![xviii Preface
5. Problem-solving suggestions including check lists to help you diagnose and
overcome problem-solving difficulties you may experience.
To provide an appreciation of what processing equipment re3.Ily looks like and
how it works, in the files on the CD disk in the worked-out problems are numerous
pictures of the equipment along with an exp]anation of their function and operation.
Problem Sets
We have included several categories problems in the books to assist in
study.
1. tests with answers (in Appendix A) foHow each section,
2. Thought and discussion problems follow the tests. Thought
problems require reflection more than calculation. Discussion problems. which
can be used as the basis of research. papers, and class discussions, pertain to
broader issues and are more open ended.
3. Homework-type problems are listed at the end of each chapter, one-third of
which have answers (in Appendix N). Each of the problems is rated 1 to 3
(using asterisks) to indicate the degree of difficulty, with 3 being the most difficult.
4. The contains more than 100 worked-out examples and another 100 prob-lems
with answers keyed to in the chapters in the text.
An of the examples and problems are designed build your problem-solving skills.
Miscellaneous Useful Features in this Book
To make the book more usable and friendly, we have incorporated a number of
beneficial features:
1. A list of contents at beginning of each chapter.
2. A list of instructional objectives at the beginning of each chapter.
3. Important terms appear in boldface type.
4. A glossary has been at the end of chapter.
S. Supplementary references that you can use to additional information are
listed at the end of each chapter.
6. Web sites containing information and links are listed at end of each
chapter.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-19-320.jpg)
![Frequently Asked Questions xxix
Many of the materials we talk about and use every day are not pure compounds.
but nevertheless you can obtain information about the properties of such
materials. Data on materials as coal, coke, petroleum products, and natural
gas-which are the main sources of this country-are available in refer-ence
books and handbooks. Tables C, 0, and E.
TABLE C Sources of Physical Property Data
American Chemical Society. 17,000 compounds, on disk. ACS, Washington, D.C. (1994).
American Chemica] L-n4e:ml,CUI Abstracts Service, ACS, Washington, D.C. (Continuing printed. mi-
J:F'.T"IIf"P. with over 20 million abstracts.)
New York (1941).
ecn'THt:.Ul Dala Book-Petroleum Refining. New York: (1970).
compounds. Also CD-Disk, ongoing,
GeseUschaft fur Chemisches Apparatwesen e. v,, Berlin,
on and thennodynamic properties.) 10.000 compounds.
Analytical Chemistry and General Scientific Data Analysis. Cambridge
(2001),
gives properties and prediction equations for over
on-line; Amer. lnst Chemical Engineers. New
International Data Series. "London. (Continuing series of data and
Environmental Data Information Network. Ecdin, data on 25,0CXl substances,
103.000. Distributed by Technical Database Services (IDS).
Chemistry, eRe Press, Boca Raton, FL., a.n.rlual editions. Also on
Lange's Handbook of Chemistry and Physics, McGraw-Hill, New York; issued periodically.
Web.
for Synthetic Fuels"; Hydrocarbon Process. p. 229 (May 1980).
Laboratories, PPDS2, 1,600 compounds; Glasgow, UK, ongoing.
Narural Processors Suppliers Association, Engineering Data Book, Okla. (Continlling editions.)
R. H., D. W. and J. O. Maloney. Perry's Chemical 7th McGraw-
Hill, New York (2000). Also on the Web.
B. D. 11. M. Prausnitz, and 1. O'Connell. The Properties 5th ed., McGraw-
York (2002).
laboratory, Chemsafe, 1,600 gases, liquids. and dusts that can distributed on disk. tape. on-
Braunschweig, Germany (1995).
STN Express, provides access to many databases Chemical Abstracts Service..
Ohio, continuing.
Research Center, Texas A&M University, Vapor Data Profilefor 5,500
on (1994).
L. Chemical Properties Handbook. McGraw-Hill. New York (1999).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-30-320.jpg)
![/
Sec. 1 Operations with Units 11
1 .. 2
c. the unit all types of power
of the above
e. none of the above
3. kg/s a basic or derived unit in SI?
4. In the IEEE Spectrum (Jan. 2001~ pp. 14-16) an article on building out the wireless Inter-net
proposed a each 0.05 km2. Does this seem reasonable?
Problems
1. Prepare a table in which the rows are: length, mass, time. two
columns. one for the the the AE system of units. Fill in each row with the
name of column show the numerical equivalency (Le .•
I ft == 0.3048 m).
2. Classify following units as correct or incorrect units in the SI svstenn:
a. om
b. OK
c. sec
N/mm
e. kJ/(s)(m3)
Thought Problem
1. What volume of material will a barrel hold?
Discussion Problem
1. In a letter to the editor, the letter says:
I believe notation be improved so as to make it mathematically
more useful by SI-sanctioned prefixes in boldface Then one WOUJG
1 c;:::; 10 m without any ambiguity [c ~ m == the meaning
of Hmmlt would at once clear to literate, if scientifically
illiterate. citizen, namely m [mm], 10-6 [mm], or (after Gauss and
(mmJ.
With respect to the "mm" problem and remarks regarding the difference
between "one square millimeter" [(mm)2] and mili squaremeter" [m(m2)].
these difficulties are analogous to confusion a "earners-hair brush"
and a camel's hair-brush."
What do you think of author's proposal?
Operations with Units
Answers a question such as: how much is 2 + 2 can sometimes be debatable.
might state 4. A bad calculator might show 3.99999. What about 9 + 5? Can the
answer for 9 + 5 = 2 possibly be correct? Look at a wall clock.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-42-320.jpg)
![12 Dimensions, Units, and Their Conversion Chap. 1
Every freshman knows that what you get from adding apples to oranges is fruit
salad! The rules for handling units are essentially quite simple:
1 .. 2-1 Addition, Subtraction, Equality
You can add, subtract, or equate numerical quantities only if the associ·
ated units of the quantities are the same. Thus, the operation
5 kilograms + 3 joules
cannot be carried out because the units as well as the dimensions of the two terms
are different. The numerical operation
10 pounds + 5 grams
can be performed (because the dimensions are the same, mass) only after the units
are transformed to be the same, either pounds, grams, or ounces, or some other mass
unit.
1.2 ... 2 Multiplication and DiviSion
You can multiply or divide unlike units at will such
50(kg)(m)/(s)
but you cannot cancel or merge units unless they are identical. Thus~ 3 m2/60 ern
can converted to 3 m2/O.6 m, and then to 5 m, but mJs2, the units cannot
cancelled or combined. In summary, units contain a significant amount of infonna·
tion that cannot be ignored. They also serve as guides in efficient problem solving,
as you will see shortly.
Frequently Asked Question
How should you handle mathematical operations or units such as sine. log. or
exponential? To be specific, if you take the log of 16 m2 and treat the number and
units as a product, then you would have
log (16m2) = log (16) + 2 log (m)
Various awkward ways and tricks of handling quantities such as 2 log (m) have been
proposed (see, M. Karr and D. B. Loveman, "Incorporation of Units into Programming
Languages," Comma. ACM, 21, 385-391 [1978]). We prefer for simplic.
ity to require that a variable be transformed or scaled to be dimensionless before you
apply nonlinear operations such as log. For example, for a pipe of radius R with](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-43-320.jpg)
![Sec. 1.3 Conversion of Units and Conversion Factors 19
mass (consult some of the at the end of this chapter for a more precise
finition of weight)? To avoid confusion. just interpret the statement as follows: a
person or object weighs as much as a mass 100 or 200 pounds, would weigh.
if by a force scale.
Some Useful Trivia Concerning Conversion
A U.S. frequent-flier mile is not the same as a U.S. mile-the former a nauti-mile
(1.85 km), whereas latter 1.61 km. In AE 1 m ;;;;; 39.37 in.,
whereas for U.S. land survey applications it 2 x 10-6 in. shorter.
EXAMPLE 1.4 Conversion Involving Both Ibm and Ibf
What is the potential energy in (ft)(lbf) of a 100 Ib drum hanging 10 ft above
surface of the earth with reference to the surface the
Solution
The [lIst thing to do is read the problem carefully. What are the unknown quanti-
The potential (FE) is unknown. What are the known quantities? The
mass and the height the drum are known. How are they related? You have to
look up the relation unless you it from physics:
Potential energy;;: P ::::: mgh
Assume that 100 means 100 mass; g = ........ ""''"',n.''. of gravity =
EL4 is a sketch of system.
-lL-___ ..I..-._ ro:.f",rAl"lt'.:o plane
Figure El.4
Now substitute the numerical values of the variables into
perfonn the necessary unit conversions.
equation and
100 Ibm ftltOnl (s2)(lbr)
P = -- 174(ft)(lb
m
) 1000 (ft)(lb()
Notice that in the ratio of ftJs2 divided by 32. 174[(ft)(lbm)]/[(s2)(1bf)), the nu-mencal,
values are almost Many engineers would solve the problem by say-ina
that 100 lb x 10 ft:: 1000 (ft)(lb) without realizing that, in they are canceling
out the numbers in the glgc ratio, and (hat the Ib in the solution means lhf.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-50-320.jpg)
![22 and Their Conversion Chap. 1
(p (V - b) = RT
Inspection of the equation shows that the constant a must have the units of [(pres-sure)(
volume)2] for the expression in the first set of parentheses be consistent
throughout. If the units of pressure are and those of volume are a will have
the units [(atm)(cm)6]. Similarly, b must have the same units as V, or in this par-ticular
case units of cm3• If T is in what must be the units Check your
answer by up R inside the front cover of the book. An must ex-hibit
dimensional consistency.
EXAMPLE 1.6 Dimensional Consistency
Your handbook shows that microchip etching roughly follows the relation
d = 16.2 - l6.2e-O·02It t < 200
where d is the depth of the etch in microns JLm) and t is the
the etch in What are the units a;:l>;""",",uu#U with the numbers 16.2 and
Convert the
minutes.
Solution
so d becomes expressed in inches and t can be
of
I?
After you the equation that d as a function of t, you should be
able to reach a decision about the units associated with
side of the equation. Both values of 16.2 must units of microns
(J,tm). The exponential must dimensionless so that 0.021 must have the "'''''''' ........ 1-
aled units of out the conversion, look up suitable conversion factors
inside the front cover book and multiply so that units are converted from
16.2 JLm to inches, and tIs to tlmin.
39.27 in. [ -0.021 60s tmin 1 d· == ----:...- --- 1 - exp---
m I m s 1 min
As you proceed the study of you will find that
groups of symbols together, either by theory or based on experiment.
that no collections of variables or parameters are called dimen-sionless
or nondimensiona1 groups. One example Reynolds number (group)
arising in fluid mechanics.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-53-320.jpg)
![Sec. 1.4 Dimensional Consistency (Homogeneity)
Dvp
Reynolds number = --= N RE
JJ..
23
where D the pipe diameter, say in cm; 11 the fluid ve]ocity, say in cmJs; p is
fluid density. say in g/cm3; and /L is the viscosity, say in centipoise, units that can be
converted to g/(cm)(s). Introducing the consistent of units for D. v. p, and f.L into
Dvp I JL. you will find that all the units cancel out so that the numerical value of 1 is
the result of the cancellation of the units.
EXAMPLE 1.7 Interesting Example of Dimensional Consisten'cy
Explain without differentiating why the foHowing differentiation cannot be
correct:
d
dx
where x is length and a is a constant.
Solution
- ---:;::::====
Observe that x and a must have the same units because the ratio 2 must be
dimensionless (because 1 is dimensionless), a
Thus, the lefthand side of the equation has units of 1. (from dJdx). However.
x
the righthand side the equation has units of x2 (the product of ax).
Consequently, something is wrong as the equation is not dimensionally consistent.
S LF .. ASSESSMENT T ST
Questions
1. Explain what dimensional consistency means in an equation.
Explain why the so-called dimensionless group no net dimensions.
3. If you divide aU of a series of terms in an equation by one of the terms. will the resulting
series of terms be dimensionless?
4. How might you make the following variables dimensionless:
a. Length (of a pipe).
b. Time (to empty a tank full of water).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-54-320.jpg)
![1 Significant 21"
What do you think about applying rule that states the number of
digits in the least precise number (two significant here because
two significant should be the number of significant digits retained in an-swer?
If you apply rule, the calculated answer is 1.1, clearly a distortion
error in the because 98 1 has an error of only about 1 %. the
1.1 ± 0.1, an error of about 10%! Certainly 1 indicates too a
precision so that choice should 1.05 or I Which do you think is
better?
2. Perhaps the use of relative error can often be a better way to how
many significant figures to retain in your answers. Suppose you divide one number
by another close to it such as 1.0111.09 = 0.9266, and select 0.927 as the
answer. The in answer based on absolute error analysis is
0.001/0.927, or about 0.1 %, 1/1.09)100, or about a 1 % uncertainty, ex-isted
in the numbers. Should relative the answer be fixed
at about 1 %, that is, truncate the answer to 0.93 rather than 0.927? would be
the case if you applied the concept of relative error. decision is up to you. In
any case, avoid increasing the of your answer very much over the preci-sion
in your measurements or when presenting results of calculations. You do
have to use some common sense applying the of relative error to scales
[hat use both relative and units. For suppose the error
in a temperature of 25°C is 1 or 4%. Can you the error by changing the
temperature kelvin, so that error becomes (1/298)]00 == 0.33%1 Of course
not.
3. A more rigorous and
in numbers apply
more complicated third way to treat uncertainty
in the calculations. What is involved is the con-cept
of confidence limits for
gation of errors step by step through each stage
suIt. Bm even a statistical is not
ratios of numbers. Refer to a book on statistics
approach.
a calculation, the propa-calculations
to the final re-because
we nonlinear
further information about this
In book we base answers on absolute error because such a choice
convenientt but will often show one or two e~gures in intennediate calculations
as you should. (The numbers in your calculator are not a Holy Writ!). Keep in mind
that some numbers are such as the 12 = 1/2 mv2 2 in the super-script
the operation of You will encounter as 1, 2, 3,
and so on, which in some cases are exact (2 reactors, 3 input streams) but in other
cases are shortcut substitutes for presumed very accurate in probJem
moles, 10 kg).
a mass as to kg, in which the number does not have a decimal
point, despite our remarks above about you can infer that quite a few
significant figures apply to the mass, particularly in other values of](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-58-320.jpg)
![28 Dimensions, Units, and Their Conversion Chap. 1
the parameters in an example or problem, because you can easily measure a
mass to a level mg. You will also occasionally encounter fractions such as
which can be as 0.6667 in relation the accuracy of other values in a prob-lem.
In this convenience we use 273 K for the temperature equivalent to
O°C instead of 273.15 K, thus introducing an absolute error of 0.15 degrees. This
a smaIl error relative to the other known or presumed errors your calcula-that
it can be neglected in almost all instances. Keep in mind, however,
addition. subtraction, multiplication, division~ all the errors that you introduce
propagate into the final answer.
Feel to round off parameters such as 1'1' = 1416, = 1.414. or Avo-gadro's
number N = 6.02 X 1()23. In summary, be sure to round off your answers to
problems to a reasonable number of significant figures even though numbers are
carried out to 10 or more digits your computer or calculator in the intermediate
calculations.
EXAMPLE 1.8 Retention of Signific;-nt Figures
100 is subtracted from 22,400~. is the answer 2.300 kg to
four significant
Solution
If you note that 22.400, 20.100, and have no decimal points the
righthand zero. how many significant can you attribute input to 22,400 and
20,1 DO? By applying the absolute error you can conclude that number
of significant figures is Scientific notation makes decision cle,ner
2.24 X let kg
and the result retains two significant figures.
On the other hand if a decimal point were p]aced in each number thus,
22,400. and 20,100., indicating that the last zero was significant. then the answer of
2,300. would be valid to four significant figures.
From the viewpoint of relative error, 22,400 has an error of about
0/224) as does 20,100 (1201), whereas 2,300 has an error of about
5% (1/23). Should relative error have been to e~tablish the number of significant
figures to be retained? you add a 0 to the right of 0.23 to give a
relative error of (11230) or about 112%? No. what about giving (he 3Mwer as
x 10?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-59-320.jpg)
![34
, "
Dimensions, Units, and Their Conversion Chap. 1
$1.5 following test win measure your SIQ. List the correct answer.
(a) Which the correct symbol?
(1) nm (2) oK
(3) sec (4) N/rnm
(b) Which is the wrong symbol?
(1 ) lvfN/m2 (2) GHzls
(3) kl/(s)(m3) (4) °ClMJs
(c) Atmospheric pressure is about:
(1) 100 Pa (2) 100 kPa
(3) 10 MPa (4) 1 GPa
(d) The temperature O°C is defined as:
(1) 273.15°K (2) Absolute zero
(3) 273.15 K (4) The freezing point of water
(e) Which height and mass are those of a petite woman?
(1) ] .50 m, kg (2) 2.00 m, kg
(3) 1.50 m, 75 kg (4) L80 m. 60 kg
(0 Which a recommended room temperature winter?
(I) 1 (2)
(3) (4)
(g) The watt is:
(1) One joule second (2) Equal to 1 . m2/s3
(3) The unit an types power (4) All of above
(h) What force be needed to lift a heavy suitcase?
(1) 24 N (2) 250 N
(3) (4) 250 kN
*1.6 A technical publication describes a new model 20-hp Stirling (air cycle) engine that
drives a 68-kW generator. this possible?
**1.7 Your boss announced that the of the company Boeing 737 is to cut from
milhr to 475 milhr to "conserve fuel," thus cutting consumption from 2200 gaJlhr to
2000 gal/hr. How gallons are saved in a 1000-mi trip?
··1.8 From Parade Magazine, 31, 1997. page 8 by Marilyn Voss·Savant:
Can help with this problem? Suppose it takes one man 5 hours to paint a house, and it
another roan 3 hours to paint the same house. If two men work together, how many
hours would it take them? This is driving me nuts. Calculate the answer.
"'1.9 Two scales are shown, a balance (a) a spring scale (b)
a b](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-65-320.jpg)
![, .
.'
Chap. 1 Problems 37
Was the calculation correct? (Answer Yes or No and explain briefly the reasoning
underlying your answer.)
··"1.21 The density of a certain liquid is given an equation of the following form:
where p = density in g/cmJ
t = temperature in °c
P = pressure in atm
p = (A + B/)eCP
(a) The equation is dimensional1y consistent. What are the units of A, B, and C?
(b) In the units above,
A = 1.096
B = 0.00086
C = 0.000953
Find A, 8, and C if p is expressed in Ib/ft3, t in oR, and P in Iblin.2
···1.22 A relation for a dimensionless variable called the compressibility (z) is z = 1 + pB +
p2 C + p3 D where p is the density in g mol/em3. What are the units of B, C, and D?
Convert the coefficieflts in the equation for z so that the density can be introduced
into the equation in the units of lbm/ft3 thus: z = 1 + p* B* + (p")2 C· '+ (pll)3 D* where
p. is in Ibm/ft3. Give the units for B·, C·, and D", and give the equations that relate
B''' to B, C· to C, and D'" to D.
···1.23 The velocity in a pipe in turbulent flow is expressed by the following equation
u=k-[p" ]'/2
where l' is the shear stress in N/m2 at the pipe wall, p is the density of the fluid in kg/m3,
u is the velocity, and k is a coefficient. You are asked to modify the equation so that the
shear stress can be ihtroduced in the units of 1" which are Iblft2, and the density be p' for
i,i "': . which the units are lbn/ft3 so that the velocity u' comes out in the units off tis. Show all
calculations, and give the final equation in tenus of u', 1"', and p' so a reader will know
that American Engineering units are jnvol ved in the equation.
·1.24 Without integrating, select the proper answer for
a
a
(1Ia)
( 1/a)
where x = length and a is a constant.
arctan (ax)
arctan (x/a)
arctan (x/a)
arctan (ax)
+ constant
*1.25 In many plants the analytical instruments are located some distance from the equipment
being monitored. Thus, some delay exists before detecting a process change and
the activation of an alarm.
/](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-68-320.jpg)
![~-.
2.1 Mole 47
ELF-A SESSM NT TEST
Questions
1. Answer the foHowing questions true or false:
a. The pound mole is comprised of x 1 Q26 molecules
h. kilogram mole is comprised of 6.022 x 1026 molecules.
c. Molecular weight is the mass of a compound or element per mole,
1. What is the molecular weight of acetic acid (CH3COOH)?
Problems
1. Convert the foHowing:
a. 120 g mol of NaCI g.
b. 120 g of NaCI to g mol.
c. 120 lb mol of NaCl to lb.
d. 120 Ib of NaCI to Ib moL
2. Convert 39.8 kg of NaCI per 100 kg of water to mol of NaCI per mal of water.
How many Ib mol NaN03 are there in 100 Ib?
Thought Problem
1. There is twice as much copper in 480 g of copper as there is in 240 g of copper, but is
twice as much copper in 480 g of copper as there is silver in 240 g of silver?
2. If a mole of doughnuts 6.022 x 1 doughnuts, can you also say that the mole of
doughnuts contains the same number of moles of doughnut holes when holes are not
doughnut material?
Discussion Problems
In journal Physics Education (July 1 p. 276) McGlashan suggested that the physical
quantity we call the mole is not necessary. Instead. it would be quite feasible to use
molecular quantities. that number of molecules or atoms, directly. Instead of pV ::::: nRT
where n denotes the number of moles of a substance. we should write pV ::::: NkT where
denotes the number of molecules and k is the Boltzmann constant [1.380 x lo-23JJ(motecule)(
K)]. Thus, example. 3.0 x 10-29 m3/molecule would the molecular volume for
water, a value used instead 18 cm3Jg mol. Is the proposal a reasonable idea?
2. In asking the question "what meant by a mole," the following answers were obtained.
Explain which are correct and which are not. and why.
a. A mole is the molecular weight expressed in grams.
b. A mole is the quantity of material one gram.
c. A mole is a certain number of of one substance or another.
d. A mole is the weight of a molecule expressed in grams.
e. A mole is the number of molecules one gram of a substance.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-78-320.jpg)
![62 Moles, Density, and Concentration Chap. 2
c. The mole percent of each component.
d. The average molecular weight of the mixture.
2.7 Concentration
Concentration generally refers to the quantity of some substance per unit volume,
but other related measures of the amount of material frequently occur, as indicated
by some of the terms in the following list of ways to express concentration:
3. Mass per unit volume (lb of solute/ft3 of solution. g of solute/L, ]b of
solutelbarrel, kg of solute/m3).
b. Moles per unit volume (lb mol of solutelft3 of solution. g mol of solutelL, g
mol of solute/cm3).
c. Parts per million (ppm); parts per billion (Ppb), a method of expressing the
concentration of extremely dilute solutions; ppm is equivalent to a mass
(weight) fraction for solids and liquids because the total amount of material is
of a much higher order of magnitude than the amount of solute; it is a mole
fraction for gases. Why?
d. Parts per million by volume (ppmv) and parts per billion by volume (ppbv)
e. Other methods of expressing concentration with which you may be familiar are
molarity (g mollL) , molality (mole solutelkg solvent), and normality (equivalentslL).
A typical example of the use of some of these measures of concentration is the
set of guidelines by which the Environmental Protection Agency defines the extreme
levels at which the five most common air pollutants could hann people over
stated periods of time.
a. Sulfur dioxide: 365 f.,Lg/m3 averaged over a 24-hr period
b. Particulate matter (10 JLm or smaller): 150 JLglm3 averaged over a 24-hr period
c. Carbon monoxide: 10 mg/m3 (9 ppm) when averaged over an 8-hr period; 40
mg/m3 (35 ppm)' when averaged over 1 hr
d. Nitrogen dioxide: 100 f..Lg/m3 averaged over 1 year
e. Ozone: 0.12 ppm measured over 1 hour
Note that the gas concentrations are mostly mass/volume except for the ppm.
In the European Union (EU), aU gasoline marketed after January 1, 2000, was
to be lead-free. Other components are limited as follows: sulfur, less than 150
mglkg; benzene, 1 vol. %~ and aromatics, 42 voL %. Sulfur in diesel fuel is limited
to 350 mg/kg. By 2005, the maximum sulfur content of both gasoline and diesel will
I](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-93-320.jpg)
![l
Sec. 2.7 Concentration
drop to 50 mg/kg. The aromatics limit for gasoline wil1 be volume percent, and
diesel will have an 11 vol. % limit for po]yaromatics.
Parts per million is a very small number. One ppm is equivalent to I inch in 16
miles, or a 1.0 g needle in a metric ton of hay.
EXAMPLE 2.8 Use of ppm
The current OSHA 8-hour limit for HeN in air is 10.0 ppm. A lethal dose of
HCN in air is (from the Merck Index) 300 mglkg of air at room temperature. How
many mg HCNlkg air is 10.0 ppm? What fraction of the lethal dose is 10.0 ppm?
Solution
In this problem you have to convert ppm in a gas (a mole ratio, remember!) to
a mass ratio.
Basis: 1 kg mol of the airIHeN mixture
Draw a simple picture. Put the data in the figure.
HeN 10 ppm
Air
Figure E2.8
You can treat the 10.0 ppm as 10.0 g mol HCN/106 g mol air because the
amount of HCN is so small when added to the air in the denominator of the ratio.
10.0 g mol HeN 10.0 g mol HCN
The 10.0 ppm is - ----=~---
l06(air + HCN)g mol 106 g mol air
Next get the MW of HCN so that it can be used to convert moles of HeN to
mass of HCN; the MW;;;; 27.03.
Then
10.0 g mot HeN 27.03 g HeN [t g mol air 1000 mg HeN
a. 6 109 mol air 1 g mol HeN 29 g air 1 g HCN
1000 g.air
X 1 k . = 9.32 mg HCNlkg air
g air
9.32
b. 300 = 0.031
Does the answer seem reasonable? At least it is not more than I!](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-94-320.jpg)
![64 Moles, Density. and Concentration
EXAMPLE 2.9 CalCulation of Mole Fraction and ppm
from a Concentration in gIL
Chap. 2
A solution of HN03 in water has a specific gravity of 1.10 at 25°C. The concentration
of the HNO) is 15 gIL of solution. What is the
a. Mole fraction of HN03 in the solution?
b. ppm of HN03 in the solution?
Solution
Let the value of the specific gravity be the value of the density (ignoring any
minor effects related to the density of water),
Basis: 1 L solution
15 HN03i 1 L 1 == 0.01364 HNO)
1 soln 1000 cm3 1.10 g soln g soln
Basis: 100 g so]ution
The mass of water in the solution is: 100 - 0.0134 = 99.986 g H20.
g MW gmol mol fraction
a. HNO:] 0.01364 63.02 2.164 X 1Q-4 3.90 x 10-5
H2O 99.986 18.016
Total 1.00
b.
0.01364 13,640
= or 13,640
106
EXAMPLE 2.10 Evaluation of Alternate Processes for the Production
of Methyl Methacrylate
Methyl methacrylate can be made using as feed (a) acetone and hydrogen
cyanide, or (b) isobutlyene. Table 10 lists some data assembled from book
by D. Allen and D. R. Shonnard (Green Engineering, Prentice Hall, Upper Saddle
River, N. 2002) relating to these two processes. The mass values are on the
basis of the production of one pound of methyl methacrylate, Oxygen also involved
but neglected.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-95-320.jpg)
![68 Moles, Density, and Concentration 2
Compound A species composed more than one element.
Concentration Quantity of a per unit volume.
Density Mass unit volume of a compound~ molar density is the number of
moles divided by the volume.
Flow Mass, moles, or of a material that is in motion divided by
unit
Gram mole 6.022 x 1023 molecules.
Mass fraction Mass a particular compound in a mixture or solution divided
the total amount of mass present.
Mole Amount a substance containing 6.022 x 1023 entities.
Mole fraction Moles of a particular compound in a mixture or solution divided by
total number of moles preSel1lL
Molecular weight Mass of a compound mole.
Parts per million Concentration expressed teons of millionths units (ppm).
Pound mole 6.022 x x 453.6 molecules.
Solution Homogeneous mixture of two or more compounds.
Specific gravity Ratio of density a compound to the density of a reference
compound.
Specific volume Inverse of the density (volume per unit mass).
Weight fraction historical for mass fraction.
UPPL MENTARY R FER NCES
In addition to the genera] N'T'I"t" ... "'r .. "
ing are pertinent.
in the FAQ in the front material, the foHow-
AHsoP1 "The Place and Importance of the Mole in Chemistry Courses:' Phys. Educ.
285 (July 1
2. B. 1. and S. M. Vora. Stoichiometry (SI Units), McGraw-Hill, New Delhi
(998).
3. R. "SI for : Persistent Problems. Solid Solutions,H J. Chem. &Juc.
80, 16-21 (2003).
4. D. M., and J. M. "Understanding the Mole and Its Use in Chemical
neering," Che. Eng. Educ., 332-335 (Fall 1999).
Gorin, "Mole Chemical Amount: A Discussion of the Fundamental Measurements
in Chemistry," }, Chem. Educ., 71, 114-116 (1994).
6. Gorin, "Mole, Mole Liter, and Molar," J. Chem. 80, 103-104 (2003).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-99-320.jpg)
![70 Moles, Density, and Concentration Chap. 2
" Objective 1. The student will know that the mole is a counting unit and that one
mole of any substance contains the same number of units as one mole of any
other substance.
" Objective The student will know that the mole is defined as the amount of
substance containing Avogadro's number of units or particles of that substance.
" Objective 3. The student will be able to calculate the atomic or molecular masses
in grams from the molar masse.. ... of the respective atoms or molecules, and viceversa.
"'2.3 Explain the differences between mole, molecule, and molecular weight.
l1li1.4 What is wrong, or correct, about each of the following answers to the question: What
is a mole?
(a) A mole is found in a certain number of cm3 of one substance or another.
(b) A mole is the weight of a molecule expressed in grams.
(c) A mole is the number of molecules in one gram of a substance.
(d) A mole is the sum of atomic weights.
(e) mo]e is the molecular weight of an element.
·2.S A textbook states: "A mole is a quantity of material whose weight is numerically
equal to the molecular weight." State whether this statement is correct, incorrect, or
partially correct, and explain in no more than three sentences reasoning behind
your answer.
*1.6 What does unit mol-1 mean? Can a unit moll/3?
·2.7 Select the correct answer(s):
1. A mole of H20 and a mole of O2
(a) have the same mass
(b) contain one molecule each
(c) have a mass of 1 g each
(d) contain the same number of molecules
2. One mole of oxygen molecules contains more independent units (02) than one
mole of oxygen atoms (0).
(a) True, because there are two atoms a for every molecule of O2,
(b) because one mole of 02 weighs more than one mole of
(c) because both them have the same number of particles.
(d) False, one mole of ° has the same mass as one mole of 02'
*1.8 Calculate the mass of one mole of chlorophyll (CssHnMgN40s)
*2.9 Convert the following:
(a) 120 Ib moles of NaCI to g.
(b) t 20 g moles of NaCI (0 lb.
(e) 120 Jb of NaCl to g moles.
(d) 120 g of NaCl to lb moles.
·1.10 (a) What is the molecuLar weight of CaCO)?
(b) How many g mol are in 109 of CaC03 ?
(c) How many Ib mol are in 20 lb of CaC03?
(d) How many g are in 2lb mol of CaC03?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-101-320.jpg)
![12 Moles, Density, and Concentration Chap. 2
Set 1 2 3
a b c a b c a b c
c Pb Pb c
Figure n.!7
*2.18 Inserting genetic material into cells to create disease~resistant or herbicide-strains
of crops is now possible with a technique that requires no special
equipment. Known as "transfonnation," method differs microtnJection or
the ballistic method, which uses a "gun" to fire gold particles coated with the "" ....... "' ... ,'"'
material into cells. Transformation only requires commercially available silicon carbide
"whiskers/' a test tube and a mixer. Suppose that 10,000 SiC crystals, with an
average diameter of ;.tm and length of 20 /Lm. are added to a test tube containing
1,000 plant cells and the genetic material to be inserted. The tube is agitated in a simple
mixer for about a minute. Collisions between the cells and needle· like SiC crys-tals
pores through which the genetic material into the cell. How many
grams of SiC were added? The specific gravity of SiC 17.
"*2.19 A cylindrical rod of silica (Si02) heated to its melting point and a thread of silica
0.125 mm in diameter is drawn from melt. By careful control of the temperature
and the tension on the thread being drawn, long of uniform diameter can be ob·
tained that make optical
(a) Calculate how much fiber (in can drawn from one cylinder of 1.0m
in length and cm in diameter. Also estimate mass. The specific gravity of
silica is
(b) optical fiber is it is typically covered with a thin protective
a polymer. fiber discussed above is with a 50 /Lm layer of a poly-mer
with density 1 kg/m3. How much polymer is needed coat the entire
fiber?
*2.20 In fully vau]ted storage systems (see P2.20) primary container is installed
in a completely reinforced concrete vault of monolithic construction. Leak
tection is normally achieved by the installation of monitoring wells or sensors in the
backfill area. vaults must be constructed by pouring the floor and walls as continuous
structural elements. Chemical resistant coatings on interior surfaces and
water coatings on exterior surfaces are required.
If a 10,000 gallon cylindrical tank is to be buried in an approved containment
system, the volume of the containment system must by law hold 10% of the volume
of tank. If the void fraction of the backfill material (after compaction) is 20%,
what must be the volume in ft3 of the excavation before the concrete vault is
poured? What is the weight of the backfill it is (density 2.2g1cm3)?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-103-320.jpg)
![82 Choosing a Basis
EXAMPLE 3.4 Calculation of the Mass Fraction of Components
in N anoparticles
Chap. 3
The microstructure of nanosized particles has proved to be important in nanotechnology
in developing economic magnetic performance of nanocomposites. In a
ternary alloy such as Nd45 Fe17BI8.5 the average grain size is about 30 nm. By replacing
0.2 atoms of Fe with atoms of Cu, the grain size can be reduced (improved)
to 17 nm.
(a) What is the molecular formula of the alloy after adding the Cu to replace
the Fe?
(b) What is the mass fraction of each atomic species in the improved alloy?
Solution
Pick a convenient basis. Because the atoms in the chemical fonnula of the
alloy total to 100, pick a basis of
Basis: 100 g mol (or atoms) of Nd4.5Fe77BI8.5
(a) The final alloy is Nd4.5F~6. 8BI8.5CUo. 2'
(b) Use a table to calculate the respective mass fractions.
Component Original g mol Final gmol MW g
Nd 4.5 4.5 144.24 649.08
Fe 77 76.8 55.85 4289.28
B ~ 18.5 lO.81 199.99
Cu --.JU 63.55 12,7}
Total 100.0 100.0 5151.06
Mass fraction
0.126
0.833
0.039
rum
1.000
In summary, be sure to state the basis of your calculations so that you will keep
clearly in mind the real nature of your calculations, and so that anyone checking
your problem solution will be able to understand on what basis your calculations
were perfonned.
You no doubt have heard the story of Ali Baba and the 40 thieves. Have you
heard about Ali Baba and the 39'camels? Ali Baba gave his four sons 39 came]s to be
divided among them so that the oldest son got one-half the camels; the second son a
quarter, the third an eighth, and the youngest a tenth. The four brothers were at a loss
as to how they should divide the inheritance until a stranger came riding along on his
came1. He added his own camel to Ali Baba's 39 and then divided the 40 among the
sons. The oldest son received 20~ the second, 10; the third, 5; and the youngest, 4. One
camel was left. The stranger mounted it-for it was his own-and rode off. Amazed,](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-113-320.jpg)
![90 Temperature Chap. 4
system to transfer energy (as heat). When you get to Chapter 21 you will find out. if
you do not already know. that a big difference exists between these two concepts.
In this book we use four types of temperature, two based on a relative scale,
Fahrenheit (OF) and Celsius (OC), and two based on an absolute scale.
gree Rankine (OR) and kelvin (K). Note that the degree symbol (0) not used with
the kelvin temperature abbreviation K. Relative scales are the ones you hear the TV
or radio announcer give. and are based on a specified reference temperature (32°F or
O°C) that occurs in an ice-water mixture (the freezing point of water).
Absolute temperature scales have their zero point at the lowest possible temperature
that we believe can exist. As you may know, this lowest temperature is related
both to the idea] gas laws and to the laws of thermodynamics. The absolute
scale. based on degree units the size of those in the Celsius scale, is caned the kelvin
scale in honor of Lord Kelvin (1824-1907) who reconciled the divergent views of
principle of conservation of energy. The absolute scale, which corresponds to
the Fahrenheit degree units, called the Rankine scale in honor ofW. J. M. Rankine
(1820-1872), a Scottish engineer. Figure 4.1 illustrates the relationship between rel~
ative temperature and absolute temperature. We shall usually round off absolute
zero on the Rankine scale of -459.67°F to -460°F; similarly, -273.1SoC frequently
will be rounded off ~o -273°e. In Figure 4.] all of the values of the temperatures
have been rounded off, but more significant figures can be used. O°C and its equivalents
are known as the values of the standard conditions of temperature.
Now we turn to a topic that causes endless difficulty in temperature conversion
because of confusing semantics and notation. To start, you should recognize that the
unit degree , the unit temperature difference or division) on the Kelvin-Celsius
scale is not the same as that on the Rankine-Fahrenheit scale. If we AOF represent
the unit temperature difference in the Fahrenheit scale and AOR be the unit
temperature difference in the Rankine scale, and AOC and AK be the analogous units
in the other two scales, you probably are aware that
AOF = AOR
Also. if you keep in mind that the A °C is larger than the
Aoe
~= 1.8 or
or
Thus, when we cite the temperature of a substance we are stating the cumulative
number of units of the temperature scale that occur (an enumeration of
measured from the reference point. Reexamine Figure 4.1.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-121-320.jpg)
![92 Temperature Chap. 4
Unfortunately, the units for b are usually ignored; just the value of b 0.8) is employed.
When you reach Chapter 21 you will note that the heat capacity in the SI sys·
tem has the units of 1/(g mol)(K). Does the K the heat capacity designate the temperature
in degrees K or the unit interva] ilK? this case K represents a temperature
difference. Consider gas constant, R. which can have units of (kg)(m2)/(kg
mol)(s2)(K). Does K represent the temperature in degrees K or the interval AK? You
should recall for the ideal gas that you use the absolute temperature.
If you are not well acquainted with temperature conversion. be sure to practice
conversions until they become routine. Many calculators and computers make the
conversions automaticaHy, but you should know that
Now Jet us look at some examples.
EXAMPLE 4.1 Temperature Conversion
Convert 100°C to (a) K. (b) OF, and (c) OR.
Solution
(a) (100 + 273)OC]1 :o~ = 373K
or with suppression of the ~ symbol.
(100 + 273)OC I K = K
lOC
0.6)
273 K (1.7)
(1.8)
(1.9)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-123-320.jpg)
![Chap. 4 Probrems 97
Pellicer. 1., M. Ampano Gilabert. and E. Lopez-Baeza. 'The Evolution of the Celsius and
Kelvin Scales and the State the Art:' J. Chem. Educ.; 76, 911-913 (1999).
Quinn, J. Temperature. Academic Press, New York (1990).
Romer. R. H. ''Temperature Scales," The Physics Teacher. 4S0 (October 1982).
Thompson. H. B. "Is goC Equal to 500P? 1. Chern. Educ., 68, 400 (1991).
Web Pages
http://www.sant.esson.comJengtemp.html
http://www.scieneemadesimple.netltemperature.html
http://www.inidata.ucar.edulstafflblyndsltmp.htrnl
http://www.temperatures.com
PROBLEMS
"4.1 "Japan, Aim for Better Methanol-Powered Cant read the headline in the
Wall Street Journal. Japan and the plan to join in developing technology to
prove cars that run on methanol, a fuel that causes less air pollution than gasoline. An
unspecified number of researchers from Japanese companies will work with the EPA
to develop a methanol car that win start in temperatures as low as minus 10 degrees
Celsius. What is this temperature degrees Rankine. kelvin~ and Fahrenheit?
·*4.2 Can negative temperature measurements exist?
·'4.3 The heat capacity Cp of acetic acid in J/(g rnol)(K) can be calculated from the equation
Cp :: 8.41 + 2.4346 X 1O-5T
where T is in Convert the equation so that T can be intrOduced into the equation in
OR instead K. Keep the units of Cp the same.
·4.4 Convert the following temperatures to the requested units:
(a) 10°C to OF
(b) IOoe to OR
(c) -25°F to K
(d) ISOK to OR
*4.5 Heat capacities are usually given in terms polynomial functlons of temperature.
The equation for carbon dioxide is
Cp :: 8.4448 + 0,5757 X 10-2 T - 0.21S9 X 100ST2 + 0.3059 X 10-9']'9
where T is in OF and Cp is in Btu/(lb mol)(°F). Convert the equation so that Tean be
in °C and Cp will be in 11(g mo1)(K).
·4.6 In a report on the record low temperatures in Antarctica. Chemical and EngiMering
News said at one point that "the mercury dropped to -76t:.C." In what sense is that
possible? Mercury at](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-128-320.jpg)
![98 Temperature Chap. 4
··4.7 "Further, the degree Celsius is exactly the same as a kelvin. The only difference is
that zero degree Celsius is 273.15 kelvin. Use of Celsius temperature gives us one
Jess digit in most cases", from [Eng. Educ., (April 1977):678]. Comment on the quotation.
Is it correct? If not, in what way or sense is it wrong?
-4.8 Calculate all temperatures from the one value given:
(a) (b) (e) (d) (e) (f) (g) (h)
140 1000
500 1000
298 toOO
-40 1000
"4.9 The emissive power of a blackbody depends on the fourth power of the temperature
and i§ given by
W=Af4
where W = emission power, BtuI(tt1) (hr)
A = Stefan-Boltzman constant. 0.171 X 10-8 Btu/(ft2)(hr)(OR)4
T = temperature, OR
What is the value of A in the units J/(m2)(s)(K4)?
·4.10 Suppose that an alcohol and a mercury thermometer read exactly ooe at the ice point
and 100°C at the boiling point. The distance between the two points is divided into
100 equal parts in both thennometers. Do you think these thermometers will give exactly
the same reading at a temperature of, say, 60°C? Explain.
·*4.11 For the following data plot p vs. lIT. Do you get a straight line? Try In(p) vs. liT and
repeat. Use curve fitting software (refer to the CD accompanying this book) to estimate
the values of the coefficients in the relation
In (p) = a + b (1/7)
T is in K and p is in nun Hg absolute.
T (K) p (rnm Hg)
273
298
323
373
Predict the value of p at 1340 mm Hg.
5
24
93
760
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-129-320.jpg)
![Sec. Pressure and Its Units
Some common nonstandard variations of pressure measurement
SI system are
8. Bars (bar): 100 kPa = 1 bar
10f
with the
b. Kilograms (force) p~r square centimeter (kgpcm2Y-'-a very common measure
of pressure but not standard in (often called just "kilos")
c. Torr (Torr): 760 Torr :: 1 atm
In the system can expressed a variety including
8. Millimeters of mercury (mm Hg)
b. Inches of (in. Hg)
c. Feet water (ft H2O)
d. Inches of water (in. H2O)
c. Atmospheres (atm)
f. Pounds (force) per square inch (often just "pounds") (psi)
Keep in mind the confusion that can be caused by
"pounds/' such as the caption on a cartoon that
nontechnical use of the word
"It says inflate to 12 pounds. How can I throw a 12-pound football?"
You can calculate the force exerted at the bottom a static fluid by applying
Equation I). For example, suppose that the cylinder fluid in 5.1 is a
umn of mercury that an area of 1 cm2 and is em high. From Table D 1 in the
Appendix you can find the specific gravity of mercury at 20°C. and hence the den-sity
the Hg, 13.55 g/cm3. Thus, the exerted by mercury alone on
l-cm2 section of the bottom plate by the column of mercury
980 cm 50 cm I cm2
] 1 m 1 (N)(S2)
F=--..;;..
S2 1000 g 100 em l(kg)(m)
= 6.64 N
The on the of the plate covered by mercury is the force per unit
area of the mercury plus the pressure of the atmosphere
= 6.64 N (100 cm)2 1 kPa
p 11m -'-----'----'--- 1000 Pa Po = 66.4 Po
.. Although
quently denoted in
SI system are in common pressures are
kglcm2), units that have 10 multiplied by
9.80 X (cm2) - to pascal](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-132-320.jpg)
![L
Sec. 5.3 Differential Pressure Measurements
If fluids 1 and 3 are gases, and fluid 2 is mercury. the density of the gas is so
much less than that of mercury that you can ignore the term involving the gas in
Equation (5.3) for practical applications.
However, if fluids 1 and 3 are liquids and fluid 2 is an immiscible fluid, the
terms invo]ving the densities of fluids 1 and 3 cannot be neglected in Equation (5.3).
As the densities of fluids 1 and 3 approach that of fluid 2, what happens to the differentia1
reading d2 in Figure 5.9 for a given pressure difference?
Can you show for the case in which Pl ::::: P3 = P that the manometer expression
reduces to the we11-known differential manometer equation?
(5.4)
As you know, or will learn in Chapter 27, a flowing fluid experiences a pressure
drop when it passes through a restriction such as the orifice in a pipe, as shown
in Figure 5.10.
The pressure difference can be measured with any instrument, such as a
manometer. connected to the pressure taps, as illustrated in Figure 5.10. Note that
the manometer fluid is static if the flow rate in the pipe is constant.
Fluid
Flow
~--
Differential
Pressure Manome1er
Figure 5.10 Concentric orifice used to restrict flow and measure the fluid flow rate
with the aid of a manometer.
EXAMPLE 5.4 Calculation of Pressure Differences
In measuring the flow of fluid in a pipeline as shown in Figure E5.4. a differential
manometer was used to determine the pressure difference across the orifice
plate.
The flow rate was to be calibrated with the observed pressure drop (difference).
Calculate the. pre.ssure drop P CP1 in pascals for the manometer reading in
Figure E5.4.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-146-320.jpg)
![120 Pressure Chap. 5
*5.3 A centrifugal pump is to be used to pump water from a lake to a storage tank that is
148 ft above the surface of the lake. The pumping rate is to me 25.0 gal/min, and the
water temperature is 60°F. The pump on hand can develop a pressure of 50.0 psig
when it is pumping at a rate of 25.0 gaUmin. (Neglect pipe friction, kinetic energy effects,
or factors involving pump efficiency.)
(a) How high (in feet) can the pump raise the water at this flow fate and temperature?
(b) Is this pump suitable for the intended service?
"'5.4 A manometer uses kerosene, 0.82. as the fluid. reading of 5 in. on the
manometer is equivalent to how many millimeters of mercury?
*5.5 manufacturer of large tanks calculates the mass of fluid in the tank by taking the
pressure m~asurement at the bottom of the tank in psig. and then multiplies that value
by the area of the tank in square inches. Can this procedure be correct?
·5.6 A letter to the editor says
"An error in units was made in the article "Designing Airlift Loop
Fermenters." Equation (4) is not correct
Lip = 4fp [(v2/2g)(UD)]
Is the author of the letter correct? if is dimensionless.)
(4)
Leaking au tanks have become such environmental problems that the Federal Government
implemented a number of rules to reduce the problem. A leak from a
small hole a tank can be predicted from the following relation:
Q = 0.61 Sv'(2ilp)p
where Q = the leakage rate
S = crossectional area bf the leak
il p ::::: pressure drop
p = fluid density
To test the tank, the vapor space is pressurized with N2 to a pressure of 23 psig. If the
tank is filled with 73 inches of gasoline (sp. = 0.703) and the hole is 114 in. in diameter,
what is the initial value of Q (in ft3Jhr)?
- -----------~-t-·------- -
Gasoline 73 in
t
Leak
Figure PS.7
·5.8 Suppose that a submarine inadvertently sinks to the bottom of the ocean at a depth of
1000 m. It is proposed to lower a diving bell to the submarine and attempt to the
conning tower. What must the minimum air pressure be in the diving bell at the level](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-151-320.jpg)
![Sec. 6.5 Accounting for Chemica! Reactions in Material Balances 149
followed the processing of (he orange juice from squeezing, through concentration, and
cooling to slush. She watched the waiting being steam cleaned and sanitized, and
then the slush being pwnped into the trucks prior to being sent off to the warehouse.
The trucks were weighed before and after filling. She placed a seal on one or two
trucks, and flew to the warehouse district to wait for "her>~ trucks to come in. On arrival
she noted that the seals were undisturbed. A pump was attached to the bottom drain of a
truck, and the slush pumped out into a holding tank until the pump started sucking air, at
which time the pump was disconnected.
Then she watched the cans being fined from the holding tank, counted the cans, had
some weighed, and noted a minor amount of spillage, but the spillage and filling operation
amounted to less than 1 % of the overall loss. At this point she concluded that either (a)
ing was going on in Florida, and/or (b) someone was stealing cases of concentrate from the
warehouse. She spent a couple of nights in the warehouse and did note someone take a couple
of cases of concentrate, but that amount had negligible effect on the overall loss.
Two weeks later while drinldng orange juice at breakfast, she had sudden idea as to
what the company's problem was. What was it?
Discussion Problems
1. Isotope markers in compounds are used to identify the source of environmental pollutants,
investigate leaks in underground tanks and pipelines, and trace the theft of oil and other
liquid products. Both radioactive and isotopic markers are used.
. Deuterium is typicalIy used as a marker for compounds, Three or more hy-drogen
atoms on the organic molecule are replaced by deuterium. However, isotopes of
carbon and oxygen can be used. The detection limit of using a combination of
gas chromatography and mass spectrometry is about 100 ppb in crude oil and about 20
ppb in refined products.
Explain how such markers can be used in chemical processes.
2. Projects to avoid climatic changes engendered by man's activities and in particu-lar
the increase in CO2 in the atmosphere include dispersal of sulfate particles in the stratosphere
to reflect sunlight and fertilizing the southern oceans with iron to stimulate phytoplankton
growth. It is believed that low levels of iron limit the biological productivity of
nutrient-rich southern oceans. Adding iron to these waters would increase the growth of
phytoplankton, thus reducing CO2 levels in the seawater and thereby altering the CO2 balance
between sea water and the atmosphere. What do you think of such a suggestion?
6.5 Accounting for Chemical Reactions
in Material Balances
Chemical reaction a system requires the augmentation of Equation (6.4) to
take into account the effects of the reaction(s). To illustrate this point, look at Figure
6.10, which shows a steady-state system in which HC] reacts with NaOH by the following
reaction: NaOH + Hel ~ NaCl + H20.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-180-320.jpg)
![Sec. 6.6 Material Balances for Batch and Semi·Batch Processes
6.6 Material Balances for Batch
and Semi-Batch Processes
1St'
A batch process is used to process a fixed amount of material each time it is
operated. Initially, the material to be processed is charged into the system. After pr0-
cessing of the material is complete, the products are removed. Therefore, if you include
the material originally charged into the vessel as well as the material remaining
in the vessel after the processing has been completed as part of the system, batch
processes fall into the category of closed systems. Batch processes are used industrially
for speciality processing applications (e.g., producing pharmaceutical products),
which typically operate at relatively low production rates. Look at Figure 6.11a that
illustrates what occurs at the start of a batch process.
Two separate quantities of material are put into a mixing vessel at the start of
the process, 9000 lb of H20 and 1000 lb of NaOH (Figure 6.11a), and after thorough
mixing, the final so]ution remains in the system (Figure 6.11 b).
From the viewpoint of Equation (6.4), the flows by definition are zero because
material does not cross the system boundary. Would you conclude that the accumulation
of the total mass in the system is zero? Yes. All that happens in the batch system
(in the absence of chemical reaction) insofar as the total mass or moles of material,
or the mass or moles of the individual componentst is that the final conditions
----------------------------------------~~~-~---
boundary
1
----------------------------------------------- ______ 1
Figure 6.118 The initial state of a batch mixing process.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-182-320.jpg)
![Sec. 6,6 Material Balances for Batch and Semi-Batch 157
Rate Flow unit
Semi-batch process A process in which material enters the system but product is
not removed during operation.
Steady-state system A system for which all the conditions (e.g., temperature,
pressure, amount of material) remain constant with time.
System Any arbitrary portion of or whole process that is considered for analysis.
System boundary The closed line that encloses the partion of the process that is
to be analyzed,
Transient system A system for which one or more the conditions (e.g., temper-ature,
pressure, amount of material) of the system vary with time. Also known
as an unsteady-state system,
Unsteady-state system system for which one or more of the conditions (e.g.,
pressure, amount of material) of the system vary with time. Also
known as a transient system ..
UPPLEMENTARV R F R NCES
In addition to the general references listed in the Frequently Asked Questions in the
front materia], the following are pertinent:
CACHE Corp. Material and Energy Balances 2,0 (CD). Austin, TX (2000).
Fogler, S., and M. Montgomery. Interactive Computer Modules for ChE. Instruction,
Corp., Austin. TX (2000).
Veverka, V. V" and F. Madron. Material and Energy Balances in the Process Industries:
From Microscopic Balances to Large Plant, Elsevier Science, Amsterdam (1998).
Web Sites
http://www.outokumpu.filhsclwhats_new_heat.htm
http://voyager5.sdsu.edultestcenter/features/cgprocessJindex
www.chemeng.ed.ac.uV-jwpIMSO/sectionllbalance.html
Answer to the puzzle
The additions and subtractions made in the puzzle are invalid. A diagram of
the flow of money shows the transfers (amounts on the arrows) and final conditions
(the amounts within the boxes):
You can see that the money balance should be made for the net resul among
the five individuals in the boxes, and not of flows between the boxes:
$55 + $2 - ($19) (3) = 0](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-188-320.jpg)
![168 A General Strategy for Solving Material Balance Problems Chap. 7
as an science. instance, suppose that it takes one man 10 days to build a
brick waH; then 10 men can finish it in one day. Therefore, 240 men can finish the
wall in 1 hr, 14AOO can do the job in 1 min$ and with 864,000 men the waH will be
up before a single brick is in place! Your password to success to use some common
sense in problem solving and always maintain a mental picture of system
that you are analyzing. Do not allow a problem to become abstract and unrelated to
physical behavior.
7.2 The Strategy for Solving Problems
Howe's Law: Every person has a scheme that will not work.
Gordon's Law: If a project is not worth doing, it is not worth doing well.
You neither have to follow the steps in the list below in any particular sequence
nor formally employ every one them. You can go back several steps and
repeat steps at will. As you might expect, when you work on solving a problem you
will experience fa]se starts, encounter extensive preliminary calculations, suspend
work for higher priority tasks, look for missing links, and make foolish mistakes.
The strategy outlined below is designed to focus your attention on the main path
rather than the detours. I
1. Read and understand the problem statement
This means read the problem carefully so that you know what given
and what to be accomplished. Rephrase the problem to make sure you
understand it. An anecdote illustrates the point of really understanding the
problem:
An English family visiting Khartoum in the Sudan took their young son
day by the statue of General Gordon on a camel. On the day of
their visit to the statue as the family was leaving. the boy asked, "Who was
that man sat on General Gordon?"
is a question to answer:
How many months have 30 days?
Now you remember the mnemonic: 30 days hath September ... and give
the answer as four, but is that what the question concerns-how many months
have exactly 30 days? does the question ask how many months have -at
least 30 days (with the answer being II)?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-199-320.jpg)
![• ·t' >.'.
I·
I •
7.2 The Strategy for Solving Problems
TABLE 7.3 A List of of the Variables Involved in the Example Problem
Feed Water Product
Let's analyze Example 7.2 again taking into account of the variables.
You can count the number variables in Example 7.2 simply by multiplying
the number of rows by the number of columns table.
Then for one stream
Nsp = the number species
Ns = the number of streams
Nu = the total number variables
and for the whole
The value of] in (Nsp + () comes from last row the table listing the totals.
What kinds of independent specifications and constraints in general must
you consider in arriving at a value for NE? are some common possibilities:
a. Specifications and values variables that are given in the problem
statement such as
I. The ratio of two flow rates is some specific value.
The conversion in a IS gwen.
3. The value of a concentration, flow temperature, pressure, den-sity,
volume, and so on is given.
4. A variable not present in a stream, hence its value is zero.
For the case of the problem in Example 7.2 we have
(1) specifications the mass fractions and one specification of the mass
flow:
F NaOH = 1000 kg (the basis) (1)
1 the variables in Table
w~20 = 0, hence FH20 = 0 (2)
or. the variable in Table 7.3,
~aOH 0, hence WNaOH 0 (3)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-210-320.jpg)
![Sec. 7.2 The Strategy for Solling Problems
must be made. so that you can determine the respective contributions from each
cylinder to get the desired composition in the cylinder with the three components?
Solution
A sketch of the process greatly helps in the analysis of the degrees of freedom.
Look at Figure E7.4.
F2
C2H6 0.10
N2 0,90
1.00
,
F, F4 F;j
CH4 0.20 CH, XC~ N2 1.00
N2 0.80 ~ C2HS Xc.zHe
1.00 N2 xt.I2
Figure E7.4
Do you count seven unknowns-three values of xi and four values of F/! How
many independent equations can be written?
Three material balances: CH4, C2H6• and N2
One specified ratio: moles of CH4 to C2H6 equal 1.5 or (xCH !xc H ) == 1.5
• ~ F'. 4 2 6
One summatIon of mole fractions: ""X;4 = 1
Thus, there are seven minus five equals two degrees of freedom. If you pick a
basis, such as F4 = 1. one other value has to be specified to solve the problem to calculate
composition of F4. Keep in mind that you must be careful in making any
specifications to maintain only independent equations. Avoid transforming one or
more independent equations in a set such that the resulting set contains redundant
(dependent) equations. Did you notice in the problem formulation that the LX; = 1 equations for F l' F2• and F] were redundant because of the way the spec·
ification of the mole fractions was made?
181](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-212-320.jpg)
![, 1 A General Strategy for Solving Material Balance Problems Chap. 7
EXAMPLE 7.S Analysis of the Degrees of Freedom
in the Production of Biomass
[n the gTowth of biomass CH1.gOO.5No.16S0.004SPO.OO.55' with the system comprised
of the biomass and the substrate, the substrate contains the carbon source for
growth, Ca Hp 0,., plus NH3• °2, H20. C021 H3P04• and H2S04- The relations between
the elements and the compounds in the (refer to Appendix for more
details) are:
CHt.sOo.sNo.t ~O.OO4SP 0.0055 NH) Oz CO2 H2O H2SO4 H3P04
C I ct 0 0 1 0 0 0
H 1.8 ~ 3 0 0 2 2 3
0 0.5 'Y 0 2 2 I 4 4
N 0.16 0 I 0 0 0 0 0
S 0.0045 0 0 0 0 0 I 0
P 0.0055 0 0 0 0 0 0
How many degrees of freedom exist for this system (assuming that the values
of ct, ~, yare specified)?
Solution
Based on the given data six element balances exist for the 8 species present,
hence the system has two degrees of freedom. However, it turns out for this type of
system that experiments show that the change in CHLSOO.sNo.16S0.004SPO.OO55 and
the change in CaHpOy prove to related by the amount of biomass present and the
maintenance coefficient (the moles of substrate per mole of biomass second) so
that the respective quantities cannot be chosen independently. Consequently, with
this extra constraint, only one degree of freedom remains to be specified. the basis.
8. Write down the equations to be solved in terms of the knowns and unknowns.
Thus [Beatrice] began: "You dull your own perceptions with false imaginings and do
not grasp what would be clear but for your preconceptions . ..
Once you have concluded from the degree of freedom analysis that you
can solve a problem, you are prepared to write down the equations to be solved
(if you have not already done so as part of Step 7), Bear in mind that some formulations
of the equations are easier to solve by hand, and even by using a
computer, than others. In particular, you should attempt write linear equations
rather than nonlinear ones. Recall that the product of variables, or the ratios
of variables, or a logarithm or exponent of a variable, and so on, in an
equation causes the equation to be nonlinear.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-213-320.jpg)
![7.2 The Solving Problems 183"
In many instances you can easily transform a nonlinear equation to a linear
one. For instance, in the problem posed in Example 7.2, one constraint
was that WIP = 0.9, a nonlinear equation. If you multiply both sides of
the equation by P, you a linear equation W =:
Another of judicious formulation equations occurs in
same problem. chose as the variables listed Table 7.3 the mass flows
such as WH20 and PHlO for the water in the respective streams. If, instead
of PH10, you use as variables the product of the mass fraction of in P
times P:
and substitute the ~20P for PH20 in the
ance for water, instead of having a linear equation
and in the material balwater
balance
F(O) + W(1.000) PH20
you would
F(O) W(LOOO) = W~lO P
a nonlinear equation (which is why we didn't use W~20 as a variable).
With these in mind, you can formulate the set of equations to be
used to solve the problem in Example 7.2 as follows. First. you introduce
ifications 0), (2), and (3) into the material balances (5) and (6), and into the
summation of mass fractions or their equivalents, (7), (8), and (9). Then you
will a set of four independent equations in four unknowns, which are the
same set of four we introduced at the of Step 7.
basis is stiB 1 hr (FNaOH = 1000 kg) and the process been assumed to
be at steady state. Recall from Chapter 6 that in such circumstances a material
balance simplifies to in = out or in - out = O. The equations are
NaOH balance: 1000 = PNaOH or 1000 - P NaOH = 0 (1)
H20 balance: W = Pu,o or W - PHlO = 0 (2)
Given ratio: W = O.9P or W - D.9P = 0 (3)
of components in P: PH10 = Por PNaOH + PH10 - P = 0 (4)
Could you substitute total mass balance ] 000 + W = P for one of the two
component balances? Of course. fact, could calculate P by
just two equations:
Total baJance: 1000 + W = P
W = O.9P](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-214-320.jpg)
![Sec. The Strategy for Sotving Problems
Then
1000 NaOH
p - --....;:::;......-- = 0.1 WNaOH - 10,000 kg Total
P 9,000
W H - ----------- 20 - 10,000
Examine the set of four equations listed in Step 8. Can you find a shorter or
"" ........ ~A series of calculations get a solution for the problem?
10. Check your answer(s)
Error a hardy plant; it flourishes in every soil.
Martin Tupper
Everyone makes mistakes. What distinguishes a good engineer is that he
or she is able to find their mistakes before they submit their work. In Chapter 1
we listed several ways to validate your solution. We will not them here.
Refer to Section 1 A good engineer uses his or her accumulated knowledge
as a primary tool to make sure that the results obtained for a problem (and the
data in the problem) are reasonable. Mass fractions should faU between
zero and one. Flow rates should nonnegative.
In any collection of data, the figure that is most obviously corfeetbeyond
all need of checking-is the mistake.
Unknown
To the list of validation techniques that appeared in Chapter 1, we want to
add one more very useful one. After solving a problem, use a redundant equation
to check your values. In the problem in Example 7.2 that we have been
using in the presentation of the problem-solving strategy. one of the three material
balances is redundant (not independent), as pointed out severa] times.
Supposed you had solved the problem using the N aOH and H20 balances.
Then the total balance would have been a redundant balance, and could be
used to check the answers
Insert the numbers
1)000 + 9,000 = 10,000
Table 7.4 summarizes the set of 10 steps we have discussed.
H you use the steps in Table 7.4 as a mental checklist each time you start
to work on a problem, you will have achieved the major objective of this chap-](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-216-320.jpg)
![198 Solving Material
Solution
Stept
Problems for Single Units without Reaction Chap.S
This is an open (flow), steady-state process without reaction. Assume because
of the low concentration of Strep. in the aqueous and organic fluids that the flow
rates of the entering fluids equal the flow rates of the exit fluids.
Steps 3, and 4
AH of the data has been placed on Figure ES.l.
Organic solvent S
10Umln p::: 0.6 glcm3
No Strep
Aqueous solution Aqueous solution
A Extraction
200 Umin Process 0.2 9 Strap Il
10 9 Strep IL
p:: 1 glcm3
Organic solvent
Extracted Strep
Figure £.8.1
StepS
Basis: 1 min
Steps 6 and 7
The degree-of-freedom analysis
Number of variables (8): 4 flows (in L) plus 4 concentrations (in gIL)
Number of equations (8):
Basis: :: 200 L (flow of aqueous entering aqueous solution) 1
Specifications: Concentration Strep in entering aqueous solution 1
Concentration of Strep in aqueous solution ]
Concentration of Strep in entering organic solvent 1
Flow exiting aqueous solution (same as existing flow) 1
Flow of entering organic solution 1
Flow of exiting organic solution (same as existing flow) 1
material balance j
Total 8
degrees of freedom are O.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-229-320.jpg)
![Chap. 8 Solving Material Balance Problems for Single Units without Reaction
-""""------------. .......
,~ "
System' /' ...
8oundory~
1000 kg Fee
F 10% EtOH
90"1) H20
,
I
I
,I
,I
dl
I
I
I
I
t
EllchO/1ger ,
I Vapor Heal
I
I
Reflux I
I
/
,/
OistiliotiOfl ,/
/'
Column /'
/'
/'
/
/
/
I
J
I Bottoms (Woste) 8
/
Figure E8.3
Oi stillote (Product) p::c kg
60,"0 E.tOH
40,.., H20
1/10 feed
"" kg
this problem. Let m designate the mass of a component. Clearly the process is an open
system, and we assume it is in the steady state. No reaction occurs. Thus, the material
balances reduce to In;:: Out in kg. The cooling water enters and leaves the heat exchanger
without mixing with the components being separated, and can be ignored for the material
balances. In addition, the heat added at the bottom of the column does not involve
mass entering or leaving the system, and can be ignored for the material balances.
Steps 1, 2, 3, and 4
All of the symbols and known data have been placed on Figure E8.3.
StepS
Select as the basis the given feed. Pick ] hour so that you can suppre~s the
time variable in all of the calculations.
Basis: I hour so that F = 1000 of feed
Step 4
are gI.V en that p'1 8 -1 of
10
so that P ::: 0.1 (l000) = 100 kg
Steps 6 and 7
The next step is to carry out a degree of freedom analysis.
Number of variables: 9
F F P P B B FPB mEtOH. mHzo, mEtOH. mHzo- mBtOH. mH20. , ,
203](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-234-320.jpg)
![Chap.S Solving Material Balance Problems for Single Units without Reaction 209"
Example 8 the BDC in the wet and dry fish cake is known as a tie component
because the BDC goes from a single stream in the process to another single
stream without loss, addition, or splitting. The BDC ties two streams together. Con-sequently,
you pick BDC as the basis in the you can add and subtract ra-tios
the amount of a compound per unit amount BDe in a material balance.
Here is what you could have done in Example
H20 balance:
In
0.80 H20 inA
0.20 kg BDC in A
Basis: 1 BDC
Out
0.40 kg H20 in B
0.60 BDC in B
=
H20 removed
3.33 kg H20
kg BDC (in A or B)
Next, shift to a basis of 100 kg of H20 in W. The H20 removed is H20 in W:
BDC in A
0.20
;;; ] 50 A
Sometimes the use of a tie element reduces the extent of the calculations in a
problem.
EXAMPLE 8.6 Crystallization
A tank holds 10,000 kg a saturated solution of Na2CO) at 30°C, You want
to crystallize from this solution 3000 kg NazCO} . IOH20 without any accompa-nying
water. what temperature must the splulion be cooled?
Solution
This problem is a little more complicated to analyze than the previous problems
because it not only requires a decision as what the are in the
problem, but also it implies without specifically stating so that the final solution is
saturated at the final temperature. No reaction occurs. Although problem could
set up as a steady-state problem with flows in and out of system (the tank), it
is equally justified to treat the process as an unsteady-state process. The major diffiCUlty
posed in this problem is to all the necessary information about the compo·
sitions of the solutions and solid precipitate. If you can calculate the final concen-tration
the N~C03 the tank, you can look up corresponding temperature
a handbook containing solubility data. The components mentioned the problem
statement are Na2C03' H20, and Na2C03 . 10H20. Although these components can](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-240-320.jpg)
![212 Solving Material Balance Problems for Single Units without Reaction Chap. 8
Because we are treating problem as an unsteady-state problem,
ance reduces to (the flow in = 0)
accumulation = out
Steps 6 and 7
The analysis of the
initial, F final and
freedom yields a value of zero (l stands for the
the crystals).
Number variables: 9
I J F FCC I
mN.a2C03' mHzO. mNa2C03' mHlO. mNa2C03' mHzO. ,
Number of equations: 9
Basis: 1 == 10,000 kg
Specifications:
I 1 F FCC
WNa2C03' WHlO. WNs2c03, WA20, WNa2CO), wH20
Material balances:
c
Note that w/I = m{, w/F = m{, and {dice = mf are redundant equations.
For example, m~a2 could be replaced with 0.2801 by using the specification for
~a2 Also redundant are equations su?h as = ] :Lm; mlotal.
Steps 8 and 9
After substituting the specifications and basis into the material balances, (only
two are independent) get (in kg).
Accumulation in Tank
Final Initial
N~C03 F
mNa2CO) 10,000(0.280) ;;;; -3000(0.37 J)
F
mHlO 10,000(0.720) = -3000(0.629)
Total F 10,000 == -3000
solution for the composition and amount of the final solution is
Component kg
1687
F (total) 7000](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-243-320.jpg)
![. 220 Solving Material Balance Problems for Single Units without Reaction Chap.B
order for 3000 pounds of an adhesive solution containing 13% polymer by weight.
On hand is 500 pounds of 10% solution and very large quantities of 20% solution and
pure solvent. Calculate the weight of each that must be blended together to fill this
order. Use all of the 10% solution.
-8.7 A lacquer plant must deliver 1000 lb of an 8% nitrocellulose solution. They have in
stock a 5.5% solution. How much dry nitrocellulose must be dissolved in the solution
to fill the order?
J*8.8 A gas containing 80% CH4 and 20% He is sent through a quartz diffusion tube (see
Figure PS.8) to recover the helium. Twenty percent by weight of the original gas is
recovered, and hs composition is 50% He. Calculate the composition of the waste gas
if 100 kg moles of gas are processed per minute.
BO%CH4 __
20% He ..
• L
Figure PS.S
-.~ Waste gas
.. 50% He
recovere d gas
*8.9 In many fennentations, the maximum amount of cell mass must be obtained. However,
the amount of mass that can be made is ultimately limited by the cell volume.
Cells occupy a finite volume and have a rigid shape so that they cannot be packed beyond
a certain limit. There wiU always be some water remaining in the interstices between
the adjacent cells, which represents the void volume that at best can be as low
as 40% of the fermenter volume. CaJculate the maximum cell mass on a dry basis per
L of the fennenter that can be obtained if the wet cell density is 1.1 glcm3. Note that
cells themselves consist of about 75% water and 25% solids, and cell mass is reported
as dry weight in the fennentation industry.
"'8.10 A polymer blend is to be formed from the three compounds whose compositions and
approximate formulas are listed in the table. Determine the percentages of each compound
A, B, and C to be inh-oduced into the mixture to achieve the desired composition.
Compound (%)
Composition A B C Desired mixture
(CH4).r 25 35 55 30
(C2H6)x 35 20 40 30
(C3Hg)x 40 .-&. _5 40
Total 100 100 100 100
How would you decide to blend compounds A, B. C. and D f(CH4)x = 10%, (C2H6)x
= 30%, (C3Hg;: 60%] to achieve the desired mixture?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-251-320.jpg)
![Sec. 9.1 Stoichiometry 221
tar reaction equation. If you multiply each term in a chemical reaction equation
by the same constant, say two, the absolute stiochiometric coefficient each
tenn doubles. but the coefficients still exist in the same relative proportions.
2. Use the proper degree of completion for the reaction. If you do not know how
much reaction has occurred, you have to assume some amount, such as complete
reaction.
3. Use molecular weights to convert mass to moles for the reactants and products,
and vice versa.
4. Use the coefficients in the chemical equation to obtain the molar amounts of
products produced and reactants consumed by the reaction.
Steps 3 and 4 can be applied in a fashion similar to that used in carrying out the
conversion units as explained in Chapter 1. As an example, the combustion of
heptane takes place according to the fonowing reaction equation
(Note that we have put the states of the compounds in parentheses after the species
formula) information not needed for this chapter, but that will be vital in Parts 3 and
4 of this book.)
What can you learn from a chemical reaction equation? The stoichiometric
coefficients in the chemical reaction equation (l for C7H16, 11 for 02' and so on)
tell you the relative amounts of moles of chemical species that react and are produced
by the reaction. The units of a stoichiometric coefficient for species i are the
change in the moles of species i divided by the moles reacting according to the specific
chemical equation. We will abbreviate the units simply as mol vmoles reacting
when appropriate, but frequently in practice the units of moles reacting are ignored.
You can conclude that 1 mole (not Ibm or kg) of heptane wi]) react with 11 moles of
oxygen to give 7 moles of carbon dioxide plus 8 moles of water. These may be Ib
mol, g mol, kg mol, or any other type of mole. Another way to use the chemical reaction
equation is to indicate that 1 mole of CO2 is formed from each ~ mole of
C7H16, and I mole of H20 is formed with each ~ mole of CO2- The latter ratios indiuse
of stoichiometric ratios determining the relative proportions of products
and reactants.
Suppose you are asked how many kg of CO2 will be produced as the product if
10 kg of C7H I6 react completely with the stoichiometric quantity of °21 On the
basis of 10 kg of C7H6
10 kg mol CO2 44.0 kg
= 30.8 kg CO2 g rno 16 1 g mo 2
1 k I k 1 CO](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-258-320.jpg)
![Sec. 9.1 Stoichiometry
EXAMPLE 9.2 Use of the Chemical Equation to Calculate the Mass
of Reactants Given the Mass of Products
In the combustion of heptane. CO2 is produced. Assume that you want to produce
500 kg of dry ice per hour. and that 50% of the CO2 can be converted into dry
ice, as shown in Figure E9.2. How many kilograms of heptane must be burned per
hour?
Other ProGJcts
COt Gos
150%)
............ ---"'t"""l· CO2 Solid
C11-, Gos (50~
R.odor
0t GO& 500 kg/II Figure E9.1
Solution
In solving a problem of this sort. the grand thing is to be
able to reason backward. This is a very useful accomplishment,
and a very easy one, but people do not practice it
much.
Sherlock Holmes, in Sir Arthur Conan Doyle's
a Study in Scarlet
From the problem statement you can conclude that you want to use the product
mass of CO2 to calculate a. reactant mass, the 16' The procedure is flrSt to convert
kilograms of CO2 to moles, apply the chemical equation to moles of C,HI6 > and
finally calculate the kilograms of C7Hu1' We will use Figure E9.2 the analysis.
Look in Appendix Dl to get the molecular weights of CO2 (44.0) and C;H!6
(100.1), The chemical equation
tions.
C7Hr6 + 1102 ~ 7C02 + 8H20
The next step is to select a
Basis: SOO kg of dry ice (equivalent to 1 hr)
The calculation of the amount of C7HI6 can be made in one sequence:
500 mol CO2 1 mol
0.5 kg dry ice 44.0 kg CO2 7 kg rno] CO2
100.1 -----=
1 kg mol C7H16
Finally. you should check your answer by reversing the sequence of calcula-
229 ,.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-260-320.jpg)
![Sec. 1
StepS
Stoichiometry
Mol. Wt.~
CaCOl
100.1
next step is to pick a
Basis: 100 Ib of limestone
This was selected because pounds will be equal to percent. You could also
pick 1 of limestone if you wanted, or 1 ton.
Steps 6, 7, 8. and 9
Calculations of the percent composition and lb moles of the and
products the fonn of a table will serve as an adjunct to Figure E9.3, and will
prove helpful in answering posed.
MgCOJ
Inert
Total
The q~antities
exam~le, for
92.89
5.41 lb
Limestone Solid Products
Ib == percent lbmol Compound Ib mol Ib
92.89 0.9280 0.9280 52.04
5.41 0.0642 0.0642
1.70
100.00 0.9920 0.9920
from the chemical equations.
lIb mol CaCOl lIb mol CaO 56.081b CaO = 52.041b CaO
100.1 Ib CaCOl 1 Ib mol CaC03 1 lb mol CaO
lIb mol MgO
1 Ib mol MgCO]
The production of
0.9280 Ib mol CaO is equivalent to 0.9280 Ib mol CO2
0.0642 MgO is equivalent to 0.0642 lb mol
Total 0.992 lb C02
O.9921b mol 44.0 lb CO2 1
1 lb mol CO
2
= 44.65 b
Alternately, you could have calculated the Ib CO2 a total balance: 100 -
56.33 = 44.67. Note that total pounds of an of the products 100 lb of
entering limestone. If it did not. what would you do? Check weight
values and your calculations.
231](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-262-320.jpg)
![Sec. 9.2 Terminology for Applications of Stoichiometry 237 .-
iCing reactant is the species in a chemical reaction that would theoretically run out
first (would be completely consumed) if the reaction were to proceed to completion
according to the chemical equation-even jf the reaction does not proceed to
completion! An other reactants are caUed excess reactants.
amount of the excess reactant fed - amount of the
..,"' .... .., .. '" reactant required to react with the limiting reactant
% excess reactant = 100 --'-----,-------~-------~
amount of the excess reactant required
to react with limiting reactant
For example. using the chemical reaction equation in Example 9.2.
C7HI6 + ] 102 """, 7C02 + 8H20
if 1 g mol of C,H16 and 12 g morof 02 are mixed, C7Hl6 would be the limiting reactant
even if the reaction does not take place. The amount of the excess reactant
would be 12 g mol less the 11 g mole needed to react with 1 g mol of C7H16; or 1 g
mol of 02' Therefore. if the reaction were to go to completion, the amount of product
produced would be controlled by the amount of the limiting reactant.
As a straightforward way of determining the limiting reactant, you can determine
the maximum extent of reaction, ~ax, for each reactant based on the complete
reaction of the reactant. The reactant with the smallest maximum extent of reac ..
tion is the limiting reactant. For the example. for 1 g mol of <;H16 plus 12 g mole
of 02' you calculate
o mol O2 - 12 g mol O2
~max (based on 02) = 10 1 . = 1.09 moles reacting
- 11 g mo 2/mo es reacting
o g mol -1 g mol C,HI6 emax (based on C,H16) = 1 I C H _I 1 . = 1.00 moles reacting
- g rno 7 16'mo es reacting
Therefore, heptane is the limiting reactant and oxygen is the excess reactant.
Consider the foHowing reaction
A + 3B + 2C ......, Products
If the feed to the reactor contains J.l moles of A, 2 moles of B, and 2.4 moles of
C. The of reaction based on complete reaction of A, B. and C are
gmn (based on A) = -1.1 mol A 1.1
-1
-3.2 mol B
~JruU( (based on B) = = 1.07
,max (based on C) = -2.4 mol C = 1.2](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-268-320.jpg)
![Terminology for Applications of Stoichiometry
Solution
You arc asked to calculate the limiting reactant, and use a chemical reaction
equation to calculate the NH3 produced. At room temperature and pressure no reaction
will occur, but you are asked to calculate what would result if the reaction were
to occur (as it does under other conditions of temperature and pressure),
Look at E9.5.
N2 (g) ___. ...... ,..j
10 9
Reactor 1-+---- H2 (g)
10 9
Next. write down
Given g:
MW:
Cal cd . g mo]:
Figure E9.5
chemical equation. and
N2{g)
10
28
0.357
+ 3H2(g)
10
2.016
4.960
the molecular weights:
2NH3(g)
o
17.02
o
The next step is to determine the limiting reactant by calculating the maximum
extent of reaction based on the complete reaction of N2 and
-0.357 g mol N2
~m!lX (based on N2) = 1 N I I . = 0.357 moles reacting
- g mo] 2 mo es reactmg ,
-4.960 mol H2 .
1 HI' = 1.65 moles reacting
g rno 2 moles reactmg
You can conclude that (b) N2 is the limiting reactant, and that (c) H2 is the excess
reactant. The excess H2 is 4.960 - 3(0.357) = 3.89 g mol. To answer question (a),
the maximum amount of NH3 that can be produced is based on assuming complete
conversion of the limiting reactant
Finally, you should check your answer by working from the answer to the
given reactant, or, alternatively. by adding up the mass of the NH3 and the mass of
excess H2• What should the sum be?
239 '](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-270-320.jpg)
![Sec. 9.2 Terminology for Applications of Stoichiometry 241
product produced in a set of reactions. For example, methanol (CH30H) can be
converted into ethylene (C2H4) or propylene (C3H6) by the reactions
2 CH30H -+ C214 2H20
3 CH30H -+ C314 + 3HzO
Of course, for the process to be economical, the prices of the products have to
significantly greater than the reactants. Examine the data in Figure 9.1 for the
,concentrations of the products of the reactions. What is the selectivity of C2H4 rela'
live to the C3H6 at 80% conversion of the CHlOH? Proceed upward at 80% conversion
to get for C2H4 :; 19 mole % and for C3H6 :; 8 mole %. Because the basis for
both values is the same, you can compute the selectivity 19/8 == 2.4 mol c;H4 per
mol C3H6.
- C:]H6
10 & ~, .. c
~ •
~ ~ CI) 5 5
0.
0 0
40 60 80 40 60 ao
Percent conversion of CH:PH Percent conversion of CH30H
Figure 9.1 Products from me conversion of ethanol.
9.2-5 Yield
No universally agreed~upon definitions exist for yield-in fact~ quite the con,
,trary. Here are three common ones:
.. yield (based on feed}-the amount (mass or moles) of desired product ob-
/ tained divided by the amount of the key (frequently the limiting) reactant fed.
• yield (based on reactant consumed)-the amount (mass or moles) of desired
product obtained divided by amount of the key (frequently the limiting) rectant
consumed.
• yield (based on theoretical consumption of the limiting reactant)-the amount
(mass or moles) of a product obtained divided by the theoretical (expected)
amount of the product that would be obtained based on the limiting reactant in
the chemical reaction equation(s) if it were completely consumed.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-272-320.jpg)
![242 The Chemical Reaction Equation and Stoichiometry Chap. 9
EXAMPLE 9.6 Yields in the Reaction of Glucose to Produce Ethanol
Yeasts are living organisms that consume sugars and produce a variety of
products. For example, yeasts are used to convert malt to beer and corn to ethanol.
The growth of cerevisiae (a specific type of yeast) on glucose (a sugar) under
anaerobic conditions (in the absence of oxygen) proceeds by the following overall
reaction to produce biomass. glycerol, and ethanol
C6H120 6(gIucose) + 0.118 NH) -+ 0.59 CHL74No.200.45 (biomass)
+ 0.43 C3Hs0 3(glycerol) -+- 1.54 CO2 + 1.3 C2HsOH (ethanol) + 0.03 H20
Calculate theoretical yield of biomass in g of biomass per g of glucose. Also,
calculate the yield of ethanol in g of ethanol per g of glucose.
Solution
Basis: 0.59 g mol of biomass
0.59 mol biomass g biomass
---'--- = 0.0778 g biomass/g glucose
1 g mol glucose I g mol biomass
1.3 g mol C2HsOH --=---=------ ~--"'--- = 0.332 g C1HsOHJg glucose
] g mol glucose
Why doesn't the actual yield in a reaction equal the theoretical yield predicted
from the chemical reaction equation? Several reasons exist:
., impurities among the reactants
'" leaks to the environment
II side reactions
it reversible reactions
As an illustration) suppose you have a reaction sequence as follows:
A~B~C
~
with B being the desired product and C the undesired one. The yield B according
to the first two definitions is the moles (or mass) of B produced divided by the respective
moles (or mass) of A fed or consumed. The yield according to the third definition
moles (or mass) of divided by the maximum amount of B that could](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-273-320.jpg)
![244 The Chemical Reaction Equation and Stoichiometry
EXAMPLE 9.8 Calculation of Various Terms
Pertaining to Reactions
Chap. 9
Semenov (Some Problems in Chemical Kinetics and Reactivity, Princeton
Univ. Press (1959), Vol II. 39-42) described some of the chemistry of allyl
chlorides. The two reactions of interest for this example are
CI2(g) + C3H6(g) -;+ C3HsCI(g) + HCl(g)
C12(S) + C3H6(g) ~ C3H6C12(g)
C]H6 is propylene (propene) (MW;;;:; 42.08)
allyl chloride (3-chloropropene) (MW ;;;:; 76.53)
C3H6CI2, is propylene chloride (l,2-dichloropropane) (MW;;;:; 112.99)
(a)
(b)
The species recovered after the reaction takes place for some time are Hsted in
Table
Species
C12
C)H6
C:lHSCl
C3H6Cl2
HCl
TABLEE9.8
gmat
141.0
1.0
4.6
24.5
4.6
Based on the product distribution assuming that no aUyl chlorides were
sent the feed, the following:
a. How much CI2 and C3H6 were to the reactor in g mol?
b. What was the limiting reactant?
c. What was the excess reactant?
d. What was the fraction conversion of C3H6 to C)HSCI?
e. What was the selectivity of C)HSCI relative to C3~C12 ?
f. What was the yield of C3HsCl expressed in g of C)HSCI to the g of C3H6
fed to the reactor?
g. What was the extent of reaction of the and second reactions?
h. In the application of green chemistry. you would like to identify classes
of chemical reactions that have the potential for process improvement.
particularly waste reduction. In this example the waste is HCl (g). The
C12 is not considered to be a waste because it is recycled. What is the
mole efficiency. fraction an element in the entering reactants
that emerges exiting products, for chlorine?
,
'.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-275-320.jpg)
![Sec. 9.2 Terminology for Applications of Stoichiometry 249"'
The table lists some of the compounds discharged into the river along with the LC50
value (the concentration that will kit] 50% of rainbow trout).
Compound
(1)
Thiometon (I)
Ethoxyethyl mercury
hydroxide (P)
DNOC(P)
Endosulfan (P)
! = insecticide; P "" pestl:CII;te
Estimated
discharge (kg)
3000-9000
1200-4000
18-200
600-2000
20-60
Estimated concentration
near discharge point (,...gIL)
600
500
100-430
LC50
(,...gIL)
6000
8000
3 to 1000
66 to 1250
1.4
What were the probable consequences of the discharge along the river to the
biota. drinking water, and benthic organisms? Note that the Rhine several dams to
provide water for navigation. What would concentration these compounds be as a
function of time at various towns downstream of Basel?
Looking Back
In this chapter we explained how the chemical reaction equation can be used to
calculate quantitative relations among reactants and products. We also defined a
number of terms used by engineers in making calculations involving chemical re~
actions.
GLOSSARY OF NEW WORDS
Con version The fraction of the or some material in the feed that is con-verted
into products.
Degree of eompletion The percent or fraction of the limiting reactant converted
into products,
Exeess reactant All reactants other than limiting reactant.
Extent of reaction € == nj
Vi
'<"'"FATFA ni is the number of moles of species i present in the system after the reaction
occurs, nio is the number of moles of species i present in the system
when reaction starts, and Vi the stoichiometric coefficient for species i in
chemical equation.
Limiting reactant The species in a chemical reaction that would theoretically run
out (would completely consumed) the reaction were to proceed to](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-280-320.jpg)
![264 The Chemical Reaction Equation and Stoichiometry Chap. 9
What would another company have to bid. to match the VWR bid if the bid they submitted
was for ferrous sulfate (FeS04 . H20)? For (FeS04 . 4HzO)?
*9.13 Three criteria must be met if a fire is to occur: (1) There must be fuel present; (2)
there must be an oxidizer present; and (3) there must be an ignition source. For most
fuels, combustion takes place only in the gas phase. For example. gasoline does not
bum as a liquid. However, when is vaporized. it bums readily.
A minimum concentration of fuel in air exists that can be ignited. If the fuel
concentration is less than this lower flammable limit (LFL) concentration, ignition
will not occur. The can be expressed as a volume percent. which is equal to the
mole percent under conditions at which the LFL is measured (atmospheric pressure
and 25° C). There is also a minimum oxygen concentration required for ignition of
any fuel. It is closely related to the La and can calculated from the LFL. The
minimum oxygen concentration required for ignition can be estimated by multiplying
the LFL concentration by the ratio of the number of moles of oxygen required for
complete combustion to the number of moles of fuel being burned.
Above the LFL. the amount of energy required for ignition is quite smalL For
example, a spark can easily ignite most flammable mixtures. There is also a fuel con~
centration called the upper flammable limit (UFL) above which the fuel-air mixture
cannot be ignited. Fuel-air mixtures in the flammable concentration region between
the LFL and the can be ignited. Both the LFL and the have been measured
for most of the common flammable gases and volatile liquids. The LFL is usually the
more important of the flammability concentrations because if a fuel is present the
atmosphere in concentrations above the UFL, it will certainly be present within
the flammable concentration region at some location. LFL concentrations for many
materials can be found in the NFP A Standard 325M, "Properties of Flammable liquids,"
published by the National Fire Protection Association.
Estimate the minimum permissible oxygen concentration for n-butane. The LFL
concentration for n-butane is 1.9 mole percent. This problem was originally based on a
problem in the text Chemical Process Safety: Fundamentals with Applications. by
D.A. Crowl and J.F. Louvar, published by Prentice Han, Englewood Cliffs. NJ, and has
been adapted from Problem 10 of the AIChE publication. Safety, Health, and Loss Prevention
in Chemical Processes by l.R. Welker and Springer, New York (1990).
·9.14 Odors in wastewater are caused chiefly by the products of the anaerobic reduction of
organic nitrogen and sulfur.containing compounds. Hydrogen sulfide is Ii major component
of wastewater odors; however. this chemical is by no means the only odor
producer since serious odors can also result in its absence. Air oxidation can be used
to remove odors, but chlorine is the preferred treatment because it not only destroys
H2S and other odorous compounds but it also retards the bacteria that cause the compounds
in the first place. As a specific example. HOC} reacts with H2S as foHows in
low pH solutions
HOCl+
If the actual plant practice calls for 100% excess HOCI (to make sure of the destrucrion
of the H2S because of the reaction of HOC] with other substances); how much
HOCI (5% solution) must be added to of a solution containing 50 ppm](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-285-320.jpg)
![262
m
18
15
Material Balances for Processes Involving Reaction Chap. 10
Reactor
9 H2
1----....... 12N2
8 NHa
Figure 10.1. A reactor to produce
NH3·
the respective three generation and consumption terms are related by the reaction
equation. Given the value for the generation NH3, you can calculate the values for
the consumption of and using the reaction equation. the generation and
consumption tenns are not independent, and you cannot specify more than one of
the values without introducing a redundant or inconsistent specification. In general,
if you specify the value for the generation or consumption of one of the N species in
a reaction, you are to calculate the values of the N - I other species from the
chemical reaction equation.
Here is where the extent of reaction S becomes useful. Recall that an open
system
where Vi is the stoichiometric coefficient of species i in
to Section 9.2). the reaction
VNH3 = 2
VH2 = -3
VN2 = - I
, and the extent of reaction can be calculated any species:
in
~=
- nNH3
vNHJ
n out - n~2
~=
H2 -
VHl
6 - 0
- =3 2
9 - l8
3
-3
12 - 15 = 3
-1
(10.2)
reaction equation (refer
You can conclude for the case of a chemical reaction that specifica-tion
of the extent of reaction provides one independent quantity that accounts for
all of values of the generation and consumption tenns the various species in
the respective implementations of Equation (10.]). The three species balances corresponding
to process in Figure 10.1 are](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-293-320.jpg)
![· l
~ t
lSec. 10.1 Species Material Balances
Solution
Steps 1, 2, 3, and 4
Assume the reactor is an open. steady-state process. Figure E 1 0.1 is a sketch
of the process with the known information placed on it.
Reactor
Feed 67% Converalon Product
100g mol nc~
4{)% CHe F.[><]) p "a2
!ro%C12 ... 'tiel
10%~
Figure EIO.l
StepS
Select as a basis 100 g mol feed
Step 4
~
~2
Yau have to determine the limiting reactant if you are to make use of the in~
formation about the 67% conversion. By comparing the maximum extent of reaction
(refer to Chapter 9) for each reactant, you can identify the limiting reactant.
-~~ -40
fJ1W:(CRa) = = - = 40
vCHc (-1)
-n~12 -50
~(CIl) = v0
2
= (-1) = 50
Therefore. CI4 is the limiting reactant. You can now calculate the extent of
reaction using the specified conversion rate and Equation (10.5).
- f nj~ (-0.61)( 40) .
, = = = 26.8 g moles reacting
Vir -1
Steps 6 and 7
The next step is to carry out a degree-of-freedom analysis
Number of variables: 11
~U..·na2' n~2' n~~4' ~i, nff6. n~t,Cl. n~ F. P, f
Number of equations: 11
Basis: F = 100
Species material balances: 5
CH4• C~. HO. CH3CI, N2](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-296-320.jpg)
![Material Balances for Processes Involving Reaction Chap. 10
Specifications: 4 (3 independent)
X~2S = 0.20 or xtH4 0.80i (nEo/nk1s) = 3. (nk1olnk2s) = 10
Implicit equatio~s: 2
'Lnf = P 'Znf = (redundant if you use both specifications in F)
The degrees of freedom are and the problem is exactly specified.
StepS
The species balances in pound moles
cations are:
introduction of most of the specifi-s:
156.3 = 0 3 §
H2S: n~2S = O.20F - 2 ~
S02: n§o2 = Fso:z - 1 e
H20 : nk:zo = 0 + 2f
CH4: ntH.. = O.80F + 0 (f)
remaining specifications are
n~o2 = 3nh2s
n~2o = 1 On~2s
Equations (a) through (g) comprise seven
unknowns.
Step S-
(a)
(b)
(c)
(d)
(e)
(0
(g)
equations and seven
If you solve the equations without using a computer, you shou]d by
culating ~ from (3)
Then Equation (d)
np
H20 --
Next. Equation (g)
mol
3 = 52.1 mol nn
I) = 104.21b mol H20](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-299-320.jpg)
![Sec. 10.1 Species Material Balances
and Equation (f) gives
n~02 :: 3(JO.4) = 31.2lb mo] S02
If you solve the rest of the equations in the order (b). (c), and (e), you find
F = 5731b mol
F S02 = 83.3 Ib mol
ntH4 = 458 Ib mol
Finally, you can identify H2S as the limiting reactant because the molar ratio of S02
to H2S in the product gas (3/]) is greater than the molar ratio in the chemical reaction
equation (211). The fractional conversion from Equation (10.5) is the consumption
of H2S divided by the total feed of H2S
-(-2)~ (2)(52.1)
f;:::: O.2F ;:::: (0.2)(573) = 0.91
Step 10
Because of the coincidence of the equality of moles of reactants and products
for this particular reaction, you cannot use the overall mole balance for this process
as a consistency check.
10.1-2 Processes Involving Multiple Reactions
269
In practice reaction systems rarely involve just a single reaction. A primary reaction
(e.g., the desired reaction) can occur, but there are invariably additional or
side reactions. To extend the concept of the extent of reaction to processes involving
multiple reactions, the question is do you just include a ~ for every reaction. The answer
is no. You should include in the species material balances only the ~i associated
with a (nonunique) set of independent chemical reactions called the mini·
mal* set of reaction equations. What this term means is the smallest set of chemical
reactions equations that can be assembled that includes all of the species involved in
the process. Analogous to a set of independent linear algebraic equations, you can
fonn any other reaction equation by a linear combination of the reaction equations
contained in the minimal set.
For example, look at the following set of reaction equations:
c + 1/20 2 ~CO
CO + 1/20 2 ~ CO2
·Sometimes called the maximal set.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-300-320.jpg)
![Sec. 10.1 Species Material BaJances
Step 8
Because the variables in Figure EIO.3 are yf and not nr. direct use of yf in
the material balances will involve the nonlinear tenns yf P. Consequently, to avoid
this situation, let us first calculate P using Equation (10.7):
S S R
P = :Ln}n + ~- ~ViJ
1'= 1 t:=1 J= I
6 1
= 1 + 4.76 + 2: :LVi- ~j
i=I)==1 J
= 5.76 + [( -1) + (_1/2) + (1) + 0 + (1) + 0] 0.9
+ [0 + ( ~2) + (-I) + 0 + (1) + (1)]0.15 =:; 6.28gmol
The material balances after entering the values calculated in Step 4 are:
n~~30H =:; YeHloH (6.28) = 1 - (0.9) + 0 = 0.10
Step 10
n~t = Y02 (6.28) = 1.0 - (V2)(O.9) - (~)(0.15) = 0.475
= YCH20 (6.28) = 0 + 1 (0.9) - 1 (0.15) = 0.75
= YH20 (6.28) = 0 + 1 (0.9) + 1 (0.15) = LOS
== yco (6.28) = 0 + 0 + 1 (0.15) = 0.15
= YN2 (6.28) ;;::: 3.76 0 - 0 == 3.76
You can check the value of P by adding all of the n?ut above.
Step 9
The six equations can be solved for the >'1':
YeHloH = 1.6%, Y~ = 7.6%, YN2 = 59.8%,
YeH20 = 11.9%, Y"20 = 16.7%, Yeo = 2.4%.
EXAMPLE 10.4 Analysis of a Bioreactor
A bioreactor is a vessel in which biological conversion is carned out involving
enzymes, microorganisms, and/or animal and plant cells. In the anaerobic fermentation
of grain. the yeast Saccharomyces cerevisiae digests glucose (C6HI20 6)
from plants to fonn the products ethanol (C2HsOH) and propenoic acid
(C2H3C02H) by the following overall reactions:
273 "](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-304-320.jpg)
![,, ~,~,. '
sec: to'.t Species Material Bal~nces
J .. I 4000 ( 0.88)
n nJlla = = 195.3
H20 18 .02
J"j iaJ _ 4000 ( 0.12) _
nC6~llo6 - 18 0 .1 - 2.665
Steps 6 and 7 .
The degree-of-freedom analysis is ¥ follows:
, Number of variables: 9
Number of equations: 9
Basis: 4000 kg of initial solution (equivalent to initial moles of
H20 plus moles of sucrose)
Species material balances: 5
H20. C6H120 6• ~H50H. C2H]C02H, CO2
Specifications: 4 (3 independent)
nlf;8a1 = 195.3 or nb~a~;06 = 2.665 (one is independent, the sum is F in mol)
F " J 120 ncD; = 44 = 2.727.
The degrees of freedom are zero.
Step 8
. .
The material balance equations. after introducing the known values for -the
variables, are: " ,. ,
(a)
(b)
C2HsOH: n~1f;oH = 0 + 2g1 + (O)f2 (c)
, ' , . F'rud ' . .
C2H3C02H~ ,' . nC~HJcolH = 0 + (O)~1 + (2)~2 (d)
CO2 2.727 = 0 + (2) €l + (O)~2 (e)
If yo~ do not use ~ computer to solve the equations. the sequence you should
use to solve t:l).em would be
(e) (b) simultaneously, and then solve, (a), (c), and (d) in order.
275](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-306-320.jpg)
![1 1 Species Material Balances 2n
Without With
Type of balance chemical reaction chemical reaction
Total ,balances
mass [ J []
Total moles [ ] [ J
Component balances
Mass of a pure compound [ ] [ ]
Moles of a pure compound { J fJ
Mass of an atomic species [ J [ ]
Moles of an [1 []
Explain how the extent of reaction is related to the fraction conversion of the limiting reactant.
Problems
Hydrofluoric acid can manufactured by calcium fluoride (CaF2) with sulfuric
acid (H2S04), A sample of f1uorospar (the raw material) contains 75% by weight
CaF2 and inert (nonreacting) materials. The pure sulfuric used in the process is
in 30% excess of theoretically required. Most of the manufactured HF leaves the re-action
chamber as a but a solid that contains 5% of the HF fonned. plus
CaS04, inerts. unreacted sulfuric acid is also removed from the reaction chamber. Assume
complete conversion of the occurs. How many kilograms of cake are produced
per 100 kg fluorospar charged to the process?
2. Corrosion of pipes in boilers by oxygen can be alleviated through the use of sodium sulfite.
Sodium sulfite removes oxygen from boiler feedwater by the foHowing reaction:
2Na2S03 + 02 ~ 2NaS04
How pounds of sodium sulfite are theoretically required (for complete reaction) to
remove the oxygen from 8.330,000 Ib of water (106 gal) containing 10.0 parts per million
(ppm) of dissolved oxygen at the same time maintain a 35.% excess of sodium sulfite?
3. Consider a continuous. steady-state process in the following reactions take place:
C6HI2 + 6H20 -4 6CO + 12H2
C6H12 + H2 -4 C6HI4
In process moles of C6H12 and 800 moles of H20 are fed into the reactor each
hour. The yield of H2 40.0% and selectivity of H2 relative to C6H14 is 12.0. Calculate
the molar flow rates of all five components in the output stream.
Discussion Problems
1. A scheme has proposed for use on the moon to generate oxygen. The idea is to pass
an electric current through molten silica to oxygen at the anode. is pr0-
duced at the cathode. What problems you anticipate might occur with the proposed
.-](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-308-320.jpg)
![278 Material Balances for Processes Involving Reaction Chap. 10
process if the silica contains contaminants such as iron? Will the process have to take
place at a high temperature? Where might the electric energy come from?
2. A liquid-emulsion membrane process that offers the promise of cleaning up waste streams
contaminated with metals was demonstrated by the Copper Co. on June 1. The process
was reported to recover 92.,-98% of the copper from solutions containing 320-1400 ppm
of copper at a feed rate of up to 4 Umin. The emulsion consists of globules of aqueous so~
lution surrounded by an organic phase that contains a reagent selective for copper. The
metal loaded emulsion is removed by flotation, and the copper collected by a high voltage
electrical field.
Does this process involve a reaction(s)? How would you involve the extent of reaction in
making material balances on the process?
10.2 Element Material Balances
In Section 10.1, you learned how to use species mole balances in solving problems
involving reacting systems. Equation (10.1), which included tenns for the generation
and consumption of each reacting species, was the basic equation from
which the other equations were derived.
As you know, elements in a process are conserved, and consequently you can .
apply Equation (10.1) to the elements in a process. Because elements are not generated
or consumed, the generation and consumption tenns in Equation (10.1) can be
ignored. Why not use e1ement balances to solve material balance problems rather
than species balances?
How wonderful that we have met with a paradox.
Now we have some hope oj making progress.
Niels Bohr
You can, but you must first make sure that the element balances are independent.
Species balances are always independent. Here is an illustration of the issue.
Carbon dioxide is absorbed in water in the process shown in Figure 10.2. The reac~
tion is
CO2(g) + H20Cf) --t H2C03Cf)
Three unknowns exist: W, F, and P, and the process involves three elements:
C, H. and 0. It would appear that you can llse the three element balances (in moles)
C: W(O) + F(l) = O.05P(1)
H: W(2) + F(O) = [0.05(2) + 0.95(2)]P = 2P
0: W(l) + F(2) = [0.05(3) + 0.95(1)]P = 1.10P](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-309-320.jpg)
![280 Material Balances for Processes Involving Reaction
In Steps 8 and 9 the element material balances are:
100 (0.40) = n~~(l) + n~ljtcl(1)
H: 100 (0.40)(4) = ~~(4) + n~61(1) + n~'ii3Cl(3)
Cl: 100 (0,50)(2) = n~Y;(2) + nrftl(l) + n~3Cl(1)
2N: 100 (0.10)(1) = n~~t(1)
Chap. 10
Substitute these equations for the species balances used in Example 10.1. As
extIeCI:ea. the solution of the probJem will be same as found in Example 10.1.
Example 10.3
degree-of-freedom analysis Steps 6 and 7
Variables: 9 (e1 and are not involved)
Equations: 9
~ ...... ,.". F;;:; 1
Element balances: 4 (independent)
C,H,O,N
Specifications: 3 (the same as in Example 10.3)
Implicit equations: 1 (the same as in Example 10.3)
of freedom are zero.
In Steps 8 and 9 the element balances are:
c: 1(1) + 4.76(0) = P[y~30H(1) + Y&20(1) + yto(l)]
H: 1(4) + 4.76(0) = P[YtH30H(4) + ytHzo(2) + Yk2o(2)]
/ ,-,
l(l) + LO().~~J= P[YtHJOH(1) + y&(2) + YtH20 (1)
+ Yk2o(1) + yto(l)]
1(0) 3.76
Substitute these equations for the species balances used in Example 10.3. The
solution of the problem not change.
It would be easier to use term yf' P = nf in the equations above in place
of the product of two variables, /; and](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-311-320.jpg)
![282
l
Material Balances for Processes Involving Reaction
StepS Basis: P = 100 g mol
Steps 6 and 7
The degree-of-freedom analysis is
Variables: 3
F,G,P
Equations: 3
Element balances: 2
H,e
Basis: P = 100
Chap. 10
If you calculate the degrees of freedom based on species balances, you have
as unknowns five species molar flows, P, and R (the number of independent reaction
equations which equals the number of unknown extents of reaction), and as
equations five species balances, one specification (the basis). and one implicit equation
so that
degrees of freedom = (6 + R) - (5 + 2) :;: R - 1
Thus, although you do not know what R is from the problem statement, for zero degrees
of freedom to occur, R = I, that is, one independent reaction equation exists.
You can be assuaged that if you do not know R and a minimum reaction set, and
thus you do not know how to involve the respective ~ in the species equations. you
can fall back on element balances.
StepS
The element balances after introducing the specification and basis are:
C: F(8) + G(O) = 100[(0.195)(3) + (0.594)(4) + (0.211)(5)]
H: F(l8) + G(2) = 1 OO[ (0.195)(8) + (0.594)( 10) + (0.211 )(12)]
and the solution is
F = 50.2 g mol G=49.8gmol
The ratio
H2 consumed 49.8 g mol
= = 0.992
CgHI8 reacted 50.2 g mol
You will find that employing element material balances can be simpler than
employing the extent of reaction for problems in which the reaction equations are
not specifically known or must be inferred. as shown in the examples in the next
section.
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-313-320.jpg)
![288 Material Balances for Processes Involving Reaction Chap. 10
and air (02 plus N2) produce electricity p]us CO2 and H20. Special membranes and
catalysts are needed to promote the reaction of CH4•
Based on the given in Figure EIO.8, arc asked to the COffi-position
of the products in
Solution
Steps 1, 3, and 4 This is a steady-state process with reaction. Can you as-sume
a complete reaction occurs? Yes. No CH4 appears in p, The system is the fuel
cell (open, steady state). Because the process output is a composition will
be mole fractions or moles, hence it more convenient to use moles rather than
mass in this problem even though the quantities of CH4 are stated in kg. You
can carry out the necessary preliminary conversions as fol lows:
300 kg A 1 mol A
----=---= -..;;;.......-- = ]
29.0 kg A
kg mol A in
16.0 kg 1 kgmolC~ .
1 k CH
= 1.OOkgmolCH4 ln
6.0 g 4
10.35 kg mol 0.21
----'''---...=. = 2.17 kg mol O2 in
10.35 kg mol 0.79 kg mol N2 .
1 kg mol A = 8.18 kg mol N2 In
Step S Since no particular basis is designated we will pick a convenient basis
Basis: 16.0 CH.t F 1 mol CH4
Steps 6 amd 7
The degree-of-freedom analysis is (A has calculated):
Variables: 8
FP P P P P A A , ,neol , nN1, no2, nH201 nOl' nN2,
Equations: 8
Basis: F = 1 kg mol
Element material balances: 4 (independent)
C,H.O,N
Specifications and calculated quantities: 2
n~l = 2.17. n~2 = 8.18
Implicit equation: 1
Inr =
The degrees of freedom are zero,
j](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-319-320.jpg)
![298 Material Balances for Processes Involving Reaction 10
ROBl MS
*10.1 Pu re I. n ph a se a reactor. of this is converted to B through the re-action
A ~ 3 What the mole fraction of in the exit "'~ .. "." .....
What is extent of "' ... "'."' .... "
"'10.2 One of the most common commercial methods for the production pure silicon that
is to be for the manufacture of semiconductors is the Si~mens process
ure PIO.2) of chemical vapor deposition (CVD). A chamber contains a heated silicon
rod, and a mixture of high purity trichlorosilane mixed with high purity hydrogen that
is passed over the Pure silicon (EGS-electronic grade silicon) deposits on the rod
as a polycrystalline solid. (Single crystals of Si are made by subsequently' melt-ing
the EGS and drawing a single from melt.) The reaction is: Hig) +
SiHC]3(g) ~ SICs) + 3HCI(g). .
The rod initially has a mass of 1,460g, and the mole fraction of Hi in exit"
is 0.223. The mole fraction of in the to the reactor 0.580, and feed enters
at the rate kg mollhr. What will the mass the rod at the of 20 ..... u .. ,..,'"
Gu Gu
SiHCI3 --~ ',...--....... Hel
H2 L-'====--1 SiHCla
H2
Figure PI0.l
"10.3 A low-grade pyrite containing 32% S is mixed with 10 Ib of pure sulfur per 100 lb of
pyrites so the mixture will bum readily with fonning a gas analyzes
13.4% S02' 2.7% °2, and 83.9% N2• No sulfur left in cinder. Calculate the percentage
of the sulfur fired burned to S03' (The S03 is not detected by the analysis.)
"'10.4 Examine the reactor in Figure PlO.4. Your boss something has gone wrong with
the yield of CH20, and it is up to you to find out what the problem is. You start by
making material balances (naruralIy!). Show all calculations. Is there some problem?
Methanol
(CH30H) ---IIII-i Reactor
Air
Figure PIOA
Product
62.S N2
1 02
H20
4.6CH20
12.3CHsOH
1.2 HCOOH](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-329-320.jpg)
![300 Material Balances for Processes Involving Reaction Chap. 10
Assume the reactor is loaded with wafers baving a silicon oxide surface, a flow of
and 50% nitrogen is started, and the reactions proceed.
he reaction 10% of the is consumed. What is the composition of the ex~
haust strea ?
In the anae obic fermentation of grain, the yeast Saccharomyces digests
glucose fro plants to form the products ethanol and propenoic acid by the following
overall rea tions:
actio 1: C6HI20 6 ~ 2C2HsOH + 2C02
Reaction 2: Hl20 6 ~ 2C2H3C02H + 2H20
In an open. flow reactor 3500 kg of~ 12% glucose/water solution flow in. During fermentation,
120 kg of carbon dioxide are produced together with 90 kg of unreacted glucose.
What are the weight percenrs of ethyl alcohol and propenoic acid that exit in the
broth? Assume that none of the glucose is assimilated into the bacteria.
Semiconductor microchip processing often involves chemical vapor deposition
(CVD) of thin layers. The material being deposited needs to have certain desirable
properties. For instance, to overlay on aluminum or other bases. a phosphorus pentoxide-
doped silicon dioxide coating is deposited as a passivation (protective) coating
by the simultaneous reactions
Reaction 1: SiH4 + 02 -7 Si02 + 2H2
Reaction 4PH3 + 502 -7 2P20 S + 6H2
Determine the relative masses of SiH4 and PH) required to deposit a film of 5% by
weight of phosphorus oxide (P20S) in the protective coating.
··10.11 Printed circuit boards (PCBs) are used in the electronic industry to both connect and
hold components in place. In production, 0.03 in. of copper foU is laminated to an insulating
plastic board. A circuit pattern made of a chemically resistant polymer is
then printed on the board. Next, the unwanted copper is chemically etched away by
using reagents. If copper is treated with Cu(NH)4C12 (cupric ammonium
chloride) and NH40H (ammonium hydroxide), the products are water and
Cu(NH3)4C1 (cuprous ammonium chloride). Once the copper is dissolved, the polymer
is removed by solvents leaving the printed circuit ready for further processing. If
a single-sided board 4 in. by 8 in. is to 75% of the copper layer removed using
the reagents above, how many grams of each reagent will be consumed? Data: The
density of copper is 8.96 glcm3.
··10.12 The thermal destruction of hazardous wastes involves the controlled exposure of
waste to high temperatures (usually 9000c or greater) in an oxidizing environment.
Types thermal destruction equipment include high-temperature boners~ cement
kilns. and industrial furnaces in which hazardous waste is burned as fueL In a properly
designed system, primary fuel (100% combustible material) is mixed with waste
to produce a feed for the boiler.
(a) Sand containing 30% by weight of 4,4'-dichlorobiphenyl [an example of a poly·
chorinated biphenyl (PCB)] is to be cleaned by combustion with excess hexane](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-331-320.jpg)
![302 Material Balances for Processes Involving Reaction Chap. 10
*10.21 In underground coal combustion in the phase
CO + 1/202 ~ CO2
reactions take place including
+ 1/2°2 ~
CH4 + 312°2 ~ CO + 2H20
where the CO, and CH4 come from coal pyrolysis.
If a gas phase composed of CO: 13.54%, CO2: 15.01 %, CH4:
3.20%, and the balance N2 burned with 40% excess air, (a) how much is needed
per 100 moles of gas, (b) what will the analysis of the product gas on a wet basis?
*10.22 Ethanol (CH3CH20H) is dehydrogenated in the presence of air over a catalyst. and
the following reactions take place
CH3CH20H~ +
2 CH3CH20H + 302 ~ 4C02 + 6H2
2 CH3CH20H + 2H2 -4 4CH4 + 02
Separation of product, (acetaldehyde), as a liquid leaves an output gas
with the following Orsat analysis:
CO2:O.7%. °2:2.1 %, CO:2.3%. H2:7.1 %, CH4:2.6%, and N2:85.2%
How many kg acetaldehyde are produced per of ethanol fed into the process?
"10.23 So]vents emitted from industrial operations can become significant pollutants it: not
disposed of properly. A chromatographic study of the waste exhaust gas from a synthetic
fiber plant has the following analysis in mole percent:
CS2 40%
S02 10
50
It has been suggested that the gas be disposed of by burning with an excess of air.
The gaseous combustion products are then emitted to the air through a smokestack.
The local air pollution regulations say that no stack gas is to analyze more than 2
S02 by an Orsat analysis averaged over a 24-hr period, Calculate the minimum
percent excess air that must be used to stay within this regulation.
*10.24 The products and byproducts from coal combustion can create environmental prob-lems
if the combustion process is not out properly. boss asks you to
out an analysis of the combustion in boiler No.6. carry out the work as-signment
using existing instrumentation, and obtain following
Fuel analysis (coal): 74% 14% H, 12% ash
Flue gas analysis on a dry 12.4% C02t 1.2% CO, 5.7% 02 and 80.7% N2
Wbat are you going to report to your boss?
"'10.25 The Air Act requires automobile manufacturers to warrant their control sys-tems
as satisfying the emission standards for 50,000 mL It requires owners to have](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-333-320.jpg)
![Chap. 10 Problems
their engine control systems serviced exactly according to manufacturers' specifica~
tions and to always use the correct gasoline. In testing an exhaust having a
known Orsat analysis of 16.2% CO2, 4.8% 02' and 79% N2 at the oudet, you find to
your that at the end of the muffler the Orsat analysis is 13.] % CO2- this
discrepancy caused by an air leak into the muffler? (Assume that the analyses are
satisfactory,) If so, compute the moles of air leaking in per mole of exhaust gas leaving
the engine,
/*"'10.26 One of the products of sewage treatment is sludge. After microorganisms grow in the
activated sludge process to remove nutrientc.; and organic material, a substantial
amount of wet sludge is produced. sludge must dewatered. one of the most
expensive parts of most treatment plant operations.
How to dispose of the dewatered sludge is a major problem. Some organizalions
seH dried sludge for fertilizer, some spread the sludge on farmland, and in some
places it is burned. To bum a dried sludge. fuel oil is mixed with it, and the mixture is
burned in a furnace with air. If you collect the foHowing analysis for the sludge and
for the product gas
Sludge(%) Product Gas (%)
S 32 S02 1.52
C 40 CO2 10.14
H2 4 °2 4.65
°2 24 N2 81.67
CO 2.02
(a) Determine the weight percent of carbon and hydrogen in the fuel oil.
(b) Determine the ratio of pounds of dry sludge to pounds of fuel oil in the mixture
to the furnace.
*10.27 Many industrial processes use acids to promote chemical reactions or produce acids
from the chemical reactions occurring in the process. As a result, these acids many
time end up in the wastewater stream from the process and must be neutralized as
part of the wastewater treatment process before the water can be discharged from the
process, Lime (CaO) is a cost effective neutralization agent for acid wastewater.
Lime is dissolved in water by the following reaction:
CaO + 1/202 ~ Ca(OHh
which reacts directly with acid. for H2S04,
H2S04 + Ca(OHh ~ CaS04 + 2H20
Consider an wastewater steam with a flow rate of lOOO gaVrnin with an acid
concentration 2% H2S04, Determine the flow rate of lime Ib/rnin necessary to
the acid in this stream if 20% excess lime is used. Calculate the production
rate of CaS04 from this process . ~ons/yr, Assume that the specific gravity of the
acidic wastewater stream is 1.05,](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-334-320.jpg)
![322 Material Balance Problems InvoMng Multiple Units
EXAMPLE 11.4 An.!lysis of a Sugar Recovery Process Involving
Multiple Serial Units
Figure E 11.4 shows the process and the known data. You are asked to calculate
the compositions of every flow ~tream •. and the fraction of the sugar in the cane
that is recovered. .
F Cone
16% Sugar
25% Water
59% Pulp
.--"-........
M Sugar,
10001b/l1r
t
. L--....- L
Crystallizer '-----Water
[ H
Mill
J
Screen 1-----1001 Evaporab f--- Water
13% Sugar
1.....-""""11""'---' 14% Pulp 1.....-""""11""'---'
D Boga5&e
80% Pulp
Solution
Steps 1, 2, 3, and 4
Solids G Conloin
95% Pulp Ftgure Ell."
An of the known data have been placed on Figure E 11.4. If you examine the
figure, two questions naturally arise~ What basis should you pick, and what system
should you pick to start the analysis with? Some potential bases and systems lead to
more equations to be solved simultaneously than others. You could, pick as a basis
F = 100 lb, M::::: 1000 Ib (the same as I hour), or the value of any of the intermediate
flow streams. You could pick an overall process as the system with which to start,
or any of the individual units, or any consecutive combinations of units.
StepS
Basis: 1 hour ( M = 1000 Ib)
Another important decision you must make is: what are the comppsitions of
streams D, E. G. and H? Stream F has three components, and presumably stream K
contains only sugar and water. Does stream H contain pu]p? Presumahly not, be·
cause if you inspect the process flowsheet you will not find any pulp exiting anywhere
downstream of the evaporator. Presumably streams D and G contain sugar
and water because the problem implies that not all of the sugar in stream F is recovered.
What happens if you assume streams D and G contain no water or no sugar?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-353-320.jpg)
![324 Material Balance Problems Involving Multiple Units Chap. 11
lb mass fraction
D;;;;;. 16,755 wf = 0.174
£=7,819 lLI~ = 0.026
F;;;;;. 24,574 wij, = 0.73
G= 1,152 lJ)~ = 0.014
H= 6,667 w~ = 0.036
J = 4,167 w!(, = 0.85
K= 2,500 It)~ = 0.60
L= 1,500
M=lOOO
The fraction of sugar recovered is [1000/(24.574)] (0.16) = 0.25.
EXAMPLE 11.5 Production of a Hormone in Connected Reactors
Figure E 11.5 shows two different reactor configurations to produce a honnone.
Option 1
Option 2
F J
F = 15 Uhr
:.: I :z I
F F
~
csln = 10 gil X1 x2
1 2
Figure E 11.5
The relation that gives the growth (generation in Equation (10.1» of the hormone in
each vessel in the steady state is the Monod equation
J-Lmu.Cs
1-'=
Ks + C!
where p, = the specific growth rate, llhr
Ikrrw. ;: maximum specific growth rate (the state at which the concentration
of nutrient becomes the limiting factor in the growth of cells).
IIhr
c of = nutrient concentration in the vessel, g nutrientIL substrate
~ = Monad constant, g nutrientIL substrate](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-355-320.jpg)
![328 Material Problems Involving Multiple Units Chap. 11
3. In for the production of pure acetylene, C2H2 (see SATIIP3), pure
methane (CH4), and pure oxygen are combined in the burner, .. lh ......... the following reactions
occur:
C~ + 202 --;. 2H20 + C02
C~ + I! --;.2H20 + CO
2C14 --;. + 3H2
(1)
(2)
(3)
a. the ratio of £he moles of to moles of CH4 fed (0 the burner.
b. On of 100 Ib rna] of gases leaving the condenser, calculate how many pounds
of water are removed by the COI1tOellser
c. What is overall percentage yield product (pure) C2H2• on the carbon in
the natural entering the burner?
L-
.ca.>. ::I
ttl
tJ,
CH4
Waste
H2O
Solvent
CO2 and C2H2
Solvent
and
Figure SA TllP3
The gases from the burner are cooled in the condenser removes all of the
analysis of the gases leaving the condenser is as follows:
These gases are sent to an
removed with the solvent.
%
8.5
CO 58.3
CO2 3.7
4.0
Total 100.0
where 97% of the C2~
solvent from the ~h(e,nrhpr
essentially all the CO2 are
sent to the CO2 stripper,
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-359-320.jpg)
![-346 Recycle, Bypass, Purge, & the Industrial Application Material, Balances Chap. 12
Problems
1. How many recycle streams occur in Figure SAT12.lP I?
P1 -
! XI cR4~a
-F-1 ---1:1 ~-------.
P2
X2
Figure SAT12.1Pl
2. The Hooker Chemical Corporation operates a process' in Michigan for the purification of
Hel. Figure SATl2.lP2 shows the flow sheet for the Hooker ,process. The streams from
the bottoms of the five towers are liquid. The streams from ', the tops of the towers are
gases. Hel is insoluble in the HCB (hexach1oro~utadiens), The various stream compositions
are shown in Figure SAT12.1 P2.
How many recycle streams are there in the Hooker process?
wt.% Helga.7 t
CI2 9.3 , (Feed)
Pure
wt.% eCI.
CC1490.7
HCI 5.6
Cli" 18.2
wt.%
CCISO.5
HC149.5
~ II
IV,
289~,
.L.3.40atm+
.Q-.....
Compressor
. wt.'%
HCS68.S
CC1431.S
v
~
wt.% "
HCS88.S
eCI. 11.5 ~ Pure HCB
"'"---~~ HCI and CI2 recycled to chlorination units
Figure SA T12.~ P2
]](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-377-320.jpg)
![350 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12
Solution
Open, steady-state process
a. Steps 1, 2, 3, and 4
Figure E12.la contains the information needed to solve the problem.
Step 5
Basis: 10,000 lb fresh feed (equivalent to 1 hour)
Steps 6 and 7
The unknowns are W, G, P, and R. You can make two component balances
about three systems: the mixing point A, the evaporator, and the crystallizer as well
as two overall component balances. You can also make total balances for the same
selection of systems. What balances should you choose to solve the problem? If you
plan to put four equations in an equation solver, it does not make any difference as
long as the equations are independent. But if you solve the problem by hand, you
should count the number of unknown variables involved for each of the three subsystems
and the overall system as follows:
Component balances
Mixing point
Evaporator
Crystallizer
Overa11
Unknowns
R plus feed (and compositions) to evaporator
(not labeled)
W, 0, and feed to evaporator
0, P, andR
WandP
You can see that by using just two overall component balances (you can substitute
the overall total balance for one component balance) you can determine the values
of Wand P. Consequently, you should start with overall balances.
Steps 8 and 9
Overall NaOH balance
(0.4)(10,000) = [0.95 + (0.45) (0.05)]P
P = 41131b
Overall H20 balance
(0.6) (10,000) = W + [(0.55)(0.05)](4113)
W= 58871b
(or use the overall total balance 10,000 = 4113 + W)
The total amount of NaOH exiting with P is
[(0.95) + (0.45)(0.05)](4113) = 4000 Ib
I
.J.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-381-320.jpg)
 + 0.45H
H20 balance on the crystallizer
0.5G == [(0.05)(0.55)(4113) + 0.55H
H== 38.870 lb
An overall NaOH balance gives the new F
Overall NaOH balance
O.40F == 0.45(38.870) + 4000
F= 53,730 Ib
Note that without recycle, the feed rate must be times larger than with recycle
to produce the same amount of product, not to mention the fact that you would have
to dispose of a large volume of filtrate.
SELF .. ASS SSM NT T ST
QuestIons
Why have we not considered the buildup of material in recycle streams this chapter?
2. Under what circumstances might material be accumulated or depleted in a recycle stream?
3. Can you make material balances in both steady-state and unsteady-state flow processes
that involve recycle?
4. Can you fonnulate sets of equations that are not independent if recycling occurs in a system?
Problems
1. ball mill grinds plastic to make a very fine powder. Look at Figure SAT12.2Pl.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-383-320.jpg)
![Sec. 12.2 Recycle without Chemical Reaction
;- ~L.. ___P_ a __ rtl_ c_l e_- rCo~l1ect_o_r_...J
I
I
t
I Uncollected powder 1-------.... to waste
Ball
Mill
Product (10,000 kg) fine powder
Figure SAT12.2Pl
353
At the present time 10,(X)() kg of powder are produced per day. You observe that the
process (shown by the solid lines) is inefficient because 20% of the feed is not recovered
as powder-it goes to waste.
You make a proposal (designated by the dashed lines) to recycle the uncollected materia]
back to the feed so that it can be remiUed. You plan to recycle 75% of the 200 kg of uncollected
material back to the feed stream. If the feed costs $ L201kg, how much money
would you save per day while producing 10,000 kg of fine powder?
2. Sea water is to be desalinized by reverse osmosis using the scheme indicated in Figure
SAT12.2P2. Use the data given in the figure to detennine: (a) the rate of waste brine removal
(B); (b) the rate of desalinized water (called potable water) production (P); (c) the
fraction of the brine leaving the reverse osmosis cen (which acts in essence as a separator)
that is recycled.
1000lblhr
Sea water
3.1% salt
Brine Recycle
4.0%
salt
Reverse
osmosis
cell
P
Desalinized Water
500 ppm salt
Figure SATll.2P2
Brine waste (B)
5.25% salt
3. A material containing 75% water and 25% solid is fed to a granulator at a rate of 4000
kglhr. The feed is premixed in the granulator with recycled product from a dryer. which
follows the granulator (to reduce the water concentration of the overall material fed into](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-384-320.jpg)
![.12.3 Recycle with Chemical Reaction 355
Discussion Problems
1. of limitations in supply as well as economics, many industries reuse their water
over and over again. example~ recirculation occurs in cooling towers, boilers, powdered
coal transport, multistage evaporation~ humidifiers. and many devices to wash agri.
cultural products.
Write a report discussing one of these processes, and include in the report a descrip-tion
of the process, a simplified flow problems with recycling, the extent of purge.
and, if you can find the information, the made by recycling.
A of with forward and reverse flow, and among is known as a
cascade. The enrichment of natural uranium by of uranium hexafluoride is a
well-known example of a process involving a cascade. Deve]op a flowsheet of a cascade.
and indicate what material balances are for individual units at the beginning, middle,
and end of the cascade,
The Celanese company appealed to the Court the application an rule
concerning benzene leakage to the atmosphere, Four of their plants used benzene in their
processes. recycling the benzene over and over the'plants. The company believed that
EPA rule on use applied to inventory (storage) of benzene. 'which was tess than
the lOoo-metric-ton regulatory threshold "use," whereas the EPA interpretation of the
"use' the benzene was it should counted each it circulated through a plant. The
EPA argument was that every time the benzene circulated through a plant, it had the same
chance of leaking into the as new benzene would. What is your opinion about this ar-gument?
(The U.S. Court of Appeals with EPA.)
12.3 Recycle with Chemical Reaction
The most common application of recycle for systems involving chemical reaction
is the recycle of reactants, an application that is used to increase th~ overall con-in
a reactor. Figure 12.9 shows a simple example for reaction
A-tB
From data in 12.9 you can see that the steady-state material balances for
the mixer, reactor, and separator are satisfied. Also, that for the overall material
balance if you calculate the extent of reaction you will find it 100 reacting g molls.
Reactor
100 gmalls
I--~ B
900 gmolls A
Figure 12.9 A simple recycle system with chemical](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-386-320.jpg)
![364 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12
Accumulation Input Output Generation Consumption
o :: {l00 (0.10) + Rw] - Rw + 0 -DAD [100 (0.10) + Rw]
Rw :: 15 kg of component in the recycle stream
Next, pick the mixer as the system.
Component balance: 100 (0.10) + 15 :: 0.03 FI
Total balance: R= - 100
15
w = - = 0.0205
SELF-ASSESSMENT TEST
Questions
1. If the components in the feed to a process appear in stoichiometric quantities and the subsequent
separation process is complete so that all of the unreacted reactants are recycled,
what is the ratio for reactants in the recycle stream? " .
2. Answer the following questions true or
a. The general material balance applies for processes that involve recycle with reaction
as it does for other processes.
b. The key extra piece of information in material balances on processes with recycle in
which a reaction takes place is the specification of the fraction conversion or extent of
reaction.
c. The degrees of freedom for a process with recycle that involves chemical reaction are
the same as for a process without recycle.
Cite two reasons using recycle in a process.
Problems
A catalytic dehydrogenation process shown in SATI2.3Pl, produces 1,3 butadiene
(C4H6) from pure normal butane (C4HIO). The product stream contains 75 mollhr of
H2 and 13 mollhr C4HIO as well as C4H6. The recycle stream is 30% (mol) C4HlO and
70% (mol) C4H6• and the flow is 24 mollhr.
Pure ~---to~r--""
C4H10
F :::: ? moles/hr
Recycle xC.Ho::::: 0.30
XC,He :: 0.70
Figure SA TI2.3Pl
1---100- nH~ =: moles/hr
"C,H10 :: 13 moles
"C4H6 ==?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-395-320.jpg)
![12.4 Bypass and Purge
In Out Consumed co:
[32.5 + Ry] - [y(R + P)J = (32.5 + Ry)(O.lS)
(f)
Equation (8) can be substituted into Equations (b) through (0. and the resultfive
equations solved by successive substitution or by using a computer
The resulting values obtained are (in moles)
E CH]OH 31.25
p purge 6.25
R recycle 705
x H2 0.768
Y CO 0.200
Z CH4 0<032
Step 10
Check to see each of the balances (b)-(f) is satisfied.
If you want use the. extent of reaction to make the calculations. you must
first calculate gmax for CO H2• and then find that was the limiting reactant.
You could reach the same conclusion by inspection. Then
0.18 E = 0 + (1)~ would equations to use.
SELF .. ASS SSM NT T ST
Questions
1. Explain what bypass means in words and also by a diagram.
2. Answer the following questions true or
371 -'
a. Purge is used to maintain a concentration of a minor component of a process stream
below some set point so that it not accumulate the
b. Bypassing means that a process stream the process in advance of the to the
process.
c. A trace component in a or produced a reactor has negligible on the
overall material balance when occurs.
3. the waste stream the same as a purge stream a process?
Problems
1. In the famous process (Figure SAT12.4PI) to manufacture ammonia, me reaction
is carried out at pressures of 800 to 1000 atm and at 500 to 600°C using a suitable catalyst
Only a small fraction of the material entering the reactor reacts on one pass, so recy-is
needed. Also, because nitrogen is obtained from air, it contains almost 1 %
rare (chiefly that do not react. The rare gases would continue to build up in](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-402-320.jpg)
!['-.
f "
, , " . " ~
386 Recycle, BYRassJ P.tlrg~,··& the· lndustria~ Application Material Balances Chap. 12
.. , I r* j.~' .
, '~." ,'1' "
···12.13 B~nzene. !o]uene -and~.~~r:aroma~c compounds can be recovered by solvent extraction
witli~lfui:!dioxide~;As,an example. a catalytic refonnate stream containing 70%
by ~ejg~t ~nzene and 30% non-benzene material is passed through the countercurrent
~~~tivS' recovery scheme shown in the diagram in Figure P12.13. One
thot,sana kg-of. the reform ate stream and 3000 kg of sulfur dioxide are fed to the system'
'hour. The benzene product steam contains O.IS.kg of sulfur dioxide per of
benzene. the waste steam contains all the initially charged non-benzene material as
weB ~ O~5 kg of benzene per kg of the non-benzene material. The remaining component
in'the waste steam is the sulfur dioxide.
(a) How many kg of benzene are extracted per hour (are in the product stream)?
(b) If 800 kg of benzene containing 0.25 kg of the non-benzene material kg of
benzene are flowing per hour at paint A and 700 kg of benzene containing 0.07
of the non-benzene material per kg of benzene are flowing at point B. how
many kg (exclusive the sulfur dioxide) are flowing at points C and D?
O.
Product benzene includes --.;;;......--.;;;;.
kgBz
B -
Sulfur dioxide
3000 kglhr
3
Product stream
- en
Figure P12.13
*12.14 Examine Figure PI 14 (data for I hour),
f
2
A
"-""
~
00
1000 kglhr
catalytic reformate
70%
a. What the single pass conversion H2 in the reactor?
b. What is the single pass conversion of CO?
c. What the overall conversion of H2?
d. What is the overall conversion of CO?
1
Waste
0.25 kg Bz
kg the non-Bz](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-417-320.jpg)
![Sec. 13.1 The Ideal Gas Law
EXAMPLE 13.5 Calculation of the Specific Gravity of a Gas
What is the specific gravity of N2 at 800 P and 745 mm Hg compared to air at
80°F and 745 mm Hg?
Solution
One way to solve the problem is to calculate the densities of N2 and air at
their respective conditions of temperature and pressure, and then calculate the specific
gravity by taking a ratio of their densities. Example 13.4 covers the calculation
of the density of a gas, and therefore, to save space, no units will appear in the intermediate
calculation here:
Basis: 1 ft3 of air at Soop and 745 mm Hg
1 492 745 29 3
- 540 760 359 - = 0.0721 Ih/ft at 800P and 745 mm Hg
Basis: I ft3 of N2 at Soop and 745 mm Hg
1 492 745 28
- - - - - = O.06971b/ft3 at 80°F and 745 mm Hg
540 760 359
)
0.0697 Ib N2/ft3 at BO°F, 745 mm Hg
(sp gr N2 = = 0.967-----------.;;;-
0.0721 Ib air/ft3 air at 80°F. 745 mm Hg
,-
409
Did you note from Example 13.5 that for a gas and reference at the same temperature
and pressure, the specific gravity is just the ratio of the respective molecular
weights? You can gerive this result as foHows. Let A be one gas and B be another.
Keep in mind that V for a gas is the volume per mole and not the volume per mass.
Thus
...
pV = RT or
p=
" 1
V = = [RTlp]
p/MW
(p)(MW)
RT
sp gr - :; - (:::.' ::.:)(~:)(;;) (13.3)
and at the same temperature and pressure for A and B the specific gravity is just the
ratio of the respective molecular weights.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-440-320.jpg)
![410 Chap. 13
SELF .. ASSESSMENT TEST
Questions
1. What is the idea] gas law? Write it down,
2. What are the dimensions of T. p, V, n, and R?
3. List the standard conditions for a gas in the SI and AE systems of units.
4. How do you calculate the density of an ideal gas at S.c.?
5. Can you use the respective specific molar densities (mole/volume) of the gas and the reference
gas to calculate the specific gravity of a gas?
6. Does the value of R depend on pressure, temperature, density, or number of moles of a
gas?
7. If a gas meets these three conditions, will it behave like an ideal
a. The volume occupied by molecules is so sman compared to total space that essentially
the molecules occupy no
b. The collisions are so few that it can be said that no energy is lost due to collisions.
c. The average distance between the molecules is great enough to neglect the effect of
the intermolecular forces.
Problems
1. Calculate the volume in ft3 of 10 lb mol of an ideal gas at 68°F and 30 psia.
2. A steel cylinder of volume 2 m3 contains methane (CHJ at 50°C and 250 kPa
solute. How many kilograms of methane are in the cylinder?
3. What is the value of the ideal gas constant R to use if the pressure is, to expressed in
atm, the temperature in kelvin, the volume in cubic feet, and the quantity of material in
pound mo]es?
4. Twenty-two kilograms per hour of CH4 are flowing in a gas pipeline at 30°C and 920 mm
What is the volumetric flow rate of the CH4 in m3 per hour?
5. What is the density of a gas that has a molecular weight of 0.123 kglkg mol at 300 K and
1000 kPa?
6. What is the specific gravity of CH4 at 70°F and 2 atm compared to air at S.c. ']
7. An automobile tire is j nflated to a pressure of 35 psig at a temperature of OOE Calculate
the maximum temperature to which the may be heated without the gauge pressure ex-ceeding
50 psia. (Assume that the volume of the tire does not change).
Thought Problems
1. A candJe is placed vertically in a soup plate, and the soup plate filled with water. Then the
candle is lit. An inverted water is carefully placed over the candle. The candle soon
goes out, and the water rises inside the It is often said that this shows how much
oxygen in the air has been used up. Is this conclusion correct?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-441-320.jpg)
![Sec. 13.1 The Ideal Gas Law 411- -
2. A scientific supply house markets aerosol-type cans containing compressed helium for
filling balloons, doing demonstrations, and the Like. On the la~1 there appears the notice:
"Because the can contains helium, it quite naturally feels empty. It is actually lighter
full than empty." Is this statement correct? If so, why? If not, why not?
3. Some reviewers of books and articles have suggested that the gas constant R in the ideal
gas equation be forced to take the value of unity (1). What would this step require as far
as using the ideal gas equation?
4. In an article by P. Hickman in Physics Teacher, vol. 25, p. 430 (1987) it was suggested
that the density of air be determined using marshmallows as follows:
1. Put the marshmallows in a vacuum so they expand and release trapped air. Repressurize
the vessel and compress the marshmaUows to a fraction of their original
volume. "
2. Weigh the marshmallows on a balance before and after the vacuum treatment. The difference
should be the mass of air trapped inside.
3. Measure their volume before and after the vacuum treatment. The difference should be
the volume of the air trapped inside.
Is this a sound way of getting the density of air?
Discussion Problems
1. Three identical glasses are arranged as in Figure DP13.1(pl). Hollow stirrers are also
needed. Glasses A and B are comp]etely filled with water (by submerging them jointly in
a bucket or sink and joining the mouths before removing), and C is empty. Glasses A and
B are carefully placed on a few hollow stirrers as shown in the figure. How can most of
the water in glass A be transferred to glass C without ever touching or moving the glasses
or their supporting stirrers?
Glasses filled with water
==~=T==J4-- Supporting hollow straws
C Empty glass
Additional hollow straws ~~
Figure DP13.IPl](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-442-320.jpg)
![414 Ideal Gases Chap. 13
A + B at 700 kPa and 20°C for this mixture. You could say that gas A exerts a partial
pressure of 300 kPa and gas B exerts a partial pressure of 400 kPa in tank C.
Of course you cannot put a pressure gauge on the tank and check this conclusion
because the pressure gauge will read only the total pressure. These partial pressures
are hypothetical pressures that the individual gases would exert if they were
each put into separate but identical vo]umes at the same temperature. In tank C the
partial pressures of A and B are according to Equation (13.5)
PA = 7oo(~) = 300 kPa
PB = 700( ~) = 400 kPa
EXAl1PLE 13.6 Calculation of the Partial Pressures of the
Components in a Gas from a Gas Analysis
Few organisms are able to grow in solution using organic compounds that
contain just one carbon atom such as methane or methanol. However, the bacterium
methylococcus capsulates can grow under aerobic conditions (in the presence of
air) on C-l carbon compounds. The resulting biomass is a good protein source that
can be used directly as feed for domestic animals or fish.
In one process the off-flue gas analyzes 14.0% CO2, 6.0% 02' and 80.0% N2.
It is at 400°F and 765.0 mm Hg pressure. Calculate the partial pressure of each
component.
Solution
Use Equation (13.5): Pi ;;; P'Oio./ Yi
Basis: 1.00 kg (or 1b) mol flue gas
Component kg (or Ib) mol p(mmHg)
CO2 0.140 107.1
°2 0.060 45.9
N2 0.800 612.0
Total 1.000 765.0
On the basis of 1.00 mole of flue gas, the mole fraction Y of each componenr, when
multiplied by the total pressure, gives the partial pressure of that component. If you
find that the temperature measurement of the flue gas was actually 437°F but the
pressure measurement was correct, would the partial pressures change? Hint: Is the
temperature involved in Equation 13.5?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-445-320.jpg)
![Sec. 13.3 Material Balances Involving Ideal Gases
100 1b mol entering ttl at SC 5500R -29-.9-2- -=- -
1 1b mol 492°R 35.0 in. Hg
3 i 0 lb mol air 359 at SC 5300R 29.92 in.
I Ib mol 492QR 29.4 Hg
1220 X 102
3 366.61b mol P 359 ft3 at SC 8600R 29.92 in. Hg 2
ft of product: 1 ib mol 4290R 35.0 in. Hg = 2255 X 10
The answers to questions are
-] 2-2-0- X- :1:0-2 = ft3 air at and 29.4 in. Hg 3.56----------"-
343 X 1 ft3 at 5500R and .0
2255 X 102 = 6.57_=--__at_ 8_6_00_R_a_n_d_2_9_.4_in_,_H-'-g
x and 35.0 in. Hg
EXAMPLE 13.8 Material Balance without Reaction
at and ]05 kPa is flowing through an irregular duct. To determine
the rate of flow of the gas, CO2 from a tank is passed into the gas stream. The gas
analyzes 1.2% CO2 by volume before 3.4% CO2 by volume after the addition.
the CO2 that wa.~ injected left the tank, it was passed through a rotameter, and
found to flow at the rate of 0.0917 m3 fmin at and 13] kPa. What was the rate
of flow of the entering in the duct cubic meters minute?
Solution
This is an open, steady-state system without reaction. The system is the ·duct.
Figure 3,8 is a sketch of process.
Steps 1, 2, 3, and 4
The data are presented in Figure 3.8.
F ..
tSGC and 105 kPa
%
C(h 1.2
Other 98.8
100.0
Figure E13.8
15°C and 105 kPa
%
CO2 3.4
Other 96.6
100.0
419](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-450-320.jpg)
![Sec. 13.3 Material Balances Involving Ideal Gases 421
2. Does including the ideal gas law together with material balances add an additional equation
to the analysis of the degrees of freedom?
Problems
1. A furnace is fired with 1000 ft3 per hour at 60°F and 1 atm of a natural gas having the following
volumetric analysis: 80%, C2H6: 1 02: 2%. CO2: 1%, and N2: 1 %. The
exit flue gas temperature 800°F and the pressure is 760 mm Hg absolute; 15% excess
air is used and combustion is complete. Calculate the (a) volume CO2 produced per
hour; (b) volume of H20 vapor produced per hour; (c) volume of N2 produced per hour~
and (d) total volume of flue gas produced per hour.
2. A flue gas contains 60% N2 and mixed with air to cool it. If the resulting mixture flows
at a rate of 250,000 ft3 Ihr and contains 70% Nz, what is the flow rate of the flue gas?
State all your assumptions concerning the temperatures and pressures of the streams.
3. Two tanks containing N2 at the following conditions sit next to each other
TankA TankB
Volume (m3) 1 5
Temperature (ae) 25 40
Pressure (kPa) 300 ?
AmouDr of (g mol) '] ?
After the two tanks are connected and reach equilibrium, the conditions in the combined
tanks are 700 kPa and 35°C. What was the pressure in Tank B?
Thought Problems
1. In a test of the flow of gases through a pipe, pure hydrogen was found to flow at a volumetric
flow rate 22 times that of carbon dioxide. When the hydrogen was diluted with carbon
dioxide entering the pipe midway from the ends, the exit flow was less than that
of pure hydrogen. Explain the observed differences.
2. A pair of identical balloons are inflated with air to the same pressure, and tied to a stick
that is held in the center by a balloons are each same distance from the
center of the stick so that the stick remains horizontal to the ground. When the lefthand
baUoon is carefully punctured, will the stick rotate down from the left, up from the left, or
remain horizontal?
Discussion Problem
1. In a demonstration. a 30 em diameter balloon was filled to two-thirds of its maximum
pressure with SF6, a gas. Students measured the balloon's diameter for 10 days. at which
time the balloon burst. No one ever touched it. Explain how this could happen. (Note: The
baUoon was not defective.)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-452-320.jpg)
![436 Real Gases: Compressibility Chap. 14
Main Concepts
14.1 Corresponding States
When gases do not confonn to the assumptions underlying ideality that were explained
in Chapter 13. we call them real gases. For example, if you compress argon to
1000 atm in a chamber with a viewport, the argon will have about the same density as
water (but it is still a gas), and you can float a material in the argon that floats on water.
You cannot apply the idea] law under such c9nditions, Figure 14.1 shows the error
involved in using the ideal gas law to predict p-V-T properties for steam.
In the attempt to devise some truly urn versal gas law that predicts well at low and
high pressures, the idea of corresponding states was developed. Early experimenters
found that at the critical point all substances are in approximately the same state of
molecular dispersion. Consequently. it was felt that their thermodynamic and physical
properties should be similar. The law of corresponding states expresses idea that
in the critical state all substances should behave alike.
What does the critical state (point) mean? For a pure component it means
the maximum temperature and corresponding pressure at which liquid and vapor
can coexist. You can fmd many definitions, but the one most suitable for general
use with pure component systems as well as with mixtures of gases is the followmg:
The critical state for the gas-liquid transition is the set of physical conditions
at which the density and other properties of the liquid and vapor become
identical.
1
O'----.L----""------l
o 40 80 120
Pressure (AtM)
s.
2....---------........,
1
o ~ ______ -4 ___ ~~
250 350 450 550
Temperature (K)
b.
Figure 14.1 Graph of the measured specific volume V of stearn as It function of (a) the
pressure and (b) the temperature contrasted with predictions by the ideal gas law, Note
that rhe deviations at high pressure and low temperature are relatively quite large.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-467-320.jpg)
![Sec. 14.2 Critical State 437
14 .. 2 Critical State
One way to determine the critical is by acoustic because
,>1..1 ... ,""' .... of sound in a fluid dIOpS to a minimum wh.en the fluid in the criticru.
fact. in 1822 acoustic effects were exploited unwittingly by Cagniard de 1a
Tour when heated alcohol in a sealed gun barrel and to the musket ban
roIling Figure 14.2 shows measurements by Andrews 1863 of the
pressure versus the volume for CO2,
Note the point C at 31 the highest temperature at which liquid and
gaseous CO2 can coexist in equilibrium. Above 31°C only fluid exists, so that the
critical temperature for is 31°C (304K). The corresponding pressure 72.9 atm
(7385 Also note at higher temperatures, such as 50°C. the data can be rep-by
the ideal law because p V = constant, a hyperbola.
You can experimental values of the temperature (Tc) and the,
cal pressure (Pc) for various compounds in Appendix D on the CD that accompanies
book. If you cannot find a desired critical value this text or a hand-book,
you can always consult Poling a]. (refer to references at end for this
chapter), which and evaluates methods estimating constants
various compounds.
p(atm)
80
72.9
Liquid .. •• - ••
I
4O"C
',-----31"C
~------- 20"C
40 -------:---......-....- --10°C ~ _ _L __ ~ ____________ ~ ________________ ~ __ •
100 94 75 50
Specific volume. V (crrt3/g mol)
Figure 14.2 Experimental measurements of carbon dioxide by Andrews. The
solid lines smoothed is the highest temperature which any
liquid exists, At the solid dots Hquid and vapor coexist.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-468-320.jpg)
![Real Gases: Compressibility Chap. 14
A supercritical fluid is a compound in a state above critical point. Super-critical
fluids are used to replace solvents such as trichloroethylene and methylene
chloride, the emissions from which, and the contact with, have severely limited.
For example. coffee decaffeination, the remova] of cholesterol from egg yolk
with CO2! the production of vanilla extract, and the destruction of undesirable organic
compounds all can take place using supercritical water. Oxidation in supercritical
water has been shown to destroy 99.99999% of all of the major types of toxic
agents present in chemical weapons, which are being eliminated.
14 .. 3 Reduced Variables
Other terms with which you should become familiar are the reduced variables.
These are corrected, or normalized, conditions of temperature. pressure, or
volume, normalized (divided) by their respective critical conditions. as follows:
T
Tr =-
p
p, = Pc
A
V
V, = -;;:-
Vc
In theory, the law of corresponding states indicates that any compound should have
the same reduced volume at the same reduced temperature and reduced pressure so
that a universal gas law might be
(14.1)
Unfortunately Equation (14.1) does not make accurate predictions universally. You
can check this conclusion by selecting a compound such as water, applying Equation
(14.1) at some low temperature and ~igh pressure to calculate V, and comparing
your results with the value obtained for V for the corresponding conditions from the
steam tables that are the folder in the back of this book.
14.4 Compressibili
How can the ideas presented above be used? One common way to modify
the ideal gas law by inserting an adjustable coefficient z, the compressibility factor,
a factor that compensates for the nonideality of the gas, and can be looked at as a
!](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-469-320.jpg)
![442 Real Gases: Compressibility Chap. 14
charts is of considerable practiool as well as pedagogical value because their existence
enables you to make engineering calculations with considerable ease~ and also
pennits the development of thermodynamic functions for gases for which no experimental
data are available.
Frequently Asked Questions
" 1. What is in the blank region in Figure 14.4a below the curves for Tr and V 7,? The blank re-gion
corresponds to a different phase-a liquid.
A 2. Win pV = zRT work for a liquid phase? Yes, but relations to calculate z accurately are
more complex than those for the g~s phase. Also, liquids are not very compressible so that
for our purposes we can ignore p-V-T relations for liquids.
A
3. Why should I use pV = zRT when I can look up the da,!a needed in a handbook or on the
Web? Although considerable data exists, you can use pV = zRT to evaluate the accuracy
of the data and" interpolate within data points. If you do not have data in the range you
want. use of pV = zRT is the best method of extrapolation. Finally, you may not have
any data for the gas of interest.
EXAMPLE 14.1 Use of the Compressibility Factor to Calculate
a Specific Volume
In spreading liquid ammonia fertilizer. the charges for thaemEt of NH3
used are based on the time involved plus the pounds of NH3 inject nto the soil.
After the liquid has been spread, there is still some ammonia left in e source tank
(volume:::: 120 ft3), but the fonn of a Suppose that your weight tally, which
is obtained by difference, shows a net weight of ] 25 Ib of NH3 left'in the tank at
292 psig. Because the tank is sitting in the sun, the temperature in the tank is 125°E
Your boss complains that his calculations show that the specific volume of
the NH3 is 1.20 ft3 lIb, and hence that there are only 100 Ib of NH3 in the tank.
Could he be correct? See Pi gure E 14.1
Figure E14.1
Solution
Basis: 1 Ib of NH)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-473-320.jpg)
![l
Chap. 15 Equations of State
8.
-2
b.
4
x
463
Figure IS.2 Gra.ph (a) shows the existence of one positive real root whereas graph (b)
shows the of two rea] roots.
One feature of the cubic equations of state listed in Table 15.1 com-ment.
In solving for n or VI solve a cubic equation that more
than one real root, as indicated in 15.2.
For example. the van der Waals can easily solved explicitly for p
as foHows.
p=
However, if you
cubic in V (or n):
V (or n). you can see that the equation becomes
f(V) = V3 - (nb
and can have multiple roots (see Figure 15.2), You want a positive real root in the
gas phase. In Figure 15.2b, the smaller positive root figuratively pertains to the pressure
of the liquid. See Pratt'll for a good about the meaning of the multiple
roots of an equation of state. On the disk that accompanies this book you will find a
computer program caned Polymath that can nonlinear equations, and can be
used to solve for V (or n) if you have a reasonable initial guess for V (or n). say from
"Pratt, R.M. "Beware of Bogus Roots with
278-281 (Fall 1999).
,. Chern Eng. Educ ..](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-494-320.jpg)
![J
Chap. 15 Real Gases: Equations of State
p = (4790 + 101.3) kPa IO:.~~a = 48.3 atm
, . 82.06(cm3)(atm)
R the proper Units = (g mol)(K)
22.7
n = = mol propane
44 kglkg mol
[
(0.516 X loJ)2(9.24'X 106)] 6
nRT = 48.3 + [0.150 X 10
, (0.150 X 1
- (0.516 X 103)(90.7)] = (0.516 X 103)(82.06)(T K)
T=384 K
Suppose you want to apply the PR or SRK equation in solving problem
posed in Example 15.1. It may not be immediately obviou$, but both of these equa-tions
a value for T in to obtain values for the coefficients in respec-tive
and you do not know T. Look at Table I 1. Can you locate where T
is involved in calculating the values of the coefficients? You might augment the PR
or equation with the equation for A. and solve pair for T to avoid a trial-am:
l-eJror solution. Because A does not change with temperature, one or two
iterations using assumed values of A should suffice.
EXAMPLE 15.2 Solution of van der Waals's Equation for V
Given the 'UlU' ........ " a vessel of
p = 679.7 psia a = 3.49 X 10·
n = L1361b mol
b=L
for the volume [he vessel using Van der Waals's equation.
Solution
Wrice van
V.
Waals's equation as a cubic equation in one unknown
f(V) = V3 _ -,-pn_h_+_n_RT_
p
/
(a)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-496-320.jpg)
![410 Real Gases: Equations of State Chap. 15
Looking Back
We provided a condensed view of equations of state for pure components in
this chapter, citing their advantages and disadvantages, expected accuracy, and examples
of application. We also mentioned briefly one way to treat gas mixtures.
GLOSSARY OF NEW WORDS
Benedict-Webb-Rubin An eight-parameter equation of state that relates the physical
properties p. V, T. and n for a gas.
Group contribution method A technique of estimating physical properties of
compounds by using properties of molecular groups of elements in the compound.
Holborn A multiple parameter equation of state expanded in p.
KammerHngh.Onnes A multiple parameter equation of state expanded in
Peng .. Robinson A three-parameter equation of state.
Soave-Redlich-Kwong A three-parameter equation of state.
UNIFAC A group contribution method of estimating physical properties.
UNIQUAC An of the UNIFAC method of estimating pbysical proper-ties.
Van der Waals A two-parameter equation of state.
Vinal equation of state An equation of state expanded in successive terms of one
of the physical properties.
UPPLEMENTARY REF R NCES
In addition to the genera] references listed in the Frequently Asked Questions
in the front material, the following are pertinent.
Barrufet. M. A., and Eubank. "Generalized Saturation Properties of Pure Fluids via
Cubic Equations of State," Chem. £.ng. Educ .. 168-1 (Summer 1989).
Benedict, P .• and Olti. Computer Aided Chemical Thermodynamics of Gases and Liquids,
Wiley-interscience, New Yark (1985).
Chaa~ K. C., and R. L Robinson. Equations of State in Engineering and Research, American
Chemical Society, Washington, D.C. (1979).
Copeman. T. W., and P. M. Mathias. "Recent Mixing Rules for Equations of State:' ACS
Symposium Series 300~ 352-369. American Chemical Society. Washington, D.C.
(1986).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-501-320.jpg)
![f
Chap. 15
"""$15.10
"'*15.12
Problems
acid. That generates more carbon dioxide, giving more pressure in the pouch, which
expands and helps push out more product. (The CO2 does not escape from the can,
just the product.)
How many grams of NaHC03 are needed to generate a residual pressure of
81.0 psig in the can to deliver the very last em3 of product if the cy Hndrical can is
8.10 cm in diameter and 17.0 em high? Assume the temperature is 25°C. Use the
Peng-Robinson equation.
Pi rst commercialized in the 1970s as extractants in "natural" decaffeination
processes, SCFs (supercritical fluids)-panicularly carbon dioxide and water-are
finding new applications, as better, less-expensive equipment lowers processing
costs, and regulations drive the chemical industries away from organic
vents.
extraction capabilities are now being exploited in a range of new pharmaceutical
and environmental applications. while supercritical extraction, oxidation
and precipitation are being applied to waste cleanup challenges.
A compressor for the carbon dioxide compresses 2,000 m3/min at 20°C and
500 kPa to llOoC and 4800 kPa. How many m3/min are produced at the high pressure.
Use Vander Waals' equation.
A tank of H2 is left out overnight in Antarctica. You are asked to determine how
many g moles of H2 are in tank. The pressure gauge reads 39 atm gauge and the
temperature is - 50°C. How many g moles of H2 are in the tank?
Use the van der Waals and Redlich-Kwong equations state to solve this
problem. (Hint: The nonlinear-equation-solving program on the CD in the pocket at
the back of this book win make execution of the calculations quite easy.)
Find the molar volume (in cm3/g mol) of propane at K 21 atm. Use the
Redlich-Kwong and Peng-Robinson equations, and solve for the molar volume using
the nonlinear equation solver on the CD in pocket at the back of this book. The
acentric factor for propane to use in the Peng-Robinson equation is 0.1487.
4.00 g mol of is contained in a 6250-cm3 vessel at 298.15 K and 14.5 atm. Use
the nonlinear equation solver on the in the back of the book to solve the RedlichKwong
equation for molar volume. Compare the calculated molar volume of the
CO2 in the vessel with the experimental value.
··15.14 The tank cited in problem 15.5 is constructed and tested, and your boss informs you
that you forgot to add a safety factor in the design of the tank. It tests out satisfacto- t rily to 3500 kPa, but you should have added a safety factor of 3 to the design. that
the tank pressure should not exceed (3500/3) 1167 kPa. say 1200 kPa. How many
kg of CO2 can be stored in the tank if the safety factor is applied? Use the RedlichKwong
equation. Hint: 'Polymath will solve the equation for you.
..
I
J
··15.15 A graduate student wants to use van der Waa]s· equation to express the pressurevolume-
temperature relations for a gas. Her project required a reasonable degree of
precision in the p- V-T calculations. Therefore, made the following experimental
measurements with her setup to get an idea how easy the experiment would be:](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-504-320.jpg)
![480 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16
two-phase region
vaporization
Conditions of T and p in which two phases
can coexist at equilibrium (the region com-prising
vapor and liquid compressed
into curve N-H-E-B in Figure 16.3)
The change of phase from liquid to vapor,
that boiling (D to F).
In Figure 16.3 process of evaporation and condensation of water at 1 atm is
represented by the ABC with the phase transformation occurring at 100°C. Sup-pose
that you went to the top of Pikes and repeated the of evaporation
and condensation in the open What would happen then? The process would be
the same (points D-E-F) with the exception of the temperature at which the water
would begin to boil, or condense. Since the pressure of the at the top of
Pikes Peak lower than 101.3 kPa, the water would start to displace the or boil,
at a lower temperature. Some unfortunate consequences might result if you expected
to kill certain of disease-causing bacteria by boiling!
To conclude, you can see that (a) at any given temperature water exerts its
vapor pressure (at equiHbrium)~ (b) as the temperature increases, vapor pressure
increases as wen; and (c) it makes no difference whether water vaporizes into
into a cylinder closed by a piston, or into an evacuated cylinder-at any temperature
it still exerts same vapor pressure as long as the liquid water is in equilibrium
with its vapor.
A pure compound can change at constant volume from a liquid to a vapor. or
reverse~ via a constant temperature process as well as a constant pressure
process. A process of vaporization or condensation at constant temperature is
illustrated by the lines G-H-/ or JaH-G, respective]y, Figure 16.3. Water would
vaporize or condense at constant temperature as the pressure-reaches point H on
the vapor-pressure curve. TQe change that occurs H the increase or
in the fraction vapor or 1iquid~ respectively. at the fixed temperature. The
sure does not change until all of the vapor, or liquid, has completed the phase transition.
Now let's go back to th~ experimental apparatus and collect data to prepare a
graph of values of p and V as a function of temperature. This time you want to
hold the temperature in the chamber constant and adjust the volume while measuring
the pressure. Start with compressed liquid water (subcooled water) rather than
ice. and raise the piston so that water eventually vaporizes as the pressure drops.
Figure 16.4 illustrates by dashed lines the measurements two different temperatures,
T) and T2. the pressure is reduced at constant Ttt V increases very slightly
(liquids are not very compressible) until the liquid pressure reaches p., the vapor
pressure, at point A.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-511-320.jpg)
![Sec. 1 Modeling and Predicting Vapor Pressure as a Function of Temperature 485
and dissolved gases come up through the annular space between the first tube and
the closed second tube (34 cm dia.), About 400 m down. pure oxygen is sparged
into the sludge to promote oxidation of the suspended organic matter. The static
head of the j ,200-m column prevents boiling at about 280°C and 100 bars. The alternati
yes: incineration, composting. drying, or conventional wet oxidation have per
m.t. costs DGL 790. DGL 790, DGL 700, and DGL 600, respectively. versus
DGL 580 for the VerTech process.
Check the consistency of the data. and offer an opinion as to the problems that
might occur if the process operated on a 20% sludge solution in water.
2. The term (boiJing·liquid expanding-vapor explosion) refers to an explosion that
occurs when the pressure on a liquid is suddenly reduced substantiaHy below its vapor
pressure. Explain how such an explosion can occur. Might it occur if a tank car ruptures
in an accident even if no occurs? Give one or two other examples of vessel failures
that might cause a BLEVE.
16.2 Modeling and Predicting Vapor Pressure
as a Function of Temperature
How can youfind the answer if you don't know the question?
M.C. Sonby. Shell Development Co.
In this section we are going to explain how to determine values for the vapor
pressure given the temperature, or the temperature the vapor pressure. Severa]
aids exist for you to use: (a) Equations of as a function of T; (b) charts of versus
(c) tables of p" versus T;
Dr. Sonby reported at one of the A.I.ChE. meetings that
"In one of my first projects at Shen. I was trying to determine the rate of
emission of sulfur vapor from a pit full of molten sulfur. I was able to locate four
different references on the vapor pressure of molten sulfur. Unfortunately, all
four references different vapor pressures. with the highest and lowest being
an order of magnitude apart! As it turned out, the experimental methods used to
determine the vapor pressures varied greatly, as did the purity of the sulfur in the
experiments. "
Thus, you need to be able to appraise the accuracy of the predictions values of the
vapor pressure of even old standbys.
16.2-1 Prediction via Equations
You can see from Figure 16.2 that the function of p* versus T is not a linear
function (except as an approximation over a very small temperature range). Many
functional forms have been proposed to predict p* from T (refer to Ruzicka, K., and](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-516-320.jpg)
![486 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16
V. Majer, "Simple and Controlled Extrapolation of Vapor Pressures toward the
Triple Point," A.I.ChEJ., 42, 1723-1740 [J996] for an evaluation of numerous
equations) and the reverse, some with numerous coefficients. We wiH use the Antoine
equation in this book-it has sufficient accuracy for our needs, and coeffi-
"' ......... ., for equation for over 5,000 compounds in the literature:
B lnp '" = A - --T-where
A. B, = constants for each substance
= temperature, K
( 16.2)
Refer to Appendix for values of A, B, and C for a small set of compounds.
The CD in back this book~ based on data provided by Yaws,'" will enable you to
retrieve vapor pressures for over 700 compounds. The National Institute of Standards
and Technology a program (www.nist.gov/srdinist87.htm) that calculates the
vapor pressures and displays tables for approximately 6,000 pure compounds.
You can estimate the values of A, B. and C in Equation (16.2) from experimental
data by using a regression program such as the one in Polymath on the CD
that accompanies this book:. With just experimental values the vapor pres-sure
versus temperature you can Equation (16.2),
EXAMPLE 16.1 Vaporization of Metals for Thin Film Deposition
Three methods of providing vaporized metals for thin film deposition are
evaporation from a boat or from a filament transfer via an electronic beam. Fig-ure
1 illustrates evaporation from a boat placed in a vacuum chamber.
Vacuum chamber
To pump
Electrode Electrode
Figure E16.1
ilYaws. C. and Yang. "To ..... "'''' ............ Vapor Pressure Easily," Hydrocarbon ProceSSing,
p. 65 (October 1989).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-517-320.jpg)
![Sec. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature 493 "
16.2 .. 3 Predicting Vapor Pressures from Reference
Substance Plots
Because of the curvature of the vapor pressure versus temperature data (see
16.2), no simple equation with two or three coefficients will fit the data accurately
from the triple point to the critical point. Othmer proposed in numerous articles
(see. for example. Othmer, D.F., Ind. Eng. Chern. 32, 841(1940) and Perry, J.H.,
and E.R. Smith, Ind. Eng. Chem., 25, 195 (1933» that reference substance plots
(the name will become clear in a moment) could convert the vapor pressure versus
temperature curve into a straight line. One well-known example is the Cox chart
(Cox, E.R., Ind. Eng. Chern .• 15, 592[ 1923]). You can use the Cox chart to retrieve
vapor pressure values to test reliability of experimental data, to interpolate, and
to extrapolate. Figure 16.8 is a Cox chart. Here is how you can make a Cox chart.
1. Mark on the horizontal values of Jog p* so as to cover the desired range
of p* for the compound of interest.
2. Next draw a straight line on the plot at a suitable angle. say 45° I that covers the
range of T that is to marked on the vertical axis.
Vapor pressure, kPo /log 10 scale)
tOO 1000 47
200
150
_ 2501---1---
.-
100 .!!
0 "~"
11:1 "8 2001--+---+-
VI
fOO
LI...
'" 75
50
25 ~
IJ-
"- 50 3
.a.
25 0 ... c.. .e.. I-
-25
-25
1.0 10 100 1000 10,000
Vapor pressure psio lIoQlo scale)
Figure 16.8 Cox chart. The vapor pressure of compounds other than water is observed
to fall on straight lines when plotted on the special developed for the Cox chart by
using a reference substance water).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-524-320.jpg)
![502 Single-Component Two-Phase Systems (Vapor Pressure)
Web Sites
http://antoine.fsu.umd.edu⪙liquids/faq/antoine-vapor-pressure.shtml
http://chemengineer.abouLcomllibrary/weekl y laa082 800a.h tm
http://w ww .questcons u It.comJ -j rmJdew bu b.html
http://www.taftan.comJsteam.htrn
PROBLEMS
-16.1 Select the correct answer(s) in the following statements:
Chap. 16
] . In a container of 1.00 L of toluene, the vapor pressure of the toluene is J 03 mm
Hg. The same vapor pressure will be observed in a container of (1) 2.00 L of
toluene at the same temperature~ (2) 1.00 L of toluene at one-half the absolute
temperature~ (3) 1.00 L of alcohol at the same temperature~ (4) 2.00 L of alcohol
at the same temperature.
2. The temperature at which a compound melts is the same temperature at which it
(1) sublimes; (2) freezes; (3) condenses; (4) evaporates.
3. At what pressure would a liquid boil first? (1) 1 atm; (2) 2 atm; (3) 200 mrn Hg;
(4) 101.3 kPa.
4. When the vapor pressure of a liquid reaches the pressure of the atmosphere surrounding
it, it will (I) freeze; (2) condense; (3) melt; (4) boil.
5. Sublimation is the phase change from (1) the solid phase to the liquid phase; (2)
the liquid phase to the solid phase; (3) the solid phase to the gas phase; (4) the
gas phase to the solid phase.
6. A liquid that evaporates rapidly at ambient conditions is more like than not to
have a (1) high vapor pressure; (2) a low vapor pressure; (3) a high boiling point;
(4) strong attraction among the molecules.
-16.2 Based on the following phase diagrams, answer the questions below, and explain
your answers.
(a) What is the approximate normal melting point for compound A?
(b) What is the approximate nonnal boiling point for compound A?
Compound A Compound B
1.5
10
E..- -E a ~ 1.0 ~
a. Q. 6
4
2
o 40 80'20160200 20 40 60 ao 1 00 120
T (0C) T (DC)](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-533-320.jpg)
![1
506 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16
*16.16 Take 10 data points from the steam for the vapor pressure of water as a func-tion
of temperature from the freezing point to 500 K. and fit file following function:
p* = exp [a+b T + c (in Til. + d (1n n3)
j where p is in kPa and T is in K.
*16.17 For each the conditions of temperature and pressure listed below for water, state
whether the water is a solid phase. liquid 'phase1 superheated, or is a saturated mix-ture.
and if the latter, indicate how you would ,calculate the quality. Use the steam ta- f
bles (inside the back to assist the calculations. i
i .
State p (kPa) T(K) if (mJlkg)
1 2000 475
2 1000 500 0.2206
3 101.3 200
4 245.6 400 0.7308
5 ]000 453.06 0.001127
6 200 393.38 0.8857
*16.18 Repeat Problem 16.17 for the fonowing conditions:
State p (psia) T(CF) if (ttlJ1b)
1 0.3388 927.0
2 1661.6 610 0.0241
3 308.82 420 0.4012
4 180.0 440 2.812
·16.{9 Calculate the specific volume for water that exists at the following conditions:
(a) T= 100°C, p:;; 101,4 kPa. x= 0.5 (in m3/kg)
(b) T::; 406.70 K, p == 300.0 kPa. x ::; 0.5 (in m3/kg)
(c) T= loo.OoP, p:::.: 0.9487 psis, x:: 0.3 (in ff1llb)
Cd) T= 860.97"R. p ::; psia, x;;;:; 0.7 (in ft3l1b)
"'16.20 Answer the following questions true or false:
(a) A pot full of boiling water is tightly closed by a heavy lid. The water will stop
boiling.
(b) Steam quality is the same thing as steam purity.
(c) Liquid water that is in equilibrium with its vapor is saturated.
(d) Water can exist in more than three different phases.
(e) Superheated steam at 300°C means steam at 300 degrees above boiling point.
(f) Watercan be made to boil without heating it. '
t
1](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-537-320.jpg)
![Sec. 17.1 Saturation 511
balances. From Chapter 13 you should that conditions the ideal
gas law applies to both air and water vapor with excellent precision. Thus, we can
say that the following relations hold saturation:
nH20RT
-
PairV nairRT
or ,.
PH20 P H2O nH20 Protal - Pair --= - -
Pair Pair
because V and Tare the same for the air and water vapor.
Also
PH20
Ptotal - PH20
PH20
---=--- = 1 - Yair
Prur PH20
(17.1)
(1
(1
Asa
51°C. and the
pressure of
water vapor is
find that pit: =
suppose you have a saturated gas) say water in air
....... "u .. ...., on the system is 750 mm Hg absolute. What is the partial
Furtbennore.
saturated, you know that the partial oreSSUlre
You can look in a handbooks or use the steam
Then
Pair = 750 - 98 = 652 mm Hg
vapor air mixture has the following composition
PH20 98
YH20 = -- = - = 0.13
Ptotal 750
Pair 652
Yrur = -- = 750 = 0.87
Ptotal
You can use these compositions when applying material balances.
Calculation of the Dew Point of the Products
of Combustion
burned at atmospheric
so that of the carbon burns to CO2, the dew
So]ution
Examine Figure E17. L To get the dewpoint, a
mine P~20' To get P~20 you have to calculate P~20 =
with 248% excess air
product gas.
you have to deter.
The solu-](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-542-320.jpg)
![Sec. 17.2 Condensation
Components
Total
kg mol
1.0
2.0
Composition of combustion product gases (kg
LO
2.0
1.0 2.0 0.80
10.5
10.5
total kilogram moles of produced are 14.3 and mole fraction is
2.0
14.3 = 0.14
Coal
is
Excess
Components
C
2H
Air
Total
1 + 0.47(112) = 1 kg mol
( 1.24 )(0.40) = 0,49 mol
1.40(79/21)[1 + 0.47(1/2)] = 6.50 kg mol
kg mol
1
0.47
CompMition of combustion product
1
1
0.47
0.47
0.49
0.49
(kg mol)
6.5
6.5
The total kilogram moles of produced are 8.46 and the mole fraction H20
0.47
8.46 = 0.056
the barometric pressure is, 100 kPa. and if the stack became satu-rated
so that water vapor would start to condense at PH20. condensed vapor could be
photographed. The respective partial pressures of the water vapor in combustion
gases are:
Partial "' ...... ,,', ..
Equivalent temperature:
Natural gas
100(0.14) = 14 kPa
52.SoC
Coal
100(0.056) = 5.6 kPa
35QC](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-552-320.jpg)
![522 Two-Phase Gas-Liquid Systems Chap. 17
Thus. the stack emit condensed water vapor at higher ambient temperatures for
a boiler natural than for one burning coal. The public. unfortunately.
sometimes concludes all the emissions they perceive are pollution. Natural
could appear to the public to a greater pollutant than oil or coal, whereas.
fact, the emissions are just water vapor;The sulfur content coal and oil can be
released as sulfur dioxide to atmosphere. and the capacities of mercury
heavy metals in coal and oil are much greater than natural gas when all
three are being burned properly. The sulfur contents as to the consumers
are as follows: natural gas, 4 X )0-4 mol (as added mercaptans to provide smell
for safety); number 6 fuel oil, up to 2.6%; and coal. from 0.5 to 5%. In addition.
coal release particulate matter into the stack plume. By mixing the stack gas
with air, and by convective above the stack. the water
vapor can reduced. and hence the condensation temperature can be reduced.
However, equivalent dilution, the coal·burning plant will always a iewer
condensation temperature.
With the information calculated above, how would you resolve the questions
that were originally posed?
EXAMPLE 17.4 Material Balance Involving Condensation
Contamination from accidental discharges of compounds used by
industries, airports. businesses, and homeowners a challenging
problem for amelioration. contaminated with poly aromatic hydrocarbons can
be with hot air and steam drive out the contaminants. 30.0 m3 of air at
100°C and 98.6 kPa with a point of 30°C is introduced into the contaminated
soil. and in the soil the gas cools 14°C at a pressure of 109.1 what fraction of
the vapor in the gas at 100°C wi]] condense out in the soil if the gas does not
esc:ape too rapidly from the soil?
Solution
Assume the system at 14°C is at equilibrium. and select a fixed volume of initial
gas as the SVS:tCt1I1.
Step S
Basis: F;::;; 30 wet air at lOOOC and 98.6 kPa.
Steps 1,2,3, '"
Some of the data been placed in Figure 7.4](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-553-320.jpg)
![Chap. 17 Problems 533
'*11.8 One ganon of benzene (C6H6) vaporizes in a room that is 20 ft by 20 ft by 9 ft in
at a constant barometric pressure of 750 mm Hg abs01ute and 70°F. The lower explolimit
for benzene in air 1S 1.4%. Has this value been exceeded?
"'*]7.9 mixture of acetylene (C2H2) with an excess of oxygen measured ft3 at
and 745 mm Hg absolute pressure. After explosion the volume of the dry gaseous
product was 300 ft3 at 60°C. and the same pressure. Calculate the volume of actylene
and of oxygen in the original mixture. The final was saturated. Assume that all of
the water resulting from the reaction was in gas phase after the reaction .
. ··17.10 In a science question and answer column, the foHowing question was
On a trip to see the elephant seals in California, we noticed that when the male
elephant seals were benowing, you could see their breath. But we couldn't see our
own breath. How come?
"'17.11 One way that safety enters into specifications is to specify the composition of a vapor
in air that could burn if ignited. If the range of concentration of benzene in in
which ignition could take place is 1.4 to 8.0 what would the correspond-ing
temperatures for saturated with benzene of a storage tank?
The lOtal pressure in the vapor is 100 kPa.
··17.12 In a dry cleaning establishment warm dry is blown through a revolving drum in
which clothes are tumbled until all of the Stoddard solvent is removed. The solvent
may be assumed to be n-octane (CgH 1s) and have a vapor of 2.36 Hg at
120°F. J f the air at 120°F becomes saturated with octane, calculate the:
(a) Pounds of air required to evaporate one pound octane;
(b) Percent octane by volume in the the drum~
(c) ft3 of inlet air required per lb of octane. The barometer reads 29.66 in. Hg.
"'17.13 When people are exposed to chemicals at relatively low but toxic con centra-tion~,
the toxic effects are only experienced prolonged exposures. Mercury is
such a chemical. Chronic exposure to low concentrations of mercury can cause permanent
menta! deterioration, anorexia. instability, insomnia, pain and numbness in
the hands and feet. and several other symptoms. The level of mercury that can cause
these symptoms can be present in the atmosphere without a being aware of it
because Jow concentrations of mercury in the air cannot be seen or smelled.
Federal standards based on the toxicity of various chemicals have been sel for
the "Pennissible Exposure Limit," or PEL. These limits are set by the Occupational
Safety Health Administration (OSHA). The PEL is [he maximum level of exposure
permitted in the workplace based on a time weighted average (TWA) exposure.
TW A is the average concentration permitted for exposure day after day
without causing adverse effects. It is based on exposure for 8 hours day for the
worker's lifetime.
The present Federal standard (OSHNPEL) for exposure to mercury in air is
0.] mg/m3 as a ceiling value. Workers must be protected from concentrations greater
than 0.1 mglm3 if they are worldng in areas where mercury is being used.
Mercury manometers are filled and cali braled in a small store room that has no
ventilation. Mercury has been spil1ed in the storeroom and is not completely cleaned](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-564-320.jpg)
![· 1 1 Terminology Involved for Partial Saturation 539
18.1 .. 1 Relative Saturation (Relative Humidity)
Relative saturation defined as the partia1
vapor pressure of the vapor at the temperature of the
of the vapor divided by the
'RS ::= --. - = relative saturation
p
(18.1 )
where Pvapor = partial pressure of the vapor in the mixture
p II< = partial pressure of the vapor in the mixture if the gas were
saturated at the given temperature of the mixture (i.e., the vapor
pressure of the vapor component)
Then, for brevity, if the subscript 1 denotes the vapor
ns ::= !?L = p)/PtO[ = V1/VtOl __ -'--- __ m_a_s_s,,--
p ~ P; / Plot VI, satdlVtot n 1. satd mass 1. satd
(18.2)
You can see that relative saturation, in effect, represents the fractional approach to
the tota1 saturation. If you listen to the radio or TV and hear the announcer say that
the temperature is 25°C (77°F) and the relative humidity is 60%, he or she implies
that
P.H20 (100) = %'R1i = 60
P H20
with both the PH20 and the P * H20 being measured at 25°C. Zero percent relative saturation
means no vapor in the gas. What does 100% relative saturation mean? It
means that the partial pressure of the vapor the IS same as the vapor pressure
of the substance.
EXAMPLE 18.1 Application of Relative Humidity to Calculate
the Dewpoint
The weather report on the radio this morning was that the temperature this afternoon
would reach 94C1F, the relative humidity would be 43%, the barometer was
29.67 in. Hg, part]y cloudy to clear. with the wind from SSE at 8 milhr. How many
pounds of water vapor would be in 1 mil of afternoon air? What would be the dew
point of this air?
Solution
The vapor pressure of water at 94°P is 1.61 in. You can ca1culate the
tia} pressure of the water vapor in the air from the given percent relative humidity.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-570-320.jpg)
![Sec. 18.2 Material Balance Problems Involving Partial Saturation
StepS
Either the Wor the A stream can serve as basis. Let's pick
Basis: 1000 bone-dry air (BOA) at 200e and 108.0 kPa
Steps 3 and 4
You first need to calculate the amounts (in kg mol) of water vapor and dry air
in the original air. W = 0.93 kg or
0.93
--=..........;;;...... ------= 0.05 kg mol H20
18 kg H20
As for the bone dry air (BDA)
1000 BDA 273 K 1108.0 kPa 11 kg mol
293 K 101.3 kPa 22,4 m3 = 44.35 kg mol BOA
Steps 6, 1, 8 and 9
The materia] balances are trivial. All of the water and all of the air in the orig-inal
moist are accounted for in the above calculations. The problem has zero. de-grees
of freedom.
To get the relative humidity the moist air, you first have to calculate the
partial pressure of the water vapor in the moist
PH20 nH20 0.0517 -3
·PCOI = ntot = 0.0517 + 44.35 = 1.164 X 10
PHzO = (1.164 X W-3}(108.0) = 0.126 kPa
Next you look up the vapor pressure at 15°C for water, namely 1.70 kPa. Conse-quently,
fractional relative humidity of the original air was
0.126 = 0.074
1.70
EXAMPLE 18.4 Humidification of Air
To condition the in an office building in the winter, 1000 m3 of moist air
hour at 101 kPa 22°C with a dew point 11°C enter the system. The air
leaves the system at 98 kPa with a dew point of 58°C. How many kilograms
water vapor are added to each kilogram of wet entering the process?
Solution
Steps 1, 2, 3, and 4
The known data appear in Figure E18A .
. "
545
..](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-576-320.jpg)
![552 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18
Looking Back
In this chapter we explained a number of new tenns, all of which are used to
identify partially saturated conditions in a noncondensable gas. We also solved material
ba]ance problems in which these tenns were used to get the stream compositions.
GLOSSARY OF NEW WORDS
Humidity The mass of water vapor per mass bone-dry
Molal saturation Moles of vapor divided by the moles vapor-free gas.
Partial saturation A noncondensable containing a vapor whose partial
sure less than the saturation pressure.
Relative humidity The partial pressure of water vapor in a non condensable gas
vided by the vapor pressure of the water at the temperature of the
Relative saturation The partial pressure of a vapor in a noncondensable gas
vided by the saturation pressure the vapor at the temperature of gas.
SUPPL M NTARY R FERENCES
American Institute of Chemical Saturation and Material Balances. Molular In-struction
Series AIChE. New York (1981).
r/--",
Instrument Society
( 1985).
America, Moisture and HumiditY. Instrument Society of America
Shallcross, C. Handbook of Psychrometric Charts-Humidity Diagrams for Chemical
gineers, Kluwer Academic, New York (1997).
Web Sites
http://members.aol.coml_hCalRogluthierlhumidity.html
http://w3.one.netJ-jwclyrner/wet.html
http://www.agsci.kvl.dkJ-beklrelhum.htrn
http://www.crh.noaa.gov/mkfsoo/docuJhumidity.htm
ROBLEMS
at 60.0°C and 1 ° 1.6 abs. has a molal humldity of 0,030, detennine:
(a) the relative humidity
(b) the dewpoint of the (in °C)
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-583-320.jpg)
![Chap. 18 Problems 553
"'18.2 What is the relative humidity of 28.0 m3 of wet at 27.0°C that is found to contain
-18.3
·18.4
0.636 kg of water vapor?
Air at 80°F and I atm has a dew point of 40°F. What the relative humidity of this
air? If the air is compressed to 2 atm and what the relative humidity of the resulting
air?
If a gas at 140°F and 30 in. Hg abs. has a molal humidity of 0.03 mole of H20 per
mole of dry air. calculate:
(a) The relative humidity (%)
(b) The dew point of the gas (OF)
The Weather Bureau reports a temperature of 90°F, a relative humidity of 85%, and a
barometric pressure of ] 4.696 psia.
(a) What is the molal humidity?
(b) What is the humidity (weight basis)?
(c) What is the saturation temperature or the dew point?
(d) What is the number of degrees superheat of the water vapor?
(e) Determine the molal humidity and dew point if the air is heated to 105°P. the
pressure remaining steady.
(f) Determine the molal humidity and dew point if the air is cooled to 60°F, the pres-sure
remaining steady.
(g) What fraction of the original water is condensed at 60°F?
The Envirorunental Protection Agency has promulgated a national ambient air quality
standard for hydrocarbons: 160 J.Lglrn3 is the maximum 3-hr concentration not to be
exceeded more than once a year. It was arrived at by considering the role of hydrocarbons
in the formation of photochemical smog. Suppose that in an exhaust gas benzene
vapor is mixed with air at 25°C such that the partial pressure of the benzene
vapor is 2.20 mrp Hg. The total pressure is 800 mm Hg. Calculate:
(8) The moles ofoenzene vapor per mole of gas (total)
(b) The moles of benzene per mole of benzene free gas
(c) The weight of benzene per unit weight of benzene-free gas
(d) The relative saturation
The micrograms of benzene per cubic meter
(f) The grams of benzene per cubic foot
Does the exhaust gas concentration the national quality standard?
A constant volume bomb contains air at 66°F and psia. One pound of liquid
water is introduced into the bomb. The bomb is then heated to a constant temperature
of 180°F. After equilibrium is reached. the pressure in the bomb is 33.0 The
vapor pressure of water at 1800P 7.51 psia.
(a) Did all of the water evaporate?
(b) Compute the volume of the bomb in cubic
(c) Compute the humidity of the air in the bomb at the final conditions in pounds of
water per pound of air.
·18.8 A mixture of ethyl acetate vapor and air has a relative saturation of 50% at 30°C and
a lotal pressure of 740 mm Hg. Calculate (a) the analysis of the vapor and (b) the
molal saturation.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-584-320.jpg)
![554 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18
a gas mixture there are 0.0083 Ib mol of water per Ib mol of dry at a
temperature of 80°F and a total of 2 atm.
(a) Calculate the relative saturation of this mixture.
(b) Calculate the temperature to which the mixture must be heated in order that
relative becomes 20%.
A drier must remove 200 kg of H20 per hour from a certain material. Air at 22°C and
50% relative humidity enters the and leaves 72°C and 80% relative humidity.
What is the weight (in kg) of dry air used per hour? The barometer reads
103.0 kPa.
··18.11 thousand (1 metric ton) of a slurry containing 10% by weight of CaCOJ are to
filtered in a rotary vacuum The filter from filter contains 60%
water. This cake is then placed into a drier and to a moisture content 9.09 kg
H20/1OOkg CaC03• the humidity of the entering the is 0.005 of water
of dry and the humidity of the leaving the is 0.015 kg
of dry air, calculate:
(a) the kg of water removed by the filter
(b) the wet air entering the drier
"18.12 In the solid electronic chips, concern over contamination pre~
dominates concern over cost In specification of nitrogen, control hydrocar~
bon content is essential. Cryogenic separation is a preliminary in the purifica-tion
process.
If nitrogen containing a hydrocarbon at a relative"{aturation of 3.8% at
750mm Hg absolute and 300K is to be cooled at 750mm Hg to remove 90% of the
hydrocarbon as liquid, what must temperature of the outlet gas be? The vapor
sure of the hydrocarbon the range of interest is given by
1 ,.. = 159008 _ 2788.5]
n p , T - 52.36
where p* is in mm absolute and T is in K.
"18.13 wet gas at 30°C and 100.0 kPa with a relative humidity of 75.0% was compressed
to 275 kPa, then cooled to 20°C. How many m3 the original was com~
pressed if 0.341 kg of condensate (water) was removed from separator that was
connected to the
*18.14 An absorber receives a mixture air containing 12 percent carbon (CS2).
absorbing solution is and gas exits from the absorber with a
content of 3 percent and a benzene content of 3 percent (because some of the ..,"' .. "' ........
evaporates). What fraction of the was recovered?
***18.15 If a liquid with a fairly high vapor pressure at room conditions is stored in a fixed size
tank that breathes, that is, has a vent to the atmosphere. because of ambient tempera-changes,
how much Joss per day occurs in g mol octane m3 of free space
under the following conditions, name]y the material stored is n-octane at the 50°C
during day and at at night. The space the octane consists of and
octane vapor that ex.pands and contracts. Ignore changes the liquid density.
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-585-320.jpg)
![Chap. 18
i: l
Problems 559'
before entering the reactor. All of the maleic acid produced the reactor is removed
with water in the bottom stream from the water scrubber. All of the C6H6• 02' CO2,
and N2 leaving the reactor leave in the stream from the top of the water scrubber, saturated
with H20. Originally. the benzene contains trace amounts of a nonvolatile
contaminant that would inhibit the reaction. This contaminant is removed by steam
distillation in the steam still. The still contains liquid phases of both benzene
and water (benzene is completely insoluble in water). The benzene phase is 80% by
weight, and the water phase is 20% by weight of the total of the two liquid phases in
the stilL Other process conditions are given in the flow sheet. Use the foHowing
vapor-pressure data:
Temperature (oF)
110
120
130
140
150
160
170
180
190
200
The reactions are
Benzene (psla)
4.045
5.028
6.195
7.570
9.178
11.047
13.205
15.681
18.508
21.715
o
CH-C(
OH
c<O
OH
Water (psis)
1.275
1.692
2.223
2.889
3.718
4.741
5.992
7.510
9.339
1] .526
+ 2~ + H20
~~ + 7!Ch ---+ 6CO:z + 3H20
Calculate:
(1)
(2)
(a) The moles of "' .... &li .... "" •.• '" undergoing reaction (2) per mole of benzene feed to the
reactor
(b) The pounds of H20 removed in the top stream from the dehydrator per pound
mole of benzene feed to the reactor
(c The composition (mote percent, wet basis) of the gases leaving the top of the
water scrubber
(d) The pounds pure liquid H20 added to the top of the water scrubber per pound
mole of benzene feed to the reactor](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-590-320.jpg)
![562 The Phase Rule and Vapor-Liquid Equilibria Chap. 19
where = number of degrees of freedom (Le" the number of independent
properties that have to specified to an of the intensive
properties of each of the system of interest)-not to be confused
with the degrees of freedom calculated in solving material
balances that can involve both intensive and variables.
P = number of phases that can exist in the system; a IS a .. '-'.uv""','"'
neous quantity of material such as a gas, a liquid, a solution,
or a homogeneous solid.
C = number of (independent chemical species) in the sys-
Without reaction the system, the number of components
(species) is equal to With reaction in the system, the value for C
may be reduced to a value than the number of species, as explained
in Appendix
In particular cases additional constraints must be included Equation (19.1)
that reduce the number of degrees of freedom (such as equilibrium and e]ectroneutrality
among ions solution), but we do not have space to discuss them. Refer to
the references at the of this chapter. such as Rao (1985), for "-
Look at Figure 19.1, the three-dimensional representation of the p-V-T proper-of
water that you previously saw in 16.6 in Chapter 16, Consider the
vapor phase.
Three-dimensional
of the properties of
water as a surface with
dimensions p. V, and T.
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-593-320.jpg)
![512 The Phase Rule and Vapor-liquid Equilibria Chap. 19
10
Total pressure == 1 atm
K
Kwaler
0.1 L-i--L--I----1-....L-..L.......I'--1..---L--'-....I..--'---lI....-L--'--'-.....L-.J.....",;L........I
o 0.2 0.4 0.6 0.8
Mol fraction acetone in the liquid phase.
1. Empirical equations such as·
Figure 19.7 Change of K of water
with composition at p == 1 atm.
K
. = (pc,i)exp [7.224 - 7.534rrr.i - 2.598 In Tr,i]
If Tc,/f > 1.2: I
Ptotal
2. Databases (refer to the supplementary references at the end of the chapter).
3. Charts such as Figure 19.8.
4. Thennodynamic relations (refer to the references at the end of the chapter).
Typical problems you should be able to solve that involve the use of the equilibrium
coefficient Kj together with material balances are:
1. Calculate the bubble point temperature of a liquid mixture given the total pressure
and liquid composition.
2. Calculate the dew point temperature of a vapor mixture given the total pressure
and vapor composition.
3. Calculate the compos~tion of the vapor and liquid streams, and their respective
quantities, when a liquid of given composition is partially vaporized (Dashed)
at a given temperature and pressure (the temperature must He between the bubble
and dew point temperatures of the feed).
4. Calculate the re]ated equilibrium vapor-liquid compositions over the range of
mole fractions from 0 to 1 as a function of temperature given the total pressure,
Analogous problems occur with respect to calculating the total pressure given a
fixed temperature.
Let's next outline the procedure to solve the first three problems cited above.
·SJ. Sandler, in Foundations o/Computer Aided Design, VoL 2, R.H.S. Mah and W.O. Seider,
eds .• p. 83, American Institute of Chemical Engineers, New York (1981).](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-603-320.jpg)
![574 The Phase and Vapor-Liquid Equilibria Chap. 19
an ideal solution, Equation (19.6) becomes
(19.7)
LYi ~ ---'-- + --.....::.-.. = 1
Ptotal Ptota!
(19.7a)
No. te that this equation is nonlinear and requires an iterative solution to deter- mme
You might use Antoine's equation for to formulate an equation ex-plicit
in T. Solve it as described above in connection with Equation (19.6).
Once the bubble point temperature is determined, you can calculate the vapor
composition from
..
X· I Yi =
, Ptotal
A degree-of-freedom analysis for the bubble point temperature for a bi-nary
mixture shows the degrees of freedom are zero:
Total variables = 2 2 + 2: Xl' X2; Yl' Y2; Ptotal; T
Prespecified values of variables = 2 + 1: xl' Ptotal
Independent equations::.: 2 + I: Y1 = K1x., Y2 = KzX2; Y1 + Y2 = 1
2. To calculate the dew point temperature (given the total pressure and vapor
composition), you can write Equation (19.5a) as Xi = y/Ki• and you know
k = 1 in the liquid phase. Consequently, you want to solve the equation:
1 = lL +
Kl
(19.8)
in which s are a function of as explained for bubble point
calculation. For an ideal binary solution, Equation (19.8) becomes
Equation (19.8a)
To calculate
1 = YI (19.8a)
Y2 ]
P~(Tdp)
nonlinear requiring an iterative numerical solution.
composition of the liquid phase, you use
The degree-of-freedom analysis is similar to that for the bubble point temperacalculation.
J](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-605-320.jpg)
![Sec. 19.2 Vapor-Liquid Equilibria in Systems
Basis: 1 mol of liquid
You want to solve the following equation to the bubble point tenape:ratlJre usmg
a nonlinear equation solver:
(
2697.55
760 == exp 15.8366 - -48.784 + + exp( 15.9798 - ---polymath
is T for which the vapor pressure of hexane is
mm the vapor Dlc:ssmrc octane is 661 rom Hg. respective mole
fractions in the vapor phase are
31
== 760 (0.04) = 0.164
Yea = 1 0.164 = 0.836
EXAMPLE 19.4 Flash Calculation for a Binary Liquid Mixture
Calculate the fraction of liquid that will remain at equilibrium when a mixture
of 68.6% hexane and 31.4% toluene is vaporized at 80°C and I atm.
Solution
You can treat the hexane toluene mixture as an solution, and use
(t 9.11) in solving problem.
3and4
Remember given in the problem statement are in mass. so
concentrations to fractions.
Component
Hexane
Toluene
The next
Basis: 100 g
Grams MoL wt.
68.6 86.17
31.4 92.13
100.0
data at 80°C are:
p'" (mm Hg) ]020
to calculate the values of
K = 1020 = I
hexane = 760
Ptotal
gmol Mol fr.
0.796 0.70
0.341 0.30
TTI7 r:oo
Toluene
290
1
K ;;::;; 0.745
hexane
'" 5n](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-608-320.jpg)
![586 The Phase Rule and Vapor-Liquid Equilibria Chap. 19
dioxide. The sulfur dioxide is to be removed by a stream of water. The entering water
contains no sulfur dioxide. The temperature 290 and the pressure on the process
is one atmosphere. Find the kilograms water per minute need to remove the sulfur
dioxide assuming that the exit water is in equilibrium with the entering gas, and
(b) the ratio of the water stream to the gas stream. The Henry's Law constant for sulfur
dioxide at 290 K 43 atmlmol fraction.
*19.14 A tank contains a liquid composed of 60 mole percent toluene 40 mole percent
benzene in equilibrium with vapor phase and air at 1 atro and 60oP.
What is the concentration of hydrocarbons in the vapor phase?
If the lower flarrunabity limit for toluene in air is 1.27% and benzene is 1.4%,
is the vapor phase flammable? ~
"'19.15 Fuel tanks for barbeques contain propane and n-butane. At 120oP, if an essentially full
tank liquid that contains liquid and vapor in eqUilibrium and exhibits a pressure of
100 psia, what is the overall (vapor plus liquid) mole fraction of butane in the tank.
"'19.16 Based on the following vapor pressure data, construct the temperature-composition
diagram at 1 atm for the system benzene-toluene, assuming ideal solution behavior.
Vapor mmHg
Temperature °c Benzene Toluene
80 760 300
1078 432
100 1344
1 lOA 1748 760
·19.17 Sketch a T-x-y diagram that shows an azeotrope locate and label the bubble and
dew lines and the point.
·19.18 What is (a) the pressure in the vapor phase, and (b) the composition of the vapor
phase in equilibrium with a liqUid mixture of 20% pentane and 80% heptane at 50°F?
the mixture is an ideal one at equilibrium.
*19.19 Methanol has a flash point at at which temperature its vapor pressure is 62 mm
Hg. What is the flash point (temperature) of a mixture 75% methanol and 25%
water. Hint: The water does not burn.
11119.20 Two kiJograms of a mixture of 50-50 benzene and toluene is at 60°C. As the total
pressure on the system is reduced. at what pressure will boiling commence? What
will be the composition of the first bubble liquid?
*"'19.21 The normal boiling point of propane is -42.1°C and the normal boiling point of nbutane
is -O.soC. (a) Calculate the mole fraction of the propane in a liquid mixture
that boils at - 31.2°C and ] atm. Calculate the corresponding mole fraction of the
propane in the vapor at - 31 (c) Plot temperature vs. propane mole fraction
for the system of propane and butane. -
*19.22 In the system n-heptane, n-octane at 200oP, determine the partial pressure of each
component in the vapor phase at liquid mole fractions of n-heptane of 0, 0.2~ OA. 0.6,
0.8 and] Also calculate the total pressure above each solution.](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-617-320.jpg)
![Chap. 19 Problems 587
1
;
/ ·"'19.25
"·19.27
Plot your results on a P-x diagram with mole fractions of C7 to the
right and mole fractions of Cg increasing to the The ordinate should pressure
10
Read from the plotted graph the following information:
The total and partial pressure of each at a mole frac-tion
of C, = 0.47 in the liquid.
Calculate the bubble point of a liquid mixture of 80 mole% n-hexane and 20 mole%
n-pentane at 200
Calculate the dew point a vapor mixture of 80 mole% n-hexane and 20 mole% n-pentane
at 100
A mixture of 50 mole% benzene and 50 mole% toluene is contained in a cylinder at
39.36 in. absolute. Calculate the temperature range in which a two phase system
can eXIst.
A liquid mixture of n-pentane and containing 40 mol per cent n-pentane is
fed continuously to a flash separator operating at 250°F and 80 psia. Determine:
(a) quantity of and liquid obtained the separator mol of feed.
(b) The composition of both the vapor the liquid leaving the separator.
hundred moles per minute of a binary mixture of A and B are separated in a two
stage (serial) process. In the first the liquid vapor flow rates exiting from
the are each 50 moles per minute. liquid then into a sec-ond
separator that operates at same temperature as the first and the respec-tive
exit streams of liquid and vapor from the second stage are each moles per
minute. temperature is same for each stage. and at that temperature, the vapor
pressure of A is 10 kPa while vapor pressure B is 100 kPa. Treat the liquids
and vapors as ideal.
Calculate the compositions of an the streams in the process, and calculate
the pressure in each stage.
·"'19.28 Most combustible reactions occur in the phase. any flammable material to
burn both fuel oxidizer must be present., and a minimum concentration of the
flammable gas or vapor in the phase must also exist. mlnlmum concentration
at which ignition will occur called the lower flammab1e limit (LFL). liquid
temperature at which the vapor concentration reaches the LFL can be found experimentally.
is usually measured a standard method called a "dosed cup flash
point" The Hflash of a liquid fuel is thus the liquid temperature at which
the concentration of fuel vapor in air is large enough for a flame to flash across the
surface of the fue] if an ignition source is present
The point and the LFL concentration are dosely related through the vapor
pressure of the liquid. Thus, if the flash point is known, the concentration can be
estimated, and if the concentration is known, the flash point can be estimated.
Estimate the point (the temperature) of liquid n-decane that contains 5.0 mole
percent pentane. The for pentane is 1.8% and that for n-decane is 0.8%. Assume
the propane-n-decane mixture is an liquid. Assume the ambient pressure is 100
kPa. This problem has adapted from Safety, Health, and Loss Prevention in](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-618-320.jpg)
![588
Chemical Processes,
cal Engineers. New
The Phase Rule and Vapor-Liquid Equilibria Chap. 19
lR. Welker and C. Springer, American Institute of Chemi(
1990), with permission.
"'19.29 You are asked to maximum pressure at which steam distillation
naphtha can be out at ] (the maximum allowable temperature). IS
injected into the liquid naphtha to vaporize it If (1) the distillation is out at
160°F, (2) the liquid naphtha contains 7.8% (by weight) nonvolatile impurities. and
(3) if the initial charge to the distillation equipment is 1000 Ib water and 5000 Ib of
naphtha. how much water wiI1 left in the still when last drop of naphtha
vaporized? Data: naphtha the MW is about 107, and p" (180°F) = 460 mm Hg,
(l60°F) = 318 mm .
"19.30 Late in evening of2l August 1986 a large volume toxic gas was released from
beneath and within Lake Nyos in the Northwest Province of Cameroon. Ari aerosol
of water mixed with toxic swept down the valleys to the north of Lake Nyos,
leaving more than 1.700 and dying people in its wake. The lake had a surface
area of 1.48 km2 and a depth of 200-250 m. It took 4 days to refill the lake, hence it
was estimated to have lost about 200,000 tons of water during the gas emission. To
the south of tbe lake and in the small cove immediately to the east of spillway a
wave rose to a height about 25 m.
conclusion of investigators studying this incident was that the waters of
Lake Nyos were saturated with CO2 of volcanic origin. in the evening of 21 August
a pulse of vo1canic gas-mainly CO2 but containing some H2S-was released
above a volcanic vent in the northeast corner of the lake. The stream of bubbles rising
to surface brought up more bottom waters highly charged with CO2 that gushed
out increasing the gas flow and hence the flow of water to surface much as a
warm soda overflows on release of pressure. At the surface, the of
transformed the accompanying water into a mist and sent a wave of water ....... "''''"-
ing across the lake. The aerosol water and CO2 mixed with a trace of H2S
down valleys to the north of lake leaving a terrible toll of injury and death in
its wake.
If the solution at the bottom of the lake obeyed Henry' slaw, how much was
....... ,...." ........ with the 200.000 tons of water, and what would the volume of the
at cubic meters? At 25°C Henry's law constant is I X lQ3 atmlmol fro
"19.31 Examine the statements below:
(a) "The vapor pressure of gasoline is about ]4 psia at 130oP."
(b) "The vapor pressure of system, water-furfural diacetate, 760 mm Hg at
99.96°C."
Are the statements correct? If not, correct them. Assume the numerical va]ues are
correct.
"'19.32 If pressure in the head space (gas space) in a bioreactor is ItO kPa and and
the oxygen concentration in the head space is enriched to 39.7%. what is the mole
fraction of the dissolved oxygen in liquid phase? What is the percent excess oxydissolved
in the liquid phase compared with the saturation value that could
obtained from air alone dissolved the liquid?](https://image.slidesharecdn.com/4rx73ugssik2wkqxrhad-signature-cf3c694303469e0b508378f724933067112ce89184f38800e9f0997500fe86e4-poli-140907101232-phpapp02/85/Basicprinciplesandcalculationsinchemicalengineering7thedition2-140607171816-phpapp02-619-320.jpg)
Basicprinciplesandcalculationsinchemicalengineering7thedition2 140607171816-phpapp02
- 1. Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com David M. Himmelblau / James B. Riggs Seventh Edition Prentice Hall International Series In the Physical and Chemical Engineering Sciences ,
- 3. IN PRENTICE HALL INTERNATIONAL t..JI~.............., PHYSICAL AND ......................................... ENGINEERING SCIENCES NEAL R. AMUNDSON, SERIES EDITOR, University oj Houston ADVISORY EDITORS ANDREAS ACRlVOS, Stanford University JORN DAHLER, University of Minnesota SCO'IT University of Michigan THOMAS 1. HANRATIY. University of Illinois JOHN M. PRAUSN1TZ. University of California L. SCRIVEN, University of Minnesota SAMUELS, AND Chemical Thermodynamics BEQUEITE BEQUETIE BIEGLER. Design Process Control: Modeling, Design, and Simulation Process Dynamics AND WESTERBERG Systematic Methods Chemical Process BROSlLOW AND Techniques of Model-based Control CROWL AND Louv AR Chemical Process Safety: Fundamentals with Applications, edition CONSTANTINIDES AND Numerical Methodsfor Chemical Engineers with MATLAB Applications CUTUP AND SHACHAM Problem Solving in Chemical with Numerical Methods DENN Process Fluid Mechanics ELLlOT AND Introductory Chemical Engineering Thermodynamics FOGLER Elements of Chemical Reaction Engineering. 3rd edition , , , ,"" ' HIMMELBLAU AND R1GGS Basic Principles and Calculations in Chemical Engineering, 7th edition AND MADDOX Mass Transfer KYLE Chemical and Thermodynamics, 3rd edition PRAUSNm. LICHTENTHALER, AND DE AzEVEDO Molecular Thermodynamics of Fluid·Phase Equilibria, 3rd edition PRENTICE Electrochemical Engineering Principles SHULER AND KAROl Bioprocess Engineering. 2nd edition STEPHANOPOULOS Chemical Process Control TEsTER AND MODELL Thermodynamics and Applications. 3rd edition TuRTON, BAll.JE, WHITING, SHAElWm Analysis, Synthesis, and Design of Chemical Processes, edition Fluid Mechanics for Chemical Engineering
- 4. SETH EDITI I I I L I I E L I I David . Himmelblau University of Texas James B. Riggs Texas h University PRENTICE HAll PTR PRENTICE HALL Professional Technical Reference Upper Saddle River, New Jersey 07458 www.phptr.com
- 5. Himmelblau. Basic and in chemical engineeri.ng.- 7th ed. I David M. Himmelblau and James B. Riggs. t, engineering·· Tables. I. TPISI .HS 2004 660'.2-«22 Publisher: Bernard Goodwin James B. II. Title. 2003016672 Editorial/production supervision: Patty Donovan Cover design dilUtor: Jerry Votta Compositioll, Inc.) Art director: Gail Coclcer-Bogusz Manu!ac:turing manager: Maura Zaldivar Marketing manager: Dan DePasquale Editorial assistant: Michelle Vincenti Full-service production manager: Anne R. Garcia ~ PRENTICE HAll PTR C 2004. 1996,1989.1982 by David M. HiIlnm~~bJJw to 1974.1967, and 1962 by Pearson Education, Publishing as Prentice Hall Profes~ljona1 Techmc.al.R.eiC1I'eD1:e Upper Saddle River, NJ 07458 The publisber offers exeeDent disooUBts 00 this book wbeD nl'fll",wlfl For more infonnation, please contact: U.S. Corporate and Government Sales 1-800-382-3419 corpsa1e.s@peacsonr.echgroup.oom For sales outside of the U.S .• please contact: International Sales 1-317-581-3793 in(ernational@pearsollr.echgroop.com Printed in the United States of America First printing ISBN: 0-13-140634-5 Pearson Education Ltd., London ........ ,,..,,..., Education Australia Pty, Limited, Sydney ....... '''''''n Education Singapore, Pte. Ltd. P"'IIIr"lMl .t::.aIJClllIl0D North Asia Ltd .• Hong Kong Pt"A1r!tnn Education Canada, Ltd., Toronto ................ ., 1::!(!~leac:lon de Mex.ico, S.A. de C.V. Pearson t'.OUcabJon-J Pearson Education Malaysia. Pte. Ltd. qmmtfity tor bulk p1.l!"clutsa or spedaI SIIIe:s.
- 6. To Betty for the 7th time DMH To Brenda, Michelle, J. Michael, and Matt JBR
- 7. About Prentice Hall Professional Technical Reference With origins reaching back to the industry's first computer science publishing OF02ram in formally launched as its own imprint in 1986, Prentice Halt proreS:ilonial Technical (PH PTR) has developed into the leading provider of rOI'I,nll'::l1 in the world today. editors now publish over 200 books annually. by 1&:I;II(I&:Irc in of computing, engineering, and business, Our roots are firmly planted in the soil that gave rise to the technical revolution. Our bookshelf contains many of the industry's computing and engineering classics: Kernighan and Ritchie's C Programming Language. Nemeth's UNIX System Adminstration Handbook, Horstmann's Java, johnson's High-Speed Digital Design. PH PTR acknowledges its oU"I-"""vu,;> beR~inning:s,wlhile it looks to the future for inspiration. We continue and break new ground in publishing by providing today's with tomorrow's solutions.
- 8. CONTENTS PREFACE README FREQUENTl V ASKED QUESTIONS PART! INTRODUCTION 1 DIMENSIONS, UNITS, AND THEIR CONVERSION 1.1 Units and Dimensions 1.2 Operations with Units 1.3 Conversion of Units and Conversion Factors 1.4 Dimensional Consistency (Homogeneity) 1.5 Significant Figures 1.6 Validation of Problem Solutions 2 MOLES, DENSITY" AND CONCENTRATION 2.1 Mole Density xv xxi xxv 1 5 6 11 14 21 24 30 .. 43 48 vii
- 9. I viii 2.3 Specific Gravify 2.4 F10w Rate Mole Fraction and (Weight) Fraction 2.6 Analyses of MuHicomponent Solutions and Mixrures 2.7 Concentration 3 CHOOSING A BASIS 4 TEMPERATURE 5 PRESSURE Pressure and Its Units ~easurement Pressure 5.3 Differential Pressure Measurements PART 2 MA TERJAL BALANCES 6 INTRODUCTION TO MATERIAL BALANCES 6.1 The Concept of a Material Balance Open and Closed Systems 6.3 Steady-State and Unsteady-State Systems ~ultiple Component Systems 6.5 Accounting for Chemical Reactions in Material Balances 6.6 Material Balances for Batch and Semi-Batch Processes Contents 51 56 57 59 18 89 99 100 114 129 133 134 136 138 144 149 151 1 A GENERAL STRATEGY FOR SOLVING MATERIAL BALANCE PROBLEMS 166 7.1 Problem Solving 7.2 Strategy Solving Problems 8 SOLVING MATERIAL BALANCE PROBLEMS FOR SINGLE UNITS WITHOUT REACTION 167 168 196
- 10. Contents ix" 9 THE CHEMICAL REACTION EQUATION AND STOICHIOMETRY 225 9.1 Stoichiometry 226 9.2 UU.IIV!V)o;y for API)llClltlO!nS of Stoichiometry 10 MATERIAL BALANCES FOR PROCESSES INVOLVING REACTION 260 to. 1 ~peCles; Material Balances 10.2 Element Material Balances 10.3 Material J:i8.1anCles Involving Combustion 11 MATERIAL BALANCE PROBLEMS INVOLVING MULTIPLE UNITS 12 RECYCLE, BYPASS, PURGE, AND MATERIAL BALANCES Introduction Recycle without cne:m1cal Reaction t<eCVCle with Che~mi(;al K,eacltlon and Purge INDUSTRIAL APPLICATION Industrial Application of Material Balances PART 3 GASES, VAPORS, LIQUIDS, AND SOLIDS 13.1 The Ideal Law 13.2 Ideal Mixtures and Partial Pressure 261 278 283 341 342 347 355 365 373 396 401 402 4 13.3 Material Balances Involving Ideal 416 14 REAL GASES: COMPRESSIBILITY 435 15 REAL GASES: EQUATIONS OF STATE 459 16 SINGLE COMPONENT TWO-PHASE SYSTEMS (VAPOR PRESSURE) 415 Diagrams 476 Modeling and Predicting Vapor Pressure as a FUDlcticm of Temperature 485
- 11. x Contents 17 TWO·PHASE GAS-UQUID SYSTEMS (SATURATION, CONDENSATION, AND VAPORIZATION) 509 17.1 Saturation 17.2 Condensation 17.3 Vaporization 18 TWO-PHASE GAS-LIQUID SYSTEMS (PARTIAL SATURATION AND HUMIDiTY) 18.1 Terminology Involved for P~rlll~1 Saturation 18.2 Material Balance Problems Involving l-'!:In"I~1 Siatuf'ati{)O 19 THE PHASE RULE AND VAPOR-LIQUID EQUIUBRIA 19.1 The Gibbs Phase Rule Vapor-Liquid Eqvllibrja in Binary Systems , 20 LIQUIDS AND GASES IN EQUILIBRIUM WITH SOLIDS PART 4 ENERGY BALANCES 21 ENERGY: TERMINOLOGY, CONCEPTS, AND UNITS 21.1 21.2 Terminology ASSOCHttect of Energy Energy Balances 22 INTRODUCTION TO ENERGY BALANCES FOR PROCESSES WITHOUT REACTION 22.1 Concept of the Conservation of .... np',rlJv 22.2 Energy Balances for '-lU ..... " •• , uns,tea'ly-~;rate Systems Energy Balances for Closed, Steady-State Systems 22.4 Energy Balances for """ .......... Unsteady-State Systems Energy Balances for Open. Steady-State Systems 510 514 537 538 544 560 561 565 590 603 601 608 613 645 646 648 655 666
- 12. Contents 23 CALCULA nON OF ENTHALPY CHANGES 23. i Phase Transitions .2 Capacity Equations Tables and ChartS to 23.4 Computer Databases ... Tn .. ", ... Enthalpy Values 24 APPLICATION OF ENERGY BALANCES IN THE ABSENCE OF CHEMICAL REACTIONS 1 Simplifications of the General Energy Balance 24.2 The Strategy Solving Energy Balance Problems 24.3 Application of the Energy Balance to Closed Systems Application the Energy Balance to Systems xi 681 682 690 699 705 111 718 723 728 26 ENERGY BALANCES: HOW TO ACCOUNT FOR CHEMICAL REACTION 763 1 The Standard Heat (Enthalpy) of Fonnation 25.2 The Heat (Enthalpy) of Reaction 769 25.3 Merging Heat of Fonnation with Sensible Heat of a Compound in Making an Balance 780 25.4 The of Combustion 26 ENERGY BALANCES THAT INCLUDE THE EFFECTS OF CHEMICAL REACTION 26.1 Analysis of the Degrees of Freedom to Include the Energy Balance with Reaction 26.2 Applications of Energy Balances in Processes that Include Reactions 21 IDEAL PROCESSES, EFFICIENCY, AND THE MECHANICAL ENERGY BALANCE 27.1 Ideal Reversible Processes 27.2 Efficiency 27.3 The Mechanical Energy Balance 785 802 803 806 836 837 843 848
- 13. xii 28 HEATS OF SOLUTION AND MIXING 28.1 Heats of Solution. Dissolution, and Mixing 28.2 Introducing the Effects of Mixing into the "'"""'''O''J Balance 29 HUMIDITY (PSYCHROMETRIC) CHARTS AND THEIR USE 29.1 TenninoJogy 29.2 The Humidity (psychrometric) 29.3 Applications of the Humidity PART 5 SUPPLEMENTARY MATERIAL (ON THE ACCOMPANYING CD) 30 ANALYSIS OF THE PROCESS " ................ " IN A STEADY -STATE 31 SOLVING MATERIAL AND ENERGY BALANCES USING PROCESS Contents 864 865 872 884 885 888 897 913 SIMULATORS (FlOWSHEETING CODES) 938 32 UNSTEADY-STATE MATERIAL AND ENERGY BALANCES 910 APPENDICES 997 A ANSWERS TO SELF-ASSESSMENT TESTS 001 B ATOMIC WEIGHTS AND NUMBERS 1030 C TABLE OF THE zO AND Z' FACTORS 1031 D PHYSICAL OF VARIOUS ORGANIC AND INORGANIC SUBSTANCES 1036 E HEAT CAPACITY 1048 F HEATS OF FORMATION AND COMBUSTION 1052 G VAPOR 1051 .. «
- 14. Contents kiii H HEATS Of SOLUTION AND DILUTION 1058 ENTHALPV..cONCENTRATION DATA 1069 J THERMODYNAMIC CHARTS 1065 K PHYSICAL PROPERTIES Of PETROLEUM fRACTIONS 1067 L SOLUTION OF SETS OF EQUATIONS 1069 M FlrnNG FUNCTIONS TO DATA 1085 N ANSWERS TO SELECTED PROBLEMS 1089 INDEX 1106
- 16. E This book is intended to serve as an introduction to the principles and techniques used in the field of chemical, petroleum. and environmentaJ engineering. Although the range of subjects deemed to be in the province of "chemkal engineering" has broadened over the last decade, the basic principles involved in chemical engineering remain the same. This book lays a foundation of certain information and skills that can be repeatedly employed in subsequent courses as well as in professionallife. Our Motivation for Writing This Book Far too many chemical engineering textbooks have become difficult, dry, and demoralizing for their readers. With this book, we have maintained a conversational style and detailed explanation of principles both in the text and examples to provide a readable yet comprehensive text. We have strived to maintain a suitable balance between understanding and developing skills. Our vision is to avoid comments (from a student about a different text) such as: "My text is useless, well not really, I use it to kill roaches in my room." Piaget has argued that human intelligence proceeds in stages from the concrete to the abstract and that one of the biggest problems in teaching that the teachers are fonnal reasoners (using abstraction) while many students are still concrete thinkers or at best in transition to formal operational thinking. We believe that there is considerable truth in this viewpoint. Consequently, we initiate most topics with simple examples that illustrate the basic ideas. In this book the topics are presented xv
- 17. xvi Preface in order of assimilation. We start with easy material followed by more difficult ma-terialro readers a "breather' before passing over each hump. Assumed Prerequisites The level the book is directed to the first course in chemical engineering, which usually occurs in a student's sophomore year. We have assumed that you as a reader to have completed the second part calculus and started chemistry. Familiarity with hand-held calculators is essential, but computer programming is not Familiarity with software would be helpful, but is not criticaL Intended Audience We believe that the main category of individuals who will use this book will students of chemical engineering. However, the book is wen designed for courses for nonchemical engineers as wen as independent study. long-distance learning. and review licensing examinations through its features. Our ·Objectives This book is not an introduction to chemical engineering as a profession. We have focused instead on five general objectives in writing this book: 1. To introduce you the principles and calculation techniques used in chemical engineering. 2. To acquaint you with what material and energy balances and how to for~ mulate and solve 3. To you efficient and consistent methods of problem solving so that you can effectively solve problems you will encounter after leaving school. 4. To offer practice defining problems, collecting data, analyzing the data breaking it down into basic patternst and selecting pertinent information for application. To review certain principles applied physical chemistry. In addition to focusing on the above objectives, we expose you to background infonnation on units and measurements of physical properties; basic laws about the behavior of liquids, and solids; and some basic mathematical tools. Other ob~ 1
- 18. .""""Ih" ... .:- that an instructor may want to include a such as and communication skills, information about professional activities, developing a pro-fessional attitude~ establishing personal goals, developing social and so on, must implemented sources. Economic feasibility, a major in engineering making. costing, and optimization, have been omitted because of tack We have not focused on solve problems even though it is questions as well as removing some of process simulation software to ~n~ll,,~r ... good in and handling "what drudgery in problems bec:am:;e 1. the too closely to cookbook-style problem solving; 2. learning to use the software with ease takes some and 3. development a problem-solving strategy is taken out the hands of the user by the software programmers. software provides too much ~m.daJtlce neophytes. Organization and Scope of this Book major portion book comprises fOUf parts: L Background information (Chapters 1-5) Material balances (Chapters 6-12) 3. Behavior of gases, liquids, and solids Part 4. Energy balances (Chapters 21-29) Ch~lpters 13-20) In addition, on the accompanying CD, Chapter 30 treats the degrees of dom, Chapter 31 process simulatorst Chapter 32 state material ergy balances. en- A series appendices foHow that include. in addition to tables and charts of physical prc.pelr'tles. miscellaneous information you will Look at Table of Contents details. In the CD that accompanies this valuable tools: 1. Polymath: "f"I1~hlJ!>I"p that solves equations, and can be used without reading any ins tructi ons. to physical property for over 740 compounds. A Supplementary Problems Workbook containing 100 problems with complete ................ 1£ .. ' .... solutions, and another 100 problems with answers. 4. Descriptions of process equipment, and animations that illustrate functions equipment
- 19. xviii Preface 5. Problem-solving suggestions including check lists to help you diagnose and overcome problem-solving difficulties you may experience. To provide an appreciation of what processing equipment re3.Ily looks like and how it works, in the files on the CD disk in the worked-out problems are numerous pictures of the equipment along with an exp]anation of their function and operation. Problem Sets We have included several categories problems in the books to assist in study. 1. tests with answers (in Appendix A) foHow each section, 2. Thought and discussion problems follow the tests. Thought problems require reflection more than calculation. Discussion problems. which can be used as the basis of research. papers, and class discussions, pertain to broader issues and are more open ended. 3. Homework-type problems are listed at the end of each chapter, one-third of which have answers (in Appendix N). Each of the problems is rated 1 to 3 (using asterisks) to indicate the degree of difficulty, with 3 being the most difficult. 4. The contains more than 100 worked-out examples and another 100 prob-lems with answers keyed to in the chapters in the text. An of the examples and problems are designed build your problem-solving skills. Miscellaneous Useful Features in this Book To make the book more usable and friendly, we have incorporated a number of beneficial features: 1. A list of contents at beginning of each chapter. 2. A list of instructional objectives at the beginning of each chapter. 3. Important terms appear in boldface type. 4. A glossary has been at the end of chapter. S. Supplementary references that you can use to additional information are listed at the end of each chapter. 6. Web sites containing information and links are listed at end of each chapter.
- 20. Preface xix 7. The examples are simple and concrete so that the book is both teachable and useful for self instruction. S. The chapter topics are independent but linked through a few·principles. 9. The examples demonstrate a proven problem-solving strategy. New Features in the Seventh Edition The seventh edition is a completely rewritten and revised version of Basic Principles and Calculations in Chemical Engineering. Instead of five long chapters, the book is now comprised of 32 short chapters. each typically corresponding to one class session in a schedule of three meetings a week. New features include: 1. A consistent, sound strategy for solving material balance and energy balance problems, one can be used again and again as a framework for solving word problems, which 1s explained in Chapter 7. All of the examples in this book showing how to solve material and energy balances have been fonnulated according to this strategy. 2. The examples and problems in each chapter have been augmented to include expanded areas of importance to chemical engineers such as safety, semiconductor processing, and biotechnology. 3. The chapters on material balances have been revised to offer practice in find· ing out what the problem is, defining it. collecting data to be used in the problem, analyzing the information pertaining to the problem in order to relate it to what you know about similar problems, and, in effect, doing everything but testing the solution experimentally. 4. The extent of reaction has been added to the tools used to solve problems involving chemical reactions. 5. The degree of freedom analysis in solving problems has been emphasized and simplified. 6. A glossary has been added to each chapter. On the CD that accompanies this book is 7. A new version of Polymath, a self-documented, widely used software package that runs on pes and can solve linear, nonlinear, and differential equations as well as regression problems. 8. A new physical properties database that contains retrievable physical properties (such as vapor pressures and heat capacities and enthalpies for 740 compounds plus the steam tables).
- 21. Preface ACKNOWLEDGMENTS We are indebted to many former teachers, colleagues, and students who directly or indirectly helped in preparing this book, in particular the present edition of it. Special thanks go to Chris' Bailor for getting the manuscript to its final form.. and to H. R. Heichelheim and Dale Slaback for their reviews of the manuscript. We also want thank Professor C. L. Yaws for his kindness in making available the physical properties database that is the basis of the physical properties packages in the CD in back of this book, and also thanks to Professors M. B. Cutlip and M. Shacham who graciously made the Polymath software available. Far too many instructors using the text have contributed their corrections and suggestions to list them by name. However, we do wish to express our appreciation for their kind assistance. Any further comments and suggestions for improvement of the book would be appreciated. David M. Himmelblau Austin, Texas James B. Riggs Lubbock. Texas
- 22. READ ME Welcome to our book Basic Principles and Calculations in Chemical Engineering. Several tools exist in the book in addition to the basic text to aid you in learning its subject matter. Don't neglect to use them. Learning Aids 1. Numerous examples worked out in detail to illustrate the basic principles. 2. A consistent strategy for problem solving that can be applied to any problem. 3. Figures, sketches, and diagrams to provide reinforcement of what you read. 4. A list of the speci(ic objectives to be reached at the beginning of each chapter. Self assessment tests at the end of each section, with answers so that you can evaluate your progress in learning. 6. A large number of problems at the end of each chapter with answers provided in Appendix N for about a third of them. 7. Thought and Iliscussion problems that involve more reflection and consideration than the problem sets cited in #6 above. 8. An appendix containing data pertinent to the examples and problems. 9. Supplementary references for each chapter. 10. A glossary following each section. 11. A CD that includes some valuable accessories:
- 23. xxii Read Me a. Polymath-an equation-solving program that does not require training to use. b. Software that contains a physical properties database of over 700 compounds c. Supplementary Problems Workbook with over 100 completely solved prob-and another 100 problems with answers. Workbook indexed descriptions of process equipment, and ani-mations that the functions equipment. You can instantly access these pages want to look something up by clicking on the page number. e. Problem-solving suggestions including check lists to and overcome problem-solving difficulties that you experience. 12. In the pocket the back of the book is a set of tables (properties of water) in SI and American Engineering units. Scan through book now locate these features. Good Learning Practices (Learning How to Learn) You cannot put the same shoe on every foot. PubliUus Syrus Those who study characteristics educational psychologists say people learn by and reflecting, and not by watching listening to someone else telling them what they are to learn. is not teach-and listening is learning." You learn by doing. Learning involves more than memorizing. Do not memonzmg Recording. copying, and outlining notes or the text to memorize problem will help in really lInr'<!Iot"_ standing how to solve material and energy balance Practice will help to be able apply your knowledge to problems that you have not seen before. Adopt good learning practices. You will find that skipping the text and to equations or examples to solve problems may work but in the long run will lead to frustration. Such a strategy is called "formula centered," is a very poor way approach a problem-solving subject. adopting it. you will not be able to each proble:m will be a new challenge, and the interconnections among similar be llUi~;)vIU.
- 24. Read Me xxiii"' Various appropriate learning (information processing) hence you should reflect on what you do to and adopt techniques best suited for you. students leam through thinldng things out in solitary study. Others prefer to talk things through with peers or Some focus best on examples; oth-ers abstract ideas. Sketches used in explanation usually appeal to Do you get bored by going over the same ground? You might want to take a battery of tests to assess style. Students find such invento-and helpfuL CD that accompanies trus book to read about learning styles. Whatever your learning style, here are some to enhance learning that we feel are appropriate to on to you. Suggestions to Enha Learning 1. Each chapter in this book will require three or more hours to read, assimilate, and practice your skins in solving pertinent problems. Make aHowance in your schedule so that you will have read the pertinent before coming to class. 2. If you are enrolled in a work with one or more classmates. if permitted. to exchange ideas. But do not rely on someone to do your work for you. 3. Learn everyday. up with the scheduled assignments--don't get behind because one topic bunds on a previous one. 4. Seek answers to unanswered questions right away. s. Employ that is, every 5 or 10 minutes stop for I or 2 minutes. and summarize what you have learned. Look for connecting ideas. Write a summary on paper it helps. Suggestions as to How to Use Th Book Effectively How can you best use of this book? Read the objectives ... "" ... "... .. .0 studying each section. Read the text, and when you to an example, first cover the solution and tty to solve the stated problem. Some people. those who learn by reading examples. might look at the examples first and then the text. After reading a solve the self·assessment problems at the end of the section. The answers are in Appendix A. After completing a chapter, solve a of the problems listed at end of the chapter. Feynman, the Nobel laureate in physics. made point "You do not know anything until you have practiced." Whether you solve the problems using hand calculators or computer proarams is nn to you, but use a systematic approach to formulatino- tJ,. .... ~_1"
- 25. xxiv Read Me proper solution. Use the supplement on the CD in the back of the book (print it out if you need to) as a source of examples additional solved problems with which to practice solving problems. This book functions as a savings account-what you put in you get out, with interest I
- 26. FREQUENTLY ASKED QUESTIONS What Do Chemical Engineers Do? Chemical engineering is an intriguing, challenging, and flexible profession. Chemical engineering graduates work in a wide variety of industries, as indicated in Table A. TABLE A Initial Job Placement of Chemical Engineering Graduates In 2000-2001 In Percent (Source: AIChE, NY, NY). Industry BS MS PhD Chemical 23.3 1.8 21.3 Fuels 15.7 7.6 . 10.6 Electronics 15.9 27.4 29.S Food/Consumer Products 10.6 6.6 4.3 Materials 3.1 2.5 3.4 Biotechnology & Related Industries (Pharmaceuticals) 9.3 14.7 15.9 Pulp & Paper 2.t 1.5 1.5 Engineering Services Design & Construction 5.6 6.6 1.9 Research & Testing 1.8 4.1 3.4 Environmental Engineering 2.4 2.5 1.5 Business Services 5.8 2.0 2.9 Other Industries 3.9 2.5 3.9
- 27. xxvi Frequently Asked Questions In the industries listed in Table A, chemical engineers focus on design, operation, control, troubleshooting, research, management, and even politics, the latter because of environmental and economic concerns. Some chemical engineers design processes and solve problems using their computing skills and specialist knowledge of reactions, separations, heat transfer, fluid flow, control, and economics. Others lead teams of experts from various disciplines in managing installations and directing plant operations. You will find chemical engineers use their expertise in management, marketing, infonnation technology, business, and financial planning. If you want additional infonnation, look at some of the web sites listed in Table B. TABLE B Web Sites Providing Information on Chemical Engineering http://www.aiche.org http://www.chemspy.com h up://www.j-edainc.comlGroups/ChemicalEngineering.htm http://www.monstertrak.co'm http://www.umin.che What Other Text Books Can I Read That Cover the Same Topics as This One? Here are some of the more recent ones; many others were published in the period 1950-1980: Felder, R.M .. and R. W. Rousseau. Elementary Principles of Chemical Processes. 3rd ed., John Wiley, New York (2000). Luyben, W.L., and L.A. Wenzen. Chemical Process Analysis: Mass and Energy Bal~ ances. Prentice-Hall, Englewood Cliffs, N.J. (1988). Reklaitis, E. V., and D. R. Schneider. Introduction to Material and Energy Balances. John Wiley, New York (1983). Shaheen, E. I. Basic Practice of Chemical Engineering. Houghton Mifflin, Palo Alto, CA (1975). What Computer Codes Can Be Used to Solve the Equations Formulated in Homework Problems (and Examples)? Software packages involving symbolic and numerical calculations along with graphics have over the last decade become essential tools for all enpiopp ":- - .--.
- 28. Frequently Asked Questions xxvii great potential of computers is their capacity to do anything that can be described mathematically as a series of operations logical decisions-theoretically. From a practical viewpoint, you should not merely whether it is feasible for a task to be performed on a computer. but whether it is sensible. Two questions should asked in reaching a decision: (1) can the be performed (or prob~ lem solved) at all without the use of a computer; and (2) is it better or cheaper or faster to use a computer to solve a problem than a hand-held calculator (or no machine all)? Some commercial software that solves equations and provides graphics (and much more in many cases) in order of increasing difficulty in learning how to use is: Polymath TK solver Mathe ad Matlab Mathematica Maple You can find many of these codes installed on university computers, or purchase them at reasonable prices with an educational discount. On the CD that accompanies this book you will find a software program called Polymath. This program solves linear equations, nonlinear equations, differential equations, and carries out linear and nonlinear regression fitting). significant advantage of Polymath is that you do not have to read an instruction manual to use You just look at the sample equations displayed on screen and follow their format. In addition, process simulators solve equations as part of their many other functions. Refer Chapter 31 for examples of such codes. Most departments of chemical engineering have licenses for one or more of these process simulators~ but they take some effort to learn how to use. Where Can I Find More Examples of Problems and Their Solutions? The CD that accompanies this book contains more than 100 additional examples of problems with detailed solutions, and another 100 problems with answers. In addition, you can find numerous examples problems with answers in the references cited above.
- 29. xxviii Frequently Asked Questions Where Can I Get Information and Data To Solve the Homework Problems If the Appropriate Data is Not in the Problem Statement, Appendix, or CD? Accurate values of physical properties are needed in almost aU phases of chemical engineering design and analysis. Various ways to obtain data for the physical properties of components besides the Internet are: 1. Employer's database 2. Design software (such as flowsheeting codes) 3. On-line databases 4. On~line bulletin boards/e-roml 5. Personal files and books 6. Departmental library 7. Employer's main library 8. Outside library 9. Technical magazines/newsletters 10. Professional society meetings 11. Trade association meetings 12. Continuing education courses 13. Other engineers in department 14. Outside consultants 15. Regulatory agencies 16. Raw material/equipment vendors 17. Clientslcustomers 18. Direct experimentation Much of the data you want is available with little or no cost, particularly over the Internet. You win be interested in using physical property databases in one of three ways: 1. Retrieve an isolated value to be used in a calculation or in the calculation of other property values. Often a value is to be employed in hand calculations, or perhaps fed as input data to a computer program for further calculation. 2. Serve as a subroutine (such as a physical properties library) to another computer program to provide physical property data for process calculations. 3. Provide interactive capabilities for the rapid rendering of physical properties of substances of interest for parametric studies of process units, that is, "ask what if."
- 30. Frequently Asked Questions xxix Many of the materials we talk about and use every day are not pure compounds. but nevertheless you can obtain information about the properties of such materials. Data on materials as coal, coke, petroleum products, and natural gas-which are the main sources of this country-are available in refer-ence books and handbooks. Tables C, 0, and E. TABLE C Sources of Physical Property Data American Chemical Society. 17,000 compounds, on disk. ACS, Washington, D.C. (1994). American Chemica] L-n4e:ml,CUI Abstracts Service, ACS, Washington, D.C. (Continuing printed. mi- J:F'.T"IIf"P. with over 20 million abstracts.) New York (1941). ecn'THt:.Ul Dala Book-Petroleum Refining. New York: (1970). compounds. Also CD-Disk, ongoing, GeseUschaft fur Chemisches Apparatwesen e. v,, Berlin, on and thennodynamic properties.) 10.000 compounds. Analytical Chemistry and General Scientific Data Analysis. Cambridge (2001), gives properties and prediction equations for over on-line; Amer. lnst Chemical Engineers. New International Data Series. "London. (Continuing series of data and Environmental Data Information Network. Ecdin, data on 25,0CXl substances, 103.000. Distributed by Technical Database Services (IDS). Chemistry, eRe Press, Boca Raton, FL., a.n.rlual editions. Also on Lange's Handbook of Chemistry and Physics, McGraw-Hill, New York; issued periodically. Web. for Synthetic Fuels"; Hydrocarbon Process. p. 229 (May 1980). Laboratories, PPDS2, 1,600 compounds; Glasgow, UK, ongoing. Narural Processors Suppliers Association, Engineering Data Book, Okla. (Continlling editions.) R. H., D. W. and J. O. Maloney. Perry's Chemical 7th McGraw- Hill, New York (2000). Also on the Web. B. D. 11. M. Prausnitz, and 1. O'Connell. The Properties 5th ed., McGraw- York (2002). laboratory, Chemsafe, 1,600 gases, liquids. and dusts that can distributed on disk. tape. on- Braunschweig, Germany (1995). STN Express, provides access to many databases Chemical Abstracts Service.. Ohio, continuing. Research Center, Texas A&M University, Vapor Data Profilefor 5,500 on (1994). L. Chemical Properties Handbook. McGraw-Hill. New York (1999).
- 31. Frequently Asked Questions TABLE D Proressional Journals Archival journals (in English) AlChE Journal Canadian Journal of Chemical Engineering Chemical Engi~ering Comnrunications Chemical Engineering Journal (Lausanne) Chemical Engineering Science Computers and Chemical Engineering Industrial &: Englneering ChemJ,srry Research, Journal of Chemical and Engineering Data Journal of Chemical Engineering of Japan . Journal of Chemical Technology and Biotechnology Other journals and magazines Chemical Engineer (London) ChLmical Engineering Chemical Engineering Progress Chemical Processing Chemical Technology Chemistry and Industry (London) Table E lists some valuable web sites that have links to,hun<lre<is other sites. TABLE E Data Source on the Web : ," http://www.chemicalonline.com http://www.chempute.comlmain.htm http;//www.cheresources.com http;//www.che.ufl.edulwww-che http://www.deb.uminho.ptlfontesJchem_engleduC8.tionlchee_UnkS..htfu httpJ/www.knovel.com http://www.mwsofiware.comldragonldesc.html http://myplant.com http://www.retallick.comlresourceslnetresrc.html http://www.shef.ac.uklunil8cademiclA-C/cpe/mpitt/chemengs.html ..
- 32. I 1 I CHAPTER 1 Dimensionsl Units, and Their Conversion 2 Moles, Density, and Concentration 3 Choosing a Basis 4 Temperature 5 Pressure PAGE 5 42 18 89 99 Part 1 begins your introduction to chemical engineering calculations by reviewing certain topics underlying the main principles to be discussed. You have already encountered most of these concepts in your basic chemistry and physics courses. Why, then. the need a review? First, from experience we have found it necessary to restate these familiar basic concepts in a somewhat more precise and ,""U",';U...,J, fashion; second, you will need practice to develop your ability to analyze and work engineering problems. If you encounter new material as you go through these chapters, or if you flounder over little gaps in your skills or knowledge of old material. you should devote extra attention to the chapters by solving extra prob1ems in the set that you will find at the end of each chapter. To read and understand the principles discussed in these chapters is relatively easy; to apply them to different unfamiliar situations is not. An engineer becomes competent in his or her profession by mastering the techniques developed by one's predecessors-thereafter comes the time to pioneer new ones. What I hear, I forget; What I see, I remember; What J do, I understand. Confucius Part 1 begins with a discussion of units, dimensions. and conversion factors, and then goes on to review some terms you should already be acquainted with, including: 1
- 33. 2 Part 1 You are FJgure Part 1.1 The bridge to success. a. Mole and mole fraction b. Density and specific gravity c. Measures of concentration d. Temperature e. Pressure Part 1 Introduction You wan1 to get here + A finn grasp of this information as presented in the next five chapters wi 11 heJp guarantee "plug-and-play" acquisition of the information in the remaining chapters. You will find that adding new ideas and techniques will be as easy as copying and pasting images in a computer. Consider the following story. One night a group of nomads were preparing to retire for the evening when suddenly they were surrounded by a great light. They knew they were in the presence of a celestial being. With great anticipation, they awaited a heavenly message of great importance that they knew must be especially for them. Finally, the voice spoke. "Gather as many pebbles as you can. Put them in your saddle bags. Travel a day's journey and tomorrow night will find you glad and it will find you sad." After the light departed, the nomads shared their disappointment and anger with each other. They had expected the revelation of a great universal truth that would enable them to create wealth, health. and purpose for the world. But instead they were given a menial task that made no sense to them at all. However, the memory of the brilliance of their visitor caused each one to pick up a few pebbles and deposit them in their saddle bags while voicing their displeasure. They traveled a day's journey and that night while making camp. they reached into their saddle bags and discovered every pebble they had gathered had become a diamond. They were glad they had diamonds. They were sad they had not gathered more pebbles.· *Schlatter. 1. W., quoted in A Second Helping of Chicken Soup for [he Soul, 1. Canfield and M. Hansen (eds.). Healrh Communications, Deerfield Beach, FL (l995). ~ l .
- 34. Part I Introduction SUPPLEM NTARV REFERENC S M., Wiley, W. Rousseau. Elementary Principles (2000). Chemical Processes. 3rd John Luyben. W. L.. and Chemical Process Analysis: Mass and Energy Balances. Prentice-Hall, ..., un.nn Cliffs, N.J. (1988). Reklaitis, E. V., and D. R. .. ylJ' ... l Introduction 10 Material and Energy Balances. John Wiley, N.Y. (1983), I. Basic Practice of Chemical Houghton Mifflin, Palo Alto, CA (1915).
- 36. CHAPTER 1 DIMENSIONS, UNITS, AND THEIR CONVERSION 1.1 Units and Dimensions 1.2 Operations with Units 1.3 Conversion of Units and Conversion Faciors 1.4 Dimensiona' Consistency (Homogeneity) 1.5 Significant Figures 1.6 Validation of Problem Solutions Your objectives In studying this chapter Bre to be able to: 1. Understand and explain the difference between dimensions and units. 2. Add, subtract multiplYl divide units associated with numbers. 3. Specify the and derived units in SI and American ing (AE) systems for mass. length, volume. density, and time, and their equivalents, 4. Convert one set of units in a function or equation into another equi-valent for mass, length, area, volume, time, and force. Explain the difference between weight and mass. 6. Define and know when to use the gravitational conversion factor 9c' 7. Apply the concepts of dimensional consistency to determine the validity of an equation or function. S. Employ an appropriate number of significant figures in your calculations. ''Take care of your units and they will take care of you. " Anonymous 6 11 14 21 24 30 At some time every engineer's life comes the exasperating sensation frustration in problem solving. Somehow, the answers or the calculations do not come out as expected. Often this outcome arises of errors in the handling units. 5
- 37. Dimensions, Units, and Their Conversion Chap. 1 The use of units along with the numbers in your calculations requires more attention than you probably have been giving to your computations in the past. In addition. you will discover that checking the consistency of units U, yO'UI' equations win prove to be a valuable tool that will reduce number errors you commit when performing engineering calculations. Looking Ahead In this chapter we review the SI and American" Engineering systems of units, show how conversions between units can be accomplisl)ed efficiently. and discuss the concept of dimensional homogeneity (consistency). V:lt;. a~so provide some com- .u""' .. '"'" with to the number of significant to use your calculations. 1.1 Units and Dimensions Engineers and scientists have to be able to communicate not only with words but also by carefully defined numerical descriptions. Read ~<; folJ.owing news report that appeared in Wall Journal, June 6,2001, on pa~ All: SEOUL, South Korea-A mix up in the cockpit over whether' altitude guidance was measured in feet or meters led to the crash of a Korean Air Lines McDonnell Douglas MD-ll freighter soon after takeoff in Shanghai in April 1999. investigators ' The crash killed all crew-members. Five people on the ground were kuted and 40 more were injured when the plane went-down in tight rain onto a construction near Shanghai's Hongqiao Airport. According to a of the report released by South Korean au-thorities, a Chinese air-traffic controller directed the pilots to an altitude of 1.500 meters (4.950 feet). plane was climbing rapidly to that level when the co-pHot told the pilot thought the instructed height was 1.500 feet, equivalent to 455 meters. The international aviation industry commonly measures rutitude in feet. and the confusion led pilot to conclude the jet was almost J ,000 meters too high, so he quickly moved controls to lower the plane. As the plane descended, the pilot realized the error but couldn't correct the mistake in time. South Korea's Ministry of Construction and Transportation said Korean Air Lines would lose right to serve the Seoul-Shanghai cargo route for at least two years because of errors by the pilots. Korean Air said it would appeal decision . . . Now you can understand the point of defining your quantities carefully so that your communications are understood.
- 38. 1.1 Units and Dimensions 1.1 .. 1 What Are Units and Dimensions and How Do They Differ? " 7 Dimensions are our basic concepts measurement such as length, time, mass, temperature, and so on; units are the means of expressing the dimensions, such as feet or centimeters for length, and hours or seconds for time. By attaching units to all numbers that are fundamentally dimensionless, you get following very practical benefits: a. diminished possibility of errors in your calculations, b. reduced intermediate calculations and time in problem solving. c. a logical approach to the problem than remembering a formula and sub-stituting numbers into the fOffilu1a, d. interpretation of the physical meaning of numbers you use. In this book you will use two most commonly used systems of units: 1. SI, fonnally called Le Systeme Internationale d'Unites, and informally called S1 or more often (redundantly) the SI system of units. 2. or American Engineering system of units, not be confused with what is called the U.S. Conventional System (USCS) nor the English system of units. The 81 system has certain advantages over the AE system in fewer names are associated with the dimensions, and conversion one of units to another is easier. but in the United States the AE system has deep roots. Most modern computer pro-grams t process simulators} allow use either or mixed sets of units. Dimensions and their respective units are classified as fundamental or derived: " Fundamental (or basic) dimensions/units are those that can be measured independently and are sufficient to describe essential physical quantities. II Derived dimensions/units are those that can be developed in terms of the fundamental dimensions/units. Tables 1.1 and 1 list both basic. derived. and alternative units in the S1 and AE systems. Figure 1.1 illustrates the relation between the basic dimensions and some of the derived dimensions. example, squaring length results in area, cubing length results in volume. and dividing volume by time gives the volumetric flow What are the dimensions of the mass flux (mass flow rate unit area)? Can you add the appropriate lines in Figure 1. I? The distinction between and lowercase letters should be followed even if the symbol appears in applications where the other lettering is in uppercase style. Unit abbreviations have the same fonn for both the singular and plural, and they are not followed by a period (except the case of inches). One of the fea-
- 39. 8 Dimensions, Units. and Their Conversion Chap. 1 TABLE 1.1 SI Units Encountered in This Book Physical Quantity Name of Unit Lenglh Mass Time Molar amount Energy Power Density Velocity Acreleration Heat capacity Time Temperature Volume Mass Basic Sl Units mette, meLer kilogramme, kilogram second kelvin mole joule newton walt SI Units kilogram per cubic meter meter second meter second newton per square meter, pascal joule per (kilogram· Icelvin) Altemalive Units minute. hour. day. year aeJl:ree Celsius liter (dm3) tonne. ton (Mg), gram Symbol for Unit'" m kg s K mol J N W min,h,d,), L t, g Definition of Unit kg . m2 • -* Pa . m3 kg . m • s-2 .....f> J . m-I . m2 . -t J . s-1 kg. m-3 m' 5-1 m· N· m-2, Pa J. ... Symbols for units do not take a plural form. but plural forms are used for the unabbreviated names. Non·SI units such as day (d), liter or litre (L). and ton or tonne (t) are legaJly for use with Sl. of the S1 system that (except for time) units and their multiples and submultiples are related by standard factors designated by the prefix indicated in Table 1 When a compound unit is formed by mUltiplication of two or more other units, its symbol consists of the symbols for the units joined by a centered dot , N . m for newton meter). The dot may be omitted in the case of familiar units such as watt-hour (symbol Wh) if no confusion will result, or if the symbols are separated by exponents, as in N . m2kg-2. Hyphens should not be used in symbols for compound units. Positive and negative exponents may be used with the symbols for the separate units either separated by a solidus or multiplied by using negative powers (e.g., m/s or m . 8-1 for meters per second). However, we do not use the center dot for multiplication in this text. A dot can easily confused with a period or '*
- 40. Sec. 1.1 Units and Dimensions TABLE 1.2 American Engineering (AE) System Units Encoontered This Book Physical Quantity Name of Umt Mass Time Temperature MoLar amount Force Energy Power Density Velocity . Acceleration Pressure Heat capacity Basic Units foot pound (mass) second. minute, hour. day degree or degree Fahrenheit pound Derived Uni,s pound (force) British thermal horsepower foot pound (force) pound (mass) per cubic foot feet per second feet per second SQUI!lred (force) per square inch per pound (mass) per degree F Symbol ft Ibm s, min, h (hr). day °RoroP lbmol Ibf Btu, (ft)(lbr) hp Ibn/ttl ftls2 Ibfin.2. psi BtuI(lbm)(°F) Figure 1.1 Relation between the basic dimensions On boxes) and various derived dimensions (in eUipses).
- 41. 10 Dimensions, Units, and Their Conversion Chap. 1 TABLE 1.3 SI Prefixes Factor PrefIx Symbol Factor Prefix Symbol 109 giga 10-1 deci d 1()6 mega M 10-2 c 103 kilo k 10-3 miUi m 102 hecto h 10-6 micro J.L 101 da 10-9 nano n missed entirely in handwritten calculations. Instead, we will use parentheses or vertical ruJes, whichever is more convenient, for multiplication and division. Also, the SI convention of leaving a space between groups of numbers such as 12 650 instead of inserting a comma. as in 12.650, will be ignored to avoid confusion in handwritten numbers. Frequently Asked Questions 1. Is the SI system of units the same as the metric system? The answer is no. SI differs from versions the system (such as CGS) in the 'number of basic units and in the way the basic units are defined. 2. What is the major difference between the and uses systems? In the uses system the pound force is a basic unit and the pound mass a derived unit 3. What does ms mean: millisecond or meter seconds? Mind your use of meters! The letters ms mean millisecond; the combination (m) (s) or rn . s would mean meter seconds. I Mm is not 1 mm! Notation such as emi , meaning square centimeters, frequently has to be written as (cm)2 to avoid confusion. SELF-ASSESSMENT T ST (Answers to the self-assessment tests are listed in Appendix A.) Questions 1. Which of the foHowing best represents the force needed to a heavy suitcase: a. N b. leN c. 250 N d. kN? 1. Pick the correct answer(s); a watt is a. one joule second b. equaJ to 1 (kg)(m2)/s2 r l l' I l t I I f . r ' ; 1.
- 42. / Sec. 1 Operations with Units 11 1 .. 2 c. the unit all types of power of the above e. none of the above 3. kg/s a basic or derived unit in SI? 4. In the IEEE Spectrum (Jan. 2001~ pp. 14-16) an article on building out the wireless Inter-net proposed a each 0.05 km2. Does this seem reasonable? Problems 1. Prepare a table in which the rows are: length, mass, time. two columns. one for the the the AE system of units. Fill in each row with the name of column show the numerical equivalency (Le .• I ft == 0.3048 m). 2. Classify following units as correct or incorrect units in the SI svstenn: a. om b. OK c. sec N/mm e. kJ/(s)(m3) Thought Problem 1. What volume of material will a barrel hold? Discussion Problem 1. In a letter to the editor, the letter says: I believe notation be improved so as to make it mathematically more useful by SI-sanctioned prefixes in boldface Then one WOUJG 1 c;:::; 10 m without any ambiguity [c ~ m == the meaning of Hmmlt would at once clear to literate, if scientifically illiterate. citizen, namely m [mm], 10-6 [mm], or (after Gauss and (mmJ. With respect to the "mm" problem and remarks regarding the difference between "one square millimeter" [(mm)2] and mili squaremeter" [m(m2)]. these difficulties are analogous to confusion a "earners-hair brush" and a camel's hair-brush." What do you think of author's proposal? Operations with Units Answers a question such as: how much is 2 + 2 can sometimes be debatable. might state 4. A bad calculator might show 3.99999. What about 9 + 5? Can the answer for 9 + 5 = 2 possibly be correct? Look at a wall clock.
- 43. 12 Dimensions, Units, and Their Conversion Chap. 1 Every freshman knows that what you get from adding apples to oranges is fruit salad! The rules for handling units are essentially quite simple: 1 .. 2-1 Addition, Subtraction, Equality You can add, subtract, or equate numerical quantities only if the associ· ated units of the quantities are the same. Thus, the operation 5 kilograms + 3 joules cannot be carried out because the units as well as the dimensions of the two terms are different. The numerical operation 10 pounds + 5 grams can be performed (because the dimensions are the same, mass) only after the units are transformed to be the same, either pounds, grams, or ounces, or some other mass unit. 1.2 ... 2 Multiplication and DiviSion You can multiply or divide unlike units at will such 50(kg)(m)/(s) but you cannot cancel or merge units unless they are identical. Thus~ 3 m2/60 ern can converted to 3 m2/O.6 m, and then to 5 m, but mJs2, the units cannot cancelled or combined. In summary, units contain a significant amount of infonna· tion that cannot be ignored. They also serve as guides in efficient problem solving, as you will see shortly. Frequently Asked Question How should you handle mathematical operations or units such as sine. log. or exponential? To be specific, if you take the log of 16 m2 and treat the number and units as a product, then you would have log (16m2) = log (16) + 2 log (m) Various awkward ways and tricks of handling quantities such as 2 log (m) have been proposed (see, M. Karr and D. B. Loveman, "Incorporation of Units into Programming Languages," Comma. ACM, 21, 385-391 [1978]). We prefer for simplic. ity to require that a variable be transformed or scaled to be dimensionless before you apply nonlinear operations such as log. For example, for a pipe of radius R with
- 44. 1 Operations with Units 13 units of rn, we would develop a dimensionless variable ,., a fraction, for a distance, from the also in m, to operate on rm =- Rm so that log r = , + log m - log R - log m = log r - log R = r Can you suggest what the scaling could be for a square duct? What if the units of , are not in meters? EXAMPLE 1.1 Dimensions and Units Add the following: (a) 1 foot + 3 seconds (b) 1 horsepower + 300 watts Solution The operation indicated by Ift+3s has no meaning since the dimensions of the two terms are not the same. One foot has dimensions of length, whereas 3 seconds has the dimensions of time. In the case of I hp + 300 watts the dimensions are the same (energy per unit time). but the units are different You must transform the two quantities into like units, such as horsepower or watts, fore the addition can be carried out. Since 1 hp = 746 watts, 746 watts + 300 watts = 1046 watts ELF .. ASS SSMENT TEST Questions 1. Answer the following questions yes or no. Can you a. divide ft by b. divide m by c. multiply ft by
- 45. 14 d. divide ft by em? e. divide m by (deg) K? f. add ft and g. subtract m and (deg) h. add em and ft? 1. add em and m2? j. add 1 and 2 cm? Dimensions. Units. and Their Conversion 2. Why it not po~sible to add 1 ft and 1 ft2? Chap. 1 3. Explain how to accommodate operations such as exp and In on a number accompanied by units. Problems 1. Add 1 cm and I m. 2. Subtract 3 ft from 4 yards. 3. Divide 3 mLS by 2 mO.s, 4. Multiply 2 h by 4 lb. Discussion Problem 1. There seems to be two schoolS of thought concerning how to take the logarithm of a number that has associated dimensions. The proponents of the first school hold that laking the logarithm of a dimensioned variable is a perfectly acceptable procedure, one that leads to a dimensionless result regardless of the dimensions of the original variable. The opposing school is that taking the logarithm of a dimensioned variable is improper, and even meaningless. and the variable should be in dimensionless form before the logarithm is taken. What side do you believe is correct? Explain the reasons for your choice. 1.3 Conversion of Units and Conversion Factors Mistakes are the usual bridge between inexperience and wisdom. Phyllis Theroux, Night Lights Columbus had many of the qualities that would appeal to today's venture capitalists. He was an experienced seafarer, prepared detailed written proposals for his ventures, and was dedicated and sincere. King John of Portugal, who rejected his first proposal in 1484, regarded him as boastful, fanciful, and overimaginative. His Portuguese experts believed that the distance to the Indies was 10,000 (U.S,) miles, four times Columbus's estimate of 2,500 (U.S.) mnes. Both the experts and Columbus knew he had to travel about 68° of longitude, but Columbus apparently interpreted the Arabic literature in which the measure for 10 was 56 2/3 miles (U.S.) as
- 46. Sec. 1.3 Conversion of Units and Conversion Factors ancient Italian miles, which are equal to modem 37 U.S. miles. Consequently, he thought that 68° was about 2~500 U.S. miles, whereas the correct distance was about 3900 U.S. miles. As another example of a serious conversion error, in 1999 the Mars Climate Orbiter was lost because engineers failed to make a simple conversion from English units to SJ, an embarrassing lapse that sent the $125 million craft fatally close to the Martian surface. As a prospective engineer you must be carefu1 of handling all sorts of units, and be able to convert a given set of units to another set with ease. As you probably already know the procedure for converting one set of units to another is simply to multiply any number and its associated units by ratios tenned conversion factors to arrive at the desired answer and its associated units. Conversion factors are statements of equivalent values of different units in the same system or between systems of units used in the fonn of ratios. You can view a pair of (correct) conversion factors as quantities that form a ratio so that multiplying a teon by the ratio is essentially the same as multiplying the term by 1. On the inside of the front cover of this book you will find tables of commonly • used conversion factors. You can locate many others in handbooks and on the Internet. Some of the references to consult can be found at the end of the chapter. Memorize a few of the common ones to save time looking them up. It win take you less time to use conversion factors you know than to look up better ones. Some web sites do the conversions for you', In the physical property software on the CD in the back of this book you can insert almost any units you want in order to retrieve property values. Nevertheless, being able to make conversions by yourself is important. In this book. to help you foHow the calculations and emphasize the use of units, we frequently make use of a special format in the calculations, as shown below. Consider the following problem: If a plane travels at twice the speed of sound (assume that the speed of sound is 1100 ft/s), how fast is it going in miles per hour? We formulate the conversion as follows 2X 1100 ft j mi 60 s 60 min ----,- --_._ .. - --- s 5280 ft I min 1 hr ft Inl tnt S S mm Note the format of the calculations. We have set up the calculations with vertical lines separating each ratio. These lines retain the same meaning as a " or parenthesis, or a multiplication sign (x) placed between each ratio. We will use this fonnulation frequently in this text to enable you to keep clearly in mind the significance of units in problem solving. We recommend that you always write down the units next to the as-
- 47. 16 Dimensions. Units, and Their Conversion Chap. 1 sociated numerical value (unless the calculation is very simple) until you become quite familiar with the use of units and can carry them in your head. Another convenient way you can keep track of the net units in an equation is to strike through the units that can be cancelled as you proceed with the calculations. For example: 2( l100)ft 1 mile 60 g 60 mi1f 5 5280 it 1 miff 1 hr At any stage in the conversion you can determine the consolidated net units and see what conversions are still required. If you want, you can do this fonnally, as shown above, by drawing slanted lines below the dimensional equation and writing the consolidated units on these lines; or you can do it by eye, mentally canceling and accumulating the units; or you can strike out pairs of identical units as you proceed. Consistent use of units a10ng with numbers throughout your professional career will assist you in avoiding silly mistakes such as converting 10 centimeters to inches by multip1ying by 2.54: 10 em 2.54 em .. 10 cm 1 in. . 1 in. =1= 25.4 10 .• mstead of 2.54 cm = 3.94 m. By three methods we may learn wisdom: First. by ,efl~ction, which is noblest; second, by imitation, which is easiest; and third by experience. which is the bitterest. Confucius Now let's look at an example. EXAMPLE 1.2 Conversion of Units (a) Convert 2 km to miles. (b) Convert 400 in.3/day to cm3/min. Solution (a) One way to carry out the conversion is to look up a direct conversion factor. namely 1.61 km = 1 mile: -2 k-In 116 m ile = 1.24 mile 1. 1 km Another way is to use conversion factors you know 2lErrt' 105 ~ l.i:rf. ~ I mile = 1 24 il 1l61'i' 2.54 ~ 12 m. 5280 ir. . m e
- 48. Sec. 1.3 Conversion of Units and Conversion Factors 400 in.3 (2.54 cm)3 1 day 1 hr 4 55 cm) (b) - =- day 1 in. 24 hr 60 min . min In part (b) note that not only are the numbers in the conversion of inches to centimeters raised to a power, but the units also are raised to the same power. EXAMPLE 1.3 Nanotechnology Nanosized materials have become the subject of intensive investigation in the last decade because of their potential use in semiconductors, drugs. protein detectors, and electron transport. Nanotechnology the generic tenn that refers to the synthesis and application of such small particles. An example of a semiconductor is ZnS with a particle diameter of 1.8 nanometers. Convert this value to (a) dm (decimeters) and (b) inches. Solution t, 1.8 nm 10-9 milO dm = 1.8 X 10-8 dm (a) I nm 1 m (b) 1.8 nm 10-9 m 39.37 in. = 7.09 X 10-8 in. 1 nm 1m 11 In the AE system the conversion of tenns involving pound mass and pound force deserve special attention. Let us start the discussion with Newton's Law: where =Cma = force C = a constant whose numerical value and units depend on those selected for F. m, and a m= mass a = acceleration (1.1) In the SI system in which the unit of force is defined to be the Newton (N) when 1 kg is accelerated at 1 mls2• a conversion factor C = 1 N/(Kg)(m)/s2 must be introduced to have the force be 1 N: IN F = -~--,- C 1 kg 1 m -IN ,...., m (1.1)
- 49. Dimensions, Units, Their Conversion Chap. 1 Because the numerical value associated with the conversion factor is 1. the conver~ sion factor seems simple. even nonexistentt and the units are ordinarily ignored. In the AE system an analogous conversion factor required. However, to make the numerical value of the and the mass be essentially the same at the earth's surface, a mass of 1 Ibm is hypothetically accelerated g ftls2, where g is the acceleration that would caused by gravity (about ftls2 depending on the location of the mass), we can make the force be I 1 bf by choosing the proper numer-ical value and units for conversion factor F=( IIbr (1.2) ,...." g A numerical value of 174 has been chosen for numerical value in the conversion because 32.174 the numerical value of acceleration of gravity (g) (9.80665 mls2) at sea level at latitude when g expressed in ftls2. acceleration caused by gravity, you may by a few tenths of 1 % from place place on the of the but quite different on the surface of the moon. The inverse of the conversion factor with is given the special symbol gc numerical value 32.1 included that you will see induded in equations in some texts to remind you that the numerical of conversion factor is not a unity. avoid confusion~ we will not place gc in the equations in this book because we will be using both SI AE You will discover the use of gc is essential in the system when need a conversion factor to adjust units when both Ibm and Ibf are involved in a calculation. or when tbf has to be transfonned to Ibm in a unit such as psia (lbf/in. In summary, can see that the AE system has convenience that the nu-merical value of a pound mass is also that of a pound force if the numerical value of the ratio g/gc equal to 1, as it is approximately most cases. No one gets confused by the fact that a person who 6 feet tall has only two In this book, we will not subscript the symbollb with m (for mass) or f (for force) unless it becomes essential to do so to avoid confusion. win always mean by the unit Ib without a subscript quantity pound mass. But never forget that the pound (mass) and pound (force) are not the same units in the AE system even though we speak of pounds to express force, weight, or mass. What the difference between mass and weight? When someone says weigh 100 kg, or 200 pounds, how can that s"ta teme. nt correct when you know that weight is a force. not a mass, equal the opposite of the force required support a
- 50. Sec. 1.3 Conversion of Units and Conversion Factors 19 mass (consult some of the at the end of this chapter for a more precise finition of weight)? To avoid confusion. just interpret the statement as follows: a person or object weighs as much as a mass 100 or 200 pounds, would weigh. if by a force scale. Some Useful Trivia Concerning Conversion A U.S. frequent-flier mile is not the same as a U.S. mile-the former a nauti-mile (1.85 km), whereas latter 1.61 km. In AE 1 m ;;;;; 39.37 in., whereas for U.S. land survey applications it 2 x 10-6 in. shorter. EXAMPLE 1.4 Conversion Involving Both Ibm and Ibf What is the potential energy in (ft)(lbf) of a 100 Ib drum hanging 10 ft above surface of the earth with reference to the surface the Solution The [lIst thing to do is read the problem carefully. What are the unknown quanti- The potential (FE) is unknown. What are the known quantities? The mass and the height the drum are known. How are they related? You have to look up the relation unless you it from physics: Potential energy;;: P ::::: mgh Assume that 100 means 100 mass; g = ........ ""''"',n.''. of gravity = EL4 is a sketch of system. -lL-___ ..I..-._ ro:.f",rAl"lt'.:o plane Figure El.4 Now substitute the numerical values of the variables into perfonn the necessary unit conversions. equation and 100 Ibm ftltOnl (s2)(lbr) P = -- 174(ft)(lb m ) 1000 (ft)(lb() Notice that in the ratio of ftJs2 divided by 32. 174[(ft)(lbm)]/[(s2)(1bf)), the nu-mencal, values are almost Many engineers would solve the problem by say-ina that 100 lb x 10 ft:: 1000 (ft)(lb) without realizing that, in they are canceling out the numbers in the glgc ratio, and (hat the Ib in the solution means lhf.
- 51. 20 Dimensions, Units, and Their Conversion EXAl1PLE 1.5 Conversion of Units Associated with Biological MateriaJs Chap. 1 In biological systems, enzymes are used to accelerate the rates of certain biological reactions. Glucoamylase is an enzyme that aids in the cODversion of starch to (a sugar that cells use for energy). Experiments show that I JLg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 Jtg moll(mL)(min). Detennine production rate of glucose for this system in the units of Ib moll(ft3)(day). Solution Basis: 1 min 0.6 Jtg mol (mL)(rnin) mol 1 lb mol 1000 mL I L JLg mol 454 g mol 1 L 3.531 X lb mol = 0.0539 (rt3)(day) Questions 1. What I·S gc?' SELF-ASSESSMENT TEST 60 min 24 hr hr day 2. Is the ratio of the numerator and denominator in a conversion factor equal to unity? 3. What is the difference, if any, between pound force and pound mass in the system? 4. Could a unit of force in the SI system be kilogram force? 5. Contrast the procedure for converting units within the SI system with that for the AE system. 6. What the weight a one pound mass at sea level? Would the mass the same at the center of Earth? Would the weight be the same at the center of Earth? 7. What is the mass of an object that weighs 9.80 kN at sea level? Problems 1. What are the value and units of in the SI system? 2. Electronic communication via radio travels at approximately the speed of light (186,000 miles/second), The edge the sblar system is roughly at Pluto, which is 3.6 x 109 miles from Earth at its closest approach. How many hours does it take for a radio signal' from Earth to reach Pluto?
- 52. 1.4 Dimensional Consistency (Homog~nejty) 21 3. the kinetic energy of one pound of fluid moving in a pipe at the speed of 3 feet per second. 4. Convert the following from AE to units: a. 4lbm/ft kg/m b, 1.00 Ibm/(ft3)(s) to kg/(m~)(s) S. Convert the foHowing 1.57 X 10--2 g/(cm)(s) to Ibm/(ft)(s) 6. Convert 1.1 gal to ft3. 7. Convert 1.1 gal to m3, Thought Problems 1. Comment as to what is wrong with the following statements from a textbook: a. Weight is product of mass times the force of gravity. b. A 67-kg person on earth will weigh only lIon the moon. c. If you have 1 g of water at 4°C that a volume of 1.00 mL, you can use the ratio 1.00 g water/4°C as a conversion factor. 2. In the conversion tables in Perry's Handbook (5th is a row showing that factor 0.10197 converts newtons to kilograms. Can this be correct? Discussion Problem 1. In spite of the official ,adoption of SI system of units most countries, people stm buy 10 kg, of potatoes and inflate automobile tires to a value in (or kg/cm2). Why does this usage occur? 1.4 Dimensional Consistency (Homogeneity) Now that we have reviewed some background material concerning units and dimensions, we can immediately use this information a very practical and important application. A basic principle states that equations must be di" mensionally consistent. What the principle means is that each term in an equation must have same net dimensions and units as every other to which it is added, subtracted. or equated. Consequently, dimensional considerations can be used to help identify the dimensions and units of terms or quantities in an equation. /" The concept of dimensional consistency can be illustrated by an equation that represents the pressure/volume/temperature behavior of a and known as van der Waals's equation, an equation that discussed in more detail in Chaper 15:
- 53. 22 and Their Conversion Chap. 1 (p (V - b) = RT Inspection of the equation shows that the constant a must have the units of [(pres-sure)( volume)2] for the expression in the first set of parentheses be consistent throughout. If the units of pressure are and those of volume are a will have the units [(atm)(cm)6]. Similarly, b must have the same units as V, or in this par-ticular case units of cm3• If T is in what must be the units Check your answer by up R inside the front cover of the book. An must ex-hibit dimensional consistency. EXAMPLE 1.6 Dimensional Consistency Your handbook shows that microchip etching roughly follows the relation d = 16.2 - l6.2e-O·02It t < 200 where d is the depth of the etch in microns JLm) and t is the the etch in What are the units a;:l>;""",",uu#U with the numbers 16.2 and Convert the minutes. Solution so d becomes expressed in inches and t can be of I? After you the equation that d as a function of t, you should be able to reach a decision about the units associated with side of the equation. Both values of 16.2 must units of microns (J,tm). The exponential must dimensionless so that 0.021 must have the "'''''''' ........ 1- aled units of out the conversion, look up suitable conversion factors inside the front cover book and multiply so that units are converted from 16.2 JLm to inches, and tIs to tlmin. 39.27 in. [ -0.021 60s tmin 1 d· == ----:...- --- 1 - exp--- m I m s 1 min As you proceed the study of you will find that groups of symbols together, either by theory or based on experiment. that no collections of variables or parameters are called dimen-sionless or nondimensiona1 groups. One example Reynolds number (group) arising in fluid mechanics.
- 54. Sec. 1.4 Dimensional Consistency (Homogeneity) Dvp Reynolds number = --= N RE JJ.. 23 where D the pipe diameter, say in cm; 11 the fluid ve]ocity, say in cmJs; p is fluid density. say in g/cm3; and /L is the viscosity, say in centipoise, units that can be converted to g/(cm)(s). Introducing the consistent of units for D. v. p, and f.L into Dvp I JL. you will find that all the units cancel out so that the numerical value of 1 is the result of the cancellation of the units. EXAMPLE 1.7 Interesting Example of Dimensional Consisten'cy Explain without differentiating why the foHowing differentiation cannot be correct: d dx where x is length and a is a constant. Solution - ---:;::::==== Observe that x and a must have the same units because the ratio 2 must be dimensionless (because 1 is dimensionless), a Thus, the lefthand side of the equation has units of 1. (from dJdx). However. x the righthand side the equation has units of x2 (the product of ax). Consequently, something is wrong as the equation is not dimensionally consistent. S LF .. ASSESSMENT T ST Questions 1. Explain what dimensional consistency means in an equation. Explain why the so-called dimensionless group no net dimensions. 3. If you divide aU of a series of terms in an equation by one of the terms. will the resulting series of terms be dimensionless? 4. How might you make the following variables dimensionless: a. Length (of a pipe). b. Time (to empty a tank full of water).
- 55. 24 Dimensions, Units, and Their Conversion Chap. 1 Problems 1. An orifice meter is used to measure the rate of flow of a fluid in pipes. The flow rate is related to the pressure drop by the following equation u = c~Il.P . p where u = fluid velocity 6.p = pressure drop (force per unit area) p = density of the flowing fluid c = constant What are the units of c in the SI system of units? 2. The thennal conductivity k of a liquid metal is predicted via the empirical equation k=A exp (Bn) where k is in J/(s)(m)(K) and A and B are constants. What are the units of A and B? Thought Problems 1. Can you prove the accuracy of an equation by checking it for dimensional consistency? 2. Suppose that some short time after the "Big Bang" the laws of nature turned out to be different than the laws currently used. In particular, instead of pV = nRT, a different gas law arose, namely p'VT = nR. What comments do you have about such an equation? Discussion Problem 1. In a letter criticizing an author's equation. the writer said: The equation for kinetic energy of the fluid is not dimensionaHy consistent. I suggest the modification KE= mv2 /2gc in which gc is introduced. Then the units in the equation will not be (ft/s)2, which are the wrong units for energy. What do you think of the comment in the letter? , 1.5 Significant Figures Decimals have a point. Unknown You have probably heard the story about the Egyptian tour guide who told the visitors that the pyramid they beheld in awe was 5013 years old. "Five thousand and l
- 56. Sec. 1 Significant Figures thirteen said a visitor!H "How do you knowT~ "Well, said the guide, when I first began working here 13 years ago, I was told the pyramid was 5000 years old." What do you believe about the accuracy of a statement in a travel brochure in which you read that a mountain on a trip is 8000 m (26.246 ft high)? Responsible physical scientists and engineers agree that a measurement should include three pieces of information: a. the magnitude of the variable being measured b. its units c. an estimate of its uncertainty The last is likely to be either disassociated from the first two or ignored completely. If you have no idea of the accuracy of a measurement or a number, a conservative approach is to impJy that the last digit is known within upper and lower bounds, For example, 1.43 indicates a value of 1.43 + 0.005, meaning that the value can be deemed to be between 1 and 1.435. Another interpretation of 1.43 is that it means 1.43 0.01. What should you do when you add, subtract, multiply, and divide numbers that have associated uncertainty? The accuracy you need for the results of a calculation depends on the proposed application of the results. The question How close is close enough? For example. in income tax forms you do not need to include whereas in a bank cents (two decimals) are included. In calculations, the cost of inaccuracy is great (failure, fire, downtime, etc,), knowledge of the uncertainty in the calculated variables vitaL On the other hand, in determining how much fertilizer to put on your lawn in the summer, being off by 10 to 20 pounds out of 100 lb not important. Several options (besides common sense) in establishing the degree of cer-tainty in a number. Three common decision criteria are: (1) absolute (2) rela-ti ve error, and (3) statistical analysis. 1. First. consider the absolute error in a number. You have to consider two cases: a. numbers with a decimal point, and b. numbers without a decimal point. For case (a), suppose we assume that the last significant figure in a number represents the associated uncertainty. Thus. the number 100,3 carries along implication of 100.3 ± 0.05, meaning 100.3 lies in the interval between 100.25 to 100.35. Thus, 100.3 would have what is termed four significant figures. For case (a), the number 100.300, we will that adctitional significant figures of accu-racy exist so that 100.300 will have six significant figures. (Be aware that some textbooks and authors do not attribute significance to the trailing zeros on the righthand
- 57. 26 Dimensions, Units, and Their Conversion Chap. 1 side of a decimal point) The rationale behind attributing additional significant ures to the trailing zeros is that they would not be added to 100.3 unless there was a reason for displaying additional accuracy. As an example, rounding the number 100.2997 to retain only six significant figures would 100.300. For case (b), if a number is stated without a decimal point, such as 201,300, we will assume that trailing zeros (after the 3) do not imply any additional accuracy beyond four significant figures. When you multiply or divide numbers, generally you should retain in your final answer the lowest number of significant figures that occur among all of the numbers involved in the calculations even though you carry along 10 or 20 digits during the calculations themselves. For example, we will treat the product (1.47)(3.0926) := 4.54612 as having only three significant figures because 1.47 has only significant answer should be truncated to to avoid sugany greater precision in the result of the multiplication. When you add or subtract numbers, generally you should retain in your final answer the number of significant digits as determined by the error interval of the largest of the numbers. For example, in the addition llO.3 0.038 110.338 common sense would say to state the answer as 110.3. You should not have more than four significant figures in the sum. This decision reflects what revealed by a more detailed examination of the error bounds imputed to the two numbers: Upper Bound 110.3 0.05 = 110.35 0.038 + 0.0005:;:: 0.0385 110.3885 Lower Bound 110.3 - 0.05 :;:: 110.25 0.038 - 0.005 = 0.0375 110.2875 The midpoint of these two numbers is 110.338. Absolute errors are easy to track and compute, but they can lead to gross tortions in the specified uncertainty of a number. For example, let's divide 98 by 93.01. You can 98 93.01 = 1.1 = 1.05 = l.054 = 1.0537
- 58. 1 Significant 21" What do you think about applying rule that states the number of digits in the least precise number (two significant here because two significant should be the number of significant digits retained in an-swer? If you apply rule, the calculated answer is 1.1, clearly a distortion error in the because 98 1 has an error of only about 1 %. the 1.1 ± 0.1, an error of about 10%! Certainly 1 indicates too a precision so that choice should 1.05 or I Which do you think is better? 2. Perhaps the use of relative error can often be a better way to how many significant figures to retain in your answers. Suppose you divide one number by another close to it such as 1.0111.09 = 0.9266, and select 0.927 as the answer. The in answer based on absolute error analysis is 0.001/0.927, or about 0.1 %, 1/1.09)100, or about a 1 % uncertainty, ex-isted in the numbers. Should relative the answer be fixed at about 1 %, that is, truncate the answer to 0.93 rather than 0.927? would be the case if you applied the concept of relative error. decision is up to you. In any case, avoid increasing the of your answer very much over the preci-sion in your measurements or when presenting results of calculations. You do have to use some common sense applying the of relative error to scales [hat use both relative and units. For suppose the error in a temperature of 25°C is 1 or 4%. Can you the error by changing the temperature kelvin, so that error becomes (1/298)]00 == 0.33%1 Of course not. 3. A more rigorous and in numbers apply more complicated third way to treat uncertainty in the calculations. What is involved is the con-cept of confidence limits for gation of errors step by step through each stage suIt. Bm even a statistical is not ratios of numbers. Refer to a book on statistics approach. a calculation, the propa-calculations to the final re-because we nonlinear further information about this In book we base answers on absolute error because such a choice convenientt but will often show one or two e~gures in intennediate calculations as you should. (The numbers in your calculator are not a Holy Writ!). Keep in mind that some numbers are such as the 12 = 1/2 mv2 2 in the super-script the operation of You will encounter as 1, 2, 3, and so on, which in some cases are exact (2 reactors, 3 input streams) but in other cases are shortcut substitutes for presumed very accurate in probJem moles, 10 kg). a mass as to kg, in which the number does not have a decimal point, despite our remarks above about you can infer that quite a few significant figures apply to the mass, particularly in other values of
- 59. 28 Dimensions, Units, and Their Conversion Chap. 1 the parameters in an example or problem, because you can easily measure a mass to a level mg. You will also occasionally encounter fractions such as which can be as 0.6667 in relation the accuracy of other values in a prob-lem. In this convenience we use 273 K for the temperature equivalent to O°C instead of 273.15 K, thus introducing an absolute error of 0.15 degrees. This a smaIl error relative to the other known or presumed errors your calcula-that it can be neglected in almost all instances. Keep in mind, however, addition. subtraction, multiplication, division~ all the errors that you introduce propagate into the final answer. Feel to round off parameters such as 1'1' = 1416, = 1.414. or Avo-gadro's number N = 6.02 X 1()23. In summary, be sure to round off your answers to problems to a reasonable number of significant figures even though numbers are carried out to 10 or more digits your computer or calculator in the intermediate calculations. EXAMPLE 1.8 Retention of Signific;-nt Figures 100 is subtracted from 22,400~. is the answer 2.300 kg to four significant Solution If you note that 22.400, 20.100, and have no decimal points the righthand zero. how many significant can you attribute input to 22,400 and 20,1 DO? By applying the absolute error you can conclude that number of significant figures is Scientific notation makes decision cle,ner 2.24 X let kg and the result retains two significant figures. On the other hand if a decimal point were p]aced in each number thus, 22,400. and 20,100., indicating that the last zero was significant. then the answer of 2,300. would be valid to four significant figures. From the viewpoint of relative error, 22,400 has an error of about 0/224) as does 20,100 (1201), whereas 2,300 has an error of about 5% (1/23). Should relative error have been to e~tablish the number of significant figures to be retained? you add a 0 to the right of 0.23 to give a relative error of (11230) or about 112%? No. what about giving (he 3Mwer as x 10?
- 60. Sec. 1 Significant Figures EXAMPLE 1 .. 9 Micro-dissection of DNA A stretch-and-positioning technique on a carrier layer can be used for dissection and acquisition of a electrostatically positioned DNA strand. A device to do the micro-dissection consists of a gJass substrate on which a sacrificial layer. a DNA carrier Jayer, and a pair electrodes are deposited. DNA is electrostatically stretched and immobilized onto the carrier layer with one of its molecular ends aligned on the electrode edge. A cut is made through the two layers with a stylus as a knife at an aimed portion of the DNA. By dissolving the sacrificial layer, the DNA fragment on the piece of carrier can be recovered on a membrane filter. The carrier piece can then be melted to obtain the DNA fragment in solution. If the DNA is stretched out to a length of 48kb. and a cut made with a width of 3 J.Lm. how many base pairs (bp) should be reported in the fragment? Note: 1 kb 1000 base pairs (bp). and 3 kb = 1 J.Lm. Solution Superficially conversion 3 fJ,m 3 kb 1000 bp 1 J.Lm 1 kb 9000 bp However, because the measurement of the number of molecules in a DNA fragment can be determined to 3 or 4 significant figures in a thousand, and the 3 f.Lm reported for the cut may wen have more than 1 associated significant figure, the precision the 9000 value may actually be better if the cut were determined to have a value of 3.0 or 3.00 J,Lm. S LF .. ASSE SMENT T S Questions 29 1. Why can the use of absolute error in determining the number of significant digits be misleading? 2. How can you avoid a significant loss of precision in out calculations involving many repetitive operations as addition, multiplication. and so on)? 3. Will adding a decimal point to a reported number that does not have a decimal point, such as replacing 12.600 with 12,600., improve the precision of the number? Problems 1. Identify 3.0 0.353 1,000. number of significant figures for each of the following numbers: 23 1,000 1,000.0
- 61. 30 Dimensions, Units. and Their Conversion Chap. 1 2. What is the correct sum and the number of significant digits when you add (a) 5750 and 10.31 (b) 2.000 and 0.22? 3. Convert the water flow rate of 87.0 kg of water having a density of I OOOkglmJ per minute to the units of gal/br. giving the answer in the proper number of significant figures. 4. A oomputer chip made in Japan presumably costs $78. The calculation to convert from yen to dollars was made as follows: ( 10,000 yen )( $ 1.00 ) 1 computer chip 128 yen = $ 78/computer chip Is the number of significant digits shown in the answer correct? What is the answer to: 78.3 - 3.14 - 0.3881 Thought Problems 1. Is 6 5/8 inches equivalent to (a) S1/8? (b) 6.375 inches? 2. When you want to calculate the weight of 6 silicon chips each weighing 2.35 g. is the answer good only to one significant figure, i.e., that of 61 3. A textbook mentions the quantity of reactant as being 100 mL. How would you decide on the number of significant figures to associate with the quantity of reactant? Discussion Problem 1. In a report of the crew laying fiber optics cable, the results for the month were listed as foHows: 3OO0ft 4120 ft 1300 ft 2100 ft 10.520 ft How many significant figures would you attribute to the sum? 1.6 Validation of Problem Solutions If a mistake is not a steppingstone, it is a mistake. Eli Siegel Validation (sometimes referred to as verification) means checking that your problem solution is satisfactory, and possibly assessing to some extent your prOblem-solving procedures. By satisfactory we mean correct or close enough. Since presumably you do not know the solution before you solve the problem. trying to
- 62. 1.6 Validation of Problem Solutions 3f check your result with the unknown makes severe demands on your problemsolving skills. Unless you can compare your answer with a known such as the answers in the Appendix to this and other books, what can you do? Here is a list of suggestions. (We will not consider statistical analysis.) The extent to which you can pursue a validation depends on time you have available and the cost. 1. Repeat the calculations, possibly in a different order. 2. with the answer and perfonn the calculations in reverse order. 3. Review your assumptions and procedures. Make sure two errors do each other. cancel 4. Compare numerical values with experimental data or data in a database (handbooks~ the Internet. textbooks). 6. Examine the behavior of the calculation procedure. For example, use another starting value and that the result changed appropriately. L'"!l.iJliJI",.»i:I whether the answer is and its background. given what you know about prob- The moment you have worked out an answer, start checking probably i.m 'r right. Right Answers, Computers and Automation., p. 20 (September 1969) SElF-A SES MENT T ST Questions 1. Will using a calculator or computer help numerical errors your calculations? 2. What other ways of validating your answers to a problem can you suggest in addition to the one cited in Section 1.6? 3. Suppose you convert the amount of solid CaCl2 in a 100 mL with a net weight in grams to pounds, and 2.41 lb. How would you go about checking the va-lidity of this result? Problems 1. Check the answer in the following calculation by starting with answer to get the value for the original starting quantity. B is the molar density em3 gram mole of a com-pound, MW is the molecular weight of the compound, and p is the mass density the compound in grams per em3, B I p mol 1 nJ plbm B is the value of the variable has the units em3, )(MW)B you B?
- 63. Dimensions. Units, and Their Conversion Chap. 1 Looking Back In this we have reviewed the essential background you to l:)eC:Oltie skilled in converting units, applying the concept of dimensional In your work, numerical values with an appropriate number significant UAj::;,J.L-'>. GLOSSARY OF NEW WORDS Absolute error a u ...... u ......... " that a value. AE American tmjglneer.mg .I::v ... ,rp.m Conversion of units Change of units Derived units Units developed one set to another. fundamental units. Dimensional consistency in an equation must have the same set of net dirnensi on s. Dimensionless group A collection of or parameters that has no net di-mensions (units). Dimensions The basic concepts of measurement such as length or time. A unit for the product of the mass and the ac(~elC~raIUOli1. Fundamental units Units that can be measured independently. dimension for the amount of materiaL Nondimensional group See Dimensionless group. Pound force unit of force in the AE system. Pound mass unit of mass in the AE system. Relative error Fraction or percent error for a number. SI Le Systeme Intemationale d'Unites (51 system of units). Units Method of expressing a dimension such as ft or hour. Validation Determination that Weight A force opposite to the itational field). ..:>v" .......... 'u to a problem is correct. reQum;~ to support a mass (usually in a SUPPLEMENTARV REF RENCES In addition to the general references listed in the front material, the follow-are pertinent: L Bhatt, B. 1., and S. M. Vora. Stoichiometry (SJ Units). Tata McGraw-Hill, New Delhi (1998),
- 64. l Chap. 1 Problems 2. Horvath, A. L. Conversion Tables in Science and Engineering, Elsevier. New York (1986). 3. Luyben, W. and A. Wentzel. Chemical Process Analysis: Mass and Energy Energy Balances, Prentice-Hall, Englewood Cliffs, N. J. (1988), 4. National Institute Standards. The International System of Units (SI), NIST Special Publ. No. 330, U.S. Department of Commerce, Gaithersburg, MD 20899 (1991). Reilly. M. Statistical Look at Significant Figures," Chem. Eng. Educ. 152-155 (Summer (992). 6. Vatavuk, W. M. "How Significant Are Your Figures," Chern. Web Sites hup:J Ichemengineer .aboutcom http://www.chemistrycoach.comJrutorials-2.html http://www.ex.ac.uklcimtldictunitldictunithtm http://mcgraw-hill.knoveLcom/perrys http://www.retaHick.comlresources/netresrc .html http:.! Iwww.shef.ac. uk/uniJacademicl A ~CI cpe/mpittlchemengs .htrnl PROBLEMS 97 (August 18. 1986), (The denote the degree of difficulty. "III being the most difficult.) ·1.1 out the roHowing conversions: (a) How many m3 are there in 1.00(mile)3? (b) How many gal/min correspond to 1.00 ft3/s? $1.2 Convert (a) 0.04 gI(min)(m3) to Ibrrl(hr)(ft3). (b) 2 to n3/day. 6(in)(cm2) (c) (yr)(s)(lbrn)(tr) to all SI units. $1.3 In a article describing an oil-shale retorting process. the authors say the retort: "could be operated at a solids mass flux wen over COOO Ib/(b)(ft2) (48k Path) .. ,.. In several places they speak the grade of their shale in the mixed units "34 gal (129 L)/ton.'· Does their report make sense? ·1.4 Convert the following: (a) 60.0 milhr to ftlsec. (b) 50.0 Ib/in.2 to (c) 6.20 cmlhr2 to nmlsec2.
- 65. 34 , " Dimensions, Units, and Their Conversion Chap. 1 $1.5 following test win measure your SIQ. List the correct answer. (a) Which the correct symbol? (1) nm (2) oK (3) sec (4) N/rnm (b) Which is the wrong symbol? (1 ) lvfN/m2 (2) GHzls (3) kl/(s)(m3) (4) °ClMJs (c) Atmospheric pressure is about: (1) 100 Pa (2) 100 kPa (3) 10 MPa (4) 1 GPa (d) The temperature O°C is defined as: (1) 273.15°K (2) Absolute zero (3) 273.15 K (4) The freezing point of water (e) Which height and mass are those of a petite woman? (1) ] .50 m, kg (2) 2.00 m, kg (3) 1.50 m, 75 kg (4) L80 m. 60 kg (0 Which a recommended room temperature winter? (I) 1 (2) (3) (4) (g) The watt is: (1) One joule second (2) Equal to 1 . m2/s3 (3) The unit an types power (4) All of above (h) What force be needed to lift a heavy suitcase? (1) 24 N (2) 250 N (3) (4) 250 kN *1.6 A technical publication describes a new model 20-hp Stirling (air cycle) engine that drives a 68-kW generator. this possible? **1.7 Your boss announced that the of the company Boeing 737 is to cut from milhr to 475 milhr to "conserve fuel," thus cutting consumption from 2200 gaJlhr to 2000 gal/hr. How gallons are saved in a 1000-mi trip? ··1.8 From Parade Magazine, 31, 1997. page 8 by Marilyn Voss·Savant: Can help with this problem? Suppose it takes one man 5 hours to paint a house, and it another roan 3 hours to paint the same house. If two men work together, how many hours would it take them? This is driving me nuts. Calculate the answer. "'1.9 Two scales are shown, a balance (a) a spring scale (b) a b
- 66. r Chap. 1 Problems 35 [n the balance calibrated weights are placed in one pan to balance the to be weighted in the other pan. In the spring scale, the object to be is on pan and a spring is compressed that moves a dial on a State for each device whether it directly measures mass or Underline your answer. State in one sentence for each the reason for your answer. '1.10 In the American Engineering system of units, the can have the units of (lb f) (h r)/ft2 , while in a handbook the units are a of 20.0 (g)/(m)(s) to the given American Engineering units. **1.11 Thermal conductivity in the American ~nlglOleermg <:v<:I~ .... m of units is: Btu k =------ (hr) ..... "',,."'''' this to: kJ ··1.12 Water is through a 2-inch diameter pipe with a velocity of 3 ftls. , (ft) (lbf) (a) What is energy of the water m b ? (l m) (b) What is in gal/min? contents of are often labeled in a fashion such as grams." Is it correct to so a package? ''1.14 What is meant by a scale that shows a weight "21 A tractor pulls a load with a force equal to 800 Ib (4.0 kN) with a velocity of 300 ftlmin (1.5 mis). What is the power required using the given system data? The SI data? "'1.16 What is the kinetic energy of a vehicle a mass of 2300 mOivUllll at the rate of 10.0 ftlsec in Btu? 1 Btu = 778.2 (ft)(Ibf), *1.17 A pallet of boxes weighing 10 tons is drOPpf~d a tift truck from a height of 10 feet. The maximum velocity the pallet .............. ., "' .. ion ...... nmmg the ground is 6 ftls, How much kinetic energy does the pallet have *·"'1.18 The efficiency of cell growth a !:lIl"l!;:tr",tp a l)1otecltmcllo~:y Olrocess was given in a report as 11= AHcat In .... ni'·f'o-~"t.(' ettlclelocy of cell metabolism (energy!energy) yield, carbon basis (cells produced/substrate consumed) JIb = det!:ree of reductance of biomass (available electron equivalents! g carbon. such as 4.24 e-equiv.lmol ceH carbon)
- 67. 36 Dimensions, Units, and Their Conversion I1H{)e- = biomass heat of combustion (energy/available electron equiv.) 11 H~3t = available energy from catabolism (energy/mole substrate carbon) Chap_ 1 Is there a missing conversion factor? If so, what would it be? The author claims that the units in the numerator of the equation are (mol cell carbonlrnol substrate carbon) (mol available e-/mol cell carbon) (heat of combustionlmol available e-). Is this correct? "1.19 Leaking oil tanks have become such environmental problems that the Federal Government has implemented a number of rules to reduce the problem. A leak from a small hole in a tank can be predicted from the following relation: Q = O.6ISV(211p)Jp where Q = the leakage rate S = crossectional area of the leak 11 p ::: pressure drop p = fluid density To test the tank, the vapor space is pressurized with N2 to a pressure of 23 psig. If the tank is filled with 73 inches of gasoline (sp. gr. = 0.703) and the bole is 114 in. in diameter, what is the value of Q (in ft3/hr)? ·"1.20 In an article on measuring flows from pipes, the author calculated q = 80.8 m3/s using the formula where q = volumetric flow rate, m3/s C = dimensionless coefficient, 0.6 AI = area, 2 m2 A2 = area, 5 m2 V = specific volume, 10-3 m3/kg p = pressure; PI - P2 is 50 kPa g = acceleration of gravity
- 68. , . .' Chap. 1 Problems 37 Was the calculation correct? (Answer Yes or No and explain briefly the reasoning underlying your answer.) ··"1.21 The density of a certain liquid is given an equation of the following form: where p = density in g/cmJ t = temperature in °c P = pressure in atm p = (A + B/)eCP (a) The equation is dimensional1y consistent. What are the units of A, B, and C? (b) In the units above, A = 1.096 B = 0.00086 C = 0.000953 Find A, 8, and C if p is expressed in Ib/ft3, t in oR, and P in Iblin.2 ···1.22 A relation for a dimensionless variable called the compressibility (z) is z = 1 + pB + p2 C + p3 D where p is the density in g mol/em3. What are the units of B, C, and D? Convert the coefficieflts in the equation for z so that the density can be introduced into the equation in the units of lbm/ft3 thus: z = 1 + p* B* + (p")2 C· '+ (pll)3 D* where p. is in Ibm/ft3. Give the units for B·, C·, and D", and give the equations that relate B''' to B, C· to C, and D'" to D. ···1.23 The velocity in a pipe in turbulent flow is expressed by the following equation u=k-[p" ]'/2 where l' is the shear stress in N/m2 at the pipe wall, p is the density of the fluid in kg/m3, u is the velocity, and k is a coefficient. You are asked to modify the equation so that the shear stress can be ihtroduced in the units of 1" which are Iblft2, and the density be p' for i,i "': . which the units are lbn/ft3 so that the velocity u' comes out in the units off tis. Show all calculations, and give the final equation in tenus of u', 1"', and p' so a reader will know that American Engineering units are jnvol ved in the equation. ·1.24 Without integrating, select the proper answer for a a (1Ia) ( 1/a) where x = length and a is a constant. arctan (ax) arctan (x/a) arctan (x/a) arctan (ax) + constant *1.25 In many plants the analytical instruments are located some distance from the equipment being monitored. Thus, some delay exists before detecting a process change and the activation of an alarm. /
- 69. 38 Dimensions. Units, and Their Conversion Chap. 1 In a chemical plant, air from a process area are continuously drawn through a 1/4 diameter tube to an analytical instrument located 125 ft from the process area. The 1/4 in. tubing has an outside diameter of 0.25 in. (6.35 mm) and a wall thickness 0.030 in. (0.762 mm), The sampling rate is 10 cm3/sec under ambient conditions of 22°C and 1.0 atm. The pressure drop in the transfer line can be considered negligible. Chlorine gas is used in the process. and if it leaks from the process, it can poison workers who might in the area. of the leak. Determine the time required to detect a leak of chlorine in the process area with the equipment currently installed. You may assume the analytical equipment takes 5 ~ec to respond once the reaches the inlitfument. You may assume that samples travel through the instrument sample tubing without dilution by mixing with the air ahead of the sample. the time excessive? How might the delay reduced? (Adapted from Problem 1 in Safety Health and Loss Prevention in Chemical Processes published by the American Institute of Chemical Engineers. New York (1990). "'1.26 In 1916 Nusselt deriVed a theorerical relation for predicting the coefficient of heat transfer between a pure saturated vapor and a colder sudace: _ (k3p1gA)"4 h - 0.943 Lp.AT where h = mean heat transfer coefficient. Btu/(hr) (ft2) (A. OF) k = thermal conductivity, Btul(hr) (ft)( p = density.lb/ft3 g = acceleration of gravity, 4.17 X lOll ftJ(hr)2 A = enthalpy change. BtuIJb L = length of tube, ft J.L = viscosity. Ibm/(hr) (ft) :;;::; temperature difference, A. OF What are the units of the constant: 0.9437 "'1.27 Explain detail whether the following equation for flow over a rectangular weir is dimensionally consistent. (This is the modified Francis formula.) q = 0.415 (L - 0.2 ho)h!-S v'2K where q = volumetric flow rate. fe/s L crest height. ft ho = weir head. ft g = acceleration of gravity, 32.2 ft/(sy' "'1.28 A useful dimensionless number called the Reynolds number is DU p J.L where D = diameter or length U = some characteristic velocity p = fluid density J.L = fluid viscosity I
- 70. 1 Problems 39" Calculate the Reynolds for the following cases: 1 2 3 4 D 2 20 ft 1 ft 2mm U to 10 miIhr 1 mls 3 cmJs p 62.41b/ft3 1 Ib/ft3 12.5 kglml 251b/ft3 jL 0.14 x 10-4 2 X to-6 1 x Ibm/(hr)(ft) Jbnl(s)(ft) centipoise (cp) ·"""1.29 are used extensively in automatic plant process control systems. The computers must signals from devices monitoring the process, evaluate the data using the programmed engineering equations, and then feed back the appropriate control adjustments. The equations must be dimensionally consistent. Therefore, a conversion factor must be part the equation to change the field into the proper units. Crude oil pumped from a storage unit to a tanker is to be eX,(:lressed in but the field variables of density and the volumetric flow rate are measured in Ib/ft3 and gaVmin. respectively. Determine the units and the numerical values of the factors to convert the field variables to the deSlLred output. ·1.30 If you subtract 1191 cm from 1201 em, each number with four significant IlI(ILUC::S. does the answer 10 cm two or four (10.00) significant hgllres -1.31 What is the sum of 3.1472 "'l.32 32.05 1234 8.9426 0.0032 9.00 to the correct number of significant figures? SU~)pOl;e you make ing out a compressed following sequence of measurements for the seg:me:nts in 1ayline: 1m 210,0 rn What should be 0.500 m reported total length of the air *1.33 Given that the width of a rectangular duct is 27.81 and the is 20.49 cm. what is the area the duct with the proper number of significant tU!lllre!,'! ·1.34 Multiply 762 by 6.3 to get 4800.60 on your ca1culator. How many significant figures exist in the product, and what should the rounded answer be? ·1.35 Suppose you 3.84 times 0.36 to get 1.3824. Evaluate the maximum relative error in (a) each number product. If you add the relative errors in the two numbers, is sum the same as the relative error in their product?
- 71. Dimensions, Units, and Their Conversion Chap. 1 ~1.36 A problem was posed as follows: The equation for the velocity a fluid stream measured with a Pitot tube is v =l!p where :v velocity. = drop p = density of fluid If the pressure drop is 15 mm Hg, and the density of the fluid· is 1.20 glcrn). calculate the velocity in ftis. The solution given was IN 2 15 mmHg 1.013 X 105 Pa m2 1 -.....;..;:..;.....;............;.. 760 mm Hg 1 Pa (l N)(s2) ~ (kg)(m) , Check that the answer is correct by (a) Repeating the calculations but carrying them out in reverse starting with the answer. . (b) Consolidating the units and making sure the final set of units are correct. (c) Repeating the calculations with a pressure drop of 30 mm Hg and a fluid density of 1 glcm3, and determining if the answer has changed in the correct proportionality. (d) Reviewing the calculation procedure and determining the powers have been calculated correctly and the conversion factors are correct and not inverted. "'1.37 Repeat Problem 1.36 for the solutions in (a) Example 1.2,'(b) Example 1 (c) Exam· pIe 1.4, and (d) Example 1 "1.38 dimensionless growth factor of a cell YS,us can be represented by an input-output relation for cell growth: where Ym=------ 1 + GATP Y XlATP G A TP mol A TP produced/mol carbon catabolized (utilized) YXlATP = mole substrate carbon/mol ATP Y~s dimensionless stiochiometric coefficient associated with the biomass produced in the reaction A TP stands for adenosine triphosphate that is involved . the catabolism.
- 72. 'r0,. l;;.~~ Chap. 1 Problems Calculate the growth factor for the anerobic fennentation of glucose (C6HI20 6) to ethanol with the N supplied by NH3 to fonn cells with the formula CHI.7S00.JRNo.25' Experiments show that YXlATP = 0.404 mol cell elmo} ATP. The literature shows that 2 moles of A TP are synthesized per mole of glucose catabolized. -1.39 Calculate the protein elongation (fonnation) rate per mRNA per minute based on the following data: (a) One protein molecule is produced from x amino acid molecules. (b) The protein (polypeptide) chain elongation rate per active ribosome uses about 1200 amino acids/min (c) One active ribosome is equivalent to 264 ribonucleotides. (d) 3x ribonucleotides equal each mRNA. Messenger RNA (mRNA) is a copy of the infonnation carried by a gene in DNA, and is involved in protein synthesis.
- 73. MOLES, DENSITY, AND CONCENTRATION 2.1 The Mole 2.2 Density 2.3 Specific Gravity 2.4 Flow Rate 2.5 Mole Fraction and Mass (Weight) Fraction Analyses of Multicomponent Solutions and Mixtures Concentration Your objectives in studying this chapter ars be able to: 1. Define a kilogram mole, pound mole, and gram mole. 2. Convert from moles to mass and vice versa for any chemical compound given molecular weight Calculate the molecular weight from the molecular formula. 4. Define density and specific gravity. 5. Calculate the density of a liquid or solid given specific gravity and the vice ve rsa. 6. Look up and interpret the meaning density and specific gravity of a liquid or solid reference tables. 7. Specify common reference material(s) used to determine the specific gravity of liquids and solids. S. Convert the composition of a mixture from mole fraction (or percent) to mass (weight) fraction (or percent) and vice versa. 9. Transform a material from one measure of concentration to another, including mass/volume, moles/volume. ppm, and molarity. 10. Calculate the mass or number of moles of each component in a mixture given the percent (or fraction) composition, and vice versa, and compute the pseudo-average molecular weight. 11. Convert a composition given in mass (weight) percent to mole and vice versa. 43 48 51 59 62
- 74. ~ ... Sec. 1 The Mole In this chapter we review a number of concepts, and procedures with which you are no doubt somewhat familiar. We believe a firm grasp of this material is essential to the proper implementation of material and energy balance calculations. If you make sure that you have a sOllnd command of this material, you will reduce the number errors that you will when formulating material and energy balances throughout the remainder of the text. Looking Ahead In this chapter we flrst discllss the mole. Then we review some of the common conventions used in physical properties; including density, specific grav-ity, measures of concentration, and flow rates. 2.1 The Mole What is a mole? For our. purposes we that a mole is a certain amount of material corresponding to a specified number of molecules; atoms, electrons, or any other specified types of particles. The word mole appears to have introduced by William Ostwald in 1896, who took it from the Latin word moles meaning "heap" or "pile." If you think of a mole as a large heap of particles, you will have general idea. more precise InttiOn was out by the 1969 International Committee on Weights and Measures. which approved the mole (symbol mol in the SI system) as being "the amount of a substance that contains as many elementary entities (6.022 x 1023) as there are atoms in 0.012 kg of carbon 1 H The entities may be atoms, molecules, ions, or other particles. Thus, you could have a mole of eggs consisting of 6.022 x 1023 eggs! If you had data amounting to 1 mol of bytes on your computer hard disk, how many gigabytes of space would be filled? It would more than 6 x 1014 gigabytes! In the SI system a mole composed of 6.022 x 1()23 (Avogadro's Number) molecules. However, for convenience in calculations and clarity, we will make use of other specifications for such as the pound mole (lb mot comprised of 6.022 x 1023 X 453.6) molecules. the kg mol (kilomole, kmol, comprised of 1,000 moles), and so on. You will find that such nonconforming (to S1) definitions of the amount of material will help avoid excess details many calculations. To keep units straight, we will use the designation of g mol for the SI mole. What would a ton mole of molecules consist of? One important calculation you should become skilled at is convert the num-ber of moles to mass and the mass to moles. do this you make use of the molecu .. lar weight-the mass per mole: mass molecular weight (MW) = - mole
- 75. 44 and Moles, Density. and Concentration Thus, the calculations you carry out are example mass in g the g mol = . molecular weIght the lb mol = mass in Lb molecular weight mass in g = (MW)(g mol) mass in lb = (MW)(lb mol) 6.0 Ib mol ----..::. ---~ = 1921b02 Chap. 2 Values of the molecular weights (relative molar masses) are built up from the tables of atomic weights based on an arbitrary of the relative masses of the ments. atomic weight of an element the mass an atom based on the scale that assigns a mass of 12 to the carbon isotope 12C. The terms atomic "weight" and molecular "weight" are universally used by chemists and engineers stead of the more accurate atomic "mass" and molecular "mass." weighting was original method for determining comparative atomic as long as they were calculated in a common gravitational field, the relative values obtained for atomic "weights" were identical with those of atomic "masses." Appendix B lists the atomic weights of elements, On this of atomic weights, hydrogen is 1.008, carbon is 12.01, and so on. (In most of our calculations we shaH round these off to I and 12. respectively. convenience.) A compound is composed of more than one atom, and molecular weight of the compound is nothing more than the sum the weights of atoms which it composed. Thus H20 consists of 2 hydrogen atoms and 1 oxygen atom, and the mol-weight water (2)(1.008) + 16.000 ;;; 18 These are all relative to the I atom being 12.0000, you can any unit of mass you desire to these weights; example. H2 can be 2,016 mol, 2.016 lbllb mol, 2.016 tonlton mol, and so on. Appendices and the CD the back book molecular You can compute the average molecular weight of a mixture even though its components are not chemically bonded if composition of the mixture is known. in Section 2.7 we show how to calculate the fictitious quantity called aver-
- 76. " Sec. 2.1 The Mole 45 age molecular weight of air. Of course, for a material such as fuel oil or coal, whose composition may not be exactly known, you cannot determine an exact molecular weight, although you might estimate an approximate average molecular weight good enough for engineering calculations. Keep in mind that the symbollb without any subscript in this book refers to Ibm unless otherwise stated. EXAMPLE 2.1 Calculation of Molecular Weight Since the discovery of superconductivity almost 100 years ago, scientists and engineers have speculated about how it can be used to improve the use of energy. Until recently most applications were not economically vIable because the niobium alloys used had to be cooled below 13K by liquid He. However, in 1987 su perc on-" ductivity in Y -Ba-Cu~O material was achieved at 90 K, a situation that permits the use of less expensive liquid N2 cooling. What is the molecular weight of the following cell of a superconductor material? (The figure represents one cell of a larger structure.) • Barium • Yttrium • Copper ~~ Oxygen Figure E2.1 Solution You can count the number of atoms of each element by examining Figure E2.1. Look up the atomic weights of the elements from the table in Appendix B. Assume that one cell is a molecule. By counting the atoms and carrying out the brief calculations below. you obtain the molecular weight of the cell. Element Number of atoms "Atomic weights Mass (g) Ba 2 137.34 2(137.34) Cu 16 63.546 16(63.546) 0 24 16.00 24(16.00) Y 88.905 1(88.2Q5) Total 1764.3 The molecular weight of the cell is 1764.3 atomic masses!l molecule, or 1764.3 gig mol. Finally, check your calculations.
- 77. 46 Moles, Density, and Concentration EXAMPLE 2.2 Use of Molecular Weights to Convert Mass to Moles If a bucket holds 2.00 Ib of NaOH. how many (8) Pound moles of NaOH does it contain? (b) moles of NaOH does it contain? Solution You want to convert pounds to pound moles, and then convert respective val-ues to the SI units. Look up the molecular weight of NaOH. or calculate it from the atomic weights. It is 40.0. ( ) 2.00 Ib NaOH 11 Ib mol = 0050 Ib I NaOH a 40.0 Ib NaOH . mo 1 Ib mol NaOH 454 mol 40.0 Ib NaGH 1 Ib mol = 22.7 g mol (hI) 2.00 Ib NaGH -=1--=------lib ........ '.AvA your answer. Convert the 2.00 Ib of NaOH to the system first. then calculation. EXAMPLE Use of Molecular Weights to Convert Moles to Mass How many pounds of NaOH are in 7.50 g·mol of NaOH? Solution The problem concerns converting g mol to lb, From 2.2. the MW of NaOH is 40.0. Basis: g mol of NaGH 7.50 g mol NaGH lIb mol 40.0 Ib NaOH 454 g mol 1 mol NaGH ;;;.: 0.661 lb NaGH Note the conversion between Ib mol and g mol converts the value of 7.50 g mol the SI to the system of units. Could you first convert 7.50 g NaGH to g of NaGH, and then use the conversion of 454 g = 1 to get lb NaGH? Of course. 2 J
- 78. ~-. 2.1 Mole 47 ELF-A SESSM NT TEST Questions 1. Answer the foHowing questions true or false: a. The pound mole is comprised of x 1 Q26 molecules h. kilogram mole is comprised of 6.022 x 1026 molecules. c. Molecular weight is the mass of a compound or element per mole, 1. What is the molecular weight of acetic acid (CH3COOH)? Problems 1. Convert the foHowing: a. 120 g mol of NaCI g. b. 120 g of NaCI to g mol. c. 120 lb mol of NaCl to lb. d. 120 Ib of NaCI to Ib moL 2. Convert 39.8 kg of NaCI per 100 kg of water to mol of NaCI per mal of water. How many Ib mol NaN03 are there in 100 Ib? Thought Problem 1. There is twice as much copper in 480 g of copper as there is in 240 g of copper, but is twice as much copper in 480 g of copper as there is silver in 240 g of silver? 2. If a mole of doughnuts 6.022 x 1 doughnuts, can you also say that the mole of doughnuts contains the same number of moles of doughnut holes when holes are not doughnut material? Discussion Problems In journal Physics Education (July 1 p. 276) McGlashan suggested that the physical quantity we call the mole is not necessary. Instead. it would be quite feasible to use molecular quantities. that number of molecules or atoms, directly. Instead of pV ::::: nRT where n denotes the number of moles of a substance. we should write pV ::::: NkT where denotes the number of molecules and k is the Boltzmann constant [1.380 x lo-23JJ(motecule)( K)]. Thus, example. 3.0 x 10-29 m3/molecule would the molecular volume for water, a value used instead 18 cm3Jg mol. Is the proposal a reasonable idea? 2. In asking the question "what meant by a mole," the following answers were obtained. Explain which are correct and which are not. and why. a. A mole is the molecular weight expressed in grams. b. A mole is the quantity of material one gram. c. A mole is a certain number of of one substance or another. d. A mole is the weight of a molecule expressed in grams. e. A mole is the number of molecules one gram of a substance.
- 79. 48 Moles, Density, and Concentration Chap. 2 2.2 Density A striking example of quick thinking by an engineer who made use of the con-cept of density was reported by P. K. N. Paniker in the June I 1970, issue of Chemical Engineering: bottom outlet nozzle of a full lube-oil storage tank kept at a temperature about 80aC suddenly sprang a gushing leak as the nozzle flange became loose. Because of high temperature of the oil, it was impossible for anyone to near tank and repair the leak to prevent further loss. After a moment of anxiety, we noticed that the engineer in charge rushed to his office to summon fire department personnel and instruct" them to run a hose from the nearest fire hydrant to the top of the storage tank. Within minutes, gushed out from the was hot water instead of va1uab~e oil. Some time later, as entering cold water lowered oil temperature, it was possible to make repairs. Density is the ratio of mass per unit volume, as examp1e, kg/m3 or Ib/ft3. Density has both a numerical value and units. To detennine the density a substance. you find both volume and mass. Densities for liquids and solids do not change significantly at ordinary conditions with pressure, but they do change with temperature, as shown in Figure 1. Usually we will ignore the effect, of temperature on liquid density. Specific volume is the inverse of density, such as cm3/g or ff~/lb. What can you do with density? You use it to determine the volume of a given mass, or the mass of a given volume of materiaL For a compound mass p density = I . voume A V = specific volume 1.0r-----H2-0 ----___ _ 0.9 E 0.8 ~0.7 ~ ,---_ 0.6 ~ 0.5 0.4 0.3 - m - V volume V - mass m 0.20 10 20 30 40 60 70 80 90 100 Figure 2.1 Densities of liquid H20 NH) as a function of temperature. J
- 80. 2.2 Density so that given the density of the compound you can calculate the volume a given mass. For example, given that the density of n-propyl alcohol is 0.804 g/cm3, what would the volume of 90.0 g of alcohol? The calculation is 90.0 1 cm3 --'---- = 112cm3 0.804 g Some quantities related to density are molar density (p/MW) and molar volume (MW/p). By analogy, in a packed bed of solid particles containing void spaces, the bulk density is total mass of solids PB = bulk density = tal to empty volume A homogeneous mixture of two or more components, whether solid, liquid, or gaseous, is called a solution. Solutions have variable composition while pure substances do not. That the relative amounts of the various components in a solution can . Thus, air, salt water, and 16-carat gold each are solutions. For some solutions, to calculate the density of the solution. you can make a linear combination of the individual components by adding the respective masses and volumes, and then dividing: 1'1 V= where n = number of components m= m Psolution = V For others you cannot. Examine Figure SELF-ASSESSMENT TEST Questions 1. For numbers such as 2 of water + 2 of ethanol, the sum equal to 4 mL of the solution? 2. Answer the foHowing questions true or false: a. The inverse of the density is the specific volume. b. Density of a substance is the mass per unit volume. c. The density of water less than the density mercury.
- 81. 50 .. Moles, Density, Concentration Chap. 2 O.1o5 0::--_-..,:~--~--~--~~-~ Figure 2.2 Density of a mixture of ethyl alcohol and water as a function of 3. A cubic ... "" ....... 11 ....... "'. of mercury has a mass ,of 13.6 g at Earth's surface. What is the density of mercury-! 4. What is approximate density of water at room temperature in Problems 1. The density of a is 2 kglm3, What is its specific volume? 2. An empty 10 gal weighs 4.5 lb. What is the total water when it is filled with 5 of water? 3. If you add 50 g of to 500 of water, how do you calculate the density of the sugar solution? Thought Problems 1. The representative of Lloyd's Shipping testified in a Houston District court relati ve to the fraud trial of a businessman. The indictment alleged that the busi-nessman stole 200,000 tons of oil from the Italian owner by delivering the oil to South Africa, and then scuttling the tanker to cover up the theft The prosecutor asked the insurance representative whether the tanker could sunk with a full load of oil. What you think? to the pl1Ttnr said: My hobby photography. I was using a gallon milk container to measure the water nt!(!oe:a to OISl;Olre some developer. and realized that the was about 3 ounces short a fun of liquid when filled to the brim. too the water? I
- 82. Sec. 2.3 Specific Gravity 51 ' was taking it directly from the tap, and perhaps should have let the air bubbles settle out. I think we are getting less milk than we think when we buy milk in these containers. How would you reply to the author explaining what the cause of his dilemma probably is? 3. A refinery tank that had contained gasoline was used for storing pentane. The tank overflowed even when the level indicator said that it was only 85% full. The level indicator measured the weight of fluid. Can you explain what went wrong? Discussion Problem 1. From Chemical and Engineering News, October 12, 1992, p. 10: Two Dutch scientists have won government and industry support to explore the possibility of raising the level of the ground in coastal areas of their low-lying country by converting subsurface limestone to gypsum with waste sulfuric acid. The scheme centers on the fact that gypsum, CaS04 • 2H20, occupies twice the volume of a corresponding amount of calcium carbonate. The project envisions drilling holes as deep as 1 km at selected sites above limestone strata for injecting the acid. The resulting gypsum should raise the surface as much as several meters. Instances of ground swelling have already occurred from sulfuric acid spillage in the Netherlands at Pernis, an industrial region near Rotterdam. What do you think of the feasibility of this idea? 2.3 Specific Gravity In the winter you should have antifreeze in your car radiator. The service station attendant checks the concentration of antifreeze by measuring the specific gravity and, in effect, the density of the thoroughly ,mixed radiator solution, with a hydrometer. The hydrometer kit contains a small thermometer to enable .the attendant to adjust the hydrometer reading to the correct density. Specific gravity is commonly thought of as a dimensionless ratio. Actually, it should be considered as the ratio of two densities-that of the substance of interest, A, to that of a reference substance--each of which has associated units. In symbols: • . (g/cm3)A (kg/m3)A (lb/ft3)A sp.gr. of A = specific graVIty of A = 3 = 3 = 3 (g/cm' ),e! (kg/m )rej (lb/ft )ref The reference substance for liquids and solids nonnally is water. Thus, the specific gravity is the ratio of the density of the substance in question to the density of water, which is 1.000 g/cm3, 1000 kg/m3, or 62.43 Ib/ft3 at 4°C. The specific gravity of gases frequently is referred to air, but may be referred to other gases. We will explain what to do in Chapter 13 . . _ _ _____ 0_ . ___ _ • _._ .
- 83. • J Moles, Density, and Concentration Chap. 2 To be precise, when referring to specific gravity. the data should be accompanied by both the temperature of the substance of interest and the temperature at which the reference density is measured. Thus, for solids and liquids the notation 20° sp.gr. == 0.73 40 can be interpreted as follows: the specific gravity when the solution is at 200e and the substance (implicitly water) is at 4°e is 0.73. In case the temperatures for which the specific gravity is stated are unknown, assume ambi¢nt temperature for the substance and 4°e for the water. Since the density of water at 4°e is very close to 1.0000 g/cm3, in the SI system the numerical values of the specific gravity and density are essentially equal. The calculations for dibromopentane (DBP), sp.gr. = 1.57. are (a) (b) (c) or 1 gH20 em3 --------== 1.57-- I Ib DBP Ib H20 1.57 ft' 62.4 ft3 _~ ______ . == 97.971b DBP lb H20 OO L 3 ft 1.57g DBP (100 cm)3 1 kg == 1.57 X 103 cm3 1 rn 1000 g kgDBP 1.57---:--- ----~~- ----- ----~-- = 1.57 X 103........:--kg H20 1.00 3 m Note how units of specific gravity are used here to clarify the calculations, and that the calculation in (a) can be done in your head. Since densities in the system are expressed in Ib/ft3, and the density of water about 62.4 Ib/ft3! you can see that the specific gravity and density values are not numerically equal in the American Engineering system. Yaws, Yang, Hopper,
- 84. Sec. 2.3 Spacific Gravity Cawley (1991)* is a source for values of liquid densities, and the software on the CD in the back of book contains portions of Yaw's database. EXAMPLE 2.4 Calculation of Density Given the Specific Gravity If a 70% (by weight) solution of glycerol has a specific gravity of 1.184 at 1 SoC, what is the density of the solution in (a) g/cm3? (b) Ibn/ft3? and (e) kg/m3? Solution Use the specific gravity to get density via a reference substance. No temperatures are cited for the reference compound (presumed to be water), hence for simplicity we assume that the temperatures of the water is and that water has a density of l.00 X 103 kg/m3 (1.00 g/cm3). The answers are: (a) 1.184 g solution/em3 (calculated in your head) (b) (L1841b glyeerol/ft3)/(1lh water/ft3) x (62.4 Ib water/ft3):: 73.91b solution/ ftl (c) 1.184 X 103 kg solutionlm3 (calculated in your head), You may acquainted with the fact that in the petroleum industry the specific gravity of petroleum products is often reported in terms of a hydrometer called o API. The equation for the API IS or 141.5 60°F sp.gr,-- 131 60° 141.5 (API gravity) sp.gr. 600 = 0 API 131.5 (2.1) (2.2) The volume and therefore the density of petroleum products vary with temperature, and the petroleum industry has established as the standard temperature for volume and API gravity. The CD in the back of the book contains data petroleum products. The organization that governs SI units is phasing out the use of 0 API for densities. Many other specialized systems of measuring density and specific gravity such as Baume (OBe) and Twaddel (OTw) exist. You can find information about those measures and relationships among them in the at the end of chapter. L.. H. Yang, 1. R. Hopper, and A. Cawley. "Equation for Liquid Density," Hy-pp. 103-106 (January 1991).
- 85. Moles, Density. and Concentration EXAMPLE 2.5 Application of Specific Gravity to Calculate Mass and Moles Chap. 2 the production a drug having a molecular weight of 192, the exit stream from the reactor flows at a rate of 10.5 Umin. The drug concentration is 41.2% (in water), and the specific gravity of the solution is 1.024. Calculate the concentration of the drug (in kgIL) in the exit stream, and the flow rate of the drug in kg moUmin. Solution Read the problem careful1y because this example is more complicated than the previous examples. You have a solution with some known properties including sp.gr. The strategy for the solution is to use the specific gravity to get the density from which you can calculate the moles per unit volume. After the problem, to make the requirements of the problem clearer, you should draw a picture, and put the given data on the picture. For the first part of the problem, you want to transform the mass fraction of 0,412 into mass per liter of the drug. Take 1.000 kg of the exit solution as a basis for convenience. Reactor . 1.000 kg solution 10.5LJmin sp.gr. == 1 Drug 0.412 kg Water 0.588 kg Figure E2.S How do you get mass per volume (the density) from the given data, which is in terms of mass of drug per mass of solution (41.2%)? Use the specific gravity of the solution. Calculate the density as fonows soln g H 0 )4';;;'-- 1.000 2 cm3 density of solution :::: ---- ---- - 1.024-- Next convert the amount of drug in 1,000 kg of solution to mass of drug per volume of solution using the density 0,412 drug 1.024 g soln 1 kg 103cmJ = 0422 1.000 kg soln 1 103 gIL . drugIL soln Could you simplify the last two by knowing that 1.000 of water is 1.000 L? Certainly. To get the flow rate, take a different basis, namely 1 minute. Basis: 1 min == 10.5 L solution
- 86. Sec. 2.3 Specific Gravity Convert the volume to mass and then to moles using the information previously calculated. -10.-5 L- so-ln -0.4-22 -kg- dru-g 1 kg mol drug -~--...:.... = 0.023 kg moUmin I min 1 L soln 192 kg drug How might you check your answers? Did you notice how we represented a reactor in Figure E2.5 by a simple box with a label? In the interest of simpJification (and economy, of course), in this book we usually use a simplified drawing of a process or any equipment that is involved in a discussion or example. Figure 2.3 shows two examples of how you might replace the real apparatus with a sketch. Even Figures 2.3a and 2.3b themselves are simplified drawings of what would look to be very complicated if you took a photograph of the equipment. System Boundary Malntenanoe Access Manway Tank c. Vapor Space 1000 kg H20 b. d. UQU'dlevel 1 Figure 2.3 Examples of possible substitutes for pictures of equipment by using a simple sketch. SELF-ASSESSMENT TEST Questions 1. For liquid HeN, a handbook gives: sp. gr. IO°CJ4°C = 1.2675. What does this statement mean? 2. Answer the following questions true or false: B. The density and specific gravity of mercury are the same.
- 87. Moles, Density, and Concentration Chap. 2 Specific gravity is the of two densities. c. If you are given value of a reference density, you can determine the density of a substance of interest by mUltiplying by the specific gravity. The gravity a dimensionless quantity. Problems .. 1. For ethanol, a handbook gives: sp. gr. 60°F ::= 0.79389. What is the density of ethanol at 600 P? 2. The specific gravity of steel is 7.9. What is the volume in cubic feet of a steel ingot weighing 4000 Lb? 3. ....... "'.1".'1' ..... gravity of a solution is 0.80 at 70°F. How many cubic feet win be occupied the solution at 70°F? 4. A solution in water contains L 704 of HN03/kg HZOt and the solution has a specific gravity of 1.382 at 20°e. What is the mass of HN03 in kg per cubic meter of solution at 200e? Thought Problem 1. The National Museum is considering buying a Maya plaque from Honduras that the seller claims to jade. Jade is either jadite (sp. 3.2 to 3.4) or nephrite (sp. gr. 3.0). What liq-uids would you recommend using to test whether or not the is jade? 2.4 Flow Rate continuous the flow rate a process is rate at which material is transported through a pipe. In this book we usually use an overlay dot to denote a for the volumetric flow rate F. mass flow rate (m) of a process stream is the mass (m) transported through a line unit time (t). m m=t The volumetric flow rate (F) of a process stream the volume (V) transported through a line per unit V F=t molar flow (n) of a process stream the number of moles (n) of a sub-stance transported through a line unit time. . n n=- t Use any consistent units in your calculation.
- 88. I Mole Fraction and Mass (Weight) Fraction SELF .. ASSESSM NT TEST Problems 1. Forty gal/min a hydrocarbon fuel having a specific gravity of 0.91 flow into a tank truck with a load limit of 40,000 Ib of fuel. How long will it take to fin the tank in the truck? Pure chlorine enters a process. By measurement it is found that 2.4 of chlorine pa.<;s into the process every 3.1 minutes. Calculate the molar flow rate of the chlorine in kg mollhr. 2.5 Mole Fraction and Mass (Weight) Fraction Mole fraction is simply the number of moles of a particular compound in a mixture or solution divided by the total number of moles in the mixture or so1ution. This definition holds for gases, liquids. and solids. Similarly, the mass (weight) fraction is nothing more than the mass (weight) of compound divided by the total mass (weight) of all of the compounds in the mixture or solution. Although the mass fraction the correct word, by custom ordinary engineering usage employs the term weight fraction. Mathematically. these ideas can be expressed as moles of total moles mole fraction of . mass A mass (weight) fraction of A = ---total mass Mole percent and mass (weight) percent are the respective fractions times 100. EXAMPLE 2.6 Conversion between Mass (Weight) Fraction and Mole Fraction An industrial-strength drain cleaner contains 5.00 of water and 5.00 kg of NaOH. What are the mass (weight) fractions and mole fractions each component in the drain cleaner container? Solution You are given the masses so that it is to calculate the mass fractions. From these values you can then calculate the desired mole fractions. A convenient way to carry out calculations in such conversion problems is to form a table, as shown below. Become skilled at doing so this type of
- 89. 58 Moles, Density. arid Concentration Chap. 2 problem and its inverse, that the conversion of mole to mass fraction. will occur quite frequently. the components, their masses. and their molecular weights columns. Basis: 10.0 kg of total Component kg Weight fraction Mol. wt. kg mol Mole fraction 5.00 H2O 5.00 10.0 ;:::; 0.500 18.0 0.278 0.278 =- 0.69 OA{)3 ' NaOH 5.00 5.00 0.500 40.0 0.125 10.00 0.1 0.403 = 0.31 Toral 10.00 1.000 0.403 1.00 '. The kilogram are as follows: 5.00 kg H20 11 mol H20 . 18.0 kg H 0 = 0.278 kg mol H20 2 5.00 kg NaOH 1 mol NaOH --~-- -~---- = O. 'kg mol 40.0 NaOH Adding these quantities together the total kil.og ram moles. . EXAMPLE Nitrogen Requirements Used for the Growth of Cells, In normal living cells, nitrogen requirement for the cells is provided from protein metabolism (Le .• consumption the protein in the ceUs). When individual cells are commercially grown, (NH4hS04 is usually used as the source of nitrogen. Determine amount of (NH4)2S04 consumed a fermentation medium in which the final concentration is 35 gIL a 500 L volume of the fermentation medium. Assume that contain 9 wt. % N, and that (NH4hS04 is the only source. Solution 500 L 135 g celli 0.09 g N L 1 g cell 500 L solution containing 35 gIL 1 g mol N 14 g N
- 90. Sec. 2 Analyses of Multicomponent Solutions and Mixtures 59 ELF .. ASSE SM NT T T Problems 1. Commercial sulfuric acid is 98% A compound contains 50% sulfur and 50% by mass. Is the empirical formula of the compound (1) (2) (3) S03' or (4) 3. How many kg of activated (a in removing trace impurities) must be with 38 kg of so the final mixture is 28% carbon? 4. A gas mixture contains 40 Ib of 251b S02' and 30 Ib of S03' What is the composi-tion of the mixture in mole fractions? Discussion Problem 1. Which of the foHowing kinds of ethanol would you regard as being purest? a. Ethanol made fermentation foHowed by fractional distillation to obtain a product composed of 95% ethanol. b. Ethanol made from 95% ethanol by distillation with a compound that gives as final result 99.8% ethanoL c. Denatured ethanol ("alcohol"). d. from ethene. e. Medical ethanol ("alcohol"), 2.6 Analyses of Multicomponent Solutions and Mixtures confusion exists when you are presented with an analysis of a solu-tion or mixture you may be uncertain as to whether the numbers presented represent mass (weight) fraction or mole fraction. In this book, the composition of gases will always be presumed to be given in mole percent or fraction unless specifically stated otherwise. In this book composition liquids and solids will be given by mass (weight) percent or fraction unless otherwise specifically stated, as is the common practice in industry. an example, look at Table 1 that lists the detailed composition of dry Do the values in the column u..,."u ... ,u "Per~ent" designate mass or percent? They are mole percent.
- 91. 60 Moles, Density, and Concentration Chap. 2 TABLE 2.1 Composition of Clean, Dry Air near Sea Level Mole percent Component Nitrogen 78.084 Xenon 0.0000087 Oxygen 20.9476 Ozone Argon 0.934 Summer 0-0.000007 Carbon dioxide 0.0350 Winter 0-0.000007 Neon O.OOHHS Ammonia O-trace Helium 0.000524 Carbon monoxide O-trace Methane 0.0002 Iodine 0-0.000001 Krypton 0.000114 Nitrogen dioxide 0-0.000002 Nitrous 0.00005 SuHur dioxide 0-0.001 Hydrogen 0.00005 useful quantity in many calculations is the approximate pseudo-average molecular weight of air, a that can be calculated on the assumption that aU air that is not 02 N2 with a psuedo-moleculat weight of 28.2. Component Moles :: percent 21.0 1M 100 Mol. wt. Lborkg 672 2228. 2900 Weight % 23.17 1U3. 100.00 The average molecular weight is 2900 IbiLOO Ib mol = 29.0, or 2900 kg/IOO kg mol == 29 What value would you by carrying out a more accurate calculation. taking into an of the components in Table 1 ? shows the analysis of various compounds. Are values in the table in mass or percent? TABLE 2.2 Chemical AD8Jyses of Various Wades by Percent Raw Charred Material paper paper Moisture 3.8 0.8 Hydrogen'" 6.9 3.1 Carbon 45.8 84.9 Nitrogen 1 Oxygen" 46.8 8,5 Sulfur 0.1 0.1 Ash 0.4 Tire rubber 0.5 4.3 86.5 4.6 1.2 Dry sewage I 6.1 28.7 2.6 26.5 0.6 34.9 Charred sewage sludge 1.2 1.4 3.7 45.7 Cbarred Garbage Garbage animal composite composite manure B 0.0 3.4 12.3 5.4 6.6 7.0 41.2 57.3 1.5 0.5 26.0 22.1 42.1 0.4 0.4 0.2 10.2 5.9 hydrogen and oxygen values reflect that due to both the presence of water and that contained widtin the moisturefree material.
- 92. Sec. 2.6 Analyses of Multicomponent Solutions and Mixtures 61 SELF .. ASSES MENT ST . Questions 1. A is reported as 15% water and 85% ethanoL Should percentages be deemed to be by mass, mole, or volume? In a recent EPA inventory of 20 greenhouse that are emitted in the U.S .• carbon dioxide comprised 1 million (Gg), which was about 70% of the U,S. greenhouse gas emissions. Fossil fuel combustion the utility sector contributed to about 34% of all the carbon dioxide emissions, the transportation, industrial. and residential-commercial sectors for 34%, 21%, and 11 % of the total, respec-tively. Are the four mole percent or mass percent? Answer following questions true or a. In engineering practice the compositions of liquids and solids are usually denoted in weight (mass) fraction or percent. In engineering practice composition of gases is usuaJly denoted in mole fraction or percent. c. A pseudo-average molecular weight can be calculated a of pure compo-nents whether solid. liquid, or gases. Problems 1. Saccharin, an artificial sweetener that is 3()(x} sweeter than sucrose, is composed of 45.90% carbon. hydrogen, oxygen, 7.65% nitrogen, and 17.49% sulfur. the molecular formula of saccharin (8) C14HIO06N2S2' (b) CSH703NS. (c) CgH90 2NS, /(d) C7Hs0 3NS? A mixture of gases is analyzed and found to have 12.0% CO 6.0% CH4 27.3% 9.9% N2 44.8% much 3 lb mol of this gas weigh? IOlliOWlD2 composition: 3. liquefied mixture of n-butane, n-pentane, and n-hexane the foHowing composition: For this mix·ture. calculate: 50% n-CSHI2 30% n-C6H14 20% a. The weight fraction of each component. b. mole fraction each component
- 93. 62 Moles, Density, and Concentration Chap. 2 c. The mole percent of each component. d. The average molecular weight of the mixture. 2.7 Concentration Concentration generally refers to the quantity of some substance per unit volume, but other related measures of the amount of material frequently occur, as indicated by some of the terms in the following list of ways to express concentration: 3. Mass per unit volume (lb of solute/ft3 of solution. g of solute/L, ]b of solutelbarrel, kg of solute/m3). b. Moles per unit volume (lb mol of solutelft3 of solution. g mol of solutelL, g mol of solute/cm3). c. Parts per million (ppm); parts per billion (Ppb), a method of expressing the concentration of extremely dilute solutions; ppm is equivalent to a mass (weight) fraction for solids and liquids because the total amount of material is of a much higher order of magnitude than the amount of solute; it is a mole fraction for gases. Why? d. Parts per million by volume (ppmv) and parts per billion by volume (ppbv) e. Other methods of expressing concentration with which you may be familiar are molarity (g mollL) , molality (mole solutelkg solvent), and normality (equivalentslL). A typical example of the use of some of these measures of concentration is the set of guidelines by which the Environmental Protection Agency defines the extreme levels at which the five most common air pollutants could hann people over stated periods of time. a. Sulfur dioxide: 365 f.,Lg/m3 averaged over a 24-hr period b. Particulate matter (10 JLm or smaller): 150 JLglm3 averaged over a 24-hr period c. Carbon monoxide: 10 mg/m3 (9 ppm) when averaged over an 8-hr period; 40 mg/m3 (35 ppm)' when averaged over 1 hr d. Nitrogen dioxide: 100 f..Lg/m3 averaged over 1 year e. Ozone: 0.12 ppm measured over 1 hour Note that the gas concentrations are mostly mass/volume except for the ppm. In the European Union (EU), aU gasoline marketed after January 1, 2000, was to be lead-free. Other components are limited as follows: sulfur, less than 150 mglkg; benzene, 1 vol. %~ and aromatics, 42 voL %. Sulfur in diesel fuel is limited to 350 mg/kg. By 2005, the maximum sulfur content of both gasoline and diesel will I
- 94. l Sec. 2.7 Concentration drop to 50 mg/kg. The aromatics limit for gasoline wil1 be volume percent, and diesel will have an 11 vol. % limit for po]yaromatics. Parts per million is a very small number. One ppm is equivalent to I inch in 16 miles, or a 1.0 g needle in a metric ton of hay. EXAMPLE 2.8 Use of ppm The current OSHA 8-hour limit for HeN in air is 10.0 ppm. A lethal dose of HCN in air is (from the Merck Index) 300 mglkg of air at room temperature. How many mg HCNlkg air is 10.0 ppm? What fraction of the lethal dose is 10.0 ppm? Solution In this problem you have to convert ppm in a gas (a mole ratio, remember!) to a mass ratio. Basis: 1 kg mol of the airIHeN mixture Draw a simple picture. Put the data in the figure. HeN 10 ppm Air Figure E2.8 You can treat the 10.0 ppm as 10.0 g mol HCN/106 g mol air because the amount of HCN is so small when added to the air in the denominator of the ratio. 10.0 g mol HeN 10.0 g mol HCN The 10.0 ppm is - ----=~--- l06(air + HCN)g mol 106 g mol air Next get the MW of HCN so that it can be used to convert moles of HeN to mass of HCN; the MW;;;; 27.03. Then 10.0 g mot HeN 27.03 g HeN [t g mol air 1000 mg HeN a. 6 109 mol air 1 g mol HeN 29 g air 1 g HCN 1000 g.air X 1 k . = 9.32 mg HCNlkg air g air 9.32 b. 300 = 0.031 Does the answer seem reasonable? At least it is not more than I!
- 95. 64 Moles, Density. and Concentration EXAMPLE 2.9 CalCulation of Mole Fraction and ppm from a Concentration in gIL Chap. 2 A solution of HN03 in water has a specific gravity of 1.10 at 25°C. The concentration of the HNO) is 15 gIL of solution. What is the a. Mole fraction of HN03 in the solution? b. ppm of HN03 in the solution? Solution Let the value of the specific gravity be the value of the density (ignoring any minor effects related to the density of water), Basis: 1 L solution 15 HN03i 1 L 1 == 0.01364 HNO) 1 soln 1000 cm3 1.10 g soln g soln Basis: 100 g so]ution The mass of water in the solution is: 100 - 0.0134 = 99.986 g H20. g MW gmol mol fraction a. HNO:] 0.01364 63.02 2.164 X 1Q-4 3.90 x 10-5 H2O 99.986 18.016 Total 1.00 b. 0.01364 13,640 = or 13,640 106 EXAMPLE 2.10 Evaluation of Alternate Processes for the Production of Methyl Methacrylate Methyl methacrylate can be made using as feed (a) acetone and hydrogen cyanide, or (b) isobutlyene. Table 10 lists some data assembled from book by D. Allen and D. R. Shonnard (Green Engineering, Prentice Hall, Upper Saddle River, N. 2002) relating to these two processes. The mass values are on the basis of the production of one pound of methyl methacrylate, Oxygen also involved but neglected.
- 96. Sec. 2;7 Concentration TABLE EZ.l0 Compounds Used (a) Acetone Hydrogen cyanide Methyl methacrylate produced (b) Isobutylene Methanol Pentane Sulfuric acid Methyl produeed Lb Value ($IJb) 0.68 0.43 0.32 0.67 0.37 0.64 1.63 0.04 1.00 0.78 1.12 0.31 0.38 0.064 0.03 0.112 0.01 0.04 1.00 . 0.78 TLV· (PPW;O) orrr" 750 1 10 1000 200 10 2 10.000 100"· to 200 7 200 10 600 ? 2 10.000 1O(f ... • 10 ""Threshold limit value in the wo.dcpls.oe by Set at a level for which no adverse are expected over a worker's lifetime. , "'Overall inhalation toxicity factor developed by the U. S. Environmental Protection Agency. "·4'PEL instead ofTLV. PEL is the permissible exposwe specified by the U. S. OccupadooaJ md Safety Administration. Based on the data in Table B2.10t which process would you recommend be used to produce methyl methacrylate? Solution 2.10 includes three different that you can use in the evalua-as several compounds involved each The easiest criterion to the net value of one pound of the product: Process(s}: Net value = 1($0.78) 0.68($0.43) - 0.32{$O.67) . ($0.064) - 1.63($0.04) = $0.18 Process{b): Net value = 1 ($0.78) - 1.12($0.31) - 0.38($0.64) - 0.03 (SO.112) - 0.01($0.04) = $0.19 Only a difference exists in value. With respect to the two environmental late an index. For 11., V. the larger values are hence a mass weighted reciprocals can used: a typical procedure is to calcuconcern than the smaller ones, TLVindex = L~ I TLVj For OITF. the the value, the more concern, hence the index calculated as: OITF index = (m,)
- 97. » Moles, Density, and Concentration Chap. 2 For (a): 0.68 0.32 0.37 1.63 1 0 86 TLV index = + + -- + -- + - = . 750 10 200 2 100 OITF index = 0.68(?) + 0.32(1000) + 0.37(10) + 1.63(10,000) + 1(10) - 16,600 For Process (b): L 12 0.38 0.03 0.01 00 lLVindex --+--+ +--=.1 200 600 2 OITFindex = 1.12(?) + 0.38(10) + 0.021(?) + 0.01(10,000) + 1(10) = 114 Other environmental indices can be calculated such as carcinogenisity, persistence in the atmosphere, aquatic persistence, and so on. Capital and operating costs of production also have to be considered in making a decision. However, from the viewpoint of the results the calculations above, which process would you recom-mend used? S LF .. ASSESSMENT T T Questions 1. Do per million denote a concentration that is a mole ratio? 2. Does the concentration of a component a mixture depend on amount of the mix-ture? 3. Pick the correct answer. How many ppm are in I ppb? (a) 1000, (b) 1 (c) 1, (d) 0.1. (e) 0.01, (0 0.001? 4. How many ppb are there in 1 S. Does 50 ppm represent an increase of fi ve ti mes a of 10 ppm? 6. Is 10 400% less than 50 ppm? Problems 1. many mg/L is equivalent to a 1.2% solution a in water? 2. If a membrane filter yields a count of 69 fecal coliform (FC) colonies from 5 what should be the reported FC concentration? of well J. The danger point in breathing is this value? dioxide for humans is j,Lg/m3, How many ppm
- 98. Looking Back 67 DiscussIon Problems 1. Certain trace elements are known to be toxic to humans but at the same time are essential for your health. For example. would you knowingly drink a glass of water containing 50 ppb of arsenic? The human body normally contains 40 to 300 ppb. Wines contain 5 to 116 ppb of arsenic. Marine fish contain 2,000 to 8.000 ppb. Should you stop eating fish? Another compound essential to humans and animals is selenium. We know that 0.1 to 0.3 ppm of selenium is essential to the diet, but that 5 to 10 ppm is a toxic dose. The Delaney clause of the Food Additives law has been interpreted as prohibiting the presence of any added carcinogen in food. Can selenium be added to your diet via vitamin pills? What about arsenic? 2. It has been suggested that an alternative to using pesticides on plants is to increase the level of natural plant toxins by breeding or gene manipulation. How feasible is this approach from the viewpoint of mutagenic and carcinogenic effects on human beings? For example, solamine and chaconine, some of the natural alkaloids in potatoes, are present at a level of 15,000 ILg per 200 g potatoes. This amount is about 1/6 of the toxic level-for human beings. Neither alkaloid has been tested for carcinogenicity. The intake of man- , made pesticides by humans is estimated at about 150 ILg/day. Only about one-half have been shown to be carcinogenic in test animaJs. The intake of known natural carcinogens is estimated I g per day from fruit and vegetables alone omitting coffee (500 J.Lg per cup), bread (185 p.g per slice). and cola (2000 M-gfbottle). Prepare a brief report ranking possible carcinogenic hazards from man-made and natural substances. List the possible exposure, source of exposure, carcinogenic dose per person, the relative potency. and risk of death. Looking Back We have reviewed the concepts of moles, density, specific gravity, flow rate, mole and mass fraction, and concentration I all of which you probably have previ. ous}y encountered, but should now be old acquaintances. Become very familiar with using these quantities so that you do not have to puzzle over them when dealing with more advanced subjects. GLOSSARY OF NEW WORDS API Scale used to report specific gravity of petroleum compounds. Average molecular weight A pseudo-molecular weight computed by dividing the mass in a mixture or solution by the number of moles in the mixture or solution. Atomic weight Mass of an atom based on 12C being exactly 12. - ---- -'-- -
- 99. 68 Moles, Density, and Concentration 2 Compound A species composed more than one element. Concentration Quantity of a per unit volume. Density Mass unit volume of a compound~ molar density is the number of moles divided by the volume. Flow Mass, moles, or of a material that is in motion divided by unit Gram mole 6.022 x 1023 molecules. Mass fraction Mass a particular compound in a mixture or solution divided the total amount of mass present. Mole Amount a substance containing 6.022 x 1023 entities. Mole fraction Moles of a particular compound in a mixture or solution divided by total number of moles preSel1lL Molecular weight Mass of a compound mole. Parts per million Concentration expressed teons of millionths units (ppm). Pound mole 6.022 x x 453.6 molecules. Solution Homogeneous mixture of two or more compounds. Specific gravity Ratio of density a compound to the density of a reference compound. Specific volume Inverse of the density (volume per unit mass). Weight fraction historical for mass fraction. UPPL MENTARY R FER NCES In addition to the genera] N'T'I"t" ... "'r .. " ing are pertinent. in the FAQ in the front material, the foHow- AHsoP1 "The Place and Importance of the Mole in Chemistry Courses:' Phys. Educ. 285 (July 1 2. B. 1. and S. M. Vora. Stoichiometry (SI Units), McGraw-Hill, New Delhi (998). 3. R. "SI for : Persistent Problems. Solid Solutions,H J. Chem. &Juc. 80, 16-21 (2003). 4. D. M., and J. M. "Understanding the Mole and Its Use in Chemical neering," Che. Eng. Educ., 332-335 (Fall 1999). Gorin, "Mole Chemical Amount: A Discussion of the Fundamental Measurements in Chemistry," }, Chem. Educ., 71, 114-116 (1994). 6. Gorin, "Mole, Mole Liter, and Molar," J. Chem. 80, 103-104 (2003).
- 100. Chap. 2 Problems 69 ~,., . , Luyben, W. and A. Wentzel. Chemical Process Analysis: Mass and Energy Bal· ances, Prentice-Hall, Englewood Cliffs, N.l (1988). 8. National Advisory Committee for Acute Guideline Levels, Applicabiliry Determination Index (AD/), published periodically on the web site at: http://es.epa.gov/oecaJadi.html. 9. Winkler, M. A. (ed), Chemical Engineering Problems in Biotechnology. Elsevier Science, New York (1990). Web Sites http://chemengineer.about.com http://www.chemistrycoach.comltutorials-2.html http://www.ex.ac.uklcimtldictunitldictunit.htm hup:llmc gra w-hi 11.k novel.com/perry s http://www.retallick.com.lresources/netresrc.html http://www.shef.ac ,uklunifacademicl A -C/cpe/mpi ttlchemengs.html PROBL MS *2.1 The following was a letter to The Chemical Engineer (a British publication). In reply to Dr 1. Monis in the February issue, of course the symbols g mole and mole used to and they still can you want them to, but not in the SI. Anyway, what is wrong with the mole (mol) and kilomole (kmol)? They are easier both to and to write. We are aU aware of the apparent inconsistency in the choice of the mole rather than the kilomole as basic S1 unit for amount of substance, but the controversy is over now and it is sterile (0 pursue the matter. Dr Morris is treading on dangerous ground when he attempts "to remind us that the mole bas the dimensions of mass." The mote certainly related to mass, but this does not confer dimensions of mass on it. The amount substance is proportional to mass divided by the relative molecular mass, a mensionless ratio formerly known as the molecular weight. If SI units are in~ volved, a dimensional constant of proportionality numerically equal to 103 is normally chosen, but there is no fundamentally compelling reason why we should do so. Explain what is correct and what is not correct about this letter. The following objectives were given at the beginning a chapter discussing the concept of a mole. the objectives describe the correct characteristics of the mole? Answer yes or no.
- 101. 70 Moles, Density, and Concentration Chap. 2 " Objective 1. The student will know that the mole is a counting unit and that one mole of any substance contains the same number of units as one mole of any other substance. " Objective The student will know that the mole is defined as the amount of substance containing Avogadro's number of units or particles of that substance. " Objective 3. The student will be able to calculate the atomic or molecular masses in grams from the molar masse.. ... of the respective atoms or molecules, and viceversa. "'2.3 Explain the differences between mole, molecule, and molecular weight. l1li1.4 What is wrong, or correct, about each of the following answers to the question: What is a mole? (a) A mole is found in a certain number of cm3 of one substance or another. (b) A mole is the weight of a molecule expressed in grams. (c) A mole is the number of molecules in one gram of a substance. (d) A mole is the sum of atomic weights. (e) mo]e is the molecular weight of an element. ·2.S A textbook states: "A mole is a quantity of material whose weight is numerically equal to the molecular weight." State whether this statement is correct, incorrect, or partially correct, and explain in no more than three sentences reasoning behind your answer. *1.6 What does unit mol-1 mean? Can a unit moll/3? ·2.7 Select the correct answer(s): 1. A mole of H20 and a mole of O2 (a) have the same mass (b) contain one molecule each (c) have a mass of 1 g each (d) contain the same number of molecules 2. One mole of oxygen molecules contains more independent units (02) than one mole of oxygen atoms (0). (a) True, because there are two atoms a for every molecule of O2, (b) because one mole of 02 weighs more than one mole of (c) because both them have the same number of particles. (d) False, one mole of ° has the same mass as one mole of 02' *1.8 Calculate the mass of one mole of chlorophyll (CssHnMgN40s) *2.9 Convert the following: (a) 120 Ib moles of NaCI to g. (b) t 20 g moles of NaCI (0 lb. (e) 120 Jb of NaCl to g moles. (d) 120 g of NaCl to lb moles. ·1.10 (a) What is the molecuLar weight of CaCO)? (b) How many g mol are in 109 of CaC03 ? (c) How many Ib mol are in 20 lb of CaC03? (d) How many g are in 2lb mol of CaC03?
- 102. Chap. 2 Problems ·2.11 Convert the fonowing: (a) 4 g mol of MgClz to g (b) 21b mol ofC3Hg to g (c) g N2 to Ib mol (d) 3Ib to g mol "'2.12 How many pounds are in each of the fonowing (a) 16.11b mol of He} (b) 19.41b mol ofKCl (c) 11.9 g mol of NaN03 (d) 164 g mol of Si02 71'" "'2.13 A solid compound was found to contain 11% 51.46% 0, and Its mol-ecular was about 341. What formula for the compound? *2.14 The structural formulas in Figure P2J4 are vitamins: Vitamin Vitamin A. (a) How many pounds of compound are contained in of the following (do for each vitamin): (l) 2.00 g mol (b) How many grams are contained each of the following (do for each vitamin): (I) 1.00 Ib mol (2) 12lb StRiCwral formula CHs I CH CH -c '" CH -CH20H Relinel FiSh liver oils, liver. fish, butler. a precursor, jk;w'otene. Is present In green vegetablea. carrots, Iomatoes. squash Ascorbic acid (vitamin C) I~ Citrus fruit, tomatoes. green peppom. C - C '" C - C - C - CH20H strawberries. potatoes 11 n I I OOH HOH Figure Pl.14 to decide what size containers to use to ship 1,000 Ib "'p ... .., ........ gravity equal to 0.926, What would be the minimwn gallons? cotton-seed oil drum expressed *2.16 The density of a solution is 8.80 Ib/gal at How many cubic feet will occupied by 10,010 lb of solution at 80°F? Which of these three containers represents respectively one mole of lead (Pb). one mol of zinc (Zn). and one mol of carbon (C).
- 103. 12 Moles, Density, and Concentration Chap. 2 Set 1 2 3 a b c a b c a b c c Pb Pb c Figure n.!7 *2.18 Inserting genetic material into cells to create disease~resistant or herbicide-strains of crops is now possible with a technique that requires no special equipment. Known as "transfonnation," method differs microtnJection or the ballistic method, which uses a "gun" to fire gold particles coated with the "" ....... "' ... ,'"' material into cells. Transformation only requires commercially available silicon carbide "whiskers/' a test tube and a mixer. Suppose that 10,000 SiC crystals, with an average diameter of ;.tm and length of 20 /Lm. are added to a test tube containing 1,000 plant cells and the genetic material to be inserted. The tube is agitated in a simple mixer for about a minute. Collisions between the cells and needle· like SiC crys-tals pores through which the genetic material into the cell. How many grams of SiC were added? The specific gravity of SiC 17. "*2.19 A cylindrical rod of silica (Si02) heated to its melting point and a thread of silica 0.125 mm in diameter is drawn from melt. By careful control of the temperature and the tension on the thread being drawn, long of uniform diameter can be ob· tained that make optical (a) Calculate how much fiber (in can drawn from one cylinder of 1.0m in length and cm in diameter. Also estimate mass. The specific gravity of silica is (b) optical fiber is it is typically covered with a thin protective a polymer. fiber discussed above is with a 50 /Lm layer of a poly-mer with density 1 kg/m3. How much polymer is needed coat the entire fiber? *2.20 In fully vau]ted storage systems (see P2.20) primary container is installed in a completely reinforced concrete vault of monolithic construction. Leak tection is normally achieved by the installation of monitoring wells or sensors in the backfill area. vaults must be constructed by pouring the floor and walls as continuous structural elements. Chemical resistant coatings on interior surfaces and water coatings on exterior surfaces are required. If a 10,000 gallon cylindrical tank is to be buried in an approved containment system, the volume of the containment system must by law hold 10% of the volume of tank. If the void fraction of the backfill material (after compaction) is 20%, what must be the volume in ft3 of the excavation before the concrete vault is poured? What is the weight of the backfill it is (density 2.2g1cm3)?
- 104. Chap. 2 Problems Undisturbed Soli Monitoring Device Storage Tank __ -:...:...:..:' Monitoring Well Placed in a Low Point or Collection Fill Final Cover Undisturbed Soli Reinforced ~... .. ~....;;....-- Concrete Vault Sump - __ -=~~ ~~~~~-! Figure P2.20 73 *2.21 A sample has a specific volume of 5.2 (m3)(kg- l) and a molar volume of 1160 (m3)(kg mol-I). Detennine the molecular weight of the material. *2.22 Five thousand barrels of 28° API gas oil are blended with 20,000 bbl of 15° API fuel oil. What is the density of the mixture in Ib/gal and lb/ft3? Assume that the volumes are additive. 1 bbl = 42 gal. The density of water at 600P is 0.999 g/cm3. 600P 141.5 Specific grav ity 600F = ° API + 131.5 "2.23 In a handbook you find that the conversion between ° API and density is 0.800 density = 45.28° API. Is this a misprint? *2.24 The specific gravity of acetic acid i~ 1.049. What is the density in Ihm/ft3? ·2.25 The specific gravity of a fuel oil is 0.82. What is the density of the oil in Ib/ft3? Show an units. ·2.26 The specific gravity of Ag20 is 7.30 with the reference H20 at 25°C. What is the specific gravity if the reference water is at 4°C? ·2.27 The Handbook of Physics and Chemistry lists in one column a 30% sulfuric acid solution at 20°C as having a specific gravity of 1.2185. In the next column it lists the grarnslliter of H2S04 in the solution as being equal to 365.6. Is this value correct? ·2.28 The Federal Water Pollution Control Act, P.L. 92-500, specifies legally acceptable methods for wastewater analysis. Analysis for cyanide is done according to the method outlined in "Standard Methods for the Examination of Water Wastewater." Mercuric chloride is used in the analytic procedure to decompose complex cyanides, and 200 mg are used per analysis.
- 105. 74 Moles, Density, and Concentration Chap. 2 The Illinois Pollution Control Board has established Water Quality Standards • that limit mercury (as Hg) to 0.0005 ppm in any effluent. Permit holders are required to submit daily reports on their effluent. wm a permit holder discharging 100,000 gal/day be in violation of the cited standard if one analysis is made? ··2.29 You are asked to make up a laboratory solution of 0.10 molar H2S04 (0.10 mol H2S041L) from concentrated (96.0%) H2S04 , You look up the specific gravity of 96.0% H2S04 and find it is 1.858. Calculate (a) the weight of 96.0% acid needed per L of solution. (b) the volume of 96.0% acid used per L of solution. (c) the density of the 0,1 molar solution. $"'$2.30 A bartender claims that his special brand of rum is so strong that ice cubes sink in it. Is this possible? $2.31 The density of benzene at 600P is 0.879 g/cm3, What is the specific gravity of benzene at 60°Fl60°F? **2.32 A liquid has a specific gravity of 0.90 at 25°C. What is its (a) Density at 25°C in kg/m3? (b) Specific volume at 25°C in ft3llbm? (c) If the liquid is placed in a 1.S-L bottle that has a mass of 232 g. how much will the full bottle weigh? ·"'2.33 Given a water solution that contains 1.704 kg of lIND/kg H20 and has a specific gravity of 1.382 at 20°C. express the composition in the following ways: (a) Weight percent HN03 (b) Pounds HN03 per cubic foot of solution at 20°C (c) Molarity (gram moles of HNO) per liter of solution at 200e) "'2.34 OSHA (Occupational Safety and Health Administration) has established limits for the storage of various toxic or hazardous chemicals (OSHA 29 CPR 1910.119, Appendix A). The maximum limit for acetaldehyde is 113,0 kg. What is the minimum size spherical vessel that can be used to store this liquid at room temperature? "'2.35 For the purpose of pennit compliance, all hazardous materials are categorized into three hazard categories: toxicity, flammability, and reactivity, and assigned numbers in each category from 0 to 4 (most severe). Methyl alcohol (methanol, CHJOH) has the code 1, 4, 0 in the liquid state. For the toxic category. any amount of stored toxic material of category lover 0.35 oz. must be reported by city ordinance. Must a onehalf liter bottle of methanol (sp.gr. = 0.792) be reported? *2.36 Oil (sp.gr. = 0.8) is flowing through a 6 inch diameter pipe with a velocity of 56.7 ft/s. What is the flowrate of the oil in m3/s? "2.37 Forty gal/min of a hydrocarbon fuel having a specific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fueL How long will it take to fill the tank in the truck? *·2.38 Calculate the empirical formula of an organic compound with the following mass analysis: carbon. 26.9%; hydrogen, 2.2%; and oxygen as the only other element pre~ sent. ·*2.39 Given the following mass percent composition, determine the empirical formula: 49.5% C; 5.2%H; 28.8%N; 16.5%0.
- 106. Chap. 2 Problems ·2.40 Calculate the mass and mole fractions of the respective components in NaCI03. ·2.41 The specific gravity of a solution of KOH at 15°C is 1.0824 and contains 0.813 Ib KOH per gal of solution. What are the mass fractions of KOH and H20 in the solu· tion? """2.42 Prepare an expression that converts mass (weight) fraction (m) to mole fracrion (x), and another expression for the conversion of mole fraction to mass fraction, for a binary mixture. *2.43 You have 100 kilograms of gas of the following composition: c~ 30% H:2 10% N2 60% What is the average molecular weight of this gas? *2.44 You analyze the gas in 100 kg of gas in a tank at atmospheric pressure, and find the foHowing: What is the average molecular weight of the gas? *2.45 Suppose you are required to make an analysis of 317 Ib of combustion gas and find it has the fonowing composition: CO2 60% CO 10% N2 30% What is the average molecular weight of this gas in the American Engineering System of units? ·2.46 You purchase a tank with a volume of 2.1 ft3. You pump the tank out, and add first 20 lb. of CO2 and then 10 Ib of N2• What is the analysis of the final gas.in the tank? "'2.47 How many ppb are there in 1 ppm? Does the system of units affect your answer? Does it make any difference if the material for which the ppb are measured is a gas, liquid. or solid? "2.48 The table lists values of Fe, Cu, and Pb in Christmas wrapping paper for two different brands. Convert the ppm to mass fractions on a paper free basis. Brand A BrandS Concentration, ppm Fe Cu 1310 2000 350 50 Pb 2750 5 ·2.49 Harbor sediments in the New Bedford. Massachusetts, area contain PCBs at levels up to 190,000 ppm according to a report prepared by Grant Weaver of the Massachusetts. Coastal Zone Management Office (Environ. Sci. Technol., 16(9) (l982):491A. What is the concentration in percent?
- 107. 16 Moles, Density, and Concentration Chap. 2 '2.50 NIOSH sets standards for CC14 in air at mglm3 of (a time weighted over 40 hr). The CC14 found in a is 4800 ppb (parts billion; billion = 109). Does the sample exceed the NIOSH standard? Be careful! "'2.51 The following table shows the annual inputs of phosphorus to Source Huron Land drainage Municipal waste Industrial waSle Outflow Retained Short tonslyr 2,240 6,740 19,090 30,100 4.500 25,600 (a) Convert the retained phosphorus to concentration micrograms per liter assum-that Lake contains 1 X 1014 of water and that the average phospbo- rus retention time is 2.60 yr. (b) What percentage of input comes from municipal water? (c) What percentage of the input comes detergents, assuming they represent 70% of the municipal (d) 10 ppb of phosphorus nuisance algal blooms, as has been reported in some documents, removing 30% of the phosphorus the municipal waste and all phosphorus in the industrial waste effective in reducing the eu-trophication the algal blooms) in Lake (e) Would removing aU the in help? -2.52 A gas contains 350 ppm of H2S in CO2, If the gas liquified, what is the weight fraction H2S? Sulfur trioxide (S03) can be absorbed in sulfuric acid solution to form more concen-trated sulfuric acid. the gas be absorbed contains S031 41 N2, 3% S02' and I % °2, how many parts per million of 02 are there in (b) What the composition the gas on a N2 free "'2.S4 If the concentration Ca is 56.4 mg/L and Mg is 8.8 mgIL in water, what is the total hardness of the water expressed in rngIL of cae03 ? 42.55 Twenty-seven pounds (27 Ib) of chlorine gas is used for treating 750,000 of water each The chlorine used up the microorganisms in water is measured to be 2.6 mgfL. What is the residual (excess) chlorine concentration the treated water? *2.56 A newspaper report says the FDA found 13-20 ppb of acrylonitrile in a soft drink .... VLLH .... and if is correct, it only amounts to 1 molecule of acrylonitrile bottle. Is this statement "'2.57 Several of global warming indicate that the concentration of CO2 in the at-mosphere is increasing by roughly 1 % year. Do we have to worry about de-crease in the concentration also?
- 108. Chap. 2 Problems *2.58 The of a biomass sample % dry of cell C 50.2 0 20.1 N 14.0 H 8.2 P 3.0 Other This compound a ratio of 10.5 g cells/mol A TP synthesized in the meta-bolic reaction to form cells. Approximately how many moles of C are in the ceBs per of ATP? "'2.59 A (radioactive labeled microorganism (MMM) decomposes to NN as foHows MMM (8) ~ NN (s) + 3 CO2 (g) the (g) yields 2 x 107 dpm (disintegrations per minute) in a detection device~ how many j.LCi (micro Curies) is this? How many cpm (counts per minute) will be noted if the counting device is 80% efficient in counting disintegrations? Data: 1 Curie = 3 x 10to dps (disintegrations second). $2.60 alternative compounds have been added [0 gasoline induding methanol. ethanol, and methy-ter-butyl ether (MTBE) to increase oxygen content of ......... "'& ... .., in order to reduce formation of CO on combustion. Unfortunately, MTBE has found ground water at concentrations sufficient to cause concern. Persistence of a compound in water can be evaluated from its half-life, tl12• the time for one-half of the compound to leave system of interest. The half-life depends on the conditions in the system, of course, but for environmental can approximated by (OH-) is the concentration of hydroxyl radical in the system that for prob-lem the of water is equal to 1.5 x 106 molecules/em3, The of k detennined from experiment are: Methanol Ethanol MTBE Calculate the half-life of their persistence. k (cm3)/(moleade)(sec) 0.15 X 10-12 1 x to- I2 0.60 x 10-12 compounds, and them according to
- 109. CHAPT R 3 CHOOSING A BASIS Your objectives In studying this chapter are to be able to: 1. State three questions useful in selecting a basis. 2. Apply the three questions to problems to select a suitable basis or sequence of bases. Have you noted in some of the examples in Chapter 2 that the word basis has appeared at start of the examples? concept of a basis vitally important to both understanding of how to solve a problem and to your solving it in the most expeditious manner. Looking Ahead In this we discuss how you choose a basis on which to problems. Main Concepts A basis is a chosen by you calculations you plan to make in any particular problem, and a proper choice basis frequently makes the problem much to solve. basis may be a period of time such as hours. or a given mass of material, such as 5 kg of CO2, or some convenient quantity. To select a sound basis (which in many problems is predetermined for you but in some prob-lems is not so ask yourself the following questions: 3. What do I have to start with? (e.g., I have 100 Ib oil; I have 46 kg of fertilizer). b. What answer caned for? (e.g., How much product is produced per hour?), c. What is the convenient basis to (For example. suppose that the mole 78 fractions of an amount of material are known. Then selecting 100 kilogram moles the material as a basis would sense. the other hand, the
- 110. 3 Choosing a Basis mass fractions of the material are known, then 100 an appropriate basis), 79 of the material would be If a of flow or production of a material is stated in a then you should usually select the time interval, such as 1 minute or 1 hour, as the basis on to make the then a time unit does not have to along during all of the caIICUliaUC)fiS. These questions and their answers will suggest suitable Sometimes when several bases seem appropriate. you may find it is best to use a unit basis of I or 100 of something, for kilograms. hours, moles or cubic liq-uids and solids in which a mass (weight) analysis applies, a convenient often I or 100 Ib or kg; similarly. 1 or 100 a good choice for a rea-son for these choices is that the in the analysis of the material au-tomatically equals the number or moles, respectively, and one step in the calculations is saved. Always state the basis have chosen for your calculations by writing it prominently on your calculation sheets (or computer screen). Choosing a Basis dehydration of the lower alkanes can be carried out using a oxide (CeO) catalyst. What is the mass fracti,on and mole fraction of and 0 in the catalyst? Solution Start the solution by a basis. Because no specific amount material is specified, the question what do I have to start with does not help dec:ide basis. Neither does the question about the desired answer. Thus. sellectJlOg nient basis becomes the for a basis. What do you know about know from the formula that one is combined with one mole of O. sequently. a basis of 2 kg mol (or 2 g or 2 Ib mol, etc.) would make sense. You can get the atomic weights for Appendix B, and then you are pared to calculate the respective masses 0 in CeO. The calculations are presented in the form of the following table: Basis: 2 kg mol kg mol Mole fraction Mol. wt. kg. Mass fraction 0.50 140.12 12 0.90 0 I 0.50 16.0 16.0 O. Total 2 1.00 156.1 156.1
- 111. 80 Choosing a Basis Chap. 3 EXAMPLE 3.2 Choosing a Basis Most processes for producing high-energy~content or gasoline from coal include some type of gasification step make hydrogen or synthesis gas, Pressure gasification is preferred because of its yield of methane and higher rate of gasification. Given that a 50.0 kg test run of gas averages 10.0% H2• 40.0% CH4• 30.0% CO, and 20.0% C021 what is the average molecular weight of the gas? Solution Let's choose a basis. The answer to question J is to select a basis of 50.0 kg ("what I have to start with"), but is this choice a good basis? A little reflection will show that such a basis is of no use. You cannot multiply the given mole percent this gas (remember that the composition of is given in mole percent unless otherwise stated) times kg and expect the result to mean anything. Try it, being sure to include the respective units. Thus, the next step to choose a "convenient basis:' which is 100 kg mol of gas, and proceed as follows: Basis: 100 kg mol or Ib mol gas Set up a table such as the following to make a compact presentation of the calculations. You do not have to-you can make individual computations for each component, but such a procedure is inefficient and more prone to errors. Percent::: kg Component mol or Ib mol Mol wt. Kg or Ib CO2 20.0 44.0 880 CO 30.0 28.0 840 40.0 16.04 642 H2 10.0 2.02 20 Total 100.0 2382 2382 Average molecular weight = 100 mol = 23.8 kg/kg mol Check the solution by noting that an average molecular weight of 23.8 is reasonable since the molecular weights of the components varies from 10 to 40.
- 112. Chap. 3 Choosing a ......... "",'" EXAMPLE 3.3 Choosing a Basis for Cell Growth [n measuring the growth rate of cens the laboratory, you inject 50,000 cells into a well in a so caned well plate. Next you add mated thymidine to the culture in the well, and 24 hours later the radiation the cells the well reads SOOO cpm (counts per minute) from the uptake of the thymidine by the cells. At the same time total ceH count is 60,000 cells. The final step in the experiment is to add a growth factor to the medium under the same initial conditions (50,()(X) cells in a well). After 24 hours the radiation count per well is 8000 cpm. You are asked to compute the increase in average growth rate by adding the growth factor. What would be a good basis on which to solve the problem? Solution This problem involves amounts (numbers of cells in the well and counts of the disintegrations of radioactive material) and times (minutes and hours). A convenient basis to would be a time, say 1 or 24 hours, because several quantities of ceHs are specified. Choose 1 hour. As a matter of interest. without the growth factor the average growth rate is the 10,000 cell increase divided by 24 hours, or 417 cellslhr the selected basis. With the growth factor added, you have to find the final count 24 hours. The assumption in practice is that the cell growth is directly proportional to the tritiated thymidine uptake because thymidine is taken up at each cell division. Thus, based on the non growth data, 417 cells = k (5000 cpm) and k = 0.0833 ceUs/cpm. When the cpm is 8000, the number of cells is (0.0833)(8000) == 667 cens (in one hour), an increase of 60% .-. ------_._---- 81
- 113. 82 Choosing a Basis EXAMPLE 3.4 Calculation of the Mass Fraction of Components in N anoparticles Chap. 3 The microstructure of nanosized particles has proved to be important in nanotechnology in developing economic magnetic performance of nanocomposites. In a ternary alloy such as Nd45 Fe17BI8.5 the average grain size is about 30 nm. By replacing 0.2 atoms of Fe with atoms of Cu, the grain size can be reduced (improved) to 17 nm. (a) What is the molecular formula of the alloy after adding the Cu to replace the Fe? (b) What is the mass fraction of each atomic species in the improved alloy? Solution Pick a convenient basis. Because the atoms in the chemical fonnula of the alloy total to 100, pick a basis of Basis: 100 g mol (or atoms) of Nd4.5Fe77BI8.5 (a) The final alloy is Nd4.5F~6. 8BI8.5CUo. 2' (b) Use a table to calculate the respective mass fractions. Component Original g mol Final gmol MW g Nd 4.5 4.5 144.24 649.08 Fe 77 76.8 55.85 4289.28 B ~ 18.5 lO.81 199.99 Cu --.JU 63.55 12,7} Total 100.0 100.0 5151.06 Mass fraction 0.126 0.833 0.039 rum 1.000 In summary, be sure to state the basis of your calculations so that you will keep clearly in mind the real nature of your calculations, and so that anyone checking your problem solution will be able to understand on what basis your calculations were perfonned. You no doubt have heard the story of Ali Baba and the 40 thieves. Have you heard about Ali Baba and the 39'camels? Ali Baba gave his four sons 39 came]s to be divided among them so that the oldest son got one-half the camels; the second son a quarter, the third an eighth, and the youngest a tenth. The four brothers were at a loss as to how they should divide the inheritance until a stranger came riding along on his came1. He added his own camel to Ali Baba's 39 and then divided the 40 among the sons. The oldest son received 20~ the second, 10; the third, 5; and the youngest, 4. One camel was left. The stranger mounted it-for it was his own-and rode off. Amazed,
- 114. Chap. 3 Choosing a Basis 83 the four brothers watched him ride away. The oldest brother was the first to start cal. culating. Had his father not willed half of the camels to him? Twenty camels are obviously more than half of 39. One of the four sons must have received less than his due. But figure as they would, each found that had more than his share. After agonizing over problem you will realize that the sum of 1/2, 141 1/8, and lAo is not 1 but is 0.975. By adjusting (normalizing) the camel fractions 0) so that they total 1, the division camels is validated: Camel fractions Normalizing 0.500 (0.500) :::: 0.975 0.250 (0.250) :::;: 0.975 0.125 (0.1 ) :::;: 0.975 100 ;:: :::;: Normalized. true fractions 0.5128 x 0.2564 )( 0.1282 x 0.1026 x 1.000 39 39 Distributed camels (integer) .= 20 10 :::;: 5 ::::: 4 What we have done is to change the calculations from a basis of 0.975 to a new basis of 1.000. More frequently than you probably would like, you will have to change from your original selection a basis in solving a problem to one or more different bases in order to put together the information needed to the problem. For example, given a 1.00 mol of gas containing 02 (20%), N2 (78%), and S02 (2%), find the composition of the on an S02-/ree basis, meaning gas without S02 in it. What you do to calculate the moles of each component, remove the S02 and just the basis for the calculations so that the gas becomes composed only of 02 and N2 with a percent composition totaling 100%: Basis: 1.00 mol Components Mol fraction Mol Mol SOl free Mol fraction SO:! free °2 0.20 0.20 0.20 0.20 N2 0.78 0.78 0.78 0.80 SOl 0.02 0.02 LOO 1.00 0.98 1.00
- 115. 84 ChoOsJng a Basis Chap. 3 The round-off in the last column is appropriate given the original values for the mole fractions. Here is a more complicated example. EXAMPLE 3.5 Changing Bases A medium-grade bituminous coal analyzes as follows: Component Percent S 2 N 1 0 6 Ash 11 Wa.ter 3 Residuum 77 The residuum is C and H, and the mole ratio in the residuum is HlC = 9. Calculate the weight (mass) fraction composition of the coal'with the ash and the moisture omitted (ash- and moisture-free). Solution To calculate the mass fractions of the components of the coal-.on an ash and moisture free basis, i.e., omitting the ash and water in coal in the' Hst of compounds, you first have to determine the respective amounts oLe "d H in the residuum. Then you remove the ash and water from the list QfcQn~, add up the remaining masses, and calculate the mass fractions of each of the components remaining. Take as a basis 100 kg of coal because then percent = kilograms. Basis: 100 kg of coal The sum of the S + N + 0 + ash + water is 2 + I + 6 + 11 + 3 = 23 kg You need to determine the individual kg of C and of H in the 71 kg total residuum. To deternrine the kilograms of C and H. you have to select a new basis. Is 17 kg or 100 leg satisfactory? No. Why? Because the HlC ratio is given in terms of moles. not weight (mass), Pick instead a convenient basis involving moles. Basis: 100 kg mol
- 116. tChap.S Choosing a Basis '-' , J' :: "". Component H C the basis the values to Mole fraction 9 --:=; 0.90 1+9 I 0.10 =- 1+9 1.00 kg Mol. wt. 90 L008 10 100 90.7 Mass fraction 0.43 0.57 210.7 1.00 kg of C+H. we have the kg of H and C, but we need to change with the basis of 100 of coaL . Basis: 100 coal The next step to calculate the kg You can use the mass fractions from and H in the 77 residuum. above. or just use the kg values directly: H: (77kg)(0.43) 15kg (77 kg)(O.S7) = 43.85 kg Finally, you can prepare a table summarizing the results on the of 1.00 of the coal ash-free water· free. Component Wi. fraction C 43.85 0.51 H 33.15 0.39 S 2 0.02 N 1 0.01 0 -L QJll Total 86.0 1.00 SELF ... ASSESSMEN ST Questions' 1. What are the three questions you should ask yourself in selecting a basis? ¥,. .'." "2. Why do you sometimes have to change bases during solution of a problem? Problems 1. What would be good initial to select in solving ................ u ...... 1.2, 1.4, 2.1, and 2.2. 85
- 117. 86 Choosing a Basis Chap. 3 Thought Problem Water-based systems are an effective and viable means of con-trolling dust and virtually eliminate the historic of flres and explosions in grain -el-evators. Water-based safety systems resulted in cleaner elevators. improved respi-ratory atmospheres for employees, and dust emissions into the environment surrounding storage facilities. However, some customers have complained that adding water to the grain causes the buyer to pay too much the grain. The grain elevator erators argue all grain shipments unavoidably contain a weight component in the form of moisture. Moistur:~ is introduced to and to grain products in a broad vari-ety of UH ..... LL~ ...... " • . What would you recommend to elevator operators and problem? dealers to alleviate this Looking Back We posed three questions for you to ask yourself as guides in selecting a good basis to use in solving problems, and some examples. We also showed how to change bases if needed during the course of solving a problem. GLO SARY OF N W WORDS Basis The reference material or time selected to use in making the calculations a problem. Changing bases Shifting the basis a problem from one value to another for con-venience in ca1culations. SUPPLEM NTARY REF RENe S In addition to the listed the "Read Me" pages the front of this book, the following discuss the choice of a basis: Fogler, Elements of Reaction Engineering, Prentice-Hall, Upper Saddle River, N.J. (1999). Henley. E. and H. Chemical Engineering Calculations; McGraw-Hill, New York (1959). I
- 118. Chap. 3 Problems PROBLEMS *3.1 Read each of the following problems. and select a suitable basis for solving each one. Do not solve the problems. (a) You have 130 kg of gas of the following composition: 40% N2• 30% CO2, and 30% CH4 in a tank. What is the average molecular weight of the gas? (b) You have 25 lb of a gas of the following composition: CH4 80% C2Iit 10% C2~ 10% What is the average molecular weight of the mixture? What is the weight (mass) fraction of each of the components in the mixture? (c) The proximate and ultimate analysis of a coal is given in the table. What is the composition of the "Volatile Combustible Matter (VCM)"? Present your answer in the fonn of the mass percent of each element in the VCM. Proximate Analysis (%) Ultimate Analysis (%) Moisture 3.2 Carbon 79.90 Volatile Combustible Hydrogen 4.85 matter (VCM) 21.0 Sulfur 0.69 Fixed Carbon 69.3 Nitrogen 1.30 Ash 6.5 Ash 6.50 Oxygen 6·16 Total 1 ()(),Q Total 100.00 (d) A fuel gas is reported to analyze, on a mole basis, 20% methane. 5% ethane, and the remainder CO2, Calculate the analysis of the fuel gas on a mass percentage basis. 6S J(e) A gas mixture consists of three components: argon, B, and C. The following analysis of this mixture is given: 40.0 mol % argon 18.75 mass % B 20.0rnol %C The molecular weight of argon is 40 and the molecular weight of C is 50. Find: (a) The molecular weight of B (b) The average molecular weight of the mixture. ~i: - ,/''/''3.2 Two engineers are calculating the average molecular weight of a gas mixture contain~~, _ ing oxygen and other gases. One of them uses the correct molecular weight of 32 for oxygen and detennines the average molecular weight as 39.2. The other uses an in-
- 119. 88 Choosing a Basis Chap. 3 of 16 and the average molecular weight as 32.8. This is the only error in his calculations. What is the percentage of oxygen in the mixture ex-a!; mol %1 Choose a to solve the problem; do solve it. ·3.3 a basis for the following problem. Chlorine a water treatment plant 134.2 Ib/day. The flow rate of the leaving the plant is 10.7 million gallons per day. What is the average chlorine concentration in the """'~'T"'r1 water leaving the plant in mgIL?
- 120. l CHAPTER 4 TEMPERATURE Your objectives In studying this chapter are to be able to: 1. Define what temperature is. 2. Explain the difference between absolute and relative temperature. 3. Convert a temperature in any of the four com mon scales (Oe, K, of, OR) to any of the others. 4. Convert an expression involving units of temperature and temperature difference to other units 01 temperature and temperature difference. 5. Know the reference points for the four temperature scales. You know what temperature is from past physics and chemistry courses, hence you might well ask: why are we discussing it again? Unfortunately, certain characteristics of temperature exist that in our experience cause confusion for introductory engineering students. We want to elucidate these issues. Looking Ahead In this chapter we first discuss temperature scales and then take up the issue of converting from one temperature scale to another. Main Concepts ... You can hardly go through a single day without noticing or hearing what the temperature is. Believe it or not, considerable controversy exists among some scientists as to what the correct definition of temperature is (consult some of the references at the end of this chapter for further information). Some scientists prefer to say that temperature is a measure of the energy (mostly kinetic) of the molecules in a system. This definition tells us about the amount of energy. Other scientists prefer to say that temperature is a property of the state of thermal equilibrium of the system with respect to other systems because temperature tells us about the capability of a 89
- 121. 90 Temperature Chap. 4 system to transfer energy (as heat). When you get to Chapter 21 you will find out. if you do not already know. that a big difference exists between these two concepts. In this book we use four types of temperature, two based on a relative scale, Fahrenheit (OF) and Celsius (OC), and two based on an absolute scale. gree Rankine (OR) and kelvin (K). Note that the degree symbol (0) not used with the kelvin temperature abbreviation K. Relative scales are the ones you hear the TV or radio announcer give. and are based on a specified reference temperature (32°F or O°C) that occurs in an ice-water mixture (the freezing point of water). Absolute temperature scales have their zero point at the lowest possible temperature that we believe can exist. As you may know, this lowest temperature is related both to the idea] gas laws and to the laws of thermodynamics. The absolute scale. based on degree units the size of those in the Celsius scale, is caned the kelvin scale in honor of Lord Kelvin (1824-1907) who reconciled the divergent views of principle of conservation of energy. The absolute scale, which corresponds to the Fahrenheit degree units, called the Rankine scale in honor ofW. J. M. Rankine (1820-1872), a Scottish engineer. Figure 4.1 illustrates the relationship between rel~ ative temperature and absolute temperature. We shall usually round off absolute zero on the Rankine scale of -459.67°F to -460°F; similarly, -273.1SoC frequently will be rounded off ~o -273°e. In Figure 4.] all of the values of the temperatures have been rounded off, but more significant figures can be used. O°C and its equivalents are known as the values of the standard conditions of temperature. Now we turn to a topic that causes endless difficulty in temperature conversion because of confusing semantics and notation. To start, you should recognize that the unit degree , the unit temperature difference or division) on the Kelvin-Celsius scale is not the same as that on the Rankine-Fahrenheit scale. If we AOF represent the unit temperature difference in the Fahrenheit scale and AOR be the unit temperature difference in the Rankine scale, and AOC and AK be the analogous units in the other two scales, you probably are aware that AOF = AOR Also. if you keep in mind that the A °C is larger than the Aoe ~= 1.8 or or Thus, when we cite the temperature of a substance we are stating the cumulative number of units of the temperature scale that occur (an enumeration of measured from the reference point. Reexamine Figure 4.1.
- 122. Chap. 4 Temperature 212 672 Boi ling point of 373 100 woter at 760 mm Hg 180 100 , 32 492 Freezing point of water 273 0 0460 255 -Ia -40 420 of : 0C 233 -40 • ~ I» ~ c;;;; II> I~ ::ll G) 'iii ..... I~ " 6) Q;j &:. .lIIIII U ~ -460 0 Absolute zero 0 -273 Figure 4.1 Temperatures scaJes. Unfortunately, the symbols 8°C, 8°F. 8K, and 8°R are not in standard usage; the 8 symbol is suppressed. few books try to maintain the difference between de-of temperature eC, of, etc.) and the unit degree by assigning the unit degree the symbol Co. po, and so on. But most journals and texts use the same symbol for the two different quantities, one the unit temperature difference and the other the temperature itself. Consequently, the proper meaning of the symbols OF, K, and '"'R, as either the temperature or tbe unit temperature difference, must be interpreted from the context of the equation or sentence being examined. What this statement means is use some common sense. Suppose you have the relation Top = a + bToe What are the units of a and b? Certainly from what you have learned in Chapter 1, the of a must be for consistency. Are the units of b equal to the units in the ratio Tor/Toe? No, because the reference points for and OF differ; Tor/Toe is not a valid converSiOn factor. The correct units for b must involve the conversion factor (1.88°P/8°C), the factor converts the of a interval on one temperature scale to the other:
- 123. 92 Temperature Chap. 4 Unfortunately, the units for b are usually ignored; just the value of b 0.8) is employed. When you reach Chapter 21 you will note that the heat capacity in the SI sys· tem has the units of 1/(g mol)(K). Does the K the heat capacity designate the temperature in degrees K or the unit interva] ilK? this case K represents a temperature difference. Consider gas constant, R. which can have units of (kg)(m2)/(kg mol)(s2)(K). Does K represent the temperature in degrees K or the interval AK? You should recall for the ideal gas that you use the absolute temperature. If you are not well acquainted with temperature conversion. be sure to practice conversions until they become routine. Many calculators and computers make the conversions automaticaHy, but you should know that Now Jet us look at some examples. EXAMPLE 4.1 Temperature Conversion Convert 100°C to (a) K. (b) OF, and (c) OR. Solution (a) (100 + 273)OC]1 :o~ = 373K or with suppression of the ~ symbol. (100 + 273)OC I K = K lOC 0.6) 273 K (1.7) (1.8) (1.9)
- 124. Chap. 4 or Temperature (b) (IOO°C) 8~~ + 32°F = 212°P 1 LloR (c) (212 + 460)OF 1 AOP = 672°R EXAMPLE 4.2 Temperature Conversion The heat capacity of sulfuric acid has the units J/(g mol)(OC), and is given by the relation heat capacity = 139.1 + 1.56 X 10-IT where T is expressed in DC. Modify the formula so that the resulting expression has the associated units of BtuI(1b mol) (OR) and T is in oR. Solution The symbol °C in the denominator of the heat capacity stands for the unit temperature difference, dOC, not the temperature, whereas the units of T in the equation are in dc. First you have to substitute the proper equation in the formula to convert T in °C to T in oR, and then by multiplication by ,conversion factors convert the units on the righthand side of the equation to BtliIOb mol) (OR) as requested. heat capacity = { 139.1 + 1.56 X 1O-{ (T·R - 460 - 32) 1~8 J} 1 I 1 1 Btu 1454 g mol 1°C X (g mol)(OC) 1055 J 1 lb mol 1.80R = 23.06 + 2.07 X IO-uroR con venion fllctonl Note the suppression of the A. symbol in the conversion between 4°C and A oR. 93
- 125. 94 Temperature Chap. 4 fAardla tlon Inormometo .. Optical pyromete .. OR OF °c K r Thermocouples (all types) Iron boils 3000 ' I 2900 5500 5000 2000 2700 3000 2600 I 5000 4500 2500 2400 t Silicon boils 2300 4500 4000 2200 2500 3500 3000 1700 I Iron melts 1600 r Thormistor resistance thermometry Aluminum boils 2100 4000 2000 (.< , 3500 1900 Platinum melts 1800 2000 I Platinum resistance thermometry 1500 r 1300 I I i Filled systems including 1200 1500 glass thermometers 3000 2500 1400 , I 1500 800 r Bimetal thermometar 2500 2000 1100 1000 II , 2000 900 700 1000 1500 1000 600 Sulphur bolls 500 400 Mercury boils 1000 500 300 200 500 ~12 100 :~, 1- Water freezes 32 0-273.15 I ~d , -100 -200 Helium boils 0 -459.58 -273.15 0 "Absolute zero" Figure 4.2. Temperature measuring instruments span the range from near absolute zero to beyond 3000 K. The chart indicates the preferred methods of thermal instrumentation for various temperature regions. Whatever temperature scale you employ. to be useful in engineering, the temperature has to be measured. Look at Figure 4.2 to see the useful range of several measuring techniques. SELF-ASSESSMENT TESTS Questions 1. What are the reference points of (0) the Celsius and (b) the Fahrenheit scales? 2. How do you convert a temperature difference,b.. from Fahrenheit to CeJsius?
- 126. , l, Chap. 4 Temperature 3. Is the unit temperature difference .6.°C a larger interval than .6.0F? 10°C higher than JO°F? 4. In Appendix the heat capacity of sulfur is Cp = 15.2 + 2.68T, where Cp is in J/(g mol)(K) and T in K. Convert this expression so that Cp is in cal/(g mol)(°F) with T In 5. Suppose that you are given a tube partly filled with an unknown liquid and are asked to calibrate a scale on the tube in How would you proceed? 6. Answer the following questions: (a) In relation to absolute zero, which is higher. 1°C or 1 OF? (b) In relation to O°C, which is higher. 1°C or 1 OF? (c) Which larger. 1 .6.°C or 1.6.°F? Problems 1. Complete the following table with the proper equivalent temperatures: K -40 77.0 698 69.8 Thought Problems 1. In reading a report on the space shuttle you find the statement that "the maximum temperature on reentry is 1482.2°C." How many significant figures do you think are represented by this temperature? What temperature-measuring devices would you recommend to make the foHowing measurements: (a) temperature of the thermal decomposition of oil shale (3000 to 500°C) (b) air temperature outside your home (-20° to 30°C) (c) temperature inside a freeze-drying apparatus 100° to O°C) (d) flame temperature of a Bunsen burner (20000 to 2500°C) 3. A vacuum tower used to process residual oil experienced severe coking (carbon formation) on the tower internals when it rained. Coking occurs because the temperature of the fluid gets to be too high. The temperature of the entering residual was controlled by a temperature recorder-controller (TRC) connected to a thennocouple inserted onto a Thermowell in the pipeline bringing the residual into the column. The TRC was operating at 700°F, whereas the interior of the column was at 740°F (too hot), What might be the problem?
- 127. 96 Temperature Chap. 4 Discussion Problems 1. In the book by Rogers (Physics for the Inquiring Mind. Princeton. University Press, Princeton. NJ, 1960), temperature is defined as: "the hotness measured in some definite scale." Is this correct? How would you define temperature? In the kelvin or Rankine (absolute temperature) scales, the used for temperature T = na where a T is the value of unit temperature and n is number of units enu-merated. When n = 0, T;,; O. Suppose that temperature is defined by the relation for In (1) ;,; nA T. Does T;,; 0 occur? What does n ;,; 0 mean? Does the equivalent of absolute zero kelvin exist? looking Back In this chapter we explained difference between the absolute and relative temperature scales. We pointed out how you must be careful in distinguishing between the unit temperature difference and the temperature itself~ especially when converting units. GLOSSARY OF N W WORDS Celsius C'C) Relative temperature scale with zero degrees being the freezing point of an air-water mixture. Fahrenheit (OF) Relative temperature scale with 32 degrees being the point of an air-water mixture. Kelvin (K) Absolute temperature scale based on zero degrees being the lowest possible temperature we believe can exist. Rankine eR) Absolute temperature scale related to degrees Fahrenheit based on zero degrees being the lowest possible temperature we believe can exist. StanJard conditions of temperature 32°F, .15 K, and 459.67°R. Temperi}, ture measure of the of the molecules in a system. Temperature interval (1lT) Size of one degree in a temperature scale. SUPPLEMENTARY REF R NCES Gilabert. M. A., and 1. Pellicer. "Celsius or Kelvin: Something to Get Steamed Phys. Educ. 31,52-55 (1996). Michalski, L. (ed.). Temperature Measurement, John Wiley, New York (2001). About," J
- 128. Chap. 4 Probrems 97 Pellicer. 1., M. Ampano Gilabert. and E. Lopez-Baeza. 'The Evolution of the Celsius and Kelvin Scales and the State the Art:' J. Chem. Educ.; 76, 911-913 (1999). Quinn, J. Temperature. Academic Press, New York (1990). Romer. R. H. ''Temperature Scales," The Physics Teacher. 4S0 (October 1982). Thompson. H. B. "Is goC Equal to 500P? 1. Chern. Educ., 68, 400 (1991). Web Pages http://www.sant.esson.comJengtemp.html http://www.scieneemadesimple.netltemperature.html http://www.inidata.ucar.edulstafflblyndsltmp.htrnl http://www.temperatures.com PROBLEMS "4.1 "Japan, Aim for Better Methanol-Powered Cant read the headline in the Wall Street Journal. Japan and the plan to join in developing technology to prove cars that run on methanol, a fuel that causes less air pollution than gasoline. An unspecified number of researchers from Japanese companies will work with the EPA to develop a methanol car that win start in temperatures as low as minus 10 degrees Celsius. What is this temperature degrees Rankine. kelvin~ and Fahrenheit? ·*4.2 Can negative temperature measurements exist? ·'4.3 The heat capacity Cp of acetic acid in J/(g rnol)(K) can be calculated from the equation Cp :: 8.41 + 2.4346 X 1O-5T where T is in Convert the equation so that T can be intrOduced into the equation in OR instead K. Keep the units of Cp the same. ·4.4 Convert the following temperatures to the requested units: (a) 10°C to OF (b) IOoe to OR (c) -25°F to K (d) ISOK to OR *4.5 Heat capacities are usually given in terms polynomial functlons of temperature. The equation for carbon dioxide is Cp :: 8.4448 + 0,5757 X 10-2 T - 0.21S9 X 100ST2 + 0.3059 X 10-9']'9 where T is in OF and Cp is in Btu/(lb mol)(°F). Convert the equation so that Tean be in °C and Cp will be in 11(g mo1)(K). ·4.6 In a report on the record low temperatures in Antarctica. Chemical and EngiMering News said at one point that "the mercury dropped to -76t:.C." In what sense is that possible? Mercury at
- 129. 98 Temperature Chap. 4 ··4.7 "Further, the degree Celsius is exactly the same as a kelvin. The only difference is that zero degree Celsius is 273.15 kelvin. Use of Celsius temperature gives us one Jess digit in most cases", from [Eng. Educ., (April 1977):678]. Comment on the quotation. Is it correct? If not, in what way or sense is it wrong? -4.8 Calculate all temperatures from the one value given: (a) (b) (e) (d) (e) (f) (g) (h) 140 1000 500 1000 298 toOO -40 1000 "4.9 The emissive power of a blackbody depends on the fourth power of the temperature and i§ given by W=Af4 where W = emission power, BtuI(tt1) (hr) A = Stefan-Boltzman constant. 0.171 X 10-8 Btu/(ft2)(hr)(OR)4 T = temperature, OR What is the value of A in the units J/(m2)(s)(K4)? ·4.10 Suppose that an alcohol and a mercury thermometer read exactly ooe at the ice point and 100°C at the boiling point. The distance between the two points is divided into 100 equal parts in both thennometers. Do you think these thermometers will give exactly the same reading at a temperature of, say, 60°C? Explain. ·*4.11 For the following data plot p vs. lIT. Do you get a straight line? Try In(p) vs. liT and repeat. Use curve fitting software (refer to the CD accompanying this book) to estimate the values of the coefficients in the relation In (p) = a + b (1/7) T is in K and p is in nun Hg absolute. T (K) p (rnm Hg) 273 298 323 373 Predict the value of p at 1340 mm Hg. 5 24 93 760 J
- 130. CHAPTER 5 PRESSURE 5.1 Pressure and Its Units 5.2 Measu rement of Pressure 5.3 Differential Pressure Measurements Your objectives In studying this chapter Bre to be able tD: 1. Define pressure. atmospheric pressure, barometric pressure. standard pressure, and vacuum. 2. Explain difference between absolute and relative (gauge pressure). 3. List four ways to measure pressure. 4. Convert from gauge to absolute pressure and vice versa. 5. Convert a pressure measured in one set of units to another including kPa. mm Hg, ft H20. atm, in. Hg, and psi using the standard atmosphere or density ratios of liquids. 6. Calculate the pressure from the density and height of a column of static fluid. 100 103 114 Why review pressure, a topic you no doubt have encountered many times fore? Our experience has shown that a number of gaps may exist in your comprew hension and use of pressure. Because effective use pressure is important in chemical engineering practice, this chapter is designed to fin these gaps. looking Ahead this chapter we review various measures of both relative and ab-solute. discuss methods of measuring and illustrate converting from one set of pressure units to another.
- 131. 100 Pressure Chap. 5 5 .. 1 Pressure and Its Units In Florentine Italy the seventeenth century, well diggers observed that, suction pumps, water would not rise more than about 10 meters. In 1642 they came to the famous Gameo help. but he did not want to be bothered. As an alternate, they sought the help of Toricelli. He learned from experiments that was not being pulled up by the vacuum, but rather was being pushed up by the local pressure, Pressure is defined as "the normal (perpendicular) force per unit area." Exam- Figure 5.1. Pressure is exerted on top of the cylinder of mercury by the mosphere, and on the bottom of the cylinder itself by the mercury, including the effect of the atmosphere. The pressure at the bottom of the static (nonmoving) column of mercury exon the sealing plate is F p = - = pgh Po A where p = pressure at the bottom of the column the fluid F = force A = area p = density of fluid g = acceleration of gravity h = height of the fluid column Po = pressure at the top of the column of fluid In the SI system force expressed in newtons, and area in square 1) then the pressure N/m2 or pascal (Pa), The value of Pa is so small that the kilo- (kPa) is a more convenient unit of pressure. Atmospheric Pressure Figure Pressure is the force per unit ares. Arrows show the exerted on the respective areas.
- 132. Sec. Pressure and Its Units Some common nonstandard variations of pressure measurement SI system are 8. Bars (bar): 100 kPa = 1 bar 10f with the b. Kilograms (force) p~r square centimeter (kgpcm2Y-'-a very common measure of pressure but not standard in (often called just "kilos") c. Torr (Torr): 760 Torr :: 1 atm In the system can expressed a variety including 8. Millimeters of mercury (mm Hg) b. Inches of (in. Hg) c. Feet water (ft H2O) d. Inches of water (in. H2O) c. Atmospheres (atm) f. Pounds (force) per square inch (often just "pounds") (psi) Keep in mind the confusion that can be caused by "pounds/' such as the caption on a cartoon that nontechnical use of the word "It says inflate to 12 pounds. How can I throw a 12-pound football?" You can calculate the force exerted at the bottom a static fluid by applying Equation I). For example, suppose that the cylinder fluid in 5.1 is a umn of mercury that an area of 1 cm2 and is em high. From Table D 1 in the Appendix you can find the specific gravity of mercury at 20°C. and hence the den-sity the Hg, 13.55 g/cm3. Thus, the exerted by mercury alone on l-cm2 section of the bottom plate by the column of mercury 980 cm 50 cm I cm2 ] 1 m 1 (N)(S2) F=--..;;.. S2 1000 g 100 em l(kg)(m) = 6.64 N The on the of the plate covered by mercury is the force per unit area of the mercury plus the pressure of the atmosphere = 6.64 N (100 cm)2 1 kPa p 11m -'-----'----'--- 1000 Pa Po = 66.4 Po .. Although quently denoted in SI system are in common pressures are kglcm2), units that have 10 multiplied by 9.80 X (cm2) - to pascal
- 133. 102 Pressure If we had started with units in the AE system. the pressure would as [the density of mercury is (13 5)(62.4)lbm/ft3 = 845.5 Ibm/ft3) 32.2 ft 50 em 1 in. 1ft (lbf ) ---= 2.54 em in. 32.174(ft)(lbm) p== 1 Ibf = 1388 ft2 Po Chap. 5 computed + Po Sometimes in engineering practice, a liquid column referred to as head of liquid, the head being the height of the column of liquid. Thus, the pressure of the column of mercury could be expressed simply as 50 em Hg, and the pressure on the sealing plate at tbe bottom of the column would be 50 cm Hg + Po (in cm of Hg). S LF-ASSESSMENT TEST Questions 1. Figure SA T5.1 Q 1 shows two coffee pots sitting on a 1evel table. Both are cy lindrical and have the same cross-sectional area. Which coffee pot will hold the most coffee? o Figure SATS.IQI 12. Figure .lQ2 shows four closed containers completely filled with water. Order the containers from the one exerting the highest pressure to the lowest on their respective bases. 1 2 T : : 1" 1 I 4 I 1'!2 o.~+ O~A, i 3 A, A, Figure SAT5.1Q2 Thought Problems 1. When you lie motiortless on a bed, the bed supports you with a force that exactly matches your weight, and when you do the same on the floor, the floor pushes up against you with the same amounc of force. Why does the bed feel softer than the floor? J
- 134. Sec. 5.2 Measurement of Pressure 103 2. If you push the tube shown in Figure TP5.1 P2 into the glass of water so that bend A is below the water level. the tube will become a siphon, and drain the water out of the glass. Why does that happen? Figure TPS.IP2 5.2 Measurement of Pressure Pressure, like temperature, can be expressed using either an absolute or a relative scale. Whether relative or absolute pressnre is measured in a pressuremeasuring device depends on the nature of the instrument used to make the measurements. For example, an open-end manometer (Figure 5.2a) would measure a relative (gauge) pressure, since the reference for the open end is the pressure of the atmosphere at the open end of the manometer. On the other hand, closing off Air a , ~h=11.0 in. Hg N2 • Vacuum b I Il h = 40.90 em Hg L Figure 5.2 (a) Open-end manometer showing a pressure above atmospheric pressure. (b) Manometer measuring an absolute pressure.
- 135. 104 Pressure Chap.S the open end of the manometer (Figure 5.2b) and creating a vacuum in that end results in a measurement against a complete vacuum, or against "no pressure"; Po in Equation (5.1) is zero. Such a measurement is called absolute pressure. Since absolute pressure is based on a complete vacuum, a fixed reference point that is unchanged regardless of location, temperature, weather, or other factors, absolute pressure establishes a precise, invariable value that can be readily identified. In contrast the zero point for a relative or gauge pressure measurement usually corresponds to the pressure of the air that surrounds us at all times, and as you know, varies slightly. You are probably familiar with ihe barometer illustrated in Figure 5.3. Does a barometer read absolute or relative (gauge) pressure? Figure 5.4 shows the workings of a dial device that measures pressure called a Bourdon gauge. Does it measure absolute or relative pressure? A Bourdon gauge normally measures relative pressure, but not always. The pressure-sensing device in the Bourdon gauge is a thin metal tube with an elliptical cross-section closed at one end that has been bent into an arc. As the pressure increases at the open end of the tube, it tries to straighten out, and the movement of the tube is converted into a dial movement by gears and levers. In any of the pressure-measuring devices depicted in Figures 5.2 to 5.4, the fluid is at equilibrium, meaning that a state of hydrostatic balance is reached stabilizing the manometer fluid. The pressure exerted at the part of the tube open to the atmosphere or vacuum exactly balances the pressure exerted at the other end of the bh Figure S.3 A mercury barometer. I
- 136. Sec. 5.2 Measurement of Pressure Closed End Link Gear ond Pinion COnnection to Pressure Source Figure 5.4 Bourdon gauge pressure-measuring devices: (a) Bourdon. 105 Bourdon Tube , Conneehon to PTessure Source Bourdon and (b) Spiral tube. Water and mercury are commonly used indicating fluids for manometers; the readings thus can be expressed in "inches or em of water," "inches or cm of mer· cwy," and so on. In ordinary engineering calculations involving pressure we the vapor pressure mercury (or water) and minor changes in the density of mer-cury (or water) due to temperature changes. What are some other examples of pressure-measuring devices? Figure 5.5 presents the names of some common instruments along with their useful ranges. Another term with which you should become familiar is vacuum. In effect; when you measure the pressure in "inches of mercury vacuum/' you are reversing the usual direction measurement, and measure from the atmospheric pressure downward to zero absolute pressure. Thus: inches Hg vacuum = barometric pressure - absolute pressure. pressure-sensing device for vacuum systems is commonly used in an apparatus that operates at pressures less than atmospheric, such as a vacuum filter. A pressure that is only slightly below barometric pressure may sometimes be expressed as a "draft!) in inches of water, as, for example, in the air supply to a furnace or a water cooling tower. Here is a news article (R.E. Sanders, Chemical Engineering Progress. September 1993. p. 54) that pertains Figure 5.6. Don't Become Another Victim of Vacuum Tanks are fragile. An egg can withstand more pressure than a tank. How was a vacuum created inside the vessel? As water was drained from the column, the vent to let in air was plugged up, and the resulting pressure difference be~ tween inside and out caused the stripper to faB. We can conclude: don't let your vacuum you down!
- 137. 106 Pressure Chap. 5 mmHg Afm 105 4) 0- 100 D CJ> i ",. G) l5 ~ t04 OJ ~ I) 10 u ~ ~a ~ 0- I'l".! Q) - 'Vi g) :3 E (5 ,03 CJ> go II> C 0- .Q ~ <[ ~ .::: c0 - '0 '6 ... ~ 'i) 102 :- s e ~ 0.1 '1ii .~ .: :I 0 5l ~ } ~ ~ >. a ~ '0u Q. 10 0.0 ~ ~ 8 2 1 Figure 5.5 Ranges of application for O.OOt pressure-measuring devices. Figure 5.6 Close-up of a failed stripper column (Reproduced through the courtesy of Roy E. Sanders),
- 138. Sec. Measurement of Pressure Always keep in mind that the reference point or zero point for the relative pressure scales or for a vacuum scale is not constant, and that the relationship between relative and absolute pressure is given by the following expression: gauge pressure + barometer pressure = absolute pressure (5.2) Examine Figure 5.7. You definitely must not confuse the standard atmosphere with atmospheric pressure. The standard atmosphere is defined as the pressure (in a standard gravitational field) equivalent to 1 atm or 760 mm Hg at O°C or other equiva~ lent value. whereas atmospheric pressure is variable and must be obtained from a barometric measurement each time you need it. Look at Figure 5.7 for clarification. The standard atmosphere may not equal the barometric pressure in any part of the world except perhaps at sea level on certain days. However. it is' extremely useful converting from one system of pressure measurement to another (as well as being useful in several other ways, to be considered in Chapter 13). In a problem, if you are not given the barometric pressure, you usually assume that the barometric pressure equals the standard atmosphere, but this assumption is only that-an assumption. Expressed in various units, the standard atmosphere is equal to 1.000 atmospheres (atro) 33.91 of water (ft H20) 14.7 (14.696. more exactly) pounds (force) per square inch absolute (psi a) ~ 1.000atm j :: 101.3 kPa ... ...... =760 mmHg .. ---- ........... ~~~---------~~ ... ----- Time F!gure 5.7 terminology. Note that the vertical scale is exaggerated for Hlust:ra-rive purposes. The dashed line illustrates the atmospheric (barometric) pressure. which changes from time to time. Point ill in the figure denotes a pressure of 19.3 psia, that is, a pressure referred to a complete vacuum. or 5 psi referred to the barometric pressure, (I) is the complete vacuum or zero pressure, any point on the heavy line such as (3) is a point corresponding to the standard pressure, and @ illustrates a negative relative pressure, that is, a pressure less than atmospheric pressure. This latter type of measurement is described in the text as a vacuum measurement. Point also indicates a vacuum measurement, but one that is equivalent to an absolute pressure above the standard atmosphere.
- 139. 108 Pressure 29.92 (29.921, more exactly) inches of mercury (in. Hg) 760.0 millimeters of mercury (mm Hg) Chap. 5 1.013 x 105 pascal (Pa) or newtons per square meter (N/m2)~ or 101.3 kPa Because the standard atmosphere is an absolute pressure, you can easily con-vert from one set of pressure units to another by using a pair of standard atmospheres as a conversion factor as follows. We will convert 35 psia to inches of mercury and kPa by using ratios of the standard atmosphere to carry out the conversions: 3S psia 29.92 in. Hg . 4 . = 71.24 10 Hg 1 .7 ps!a 35 psia 101.3 kPa = 241 kPa 14.7 psia In summary, look at Figure 5.8, which illustrates the relationships between relative and absolute pressure for two of the AE pressure units and the SI units. Figure 5.8 shows the relations in pounds (force) per square inch (psi), inches of mercury On. Hg), and pascals. Pounds per square inch absolute is normally abbreviated "psia"; "psig" denotes "pounds per square inch gauge." Sometimes you have to use Pounds per mm square inch mercury Inches mercury newtons per square meter A pressure above atmospheric 5.0 19.3 259 998 39.3 10.2 0.34)( 105 1.33 x 1()5 Standard atmosphere 0.4 14.7 20.7 76029.92 0.82 0.028 )( lOS 1.013 x 1()5 Barometric pressure 0.0 14.3 0.0 740 29.1 0 0.0 0.00 0.985 x 10S A pressure below atmospheric-2.45 11.85 2.45 24.1 -5.0 5.0 -o.17x1()S 0.82 x 1()5 0.17 x 105 I E, ~ ~ E J E ::::::I ::I ::I ~ :::J e CD :J e ::I u ~ ... u u ! ::I ~ :J ::I ~ ~ e :::J ~ ::::::I III :J (J) ~ ::I ::I CIl ~ (J) ~ ~ :z U) CIl ~ ! U) CD e ! 0. ~ Q) .... - 0. 0. '- CL 0. Q) 0. Q) J " :g, 0. 0. S ~ :::s ~r CD Q) Q) CD :l 0) ca '0 '0 en ~ '0 " (I'J C'C! C!' en CIl (/) (!J .0 ~ C!) Cl oCt ~ Perfect vacuum -14.3 0.0 14.3 0 a -29.1 29.1-O.985x 105 0.00 0.985 x 10 Figure 5.8 Pressure comparisons when barometer reading is 29.1 in. Hg. 5 -,/ (
- 140. l Sec. 5.2 Measurement of Pressure 109 common sense as to whlch is being measured if the letters psi are used. For other pressure units. be certain to carefully specify whether they are gauge or absolute units. although you rarely find people doing so. For example, state u300 kPa absolute" or "12 cm Hg gauge" rather than just 300 kPa or 12 em Hg. as the latter two can on occasion cause confusion. EXAMPLE 5.1 Pressure Conversion The structures and dislocations in ceramic material determine the strength of the material in nanotechnology. NonomateriaJs are more flexible than ordinary composites because each of the nanocrystallites (particles) can move past each other so that stretching can occur. However, nanosized particles can also enhance hardness. Crystallites pack together along the boundaries of rnacrocrystallites, and prevent the structure from unzipping. For example, a hardness in excess of 60 gigapascals has been reported for nanocrystals of titanium nitride embedded in thin films of silicon nitride. Hardness is measured by the pressure required to just indent the surface of a material (diamonds exhibit a hardness of greater than 100 Gpa). What is the equivalent pressure to 60 Gpa in (a) atmospheres (b) pSla (c) inches of Hg (d) rnm of Hg Solution For the solution, use the standard atmosphere. Basis: 60 GPa ) 60 GPa 106 kPa 1 atm 6 atm (a 1 GPa 101.3 kPa = 0.59 X 10 (b) 60 GPa 106 kPa 14.696 psia 6 psia 1 GPa 101.3 kPa = 8.70 X 10 60 GPa 106 kPa 29.92 in. Hg 7 (c) 1 GPa 101 .3 kPa = 1.77 X 10 in. Hg ( d) 60 GPa 106 kPa 760 rnm Hg 8 1 GPa 101.3 kPa = 4.50 X 10 mm Hg
- 141. 110 Pressure Chap.S EXAMPLE 5.2 Pressure Conversion pressure gauge on a tank of CO2 used to fiU soda-water bottles reads 51.0 psi. At the same time the barometer reads 28.0 in. Hg. What is the absolute pressure in the tank in psia? See Figure Figure ES.2 Solution The first thing to do is to read the problem. You want to calculate a pressure using convenient conversion factors, namely the standard atmosphere. Then examine Figure E5.2. The system is the tank plus the line to the gauge. All of the necessary data are known except what means. Is the pressure gauge reading psig. not psia? Yes. You can assume the gauge is a Bourdon gauge measuring relative pressure, Equation (5.2) states that the absolute pressure is the sum of the gauge pressure and the atmospheric (barometric) pressure expressed in the same units. Let us change the atmospheric pressure to psia. The calculation is 28.0 in. Atmospheric pressure = --------=--- The absolute pressure in the tank is 51.0 psia + 13.76 psia::: 64.8 psia Lastly. you always need to check. your answers. Try using a different conversion factor. You might first convert the barometric pressure to atm, then the gauge reading to atm. add the pressure in atm. and finally convert the result to psia. A long way to solve the problem. of course. but try it.
- 142. l Sec. 5.2 Measurement of Pressure EXAMPLE 5.3 Vacuum Pressure Reading Small animals such as mice can live (although not comfortably) at reduced air pressures down to 20 kPa absolute. In a test, a mercury manometer attached to a tank, as shown in Figure E5.3, reads 64.5 cm Hg and the barometer reads 100 kPa. Will the mice survive? 64.5 em HI) t Figure ES.J Solution First read the problem. You are expected to realize from the figure that the tank is below atmospheric pressure because the left leg of the manometer is higher than the right leg, which is open to the atmosphere. Consequently, to get the absolute pressure you subtract the 64.5 cm Hg from the barometer reading. , We ignore any corrections to the mercury density for temperature, and also ignore the gas density above the manometer fluid because it is much lower than the density of mercury. Then, since the vacuum reading on the tank is 64.5 em Hg below atmospheric, the absolute pressure in the tank is 64.S em Hg 101.3 kPa 100 kPa - - 100 - 86 = 14 kPa absolute 76.0emHg The mice probably will not survive. SELF-ASSESSMENT TEST Questions 1. Answer the following questions true or false: 1(1 a. Atmospheric pressure is the pressure of the air surrounding us and changes from day to day. b. The standard atmosphere is a constant reference atmosphere equal to 1.000 aim or the equivalent pressure in other units.
- 143. 112 Pressure Chap.S c. Absolute pressure is measured relative to a vacuum. d. Gauge pressure is measured upward relative to atmospheric pressure. e. Vacuum and draft pressures are measured downward from atmospheric pressure. f. You can convert from one type of pressure measurement another using the standard atmosphere. A manometer measures the pressure difference in terms of the height of fluid(s) in the manometer tube. 2. What is the equation convert gauge pressure to absolute pressure? 3. What are the values and units of the standard atmosphere for six different methods of expressing pressure? 4. What is the equation to convert vacuum pressure to absolute pressure? Problems 1. Convert a pressure of 800 mm Hg to the foHowing units: a. pSla b. kPa c. atm d. ft H20 2. Your textbook lists five types of pressures: atmospheric pressure, barometric pressure~ gauge pressure, absolute pressure, and vacuum pressure. a. What kind of pressure is measured by the device in Figure SA TS.2P2A? VQcuum Helium 20 in. Hg. iii. Hg Helium ® Figure SATS.2P2A b. What kind of pressure is measured by the device in Figure SAT5.2P2B? c. What would be the reading in Figure SAT5.2P2C assuming that the pressure and temperature inside and outside the helium tank are the same as in parts (a) and (b)? 3. An evaporator shows a reading of 40 kPa vacuum. What is the absolute pressure in the evaporator in kPa? Thought Problems 1. A magic trick is to fin a glass with water. place a piece of paper over the top of the glass to cover the completely. and hold the paper in place as the is inverted 180°. On the release of your support the paper. no water runs out! Many books state that the
- 144. Measurement of Pressure 113-' glass should be completely filled with water with no air bubbles present. Then the outside air pressure is said to oppose the weight of the water in the inverted glass. However, the experiment works just as weB with a half-filled glass. The trick does not work if a glass plate is substituted for the piece of paper. Can you explain why? A large storage tank was half full of a flammable liquid quite soluble in water. The tank needed maintenance on the roof, Since welding was involved, the foreman attached a flexible hose to the vent pipe on the top of the tank (in which there was a flame arrestor), and inserted the end of the hose into the bottom of a drum of water sitting on the ground to pick up any exhaust vapors. When the tank was emptied, the water rose up the and the tank walls coJlapsed inward. What went wrong this incident? 3. Can a pressure lower than a complete vacuum exist? 4. If you fill a Styrofoam™ cup with water, and put a hole in the water starts to run out. However, if you hold your finger over the hole and then drop the cup from a height, water does not run out Why? Discussion Problems 1. Safety relief valves protect against excessive pressure in pipelines and process equipment. Excessive pressure can occur because of equipment failure in the process, fire, or human error. A proper must be selected for anticipated scenarios. and the spring set for the proper relief pressure. Two different valves (A and B) are proposed for use in a pipeline, as iHustrated in Figure 1. Which valve would you recommend for use? Vented '/ P2 .............. Out (A) Th!s valve the top of the closing (8) This valve has the and bottom of disk open to the outlet (P2) the closing disk open to the outlet (1'2) Figure DPS.2Pl 2. Form a study group to investigate the possibility of raising the Titanic. The mystique of the sinking and attempts to the vessel have crept into the literature (A Night to Remember), the movie ("Raising the TitanicH ), and magazines such as National Geographic.
- 145. 114 Pressure Chap. 5 Carry out a literature search to get the basic facts (4000 m deep. 4.86 x 108 N original weight.. the density of sea water, and so on). Prepare a report giving: a. An executive sununary including an estimate of feasibility. b. The proposed method(s) of raising the ship. c. A list of steps to execute to raise the ship. d. A list of the equipment needed (include costs if possible). e. A time schedule for the entire project (including obtaining the .equipment and personnel). f. A list of (1) aU assumptions made and (2) problems that might be encountered for which answers are not known. h. As an appendix, show all calculations made and references used. 5 .. 3 Differential Pressure Measurements Have you noted in the discussion and examples so far that we have ignored the in the manometer tube above the fluid. Is this OK? Let's see. Examine Figure 5.9, which illustrates a manometer involving three fluids. When the columns of fluids are at eqUilibrium (it may take some time!), the relationship among PIt P2. P3' and the heights of the various columns of fluid is as follows. Pick a reference level for measuring pressure at the bottom of dl . (If you pick the very bottom of the U tube instead of dt as the reference. the lefthand and righthand distances up to dt are equalt andt consequently. the pressures exerted by the righthand and lefthand legs of the U-tube will cancel out in Equation (5 it and see). Fluid I, Pi J f 1 1 Fluid 3, P3 Fluid 2, P'l. (5.3) Figure 5'.9 Manometer with three fluids.
- 146. L Sec. 5.3 Differential Pressure Measurements If fluids 1 and 3 are gases, and fluid 2 is mercury. the density of the gas is so much less than that of mercury that you can ignore the term involving the gas in Equation (5.3) for practical applications. However, if fluids 1 and 3 are liquids and fluid 2 is an immiscible fluid, the terms invo]ving the densities of fluids 1 and 3 cannot be neglected in Equation (5.3). As the densities of fluids 1 and 3 approach that of fluid 2, what happens to the differentia1 reading d2 in Figure 5.9 for a given pressure difference? Can you show for the case in which Pl ::::: P3 = P that the manometer expression reduces to the we11-known differential manometer equation? (5.4) As you know, or will learn in Chapter 27, a flowing fluid experiences a pressure drop when it passes through a restriction such as the orifice in a pipe, as shown in Figure 5.10. The pressure difference can be measured with any instrument, such as a manometer. connected to the pressure taps, as illustrated in Figure 5.10. Note that the manometer fluid is static if the flow rate in the pipe is constant. Fluid Flow ~-- Differential Pressure Manome1er Figure 5.10 Concentric orifice used to restrict flow and measure the fluid flow rate with the aid of a manometer. EXAMPLE 5.4 Calculation of Pressure Differences In measuring the flow of fluid in a pipeline as shown in Figure E5.4. a differential manometer was used to determine the pressure difference across the orifice plate. The flow rate was to be calibrated with the observed pressure drop (difference). Calculate the. pre.ssure drop P CP1 in pascals for the manometer reading in Figure E5.4.
- 147. 116 Solution ~ l-0rifice Water, p = 103kg/m31 I? ------~~1 i------- PI" ~ T 32mm ~ d -.liD mm Manometer fluid, rv" UO )( 103 kg/m3 Figure ES.4 Pressure Chap. 5 In this problem you cannot ignore the water density above the manometer fluid. Thus. we apply Equation (5.3) or (5.4). because the densities of the fluids above the manometer fluid are the same in both legs of the manometer. The basis for solving the problem is the information given in Figure E5.4. Apply Equation (5.4) PI - P2 = {Pf- p)gd (LI0 - 1.00)103 kg 9.807 m (22)(10-J)rri I(N)(s2) 1 (Pa)(rn2 ) = ,3 S2 (kg)(m) 1 (N) = 21.6 Pa Check your answer. How much error would occur if you ignored the density of the flowing fluid? EXAMPLE 5.5 Pressure Conversion Air is flowing through a duct under a draft of 4.0 em H20. The barometer indicates that the atmospheric pressure is 730 mm Hg. What is the absolute pressure of the air in inches of mercury? See Figure E5.5 Air ~ Figure E5.5
- 148. Sec. 5.3 Differential Measurements Solution Can you ignore the density above manometer fluid and the air above open end of the manometer? Probably. In calculations you must emptoy consistent units, and it appears in this case that the most convenient units are those of inches of mercury, so let's convert barometer reading and the manometer reading to in. using the standard atmosphere as the conversion factor. 730 mm Hg 29.92 Atmospheric 1-'.1.",."", ... "", - ----.;;;.. = 28.7 in. Hg Next, convert 4.0 cm H20 to in. Hg: 4.0cm I in. 1 ft 29.92 in. 2.54 cm 12 in. 33.91 ft H20 0.12 in. Since the reading 4.0 em ing in uniform units is draft (under atmospheric), the absolute read- 28:7 in. - 0.12 in. Hg 28.6 in. absolute S LF .. ASSESSMENT TEST Questions 1. Answer the following questions true or 117 a. flows in a pipeline. manometer containing Hg is set as illustrated in Figure E5.4. and shows a differential pressure of 14.2 mm You can ignore the effect of the density of air on height of the columns mercury. b. Lowering the pressure Figure SAT5.2P2A (see SAT problems in Section win not cause the length of the column of Hg to decrease by 10%. Problems 1. A U-tube manometer filled. with mercury is connected between two points in a pipeline. manometer reading is mm of calculate the pressure difference kPa between the points when (a) water is flowing through the pipeline, and (b) also when at atmos-pheric pressure and 20°C with a density of 1.20 kglm3 is flowing in the pipeline. A Bourdon gauge and a mercury manometer are connected to a tank as shown in SAT5.3P2. the on the pressure is 85 kPa, what is h in Hg?
- 149. 118 Pressure Chap. 5 ........,...--.. A B Figure SA TS.3P2 Looking Back In this chapter we defined pressure, discussed some of the common ways to measure pressure, and showed how to convert from one set of pressure units to another by using the standard atmosphere. We also emphasized the difference between the standard atmosphere and atmospheric pressure. GLOSSARY OF NEW WORDS Absolute pressure Pressure relative to a complete vacuum. Barometric pressure Absolute pressure measure measured by a barometer. Bourdon gauge Pressure measuring device containing a thin metal tube that flexes and moves a dial as the pressure being measured changes. Gauge pressure Pressure measured above atmospheric pressure. Manometer A U-tube or other device containing a fluid that moves in the tube as the pressure difference across the sides of the tube changes. The difference in the height of the fluid in the two sides of the tube indicates the pressure difference. Pressure The nonnal force per unit area that a fluid exerts on a surface. Pressure difference The difference between the pressure at one point and another, usually as measured by an instrument. Relative pressure Same as gauge pressure. Standard atmosphere The pressure in a standard gravitational field equivalent to 760 (exactly) mm Hg. Vacuum A pressure less than atmospheric (but reported as a positive number).
- 150. I L Chap. 5 Problems 119 UPP EM NTARY R FER NC S In addition to the general references listed the ing are pertinent in the front material. the foHow- American Society of Heating, Refrigeration, and Air Conditioning Standard Methodfor Pressure Measurement (Ashrae Standards No. 41.3). Ashrae (1989). Benedict, Fundamentals of Temperature, Pressure, and Flow Measurement, 3rd ed., John Wiley, New York (1984) Cengel, Y. A.~ and M. A. Boles, Thermodynamics, 4th ed., McGraw-Hill, New York (2002). Gillum. R. Industrial Pressure, and Density Measurement. Instrumentation, Systems, and Automation Society (1995). Web Sites bttp:llthermal.sdsu.edultestcenterffestlsolve!basicsJpressureJpressure.html http://www.efm.leeds.ac.uklCIVElCIVE1400/Section2/Manometers.htm1 bttp://www.mas.ncl.ac.ukl-sbrooks/nish.mit.edul2006rrextbookiNodesIchapo2lnode9.html http://www .omega.comJ1iteratureltransactions/volume3Jpressure,html PROBl MS *5.1 From the newspaper: "BROWNSVILLE, TX. Lightning or excessive standing water on roof of a clothes store are as leading causes suspected building's collapse. Mayor Ignacio Garza said possibilities elude excessive caused by standing water on the 19-year-old building's roof. Up to inches of rain fen here less six hours." Flat-roof buildings are a popular architectural style dry climates because of the economy materials of construction. However. during the rainy season water may pool up on roof decks so structural considerations for the added weight must be taken into account. If 15 crn of accumulates on a lO-m by 10-m area during a heavy rain storm. detennine: (a) The total added weight the building must support (b) The force the on the roof in ··S.2 A problem with concrete wastewater treatment tanks set below ground was realized when the water table rose and an empty tank floated out of the ground. This buoyancy problem was overcome by installing a check val ve the wan of the tank so that if the water table rose high enough to the tank. it would fill with water. If the density of concrete is 2080 kglm3• detennine the maximum height at which valve should installed to prevent a buoyant force from raising a rectangular tank with in, side dimensions of 30 m by m and 5 m deep. The walls and floor have a unifonn thickness of 200 mm.
- 151. 120 Pressure Chap. 5 *5.3 A centrifugal pump is to be used to pump water from a lake to a storage tank that is 148 ft above the surface of the lake. The pumping rate is to me 25.0 gal/min, and the water temperature is 60°F. The pump on hand can develop a pressure of 50.0 psig when it is pumping at a rate of 25.0 gaUmin. (Neglect pipe friction, kinetic energy effects, or factors involving pump efficiency.) (a) How high (in feet) can the pump raise the water at this flow fate and temperature? (b) Is this pump suitable for the intended service? "'5.4 A manometer uses kerosene, 0.82. as the fluid. reading of 5 in. on the manometer is equivalent to how many millimeters of mercury? *5.5 manufacturer of large tanks calculates the mass of fluid in the tank by taking the pressure m~asurement at the bottom of the tank in psig. and then multiplies that value by the area of the tank in square inches. Can this procedure be correct? ·5.6 A letter to the editor says "An error in units was made in the article "Designing Airlift Loop Fermenters." Equation (4) is not correct Lip = 4fp [(v2/2g)(UD)] Is the author of the letter correct? if is dimensionless.) (4) Leaking au tanks have become such environmental problems that the Federal Government implemented a number of rules to reduce the problem. A leak from a small hole a tank can be predicted from the following relation: Q = 0.61 Sv'(2ilp)p where Q = the leakage rate S = crossectional area bf the leak il p ::::: pressure drop p = fluid density To test the tank, the vapor space is pressurized with N2 to a pressure of 23 psig. If the tank is filled with 73 inches of gasoline (sp. = 0.703) and the hole is 114 in. in diameter, what is the initial value of Q (in ft3Jhr)? - -----------~-t-·------- - Gasoline 73 in t Leak Figure PS.7 ·5.8 Suppose that a submarine inadvertently sinks to the bottom of the ocean at a depth of 1000 m. It is proposed to lower a diving bell to the submarine and attempt to the conning tower. What must the minimum air pressure be in the diving bell at the level
- 152. Chap. 5 Problems 121 of the submarine to prevent water from entering into ben when the opening valve at the bottom is cracked slightly? Give your answer in absolute kilopascal. Assume that seawater has a constant density of 1.024 g/cm3. ·5.9 A pressure gauge on a welder's tank a of 22.4 psig. barometric pressure is in. Hg. Calculate the absolute pressure in tank (a) Ibf lft2; (b) in. Hg; (c) newtons/(meter)2; and (d) ft ·S.10 John Long says he calculated from a fonnula that the pressure at the top of Pikes Peak is 9.75 psia. John Green says that it is 504 mm because he looked it up in a table. Which John is right? "5.11 The floor of a cylindrical water tank was distorted into bulges due to the settling of improperly stabilized soil under the tank floor. However, several consulting engineers restored damaged tank to use by placing plastic around the bottom of the tank wall and devising an flotation system to move it to an adjacent location. tank was 30.5 m in diameter 11m deep. The top. bottom. and sides of the were made of 9.35-mm-thick welded sheets. The density of steel is 86 gJcm3. (a) What was the gauge pressure in of the water at the bottom of the tank when it was completely fun of water? (b) What would the air pressure have to be kPa beneath the empty tank in to raise it up for movement? ·5.12 A pressure insaument has failed on a process Hne that requires constant monitoring. A bell-type gauge as shown Figure PS.12 is available that has oU (density of em Dill 0.800 q/cmJ t p PS.12
- 153. 122 Pressure Chap.S ~ 0.800 glcm3) as a sealant liquid. COlistruction of the gauge limits the sealant liquid's travel to 12.7 em before blowout the oil occurs. Wl)at maximum pressures can this gauge measure in kPa? -5.13 Bourdon pressure gauge is connected to a large tank ,and reads 440 kPa when the ""FIn"",.,,. .. read~ 750 mm Hg. What will the gauge reading the atmospheric pres-sure increases to mm Hg? ·"'5.14 gauge on a closed tank as shown the accompanying figure 20 pounds square The contents of the tank are air and a hydrocarbon. (a) Does the read or absolute pressure?· State which why. (b) What is the pressure at the bottom of the tank in if the specific gravity of hydrocarbon is 0.921 4.0 ft Air +f-------l 2.5 ft Hydrocarbon --L '---__ --' Figure PS.14 US.IS gas cylinder to is attached an Bourdon gage appears to be at a pressure of 27.38 in. the 101.8 kPa. student claims that the pres-sure In psia, but another student points out that this impossible-the pressure is really psia. Can 1.3 psia be Explain and show calculations to up your explanation. *5.16 The on the steam condenser for a turbine indicates 26.2 in. Hg of vac" uum. The barometer reading is 30.4 in. What the absolute pressure in the con- Y,",'I'>"'" in psia? "'5.17 A pressure gauge on a process tower indicates a vacuum of 3.53 in. Hg. The barome-reads 29.31 in. What the absolute in tower in millimeters of mercury? "(l, ·"·5.18 A student a B our do n gauge deS'l gne d to read gauge to a vacuum line on finding out that it would not give a reading. decided to adjust the instru-ment so that it read 0 for a vacuum, and 14.7 at atmospheric sure. on day when this adjustment was the barometer mm Hg, so that reading of 14.7 psi was slightly in error. This gauge was later to measure a pressure a tank alI, read 51 psig. should the proper gauge reading The barometer at that time was mm State assumptions to solve this problem. *:;;:;,. 19 vacuum gauge connecu..; ~;uA to a tank read s 31.5 kPa. What is the corresponding solute pressure if the barometer reads 98.2
- 154. Chap. 5 Problems 123 *5.20 Your boss asks you to drain a completely futI leak proof water storage tank 12 m high and 3m in diameter by opening a valve at the bottom of the tank. What response should you . to your boss? ·S.21 A centrifugal pump is to be used to pump water from a lake to a storage tank which is 148 feet above the surface of the lake. The pumping rate is to be 25.0 gallons per minute, and the water temperature is 60cF. The pump on hand can develop a discharge pressure of 50.0 psig when it is pumping at a rate of 25.0 gaVrnin. (Neglect pipe friction, kinetic energy effects, or factors involving pump efficiency.) (a) How high can the pump raise the water at this flow rate and temperature? Give answer in feet. (b) Is this pump suitable for the intended service? Examine Figure PS.22. Oil (density == 0.91 glcm3) flows in a pipe, and the flow rate is measured via a mercury (density:::;:; 13.546 glcm3) manometer. If the difference in height of the two of the manometer is 0.78 in., what is the corresponding pressure difference between points A and B in mrn At which point, A or B; is the pressure higher? temperature is 6OoP. Oil- A B - I==r078 In ~ ~~ E .o!!!!5 ches - Figure **5.23 Figure PS.23. The barometer reads 740 mm Hg. Calculate the pressure in the tank in psia. Open I Tank Ah=20In~ Figure PS.23 "'·"5.24 In Figure P5.24, Hg is used to measure the pressure between the points M and N. Water occupies the space above the i:lg. What is Ap between M and N? Which pres-sure is PM or PN?
- 155. 124 Pressure Chap,5 M I 30 em h 5 --'--2~---I Scm 3 Hg Figure PS.24 )~ """"'5.25 Express Po as a function of h based on the data given Figure PS.2S. The nonA shaded tube contains oil. PA r> 4 30 em 1 t h h 30 em 2 ~ + 2 3 L..e- Hg B Ps c.... L:. ~ (Sp.gr.= 1 Valve",, v' 2pump 3.6) ... , ' " ~, Oit (Sp. gr. =0 0.88) Figure P5.2S
- 156. Chap.S Problems 125 9g"S.26 Examine Figure PS.26. Given that a = in. and b = 12 in., what is the height of the water in the right hand 6.6 pslg CCI4 Liquid Manometer oil, p::::: O.7g/cm3 Open to atmosphere I Figure P5.26 1"S.27 A differential pressure transmitter (DPT) can be used to sense the liquid level in a tank. See Figure PS.27. How much error made in not taking into account the height of the vapor above the liquid? Vapor z liqUid Figure P5.17 "5.18 Deflagration is quick but progressive combustion (with or without explosion) as tinguished from a detonation which is the instantaneous decomposition of combustible material. (The National Fire Protection Association in Standard NFPA 68 has some more precise technical defmitions.) For dust, the rate of deflagrntion pends on particle size of the dust NFP A gives a formula to calculate the area for venting a building to prevent structural damage on deflagration Av= CA/pO.5 where A" is the necessary vent area (ft2 or m2). is a constant that must be determjned by experiment (NFPA68 lists some values). As is the surface area of the build-
- 157. 126 Pressure Chap. 5 ing, and P is the maximum internal (gauge) pressure that can be sustained by the building without damage occurring. In a proposed gTain storage elevator 8m in diameter and 10m tall, the vented area (between the roof and top walls) is l0.4rn2. Is this area sufficient if the elevator is designed to have a maximum overpressure of 7.5 kPa (7.5 kPa difference between the inside and outside)? Assume for grain dust that C = 0.41(kPa)O.5, Note the vents should be distributed in practice uniformly throughout the structure. *5.29 Pressure in a gas cell is measured with an inverted manometer, as shown in Figure PS.29. The scale on the far right-hand side of the figure shows the distances in mm (not to scale) of the interfaces of the liquids in the manometer. What is the pressure in the cell? PI ;;;; 12.50 psia, Tonk 1 'Tonk 2 Figure PS.Z9 mm Scale 505 "'5.30 The indicating liquid in the manometer shown in Figure P5.30 is water, and the other liquid is benzene. These two liquids are essentially insoluble in each other. If the manometer reading is AZ = 36.3 em water. what is the pressure difference in kPa? The temperarure is 25°C. Benzene Woter Figur.e PS.30
- 158. l Chap. 5 Problems 127 ··5.31 A U-tube differential mercury manometer is connected between two pipes. One pipe contains carbon tetra chloride (sp.gr. 1.59) under a pressure of 103 kPa. and the other pipe contains oil (sp.gr. 0.8) under a pressure of 172 kPa. Find the manometer reading h in meters. ---------t ------ Pipe 2.5m y -t---*- 1.5 m --f - - h __ t _______ _ Figure PS.31 ··5.32 Examine the pressure measuring device shown in the figure. What is the gage pressure reading in psi at point®? The density ofC7H'6;;:: 0.684 g/cm3 . ~--Pressure = 15.5 psla Gage pressure = ? r- 10Smm 63mm H20 -r Air _---t-~ ~ ---.---------------- 111 mm _____________________ t. Figure PS.32 ·5.33 Examine Figure P5.33. Water flows through an orifice.
- 159. 128 Water at 25°C A Figure PS.33 Pressure Chap. 5 B The manometer fluid has a specific gravity of 1.30. What is the pressure difference between points A and B in kPa if there is a 3.1 cm difference in the heights of the two columns of manometer fluid? 0'" ;···5,34 Examine Figure P5.34. The barometric pressure is 720 mm Hg. The density of the oil is 0.80 glcm3. The Bourdon gauge reads 33.1 psig. What is the pressure in kPa of the gas? Figure PS.34
- 160. l PART MATERIAL BALANCES CHAPTER PAGE 8 Introduction to Material Balance 133 7 General Strategy for Solving Material Balance Problems 166 8 Solving Material Balance Problems for Single Units Without Reaction 196 9 Chemical Equation and Stoichiometry 10 Material Balances for Processes Involving Reaction 260 11 Material Balance Problems Involving Multiple Units 305 12 Recycle, Bypass, and Purge and the Industrial AppUcatlon 341 of Material Balances UCo<..'u£c knew how hard I worked to my mastery, it wouldn't seem so wonderful after all. Michelangelo In Part 2 you begin to what material balances are an Material bal-ances are nothing more than the application of the conservation law for mass: "Mat-neither nor destroyed," Just what this statement means in practice, and how you can use concept to solve problems varying degrees of complexity, some fairly extensive explanation. Look at Figure Part l, Material balances were used to design the plant You will also find that material balances are carned out in the plant during its operation for time periods as a year, a month, a day, so on. The objective to improve efficiency, maintain production, reduce environmental discharges, and maintain control of the processes in the plant. Material balances allow you to understand what is in a processing plant is nothing or complicated about material balances, They are basically accounting-not for money. but for material. ", .. the long years are mostly concerned with knowledge. Fact is upon fact and little time is spent in .. , On the whole it must be more important to be in thinking than to be stuffed with II Bono, The Five Day Course in Thinking (1967) 129
- 161. 130 Material Balances Part 2 Figure Part 2.1. A section of a processing plant. Why study material balances as a separate topic? You will find that material balance equations are almost invariably a prerequisite to all other calculations in the solution of both simple and complex chemical engineering problems. Furthermore. the skills that you develop in analyzing material balances are easily transferred to other types of balances and other types of problems. In solving material balance problems, you must first develop an understanding of the problem. You must be able to visualize the problem from the problem statement, and to do so, you must be familiar with the terminology of the field. Once you understand the problem, you are ready to solve it. You will find that solving material balance problems is much easier if you develop a systematic strategy that is applicable to a wide range of problems. Consequently, in Chapter 7 we focus on the strategy of making appropriate decisions, implementing them properly. and assessing whether the implementation has been correct in solving a problem. Our aim is to belp you acquire a generalized approach to problem solving so that you may avoid looking upon each new problem, unit operation, or process as entirely new and unrelated to anything you have seen before. As you scrutinize (he examples used to i1lustrate the principles involved in each chapter, explore the method of analysis, but avoid memorizing each example by rote, because, after all, they are only samples of the myriad of problems that exist or could be devised on the subject of material
- 162. Part 2 Material Balances 131" oruan(;es. Although material balances are introduced i.)alan~:es are involved throughout the remainder of this part.of the text, material that matter, there-mainder undergraduate program of study and thereafter. LO ARY OF MISSING TERMS Check (the answer) Clear)y Incompre:hensible. that demolishes your work. Difficult problem I don't know the answer (cf. Trivial). Obvious I can't prove it. Previous At some unknown time. Similarly Too complicated to explain. Trivial problem I know the answer. SUPPLEMENTARY R FERENCES R. W. Rousseau. Elementary Principles of Chemical Processes. 3rd ed., New York (2000). Wenzen. Chemical Process Analysis: t're:nUc:e-.tlall, t:<.n~'1ewooo Cliffs, N.J. (1988). ".TEII"" V" Balances. Reklaitis, E. V., and Schneider. lmroduction to Material and ~--~--.. Balances. John Wiley, New (1983). Shaheen, L Basic t'ra~ctu.~e of Chemical Engineering. Houghton 1.11.1'1"1> .. Alto. CA (1975).
- 164. CHAPTER 6 INTRODUCTION TO MATERIAL BALANCES 6.1 The Concept of a Material Balance 6.2 Open and Closed Systems 6.3 Steady-State and Unsteady-State Systems 6.4 Multiple Component Systems 6.5 Accounting for Chemical Reactions in Material Balances 6.6 Material Balances for Batch and Semi-Batch Processes Your objectives in studying this chapter are tQ be able to 1. Understand the features of open, closed, steady-state, and unsteadystate systems, and given a process in words or pictures, select the appropriate categories for the process. 2. Express in words what the material balance is for a process involving single or multiple components. 3. Determine whether positive or negative accumulation occurs in a process. 4. Understand the manner in which a chemical reaction affects the material balance. 5. Recognize a batch or semi-batch process and write the material balance for it. 134 136 138 144 149 151 Why spend time studying an introduction to material balances? Why not just start solving problems? You will find this chapter essential to all of the subsequent chapters because it explains the specialized tenninology ("jargon") associated with making material balances. The chapter also eases you into the subject of material balances by using elementary examples so that you can proceed to more complex problems wi thout stumbling. 133
- 165. 134 Introduction to Material Balances Cha.p.6 Looking Ahead This chapter introduces the concept of material balances, and demonstrates their application different types of processes. We will use severa! simple systems to do so. Using these examples, we will construct a general material balance equation that you can apply problems in remainder of this text and in your professional career. Most of the principles we consider in this chapter are of about the same n .... nr,.p.P complexity as the law of compensation devised by some unknown, self-made philosopher who said: "Things are generally made even somewhere or some place. Rain always foHowed by a dry speU. and dry weather follows rain. I have found it an invariable rule that when a man has one ~hort leg. the other is always n 6.1 The Concept of a Material Balance Success if> "j()urney. 1I0! a destination. Jerry Baird What are material balances? A material balance nothing more than the application of the law of the conservation of mass: "Matter neither created nor strayed." Although the conservation of mass is a simple concept, detailed explanations are required to enable you to be able to apply it to a fuB range of chemical engineering systems. You can a good idea what is involved in making a material balance by looking at a bank statement such as the one below, which illustrates the entries in a checking account. BANK OF THE WEST CUSTOMER SUMMARY INFORMATION Date Notes Deposit Withdrawal Balance 3/1 $1253.89 Deposit from ABC Co. $1500.00 $2753.89 3/3 Check No. 21 $550.00 $2203.89 A TM withdrawal 3/2 $200.00 $2003.89 3/5 No. $401.67 $1602.22 3115 Check No. 21 $321.83 $1280.39 3118 ATM withdrawal 3117 $200.00 $1080.39 3120 Deposit at the bank $1250.00 $2330.39 3/23 No. 2136 $1887.72 3/31 Service charge $10.00 $)877.72 3/31 C)osing balance $1877.72
- 166. -' Sec. 6.1 The Concept of a Material 135 The account initially a balance of $1 which is the initial condition of account. Deposits (what goes in, the inputs) are added to the ac-count, and withdrawals (what goes out, the outputs) are subtracted from the account. The (mal condition of account is the balance ($1877.72). Can you an equation that the closing in terms of and the deposits and withdrawals? Would you agree that the foHowing relation in words infonnation presented in the checking account: Closing balance - balance = Sum of deposits Sum of the withdrawals or Final condition - Initial condition = Sum of inputs - Sum of outputs You can check these by substituting the from the bank statement for words. In this checking account by $633.83, ........... .., .. $1877.72 - $1253.89. indicating an accumulation in the account. cumulation is equal to the final condition minus the condition. Chemical processes are similar in many ways to the checking account Mater- '-' ......... .,"" .... .., pertain to materials rather than money as you can infer from the name, concepts of money and materials are exactly the same. The initial conditions for a the amount of material initially present in the . Deposits into the account are analogous to the flow of material into a process and withdrawals are analogous flow of material out of a ..., ........ ',;)0. Accumulation of material in a process the of material into the process is than the flow out. Here is another example of making money balances in the form of a students rent a room the night before the game, and pay the desk clerk $60. A new clerk comes on duty I finds that the discounted rate for students should have $55. The new the bellhop to return to the students, but the bellhop, not having and being slightly dishonest) returns only $1 to each student, and keeps remaining $2. Now student paid $20 - $1 = $19, and 3 X $19 = $57 paid in The bellhop kept a total of $59. What happened to the other $1 ? Apply what you have learned so far to the puzzle. It provides a good~ simple illustration of the confusion that can cleared up by making a material dollar) balance. (If the puzzle stumps you, look at the end of chapter for the answer.)
- 167. Introduction to Material Balances Chap.S SELF .. ASS SSME'Nf TeST Questions I. What is the difference between the law of the conservation of mass and the concept of the material balance? 2. Derme material balance. Problem 1. Draw a sketch of the foUowing processes, and place a dashed line appropriately to designate the system: a. a tea kettle b. a fireplace c. a swimming pool. Thought Problems 1. Name some quantities besides mass that are conselVed. 2. Mr. Ledger deposited $1000 in his bank account. and withdrew various amounts of money as listed in the following schedule: Wtthdrawals AmountleR $500 $500 250 250 100 150 80 10 SO 20 ---.-lQ ........D Total $1000 5990 The withdrawals total $1000. but it looks as if Mr. Ledger had only $990 in the bank. Does he owe the bank $10? 6 .. 2 Open and Closed Systems In this section we will explain what the terms open and closed systems mean, concepts that are quite important. and will be used extensively in the remainder of this book. . What's in a name? That which we call a rose by any other name would smell as swttt. Wm. Shakespeare in Romeo and Juliet
- 168. Sec. 6.2 Open and Closed Systems System boundary 137 Figure 6.1 A closed system. a. System. By system we mean any arbitrary portion of or a whole process that you want to consider for analysis. You can define a system such as a reactor, a section of a pipe. or an entire refinery by stating in words what the system is. Or. you can define the limits of the system by drawing the system boundary, namely a line that encloses the portion of the process that you want to analyze. The boundary could coincide with the outside of a piece of equipment or some section inside the equipment. Now. let us look at two important classes of systems. b. Closed system. Figure 6.1 shows a two-dimensional view of a three-dimensional vessel holding 1000 kg of H20. Note that material neither enters nor leaves the vessel. that is, no material crosses-the system boundary. Figure 6.1 represents a closed system. Changes can take place inside the system, but for a closed system, no mass exchange occurs with the surroundings. c. Open system. Next, suppose that you add water to the tank shown in Figure 6.1 at the rate of 100 kg/min and withdraw water at the rate of 100 kg/min, as indicated in Figure 6.2. Figure 6.2 is an example of an open system (also called a flow system) because material crosses the system boundary. SELF-ASSESSMENT TEST Questions 1. Is it true that if no material crosses the boundary of a system, the system is a closed system? 2. In an automobile engine. as the valve opens to a cylinder, the piston moves down and air enters the cylinder. Fuel follows, and is burned. Thereafter. the combustion gases are dis-
- 169. 138 Introduction .to Material Balances Chap. 6 charged as the piston moves up. On a very short time scale, say a few microseconds, would the cylinder be considered an open or closed system? Repeat for a time scale of several seconds. Problems 1. Label the materials entering and leaving the systems listed in Problem 1 in Section 6.1. Designate the time interval of reference as short or long, and classify each system as open or closed. 2. Classify the following processes as (a) open, (b) closed, (c) neither, or (d) both for a characteristic operating interval: a. Oil storage tank at a refinery b. Flush tank on a toilet c. Catalytic converter on an automobile d. Gas furnace in a horne 3. As an example of a system, consider a water heater. a. What is in the system? b. What is outside the system? c. Is the system open or closed? 6.3 Steady-State and Unsteady-State Systems In this section we will describe the characteristics of steady-state and unsteadystate systems. The fonner represents the majority of the processes you will encounter in industry, and in the examples and problems in this book. a. Steady-state system. Because the rate of addition of water is equal to the rate of removal, the amount of water in the vessel shown in Figure 6.2 remains constant at its origi- ~ 100 kg H2O min System boundary 100 kg H2O min Figure 6.2 An open steady-state system. , , f
- 170. Sec. 6.3 Steady-State and Unsteady-State systems 139 nal value 0000 kg). We call such a process or system a steady-state process or a steady-state system because 1) the conditions inside the process (specifically the amount of water in the vessel in Figure 6.2) remain unchanged with time, and 2) the conditions of the flowing streams remain constant with time. Thus, in a steady-state process, by definition all of the conditions in the process (e.g., temperature, pressure, mass of material, flow rate, etc.) remain constant with time. A continuous process is one in which material enters and/or leaves the system without intenuption. b. Unsteady-state system. . What if you make a change in the process so that the flow out of the system is instantaneously reduced to a constant 90 kg/min? Figure 6.3 shows the initial condition in the vessel. Because water accumulates at the rate of 10 kg/min (100 kg/min-90 kg/min) in the system, the amount of water present in the vessel will depend on the interval of time for which the rate of accumulation is maintained. Figure 6.4 shows the system after SO minutes of accumulation. (Fifty minutes of accumulation at 10 kg/min amounts to 500 kg of total accumulation.) Because the amount of water in the system changes with time, the process and system are deemed to be an unsteady-state (transient) process or system. For an unsteady-state process, not all of the conditions in the process (e.g., temperature, pressure, mass of material, etc.) remain constant with time, and/or the flows in and out of the system can vary with time. What else might you change about the process we have been analyzing? Suppose you make the flow out of the system 100 kg/min again, and reduce the flow into the system to 90 kg/min. Figure 6.5 shows the initial conditions for this system. Note that the amount of water in the system decreases with time at the rate of 10 kg/min. Figure 6.6 shows the system after 50 minutes of operation. Figures 6.5 System boundary Figure 6.3 Initial conditions for an open unsteady-state system with accumulation.
- 171. 140 I ntroduction to Material Balances Chap.S System boundary Figure 6.4 The condition of the open unsteady-state system with accumulation after 50 minutes. and 6.6 demonstrate negative accumulation, another illustration of an unsteadystate process. From these examples of different systems, let us generalize to obtain a simple but very important equation: In words, the material balance for a single component process is { AccumulatiOnOfmaterial} = {TOtal flow into} _ {TOtal now out} (6.1) within the system the system of the system Figures 6.2 through 6.6 show the mass flows and initial and final conditions. However, Equation (6.1) does not specifically refer to mass but to "material" and "flows." Equation (6.1) can apply to moles or any quantity that is conserved. As an example, look at Figure 6.7 in which we have converted all of the mass quantities in Figure 6.2 to their equivalent values in moles. System boundary 100 kg H~ min Figure 6.5 Initial conditions for an unsteady-state process with negative accumulation.
- 172. Sec. 6.3 Steady· State and Unsteady-State Systems 141 System boundary Figure 6.6 Condition of the open unsteady-state system with negative accumulation after 50 minutes. When you apply Equation (6.1) to a material balance problem, if the process is in the steady state, the accumulation term by definition is zero, and Equation (6.1) simplifies to a famous truism What goes in must come out (6.2) If you are analyzing an unsteady-state process, the accumulation term over a time interval can be calculated as {A ccumulati.o}n = {F.i nal material} - {In.i tial material} m the system m the system (6.3) The times you select for the final and initial conditions can be anything, but you usually select an interval such as 1 minute or 1 hour rather than specific times. When you combine Equations (6.1) and (6.3) you get the general material balance for a component in the system in the absence of reaction 5.55 kg mol H20 min System boundary 5.55 kg mol H20 min Figure 6.7 The system in Figure 6.2 with the flow rates shown in kg mol.
- 173. 142 Introduction to Material Balances Chap. 6 { Final materi~} {Initial material} {FlOW intO} { Flow out Of} in the system - in the system = the system - the system at t1 at tl from tl to ~ from to ~ (6.4) a basis of 1 hour. for example, to solve a problem. the accumulation sum (or integral) of all that has accumulated ~,{IC:tpt"n 1"11'".. ........ the time interval. the accumulation are mass or moles. mass or moles per unit the flows in and out represent the sum (or integral) of all of the flows mass or moles in and out, respectively! the interval, and not the flows which would be called rates. are given in the problem. you by your selected time interval as 1 hour) to get aU of the entering or exit mass or moles. Consequently, (6.4) can deemed to be the result of a material balance if the were fonnulated as a differential equation. to Chapter 32 for a detailed dlSClU;Sl()O material balances as differential equations. EXAMPLE 6.1 Material Balance for the Blendi~ng of lJa.SOllne Will you save money $1.269 per gallon that supreme gasoline at $1 Instead of buying premfum 89 octane gasoline at octane you want, you blend sufficient 93 octane F.'U ... .,il with 87 octane at $1.149 per gallon? Solution problem is an example of applying (6.4) to neither mass nor 1Tlr.iI",c but to ocume number (ON). ON is the engine knocks using the gasoline being tested to the number of knocks per iso-octane in a standardized ON is approximately linearly ON-enhancer concentration in linear relationship is basis a principle that states that the ON of the solution times the volume of the solution is a quantity that is conserved. Chom,e a basis of 1 gallon of 89 octane the desired product. Exam-i ne Figure E6.1. The system is the gasolirle IIfillll,n .... E 6.1
- 174. l Sec. 6.3 Steady-State and Unsteady-State Systems First you have to decide whether the system is open or closed, and it is steady-state or unsteady-state. For simplicityI assume that no gasoline exists in the tank at the start of the blending, and one gallon in the tank at the end of the blending. This arrangement corresponds to an unsteady-state Clearly it is an open system. The flows into the system are the number of (fractional) gallons each of the two of gasoline. The initial number of gallons in the system is zero and the final number gallons is one. Suppose we x the number of gallons 87 octane gasoline added, and y be the number of gallons of 93 octane added to the blend. Since x + y :; 1 is the total flow into tank, you know that y = 1 - x. According to Equation (6.4) the balance on ON is Accumulation Inputs 89 octane 1 87 octane x gal octane (1 - x) gal ------0= --+ I~I 1~ 1~ The solution is x ::: 2/3 gal and thus y 1/3 gal. The cost of blended gasoline is 2/3 ($1.149) + 113 ($1.349)::: $ 1.216 a value less than the cost of the 89 octane gasoline ($1.269). In refineries usually into account the nonlinear blending gasoline (unlike the linear blending considered here) with different octane numbers because one or two of an octane number amounts to a significant amount money because the vol-ume of gasoline they sell. ElF .. A SE SM N T T Questions 1. Is mass conserved within an open process? 143 Without looking at write down the equation that represents a material balance in (a) an open system and (b) a closed system. 3. an accumulation be negative? What does a negative accumulation 4. what circumstances can the accumulation tenn in the material balance be zero for a process? 5. Distinguish a steady-state and an unsteady-state process. 6. What is a transient Is it different than an unsteady-state process? Problem 1. Classify the systems the Self-Assessment Problem #2 Section as (a) steady-state, (b) unsteady-state. (c) or (d) both.
- 175. 144 Introduction to Material Balances Chap.S Thought Problems 1. Examine Figure TP6.3Pl. Paper ashes {1} (2) ( 3) Figure TP6.3Pl A piece of paper is put into the bell in (1), In picture (2) you set fire to the paper. Ashes are left in picture (3). If everything is weighed (the bell, the dish, and the materials) in each of the three cases, you would find: a. Case 1 would have the larger weight. b. Case 2 would have the larger weight. c. Case 3 would have the larger weight. d. None of the above. 2. Certain critical processes require a minimum fluid flow for safe operation during the emergency shutdown of a plant. For example, during normal operation in one process, the chlorine is removed safely from the processing unit along with the flowing liquids. But during an emergency shutdown, the chlorine collects in the unit and its pipeline headers. Hence a minimum flow rate is needed to remove the chlorine. If the unit and pipelines are considered. to be one system. how can a minimum flow rate be obtained for safe operation if the electric power and controner fail? Discussion Problem 1. Why is the transient analysis of a process important in the overall analysis of a process? 6 .. 4 Multiple Component Systems Now, let us examine a slightly more complicated process. AU the Figures so far have illustrated the flow of water-a single component. Suppose the input to a vessel contains more than one component, such as 100 kg/min of a 50% water and
- 176. l Sec. 6.4 Multiple Component Systems 145" 50% sugar (sucrose. C12H22011' MW 342.3) mixture. As indicated in Figure 6.8, the vessel initially contains 1000 kg of water, and the exit stream flow is 100 kg/min of water and sugar. How would the material balance for the process in Figure 6.8 differ from the ones displayed in Figures 6.1 through 6.6? After examining Figure 6.8, would you agree that because the constant total inlet mass flow of 100 kg/min equals the constant total exit mass flow of 100 kg/min, the system can be represented as a steady-state system insofar as a balance on the total mass is concerned? However, with respect to the sugar and water mass balances, balances that we call component balances, the process will be unsteadystate. Sugar starts to accumulate in the system and water is depleted. Now look at the mixer shown in Figure 6.9. an apparatus that mixes two streams to increase the concentration of NaOH in a dilute solution. The mixer is a steady-state open system. Initially the mixer is empty, and after 1 hour it is empty again. We will use the values of the components listed in Figure 6.9 to show you how the total and component balances for mass and moles are made according to Equations (6.4) and (6.1). Because we deemed the process to be in the steady state, we do not have to be concerned about the initial and fmal conditions in the tankthey remain unchanged. Furthennore. if the tank is well mixed, the concentration of a component in the output stream will be the same as the concentration of the component inside the tank during the hour of mixing. Because the flows in Figure 6.9 are expressed in kglhr, we will choose a basis of 1 hour for convenience so that the time variable does not have to be carried along in each mass balance. We multiplied each rate by 1 hour to get the values for the kg listed for each stream in Figure 6.9. As an alternate to the basis we selected you could select FI = 9000 kg/hr as the basis, or F2 = 1000 kglhr as the basis; the num- System boundary 1 00 kg solution min 100 kg min • compo Mass fro compo Mass fro H2O 0.50 H2O ffiH20 Sucrose 0.50 Sucrose ffiSucrose Figure 6.8 An open system involving two components.
- 177. 146 I ntroduction to Material Balances Chap. 6 Feed 1 '" 9000 kg/hr Feed 2 ::: 1000 kglh r Component Mass fro ~ Component Mass fro NaOH 0.050 450 NaOH 0.50 500 H2O 0.950 8550 H2O 0.50 500 Total 1.000 9000 Total 1.00 1000 Product;;;;; 1 0,000 kglhr Component Mass fro kg NaOH 0.095 950 H2O 0,905 Total 1.000 10000 Figure 6.9 Mixing of a dilute stream of NaOH with a concentrated stream of NaOH. Values below the stream arrows are on 1 hour of operation. bers for this example would not change-just the units would change. Here are the component and total balances in Flowm Balances Fl NaOH H2O Total 9,000 F'}. 500 500 1,000 Flow out Accum. =0 =0 10,000 =0 Next, we win show the application of Equation (6.4) tenns of moles for the process in Figure 6.9. We can convert the shown in Figure 6.9 to kg moles by dividing each compound by its respective molecular weight (NaOH = 40 and H20 = 18). I
- 178. Sec. 6.4 Multiple Component 450 ;:: 11.25 500 12.50 950 NaOH: - - - = 23.75 40 40 40 H2O: -- 8550 500 475 - 9050 = 502.78 18 18 18 Then the component and total balances mol are: Flow in Balances F1 Flow out Aecum. NaOH 11.25 23.75 =0 H2O 502.78 Total 486.25 40.28 536.53 EXAMPLE 6.2 Concentration of Using a Centrifuge CeIltn1tugf~s are used to sep1arate particles in the 0.1 to 100 in ameter (a a liquid using force. Yeast cells are recovered from a broth mixture containing using a tubular centrifuge (a cylindrical ~'U~'tprn rOfA!tilll~ about a cylindrical Determine the amount the cell-free dls~=harge per hour if 1000 Uhr is centrifuge, the 500 mg ceBslL, and product stream contains wt. % cells. Assume the feed has a density of Solution problem involves a steady state, open (flow) system without reac::tJon. Two components are involved: cells and broth. a convenient basis 1 hour. Feed (broth) 1000 Uh 500 mg cellsIL The material balance for the mass cells that flow out: Concentrated cells P(g) 50c/o by weight cells discharge O(g) Flllure E6.1 is mass of cells that in must equal the 147 '"
- 179. ./ I J 148 / Introduction Material Balances 1000 L feed 500 mg 1 L feed 1000 mg P==l000g 0.5 g cells P g 19P Chap. 6 The mass balance fluid is the mass fluid that flows in equals the mass fluid that flows out: 1000 em3 1 g fluid 1000 g P 0.50 g fluid ------- ---= + D g fluid 1 L 1 cm3 1 g P = (106 - 500) g ~(jG _ 0 0 S LF",ASS SMENT TEST Questions 1. Does (6.4) apply to a involving more than one component? When a chemical plant or refinery uses various feeds produces various products, does Equation (6.4) apply to each component in the plant? Problems 1. Here a report from a catalytic polymerization and butanes Production: Propane lighter Butane Polymer 15,500 5,680 2,080 What production in pounds per hour of the polymer? 2. A plant 4,000 treated wastewater that contains 0.25 mgIL of PCBs (polychloronated biphenyls) into a that contains no measurable PCBs of the discharge. If flow rate is 1,500 cubic feet per after the discharged water has thoroughly mixed with the river water, what is concentration PCBs in the river mgIL? Thought Problem 1. This story was told to by Professor Woolsey: upon a time a manufacturer of canned orange juice concentrate found that 12 to 1 of the concentrate disappeared some between Florida and the storage warehouse in Northeast. A chemical engineer was on to solve the problem. When she arrived at the plant, the she noticed was a stink caused a number of dead alligators floating on the surface of an adjacent lake. the smell, she carefully
- 180. Sec. 6.5 Accounting for Chemica! Reactions in Material Balances 149 followed the processing of (he orange juice from squeezing, through concentration, and cooling to slush. She watched the waiting being steam cleaned and sanitized, and then the slush being pwnped into the trucks prior to being sent off to the warehouse. The trucks were weighed before and after filling. She placed a seal on one or two trucks, and flew to the warehouse district to wait for "her>~ trucks to come in. On arrival she noted that the seals were undisturbed. A pump was attached to the bottom drain of a truck, and the slush pumped out into a holding tank until the pump started sucking air, at which time the pump was disconnected. Then she watched the cans being fined from the holding tank, counted the cans, had some weighed, and noted a minor amount of spillage, but the spillage and filling operation amounted to less than 1 % of the overall loss. At this point she concluded that either (a) ing was going on in Florida, and/or (b) someone was stealing cases of concentrate from the warehouse. She spent a couple of nights in the warehouse and did note someone take a couple of cases of concentrate, but that amount had negligible effect on the overall loss. Two weeks later while drinldng orange juice at breakfast, she had sudden idea as to what the company's problem was. What was it? Discussion Problems 1. Isotope markers in compounds are used to identify the source of environmental pollutants, investigate leaks in underground tanks and pipelines, and trace the theft of oil and other liquid products. Both radioactive and isotopic markers are used. . Deuterium is typicalIy used as a marker for compounds, Three or more hy-drogen atoms on the organic molecule are replaced by deuterium. However, isotopes of carbon and oxygen can be used. The detection limit of using a combination of gas chromatography and mass spectrometry is about 100 ppb in crude oil and about 20 ppb in refined products. Explain how such markers can be used in chemical processes. 2. Projects to avoid climatic changes engendered by man's activities and in particu-lar the increase in CO2 in the atmosphere include dispersal of sulfate particles in the stratosphere to reflect sunlight and fertilizing the southern oceans with iron to stimulate phytoplankton growth. It is believed that low levels of iron limit the biological productivity of nutrient-rich southern oceans. Adding iron to these waters would increase the growth of phytoplankton, thus reducing CO2 levels in the seawater and thereby altering the CO2 balance between sea water and the atmosphere. What do you think of such a suggestion? 6.5 Accounting for Chemical Reactions in Material Balances Chemical reaction a system requires the augmentation of Equation (6.4) to take into account the effects of the reaction(s). To illustrate this point, look at Figure 6.10, which shows a steady-state system in which HC] reacts with NaOH by the following reaction: NaOH + Hel ~ NaCl + H20.
- 181. 150 Introduction to Material Balances System boundary Chap. 6 100 Umin 100 Umin 1.0 molar HCI 1.0 molar NaOH 200 Umin 1.0 molar NaCI Figure 6.10 Reactor for neutralizing Hel with NaOH. The exit flow of --~--------------- 200 Urnin is a rounded value. Because 100 mol/min of He} react with the stoichiometric amount of NaOH (100 mol/min), complete reaction is assumed to occur. Therefore, neither HC} nor NaOH is present in the product stream leaving the reactor. How should you account for the gain and loss of components in the system? Equation (6.4) must be augmented to include tenns for the generation and consumption of components by the chemical reaction in the system as follows ccomu lati} Input Output on {A I hi h through througb wt nt e = - the system the system sy8te~ . boundaries boundaries { Generation} {consumptiOn} + within the - witbin the system system (6.5) We will defer to Chapter 10 the analysis of material balances for reacting systems. Even though reactors (i.e., vessels that promote chemical reactions) are the heart of most chemical plants, they comprise only a small portion of the total number of process units in most chemical plants. Therefore, a large majority of the process units in the chemical processing industries do not involve chemical reactions and Equation (6.4) can be used. The remainder of this chapter as well as Chapters 7 and 8 will focus on material ba1ances for processes without chemical reaction. SELF-ASSESSMENT TEST Questions 1. Write down the general material balance in words for a process in which reaction occurs. 2. What terms of the general material balance, Equation (6.5), can be deleted if a. the process is known to be a steady-state process. b. the process is carried out inside a closed vessel. c. the process does not involve a chemical reaction. r ! :~ ',' . ~ .
- 182. Sec. 6.6 Material Balances for Batch and Semi·Batch Processes 6.6 Material Balances for Batch and Semi-Batch Processes 1St' A batch process is used to process a fixed amount of material each time it is operated. Initially, the material to be processed is charged into the system. After pr0- cessing of the material is complete, the products are removed. Therefore, if you include the material originally charged into the vessel as well as the material remaining in the vessel after the processing has been completed as part of the system, batch processes fall into the category of closed systems. Batch processes are used industrially for speciality processing applications (e.g., producing pharmaceutical products), which typically operate at relatively low production rates. Look at Figure 6.11a that illustrates what occurs at the start of a batch process. Two separate quantities of material are put into a mixing vessel at the start of the process, 9000 lb of H20 and 1000 lb of NaOH (Figure 6.11a), and after thorough mixing, the final so]ution remains in the system (Figure 6.11 b). From the viewpoint of Equation (6.4), the flows by definition are zero because material does not cross the system boundary. Would you conclude that the accumulation of the total mass in the system is zero? Yes. All that happens in the batch system (in the absence of chemical reaction) insofar as the total mass or moles of material, or the mass or moles of the individual componentst is that the final conditions ----------------------------------------~~~-~--- boundary 1 ----------------------------------------------- ______ 1 Figure 6.118 The initial state of a batch mixing process.
- 183. 152 Introduction to Material Balances System boundary Chap. 6 Figure 6.l1b The final state of a batch mixing process. are the same as the initial conditions. Chapter 28 treats the heating effect of the solution of NaOH in the H20. Can you transfonn the analysis of a batch process, as shown in Figure 6.11, so that you can treat the system as a open flow system? You can if you take a different perspective of the process. Let us assume that the NaOH and H20 are fed at nonconstant rates (all at once) into the system that contains zero initial material. Let us also assume that the product is removed at a nonconstant rate (all at once) so that the final conditions in the tank are the same as the initial conditions-nothing exists in the tank. Then the process represents an open system. Look at Figure 6.12. The inputs are the same as the initial amounts charged to the system. In Equation (6.4). the flows correspond to the accumulated total of the rates of flow over time, not the rates themselves. Hence. it is quite correct to imagine that the numerical values of the accumulated material are the same as the values that resulted from the instantaneous injection of NaOH and "20. Similar concepts apply to the output flow of the NaOH solution. We can summarize the hypothetical operation of the batch process as a flow system as foUows: Final conditions: All values = 0 Flows out: NaOH = 1,000 lb H20 = 9,000 Ib Total = 10.000 lb Initial conditions: All value = 0 Flows in: NaOH = 1.000 lb H20 = 9,000 lb Total = 10,000 Ib
- 184. Sec. S.6 Material Balances for Batch and Semi·8atch Processes ,.. _______ • _ _ _ _ _ _ _ _ _ _ System I : / boundary , I , I I f - .. - .. - - - - - - - .. - - - - - - _I 1OO0lb 100% NaOH 10,ooOIb 90% H20 10% NaOH Figure 6.12 The batch process in Figure 6.11 represented as an open system. 153 In a semi-batch process material enters the process during its operation, but does not leave. Instead mass is allowed to accumulate in the process vessel. Product is withdrawn only after the process is over. Figures 6.13 illustrates a semi·batch mixing process. Initially the vessel is empty (Figure 6.13a). NaOH (1000 lblhr) and H20 (9000 lblhr) are fed to the vessel, and the mixture (10% NaOH) accumulates in the vessel. Figure 6.13b shows the semi-batch system after 1 hour of operation. Semi-batch processes are open and unsteady·state. Can you apply Equation (6.4) to a semi·batch system? Of course. Only flows enter the systems, and none leave, hence the system is an unsteady state-one that you can treat as having continuous flows, as follows: Final conditions: NaOH = 1,000 lb H20 = 9,000 Ib Total = 10,000 lb Initial conditions: All values = 0 Flows out: All values = 0 Flows in: NaOH = COOO Ib H20 = 9,000 lb Total = 10,000 Ib You can see that all of the numerical values we have been using in this section are the same. Only the perspective of how the values are classified with respect to the tenns in Equation (6.4) discriminate among the analyses.
- 185. , 154 Introduction to Material Balances Chap. 6 System boundary 1000 Ib/hr NaOH 100% NaOH Figure 6.138 Initial condition for the semi-batch mixing process. Vessel is empty. System boundary Figure 6.13b Condition of a semibatch mixing process after I hour of operation. EXAMPLE 6.3 Discharge of Tank Residuals to the Environment Material lost in cleaning drums and tanks after their contents have been discharged not only represents economic loss but also possible environmental problems. After emptying, before tanks are flushed with water, a certain percent of the tank contents remains on the inside surface of the tank. A measurement for water flushing of a steel tank. originally containing motor oil showed that 0.15 percent by weight of the original contents remained on the interior tank surface. What is the fractional loss of
- 186. Sec. 6.6 Material Balances for Batch and Semi-Batch Processes oil before flushing with water, -and the pounds of discharge of motor oil into the environment during cleaning of a 10,000 gal tank truck that carried motor oil. Solution Basis: 10,000 gal motor oil at an assumed 77°P The density of motor oil is about 0.80 g/cm3. The initial mass of the motor oil in the tank was . 10,000 gal 0.1337 ft3 62.41b water 0.80 g/cm3 oil ---- ---- ----- -----=.--- = 66.700 lb 1 gal 1 ft3 water 1.00 g/cmJ water The mass fractional loss is O.OlliS. The oj} material balance is Initial 66,700 unloaded = 66,700(0.9985) + residual discharged on cleaning 66,700(0.0015) Thus, the discharge on flushing is 66,700 (0.0015) = 100 lb. SELF-ASSESSMENT TEST Questions 1. What is the difference between a batch process and a closed process? 2. What is the difference between a semi-batch process and a closed process? 3. What is the difference between a semi-batch process and an open process? Problems 1. Classify the following processes as (a) batch, (b) semi-batch, (c) neither. and (d) both: a. a section of a river between two bridges h. a home water heater c. a reaction carried out in a beaker d. preparing a pot of chili e. water boiling in a pot on the stove ,,- 155 2. How would yo~ proceed in the analysis of a steady-state system as an unsteady-state system? Dlscussfon Problem 1. How much time must elapse before a batch process becomes a semi-batch or open process?
- 187. 156 Introduction to Material Balances 6 Looking Back In addition to introducing you to the concept of material balances, we <·, ... l .... -~· .. some of the terminology that you will be using in solving material balance lerns. The following list summarizes the types of material introduced in this chapter: mass balance Total mole balance Component mass balance Component mole balance Other balances win discussed in later chapters. LOSSARY OF N W WORD Accumulation An increase or decrease the material (e.g., mass or moles) in the system. Batch process A process in which material neither added to nor removed from the process during its operation. CJosed system A system that does not have crossing the system boundary. Component balance A balance on a single chemical component in a sys-tern. Conservation or mass Matter is neither nor destroyed overalL Consumption The depletion of a component a system due to chemical reaction. Continuous process A process in which .. "' ........... enters andlor continuously_ condition The amount of material (e.g., mass or moles) in process at the of the processing intervaL Flow An open with material andlor leaving. Generation The appearance of a component in a system because of chemical reaction. Initial condition The amount of a material mass or moles) in process at beginning of the processing intervaL Input Material (e.g., that ....... y"' ..... Material balance The balance equation that corresponds to the conservation of mass. Negative accumulation A depletion of material (usually mass or moles) in the system. Open system A system in which material crosses the system boundary. Output Material (e.g.'! mass. that leaves the
- 188. Sec. 6,6 Material Balances for Batch and Semi-Batch 157 Rate Flow unit Semi-batch process A process in which material enters the system but product is not removed during operation. Steady-state system A system for which all the conditions (e.g., temperature, pressure, amount of material) remain constant with time. System Any arbitrary portion of or whole process that is considered for analysis. System boundary The closed line that encloses the partion of the process that is to be analyzed, Transient system A system for which one or more the conditions (e.g., temper-ature, pressure, amount of material) of the system vary with time. Also known as an unsteady-state system, Unsteady-state system system for which one or more of the conditions (e.g., pressure, amount of material) of the system vary with time. Also known as a transient system .. UPPLEMENTARV R F R NCES In addition to the general references listed in the Frequently Asked Questions in the front materia], the following are pertinent: CACHE Corp. Material and Energy Balances 2,0 (CD). Austin, TX (2000). Fogler, S., and M. Montgomery. Interactive Computer Modules for ChE. Instruction, Corp., Austin. TX (2000). Veverka, V. V" and F. Madron. Material and Energy Balances in the Process Industries: From Microscopic Balances to Large Plant, Elsevier Science, Amsterdam (1998). Web Sites http://www.outokumpu.filhsclwhats_new_heat.htm http://voyager5.sdsu.edultestcenter/features/cgprocessJindex www.chemeng.ed.ac.uV-jwpIMSO/sectionllbalance.html Answer to the puzzle The additions and subtractions made in the puzzle are invalid. A diagram of the flow of money shows the transfers (amounts on the arrows) and final conditions (the amounts within the boxes): You can see that the money balance should be made for the net resul among the five individuals in the boxes, and not of flows between the boxes: $55 + $2 - ($19) (3) = 0
- 189. 158 Introduction to Material Balances Chap. 6 $1 Student A -$19 $20 $1 Student B $20 Room $5 Bell -$19 clerk hop +$55 +$2 ~'2.() $1 Student C $3 . r $19 " Figure 6.14 PROSl MS ·6.1 A manufacturer blends lubricating oil by mixing 300 kg/min of No. 10 oil with 100 kg/min of No. 40 oil in a tank The oil is well mixed, and is withdrawn at the rate of 380 kg/min. Assume the tank contains no oil at the start of the blending process. How much oil in the tank after one hour? ·6.2 One huncJred kg of are dissolved i,n 500 kg of water in a shallow open cylindri-vessel. After standing 10 days, 300 kg of sugar solution are removed. Would you expect the remaining sugar solution to have a mass of 300 kg? *6.3 A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the solid iodine together weigh 27.0 grams. Figure Pti.3 The tube is then heated until all of the iodine evaporates and the tube is filled with io· dine gas. The weight after heating should be: (a) less than 26.0 grams. (b) 26.0 grams. (c) 27.0 (d) 28.0 grams. (e) more 28.0 grams. Mixers can be used to mix streams with different compositions to produce a product stream with an intermediate composition. Figure P6,4 shows a diagram of such a mixing process. Evaluate the closure of the overall material balance and the component material balances this process. Closure means how closely the inputs agree with the outputs for a steady state process.
- 190. l Chap. 6 Problems Figure P6.4 39.800 kglh 11.8 wi% NaCI 88.2 wi°/o H20 159 $6.5 Heat exchangers are used to transfer heat from one fluid to another fluid. such as from a hotter fluid to a cooler fluid. Figure P6.5 shows a heat exchanger that transfers heat from condensing steam to a process stream. The stearn condenses on the outside of the heat exchanger tubes while the process fluid absorbs heat as it passes through inside of the heat exchanger tubes. The feed rate of the process stream is measured as 45,000 lblh. The flow rate of steam is measured as 30.800 Iblh. and the exit flow rate of the process stream is measured as 50,000 Iblh. Perform a mass balance for the process stream. If the balance does not close adequately, what might be a reason for this discrepancy? Stream Heated Process Stream Steam Condensate Figure P6.S "'6.6 Distillations columns are used to light boiling components from heavier boiling componentsi and makeup over 95% of the separation systems for the chemi-process industries. A commonly used distillation column is a propylene-propane splitter. The overhead product from this column is used as a feedstock for the produc. tion polypropylene, which is the largest quantity of plastic produced worldwide. Figure P6.6 shows a diagram of a propylene-propane splitter (C3 refers to propane and refers to propylene), steam is used to provide energy, is not involved in the material balance). Assume that !:he composition and flow rates listed on this diagram came from process measurements. Detennine if the overall material balance is for this system. Evaluate the component material balances as well. What can you conclude?
- 191. 160 106,000 Iblll 70 wF/o c; 30 wl-l% C3 Introduction to Material Balances _-+--r:;acl- Steam ~- Figure P6.6 34.oo0Iblll 10 wt"/., Ci 90 wF/o Ca Chap. 6 ··6.7 Examine the flow sheet in Figure P6.7 (adapted from Hydrocarbon Processing, November 1974. p. 159) for the atmospheric distillation and pyrolysis of all atmospheric distillates for fuels and petrochemicals. Does the mass in equal the mass out? Give one or two reasons why the mass does or does not balance. Note: TI A is metric tons/year. Hydrogen oil x 108 T/A 50,000 BPD Atmospheric Methane distillation Steam reforming ATM Hydro-gas oil desulfurizatlon Hydrodesulturlzatlon FigureP6.1 Pyrolysis Aromatics recovery Propylene Mixed ~ Benzene Toluene Xylenes Gasoline Heavy fuel oil 0.7%S
- 192. Chap. 6 Problems 161 **6.8 Examine the flow in Figure P6.8. Does mass in equal the mass out? Give one or two reasons why the mass does or does not balance. Naphtha 250,000 ton Steam - crocking CH4 ,H2 39,500 ton Ethylene 55,000 ton Propylene 45,000 ton C" 30,700 ton Aromatics 68,000 ton Combustibles Polyethylene (30,000 tonl Styrene Polystyrene (5,000 ton 1 P. V. C. (40,000 toni Acrylonitrile (20,000 fon) Oodecylbenzene (B,OOO ton) Phenol (10,000 ton) Acetone (5,750 ton) Rubber S.B.R. (10,000 ton) L.P.G. (24,000 ton) Pentones Extraction of aromotics Aromatics (48,000 ton) Fuel-oil '---------:--------Combustibles 3,800 ton Figure P6.S ··6.9 Examine Figure P6.9. the material balance satisfactory? (T/wk means tons per week) 416.10 Examine Figure P6.10 (adapted from Environ. Sci. Technol., ,.7 (1993, p. 1975). What would be a good system to designate for this bioremediation process? Sketch the process and draw the system boundary. Is your system open or closed? Is it steady or unsteady state? *6.11 Examine Figure 11 of a cylinder that is part of a 2.9 liter V-6 engine. Pick a system and srate it.
- 193. Munlciple solid waste Particles to the atmosphere 3.8 t/wk .... ------- I I I I I 19,000 liters of recycled water I I Quench water: undissolved solids 275 t/wk t I Quench water: dissolved solids 920 tJwk Gaseous emissions 20 tlwk 620 t/wk ---+ I I I t 114.000 liters of I jaSh I Landfill I Spray chamber water: undissolved solids recycled water O.S I I I t/wk Spray chamber water: dissolved solids 0.01 0.01 0.08 1.1 t/wk t/wk 1/wk t/wk ~'--------I.~I tuniCIPI. sewer systel~~I I---------'~ FlgureP6.9 A system for tMating soil above the water table (bloventlng). Vacuum pump Batch feed: Vacuum pump Figure P6.10 162 J
- 194. 6 Problems 163 Figure Show by a crude system boundary. whether your system is a flow or a batch why (in one sentence). *6.12 State whether the following processes open or dosed in making material balances. (a) The cycle of (b) The cycle for a forest. (c) An motor for a boat. Cd) home air conditioner with respect to the '-"'"'''', .. "6.13 each one of the following scenarios. State what the system is. Draw the picture. each as belonging to one or more of [he following: open system, sys-steady- state unsteady-state process. (a) You fill your car radiator with coolant (b) You drain your car radiator. (c) You overfill the car radiator coolant runs on the (d) The radiator is full and the water pump circulates water to and from the while the engine is running. *6.14 State whether the following processes represent open or closed systems, explain answer very briefly. (a) swimming pool view point of water) (b) home furnace State whether of a block melted by the sun the ice) is an open or system, batch or flow, and steady state or state. List the choices a vertical list, and state beside entry any assumption you make.
- 195. 164 Introduction to Material Balances Chap. 6 -6.16 Examine the processes Figure P6.16, Each box represents a system, For each, state whether: (a) process is the (1) steady state, (2) unsteady state. or (3) unknown condition (b) The system is (1) closed, (2) open, (3) neither, or (4) both The wavy line represents the initial fluid level when the flows begin. In case (c), the tank stays full. F-~ (0) *6.17 Pick the correct answer(s): For a steady state system (a) rate of input is zero (b) The rate of generation is zero (c) The rate of consumption is zero (d) The rate of accumulation is zero (b) Figure P6.l6 --... p Tank 15 initially full (c) ·6.18 Consider a hot water heater in a house. Assume that the metal shell of the tank is the system boundary, (a) What is in system? (b) What is outside system? (c) Does the system exchange material with the outside of the system? (d) Could you pick another system boundary? "'6.19 Explain why the total moles entering a process may not equal the total moles leaving, ;·6.20 SUicon rods used in the manufacture of chips can be prepared by the Czochralski (LEe) process in which a cylinder rotating silicon is slowly drawn from a heated bath. Examine Fig. P6.20, If the initial bath contains 62 kg of silicon. and a cylindrical ingot 17.5 em in diameter is to be removed slowly from the melt at the rate of 3 mm minute, how long will it take to remove one-half of the silicon? What is the accumulation of silicon in the melt? I
- 196. l Chap. 6 Problems 1-4----- ---1-- Crystal-puller Emerging --+---... shaft crystal ..... I------+----"""'rt-- Seed crystal ~====L---..:t::::====:}4~rt-- 82°3 liquid Crucible --t--f""'l'-+-l encapsulant ;,:j'---::::::t+- Melt Heater Figure P6.20 165 *6.21 A thickener in a waste disposal unit of a plant removes water from wet sewage sludge as shown in Figure P6.21. How many kilograms of water leave the thickener per 100 kg of wet sludge that enter the thickener? The process is in the steady state. 100 kg 70 kg ------------~, ~ Thickener ,~ Wet Sludge L-------r----I Dehydrated 'Sludge ,V Water =? Figure P6.21 '6.22 In making a material balance, classify the following processes as (a) batch. (b) semibatch, (c) continuous, (d) open or flow, (e) closed, (f) unsteady state, or (g) steady state. More than one classification may apply. 1. A tower used to store water for a city distribution system 2. A can of soda 3. Heating up cold coffee 4. A flush tank on a toilet 5. An electric clothes drier 6. Waterfall 7. Boiling water in an open pot *6.23 Under what circumstances can a batch process that is camed out repeatedly be considered to be a continuous process.? *6.24 How long a time should elapse before a batch or semi-batch process can be considered an open process?
- 197. CHAPTER 7 A GENERAL STRATEGY FOR SOLVING MATERIAL BALANCE PROBLEMS 1.1 Problem Solving 7.2 The Strategy for Solving Problems Your objectives in studying this chapter Bre to be able to: 1. Comprehend and execute the 10 elements of effective problem solving. 2. Understand and apply a degree of freedom analysis. 161 168 The difference between the almost right strategy and the right strategy is really a large matler-it's the difference between a lightning bug and lightning. A paraphrase of Mark Twain What is the purpose of having a chapter that discusses how to solve material balance problems? The idea is to provide you with guidelines so that you can be efficient and effective in solving material balance problems. If you begin to apply the suggested strategy at oncet as you proceed through this book you win discover by the end that you have become quite skilled at problem solving. If you get stuck in solving a problem or bungle it, you can rebound and recover by applying the proposed strategy. If you are going to learn from your mistakes, let's find out what they are first. Looking Ahead This chapter presents a comprehensive methodology that you can use to solve material balance problems. We are going to describe a strategy of analysis that win help you understand, first, how similar material balance problems are, and second, 166
- 198. l Sec. 1 Problem Solving how to solve them in the most expeditious manner. You do not have to memorize strategy. you consistently use it in developing your skills, you will learn it by absorption. 7.1 Problem Solving Most of the literature on problem solving view a "problem" as a gap between some initial information (the initial state) and the desired information (the final state). Problem solving the activity dosing the gap between these two states. If you are going to become a professional, you will have to acquire a number of skills problem solving such as: III formulating specific questions from vaguely specified problems; • selecting problem -solving strategies; III deciding when an estimate will versus an exact answer; 1 using tables, graphs, spreadsheets, calculators, and computers to organize. solve, and interpret the results from solving problems; • judging the validity of the work of others; and 1 evaluating answers .. You will find as you go through this book that routine substitution of data into an appropriate equation will not adequate to solve material (and energy) balances other the most trivial ones. You can, of course, fonnulate your own strategy for solving problems-everyone has a different viewpoint. But adoption of the welltested general strategy presented in this chapter has been found to significantly ease the difficulty students have when they encounter problems not exactly the same as those presented as examples and homework in this book, or problem~ in industrial practice. After all, the problems in this book are only samples, and simple ones at of the myriad prob1ems that exist or could be fonnulated. An orderly method of analyzing problems and presenting their solutions represents training in logical thinking that is of considerably greater value than mere knowledge of how to solve a particular type of problem. Understanding how to approach these problems from a logical viewpoint will help you to develop those fundaJ1}6ntals of thinking that will assist you in your work as an engineer long after you have this material. Keep mind old Chinese proverb: None of the secrets of success will work unless you do. When solving problems, either academic or industrial, you should always use "engineering judgment" even though much of your training to date treats problems
- 199. 168 A General Strategy for Solving Material Balance Problems Chap. 7 as an science. instance, suppose that it takes one man 10 days to build a brick waH; then 10 men can finish it in one day. Therefore, 240 men can finish the wall in 1 hr, 14AOO can do the job in 1 min$ and with 864,000 men the waH will be up before a single brick is in place! Your password to success to use some common sense in problem solving and always maintain a mental picture of system that you are analyzing. Do not allow a problem to become abstract and unrelated to physical behavior. 7.2 The Strategy for Solving Problems Howe's Law: Every person has a scheme that will not work. Gordon's Law: If a project is not worth doing, it is not worth doing well. You neither have to follow the steps in the list below in any particular sequence nor formally employ every one them. You can go back several steps and repeat steps at will. As you might expect, when you work on solving a problem you will experience fa]se starts, encounter extensive preliminary calculations, suspend work for higher priority tasks, look for missing links, and make foolish mistakes. The strategy outlined below is designed to focus your attention on the main path rather than the detours. I 1. Read and understand the problem statement This means read the problem carefully so that you know what given and what to be accomplished. Rephrase the problem to make sure you understand it. An anecdote illustrates the point of really understanding the problem: An English family visiting Khartoum in the Sudan took their young son day by the statue of General Gordon on a camel. On the day of their visit to the statue as the family was leaving. the boy asked, "Who was that man sat on General Gordon?" is a question to answer: How many months have 30 days? Now you remember the mnemonic: 30 days hath September ... and give the answer as four, but is that what the question concerns-how many months have exactly 30 days? does the question ask how many months have -at least 30 days (with the answer being II)?
- 200. '. Sec. 7.2 The Strategy for Solving Problems 16!r sure to decide a problem is some simple calculation or involves a steady· state or unsteady.state Vv.,. ... , and state your assumption the top of your calculation EXAMPLE 7.1 Understanding the Problem A train is approaching the station at 1 cm/s. man in one car is walking forward at 30 cmfs relative to the seats. He is eating a foot-long hot dog, which is entering his mouth at the rate of 2 cm/s. An ant on the hot dog is running away from the man's mouth at 1 emJs. How is the ant approaching the station? Cover up the solution below, and try to determine what the problem requests before peeking. Solution As you read the problem make sure you understand how each of infor-mation meshes with the others. Would you that the following is the correct analysis? superficial analysis would take care to ignore the hot dog length but would calculate: 105 + 30 - 2 + 1 :: 134 cmls for the answer. Howe~r, the problem states on more careful reading that the ant is moving away from the mouth at the rate of 1 emls. Because the man's mouth is moving toward the station at the rate of 135 em/s, the ant is moving toward the station at the rate of 1 em/s. 2. Draw a sketch of the process and specify the system boundary In Chapter 2 we showed how a process can be represented by simple diagrams, and you have seen numerous diagrams in subsequent chapters, It always good practice to begin solving a problem by drawing a sketch the process or physical system. You do not have to be an artist to make a sketch. A simple box or drawn by hand to denote the system boundary with some arrows to designate flows material will be fine. You can also state what the system is words or by a label. EXAMPLE 7.2 Drawing a Sketcb ofa Mixing Process A continuous mix.es NaOH with H20 to produce an aqueous solution of NaOH. Detennine the composition and flow rate of the product if the flow rate of NaOH is 1000 kg/tn, and ratio of the flow rate of H20 to the product solu· tion is 0.9. We will use this example in subsequent illustrations of the proposed strategy. For this example, just a of the process is required.
- 201. 170 A General Strategy for Solving Material Balance Problems Chap. 7 Solution The process is an open one, and we assume it to steady state (nothing spe-cific is mentioned in the problem statement, steady state implied). We pick the mixer as the system. The diagram below probably looks nicer than one you would draw by hand. Mixer Product Figure System boundary NaOH 3. Place labels (symbols, numbers, and units) on the diagram for all of the known flows, materials, and compositions. For the unknown flows, materials, and compositions insert symbols and units. Add any other useful relations or information. By putting on the diagram you will avoid having look back the problem statement repeatedly, and you will also be able to clarify what data are What kinds of information might you place on the diagram? Some specific examples of information that you might put on the diagram are: .. Stream flow rates ( = 100 kg/min) } the essential information II Compositions of each stream (XHlO = 0.40) .. Given flow ratios (FIR = 0.7) • Given identities (F = P) • Yields (Y kg/X kg:::: 0.63) .. Efficiency (40%) • Specifications for a variable or a constraint (x < 1.00) .. Conversion (78%) .. Equilibrium relationships (y/x = 2.7) .. Molecular weights (MW :::: 129.8)
- 202. Sec. 7.2 The Strategy for Solving How much data should you place on the diagram? Enough to solve the p~oblem and interpret the answer. your diagram becomes too crowded with data, make a separate table it to diagram. Be sure to include the units associated with the flows and other material when you write the numbers on your diagram or in a table. The units make a difference in your thought processes. Some of the essential data may be missing from the problem statement If you do not know of a variable to put on the figure, you can substitute a symbol such as Fl, F1, or for an unknown flowrate or XI for a mole tion. Substitution for a number will focus your attention on jng for the information needed to solve the problem. EXAMPLE Placing the Known Information on the Diagram ...... "' .... ul-" • ..., 7.2, place the known information on diagram of Solution ",,,,aUi>"-' no contrary information is provided about the cornO()SltlOn H20 and we will assume that they are 100% H20 and spectively. Figure E7.3 for a typical way the data might be System boundary System Mixer / P kg Component NaOH Figure E7.3 F", 1000 kg NaOH 100% p OONaOH ::: p
- 203. 172 A General Strategy for Solving Material Balance Problems Chap_ 7 Note that the composition of the product stream is listed along with the symbols for unknown nows. Could you have listed the mass fractions instead of or in addition to the mass flows? Of course. Look to the right of the column headed "kg." Because you know the ratio WIP = 0.9, why not add that ratio to the diagram at some conve~ nient place? You win find it convenient to use a consistent set of algebraic symbols to represent the variables whose values are unknown (caned the unknowns) in a problem. In this book we frequently use mnemonic letters to represent the flow material, both mass and moles, with the appropriate units attached or inferred, as illustrated in Figure . By flow we mean a certain amount of material, not a rate of/low (that involves time), A rate of flow denoted by an overlay dot. such as F. We usually employ m for the flow of mass and n for the flow of moles with appropriate subscripts andlor superscripts to make the meaning crystal dear. Table 7.1 lists some examples. You can use any symbols you want, but it helps to be consistent. In specific problems pick obvious or mnemonic letters such as W for water and P for product to avoid confusion. If you run out of suitable letters of the alphabet, you can always insert superscripts to distinguish be· tween streams such as FI from f11, or label streams as and In the beginning there was the symbol. David Hilbert TABLE 7.1 Some Examples of the Symbols Used in This Book Symbol F FTolal or FTm FI or FI FA Ib mA m-rolill or mTOI Flow of mass in kg Total flow of material· Flow in stream number I'" Designates Flow of component A in stream F in Ib Mass flow of component A· Mass flow of the total material" Mass flow of component A stream FI" Molar flow of component A in stream w* mass (weight) fraction of A stream F. (The superscript is not required if the meaning is otherwise clear.) mole fraction of A in stream F. a liquid. (The superscript is not required if the meaning is otherwise clear.) The mole fraction of A in stream F, usuaUy a ·Units not specified but inferred from the problem statement
- 204. 7.2 The Strategy for Solving Problems 113 4. Obtain any data you need to solve the problem, but are missing Never assume obvious is true William in the Sleeping (Random House) An costs $34,700. How much did it cost pound? Clearly. something missing from the statement. Table 7.2 a clever list of the degrees of ignorance. Look at Table 7 and decide what your level of ig-norance is for the evaporator problem. you pick I? Hopefully you are not at 2! You have to find out weight of the evapora-tor is. TABLE "....,.,n .... ·'"' Laws of Ignorance o I 2 3 4 (P.G. Armour, Commun. ACM, 44, (2001) Order of frnlI"QTI,f'.I> Lack of awareness Lack of n .. t'V~"'(,(, Meta-ignorance State of Mind You know ~~...., ........ You don't know something You don't don't know something You don't know an efficient way to find out that you don't know that you don't know something. You don't about the five orders of ignorance. Another example is: How do you pronounce the name of Kentucky: uLoo-EE-ville" or "Loo-ISS-ville"? If you pick one have demonstrated at Level2! Hint a When you a immediately notice ·that some es-sential detail, such as a physical property (molecular weight, density, etc.), is in the problem statement. You can look the values up in a physical properties database as the one on the that accompanies this book, in reference books, on Web, and many other places. Or some value be but you can calculate the value in your head. For example~ you are a stream flow that contains just two one is H20 the other NaOH. You are the concentration the NaOH as 22%. no point in writing down a symbol on the the unknown concentra-of water. Why not calculate the value of 78% in your head, and put that value on the diagram. 5. Choose a basis We discussed the topic of basis in Chapter 3 of selecting a basis: we suggested
- 205. 174 A General Strategy for Solving Material (1) What do I have (2) What do I want to find, (3) What is convenient Problems Chap. 7 Although ..... n ... "' ...... know what a is Step 5 in the proposed strategy t you to pick immediately after problem statement, and can value on your calculation page What basis would you for the problem stated in 7.27 Wouldn't you pick one of the fonowing (all are equivalent)? 1000 kg 1 hour 1000 kg/hr make the material balance in of mass (or moles), or in of a namely mass (or moles) per unit, time, makes no essential dif-ference. Mass (or moles) is balanced. If you use a rate for the balance, you do have to carry along the dangling equations. Be sure to write the word on you ca1culation page, the value and associated units so that you, and anyone who reads the can later know what you 6. Determine the number of variables whose values are unknown (the unknowns) Plan ahead Unknown If you put symbols on the diagram as described in Steps 2~ 3, and 4 above, or make a list them, determining the number of unknowns is easy. Count them. In stated in Example 7.2 from which Figure E7.3 was prepared, unknowns exist? We do not know the values abIes: 'W; PNaOH; and PH o. In light of the necessary conditions in the next step, Step you should be thinking about assembling four independent equations to solve the mixing problem. Detennine the number of independent equations and carry out a degrees of freedom analysis through a maze looks easy from above. IMPORTANT COMMENT proceeding with Step 7 I we point from mathematics related to solving mining whether you can actually solve a set to call to your attention an important 6 7 focus on deter-formulated for a material I
- 206. Sec. 7.2 The Strategy for Solving Problems balance problem. For simple problems, if you omit Steps 6 and 7 and proceed directly to Step 8 (writing equations), you probably will not encounter any difficulties. However. for complicated problems. you can easily run into trouble if you neglect the two steps. Computer-based process simulators take great care to make sure that the equations you formulate indeed can be solved. What does solving a material balance problem mean? For our purposes it means finding a unique answer to a problem. If the material balances you write are linear equations (refer to Appendix L if you are not clear as to what linear equation means) as will be the vast majority of the equations you write, then you are guaranteed to get a unique answer if the following necessary conditions are fulfilled: The nwnber of variables whose values are unknown equals the number of independent equations you formulate to solve a problem. To check the sufficient conditions for this guarantee. refer to Appendix L. Frequently Asked Questions 175 1. What does the term independent equations mean? You know that if you add two equations together to get a third one. the set of three equations is said to be not independent; they are said to be dependent. Only two of the equations are said to be independent because you can add or subtract any two of them to get the third equation. For example, you can determine by inspection that the two equations are independent. However. the following two equations 3x, + 4X2 = 0 are not independent (as you can tell by inspection). Consult Appendix L I for a fonnal definition of independence. 2. If I have several equations, how can I tell if they are independent? The best thing to do is use a software program to make the calculations. Matlab, MathCad, Mathematica. Excel. and many statistical programs will do the task for you. For information as to how to use software to determine whether or not equations are independent. refer to Appendix LI. The concept of independence of equations will become clear if you look at Figure 7.1, which shows a set of three equations, only two of which are independent.
- 207. 176 A General Strategy for Solving Material Problems Chap. 7 213 ® x1+ =2 2x1+ 3X1 + :: 4 Figure An of three equations intersecting at a unique Only two of equations are independent. Note that in Figure 7.1 Equation (C) is the sum of Equations (A) (B). Thus, although the three equations give a unique solution, only two can counted as independent equations, and any two can be solved to get the unique solution. For the problem posed in Example 7.2 you can write material balances: ... one for NaOH ... one for the H20 ... one total balance (the sum of the two component balances) Only two are independent. You can use solving the problem. combination of two the three If you are not careful, you can blunder and write independent equations that have no solution. Look at 7.2. Refer to Appendix Ll for a way to detect such a case. If you are careful in writing the component and selecting additional constraints, you win not be bothered by the difficulty inus· trated in 7.2. In Step 7 you want to preview the compilation of equations you plan to use to solve the problem. You want to make sure that you have an appropriate number of independent equations. What kinds of equations should you be thinking about? The first of equations to consider, of contains the material balances. You can write as many independent material balances as there are species involved in the system. In specific case of the problem stated in Example 7.2) you have two species~ NaOH and H20, and can thus
- 208. Sec. 1.2 The Strategy for Solving Problems 177 o 2 )(1 Figure 7.2. independent eqoations having no unique solution. write two independent equations. Step 8 pertains to actually writing the equations, but you can write them down as you name them if you want. In addition to the two species balances) to solve the problem posed in Ex-ample you will need to find two more equations. In general you look for re-lations such as 41 explicit relations specified (the specifications) in the problem statement such as W/P = 0.9 stated in Example 7.2 ., implicit relations, particularly the sum of the mass or mole fractions being unity in a stream; in Example 7.2 you have ~aOH + (()~20 == 1 or, multiplying both sides of the equation by P (the given amount of material) you get the equivalent equation PNaOH + PH~O == P • specified values of variables that are given in the problem statement .. the value of the basis you select Once you have determined the number of unknowns and independent equations, an analysis of whether a problem is solvable or not is called a
- 209. 178 A General Strategy for Solving Material Balance Problems Chap. 7 degree-of-freedom analysis. The phrase degrees offreedom has evolved from the design of plants in which fewer independent equations than unknowns exist. The difference is called the degrees of freedom available to the designer to specify flow rates, equipment sizes, and so on. You calculate the number of degrees of freedom (ND) as follows: Degrees of freedom = number of unknowns - number of independent equations ND = Nu - NE When you calculate the number of degrees of freedom you ascertain the solvability of a problem. Three outcomes exist: Case Nu=NE Nu>NE Nu<NE o >0 <0 Possibility of a solution Exactly specified (detennined); a solution exists Underspecified (detennined); more independent equations required Overspecified (detennined); in general no solution exists unless some constraints are eliminated or some additional unknowns are included in the problem For the problem in Example 7.2. so that From Step 6: Nu = 4 From Step 7: NE = 4 and a unique solution exists for the problem. Another way to calculate the number of degrees of freedom involves starting with all of the variables treated as unknowns (even though some of their values are known). This approach to getting ND avoids missing one or more variables in the analysis if carried out with just the obvious variables selected as unknowns. Look at Table 7.3 in which all of the possible variables involved in the problems in Example 7.2 are listed as symbols, a total of nine. The columns headed "Feed," "Water," and "Product" each correspond to a flow stream in the process, and each row corresponds to a component in the respective stream, or to the total flow. Because the example process we have been using is steady state, you do not have to take into account the variables inside the system. Their values are the same as those of the product stream because the internal material in the mixer is assumed to be well mixed. I
- 210. • ·t' >.'. I· I • 7.2 The Strategy for Solving Problems TABLE 7.3 A List of of the Variables Involved in the Example Problem Feed Water Product Let's analyze Example 7.2 again taking into account of the variables. You can count the number variables in Example 7.2 simply by multiplying the number of rows by the number of columns table. Then for one stream Nsp = the number species Ns = the number of streams Nu = the total number variables and for the whole The value of] in (Nsp + () comes from last row the table listing the totals. What kinds of independent specifications and constraints in general must you consider in arriving at a value for NE? are some common possibilities: a. Specifications and values variables that are given in the problem statement such as I. The ratio of two flow rates is some specific value. The conversion in a IS gwen. 3. The value of a concentration, flow temperature, pressure, den-sity, volume, and so on is given. 4. A variable not present in a stream, hence its value is zero. For the case of the problem in Example 7.2 we have (1) specifications the mass fractions and one specification of the mass flow: F NaOH = 1000 kg (the basis) (1) 1 the variables in Table w~20 = 0, hence FH20 = 0 (2) or. the variable in Table 7.3, ~aOH 0, hence WNaOH 0 (3)
- 211. 180 A General Strategy for Solving Material Balance Problems Chap. 7 (2) One specification of the ratio of two variables~ W/P=O.9 b. Independent material balances that you can write (4) For the example problem, you can write two independent material balances: one for NaOH and one H20. The total balance could be substituted for either one of the two respective component balances NaOH balance H20 balance (5) (6) c. Summation of components, or the mass or mole fractions, in an in-dividual stream In each stream (and inside the system) the sum of mass or mole fractions equals unity. or, the equivalent, the sum of mass or moles of each component in a stream equals the total material in the stream. One such equation exists each stream flowing in and out of the system, and one for the components each phase inside the system. In Example 7.2, three streams exist, hence three more independent equations exist. wf ::::::: I 1 or FNaOH FH10 F (7) w"'Y = I 1 or WNaOH WH20 = W (8) Lwr ::::::: 1 or PNaOH PH20 = P (9) If you count all the equations listed above for the example problem, you get a total of 9. You can then say that ND = Nv - NE = 9 - 9 = 01 where NE the number of independent equations. Whether you introduce some of values of the known specifications (the easy ones) into the problem prior to carrying out the degrees freedom analysis in order to the number of variables to be considered is up to you. The outcome for the degrees of freedom win be the same. EXAMPLE 7.4 Analysis of the Degrees of Freedom A cylinder containing CH4, C2H6, N2 has to prepared containing a CH4 to C2H6 ratio of I to 1. Available to prepare the mixture are (1) a cy linder containing a mixture of 80% N2 and 20% (2) a cylinder containing a mix-ture of 90% N2 and 10% and (3) a cylinder containing pure N2• What is number of degrees of freedom, Le., number of independent specifications that
- 212. Sec. 7.2 The Strategy for Solling Problems must be made. so that you can determine the respective contributions from each cylinder to get the desired composition in the cylinder with the three components? Solution A sketch of the process greatly helps in the analysis of the degrees of freedom. Look at Figure E7.4. F2 C2H6 0.10 N2 0,90 1.00 , F, F4 F;j CH4 0.20 CH, XC~ N2 1.00 N2 0.80 ~ C2HS Xc.zHe 1.00 N2 xt.I2 Figure E7.4 Do you count seven unknowns-three values of xi and four values of F/! How many independent equations can be written? Three material balances: CH4, C2H6• and N2 One specified ratio: moles of CH4 to C2H6 equal 1.5 or (xCH !xc H ) == 1.5 • ~ F'. 4 2 6 One summatIon of mole fractions: ""X;4 = 1 Thus, there are seven minus five equals two degrees of freedom. If you pick a basis, such as F4 = 1. one other value has to be specified to solve the problem to calculate composition of F4. Keep in mind that you must be careful in making any specifications to maintain only independent equations. Avoid transforming one or more independent equations in a set such that the resulting set contains redundant (dependent) equations. Did you notice in the problem formulation that the LX; = 1 equations for F l' F2• and F] were redundant because of the way the spec· ification of the mole fractions was made? 181
- 213. , 1 A General Strategy for Solving Material Balance Problems Chap. 7 EXAMPLE 7.S Analysis of the Degrees of Freedom in the Production of Biomass [n the gTowth of biomass CH1.gOO.5No.16S0.004SPO.OO.55' with the system comprised of the biomass and the substrate, the substrate contains the carbon source for growth, Ca Hp 0,., plus NH3• °2, H20. C021 H3P04• and H2S04- The relations between the elements and the compounds in the (refer to Appendix for more details) are: CHt.sOo.sNo.t ~O.OO4SP 0.0055 NH) Oz CO2 H2O H2SO4 H3P04 C I ct 0 0 1 0 0 0 H 1.8 ~ 3 0 0 2 2 3 0 0.5 'Y 0 2 2 I 4 4 N 0.16 0 I 0 0 0 0 0 S 0.0045 0 0 0 0 0 I 0 P 0.0055 0 0 0 0 0 0 How many degrees of freedom exist for this system (assuming that the values of ct, ~, yare specified)? Solution Based on the given data six element balances exist for the 8 species present, hence the system has two degrees of freedom. However, it turns out for this type of system that experiments show that the change in CHLSOO.sNo.16S0.004SPO.OO55 and the change in CaHpOy prove to related by the amount of biomass present and the maintenance coefficient (the moles of substrate per mole of biomass second) so that the respective quantities cannot be chosen independently. Consequently, with this extra constraint, only one degree of freedom remains to be specified. the basis. 8. Write down the equations to be solved in terms of the knowns and unknowns. Thus [Beatrice] began: "You dull your own perceptions with false imaginings and do not grasp what would be clear but for your preconceptions . .. Once you have concluded from the degree of freedom analysis that you can solve a problem, you are prepared to write down the equations to be solved (if you have not already done so as part of Step 7), Bear in mind that some formulations of the equations are easier to solve by hand, and even by using a computer, than others. In particular, you should attempt write linear equations rather than nonlinear ones. Recall that the product of variables, or the ratios of variables, or a logarithm or exponent of a variable, and so on, in an equation causes the equation to be nonlinear.
- 214. 7.2 The Solving Problems 183" In many instances you can easily transform a nonlinear equation to a linear one. For instance, in the problem posed in Example 7.2, one constraint was that WIP = 0.9, a nonlinear equation. If you multiply both sides of the equation by P, you a linear equation W =: Another of judicious formulation equations occurs in same problem. chose as the variables listed Table 7.3 the mass flows such as WH20 and PHlO for the water in the respective streams. If, instead of PH10, you use as variables the product of the mass fraction of in P times P: and substitute the ~20P for PH20 in the ance for water, instead of having a linear equation and in the material balwater balance F(O) + W(1.000) PH20 you would F(O) W(LOOO) = W~lO P a nonlinear equation (which is why we didn't use W~20 as a variable). With these in mind, you can formulate the set of equations to be used to solve the problem in Example 7.2 as follows. First. you introduce ifications 0), (2), and (3) into the material balances (5) and (6), and into the summation of mass fractions or their equivalents, (7), (8), and (9). Then you will a set of four independent equations in four unknowns, which are the same set of four we introduced at the of Step 7. basis is stiB 1 hr (FNaOH = 1000 kg) and the process been assumed to be at steady state. Recall from Chapter 6 that in such circumstances a material balance simplifies to in = out or in - out = O. The equations are NaOH balance: 1000 = PNaOH or 1000 - P NaOH = 0 (1) H20 balance: W = Pu,o or W - PHlO = 0 (2) Given ratio: W = O.9P or W - D.9P = 0 (3) of components in P: PH10 = Por PNaOH + PH10 - P = 0 (4) Could you substitute total mass balance ] 000 + W = P for one of the two component balances? Of course. fact, could calculate P by just two equations: Total baJance: 1000 + W = P W = O.9P
- 215. 184 A General Strategy for Solving Material Balance Problems Chap. 7 Substitute the second equation into the first equation and solve for You can conclude that the symbols you select in writing the equations and the particular equations you select to solve a problem do make a difference, and require some thought. With practice and experience solving problems. such issues should resolve themselves. 9. Solve the equations and calculate the quantities asked for in the problem "Problems worthy of attack prove their worth by hitting back" Piet Hein Industrial-scale problems may involve thousands of equations. Clearly, in such cases efficient numerical procedures for solution the set of equations are essential. Process simulators to carry out the task on a computer, as explained in Chapter 31. Because most of the problems used in this text have been selected for the purpose of communicating ideas, you will find that their solution will involve only a small set of equations. You can solve two or three equations by successive substitution. For a larger set equations or for nonlinear equatiohst use a computer program such as Polymath, Excel, Matlab. or MathCad. You will save time and effort by doing so. Learn to be efficient at problem solving. When given data the AE system of units, say pounds, do not first convert the data to the SI system, say kilograms. solve the problem, and then convert your results back to the system of units. The procedure will work, but it quite inefficient, and introduces unnecessary opportunities for numerical errors to occur. Select a precedence order for the equations you write. One choice of an order can be more effective than another. We showed in Step 8 how the choice of the total balance plus the ratio WIP = 0.9 led to two coupled equations that could easily be solved by substitution for W to P 10,000 kg W = 9,000 kg these two values you can calculate the amount of H20 and NaOH in the product { NaOH balance: From the H20 balance: I 1
- 216. Sec. The Strategy for Sotving Problems Then 1000 NaOH p - --....;:::;......-- = 0.1 WNaOH - 10,000 kg Total P 9,000 W H - ----------- 20 - 10,000 Examine the set of four equations listed in Step 8. Can you find a shorter or "" ........ ~A series of calculations get a solution for the problem? 10. Check your answer(s) Error a hardy plant; it flourishes in every soil. Martin Tupper Everyone makes mistakes. What distinguishes a good engineer is that he or she is able to find their mistakes before they submit their work. In Chapter 1 we listed several ways to validate your solution. We will not them here. Refer to Section 1 A good engineer uses his or her accumulated knowledge as a primary tool to make sure that the results obtained for a problem (and the data in the problem) are reasonable. Mass fractions should faU between zero and one. Flow rates should nonnegative. In any collection of data, the figure that is most obviously corfeetbeyond all need of checking-is the mistake. Unknown To the list of validation techniques that appeared in Chapter 1, we want to add one more very useful one. After solving a problem, use a redundant equation to check your values. In the problem in Example 7.2 that we have been using in the presentation of the problem-solving strategy. one of the three material balances is redundant (not independent), as pointed out severa] times. Supposed you had solved the problem using the N aOH and H20 balances. Then the total balance would have been a redundant balance, and could be used to check the answers Insert the numbers 1)000 + 9,000 = 10,000 Table 7.4 summarizes the set of 10 steps we have discussed. H you use the steps in Table 7.4 as a mental checklist each time you start to work on a problem, you will have achieved the major objective of this chap-
- 217. 186 A Strategy for Solving Material Balance Problems TABLE 7.4 Strategy for Solving Material Balance Problems 1. Read and understand tbe problem ct'll1r"'",An~ 2. Draw a of the process and specify system boundary. 3. Place labels for unknown variables and values for known variables on the sketch. 4. Obtain missing needed data. 5. Choose a . 6. Determine number of unknowns. 1. Determine the number of independent equations9 and carry out a degree of freedom analysis. 8. Write down the equations to be solved. 9. Solve the equations and calculate the quantities asked for. 10. Check your answer(s). 7 tel' and substantially added to your professional skills. These do not have to be out in the order listed in the table, and you may repeat as the formu-lation solution of the clearer. But each of the steps is essen-tial even carried out implicitly. Postscript Tables 7.5 and 7.6 comprise a list of traits to review to help you improve your problem-solving abilities. Also look on the CD that accompanies this book. It cona section with a more detailed discussion of how to solve problems . Techniques ... ·v.'>b ... t" to Overcome Barriers to Problem Solving Read the over several Be sure to , ... f1, ..... <lf all facets of it. Emphasize the different features each problem in your own words. assumptions. Draw a comprehensive diagram of the ..... ""~ .. CC and enter all known on diagram. Enter symbols for unknown variables FonnalJy down what you are to solve for: "I want to problem to similar you have encountered before, but note any differences. for a solution, writing it down jf necessary. Consider strategies. down an the equations and that might apply to the problem. write down everything you know about the problem and what believe is needed to execute a Talk to yourself as you ... ....-.,-.. "',.. to Ask yourself questions as you problem. concerning the data, proceOur ""'-I."''''''.'''''' involved. etc.
- 218. I Sec. 7.2 The Strategy for Solving Problems 187 TABLE 7.5 Continued Talk to other people about the problem. Break off problem solving for a few minutes and carry out some other activity. Break up the solution of the problem into more manageable parts, and start at a familiar stage. Write down the objective for each subproblem (i.e., convert mole fraction to mass fraction, find the pressure in tank 2, etc.). Repeat the calculations but in a different order. Work both forward and backward in the solution scheme. Con)';ider if the results you obtained are reasonable. Check both units and order of magnitude of the calculations. _Are the boundary conditions satisfied? Use alternative paths to verify your solution. Maintain a positive attitude-you know the problem can be solved, just how is the question. TABLE 7.6 A Comparison of the Problem-Solving Habits of a Novice and an Expert A novice: Stans solving a problem before fully understanding what is wanted and/or what a good route for solution will be. Focuses only on a known problem set that he or she has seen before and tries to match the problem with one in the set. Chooses one procedure without exploring alternatives. Emphasizes speed of solution, unaware of blunders. Does not follow an organized plan of attack; jumps about and mixes problem-solving strategies. Is unaware of missing data, concepts, laws. Exhibits bad judgment, makes unsound assumptions Gives up solving the problem because he or she does not have skills to branch away from a dead-end strategy. Unable to make approximations or makes bad ones. Cannot conceive of disagreeing. Slavishly follows instructions; proceeds "by the book." Does not know what to make of Qualitative data. Fritters limes way. An expert: Reviews the entire plan. mentally explores alternative strategies, and clearly understands what result is t<;> be obtained. Concentrates on similarities to and differences from known problems; uses generic principles rather than problem matching. Examines several procedures serially or in paraJlel. Emphasizes care and accuracy in the solution. Goes through the problem-solving process step by step. checking, reevaluating, and recycUng from dead ends to another valid pa~ . Knows what principles might be involved and where to get missing data. Carefully evaluates the necessary assumptions. Aware that a dead end may exist for a strategy and has planned alternative strategies if a dead end is reached. Makes appropriate approximations. Disagrees with other experts. Breaks rules and makes exceptions. Able to deal with qualitative data. Good management of time.
- 219. 188 A General Strategy for Solving Material Balance Problems Chap. 7 SELF ASSESSMENT TEST Questions What does the conc.ept "solution of a material balance problem" mean? 2. (a) How many values of unknown variab1es can you compute from one independent material balance? (b) From three independent material balance equations? (c) From four material balances. three of which are independent? 3. What does the concept of independent equations mean? 4. If you want to solve a set of independent equations that contain fewer unknown variables than equations (the overspecified problem), how should you proceed with the solution? S. What is the major category of implicit constraints (equations) you encounter in material balance problems? 6. If you want to solve a set of independent equations that contain more unknown variables than equations (the underspecified problem), what must you do to proceed with the solu· tion? 7. As I was going to S1. Ives, I met a man with seven wives: Every wife had seven sacks, Every sack had seven cats, Every cat had seven kits. Kits, cats, sacks. and wives, How many were going to St. Ives? Problems 1. A water solution containing 10% acetic acid is added to a water solution containing 30% acetic acid flowing at the rate of 20 kg/min. The product P of the combination leaves at the rate of 100 kg/min. What is the composition of P? For this process, a. Detennine how many independent balances can written. b. List names of the balances. c. Detennine how many unknown variables can be sol ved for. d. their names and symbols. e. Detennine the composition of P . • 2. Can you solve these three material balances for F, D. and P? Explain why not. 0.1 F + O.3D ::: O.2P 0.9F + O.7D = O.SP F + D = P .' ' 3. How many values of the concentrations and flow rates in the process shown in Figure, SATI.2P3 are unknown? List them. The streams contain two components. 1 and
- 220. Sec. 7.2 The Strategy for Solving Problems F ... wF1 = 0.2 11' P OJP2= 0.1 ... 0 wDl= 0.95 Figure SAT7.2P3 189 4. How many material balances are needed to solve Problem 3? Is the number the same as the number of unknown variables? Explain. Thought Problem 1. In the steady-state flow process shown in Figure TP7 .2PI. a number of values of c.o (mass fraction) are not given. Mary says that nevertheless the problem has a unique solution for the unknown values of w. Kelly says that four values of w are missing, that you can write three component material balances, and that you can use three relations for 2: Wi = 1, one for each stream, a total of six equations, so that a unique solution is not possible. Who is right? F", 10 kg ... - P =:::: 16 kg Process A=6kg Q)1 ::: 0.030 ~=? <0:3::: 0.20 Figure Tn .2Pl ffi1 ,",0.175 <O:l =:::: ? ~=?
- 221. 190 A General Strategy for Solving Material Balance Problems Chap. 7 Discussion Problems 1. Consider the concept of zero discharge of liquid waste. It would seem to be a good idea both for the environment and the company. What are some of the arguments for and against the zero discharge of wastewater? 2. One proposed method of eliminating waste in solid, liquid, and gas streams is incineration. What are some of the pros and cons regarding disposal of waste by incineration? Looking Back In this chapter we discussed a set of 10 steps that you should apply in solving material balance problems. Not all steps have to be formally written down, but each merits consideration if you want to solve problems efficiently and effectively. We recommend that you use them throughout the remainder of the book and thereafter in your practice of engineering. GLOSSARY OF NEW WORDS Degrees of freedom The number of variables whose values are unknown minus the number of i-odependent equations. Degree of freedom analysis Determination of the number of degrees of freedom in a problem. Dependent equations A set of equations that are not independent. Exactly specified A prob1em in which the number of degrees of freedom is zero. Implicit equation An equation based on infonnation not explicitly provided in a problem such as the sum of the mass fractions is one. Independent equations A set of equations for which the rank of the coefficient matrix fonned from the equations is the same as the number of equations. Knowns Variables whose values are known. Overspecified A set of equations (or a problem) that is comprised of more equations than unknowns. Underspecified A set of equations (or a problem) that is comprised of fewer equations than unknowns. Unique solution A single solution exists for a set of equations (or a problem). Unknowns Variables whose values are unknown.
- 222. Chap. 7 Problems 191 SUPPLEMENTARY REFERENCES In addition to the general references listed in the Frequently Ask.ed Questions in the front material, the followjng are also pertinent. Felder, R. M. "The Generic Quiz," Chem Eng. Educ., 176-181 (Fall 1985). FogJer, H. S., and S. M. Montgomery. Interactive Computer Modules for Chemical Engineering Instruction, CACHE Corp., Austin, TX (2000). Woods, D. R., T. Kourti, P. E. Wood, H. Sheardown. C. M. Crowe, and 1. M. Dickson. "Assessing Problem Solving Skills," Chem Eng. Educ., 300-307 (Fall 200t). Web Site http://169 237 .64. 14/webMathematicaJMSPlExampiesIDOF Analysis PROBLEMS ·7.1 For the process' shown in Figure P7.1, how many material balance equations can be written? Write them. How many independent material balance equations are there in the set? . Composition Pi 0.10 A 0.208 0.70C Composition Fz 0.50 A 0.50C Composition 0.35A 0.108 0.55 C pc 100 Figure P7.1 Ihr FJ ~posltlon O.20A 0.30 8 O.SOC ·'.2 Examine the process in Figure f!7.2. No chemical reaction takes place, and x stands for mole fraction. How many variables are unknown? How many are concentrations? Can this problem be solved uniquely for the unknowns? /'
- 223. 192 A General Strategy for Solving Material Balance Problems F---1 J'A =0.2 ",,=0.8 X'A:: O.t 1---- P XA =0.5 x,= ? Figure M.2 Chap,7 ·7.3 Are the foHowing equations independent? Do they have a unique solution? Explain your answers. (a) XI + 2X2 = I Xl + 2X2 = 3 (b) (XI - 1)2 + (X2 - 1 r = 0 XI + = 1 ·7.4 For one process your assistant has prepared four valid material balances 0.25 mNlICl + 0.35 mKCl + 0.55 mH20 :::; 0.30 035 mNIlCl + 0.20 mKCI + 0.40 mH20 = 0.30 0,40 mNaCI + 0.45 mKCI + 0.05 mH10 = 0.40 1.00 mNIlCl + 1.00 mKO + 1.00 mH,20 = 1.00 He says that since the four equations exceed the number of unknowns, three,'no solution exists. Is he correct? Explain briefly whether it is possible to achieve a unique solution. ·"'.S Do the following sets of equations have a unique solution? U+I)+W=O (a) U 2v 3w = 0 3u + 51) + 7w = 1 u + w = 0 (b) 5u + 41' + 9w = 0 + 4v + 6w = 0 ··7.6 Answer the following questions true or false: (a) When the flow rate of one stream is given in a problem, you must choose it as the basis. (b) If all of the stream. compositions are given in a problem, but none of the flow rates are specified. you cannot choose one of the flow rates as the basis. (c) The maximum number of material balance equations that can be written for a problem is equal to the number of species in the problem. "7.7 In the steady state process (with no reactions occuring) in Figure you are asked to determine if a unique solution exists for the values of the variables. Does Show all calculations.
- 224. Chap. 7 Problems F == 10 kg == 0.10 =1 P 16 kg A=6kg Figure P7.7 lll, .= 0.175 ~=? ~""? 193 w is the mass fraction of component i. *7.8 Three gaseous mixtures, A. B. and blended into a single mixture. with the compositions listed in the table are 25 35 40 100 B C 25 30 45 tOO 60 25 100 A new analyst reports the composition of the mixture 25% CH4, 25% ~H6' and 50% C3Hg. Without making detailed calculations, explain how you know the analysis is incorrect. "7.9 A problem is posed as foHows. It is desired to mix three LPG (Liquefied Petroleum streams denoted by A, and in certain proportions so that the mixture wiH meet certain vapor pressure specifications. These specifications will be met by a stream of composition as indicated below. Calculate the proportions in which streams A, B, and C must be mixed to a product with a composition of D. The values are liquid volume %, but the volumes are additive for these compounds. Component A B 5.0 90.0 10.0 5.0 85.0 5.0 100.0 c 8.0 80.0 12.0 100.0 D 1.4 31.2 53.4 1 1.4 The subscripts on the C's represent the number of carbons, and the + sign on Cs+ dicates all compounds of higher molecular weight as well as iso-Cs. this problem have a unique solution? ,-
- 225. 194 A General Strategy for Solving Material Balance ~roblems Chap. 7 "*7.10 preparing 2.50 moles of a mixture of three gases, S02' H2S, and CS2• from three tanks are combined into a fourth tank. tanks have the following composi. tions (mole fractions): Combined Mixture Gas 1 1 .3 4 S02 0.23 0.20 0.54 0.25 0.36 0.33 0.27 0.23 0.41 0.47 0.19 In right-hand column is listed the supposed composition obtained by analysis of the mixture. the set of three mole balances for the three compounds have a solution for the number of moles taken from each of the three tanks and to make up the mixture? If so, what the solution mean? ·"7.11 have been asked to check out the process shown in Figure .11. What will be the minimum number of measurements to make in order compute value of each of the stream flow rates and stream concentrations? Explain your answer. F--fo>-I I'---W Composi1ion Composition x~ x, )(6 p Figure P7.11 >tf xf Can any arbitrary be used: that can you measure just the three flow rates and two concentrations? Can you measure just three concentrations in stream F and two concentrations in stream W? No chemical reaction takes and x is the mole fraction component A, B. or C. "·7.12 Effluent from a fertilizer pJant is processed by system shown in Figure P7.12. How many additional concentration and stream flow measurements must be made to completely specify the problem (so that a unique solution exists)? only one unique set of specifications exist? For each ofthe/ollowing three problems (a) Draw a figure (b) Put the data the problem on figure
- 226. Chap. 7 Problems (c) a basis (d) Determine the number unknowns and independent equations (e) Write the material balances neerled solve the problem (f) Write down any other pertinent equations and specifications (g) Solve the problem possible A CaS04 31% B H2S04 H2O E 11 kg/s 1.27% H N03 1.27% 0 .. H2S04 1 Yo HNOa 2% H20 Figure P1, 12 C sSO" In arts H20 195 ;:·'.13 Tank containing 90% nitrogen is mixed with Tank B containing 30% nitrogen to get Tank containing 65% You are asked to determine the ratio of the from Tank A to that used from Tank ··',14 drier takes in wet timber (20.1 % water) and reduces the water content to 8.6%. You want to determine kg of water removed kg of timber that enters the J U 7.15 CH4• CzH6• and N2 to be prepared which th~ ratio of the 1.. ... "".",,,, of CH4 to C2~ is 1.3 1. Available are (1) a cylinder containing a mixture of 70% N2 and 30% CH4• (2) a cylinder containing a mixture of 90% N2 and 10% C2H6! and a cylinder of pure Determine proportions in which the respective gases from each cylinder should be used . • , .16 you read a problem statement, what are some of the things you should think about to solve it. them. This problem not ask the 10 described in the chapter, but for brain storming.
- 227. CHAPTER 8 SOLVING MATERIAL BALANCE PROBLEMS FOR SINGLE UNITS WIT OUT REACTION Your objectives in studying this chapter are to be able to: 1. Analyze a problem statement and organize in your mind the solution strategy. 2. Apply 10-step strategy to problems without chemical reactions. In Chapters 6 and 7 you read about solving material balance problems without reaction. Can you apply ideas now? If you are quite confident. you might go on to the next chapter. If not, hone your skills by going through the applications sen ted in this chapter. Looking Ahead In this chapter we are going to through several examples that involve the analysis 'and solution of material balances for a single If you hope de-velop some skill judgment in solving material balance problems, way to through this chapter is to first cover up solution of the problem. then the problem, then out on a piece of paper your solution step by and only afterward look at the solution that appears in the example. you j read the problem and solution, you will deprive yourself of learning activity needed improve your capabilities. You will find additional solved on the CD that ac- 196
- 228. Chap.S Solving Material Balance Problems for Single Units without Reaction 197 companies this book if you would like more practice in developing your prohlemsolving skills. Main Concepts A famous magician stood on a concrete flOOf1 and with a flourish pulled a raw egg from his hair. He heJd the egg in his outstretched hand, and said he could drop it 2 meters without breaking its shell and without the aid of any other object. Then he proceeded [0 do it. What did he do? (The answer is at the end of the chapter.) Problem solving is what you do when you don't know whallo do; otherwise not a problem. G. Bodner, J. Chern. Edu. 63, 873 (1986) The use of material balances in a process allows you (a) to calculate the values of the total flows and flows of species in the streams that enter and leave the plant equipment, and (b) to calculate the change of conditions inside the equipment. You want to find out how much of each raw material is used and how much of each product (along with some wastes) is obtained from the plant. We use simple examples in this chapter to demonstrate that no matter what the process is, the problem-solving strategy evolved in Chapter 7 can be effective for all of them. If the process involves rates of flow, you can pick an interval of time as the basis on which to solve the problem so that you can avoid carrying along time as a symbol or variable in the analysis and calculations. EXAMPLE 8.1 Extraction of Streptomycin from a Fermentation Broth Streptomycin is used as an antibiotic to fight bacterial diseases. and pro-duced by the fennentation of a bacterium in a biological reactor with a nutrient of glucose and amino acids. After the fermentation process, Streptomycin is recovered by contacting the fennentation broth with an organic solvent in an extraction process. The extraction process is able to recover the Streptomycin because Streptomycin has a greater affinity for dissolving in the organic solution than in the aqueous solution. Figure E8. 1 shows the overall process. Determine the mass fraction of Streptomycin in the exit organic solvent assuming that no water with the solvent and no solvent exits with the aqueous solution. Assume that the density of the aqueous solution is 1 g/cm3 and the density of the organic solvent is 0.6 glcm3.
- 229. 198 Solving Material Solution Stept Problems for Single Units without Reaction Chap.S This is an open (flow), steady-state process without reaction. Assume because of the low concentration of Strep. in the aqueous and organic fluids that the flow rates of the entering fluids equal the flow rates of the exit fluids. Steps 3, and 4 AH of the data has been placed on Figure ES.l. Organic solvent S 10Umln p::: 0.6 glcm3 No Strep Aqueous solution Aqueous solution A Extraction 200 Umin Process 0.2 9 Strap Il 10 9 Strep IL p:: 1 glcm3 Organic solvent Extracted Strep Figure £.8.1 StepS Basis: 1 min Steps 6 and 7 The degree-of-freedom analysis Number of variables (8): 4 flows (in L) plus 4 concentrations (in gIL) Number of equations (8): Basis: :: 200 L (flow of aqueous entering aqueous solution) 1 Specifications: Concentration Strep in entering aqueous solution 1 Concentration of Strep in aqueous solution ] Concentration of Strep in entering organic solvent 1 Flow exiting aqueous solution (same as existing flow) 1 Flow of entering organic solution 1 Flow of exiting organic solution (same as existing flow) 1 material balance j Total 8 degrees of freedom are O.
- 230. Chap. 8 Solving Material Balance Problems for Single Units without Reaction Steps 8 and 9 The material balances are in = out in grams. Let x be the g of Strep per L of solvent S. Strep. balance: 200 L of A 110 g SlrCp + 10 L of S I 0 g SlIep = 200 L of A I 0.2 g Strep + 10 L of S Ix g Strep 1 L of AIL of S 1 L of AlL of S x = 196 g StreplL of S To get the g Strep/g solvent, use the density of the solvent: 196 g Strep 1 L of S 1 em3 of S 3 0 6 f S = 0.3267 g Strep/g of S 1 L of S 1000 em of S . g 0 . 0.3267 The mass fractlon Strep = 1 + 0.3267 = 0.246 EXAMPLE 8.2 Separation of Gases Using a Membrane Membranes represent a relatively new technology for the separation of gases. One use that has attracted attention is the separation of nitrogen and oxygen from air. Figure E8.2a illustrates a nanoporous membrane that is made by coating a very thin layer of polymer on a porous graphite supporting layer. What is the composition of the waste stream if the waste stream amounts to 80% of the input stream? High-pressure I Membrane Low-pressure side side I (Input) Figure EB.2a 0 2 25% Product (Output) N2 75% 199 "
- 231. Solving Material Balance Problems for Single Units without Re.action Solution Step 1 This an process without chemical reaction. The memtlrarle as in Figure E8.2. Let the mole fraction depicted in Figure E8.2) YN2 be the mole fraction of nitrogen. and let respective moles. F(g mol) molff 02 0.21 N2 0.79 1.00 Steps 2, 3, and 4 P mol} ·W(g mol) mol fr gmol O2 Yap: n'th N2 YN:2 nrt'2 . 1.00 W 1LI"1 ............. E8.1b All of the and symbols hav.e been placed, in Figure E8.2b. StepS a convenient basis. Basis: 100 g mol = F Chap. 8 A degree of You could either use analysis that includes all of the variables comes next. or mole fractions as the unknowns. Steps 6 and 7 Number of variables: 9 P, Wand 6 ni Number of equations: 9 Basis: F= 100 n62 = 0.21 (100) ,= 21 n~l = 0.79(100) = 79 Yb2 = nb/P = 0.25 Y~2 = n~/P = W = 0.80(100) = 80 nb2 O.25P n~2 = O.75P ; I I I I 1 1
- 232. Chap. 8 Solving Materia! Balance Problems for Single Units without Reaction Material balances: O2 and N2 Implicit equations: 2:nj" = W or LY;V = 1 The problem has zero degrees of freedom because of the way we have fonnu~ lated the solution. Note that nb2 + n~2 = F is a redundant equation because it repeats some of the specifications. Also, n& + n~2 = is redundant. Divide the equation by P to get Yb1 + Y~2 := 1, a relation that is equivalent to the sum of two of the specifications. Step 8 If you introduce the known values into the balances and implicit equa-tion, you three independent equations in three unknowns In Out In OUI °2: 0.21 (l00) = 0.25P + y1j2 (80) or 0.21 (100) ;;; 0.25P + ~2 N2: 0.79 (l00) = 0.75P + y;.r2 (80) or 0.79 (100) = O.75P + n;r2 LOO = yWO2 + yWNz or 80 = n~2 + n~l Step 9 The solution of these equations is n'tf2 = 16 and n'ti2 = YNW2 = 0.80, an d p;:: 20 g mol. ' or y~l = 0.20 and An calculation involves the use of the total balance first in place of one component balance. The overaH balance is easy to solve because F;:: P + War 100 = P + 80 gives P = 20 straight off. Then, the oxygen balance would be 0.21(100) = 0.25(20) + n1j:z from which YOti can get ~2 = 16 g mol, and n~ 80 - 16 = 64 g moL Alter-nately, you could use the nitrogen balance to the same result. Step 10 Check. You can use the total balance as a check on the solution obtained from the two component balances 100;:: 20 + 80 OK 201--
- 233. Solving Material Balance Problems for Single Units without Reaction Chap.S Be careful in formulating and simplifying the equations to be solved to make sure that the set evolves into a set o/independent equations. For example, if you pre-pare the following of equations from the data given in a problem with zero of freedom, 0.25mNaCl + 0.35mKcl 0.55mH20 = 0.30 0.35mNaCI O.20mKCl + OAOmH20 = 0.30 OAOmNaCl + OA5mKCI 0.05mH2 0 = OAO 1.00mNaCI I.OOmKe! + 1.00mH2o = 1.00 it may appear that the set contains three unknowns and four equations. However, only three of the equations are independent. Do you see why? In the next problem we give an example of distillation. Distillation is the most commonly used process for separating componentst and is based on the separation that results from vaporizing a liquid (see Chapter 16), When a liquid mixture is boiled to produce a vapor, the vapor contains a higher concentration of the more volatile component. A distillation column is a collection of a number of stages that lead to the concentration of the more volatile components in the top product from the column and the concentration of the less volatile components in the bottom product of the column. Look on the CD accompanying this book for more infonnation about the specialized terminology pertaining to distillation and pictures of distillation equipment. EXAMPLE 8.3 Overall Analysis for a Continuous Distillation Column A novice manufacturer of ethyl alcohol (denoted as EtOH) for gasohol is having a bit of difficulty with a distillation column. The process is shown in Figure E8.3. It appears that too much alcohol is lost in the bottoms (waste). Calculate the composition of the bottoms and mass the alcohol lost in the bottoms based on the data shown in Figure that was collected during t hour of operation. Solution Although the distillation unit shown in Figure E8. 3 is comprised of more than one unit equipment, you can select a system that indudes an of the equipment the system boundary. Consequently, you can ignore all the internal streams for
- 234. Chap. 8 Solving Material Balance Problems for Single Units without Reaction -""""------------. ....... ,~ " System' /' ... 8oundory~ 1000 kg Fee F 10% EtOH 90"1) H20 , I I ,I ,I dl I I I I t EllchO/1ger , I Vapor Heal I I Reflux I I / ,/ OistiliotiOfl ,/ /' Column /' /' /' / / / I J I Bottoms (Woste) 8 / Figure E8.3 Oi stillote (Product) p::c kg 60,"0 E.tOH 40,.., H20 1/10 feed "" kg this problem. Let m designate the mass of a component. Clearly the process is an open system, and we assume it is in the steady state. No reaction occurs. Thus, the material balances reduce to In;:: Out in kg. The cooling water enters and leaves the heat exchanger without mixing with the components being separated, and can be ignored for the material balances. In addition, the heat added at the bottom of the column does not involve mass entering or leaving the system, and can be ignored for the material balances. Steps 1, 2, 3, and 4 All of the symbols and known data have been placed on Figure E8.3. StepS Select as the basis the given feed. Pick ] hour so that you can suppre~s the time variable in all of the calculations. Basis: I hour so that F = 1000 of feed Step 4 are gI.V en that p'1 8 -1 of 10 so that P ::: 0.1 (l000) = 100 kg Steps 6 and 7 The next step is to carry out a degree of freedom analysis. Number of variables: 9 F F P P B B FPB mEtOH. mHzo, mEtOH. mHzo- mBtOH. mH20. , , 203
- 235. 204 Solving Material Balance Problems for Single Units without Aeaction Chap.S Number of equations: 9 Basis: F =. 1000 kg Specifications: 1000(0.10) = 100 m~20 = 1 000 ( 0.90) = 900 m~tOH = O.60P mk20 = OAOP P = 0.1 F:::; 100 kg Material balances: EtOH and H20 Implicit equations: 2m? = B or LW? = 1 The problem has zero degrees freedom. What redundant equation(s) can you write? Refer to 8.2 for hints. Steps 8 and 9 Let' s substitute total mass balance F :::; P + B mass balances and calculate B by direct subtraction B = 1000 - 100 = 900 one of component The solution for the composition of the bottoms can then be computed directly from the material balances: kgfeed in kg distillate out kg bonoms out Mass fraction EtCH balance: 0.10(1000) 0.60(100) ;::: 40 0.044 H20 balance: 0.90(1000) - 0.40(100) :::: MQ Mi6 900 1,000 Step to a check s use redundant equation m{hoH + mAzo B or W~tOH + = 1 Examine the last two columns of the table above for verification. The next example represents an open system, but one that is in the unsteady state. The CD that accompanies this book shows various equipment to mix liquids.
- 236. Chap. 8 Solving Material Balance Problems for Single Units without Reaction EXAMPLE 8.4 Mixing of Battery (Sulfuric) Acid You are asked to prepare a batch of 18.63% battery acid as follows. A tank of old weak battery acid (H2S04) solution contains 12.43% H2S04 (the remainder is pure water). If 200 kg of 77.7% H2S04 is added to the tank, and the final solution is to be 18.63% H2S04, how many kilograms of battery acid have been made? See Figure E8.4. Added Solulion 200 kg = A System --- .>r-----~ --__ H20 81.37,.. Original Solution Fkg Finol Solution Pkg Figure ES.4 Solution Steps 1, 2, 3, and 4 . ., All of the values of the compositions are known and have been placed on Figure E8A. No reaction occurs. Should the process be treated as an unsteady-state process or a steady-state process? If the tank is selected as the system, and the tank initially contains sulfuric acid solution, then a change occurs inside the system so that accumulation occurs in the system: the total mass increases and the mass of each Component increases accumulation = in - out From another viewpoint, as you learned in Chapter 7. you could regard the tank as initially being empty, the original solution is introduced into the system along with the 200 kg of 77.7% solution, the solutions are mixed. and fmally the entire contents of the tank are removed leaving an empty tank.. Then, the mass balance reduces to a steady-state flow process in = out because no accumulation occurs in the tank. Let us first solve the problem with the mixing treated as an unsteady-state process, and then repeat the solution with the mixing treated as a steady-state process. 205 --
- 237. 206 Solving Material Balance Problems Units without Reaction Take 200 kg of A as the basis for convenience. The degree-of-freedom amptes 8.2 and g,3. is analogous to the ones carried out for Number of variables: 9 A A mH2S04• mH20. Number of equations: 9 Basis: A = 200 Specifications: A A F F P P lOH1S04, lOH20. i.UH2S04, lOH20. i.OH2S04• lL.IH1o Material and The degrees of freedom are zero. Note that the implicit """4",,,,u',,,,,,,," = 1 for A. F, and P are redundant specifications of the mass ""'''''''A''''''''' Is P = A + F redundant also? StepS We will insert basis and specifications into the mass balances. The bal-ances will be Type of Balance Accumulation in Tank In Out H2SO4 H2O Total Note that any Step 9 Final Initial P(0.1863) F(O.1243) = 200(0.777) -- 0 P(0.8137) F(0.8757) = 200(0.223) - 0 P F = 200 0 of the three equations is independent. equations are linear and only two independent equations occur, total mass balance. solve it for and substitute for F in the H2S04 P and get 8
- 238. Chap. 8 Solving Material Balance Problems for Single Units without Reaction P = 21to kg acid F "" 1910 kg acid Step 10 You can check the answer using the H20 balance. Is the H20 balance satisfied? The problem could also be solved by considering the mixing to be a steadystate process with the initial solutions F and A in the vessel flowing through a vessel during the time interval, and the resulting mixture flowing out from the vessel. A in 200(0.777) 200(0.223) A + + + Fin F(0.1243) F(O.8757) F = = = Pout P(O.1863) P(O.8137) p You can see by inspection that these equations are no different than the first set of mass balances except for the arrangement and labels. EXAMPLE 8.S Drying Fish caught by human beings can be turned into fish meal, and the fish meal can be llsed as feed to produce meat for human beings or used directly as food. The direct use of fish meal significantly increases the efficiency of the food chain. However, fish-protein concentrate, primarily for aesthetic reasons, is used mainly as a supplementary protein food. As such, it competes with soy and other oilseed proteins. In the processing of the fish, after the oil is extracted, the fish cake is dried in rotary drum dryers, finely ground, and packed. The resulting product contains 65% protein. In a given batch of fish cake that contains 80% water (the remainder is dry w c; 100 kg H20 _-------_/system Boundary /' ----~ ...... ...... II " I 0 kg HtO WeI A r ,.- 1 Fish Coke;; ? kg ~aurner 0.80 H20.. 0 0.2080C - ------, ""', -~--B-h Dry I ............... Fish Cake ~ ? kg I . -----------/ 0.40 H,O IL _____r,e _C_omp_on_ent ________________ 0.60 BDC _ -Bone Dry Cake Figure ES.s 207 '"
- 239. Solving Material Problems for Single Units without Reaction Chap. 8 cake), 100 kg of water is removed, and it is found that the fish cake is then 40% water. Calculate the weight of the fish originally put into the dryer. is a diagram of the process. Solution We wiU abbreviate the solution. Steps 1, 2, 3, and 4- is a steady-state process without reaction. The system is the dryer. Step 5 a basis of what is given. 100 kg of water ==w Steps (I 7 degree-of-freedom analysis gives zero degrees of freedom. There are four streams, two in (air and fish cake) and two out (air and fish cake), although the air is not shown Figure E8.5 it is not involved in the process, Only the water in the air is involved. Two independent balances can be written. We will use the mass balance plus the BDC dry cake) balance (the tie component) in kg. The balance O.BOA == 0.40B + 100 can be used as a check on the calculations. In Out Total A B + W = B + 100 BDC balance: 0.20A 0.60B The A 150 kg initial and B == (150)(0.20/0.60):: 50 kg Step 10 Check via water balance: 0.80(150) ~ 0.40(50) + 100 120 = 120
- 240. Chap.S Solving Material Balance Problems for Single Units without Reaction 209" Example 8 the BDC in the wet and dry fish cake is known as a tie component because the BDC goes from a single stream in the process to another single stream without loss, addition, or splitting. The BDC ties two streams together. Con-sequently, you pick BDC as the basis in the you can add and subtract ra-tios the amount of a compound per unit amount BDe in a material balance. Here is what you could have done in Example H20 balance: In 0.80 H20 inA 0.20 kg BDC in A Basis: 1 BDC Out 0.40 kg H20 in B 0.60 BDC in B = H20 removed 3.33 kg H20 kg BDC (in A or B) Next, shift to a basis of 100 kg of H20 in W. The H20 removed is H20 in W: BDC in A 0.20 ;;; ] 50 A Sometimes the use of a tie element reduces the extent of the calculations in a problem. EXAMPLE 8.6 Crystallization A tank holds 10,000 kg a saturated solution of Na2CO) at 30°C, You want to crystallize from this solution 3000 kg NazCO} . IOH20 without any accompa-nying water. what temperature must the splulion be cooled? Solution This problem is a little more complicated to analyze than the previous problems because it not only requires a decision as what the are in the problem, but also it implies without specifically stating so that the final solution is saturated at the final temperature. No reaction occurs. Although problem could set up as a steady-state problem with flows in and out of system (the tank), it is equally justified to treat the process as an unsteady-state process. The major diffiCUlty posed in this problem is to all the necessary information about the compo· sitions of the solutions and solid precipitate. If you can calculate the final concen-tration the N~C03 the tank, you can look up corresponding temperature a handbook containing solubility data. The components mentioned the problem statement are Na2C03' H20, and Na2C03 . 10H20. Although these components can
- 241. 210 Solving Material Balance Problems for Single Units without Reaction Chap.S exist as separate compounds. only two chemical actually in the tank: N~COJ and H20. Let's select these species as the components for which to make material balances, because it takes fewer steps to make the required calculations. Steps 1, 2, aDd 3 Figure ES.6a is a diagram of the process Initiol Sfate NozC03 • 10 HzO 3000kg Crystals Removed Figure E8.6a L ~:~m Boundary -""" " ",,- ........ "- I ,I " CO : NCle Saturated I Solution I H20 '=1 I J , I / ... ",/ ...... ---~---,.." Final State Next, you need to get the compositions of the streams insofar as possible for each solution and the solid crystals of N~C03 . lOH20 Steps 3, and 4 You definitely need solubility data for Na2C03 as a function of the tempera-ture: Temp. (OC) o 10 20 30 Solubility (g N~CO:ll00 g H20) 7 12.5 21.5 38.S Because the initial solution is saturated at 30°C, you can calculate the composition of the initial solution:
- 242. Chap.. 8 Solving Material Balance Problems for Single Units without Reaction 38.8 38.8 g Na2C03 + 100 g H 2 0 = 0.280 mass fraction Na2C03 Next, you should calculate the composition of the crystals. Basis: I g mol N~C03 . 10H2O Mol Mol wt. Mass Mass!! Na2C03 106 106 0.371 H2O 10 18 180 0.629 Total 286 1.00 StepS Select a basis. 'The following is convenient; others can be used, such as 3000 g ofNa2C03' 10 H20: .. Basis: 10,000 kg of saturated solution at 300C Steps 1 and 3 (Repeated) From the known data we have calculated the compositions of the compounds and solutions, and put them on Figure E8.6b. The problem now appears to be quite similar to the previous examples, particularly Example 8.4 . .... -....------- Initial state -_ ......... 3000 kg ,,- -..,.,..""""'...----~./ , /_--------"'' I I I F=? kg I : Na2COS mNa2c~ ; I H20 mH20 J ----- Final state ,,- _/ / ./ I / Ns2COa 0.371 H20 0.629 Crystals removed FIgure Eft.6b 211
- 243. 212 Solving Material Balance Problems for Single Units without Reaction Chap. 8 Because we are treating problem as an unsteady-state problem, ance reduces to (the flow in = 0) accumulation = out Steps 6 and 7 The analysis of the initial, F final and freedom yields a value of zero (l stands for the the crystals). Number variables: 9 I J F FCC I mN.a2C03' mHzO. mNa2C03' mHlO. mNa2C03' mHzO. , Number of equations: 9 Basis: 1 == 10,000 kg Specifications: I 1 F FCC WNa2C03' WHlO. WNs2c03, WA20, WNa2CO), wH20 Material balances: c Note that w/I = m{, w/F = m{, and {dice = mf are redundant equations. For example, m~a2 could be replaced with 0.2801 by using the specification for ~a2 Also redundant are equations su?h as = ] :Lm; mlotal. Steps 8 and 9 After substituting the specifications and basis into the material balances, (only two are independent) get (in kg). Accumulation in Tank Final Initial N~C03 F mNa2CO) 10,000(0.280) ;;;; -3000(0.37 J) F mHlO 10,000(0.720) = -3000(0.629) Total F 10,000 == -3000 solution for the composition and amount of the final solution is Component kg 1687 F (total) 7000
- 244. Chap. 8 Solving Material Balance Problems for Single UnITs without Reaction Step 10 Check using the total balance 7,000 + 3,000 = 10,000 To find the temperature of the final solution, calculate the composition of the final solution in terms of grams of Na2C03/1 00 grams of H20 so that you can use the tabulated solubility data listed in Steps 2-4 above. 1,687 kg Na2C03 31 .8 g Na2CO) = 5,313 kg H20 100 g H20 Thus, the temperature to which the solution must be cooled lies between 20°C and 30°e. By linear interpolation 300e - 38.8-31.8 (lO.00C) = 260C 38.8-21.5 EXAMPLE 8.7 Hemodialysis Hemodialysis is the most common method used to treat advanced and permanent kidney failure. When your kidneys fail, hannful wastes build up in your body, your blood pressure may rise, and your body may retain excess fluid and may not make enough red blood cells. In hemodialysis, your blood flows through a device with a special filter that removes wastes. The dialyzer itself (refer to Figure E8.7a) is a large canister containing thousands of small fibers through which the blood passes. /""- Blood Inlet Header Tube sh~ Solution outlet Fibers Solutioo inlet Figure E8.7a 213
- 245. 214 Solving Material Balance Problems for Single Units without Reaction Chap. 8 Dialysis solution, the cleansing solution, is around these fibers. The fibers aHow wastes and extra fluids to pass from your blood into the solution that them away. This focuses on the plasma components of the streams: water, uric acid (UR), creatinine (CR), urea (U). p. K, and Na. You can ignore the initial filling the dialyzer because the treatment lasts for an interval of two or three hours. Given the measurements obtained from one treatment shown in Figure ES. 7b. caI~ culate the grams liter of each component of plasma in the outlet solution. Solution This is an open steady~state system. StepS 1 minute Steps 3, 4, and S data are inserted on Figure ES.7b. SIn:::: 1700 mUm;n B'n"" 1100 mUmin Boot::: 1200 mUmin BlJR= 1.16 gIL I.....-_-_-.-......-_---J Bffl{ "" 60 mg/L BiS'R= giL BIJ ::: 18 gil ;0;: o.n giL ::::5.77 ~.a = 13.0 gil ~ater= 1100mUmin soot =? SB~:: ? =1 SBut ==? ~u1::: ? ~u1=? ~m: ? ~er =1 Figure ES.7b == 120 mg/L B8ut = 1 ert= 40 mg/l B~ut ::: 2.10 mgIL ~~I gil ~~er :: 1200mUmin You can the effect of the components of the pl:lsma on the density of the solution for this problem. The entering solution is assumed to be water. Steps 6 and 7 Number of unknowns (7): Number of (7): ""tn· .... ., of freedom = 0 7 components 7 components
- 246. Chap. 8 Solving Material Balance Problems for Single Units without Reaction Steps Sand 9 The water balance in assuming that 1 mL is equivalent to I gram, 1100 + 1700 = 1200 + S~Uiter hence: S~~ter = 1600 mL The component balances in grams are: UR: 1.1(1.16) + 0 = 1.2(0.060) + 1.6 S~~ = 0.75 CR: 1.1 (2.72) + 0 = 1.2(0.1 + 1.6 1.1 ( 1 8) + 0 = 1.2 ( 1.51) + 1 S~UI P: 1.1(0.77) + 0 = 1.2(0.040) + 1.6 K: L 1 (5.77) 0 = 1.2(0.l20) + 16 SrI Na: 1.1(13.0) + 0 = 1.2(3.21) + 1.6 s~~1 Frequently Asked Questions s~~ = 1.78 S~m = 11.2 S~llt = 0.50 SoKu r -- s~~ = 6.53 215 1. All of the ex.amples presented have involved only a small set of equations to solve. If you have to solve a large of equationsl some which may redundant, how can you teU if the set of equations you select to solve is a set independent equations? You can de' termlne if set is independent for linear equations by detennining rank of the coefficient matrix. of set of equations. Appendix L ex.plains how to obtain rank and shows some examples. Computer programs in Matlab, MathCad, Polymath. and so on provide a convenient way for you to determine the nmk of the coefficient matrix without having to carry out intennediate details of the calculations. Introduce the into the computer program, and the output from computer will provide you with some diagnostics if a solution is not obtained. For example, if the equations are not independent, Polymath returns the warning "Error-Singular matrix entered." 2. What should you do the computer solution you obtain by solving a set of equations gives you a negative value for one or more of the unknowns? One possibility to examine is that you inadvertently reversed the sign of a term in a material balance, say from + to-. Another possibility is that you forgot to include an essential tenn(s) so that a zero was entered into the coefficient set for the equations rather than the proper number. S IF ... ASSESSM NT ES Questions 1. Answer the following questions true or false: a. The most difficult part of solving balance problems is the collection and for-mulation of the data specifying the compositions of the streams into out the tem, and of the material inside the system.
- 247. 216 Solving Material Balance Problems for Single Units without Reaction Chap. 8 b. All open processes involving two components with three streams involve zero degrees freedom. c. An unsteady-state l ....... n""""" problem can be analyzed and solved as a steady-state process d. a flow rate is in kg/min, you should convert it to kg moUrnin. 2. Under what circumstances do equations or specifications become redundant? ./ Problems 1. A cellulose solution contains 5.2% cellulose by weight in water. How many kilograms of 1.2% solution are required to dilute 100 kg of 5.2% solution 4.2%? 2. A cereal product containing 55% water is made at rate of 500 kglhr. You to dry the product so that it contains only 30% water. How much water has to be evaporated per hour? [f 100 g of N~S04 is dissolved in 200 g of H20 and the solution is cooled until 100 g N~S04 . 10 H20 crystallizes out, find (a) the composition of the remaining solution (the mother liquor) and (b) the grams of crystals recovered per 100 g of initial solution. 4. Salt in crude oil must be removed before the oil undergoes processing in a refinery. The crude oil is to a washing unit where freshwater fed to the unit mixes with the oil and dissolves a portion of the contained in oiL oll (containing some salt but no water), being than water, can be removed at the top of the washer. If the "spent" wash water contains 15% salt and the crude oil contains 5% salt, determine the concentration of salt in "washed" oil product if the ratio crude oil (with salt) to water used is 1. Thought Problems I. Although modern counterfeiters have mastered the duplication the outside appearance of precious metals, some simple chemicaVphysical testing can determine their authentic-ity. Consult a reference book and detennine densities of gold, silver, copper, iron, nickel, and a. Could the density of gold duplicated by using any these metals? h. Could the density of pure silver be duplicated using any of these metals? c. Assume that volumes are conserved on mixing of the metals. What physical prop-erty makes any aHoy an unlikely candidate for deception? 2. Incineration is one method disposing of sludge from treatment plants. A lower limit for the combustion temperature to prevent odorous materials re-maining in the flue gas, an upper limit exists to avoid melting ash. High temperature operation of the incinerator will vaporize some of the heavy metals that cause air pollution. Thus, to prevent metals from vaporizing, it is preferable to operate at as low a temperature as possible. In one run the following data were before and after combustion of the sludge. J
- 248. Chap. 8 Solving Material Balance Problems for Single Units without Reaction Sludge Residual ash wt. % on a dry Ash Ig-loss C 67.0 15. t 87.0 13.0 5,8 H a S 1.1 1.7 mglkg on a dry basis Cd Cu 169 412 384 184 399 474 1554 1943 Was the objective achieved of preventing vaporization of heavy metals? Discussion Problems 217,.. 1. Considerable concern has been expressed that the CO2 generated from man's activities on earth has increased the CO2 concentration in the atmosphere from 275 ppm in the last century to about 350 ppm currently. What are some of the important sources and sinks for CO2 the atmosphere on earth? Make a list. Estimate as best you can from news articles, books, and journals what the amount involved is for each source and sink. Estimate the accumulat.ion of CO2 per year in the atmosphere. Suggested references are The Scientific American, Science, Nature, Chemical and Engineering News, and various databases that can be accessed via CD disks and computer terminals. 2. Forests pJay an--integral role in the dynamics of the global carbon cyde. Through photosynthesis, forests remove from the atmosphere and, accumulatt? some carbon over long periods of time. Decomposition of dead organic matter and fires release carbon back to the atmosphere. harvesting also transfers carbon out of a forest Prepare a report and figure(s) for the mass balance of carbon with a forest as. the system. Include at a minimum growth, fire, mortality, harvesting, litter, decomposition. oxidation, and the soi) and peat as components in forest system. Designate pools of carbon that accumulate or are and show the interface between forest system and the surroundings (the atmos· phere, sediments, usage by people. etc,). Looking Back In this chapter we explained via examples how to analyze problems involving material balances in the absence chemical By using the 10-step proce6 dure outlined in Chapter 7 we explained how the strategy can be used to each problem no matter what the process GLOSSARY OF NEW WORDS Tie component A component present in only one entering stream that exits only in one stream.
- 249. 218 Solving Material Balance Problems for Single Units without Reaction Chap.S SUPPLEM NTARY REF RENCES In addition to the references listed in the Frequently Asked Questions in the front ma-terial, foHowing are pertinent. Barat. R. B. The Compleat Engineer: Guide to Critical Thinking, KendalIJHunt, Dubuque, Iowa (1933). Felder, R.M., and R.W. Rousseau. Elementary Principles a/Chemical Processes, 3rd John WHey, New York (2000). H.S., S.M. Montgomery. Material and Energy Balances Stoichiometry (Soft- CACHE Austin, TX (1993). Frensch, A., and J. Funke. Complex Problem Solving, Lawrence ErlbauffiJ Hillsdale, N.J. (1995). Larson, L. Creative Problem Solving through Problems, Springer. Verlag, New York (l Pham. Q. "Degrees of Freedom of Equipment and Processes," Chern. Eng. Science, 49. 2507-25 (1994). Rubinstein, M. and 1. Firstenberg. Patterns 0/ Problem Solving, 2nd ed., Prentice- HaU, Upper SaddJe River, N. 1. (1994). Sommerfeld, 1. T. "Degrees of Freedom and Precedence Orders in Engineering Calcu)a- " Chemical Engineering Education, 1 (Summer 1986). Woods, D. R. Problem-Based Learning, Donald R. Woods Publisher, Watertown, ant, Canada (1994). Web Sites http://www.engin.umich.edullabs/mellMnEBooklet.html http://www .glue. umd.edu/-adomatilench215/matlabintro.pdf http://www.mapleapps.comlmapIeHnkslhtmVcpc2.html hup:/Iwww2.ncsu.eduJunityllockers/userS/f/felder/pubHc/PaperS/205- KnowledgeStucture.pdf ANSW ROTHE PU ZLE The held egg 1 meters above the floor before dropping it. PROBLEMS ·8.1 You buy 100 kg of cucumbers that contain 99% water. A few days later they are found to 98% Is it true that the cucumbers now only weigh 50 kg?
- 250. Chap. 8 Problems 219 ~ *8.2 The fern Pteris vittata has been shown (Nature. 409, 579 (200t)) to effectively ex~ tract arsenic from soils. The study showed that in nonnal soil, which contains 6 ppm of arsenic, in two weeks the fern reduced the soil concentration to 5 ppm while accumulating 755 ppm of arsenic. In this experiment, what was the ratio of the soil mass to the plant mass? The initial arsenic in the fern was S ppm. ·8.3 Sludge is wet solids that result from the processing in municipal sewage systems. The sludge has to be dried before it can be composted or otherwise handled. If a sludge containing 70% water and 30% solids is passed through a drier, and the resulting product contains 25% water. how much water is evaporated per ton of sludge sent to the drier. ·8.4 Figure P8.4 is a sketch of an artificial kidney, a medical device used to remove waste metabolites from your blood in cases of kidney malfunction. The dialyzing fluid passes across a hollow membrane, and the waste products diffuse from the blood into the dialyzing fluid. If the blood entering the unit flows at the rate of 220 mUmin, and the blood exiting the unit flows at the rate of 215 mllmin, how much water and urea (the main waste product) pass into the dialysate if the entering concentration of urea is 2.30 mglmL and the exit concentration of urea is 1.70 mg/L? If the dialyzing fluid flows into the unit at the rate of 100 mUmin, wh.at is the concentration of the urea in the dialysate? Hollow-Fiber Artificial Kidney Dialysate Out 1 Dialysate In Figure PS.4 ·8.5 A multiple stage evaporator concentrates a weak NaOH solution from 3% to 18%, and processes 2 tons of solution per day. How much product is made per day? How much water is evaporated per day? ·8.6 A liquid adhesive consists of a polymer dissolved in a solvent. The amount of polymer in the solution is important to the application. An adhesive dealer receives an
- 251. . 220 Solving Material Balance Problems for Single Units without Reaction Chap.B order for 3000 pounds of an adhesive solution containing 13% polymer by weight. On hand is 500 pounds of 10% solution and very large quantities of 20% solution and pure solvent. Calculate the weight of each that must be blended together to fill this order. Use all of the 10% solution. -8.7 A lacquer plant must deliver 1000 lb of an 8% nitrocellulose solution. They have in stock a 5.5% solution. How much dry nitrocellulose must be dissolved in the solution to fill the order? J*8.8 A gas containing 80% CH4 and 20% He is sent through a quartz diffusion tube (see Figure PS.8) to recover the helium. Twenty percent by weight of the original gas is recovered, and hs composition is 50% He. Calculate the composition of the waste gas if 100 kg moles of gas are processed per minute. BO%CH4 __ 20% He .. • L Figure PS.S -.~ Waste gas .. 50% He recovere d gas *8.9 In many fennentations, the maximum amount of cell mass must be obtained. However, the amount of mass that can be made is ultimately limited by the cell volume. Cells occupy a finite volume and have a rigid shape so that they cannot be packed beyond a certain limit. There wiU always be some water remaining in the interstices between the adjacent cells, which represents the void volume that at best can be as low as 40% of the fermenter volume. CaJculate the maximum cell mass on a dry basis per L of the fennenter that can be obtained if the wet cell density is 1.1 glcm3. Note that cells themselves consist of about 75% water and 25% solids, and cell mass is reported as dry weight in the fennentation industry. "'8.10 A polymer blend is to be formed from the three compounds whose compositions and approximate formulas are listed in the table. Determine the percentages of each compound A, B, and C to be inh-oduced into the mixture to achieve the desired composition. Compound (%) Composition A B C Desired mixture (CH4).r 25 35 55 30 (C2H6)x 35 20 40 30 (C3Hg)x 40 .-&. _5 40 Total 100 100 100 100 How would you decide to blend compounds A, B. C. and D f(CH4)x = 10%, (C2H6)x = 30%, (C3Hg;: 60%] to achieve the desired mixture?
- 252. Chap. 8 Problems 221 " *8.11 Your boss asks you to calculate flow through a natural~gas pipeline. it is in. in diameter, it is impossible to run the through any kind of meter or measuring device. You decide to add 100 lb of CO2 per minute to the gas through a small II2-in. piece of pipe, collect samples of the downstream. and analyze them CO2- Several consecutive samples after 1 hr are time CO2 1 hr, 0 min 2.0 10 min 2.2 20 min 1.9 30 min 2.1 40 min 2.0 (a) Calculate the flow of in pounds per minute the point of injection. (b) Unfortunately you, gas upstream the point of injection of CO2 already contained 1,0 percent CO2, How much was your flow estimate in error (in percent)? Note: In part (a) natural gas is aU methane. CH4• ·8.12 Ammonia is a gas for whic~ reliable analytical methods are available to detennine its concentration in other To measure flow in a natural gas pipeline, pure ammo-nia is injected into the pipeline at a constant rate of kg/min for 12 min. miles downstream from injection point, the steady-state ammonia concentration is found to 0.382 weight percent. gas upstream from the point of ammonia injection contains no measurable ammonia. How many kilograms natural are flow-ing through pipelines per hour? *8.13 Water pollution in the Hudson River has claimed considerable attention, espe-cially pollution from sewage outlets and industrial wastes. To determine accurately how much effluent enters the river is difficult because to catch and weigh the material is weirs are hard to and so on. suggestion that has been offered is to add a tracer of Br ion to a given sewage stream, let it mix wen, and sample the sewage it On one test of the propos3J you add ten pounds of NaBr hour for 24 hours to a sewage stream with essentially no Br in it. Somewhat downstream (he introduction point a sampling of the sewage stream shows 0.01 NaBr. The sewage density is 60.3 tb/ft3 and river water density is 62.4 Ib/ft3, What is the flow rate of the sewage in Ib/min? *8.14 A new process for separating a mixture of incompatible polymers, such as polyethyl-ene terephthalate (PET) polyvinyl chloride (PVC), promises to expand the ding and reuse plastic waste. first commercial plant. at Celanese's recycling facility in Spartanburg, S.c., has operating since February at a capacity of 15 million Ib/yr. Operating cost: 0.5¢llb. Targeted to replace the conventional sorting of individual botdes from PVC containers upstream of the recycling step, this process first chops the mixed waste with a rotary-blade cutter to O.S-in. chips. The materials are then suspended in water, and air is forced through to create a bubble-like froth that preferentially entraps the PVC be-
- 253. 222 Solving Material Balance Problems for Single Units without Reaction Chap. 8 "8.15 ··8.19 *"8.20 ·"8.21 cause of its different surface-tension characteristics. A food-grade surfactant is also added to enhance the separation. The froth is skimmed away along with the PVC, leaving behind the PET material. For a feed with 2 % PVC, the process has recovered almost pure with an acceptable PVC contamination level of 10 ppm How many Ib of PVC are recovered per year from the above cited process? If 100 g of Na2S04 is dissolved in 200 g of H20 and the solution is cooled until 100 g of Na2S04 . 10H20 crystallizes out, find (a) The composition of the remaining solution (mother liquor). (b) The grams of crystals recovered per 100 g of initial solution. A chemist attempts to prepare some very pure crystals of borax (sodium tetraborate, N~B407 . lOH20) by dissolving l00g of Na2B407 in 200 g of boiling water. He then carefully cools the solution slowly until some Na2B407 . 10H20 crystallizes out. culate the g of Na2B40 7 ' 10H20 recovered in the crystals per 100 g of total initial solution (Na2B407 plus if the residual soiution at 55"C after the crystals are reM moved contains 12.4% N~B407' 1000 kg of FeCl3 ' 6H20 are added to a mixture of crystals FeC13 . H20 to produce a mixture of FeCl3 . 2.5H20 crystals, How much FeC}3 . H20 must be added to produce the most FeCl3 . 2.5H20? The solubility of barium nitrate at 100°C is 34 g/IOO g of H20 and at O°C is S.O g/lOO g of H20. If you start with 100 g of Ba(N03h and make a saturated solution in water at 100°C) how much water is required? If the saturated solution is cooled to O°C, how much Ba(N03h is precipitated out solution? precipitated crystals carry along with them on their surface 4 g of H20 per 100 g of crystals. A water solution contains 60% Na2S202 together with 1% soluble impurity. Upon cooling to 10°C, Na2S202 . SH20 crystallizes out. The solubility of this hydrate is 1.4lb N~S202 . 5H20l1b free water. The crystals removed carry as adhering solution 0.06 Ib solutionllb crystals. When dried to remove the remaining water (but not the water of hydration), the final dry Na-zS202 . 5H20 crystals must not contain more than 0.1 % impurity. To meet this specification, the original solution, before cooling. is further diluted with water. On the basis of 100 Ib of the original solution, calculate: (a) The amount of water added before cooling. (b) percentage recovery of the Na2S202 in the dried hydrated crystals. Paper pulp sold on the basis that it contains 12 percent moisture; if the moisture exceeds this value, the purchaser can deduct any charges for the excess moisture and also deduct for the freight costs of the excess moisture. A shipment of pulp became wet and was received with a moisture content of 22 percent. If original price for the pulp was $40/ton of air-dry pulp and if the freight is $1.001100 Ib shipped, what price should be paid per ton of pulp delivered? A laundry can purchase soap containing 30% water for a price of SOJOfkg fob the soap manufacturing plant (i.e., at the soap plant before shipping costs which are owed by the purchaser of the soap). It can also purchase a different grade of soap that con£. ains only 5% water. The freight rate between the soap plant and the laundry is $6.0S/100 kg. What is the maximum price the laundry should pay for the soap?
- 254. Chap. 8 ;'*8.22 .,-S Problems 223 -' A manufacrurer of briquets has a contract to make briquets for barbecuing that are guaranteed to not contain over 10% moisture or 10% ash. The basic material they use has the analysis: moisture 12.4%. volatile material 16.6%, carbon 57.5%, and ash 13.5%. To meet the specifications (at their limits) they plan to mix with the base material a certain amount of petroleum coke that has the analysis: volatile material 8.2%. carbon 88.7%, and moisture 1 How much petroleum coke must be added per 100 lb of the base material? ·*8.23 In a gas-separation plant. the feed to the process has the foHowing constitutents: Component Mole % 1.9 i-C4 51.5 n·C4 46.0 0.6 Total 100.0 The flow rate is 5804 mol/day. If the overhead and bottoms streams leaving the process have the following compositions, what are the flow rates of the overhead and bottoms streams in kg mol/day? Mole % Component Overhead Total Bottoms 1.1 97.6 1.3 100.0 ··8.24 The organic fraction in the wastewater is measured in terms of the biological oxygen demand (BOD) material, namely the amount of dissolved oxygen required to biodegrade the organic contents. If the dissolved oxygen (DO) concentration in a body of water drops too low, the fish in the stream or lake may die. The Environmental Protection Agency has set the minimum summer levels for lakes at 5 mgIL of DO. (a) If a stream is flowing at ml/s and has an initial BOD of 5 mgIL before reach-ing the discharge point of a sewage treatment plant. and the plant discharges ML/day of wastewater, with a concentration 0.15 gIL of BOD, what will be the BOD concentration immediately below the discharge point of the plant? (b) The plant reports a discharge of MUday having a BOD of 72.09 mgIL. If the EPA measures flow of the stream before the discharge point at 530 MUday with 3 mg/L of BOD, and measures the downstream concentration of 5 mglL of BOD. is the report correct? J "'*8.25 Suppose that 100 L/mln are drawn from a fennentation tank and passed through an extraction tank in which the fennentatJon product (in the aqueous phase) mixed with an organic solvent, and then the aqueous phase is separated from the organic
- 255. 224 Solving Material Balance Problems for Single Units without Reaction Chap. 8 phase. The concentration of the desired enzyene (3-hydroxybutyrate dehydrogenase) in the aqueous feed to the extraction tank is 10.2 gIL. The pure organic extraction solvent runs into the extraction tank at the rate of 9.5 Umin. If the ratio of the enzyme in the exit product stream (the organic phase) from the extraction tank to the concentra~ tion of the enzyme in the exit waste stream (the aqueous phase) from the tank is ;;;;; 18.5 (gIL organic)/(gIL aqueous), what 1s the fraction recovery enzyme and the amount recovered per min? Assume negligible miscibility between the aqueous and organic liguids in each other. and ignore any change in density on removal or addi~ tion of the enzyme to either stream. ; ... ·8.26 Consider the fonowing process for recovering NH3 from a stream composed of N2 and NH3 Figure PB.26). Flowing upward through the process is the stream, which can contain NH3 and but not solvent S. and flowing downward through the device is a liquid stream which can contain NH3 and liquid S but not N2. weight fraction of NH) in the stream A leaving the process is to the weight fraction of NH3 in the liquid stream B leaving the process by the following emp"mcaI rIe a'tlO ns hi p: WAN HJ = 2wBNH 3' Given the shown in Figure PS,26. calculate the flow rates and composi. tions streams A and A G as 1000 kglhr 10 mol% N 90mol'%) N2 H3 Process Gas 1000 kg/hr 10 0% Liquid Solvent S Liq uid B Figure P8.26 *S.27 MTBE (methyl tertiary butyl ether) is added to gasoline increase the oxygen con-tent of the gasoline. MTBE is soluble in water to some extent, and becomes a contaminant when the gasoline gets into surface or underground water. The gasoline used by boats has an MTBE content 10%. The boats operate in a weB-mixed flood control pond having the dimensions 3 km long, 1 wide, and 3 m deep on the average. Suppose that each of the 25 boats on the pond spill 0.5 L of gasoline during 12 hours of daylight. The flow of water (that contains no MTBE) into the pond is 10 m3/hr, but no water leaves the water level is wen below the spillway of the pond. By how much will the concentration of MTBE increase in the pond ~ the end of 12 hours of boating? Data: The specific gravity of gasoline is 0.72.
- 256. CHAPTER 9 THE CHEMICAL REACTION EQUATION AND STOICHIOMETRY 9.1 Stoichiometry 9.2 Terminology for Applications of Stoichiometry Your objectives in studying this chapter are to be able to: 1. Write and balance chemical reaction equations. 2. Identify the products for common reactions given the reactants. 3. Determine the stoichiometric quantities of reactants and products in moles or mass given the chemical reaction. 4. Define excess reactant, limiting reactant, conversion, degree of completion, selectivity, yield, and extent of a reaction. 5. Identify the limiting and excess reactants in a reaction, and calculate the fraction or percent excess reactant(s). the percent conversion or completion, the yield, and the extent of reaction with the reactants given in nonstoichiometric proportions. 226 233 What is there about stochiometry that makes the topic worth reviewing at this time? You need a solid grasp of what the chemical reaction equations imply before applying them to material and energy balances. This chapter will help you enhance your understanding of this important area. looking Ahead In this chapter we review some of the concepts related to chemical reactions, and define and apply a number by tenns associated with complete and incomplete reactions. 225
- 257. I. ,! '/ 226 The Chemical Reaction Equation and Stoichiometry Chap. 9 9.1 Stoichiometry You are probably aware that chemical engineers in practicing their profession differ from most other engineers because of their involvement with chemistry. When chemical reactions occur, in contrast with physical changes of material such as evaporation or dissolution, you want to be able to predict the mass or moles required for the reaction(s), and the mass or moles of each species remaining after the reaction has occurred. Reaction stoichiometry allows you to accomplish this task. The word stoichiometry (stoi-kiom- e-tri) derives from two Greek words: sloicheion (meaning "element") and metron (meaning "measure'} Stoichiometry provides a quantitative means of relating the amount of products produced by chemical reaction(s) to the amount of reactants. As you already know. the chemical reaction equation provides both qualitative and quantitative infonnation concerning chemical reactions. Specifically the chemical reaction equation provides you with infonnation of two types: 1. It tells you what substances are reacting (those being used up) and what substances are being produced (those being made). 2. The coefficients of a balanced equation tell you what the mole ratios are among the substances that react or are produced. (In 1803. John Dalton. an English chemist. was able to explain much of the experimental results on chemical reactions of the day by assuming that reactions occurred with fixed ratios of elements.) A chemical reaction may not occur as rapidly as the combustion of natural gas in a furnace. such as, for example, in the slow oxidation of your food. but if the reaction occurs (or would occur). it takes place as represented by a chemical reaction equation. You should take the following steps in solving stoichiometric problems: 1. Make sure the chemical equation is correctly balanced. How do you tell if the reaction equation is balanced? Make sure the total quantities of each of the elements on the lefthand side equal those on the rigbthand side. For example, CH4 +02 ~ CO2+H20 is not a balanced stoichiometric equation because there are four atoms of H on the reactant side (lefthand side) of the equation. but only two on the product side (righthand side). In addition, the oxygen atoms do not balance. The balanced equation is given by CH4 + 2 02 ~ CO2 + 2 H20 The coefficients in the balanced reaction equation have the units of moles of a species reacting or produced relative to the other species reacting for the particu-
- 258. Sec. 9.1 Stoichiometry 221 tar reaction equation. If you multiply each term in a chemical reaction equation by the same constant, say two, the absolute stiochiometric coefficient each tenn doubles. but the coefficients still exist in the same relative proportions. 2. Use the proper degree of completion for the reaction. If you do not know how much reaction has occurred, you have to assume some amount, such as complete reaction. 3. Use molecular weights to convert mass to moles for the reactants and products, and vice versa. 4. Use the coefficients in the chemical equation to obtain the molar amounts of products produced and reactants consumed by the reaction. Steps 3 and 4 can be applied in a fashion similar to that used in carrying out the conversion units as explained in Chapter 1. As an example, the combustion of heptane takes place according to the fonowing reaction equation (Note that we have put the states of the compounds in parentheses after the species formula) information not needed for this chapter, but that will be vital in Parts 3 and 4 of this book.) What can you learn from a chemical reaction equation? The stoichiometric coefficients in the chemical reaction equation (l for C7H16, 11 for 02' and so on) tell you the relative amounts of moles of chemical species that react and are produced by the reaction. The units of a stoichiometric coefficient for species i are the change in the moles of species i divided by the moles reacting according to the specific chemical equation. We will abbreviate the units simply as mol vmoles reacting when appropriate, but frequently in practice the units of moles reacting are ignored. You can conclude that 1 mole (not Ibm or kg) of heptane wi]) react with 11 moles of oxygen to give 7 moles of carbon dioxide plus 8 moles of water. These may be Ib mol, g mol, kg mol, or any other type of mole. Another way to use the chemical reaction equation is to indicate that 1 mole of CO2 is formed from each ~ mole of C7H16, and I mole of H20 is formed with each ~ mole of CO2- The latter ratios indiuse of stoichiometric ratios determining the relative proportions of products and reactants. Suppose you are asked how many kg of CO2 will be produced as the product if 10 kg of C7H I6 react completely with the stoichiometric quantity of °21 On the basis of 10 kg of C7H6 10 kg mol CO2 44.0 kg = 30.8 kg CO2 g rno 16 1 g mo 2 1 k I k 1 CO
- 259. The Chemical Reaction Equation and Stoichiometry Chap. 9 EXAMPLE 9.1 Balancing a Reaction Equation ror a Biological Reaction The primary energy source for cells is the aerobic catabolism (oxidation) of glucose (C6HI20 6• a sugar). The overall oxidation glucose produces CO2 H20 by the following reaction C6H1206 + a02 -+ b CO2 + C H20, , Detennine tion. ' val~es of a. b. and c that balance ~is ,~hemical reaction equa- ,~. . Solution . . . ~ Basis.: TI1.e given reaction " .. , . , By insPection. th~ carbon 'balance gives j, .:" the hydrogeri balanCe gives . C ,= 6. and an oxygen balance 6+2a:6x2+6 gives a = 6. Therefore, the balanced equation'is C6HI20 6 + 602 ~ 6C02 ... 6H20 Let's now write a general chemical reaction equation as i ,I c + d D a.A bB (9.1) where a. b. Ct and d are the stoichiometric coefficients for species A, B, C, and D, respectively. Equation (9.1) can written in a general form vAA + vBB+ (9.2) . . .' . , where ,Vj..is the stoichiometric coefficient for species' Sj. The products are defined to (,. have positive values for coefficients and the reactants to have negative values for coefficients. The ratios 'are fo'r.a given reac6on~, Specifically in Equation ---.f'.J- ------- -- ---. ------------ (, vo. -d VB' b ~ ----. ..-'" . If a species not in an eqtl 8:ri on , the value of its stoichiometric coefficient is ppnrip'n to be' zero. As an' example, in the re8:ctiOIf' ': '" " ' . O2 '+ 2eO ~ 2CO~ , . , '':''; 2 .. ·· VN': 0 . , '.' " .. :2 , " ,.", "" .. " ,,- ' , ".' *
- 260. Sec. 9.1 Stoichiometry EXAMPLE 9.2 Use of the Chemical Equation to Calculate the Mass of Reactants Given the Mass of Products In the combustion of heptane. CO2 is produced. Assume that you want to produce 500 kg of dry ice per hour. and that 50% of the CO2 can be converted into dry ice, as shown in Figure E9.2. How many kilograms of heptane must be burned per hour? Other ProGJcts COt Gos 150%) ............ ---"'t"""l· CO2 Solid C11-, Gos (50~ R.odor 0t GO& 500 kg/II Figure E9.1 Solution In solving a problem of this sort. the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much. Sherlock Holmes, in Sir Arthur Conan Doyle's a Study in Scarlet From the problem statement you can conclude that you want to use the product mass of CO2 to calculate a. reactant mass, the 16' The procedure is flrSt to convert kilograms of CO2 to moles, apply the chemical equation to moles of C,HI6 > and finally calculate the kilograms of C7Hu1' We will use Figure E9.2 the analysis. Look in Appendix Dl to get the molecular weights of CO2 (44.0) and C;H!6 (100.1), The chemical equation tions. C7Hr6 + 1102 ~ 7C02 + 8H20 The next step is to select a Basis: SOO kg of dry ice (equivalent to 1 hr) The calculation of the amount of C7HI6 can be made in one sequence: 500 mol CO2 1 mol 0.5 kg dry ice 44.0 kg CO2 7 kg rno] CO2 100.1 -----= 1 kg mol C7H16 Finally. you should check your answer by reversing the sequence of calcula- 229 ,.
- 261. The Chemical Reaction Equation and Stoichiometry EXAMPLE 93 Application of Stoichiometry When More than One Reaction Occurs A limestone analyses (weight %) . CaC03 MgC03 Inert 92.89% 5.41% 1.70% By heating the limestone you recover oxides known as lime. Chap. 9 (a) How many pounds of calcium oxide can be made from I ton of this limestone? (b) How many pounds of CO2 can be recovered per pound of limestone? (c) How many pounds of limestone are needed to make 1 ton of lime? Solution Steps 1, 2, and 3 Read the problem carefully to fix in mind exactly what is required. The carbonates are decomposed to oxides. You should recognize that lime (oxides of Ca and Mg) will also include other inert compounds present in the limestone that remain after the CO2 has been driven off. Step 2 Next, draw a picture of what is going on in this process. See Figure E9.3', ." / ) I I CO2 Limestone I I CoO }Li~ MgO Hect inert Figure E9..3 Step 4 To complete the preliminary analysis you need the following chemical equa-tions: CaC03 -+ CaO + CO2 MgC03 ---io MgO + CO2 Additional data that you need to look up (or calculate) are the molecular weights of the species j
- 262. Sec. 1 StepS Stoichiometry Mol. Wt.~ CaCOl 100.1 next step is to pick a Basis: 100 Ib of limestone This was selected because pounds will be equal to percent. You could also pick 1 of limestone if you wanted, or 1 ton. Steps 6, 7, 8. and 9 Calculations of the percent composition and lb moles of the and products the fonn of a table will serve as an adjunct to Figure E9.3, and will prove helpful in answering posed. MgCOJ Inert Total The q~antities exam~le, for 92.89 5.41 lb Limestone Solid Products Ib == percent lbmol Compound Ib mol Ib 92.89 0.9280 0.9280 52.04 5.41 0.0642 0.0642 1.70 100.00 0.9920 0.9920 from the chemical equations. lIb mol CaCOl lIb mol CaO 56.081b CaO = 52.041b CaO 100.1 Ib CaCOl 1 Ib mol CaC03 1 lb mol CaO lIb mol MgO 1 Ib mol MgCO] The production of 0.9280 Ib mol CaO is equivalent to 0.9280 Ib mol CO2 0.0642 MgO is equivalent to 0.0642 lb mol Total 0.992 lb C02 O.9921b mol 44.0 lb CO2 1 1 lb mol CO 2 = 44.65 b Alternately, you could have calculated the Ib CO2 a total balance: 100 - 56.33 = 44.67. Note that total pounds of an of the products 100 lb of entering limestone. If it did not. what would you do? Check weight values and your calculations. 231
- 263. 232 1. The Chemical Reaction Equation and Stoichiometry Now, to calculate the quantities originally asked for: 52.04 lb CaD 2000 Ib (a) CaO produced :: = 10411b CaO/ton 100 Ib limestone 1 ton IbC02 recovered = = 0.437 Jb C02flb limestone 100 Ib limestone (b) (c) L' 'red - 100 Ib limestone 2000 Ib _ 3550 lb limestone! Imestone reqUl - 56.33 Ib lime I ton - lime SELF-AS ESSMENT T ST Question the following reaction Mn04" + 5Fe2+ Chap. 9 a student wrote the following to determine how many moles of MnO" would react with 3 moles of 1 mol MnOi 5 mol to 1 mol MnOi/5 mol . Divide both sides by 5 mol = 1. The number of mol of MnOi that reacts with 3 mol 1 molMn04" ---~(3 mol .5 mol Is calculation correct? Problems Fe2+ = 0.6 mol MnOr 1. balanced reaction equations for the following reactions: a. 8 and oxygen to fonn carbon dioxide and water. b. and oxygen to fonn and sulfur dioxide. 2. If 1 kg of benzene (C6H6) is oxidized with oxygen, how many kilograms of 02 are needed to convert all the benzene to CO2 and H20? 3. The electrolytic manufacture of chlorine gas from a sodium chloride solution is carried out by following reaction: 2 NaCl + 2 H20 ~ 2 NaOH + + Cll How many kilograms can be produced from 10m3 of brine solution containing 5% by ofNaCl? The gravity of the solution relative that of water at is 1.07. 4. Can you balance following chemical reaction equation? a, N03 + a2HCIO ~ ~HN03 + a4HC)
- 264. Sec. 9.2 Terminology for Applications Stpichiometry 233' Thought Problems 1. An accident occurred in which <:me worker lost his A large evaporator in magne-sium chloride service. containing internal heating was to be cleaned. It was shut down, drained, and washed. next day two employees who were involved in the tenance of the evaporator the vessel to repair the They were overcome, ap· parently from Jack oxygen. Subsequently> one employee recovered and but the other never regained consciousness, died several days later. What your opinion might have caused the accident (the lack oxygen)? 2. magazine reported that to degrade crude oil in a spill in seawater the oil-degrading bac-require the dissolved oxygen in over 300,000 gal of air-saturated seawater to break down just 1 gal of crude. Can this estimate be correct? An article advocating the planting trees explains that a tree can assimilate 13 Ib of car-bon dioxide per year, or enough to the CO2 produced by one car 26,000 miles per year. this statement be correct? Discussion Problems Pollutants diesel fuel pose a threat to respira.tory tract and are a potential cause of cancer according to the Environmental Protection Agency. The Clean Air Act 1990 required that the sulfur content of fuel used on freeways must be lowered from 0.30 percent by weight to 0.05 percent-a substantial reduction. How might this be accomplished economically? At the oil at the at the service station, in the or what? 9.2 Terminology for Applications of Stoich iometry Nothing is like it seems bur everything is Like it Berra So far we .have dl,scussed the stoichiometry :re~cti.9flsjn'.·whir;.h ·~,e:'·proper.,' stoichiometric rati.6 of ' . 'are' into: a reactor~,'an:d the feacti6n::,goes, ' ,com-:' pletion. Subsequently, no reactants remain in the What if (a) some other ratio reactants.is fed, or~(b) r~action is incomplete? In'such'drcutnstances you to be familiar with a number of terms you will encounter in solving various ofpr~b~e~s. ,. '," "., " : "" , ' , ,:.,:.',': ,~ . 9:2~.1' . ::··~:':.~~~nt:ot A_cti(,ri' .. :::. ..: ....... :.':~'.:',,: . "'i: .... .. ... ...• .•• . . .. _ ~A'" .. '.. " _ ,',.. .' " ~".' ~.. . '. ~. :' You find the extent of reaction useful ." volvi'~g' c'lle~kaJ "r~a~tibh: "'Th~ 'extent of' reactio~, • • ", • ~ 4 , ... ' . ".". -. ' . ", . . ,solving, material baJances ini~ 'based o~ a particular stoi- .. . . , . ' ' .. , '. .. ',", .' . '.
- 265. 234 Chemical Reaction and Stoichiometry Chap. 9 chiometric equation, and denotes how much reaction occurs. Its units are "moles re-n The of reaction calculated by dividing change in number moles of a species that occurs in a reaction, for a reactant or a product, by related stoichiometric coefficient. example, consider the chemical reaction equa-tion for the combustion of monoxide 2CO + O2 ~ 2C02 The signs stoichiometric coefficients to used wiU conform what is dard practice in calculating the extent of namely the products of the reac-tion have positive and the reactants have negative signs. 20 meies of CO are to a with 10 moles of 02 and form 15 of the of reaction can be calculated from amount of CO2 that produced The value of change in the moles of CO2 is: 15 0:::: 15. The value of the stoichiometric coefficient for the CO2 is 2 moIlrnol Then the extent reaction is: (15 0) mol CO2 moles reacting 2 mol CO2/moles reacting S next consider a more formal definition the extent of reaction, one that into account incomplete reaction, and involves the initial concentrations of re-actants and products. extent of reaction is defined as follows: (93) in the system after the reaction occurs nio == moles of species i present in system when the reaction starts Vi = coefficient for species i the particular chemical reaction equation (moles of i produced or consumed per moles reacting) g = extent reaction (moles reacting) The coefficients of the products a chemical reaction are assigited positive values and reactants assigned negative values. Note that (nj - n;o) equal to the generation or consumption of component i by reaction. Equation (9.3) can be rearranged to calculate number of moles of compo-nent i from the value of extent of n·:::: n·o + t:v· I ! ~ I (9.4) As shown in the example, production or consumption of one species can be
- 266. Sec. 9.2 Terminology for Applications of Stoichiometry 235,..- used to calculate the production or consumption of any of the other species involved in a reaction once you calculate, or are given, the value of the extent of reaction. EXAMPLE 9.4 Calculation of the Extent of Reaction Detennine the extent of reaction for the following chemical reaction N2 + 3H2 -) 2NH3 gi yen the following analysis of feed and product: Feed 100 g 50 g 5g Product 90g Also, detennine the g and g mol of N2 and H2 in the product, and the acid rain potential (ARP) of the NH3. The acid rain potential can be characterized by the number of moles of H+ created per number of moles of compound from which the H+ are created. For ammonia the reaction considered is NH3: NH3 + 202~W + N03" + H20 In practice, the potential for acidification is expressed on a mass basis normalized by a reference compound, namely S02' for which the reaction considered produces two H+ S02: S02 + H20 + 03 ....... 2 H+ + S042- + O2 Thus the ARP is calculated as Solution mole Ht MW· ARp· = ---,'---- I mole S02 MWS02 The extent of reaction can be calculated by applying Equation (9.3) based on NH3: 90 g NH3 1 g mol NH3 nj = = 5.294 g mol NH3 17 g NH3 5 g NH3 1 g mole NH3 nlO = = 0.294 g mol NH3 17 g NHJ nj - niO (5.294 - 0.204)g mol NH3 . ~ = = = 2.50 moles reacting Vi 2 g mol NHy'moles reacting . .
- 267. 236 The Chemical Reaction Equation and Stoichiometry Equation (9.4) can be used to detennine the g mol of N2 and H2 in the reaction N2: 100 g N2 1 g N2 = 3.57 g mol N2 11m = 28 gN2 llN2 = 3.57 + 1 )(2.5) = 1.07 g mol N2 1 g mol N2 28 gN2 - = 30 g N2 1 g mol N2 50g 1 g mol H2 11;0 = - g mol H2 2gH2 nN2 = + ( )(2.5) 17.5 g mol H2 17.5 g mol --=--=-- - 1 g mol H2 The = 0/17)/(2/64) = 1.88 products of 9 If several independent (refer Appendix L for the meaning independent) reactions occur in the reactor, say k of them, , can defined for each reaction. With Vki being the stoichiometric coefficient species i in the kth reaction, the total number of of species i R ni = niO + ~ ek k=1 where the total number of independent reactions. (9.5) To summarize. the important characteristic of the variable ~ defined in Equation (9.3) that it has the same value for each molecular species involved in a parreaction. Thus, given the initial mole numbers of all species and a value for' (or the change in the number of moles of one species from which the value of ~ can be calculated), you can easily compute all other numbers of moles in the system after the reaction takes place. 9 .. 2 ... 2 limiting and Excess Reactants In industrial reactors you will rarely find exact stoichiometric amounts of materials used. To make a desired reaction take place or to use up a costly reactant, excess reactants are nearly always used. The excess material comes out with, or perhaps separately from, the product, and sometimes can be again. The Hm ..
- 268. Sec. 9.2 Terminology for Applications of Stoichiometry 237 .- iCing reactant is the species in a chemical reaction that would theoretically run out first (would be completely consumed) if the reaction were to proceed to completion according to the chemical equation-even jf the reaction does not proceed to completion! An other reactants are caUed excess reactants. amount of the excess reactant fed - amount of the ..,"' .... .., .. '" reactant required to react with the limiting reactant % excess reactant = 100 --'-----,-------~-------~ amount of the excess reactant required to react with limiting reactant For example. using the chemical reaction equation in Example 9.2. C7HI6 + ] 102 """, 7C02 + 8H20 if 1 g mol of C,H16 and 12 g morof 02 are mixed, C7Hl6 would be the limiting reactant even if the reaction does not take place. The amount of the excess reactant would be 12 g mol less the 11 g mole needed to react with 1 g mol of C7H16; or 1 g mol of 02' Therefore. if the reaction were to go to completion, the amount of product produced would be controlled by the amount of the limiting reactant. As a straightforward way of determining the limiting reactant, you can determine the maximum extent of reaction, ~ax, for each reactant based on the complete reaction of the reactant. The reactant with the smallest maximum extent of reac .. tion is the limiting reactant. For the example. for 1 g mol of <;H16 plus 12 g mole of 02' you calculate o mol O2 - 12 g mol O2 ~max (based on 02) = 10 1 . = 1.09 moles reacting - 11 g mo 2/mo es reacting o g mol -1 g mol C,HI6 emax (based on C,H16) = 1 I C H _I 1 . = 1.00 moles reacting - g rno 7 16'mo es reacting Therefore, heptane is the limiting reactant and oxygen is the excess reactant. Consider the foHowing reaction A + 3B + 2C ......, Products If the feed to the reactor contains J.l moles of A, 2 moles of B, and 2.4 moles of C. The of reaction based on complete reaction of A, B. and C are gmn (based on A) = -1.1 mol A 1.1 -1 -3.2 mol B ~JruU( (based on B) = = 1.07 ,max (based on C) = -2.4 mol C = 1.2
- 269. 238 The Chemical Reaction Equation and Stoichiometry Chap. 9 As a result, B is identified as the limiting reactant in this example while A and are the e~cess reactants. As an alternate to detennining the limiting reactant, all you have to do is to calculate the mole ratio{s) of the reactants and compare each ratio with the corresponding ratio of the coefficients of the reactants in the chemical equation thus: 12 - 12 1 > 11 - = 11 1 If more than two reactants are present, you have to use one reactant as the reference substance. calculate the mole ratios of the other reactants in the feed relative to the reference, make pairwise comparisons versus the analogous ratios in the chemical equation, and rank each compound. For example, given the reaction A + 3B + 2C ~ products and that 1.1 moles of A. 3.2 mole::; of B, and 2.4 moles of C are fed as reactants in the reactor, we choose A as the reference substance and calculate A 3.2 = 2.91 1.1 2.4 -= 18 1.1 < > We conclude that B is the limiting reactant relative to A, and that A is the limiting reactant relative to C, hence B the limiting reactant among the set of reactants. In symbols we have B < A, C> A (i.e., A < C), so that B < A < C. EXAMPLE 9.5 Calculation of the Limiting and Excess Reactants Given the Mass of Reactants If you feed 10 grams of N2 and 10 grams of H2 gas into a reactor: a. What is the maximum number of grams of NH3 that can be produced? h. What is the limiting reactant? c. What is the excess reactant?
- 270. Terminology for Applications of Stoichiometry Solution You arc asked to calculate the limiting reactant, and use a chemical reaction equation to calculate the NH3 produced. At room temperature and pressure no reaction will occur, but you are asked to calculate what would result if the reaction were to occur (as it does under other conditions of temperature and pressure), Look at E9.5. N2 (g) ___. ...... ,..j 10 9 Reactor 1-+---- H2 (g) 10 9 Next. write down Given g: MW: Cal cd . g mo]: Figure E9.5 chemical equation. and N2{g) 10 28 0.357 + 3H2(g) 10 2.016 4.960 the molecular weights: 2NH3(g) o 17.02 o The next step is to determine the limiting reactant by calculating the maximum extent of reaction based on the complete reaction of N2 and -0.357 g mol N2 ~m!lX (based on N2) = 1 N I I . = 0.357 moles reacting - g mo] 2 mo es reactmg , -4.960 mol H2 . 1 HI' = 1.65 moles reacting g rno 2 moles reactmg You can conclude that (b) N2 is the limiting reactant, and that (c) H2 is the excess reactant. The excess H2 is 4.960 - 3(0.357) = 3.89 g mol. To answer question (a), the maximum amount of NH3 that can be produced is based on assuming complete conversion of the limiting reactant Finally, you should check your answer by working from the answer to the given reactant, or, alternatively. by adding up the mass of the NH3 and the mass of excess H2• What should the sum be? 239 '
- 271. 240 The Chemical Reaction Equation and Stoichiometry Chap. 9 9.2-3 Conversion and degree of completion Conversion and degree of are terms not as defined as are the extent and limiting and excess reactant. Rather than cite an the possi-ble usages of terms, many of which conflict, we shall them as follows. Conversion the fraction of the feed or some key material in the feed that converted into products. Conversion related to the degree of completion of a reaction namely the percentage or fraction of the limiting reactant converted into products. The numerator and denominator of the fraction same units so that the fraction conversion is dimensionless. percent conversion % conversion = ( or a compound in the feed) that react (or a component in the feed) introduced example, for the reaction equation described in Example 9.2, if 14.4 of CO2 are formed in the of 10 kg of C7H16, can calculate what percent of the C,HI6 is converted to (reacts) as follows: C7H 16 equivalent 14.4 1 kg mol CO2 mol C7H16 --~---.;;. to CO2 in the product 44.0 kg CO2 mol CO 2 = 0.0468 mol C7H 16 16 in the reactants 10 I mol ---'-------:.. 00 k C H = 0.0999 kg mol C7H 16 1 .1 g 7 16 0.0468 rna! reacted % conversion = d 100 = 46.8% of 0.0999 mol fe The conversion can calculated extent of reaction as follows: conversion equal to the extent of reaction based on formation (Le., the actual reaction) divided by the extent of reaction assuming complete of 16 (i the maximum possible extent of reaction). of reaction that ........ " ........ occurs converSlOn = f extent 0 ......... ' ............ " .. that would occur (9.5) 9.2-4 Selectivity Selectivity is the ratio of the moles of a particular (usually the desired) product produced to the moles of another (usually or by-product)
- 272. Sec. 9.2 Terminology for Applications of Stoichiometry 241 product produced in a set of reactions. For example, methanol (CH30H) can be converted into ethylene (C2H4) or propylene (C3H6) by the reactions 2 CH30H -+ C214 2H20 3 CH30H -+ C314 + 3HzO Of course, for the process to be economical, the prices of the products have to significantly greater than the reactants. Examine the data in Figure 9.1 for the ,concentrations of the products of the reactions. What is the selectivity of C2H4 rela' live to the C3H6 at 80% conversion of the CHlOH? Proceed upward at 80% conversion to get for C2H4 :; 19 mole % and for C3H6 :; 8 mole %. Because the basis for both values is the same, you can compute the selectivity 19/8 == 2.4 mol c;H4 per mol C3H6. - C:]H6 10 & ~, .. c ~ • ~ ~ CI) 5 5 0. 0 0 40 60 80 40 60 ao Percent conversion of CH:PH Percent conversion of CH30H Figure 9.1 Products from me conversion of ethanol. 9.2-5 Yield No universally agreed~upon definitions exist for yield-in fact~ quite the con, ,trary. Here are three common ones: .. yield (based on feed}-the amount (mass or moles) of desired product ob- / tained divided by the amount of the key (frequently the limiting) reactant fed. • yield (based on reactant consumed)-the amount (mass or moles) of desired product obtained divided by amount of the key (frequently the limiting) rectant consumed. • yield (based on theoretical consumption of the limiting reactant)-the amount (mass or moles) of a product obtained divided by the theoretical (expected) amount of the product that would be obtained based on the limiting reactant in the chemical reaction equation(s) if it were completely consumed.
- 273. 242 The Chemical Reaction Equation and Stoichiometry Chap. 9 EXAMPLE 9.6 Yields in the Reaction of Glucose to Produce Ethanol Yeasts are living organisms that consume sugars and produce a variety of products. For example, yeasts are used to convert malt to beer and corn to ethanol. The growth of cerevisiae (a specific type of yeast) on glucose (a sugar) under anaerobic conditions (in the absence of oxygen) proceeds by the following overall reaction to produce biomass. glycerol, and ethanol C6H120 6(gIucose) + 0.118 NH) -+ 0.59 CHL74No.200.45 (biomass) + 0.43 C3Hs0 3(glycerol) -+- 1.54 CO2 + 1.3 C2HsOH (ethanol) + 0.03 H20 Calculate theoretical yield of biomass in g of biomass per g of glucose. Also, calculate the yield of ethanol in g of ethanol per g of glucose. Solution Basis: 0.59 g mol of biomass 0.59 mol biomass g biomass ---'--- = 0.0778 g biomass/g glucose 1 g mol glucose I g mol biomass 1.3 g mol C2HsOH --=---=------ ~--"'--- = 0.332 g C1HsOHJg glucose ] g mol glucose Why doesn't the actual yield in a reaction equal the theoretical yield predicted from the chemical reaction equation? Several reasons exist: ., impurities among the reactants '" leaks to the environment II side reactions it reversible reactions As an illustration) suppose you have a reaction sequence as follows: A~B~C ~ with B being the desired product and C the undesired one. The yield B according to the first two definitions is the moles (or mass) of B produced divided by the respective moles (or mass) of A fed or consumed. The yield according to the third definition moles (or mass) of divided by the maximum amount of B that could
- 274. Sec. 9.2 Terminology for Applications of Stoichiometry 243 be produced in the reaction sequence. The selectivity of B is the moles of B divided by the moles of C produced. The terms "yield" and "selectivity" are terms that measure the degree to which a desired reaction proceeds relative to competing alternative (undesirable) reactions. As a designer of equipment you want to maximize production of the desired product and minimize production of the unwanted products. Do you want high or low selectivity? Yield? EXAMPLE 9.7 Selectivity in the Production of Nanotubes A carbon nanotube may consist of a single wall tube or a number of concentric tubes. A single wal1 tube may be produced as unaligned structures or bundles of ropes packed together in an orderly manner. The structure of the nanotubes influences its properties, such as conductance. Some kinds are conductors and some semiconductors. In nanotechnology, numerous methods (arc-discharge,- laser vaporization, chemical vapor deposition, and so on) exist to produce nanotubes. For example, large amounts of single wall carbon nanotubes can be produced by the catalytic decomposition of ethane over Co and Fe catalysts supported on silica C2H6 ---+ 2 C + 3 H2 (a) i C2H4 + H2 (b) If you collec( 3 g mol of H2 and 0.50 g mol of CZH4, what is the selectivity of C relative- to C2H4? Solution Basis: 3 g mol H2 by Reaction (a) 0.50 g mol C2H4 by Reaction (b) The 0.5 g mol of C2H4 corresponds to 0.50 g mol of H2 produced in Reaction (b). Then the H2 produced by Reaction (a) was 3 - 0.50 = 2.5 g mol. Consequently, the nanotubes (the C) produced by Reaction (a) was (2/3)(2.5) = 1.67 g mol C The selectivity was 1.67/0.50 = 3.33 g mol C/g mol C2H4 The next example shows you how to calculate all of the tenns discussed above in Section 9.2.
- 275. 244 The Chemical Reaction Equation and Stoichiometry EXAMPLE 9.8 Calculation of Various Terms Pertaining to Reactions Chap. 9 Semenov (Some Problems in Chemical Kinetics and Reactivity, Princeton Univ. Press (1959), Vol II. 39-42) described some of the chemistry of allyl chlorides. The two reactions of interest for this example are CI2(g) + C3H6(g) -;+ C3HsCI(g) + HCl(g) C12(S) + C3H6(g) ~ C3H6C12(g) C]H6 is propylene (propene) (MW;;;:; 42.08) allyl chloride (3-chloropropene) (MW ;;;:; 76.53) C3H6CI2, is propylene chloride (l,2-dichloropropane) (MW;;;:; 112.99) (a) (b) The species recovered after the reaction takes place for some time are Hsted in Table Species C12 C)H6 C:lHSCl C3H6Cl2 HCl TABLEE9.8 gmat 141.0 1.0 4.6 24.5 4.6 Based on the product distribution assuming that no aUyl chlorides were sent the feed, the following: a. How much CI2 and C3H6 were to the reactor in g mol? b. What was the limiting reactant? c. What was the excess reactant? d. What was the fraction conversion of C3H6 to C)HSCI? e. What was the selectivity of C)HSCI relative to C3~C12 ? f. What was the yield of C3HsCl expressed in g of C)HSCI to the g of C3H6 fed to the reactor? g. What was the extent of reaction of the and second reactions? h. In the application of green chemistry. you would like to identify classes of chemical reactions that have the potential for process improvement. particularly waste reduction. In this example the waste is HCl (g). The C12 is not considered to be a waste because it is recycled. What is the mole efficiency. fraction an element in the entering reactants that emerges exiting products, for chlorine? , '.
- 276. Terminology for Applications of Stoichiometry SoluUon Steps 1, 2, 3, and " Examination of the problem statement reveals that the amount of feed not given, and consequently you must first calculate the g mol fed to the reactor even if the amounts were not asked for. 'lbe molecular weights are given. Figure E9.8 illusttates the process as an open-flow system. A batch process could alternatively be used. StepS a. (g) I ~H8 (g) III Figure E9.8 A convenient basis is what is given in the product list in Table E9.8. Steps " 8t and 9 Use the chemical equations to calculate the moles of species in the feed. Reaction (a) Rt1action (b) 24.5 g mol C3H6CtZ 1 mot - 1 g mol C314Ch = 24.5 g mol C12 reacts (s> Total 29.1 g mol C12 reacts Cl2 in product T4IO Total Cl2 fed 170.1 From the chemical equations you can see that if 29. I g mol 12 reacts by Reactions (8) and (b)t 29.1 g mol of C3H6 must react Since 651.0 g mol of C3~ exist in, the product. were fed to the reactor. You can check those answers by adding up the respective g mol of C 1, C. and H in the product and comparing the values with that calculated in the feed: In product Cl 2(141.0) + 1(4.6) + 2(24.5) + 1(4.6) = 340.2 C 3(651) + 3(4.6) + 3(24.5) = 2040.3 H 6(651) + 5(4.6) + 6(24.5) + 1(4.6) = 4080.6 245
- 277. 246 Cl C H The Chemical Reaction Equation and Stoichiometry In feed 2(170.1) = 340.2 3(680.1) = 2040.3 6(680.1) = 4080.6 OK OK OK We will not go through detailed steps for the remaining calculations, but ply determine the desired based on the data prepared for Part (a), Chap. 9 (b) and (c) Since both reactions involve the same value of the respective re-action stoichiometric coefficients, both reactions win have same limiting and excess ~max (based on -680.1 mol - = 680.1 -) g mol C3H(/'mo1es reacting reacting 170.1 mole Cl2 (based on C12) = Cl 1 . = 170.1 moles reacting -1 g mol zlmo es reactmg Thus, C)H6 was the excess reactant and Cl 2 the limiting reactant. (d) The 1 was 4.6 mol C3H6 that reacted - 680.1 g mol fed (e) The selectivity was (I) The yield was (g) C3HsCl is produced only by the first reaction. the extent of reac-tion of the first reaction is e. = ni - nio = __ -_0 = 4.6 Vi 1 Because C3H6CI2 is produced only by the second reaction, the extent of reaction of the second reaction is 24.5 - 0 = 24.5 1
- 278. 9.2 Terminology for Applications of Stoichiometry (h) Mole efficiency in the waste: Cl: (170.1 ::: 340.2 g mol Exiting Cl in 4.6 g mol -mo-le o-f ch-lor-ine- in -was-te = -4.-6 = 0.0135 mole of chlorine entering 340.2 Mole efficiency of the product::: 1 - 0.0135 ::: 0.987 It would difficult to find a better reaction pathway to obtain the indicated prod-ucts. Of course, the processing of HCl (g) must be considered. SELF .. A SESSMENT TEST Questions 1. What is the symbol used to denote the extent of reaction? 2. What a limiting reactant? What is an excess reactant? 4. How do you calculate the extent of reaction from experimental data? Problems 1. For the reaction in which stoichiometric quantities of the reactants are fed 2 CjHIO + 15 02 ~ 10 CO2 + 10 H20 247 and the reaction goes to completion, what the maximum extent of reaction based on CsH IO? On 01? Are the respective values different or the same? Explain the result. 2. Calcium (CaO) formed by decomposing limestone (pure CaC03). In one kiln the reaction goes to completion. a, What the composition of the solid product withdrawn from kiln? b. What is the yield in terms of pounds of produced per pound of limestone fed into process? 3. Aluminum sulfate can be made by reacting crushed bauxite ore with sulfuric acid, accordto the following chemical equation: The bauxite ore contains 55.4% by weight of aluminum oxide, the remainder being impuThe sulfuric acid solution contains 77.7% pure sulfuric acid, the remainder being water.
- 279. 248 The Chemical ~~ .... 'tl ..... n Equation and Stoichiometry Chap. 9 To produce crude aluminum sulfate containing 1798 Ib of pure aluminum 1080 lb of bauxite ore and 2510 Ib of sulfuric acid solution are reacted. in the dehydration of ethane: C2H6 ~ C2H4 + H2 C2H6 + -? 2 the product distribution measured 1'~ ~-eofu:. hase reaction of C2~ as follows (a) What """"""' ...... "" was the limiti8g ..... ""-""' ... (b) (c) Cd) (e) (0 What SO(~le~s was the excess reactant? What was was What was C2H6? conversion of C2H6 to CH4? degree of completion of the selectivity of C2H4 relative to CH4? yield of C2H4 expressed in mol of C2H4 produced per kg mol (g) What was extent of reaction of C2H6 ? Thought Problem (a) (b) 1. OSHA use of a breathing apparatus or around tanks contain-ing traces of While demolishing an old a contractor purchased several cylinders of l"'l"nTln1!"PC1'O;:p.1'1 painted gray. After two he found that he needed more cylinders, for another cylinder. The driver returned with a black cylinder. None of including the man in charge of the breathing apparatus, noticed the into use. a not injured. attached any importance to it. When the new cylinder was brought piece caught flIe. Fortunately, he pulled it off at once and was What would the most likely cause of this ... "" .... A ...... 'U Discussion Problem 1. On November 1, 1986, a substantial number of rials being introduced fighting the fire. at a Sandoz storehouse near Switzerland. resulted in a ...... u .. "''''. pesticides, dyes. and raw and intermediate mate-the Rhine River via runoff of about 15,000 m3 water used in to the North Sea where the Rhine is about 1200 J
- 280. Sec. 9.2 Terminology for Applications of Stoichiometry 249"' The table lists some of the compounds discharged into the river along with the LC50 value (the concentration that will kit] 50% of rainbow trout). Compound (1) Thiometon (I) Ethoxyethyl mercury hydroxide (P) DNOC(P) Endosulfan (P) ! = insecticide; P "" pestl:CII;te Estimated discharge (kg) 3000-9000 1200-4000 18-200 600-2000 20-60 Estimated concentration near discharge point (,...gIL) 600 500 100-430 LC50 (,...gIL) 6000 8000 3 to 1000 66 to 1250 1.4 What were the probable consequences of the discharge along the river to the biota. drinking water, and benthic organisms? Note that the Rhine several dams to provide water for navigation. What would concentration these compounds be as a function of time at various towns downstream of Basel? Looking Back In this chapter we explained how the chemical reaction equation can be used to calculate quantitative relations among reactants and products. We also defined a number of terms used by engineers in making calculations involving chemical re~ actions. GLOSSARY OF NEW WORDS Con version The fraction of the or some material in the feed that is con-verted into products. Degree of eompletion The percent or fraction of the limiting reactant converted into products, Exeess reactant All reactants other than limiting reactant. Extent of reaction € == nj Vi '<"'"FATFA ni is the number of moles of species i present in the system after the reaction occurs, nio is the number of moles of species i present in the system when reaction starts, and Vi the stoichiometric coefficient for species i in chemical equation. Limiting reactant The species in a chemical reaction that would theoretically run out (would completely consumed) the reaction were to proceed to
- 281. 250 The Chemical Reaction Equation and Stoichiometry Chap. 9 completion according to the chemical equation-even if the reaction did not take place. Selectivity The ratio the moles of a particular (usually the desired) product pro-duced to moles of another (usually undesired or by-product) product pro-duced in a set of reactions. Stoichiometric coefficient Tens the relative amounts moles of chemical species that react and are produced in a chemical reaction. Stoichiometric ratio Mole ratio obtained by using the coefficients of the species in the chemical equation including both reactants and products. Stoichiometry Concerns calculations about the moles and masses of reactants and products involved a reaction(s). Yield (based on feed) The amount (mass or moles) of a desired product obtained divided by amount of the key (frequently the limiting) reactant fed. Yield (based on reactant consumed) The amount (mass or moles) of a desired product obtained divided by the amount the (frequently the limiting) re-actant consumed. Yield (based on theorectical consumption of the limiting reactant) The amount (mass or moles) of a product obtained divided the theoretical (expected) amount of the product that would be obtained based on the limiting reactant in the chemical reaction equation(s) being completely consumed. UPPLEM NTARY REFERENCES In addition to the general references listed in the Frequently Asked Questions in the front material, the following are pertinent. Atkins, P. W. Physical Chemistry, 6th ed., Oxford University Press, Oxford. UK (1998), Atkins, P. W., and Jones. Chemistry, Molecules. Matter, and Change. Freeman, New York (1997). Brady, 1. E. Liftoff! Chemistry. Ehrlich Multimedia, John Wiley, New York (1996). Kotz, J. and TreicheL Chemistry and Chemical Reactivity. Saunders, Fort Worth, TX (1996). Peckham, G. D. "The Extent of Reaction-Some Nuts and Bolts," J. Chern. Educ., 78, 508-510 (2001). Web Sites http://dbhs.wvusd.k12.ca.us/Stoichiometry http://www.chem.ualberta.calcourseslplambeckJplOl.new
- 282. Chap. 9 Problems 251'" http://seience.widener.edulsvb/psetIJimiting http://www.gsu.edul-mstjrhlstoichiometry http://www .shsu .edufweb/sehools/SHS U/chmltchastee/14/moduies PROBLEMS ·9.1 BaC12 + Na2S04 -4 BaS04 + 2NaCl (a) How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate? (b) How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate? (e) How many grams of barium chloride are needed to produce 5.00 g of sodium chloride? (d) How many grams of sodium sulfate are necessary for the precipitation of 5.00 g of barium chloride? (e) How many grams of sodium su1fate have been added to barium chloride if 5.00 g of barium sulfate is precipitated? (f) How many pounds of sodium sulfate are equivalent to 5.00 1b of sodium chloride? (g) How many pounds of barium sulfate are precipitated by 5.00 Lb barium chlo-ride? (h) How many pounds of barium sulfate are precipitated by 5.00 Ib of sodium sulv fate? (i) How many pounds of barium sulfate are equivalent to 5.00 Ib of sodium chloride? *9.2 AgN03 + NaCl ~ AgCl + NaN03 (a) How many grams of silver nitrate will be required to react with 5.00 g of sodium chloride? (b) How many grams of silver nitrate are required for the precipitation of 5.00 g of silver chloride? (c) How many grams of silver nitrate are equivalent to 5.00 g of sodium nitrate? (d) How many grams of sodium chloride are necessary for the precipitation of the silver of 5.00 g of silver nitrate? (e) How many of sodium chloride have been added to silver nitrate if 5.00 g of silver chloride is precipitated? (f) How many pounds of sodium chloride are equivalent to 5.00 Ib of sodium nitrate? (g) How many pounds of silver chloride are precipitated by 5.00 lb of silver nitrate? (h) How many pounds of silver chloride are precipitated by 5.00 Ib of sodium chloride? (i) How many pounds of silver chloride are equivalent to 5.00 Ib of silver nitrate?
- 283. 252 The Chemical Reaction Equation and Stoichiometry Chap. 9 ·9.3 A plant CO2 dolomitic limestone with commercial sulfuric acid. The dolomite analyzes 68.0% 30.0% MgC0J' and 2.0% Si02; the acid is 94% H2S04 and 6% H20. (a) Pounds of CO2 produced per ton dolomite rrea.t.ea. (b) Pounds acid used per ton of dolomite treated. ··9.4 following reactions (find the values of aj): (3) 31 + ~H10 + ~ 34NO + 8sH3AS04 + a6H2S04 (b) 31 KCI03 + ~HCI ~ + 84CI02 + + ~H20 ·9.5 The following reaction was carried out: Fez03 + 2X ~ 2Fe + X10 3 It was found that 79.847 g of F~03 reacted with uXn to form 55.847 g of Pe and 50.982 g Identify element X. ·-'.6 A combustion device was determine empirical formula of a compound containing only carbon, hydrogen, and oxygen. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g of H20. Detennine the formula the compound. ·9.1 A hydrate 3 crystalline compound in which the ions are attached to one or more water molecules. We can dry these compounds by heating to get rid of the water. You have a 10.407-g sample of hydrated barium iodide. sample is heated to drive off the water. dry sample has a mass of 9.520 g. What is the mole ratio between barium iodidet BaI2• and water, H20? What is the fonn'Uia of the hydrate? ·9.8 The (onnula for vitamin is as foHows: How pounds of this compound are contained in 2 g mol? ,0 V i-?=?-?-?-OLOH OOHOHHOH Figure ".8 ·9.9 Sulfuric acid can be manufactured by the contact process according to the following reactions: (1) S + 02 ~ S02 (2) 2S02. + 02 ~ 2S03 +H20~ H2S04 You are asked as preliminary design of a sulfuric plant with a produc-capacity of 2000 tons/day of 66° (Baume) (93.2% H2S04 by weight) to cal-culate the following: (a) How many tons of pure sulfur are required per day to run this plant? (b) How many tons of oxygen are required per day? (c) How many tons of water are required per day for reaction (3)1
- 284. Chap. 9 Problems 253" ·9.10 Seawater contains 65 ppm of bromine in the form of bromides. In the Ethyl-Dow recovery process, 0.27 Ib of 98% sulfuric acid is added per ton of water. together with the theoretical for oxidation; finally, ethylene (C2H4) is united with the bromine to fonn C2H4Br2' Assuming complete recovery and using a basis of 1 lb of bromine, find the weights of the 98% sulfuric acid, chlorine, seawater, and ethane dibromide involved. 2Br + CIl -+ 2Cl- + BI2 Br2 + C;H4 -? C2H4Br2 *9.11 Acidic residue in paper from the manufacturing process causes paper based on wood pulp to age and deteriorate. To neutralize the paper. a vapor-phase treatment must employ a compound that would be volatile enough to permeate the fibrous structure of paper within a mass of books but that would have a chemistry that could be manipulated to yield a mildly basic and essentially non-volatile compound. George KeUy and John Williams successfully attained this objective in 1976 by designing a massdeacidification process employing gaseous diethyl zinc (DEZ). *9.12 At room temperature. is colorless liquid. It boils at 1170 When it is combined with oxygen, a highly exothermic reaction takes place: Because liquid ignites spontaneously when exposed to air, a primary con-sideration in its use the exclusion of air. In one case a fire caused by DEZ ruined neutralization center. Is the equation shown balanced? If no~ balance it How many kg of must react to form 1 kg of ZoO? If 20 cm3 of water are fanned on reaction, and the reaction was complete, how many grams of DEZ reacted? TO: 1. Coadwell DElYf: Water Waste Water DATE: BID INVITATION: 0374-AV REQUISITION: 135949 COMMODITY: Ferrous Sulfate It is recommended that the bid from VWR of $83,766.25 for 475 tons of Ferrous Sulfate Heptahydrate be accepted as they were the low bidder for this product as delivered. It is further recommended that we maintain the option of having this product delivered either by rail in a standard carload of 50 tons or by the alternate method by rail in piggy-back truck trailers.
- 285. 264 The Chemical Reaction Equation and Stoichiometry Chap. 9 What would another company have to bid. to match the VWR bid if the bid they submitted was for ferrous sulfate (FeS04 . H20)? For (FeS04 . 4HzO)? *9.13 Three criteria must be met if a fire is to occur: (1) There must be fuel present; (2) there must be an oxidizer present; and (3) there must be an ignition source. For most fuels, combustion takes place only in the gas phase. For example. gasoline does not bum as a liquid. However, when is vaporized. it bums readily. A minimum concentration of fuel in air exists that can be ignited. If the fuel concentration is less than this lower flammable limit (LFL) concentration, ignition will not occur. The can be expressed as a volume percent. which is equal to the mole percent under conditions at which the LFL is measured (atmospheric pressure and 25° C). There is also a minimum oxygen concentration required for ignition of any fuel. It is closely related to the La and can calculated from the LFL. The minimum oxygen concentration required for ignition can be estimated by multiplying the LFL concentration by the ratio of the number of moles of oxygen required for complete combustion to the number of moles of fuel being burned. Above the LFL. the amount of energy required for ignition is quite smalL For example, a spark can easily ignite most flammable mixtures. There is also a fuel con~ centration called the upper flammable limit (UFL) above which the fuel-air mixture cannot be ignited. Fuel-air mixtures in the flammable concentration region between the LFL and the can be ignited. Both the LFL and the have been measured for most of the common flammable gases and volatile liquids. The LFL is usually the more important of the flammability concentrations because if a fuel is present the atmosphere in concentrations above the UFL, it will certainly be present within the flammable concentration region at some location. LFL concentrations for many materials can be found in the NFP A Standard 325M, "Properties of Flammable liquids," published by the National Fire Protection Association. Estimate the minimum permissible oxygen concentration for n-butane. The LFL concentration for n-butane is 1.9 mole percent. This problem was originally based on a problem in the text Chemical Process Safety: Fundamentals with Applications. by D.A. Crowl and J.F. Louvar, published by Prentice Han, Englewood Cliffs. NJ, and has been adapted from Problem 10 of the AIChE publication. Safety, Health, and Loss Prevention in Chemical Processes by l.R. Welker and Springer, New York (1990). ·9.14 Odors in wastewater are caused chiefly by the products of the anaerobic reduction of organic nitrogen and sulfur.containing compounds. Hydrogen sulfide is Ii major component of wastewater odors; however. this chemical is by no means the only odor producer since serious odors can also result in its absence. Air oxidation can be used to remove odors, but chlorine is the preferred treatment because it not only destroys H2S and other odorous compounds but it also retards the bacteria that cause the compounds in the first place. As a specific example. HOC} reacts with H2S as foHows in low pH solutions HOCl+ If the actual plant practice calls for 100% excess HOCI (to make sure of the destrucrion of the H2S because of the reaction of HOC] with other substances); how much HOCI (5% solution) must be added to of a solution containing 50 ppm
- 286. 'Chap. 9 Problems 255 ·9.15 In a paper mill, soda ash (Na2C03) can added directly in the causticizing process to fonn, on reaction with calcium hydroxide, caustic s.oda (NaOH) for pUlping. The overall reaction is Na2CO) + Ca(OHh ~ 2NaOH + CaC03. Soda ash also may have potential in the on·site production of precipitated calcium carbonate. which is used as a paper fiBer. The chloride in soda ash (which causes corrosion of equipment) is 40 times less than in regular grade caustic soda (NaOH) which can also be used, hence the quality of soda ash is better for pulp mills. However. a major impediment to switching to soda ash is the need for excess causticization capacity, generally not available at older mills. Severe competition exists between soda ash and caustic soda produced by electrolysis. Average caustic soda prices are about $265 per metric [on FOB (free on board, Le .• without charges for delivery or loading on carrier) while soda ash prices are about $130/metric ton FOB. To what value would soda prices have to drop in order to meet the price of $130/metric ton based on an equivalent amount of NaOH? ·9.16 A hazardous waste incinerator has been burning a certain mass of dichlorobenzene (C6H4CI2) hour, and the He) produced was neutralized with soda ash (Na2C03)' If the incinerator switches to burning an equal mass of mixed tetrachlorobiphenyls (C t2H6CI4), by what factor will the consumption of soda ash be increased? ·9.17 Phosgene is probably most famous for being the first toxic gas used offensively in World War I, but it is also used extensively in the chemical processing of a wide variety of materials. Phosgene can be made by the catalytic reaction between CO and chlorine gas in the presence of a carbon catalyst. The chemical reaction is CO + C12 -; COC12 Suppose that you have measured the reaction products from a given reactor and found that they contained 3.00 lb moles of chlorine, 10.00 lb moles of phosgene. and 7,00 Ib moles CO. Calculate the extent of reaction, and using the value calculated, determine initial amounts of CO and C12 that were used in the reaction. ·9.18 In the reaction in which I moles of methane and 45.0 moles of oxygen are fed into a reactor, the reaction goes to completion. calculate the extent of reaction. + + I *9.19 FeS can be U£I..:l,t"-'l.l in O2 to form FeO. 2FeS + 302 -; 2FeO + 2S02 the (solid product) contains 80% FeO and 20% and the exit gas is 100% S02' determine the extent reaction and the initial number of moles of Use 100 g or 100 Ib as the basis. ·9.20 Aluminum sulfate is water treatment and in many chemical processes. It can be made by reacting crushed bauxite (aluminum ore) with 77.7 weight percent sulfu-acid. The bauxite ore contains weight percent aluminum oxide, the remainder being impurities. To produce crude aluminum sulfate containing 2000 lb of pure aluminum sulfate. 1080 lb of bauxite and 2510 Ib of sulfuric acid solution (77.7 percent acid) are used.
- 287. 256 The Chemical Reaction Equation and Stoichiometry (a) Identify the excess reactant. (b) What percentage of the excess reactant was used? (c) What was the degree completion of the reaction? Chap. 9 ·'.21 A barytes composed of 100 percent BaS04 is fused with carbon in the form of coke containing 6 percent ash (which is infusible).-The composition of the fusion mass is BaSO,. 1l.1 % BaS 72.8 C 13.9 Ash. 100.0% Reaction: BaSO ... + 4C ~ BaS + 4CO Find the excess reactant, the percentage of the excess reactant, and the degree of completion of the reaction. ·9.22 Read Problem 9.17 again. Suppose that you have measured me reaction products from a given reactor and found that they contained 3.00 kg of chlorine, J 0.00 of phosgene. and 7.00 kg of CO. Calculate the following: (a) The percent e.xcess reactant used. (b) The percentage conversion of the limiting reactant. (c) The kg mol of phosgene formed per kg mol of total reactants fed to the reactor. ··9.23 Antimony obtained by heating pulverized stibnite (Sb2S3) with scrap iron and drawing off me molten antimony from the bottom of the reaction vesseL Sb2S3 + 3Fe ~ 2Sb + 3FeS Suppose that 0.600 kg of stibnite and 0.250 kg iron turnings are heated together to give 0.200 kg of Sb metaL Determine: (a) The Jimiting reactant . (b) The percentage of ex.cess reactant (c) The degree of completion (fraction) (d) The percent conversion based on Sb2S3 (e) The yield in kg Sb prooucedlkg Sb2S~ to the reactor. ·9.24 The specific activity of an enzyme is defined in terms of the amount of solution catalyzed under a given set of conditions divided by the product of the time interval for the reaction times the amount of protein in the sample: I.L mol of solution converted specific activity = -. --. ---------------- (time mterval minutes)( mg protein in the sample) A 0.10 mL sample of pure ~-ga1actosidase (f3-g) solution that contains l.00 mg of protein per L, hydroIized 0.10 m mol of o-nitrophenyl galactoside (o-n) in 5 minutes. Calculate the specific activity of the f3-g. ·'.25 One method of synthesizing the aspirin substitute, acetaminophen. involves a three-step procedure as outlined in Figure First, p~nitrophenol is catalytically hydro-
- 288. Chap. 9 Problems 257 " genated in the presence of aqueous hydrochloric acid to the acid chloride salt of paminophenol with a 86.9% degree of completion. Next the salt is neutralized to obtain p-aminophenol with a 0.95 fractional conversion. OH HC11H;lO HaC psig) neutrali.ze with NH.OH to OH pH6 NHCOC~ NH2 OH OH Figure P9.l5 FinaHy. the p-aminophenol is acetalated by reacting with acetic anhydryde, resulting in a yield of 3 kg mol of acetaminophen per 4 kg moL What is the overall conversion fraction of p-nitrophenol to acetaminophen? "'9.26 The most economic method of sewage wastewater treatment is bacterial digestion. As an intennediate step in the conversion of organic nitrogen to it is reported. that the Nitrosomonas bacteria cells metabolize ammonium compounds into cell sue and expel nitrite as a by-product by the following overall reaction: 5C02 + 55NHi + 7602'" ...... CSH702N(tissue) + 54NOi 52H20 + l09W If 20,000 kg of wastewater containing 5% ammonium ions by weight flows through a septic tank inoculated with the bacteria, how many kilograms of cell tissue are produced, provided that 95% of the NH! is consumed? ;:',.27 One can view the blast furnace from a simple viewpoint as a process in which the S'i·'- principal reaction is Fe20 3 + 3C ......,. 2Fe + 3CD but some other undesired side reactions occur, mainly +CO After mixing 600.0 lb of carbon (coke) with 1.00 ton of pure iron oxide, the process produces 1200.0 Ib of pure iron, 183 Jb of FeO, and 85.0 Ib of F~03' Calcu-late following items: (a) The percentage of excess carbon furnished. based on the principal reaction
- 289. 258 **9.30 The Chemical Reaction Equation and Stoichiometry Chap. 9 (b) The percentage conversion of Fe203 to Fe (c) The pounds of carbon used up and the pounds of CO produced per ton of Fe20 3 charged (d) What is the selectivity in this process (of with respect to FeO)? A common method used in manufacturing sodium hypochlorite bleach is by the reaction + 2NaOH ~ NaCl + NaOCI + H20 Chlorine gas is bubbled through an aqueous solution of sodium hydroxide, after which the desired product is separated the sodium chloride (a by·product of reaction). A water-NaOH solution that contains 1 lb of pure NaOH is reacted with 851 of ch lorine. The NaOCl formed weighs 618 lb. (a) What was the limiting reactant? (b) What was the percentage excess the excess reactant used? (c) What is the degree of completion of the: reaction, expressed as moles of NaOCI formed to the moles NaOCI that would have formed if reaction had gone to completion? Cd) What is the yield of NaOCl per amount of chlorine used (on a weight basis)? (e) What was the extent of reaction? In a process for the manufacture of chlorine by direct oxidation of HCI with over a catalyst to form C12 and H20 (only), the exit product is composed of HCI (4.4%). Cl2 (19.8%), H20 (19.8%), 02 (4.0%). and N2 (52.0%). What was (a) the limiting reactant? (b) the percent excess reactant? (c) the of completion of the reaction? (d) the extent of reaction? A well known reaction to hydrogen from steam is the so caned water shift reaction: CO + H20 +:t CO2 + H2, If the gaseous feed to a reactor of 30 moles of CO hour. 12 moles CO2 per hour, and 35 moles of steam per hour at 8000 and 18 moles of are produced per hour, calculate (a) the limiting reactant. (b) the excess reactant. (c) the fraction con version of steam to H2• (d) the degree of completion of reaction. (e) kg of yielded per of steam fed. (f) the moles of CO2 produced by the reaction per mole of CO fed. (g) the extent of reaction. *9.31 The overall yield of a product on a substrate in some bioreactions is the absolute value of the production rate divided by the of consumption of the feed in the substrate (the liquid containing the cells, nutrients, etc.) The overaIl chemical reaction for the oxidation of ethylene (C2H4) to epoxide (C:zH40) 2 + O2 ~ 2 C2H40 (a)
- 290. I l Chap. 9 Problems 259 Calculate the theoreticaJ yield (100% conversion C2H4) of C2H40 in mol pet' mol for Reaction (a). The biochemical pathway for the production of epoxide is quite complex. Cofactor regeneration is required which is assumed to originate by partiaJ further oxidation of the fanned epoxide. Thus, the amount of ethylene consumed to produce-l mole of epoxide is larger than that required by Reaction (a). The following two reactions, (b 1) and (b2), when summed approximate the overall pathway 0.67 CO2 + NADH + 1.33H+ + 0.33 F ADH + 0.67 CO2 + 0.33 H+ + 0.33 FADH Calculate the theoretical yield for Reaction (b3) of the epoxide. (bI) (b2) (b3) '9.32 In the production of m-xylene (CgHlO) from mesitylene (C9H12) over a catalyst, some of the xylene reacts to fonn toluene (C7Hg): C~12 + H2 -7 CsHIO + CH4 CgH IO + H2 -7 C7Hg + CH4 The second reaction is undesirable because m-xylene sells for $O.65I1b whereas toluene sells for $O.22/lb. The CH4 is recycled in the plant. One pound of catalyst is degraded per 500 lb of C7Hg produced, and the spent catalyst has to be disposed of in a landfill that handles low-level toxic waste at a cost of $25I1b. If the overall selectivity of CgH 10 to C7Hg is changed from 0.7 mole of xylene produced/mole of toluene produced to 0.8 by changing the residence time in the reactor, what is the gain or loss in "$/100 Ib of mesitylene reacted. - .-J
- 291. CHAPTER 10 MATERIAL BALANCES FOR PROCESSES INVOLVING REACTION 10.1 Species Material Balances 10.2 Element Material Balances 10.3 Material Balances Involving Combustion Your objectives in studying this chapter are to be able to: 1. Carry out a degree of freedom analysis for processes involving chemical reaction(s). 2. Formulate and solve material barances using (a) species balances and (b) element balances. 3. Decide when element balances can be used as material balances. 4. Determine if a set of chemical reaction equations is a minimal set. S. Understand how the extent of reaction is determined for a process, and how to apply it in material balance problems. 6. Understand the meanings of stack gas, flue gas, Orsat analysis, dry basis, wet basis, theoretical air (oxygen) and excess air (oxygen), and employ these concepts in combustion problems. 261 218 283 Why are we devoting a whole chapter to discussing material balances for systems with reaction? The heart many plants is reactor in which products and byproducts are manufactured. To be able to design and operate a reactor economically and safely you have to be able to make valid material balances for the reactor (and as~ sodated equipment) often in Of computer programs can make the calculations for you, but you have to put the right information into them, and ... to calculate is not in itself to analyze. Edgar ABen in "Murders in the Rue Morgue" 260
- 292. 10.1 Species Material Balances 261 " Looking Ahead This chapter discusses material balances for reacting systems. We begin by discussing material balances on chemical species, and then examine material balances made using chemical elements. third section focuses on combustion processes. 10.1 Species Material Balances 10.1 .. 1 Processes Involving a Single Reaction Do you recall from Section 6.5 that material balance for 'a species must be augmented include generation consumption terms when chemical reactions occur in a process? In of moles of species i: { moles Of} { moles Of} { moles of i } { moles oli } { mole! or i } j all:l - i atll = entering leaving + generated In the In the the system the system by reaction !)'Stem syslem between t 1 and t l between t 2 and '. between t 1 and t t { motesofi } consumed by reaction between 12 and r I (10.1) Note we have written Equation (10.1) in moles rather than mass because the generation and consumption terms are more conveniently represented in moles. How do you include the variables corresponding to generation and con-sumption terms in Equation (10.1) and still maintain the independence of species material balances? You have to be sure to avoid redundant equations as well as excess variables. Fortunately, you do not have to add an additional variable to account for the generation or consumption of species i each present in the system if you make use of the extent of reaction that was discussed in Chapter make the idea clear. consider wen-known reaction of N2 and to form NH3. 10. t process as an open, steady-state system operating for 1 min so that the accumulation are zero on the lefthand side of the equal in Equation (10.1). The data in Figure 10.1 are in g moL Using (10.1) you can calculate via a in g mol for the generation or consumption, as the case may be, for each the three species involved in the reaction: NH3 (generation): H2 (consumption): N2 (consumption): As you know the stoichiometry of 6 - 0= 6 g mol 9 - 18 = -9 g mol 12 - 15 == -3 g mol chemical reaction equation
- 293. 262 m 18 15 Material Balances for Processes Involving Reaction Chap. 10 Reactor 9 H2 1----....... 12N2 8 NHa Figure 10.1. A reactor to produce NH3· the respective three generation and consumption terms are related by the reaction equation. Given the value for the generation NH3, you can calculate the values for the consumption of and using the reaction equation. the generation and consumption tenns are not independent, and you cannot specify more than one of the values without introducing a redundant or inconsistent specification. In general, if you specify the value for the generation or consumption of one of the N species in a reaction, you are to calculate the values of the N - I other species from the chemical reaction equation. Here is where the extent of reaction S becomes useful. Recall that an open system where Vi is the stoichiometric coefficient of species i in to Section 9.2). the reaction VNH3 = 2 VH2 = -3 VN2 = - I , and the extent of reaction can be calculated any species: in ~= - nNH3 vNHJ n out - n~2 ~= H2 - VHl 6 - 0 - =3 2 9 - l8 3 -3 12 - 15 = 3 -1 (10.2) reaction equation (refer You can conclude for the case of a chemical reaction that specifica-tion of the extent of reaction provides one independent quantity that accounts for all of values of the generation and consumption tenns the various species in the respective implementations of Equation (10.]). The three species balances corresponding to process in Figure 10.1 are
- 294. Sec. 10.1 Species Material Balances 283 Component Out In = Generation or Consumption I n9ut I -n(n I = vi ~ NH3: 6 -0 = 2 (3) = 6 H2: 9 -18 = -3 (3) =-9 N2: 12 -15 - (3) = The v,!;, corresponds to the moles of i generated or consumed. Frequently Asked Questions Does it mak.e any difference how the chemical reaction equation is written? No. While the value for extent of reaction may change~ the relative values for the generation and consumption of species will remain same. example, write the ammonia reaction as a decomposition of ammonia 1 3 NH3-t 2 N2 + 2 Calculate ~ the process in Figure 10. L Based on NH3 ~=6-0= -1 and the balance is 6 - 0 :::: (-1) (-6) :::: 6 If you the calculation for H2• what result do you 2. Does it any difference if the process in Figure 10.1 is transfonned into a closed (batch) system? No. You can treat the process as a closed. unsteady-state by changing your viewpoint as explained in Chapter For a closed. unsteady-state system the flows in and out would zero, and Equation (10.l) would become n~nal _ I (10.3) For convenience we usually treat a reactor as an open system, and use Equation (10.2) even though reaction may occur in a batch reactor. You get the same re-sults for g either viewpoint. In making a degree-of-freedom analysis} you can write one independent material balance equation for each species present in the system. If Equation (10.2) is applied to species that reacts, the resulting set of equations will all contain the extent of reaction g. For the species that do react" O. In terms of the total molar flow in and total molar flow out r
- 295. 264 Material Balances for Processes Involving Reaction" Chap. 10 S Foul = Ln?ut ;=1 S Fin = Ln;n i=1 where S is the total number of species in the system (nj may be zero for some species). The material balance for the total molar flow is s Fout = Fin + ~L Vi ;=1 Baldy J S Law: Some of it plus the rest of it is all of it. Paul Dickson (l0.4) Equation (10.4) is not an independent equation, but can be substituted for one of the species balances. Only S independent equations can be written for the system. Does Equation (l0.4) apply to a closed system? No. For a closed system you would sum the respective nj in Equation (10.3) over the final and initial moles in the system. If you add one more unknow~. ~. in the set of independent species equations, you will, of course, have to add one more piece of information in the problem statement in order to be able to solve a problem. For example, you might be given the value of the fraction conversion f of the limiting reactant; g is related to fby ( - !)nirmiting reactant ~~------ vlimiting reactant (10.5) Consequently, you can calculate the value of ~ from the fraction conversion (or vice versa) plus information identifying the limiting reactant. In other cases you are given sufficient information about the moles of a species entering and leaving the process so that ~ can be calculated directly. Now let's look at some examples. EXAMPLE 10.1 Reaction in Which the Fraction Conversion Is Specified The chlorination of methane occurs by the following reaction CH4 + C12 -+ CH3CI + Hel Y Oll are asked to delennine the product composition if the conversion of the limiting reactant is 67%, and the feed composition in mole % is given as: 40% CH4• 50% C12, and 10% N2.
- 296. · l ~ t lSec. 10.1 Species Material Balances Solution Steps 1, 2, 3, and 4 Assume the reactor is an open. steady-state process. Figure E 1 0.1 is a sketch of the process with the known information placed on it. Reactor Feed 67% Converalon Product 100g mol nc~ 4{)% CHe F.[><]) p "a2 !ro%C12 ... 'tiel 10%~ Figure EIO.l StepS Select as a basis 100 g mol feed Step 4 ~ ~2 Yau have to determine the limiting reactant if you are to make use of the in~ formation about the 67% conversion. By comparing the maximum extent of reaction (refer to Chapter 9) for each reactant, you can identify the limiting reactant. -~~ -40 fJ1W:(CRa) = = - = 40 vCHc (-1) -n~12 -50 ~(CIl) = v0 2 = (-1) = 50 Therefore. CI4 is the limiting reactant. You can now calculate the extent of reaction using the specified conversion rate and Equation (10.5). - f nj~ (-0.61)( 40) . , = = = 26.8 g moles reacting Vir -1 Steps 6 and 7 The next step is to carry out a degree-of-freedom analysis Number of variables: 11 ~U..·na2' n~2' n~~4' ~i, nff6. n~t,Cl. n~ F. P, f Number of equations: 11 Basis: F = 100 Species material balances: 5 CH4• C~. HO. CH3CI, N2
- 297. 266 Material Balances for Processes Involving Reaction Specifications: 3 Two of {X~H4' X~12' x~:J and f(used to calculate~) Implicit equations: 2 n,?ut = P and 1: ni.n = I l The degrees of freedom are zero. Steps 8 and 9 Chap. 10 The species material balances (in moles) using Equation (10.2) give a direct solmion for each species in the product: n~~ == 40 - 1 (26.8) == 13.2 n~~ = 50 - 1 (26.8) == 23.2 n~'A3Cl :: 0 + 1{26.8) :: 26.8 niftl == 0 1(26.8) = 26.8 n~~t == 10 0(26.8) = 10.0 100.0 =p Therefore, the composition of the product stream 13.2% CH4• 23,2% C12• 26.8% CH3CI. 26.8% HCI, and 10% N2 because the total number of product moles is conveniently 100 g mol. are 100 g mol of products because there are 100 g mol of feed and the chemical reaction equation results the same number moles for reactants as products by coincidence. Step 10 The fact that the overall mole balance equation is satisfied is not a consistency check for this problem. EXAMPLE 10.2 A Reaction in Which the Fraction Conversion Is To Be Calculated Mercaptans, hydrogen sulfide, and other sulfur compounds are removed from natural by various so-called "sweetening processes" that make available other~ wise useless "sour" gas. As you know H2S is toxic in very small quantities and is quite corrosive to process equipment. A proposed process to remove by reaction with S02: 2 H2S(g) + S02(g) ~ 3S(s) + 2H20(g)
- 298. Sec. 1 1 Species Material Balances In a of the process, a stream 20% HzS and 80% CH4 was com-bined with a stream of pure 502' The produced 5000 S(8), and in product gas the ratio of S02 to H2S was to 3. and the ratio of H20 to H2S was 10. You are asked to determine the fractional conversion of the limiting reactant, and feed rates of the and S02 Solution This problem in a sense is the reverse of Example 10.1. Instead of being fraction conversion, you are asked calculate the What this ,..."""' ......... "" means is that ~ win have to be calculated from the material balance equations, and then/from Equation (l0.5). Steps 1, 2, 3, and" Figure BI0.2 is a diagram of the with the known data n~p,"pl'll StepS F 20% 80% / Reactor , S 5000lb EIO.l The obvious basis is 5000 lb S (156.3 lb mol S) Steps 6 and 7 ) The next step is to carry out a degree-of-freedom analysis. Number of variables: 11 nHF2 S, nFca .. Number of equations: 11 Basis: S = 5000 Ib (156.3Ib mol) Species material balances: 5 CII4• S02' H20. S
- 299. Material Balances for Processes Involving Reaction Chap. 10 Specifications: 4 (3 independent) X~2S = 0.20 or xtH4 0.80i (nEo/nk1s) = 3. (nk1olnk2s) = 10 Implicit equatio~s: 2 'Lnf = P 'Znf = (redundant if you use both specifications in F) The degrees of freedom are and the problem is exactly specified. StepS The species balances in pound moles cations are: introduction of most of the specifi-s: 156.3 = 0 3 § H2S: n~2S = O.20F - 2 ~ S02: n§o2 = Fso:z - 1 e H20 : nk:zo = 0 + 2f CH4: ntH.. = O.80F + 0 (f) remaining specifications are n~o2 = 3nh2s n~2o = 1 On~2s Equations (a) through (g) comprise seven unknowns. Step S- (a) (b) (c) (d) (e) (0 (g) equations and seven If you solve the equations without using a computer, you shou]d by culating ~ from (3) Then Equation (d) np H20 -- Next. Equation (g) mol 3 = 52.1 mol nn I) = 104.21b mol H20
- 300. Sec. 10.1 Species Material Balances and Equation (f) gives n~02 :: 3(JO.4) = 31.2lb mo] S02 If you solve the rest of the equations in the order (b). (c), and (e), you find F = 5731b mol F S02 = 83.3 Ib mol ntH4 = 458 Ib mol Finally, you can identify H2S as the limiting reactant because the molar ratio of S02 to H2S in the product gas (3/]) is greater than the molar ratio in the chemical reaction equation (211). The fractional conversion from Equation (10.5) is the consumption of H2S divided by the total feed of H2S -(-2)~ (2)(52.1) f;:::: O.2F ;:::: (0.2)(573) = 0.91 Step 10 Because of the coincidence of the equality of moles of reactants and products for this particular reaction, you cannot use the overall mole balance for this process as a consistency check. 10.1-2 Processes Involving Multiple Reactions 269 In practice reaction systems rarely involve just a single reaction. A primary reaction (e.g., the desired reaction) can occur, but there are invariably additional or side reactions. To extend the concept of the extent of reaction to processes involving multiple reactions, the question is do you just include a ~ for every reaction. The answer is no. You should include in the species material balances only the ~i associated with a (nonunique) set of independent chemical reactions called the mini· mal* set of reaction equations. What this term means is the smallest set of chemical reactions equations that can be assembled that includes all of the species involved in the process. Analogous to a set of independent linear algebraic equations, you can fonn any other reaction equation by a linear combination of the reaction equations contained in the minimal set. For example, look at the following set of reaction equations: c + 1/20 2 ~CO CO + 1/20 2 ~ CO2 ·Sometimes called the maximal set.
- 301. Material for Ul.il:,;:);:)t:;;:) Involving Reaction 10 By inspection you can see that if you the second equation from the you obtain the third equation. Only two of the three equations are the minimal set is comprised of two of the three equations . .... ""' ......... .n L 1 for infonnation about using a computer to detennine if a set of chemical reaction equations comprise an independent set. With these ideas in mind, we can that for open, steady-state prOiCes.ses with multiple reactions, Equation (10.1) in moles becomes for component i R VI .J. (10.6) Iii) is the stoichiometric coefficient species i jn reactionj in the minimal set. extent of reaction for the minimal set. R equations (the size of the analogous to Equation (10.6) can be written for a closed. unsteadYAstate Try it. If you get stuck, look at Example 10.4. The moles, N, exiting a reactor are s s S R N = ~n?ut = 2: n;n (10.7) i= 1 i= 1 where S the number of species in the system. What Equation (10.7) means in words add the stochiometric coefficients reaction, multi-ply the sum by t that reaction~ and then sum the products for each reac-tion to get N. Frequently Asked Question Do you value for to know aU of the reaction equations in a in order to get the independent reactions? No. All you is a list of the species in-to Appendix Ll for the procedure. l, EXAMPLE 10.3 Material Balances for a Process in Which Two Simultaneous Reactions Occur Formaldehyde methanol (CH30H) is produced industrially by to the following reaction: CH30H + 11202 -7 CH20 + H20 of (1) J
- 302. Sec. 10.1 Species Material Balances Unfortunately, under the conditions used to produce fonnaldehyde at a profitable a significant portion of the fonnaldehyde reacts with oxygen to produce CO and H20. that (2) Assume that methanol and twice the stoichiometric amount of air needed for complete conversion of the CH30H to the desired products (CH20 and H20) are fed to the reactor. Also assume that 90% conversion of the methanol results, and that a 75% yield of fonnaldehyde occurs based on the theoretical production of CH20 by Reaction 1. Determine the composition of the product gas leaving the reactor. Solution Steps 1, 2, 3, and 4 Figure ElO.3 is a sketch of the process with .vi indicating the mole fraction of the respective components in P (a gas). Step 5 Step 4 F (CH:PH) 1 9 mol Formaldehyde Reactor Product P YCH~H Y02 "'""--........ YNa YCH:!O YH2<> Yoo Figure EI0.3 Basis: 1 g mol F You can use the specified conversion of methanol and yield of formaldehyde to determine the extents of reaction for the two reactions. Let represent the extent 211 ...
- 303. 272 Material Balances for Processes Involving Reaction Chap. 10 of reaction for the first reaction, and ~2 represent the extent of reaction for the second reaction. The limiting reactant is CH10H. -0.90 Use the fraction conversion. Equation (10.5): tl = -1 (1) = 0.9 g moles reacting The yield related to €j as follows By reaction 1: n~2h = nl:'ll20 + 1 ( g d = 0 ~ I = ~ I . By reaction n~'A'2b = n~~20 - 1(€2) = n~J2h - €2 = .. QI,tt,2 The ytel . d' "CH 0 IS F 2 - "::""::"--l-="';;;'::: 0 .75 ~2 == 0.1 moles reacting You should next calculate the amount of air (A) that enters the process. The entering oxygen twice the required oxygen based on Reaction I, namely . ~ ~ 2G F ) ~ 2(D(I·00) ~ 1.00 g mol n~ 1.00 A ::: - == - = 4.76 g mol 0.21 0.21 n~2 = 4.76 1.00 =. 3.76 g mol Steps 6 and' The degree-of-freedom analysis is Number of variables: 11 A. P, YbJ)OH' yb2 • .Y&~, ytH20. yk2o. yto e,. Number of equations: II Basis: F:: 1 g mol Species material balances: 6 CH)OH, 02' N2• CH20. H20, CO Calculated values in Step 4: 3 A'€1.G2 Implicit equation: 1 'Ly( == 1
- 304. Sec. 10.1 Species Material BaJances Step 8 Because the variables in Figure EIO.3 are yf and not nr. direct use of yf in the material balances will involve the nonlinear tenns yf P. Consequently, to avoid this situation, let us first calculate P using Equation (10.7): S S R P = :Ln}n + ~- ~ViJ 1'= 1 t:=1 J= I 6 1 = 1 + 4.76 + 2: :LVi- ~j i=I)==1 J = 5.76 + [( -1) + (_1/2) + (1) + 0 + (1) + 0] 0.9 + [0 + ( ~2) + (-I) + 0 + (1) + (1)]0.15 =:; 6.28gmol The material balances after entering the values calculated in Step 4 are: n~~30H =:; YeHloH (6.28) = 1 - (0.9) + 0 = 0.10 Step 10 n~t = Y02 (6.28) = 1.0 - (V2)(O.9) - (~)(0.15) = 0.475 = YCH20 (6.28) = 0 + 1 (0.9) - 1 (0.15) = 0.75 = YH20 (6.28) = 0 + 1 (0.9) + 1 (0.15) = LOS == yco (6.28) = 0 + 0 + 1 (0.15) = 0.15 = YN2 (6.28) ;;::: 3.76 0 - 0 == 3.76 You can check the value of P by adding all of the n?ut above. Step 9 The six equations can be solved for the >'1': YeHloH = 1.6%, Y~ = 7.6%, YN2 = 59.8%, YeH20 = 11.9%, Y"20 = 16.7%, Yeo = 2.4%. EXAMPLE 10.4 Analysis of a Bioreactor A bioreactor is a vessel in which biological conversion is carned out involving enzymes, microorganisms, and/or animal and plant cells. In the anaerobic fermentation of grain. the yeast Saccharomyces cerevisiae digests glucose (C6HI20 6) from plants to fonn the products ethanol (C2HsOH) and propenoic acid (C2H3C02H) by the following overall reactions: 273 "
- 305. 274 Material Balances for Processes Involving Reaction Chap. 10 In a batch process, a tank is charged with 4000 kg of a 12% solution of glucose in water. After fermentation, 120 leg of CO2 are produced and 90 kg of unreacted glucose remains in the broth. What are the weight (mass) percents of ethanol and propenoic acid in the broth at the end of the fermentation process? Assume that none of the glucose is assimilated into the bacteria. Solution We will view this process as an unsteady-state process in a closed system. For component i, Equation (10.1) becomes analogous to Equation (10.6) "R v .. n1iruzl = n~nitial + c . I I kJ I) f:.) j=l The bioorganisms do not have to be included in the solution of the problem. Steps 1, 2, 3, and 4 Figure ElO.4 is a sketch of the process. Step 5 Step 4 Bioreactor Initial CondltJons Figure EI0.4 Basis: 4000 kg F Bior8actor FInal Conditions H20 CeHsOH C2HsC02H You should convert the 4000 kg into moles of H20 and C6H120 6. J
- 306. ,, ~,~,. ' sec: to'.t Species Material Bal~nces J .. I 4000 ( 0.88) n nJlla = = 195.3 H20 18 .02 J"j iaJ _ 4000 ( 0.12) _ nC6~llo6 - 18 0 .1 - 2.665 Steps 6 and 7 . The degree-of-freedom analysis is ¥ follows: , Number of variables: 9 Number of equations: 9 Basis: 4000 kg of initial solution (equivalent to initial moles of H20 plus moles of sucrose) Species material balances: 5 H20. C6H120 6• ~H50H. C2H]C02H, CO2 Specifications: 4 (3 independent) nlf;8a1 = 195.3 or nb~a~;06 = 2.665 (one is independent, the sum is F in mol) F " J 120 ncD; = 44 = 2.727. The degrees of freedom are zero. Step 8 . . The material balance equations. after introducing the known values for -the variables, are: " ,. , (a) (b) C2HsOH: n~1f;oH = 0 + 2g1 + (O)f2 (c) , ' , . F'rud ' . . C2H3C02H~ ,' . nC~HJcolH = 0 + (O)~1 + (2)~2 (d) CO2 2.727 = 0 + (2) €l + (O)~2 (e) If yo~ do not use ~ computer to solve the equations. the sequence you should use to solve t:l).em would be (e) (b) simultaneously, and then solve, (a), (c), and (d) in order. 275
- 307. 276 Material Balances for Processes Involving Readion Chap. 10 Step 9 The solution of Equations (a) - (e) is € 1 == 1.364 kg moles reacting €2 == 0.801 kg moles reacting Results Conversion to mass 1l.ercent kg kmol MW ~ H2O 196.9 18.01 3546.1 88.7 C2HsOH 2.728 46.05 125.6 3.1 C"H3C02H 1.602 72.03 115.4 2.9 CO2 2.277 44.0 120.0 3.0 C~1206 0.500 180.1 90.1 2.3 m Step 10 The total mass of 3977 is close enough to 4000 kg of feed to validate the results of the calculations. SElF .. ASS SSMENT TEST Questions 1. Answer the following statements true (T) or false (F). a. If a chemical reaction occurs, the total masses entering and leaving the system for an open, steady-state process are equaL h. In the combustion of carbon, all of the moles of carbon that an open, steady-state process exit from the process. c. The number of moles of a chemical compound entering a steady-state process in which a reaction occurs with that compound can never equal the number of moles of the same compound leaving the process. 2. List the circumstances for a steady-state process in which the number of moles entering the system equals the number of moles leaving the system. 3. If Equation 00.2) is to be applied to a compound. what infonnation must be given about the stoichiometry involved and/or the extent of reaction? 4. Equation (10.2) can be applied to processes in which a reaction and also no reaction occurs. For what types balances does the simple relation "the input equals the output" bold for open, steady-state processes (no accumulation)? Fill in the blanks yes or no.
- 308. 1 1 Species Material Balances 2n Without With Type of balance chemical reaction chemical reaction Total ,balances mass [ J [] Total moles [ ] [ J Component balances Mass of a pure compound [ ] [ ] Moles of a pure compound { J fJ Mass of an atomic species [ J [ ] Moles of an [1 [] Explain how the extent of reaction is related to the fraction conversion of the limiting reactant. Problems Hydrofluoric acid can manufactured by calcium fluoride (CaF2) with sulfuric acid (H2S04), A sample of f1uorospar (the raw material) contains 75% by weight CaF2 and inert (nonreacting) materials. The pure sulfuric used in the process is in 30% excess of theoretically required. Most of the manufactured HF leaves the re-action chamber as a but a solid that contains 5% of the HF fonned. plus CaS04, inerts. unreacted sulfuric acid is also removed from the reaction chamber. Assume complete conversion of the occurs. How many kilograms of cake are produced per 100 kg fluorospar charged to the process? 2. Corrosion of pipes in boilers by oxygen can be alleviated through the use of sodium sulfite. Sodium sulfite removes oxygen from boiler feedwater by the foHowing reaction: 2Na2S03 + 02 ~ 2NaS04 How pounds of sodium sulfite are theoretically required (for complete reaction) to remove the oxygen from 8.330,000 Ib of water (106 gal) containing 10.0 parts per million (ppm) of dissolved oxygen at the same time maintain a 35.% excess of sodium sulfite? 3. Consider a continuous. steady-state process in the following reactions take place: C6HI2 + 6H20 -4 6CO + 12H2 C6H12 + H2 -4 C6HI4 In process moles of C6H12 and 800 moles of H20 are fed into the reactor each hour. The yield of H2 40.0% and selectivity of H2 relative to C6H14 is 12.0. Calculate the molar flow rates of all five components in the output stream. Discussion Problems 1. A scheme has proposed for use on the moon to generate oxygen. The idea is to pass an electric current through molten silica to oxygen at the anode. is pr0- duced at the cathode. What problems you anticipate might occur with the proposed .-
- 309. 278 Material Balances for Processes Involving Reaction Chap. 10 process if the silica contains contaminants such as iron? Will the process have to take place at a high temperature? Where might the electric energy come from? 2. A liquid-emulsion membrane process that offers the promise of cleaning up waste streams contaminated with metals was demonstrated by the Copper Co. on June 1. The process was reported to recover 92.,-98% of the copper from solutions containing 320-1400 ppm of copper at a feed rate of up to 4 Umin. The emulsion consists of globules of aqueous so~ lution surrounded by an organic phase that contains a reagent selective for copper. The metal loaded emulsion is removed by flotation, and the copper collected by a high voltage electrical field. Does this process involve a reaction(s)? How would you involve the extent of reaction in making material balances on the process? 10.2 Element Material Balances In Section 10.1, you learned how to use species mole balances in solving problems involving reacting systems. Equation (10.1), which included tenns for the generation and consumption of each reacting species, was the basic equation from which the other equations were derived. As you know, elements in a process are conserved, and consequently you can . apply Equation (10.1) to the elements in a process. Because elements are not generated or consumed, the generation and consumption tenns in Equation (10.1) can be ignored. Why not use e1ement balances to solve material balance problems rather than species balances? How wonderful that we have met with a paradox. Now we have some hope oj making progress. Niels Bohr You can, but you must first make sure that the element balances are independent. Species balances are always independent. Here is an illustration of the issue. Carbon dioxide is absorbed in water in the process shown in Figure 10.2. The reac~ tion is CO2(g) + H20Cf) --t H2C03Cf) Three unknowns exist: W, F, and P, and the process involves three elements: C, H. and 0. It would appear that you can llse the three element balances (in moles) C: W(O) + F(l) = O.05P(1) H: W(2) + F(O) = [0.05(2) + 0.95(2)]P = 2P 0: W(l) + F(2) = [0.05(3) + 0.95(1)]P = 1.10P
- 310. Sec. 10.2 Element Material Balances 279 . W(H~) Figure 10.2 Schematic of the CO2 absorber. to solve for W, F, and P, but you can't! Try it. The reason is tha.t the three element balances are not independent. Only two of the element balances are independent. If you pick a basis of P = 100 mol, the degrees freedom become and then you can solve for Wand F. If you are curious as to why the three element balances are not independent. you can view H2C03 as H20 . CO2 in which the fixed ratio of HlO and C/O prevents the 0 balance from being independent of the H and C balances. Refer to Appendix L1 for infonnation as to how to determine whether a set material balance equations is independent. EXAMPLE 10..5 Solution of Examples 10.1 and 10.3 Using Element Balances All the given data for this example is the same as in Examples 10.1 and 10.3 Solution Example 10.1 Instead of the degree-of-freedom analysis in Steps 6 and 7 of Example 10.1, the fonowing applies when element balances are used: Number of variables: 10 (e is not involved) Number of equations; 10 Basis: F = 100 Element material balances: 4 (independent) C, H. CIt N Specifications: 3 as in Example 10.1 Implicit equations: 2 as in Example 10.1 The degrees of freedom are zero.
- 311. 280 Material Balances for Processes Involving Reaction In Steps 8 and 9 the element material balances are: 100 (0.40) = n~~(l) + n~ljtcl(1) H: 100 (0.40)(4) = ~~(4) + n~61(1) + n~'ii3Cl(3) Cl: 100 (0,50)(2) = n~Y;(2) + nrftl(l) + n~3Cl(1) 2N: 100 (0.10)(1) = n~~t(1) Chap. 10 Substitute these equations for the species balances used in Example 10.1. As extIeCI:ea. the solution of the probJem will be same as found in Example 10.1. Example 10.3 degree-of-freedom analysis Steps 6 and 7 Variables: 9 (e1 and are not involved) Equations: 9 ~ ...... ,.". F;;:; 1 Element balances: 4 (independent) C,H,O,N Specifications: 3 (the same as in Example 10.3) Implicit equations: 1 (the same as in Example 10.3) of freedom are zero. In Steps 8 and 9 the element balances are: c: 1(1) + 4.76(0) = P[y~30H(1) + Y&20(1) + yto(l)] H: 1(4) + 4.76(0) = P[YtH30H(4) + ytHzo(2) + Yk2o(2)] / ,-, l(l) + LO().~~J= P[YtHJOH(1) + y&(2) + YtH20 (1) + Yk2o(1) + yto(l)] 1(0) 3.76 Substitute these equations for the species balances used in Example 10.3. The solution of the problem not change. It would be easier to use term yf' P = nf in the equations above in place of the product of two variables, /; and
- 312. Sec. 10.2 Element Material "" ... 1' ....... ' ... "" .. 281-' Element balances are especially useful when you do not know what reactions occur in a process. You only know information about the input and output stream components, as illustrated by the EXAMPLE 10.6 Use of Element Balances to Solve II Hydrocracking ProbJem Hydrocracking is an important refinery process for converting low-valued heavy hydrocarbons into more valuable lower molecular weight hydrocarbons by exposing the to a at high temperature and pressure in the pres-ence of hydrogen. in this study the hydrocracking of pure compo-nents, such as to understand the behavior of cracking reactions. In one such hydrocracking of octane, the cracked produ:cts had the following composition mole percent: 19.5% <;Hs. 59.4% C4HIO, and 21.1% CSH12. You are asked to detennine the molar ratio of hydrogen consumed to octane reacted for this ..... I',~ ... "'., Solution We win use ejC~lmeJU balances to solve this problem because the reactions in-volved are not specified. Steps 1, and 4 EIO.6 is a sketch of the laboratory hydrocracker reactor together with the data for the streams. F (CaH;s) Lab Reactor Product P 19.5%~HIiI '-----til"- 59.4% C"H ' 0 21.1
- 313. 282 l Material Balances for Processes Involving Reaction StepS Basis: P = 100 g mol Steps 6 and 7 The degree-of-freedom analysis is Variables: 3 F,G,P Equations: 3 Element balances: 2 H,e Basis: P = 100 Chap. 10 If you calculate the degrees of freedom based on species balances, you have as unknowns five species molar flows, P, and R (the number of independent reaction equations which equals the number of unknown extents of reaction), and as equations five species balances, one specification (the basis). and one implicit equation so that degrees of freedom = (6 + R) - (5 + 2) :;: R - 1 Thus, although you do not know what R is from the problem statement, for zero degrees of freedom to occur, R = I, that is, one independent reaction equation exists. You can be assuaged that if you do not know R and a minimum reaction set, and thus you do not know how to involve the respective ~ in the species equations. you can fall back on element balances. StepS The element balances after introducing the specification and basis are: C: F(8) + G(O) = 100[(0.195)(3) + (0.594)(4) + (0.211)(5)] H: F(l8) + G(2) = 1 OO[ (0.195)(8) + (0.594)( 10) + (0.211 )(12)] and the solution is F = 50.2 g mol G=49.8gmol The ratio H2 consumed 49.8 g mol = = 0.992 CgHI8 reacted 50.2 g mol You will find that employing element material balances can be simpler than employing the extent of reaction for problems in which the reaction equations are not specifically known or must be inferred. as shown in the examples in the next section. J
- 314. 10.3 Material Balances Involving Combustion 283 ' Elf-AS ESSMENT TEST Questions 1. Do you have to write element material balances with the units of each term being moles rather than mass? Explain your answer. 2. Will the degrees of freedom be smaller or larger using element balances in place of I-'~~"""" balances? 3. How can you determine whether a set of element is independent? 4. Can the number of independent element balances ever be larger than the number of species balances in a problem'? Problems 1. Consider a system used in the manufacture of electronic materials (all gases except Si) How many independent element balances can you make for this system? Methane bums with to produce a gaseous product that contains CH4• 02. CO2, CO, H20, and H2. How many independent element balances can you write for this system? 3. Solve the Self-Assessment Test problems in Section 10.1 using element balances. Discussion Problem 1. Read Discussion Question #2 in Section 10.1. How would you use element balances in analyzing the process? 10.3 Material Balances Involving Combustion In this section we discuss combustion as an extension of the previous discussion about chemical reactions. Combustion is the reaction of a substance with oxygen with the associated release of energy and generation of product gases such as H20, CO2, CO, and S02' Typical examples of combustion are the combustion of coal, heating oil, and natural used to generate electricity in utility power stations, and engines that operate using the combustion of gasoline or diesel fuel. Most combustion processes use air as the source of oxygen. our purposes you can assume tnat air contains 79% N2 and 21 % O2 Chapter 2 for a more detailed analysis); neglecting the other components with a total of less than 1.0%, and can assume that air has an average molecular weight of 29. Although a small amount of N2 oxidizes to NO and N02• gases called NO,;:. a pollutant, the amount so small that we treat
- 315. 284 Material Balances for Processes Involving Reaction Chap. 10 CO Hydrocarbons I J DefiCient air (fue! rich) Stoichiometric (fuel lean) air-fuel ratio Air-fuel ratio Figure 10.3 Pollutants resulting from combustion vary with the air-la-fuel ratio and the temperature of combustion. The fuel is natural Hydrocarbons and CO increase with deficient air. Efficiency decreases with too much excess air, but so the NO". (1) Nonnal operating state. (2) Operating state driven by NO" limits. (3) Minimum excess 02 to be below CO limits. N2 as a nonreacting component of air and fueL Figure 10.3 shows how the CO, unburned hydrocarbons, and NOx vary with the air-to-fuel ratio in combustion. Combustion requires special attention because of some of the tenninology involved. You should become acquainted with these special terms: 8. Flue or stack gas-all the gases resulting from a combustion process indud~ ing the water vapor, sometimes known as a wet basis. b. Orsat analysis or dry basis-all the gases resulting from a combustion process not including the water vapor. (Orsat analysis refers to a type of gas analysis apparatus in which the volumes of the respective gases are measured over and in equilibrium with water; hence each component saturated with water vapor. The net result of the analysis is to eliminate water as a component that is measured.) Look at Figure lOA. To convert from one analysis to another, you have to adjust percentages of the components to the desired basis as explained in Chapter 2. c. Complete combustion-the complete reaction of the hydrocarbon fuel producing CO2, S029 and H20. d. Partial combustion-the combustion of the fuel producing at least some CO. Because CO itself can react with oxygen, the production of CO in a combustion process does not produce as much energy as it would if only were produced. e. Theoretical air (or theoretical oxygen)-the minimum amount of air (or oxygen) required to be brought into the process for complete combustion. Sometimes this quantity is called the required air (or oxygen).
- 316. Sec. 10.3 Material Balances Involving Combustion 285 Orsat analysts dry basts Flue gas, stack gas, or wet basis Dry flue gas on S02-free basis Figure 10.4 Comparison of a gas analysis on different bases. f. Excess air (or excess oxygen)-in line with the definition of excess reactant given in Chapter 9, excess air (or oxygen) is the amount of air (or oxygen) in excess of that required for complete combustion as defined in (e). The calculated amount of excess air does not deptruJ on how much nuzterial is actually burned but what is possible to be burned .. Even if only partial combustion takes place, as, for example, C burning to both CO and CO2, the excess air (or oxygen) is computed as if the process of combustion went to completion and produced only CO2• Do not ever forget this basic assumption! The percent excess air is identical to the percent excess O2 (often a more CODvenient calculation): excess air excess 0 2/0.21 % excess air = 100 . . = 100--, -~-- requITed aIr reqUITed 0 2/0.21 (l0.8) Note that the ratio 1/0.21 of air to 02 cancels out in Equation (10.8). Percent excess air may also be computed as . O2 entering process - ~ required % excess arr = 100 O. (10.9) 2 required or excess O % excess air = 2 l00--~.--~--- O2 entenng-excess O2 (10.10) The precision achieved from these different relations to calculate the percent excess air may not be the same because when you take differences in large numbers that involve some error, the relative error in the difference is even bigger. In calculating the degrees of freedom in a problem, if the percent of excess air and the chemical equation are specified in the problem, you can calculate how much air enters with the fuel, hence the number of equations involved is increased by one, or the number of unknowns is reduced by one. Now, let us explore these concepts via some examples.
- 317. 2 286 l fI Material Balances for t-'rc)ce:sse~s Involving Chap. 10 EXAMPLE 10.7 Excess Air other than gasoline are for motor vehicles because they gen~ erate levels of pollutants than does gaSOline. Compressed propane is one such Suppose that in a test 20 of C3HS is burned with 400 of to CO2 and 12 kg of CO. What was the percent excess Solution This is a problem involving correctly balanced?) following reaction (is the reaction equation Basis: 20 kg of C3Hg Since the percentage of excess air is based on the complete combustion of to CO2 and H20, the combustion is not complete no influence on the calculation of "excess required 02 is 5 kg mol O2 = 1 kg mol C3Hg The entering 02 is 400 kg air 1 -~-- 21 kg mol O2 aIr 100 kg mol air The percentage of excess is excess O2 O2 100 x = 100 x ---"------"'---- % excess ---------~----------~ In calculating amount of excess air, the amount of air that enters combustion process over and that required for complete combustion. Suppose there is some oxygen in material being burned. exam-ple l suppose that a containing 80% C2H6 and 20% 02 is burned in an _ .. ,...~, __ with 200% excess Eighty percent of the ethane goes to CO2, 10% to CO, and 10% unburned. What is the amount the excess air per 100 moles of the gas? First, you can ignore the infonnation the CO and the ethane because the of the calculation of excess complete combustion. cally C goes to S to S02' H2 to H20, to CO2> and so on. j
- 318. I .gec~ 10,3 Material Balances Involving Combustion Second. the oxygen in the fuel cannot be ignored. Based on the reaction / 80 moles of C2H6 require 3.5(80) :::: 280 moles of 02 for complete combustion, However. the gas contains 20 moles of 02' so that only 280 - 20 :::: 260 moles of 02 are needed in the for comp~ete combustion. Thus, 260 moles of 02 are the required 02' and the calcu1ation of the 200% excess 02 (air) is based on 260. not 280, moles of °2: with air required °2: excess 02 (2)(260): total 02 (3)(260): Moles 260 780 In the following problems each step cited in Chapter 7 will be identified so that you can fonow the strategy of the solution. Example 10.8 A Fuel Cell to Generate Electricity From Methane "A Fud Cell in Every Car" is the headline of an article in Chemical arul Engineering News. March 5, 2001. p. 19. essence. a fuel is an system into which fuel and air are fed, and out of which comes electricity and waste products. E 10.8 a sketch of a fuel cell in Which a continuous flow of methane (CH4) F"" 16.0 kg Electric Load Fuel Cell CO2 ::::7 N2 =7 O2 iIII? H20 ==? p ::: ? (kg mol) Figure EtO.8 A:: 300 kg Mol % O2 21.0 N2 79.0 100.0
- 319. 288 Material Balances for Processes Involving Reaction Chap. 10 and air (02 plus N2) produce electricity p]us CO2 and H20. Special membranes and catalysts are needed to promote the reaction of CH4• Based on the given in Figure EIO.8, arc asked to the COffi-position of the products in Solution Steps 1, 3, and 4 This is a steady-state process with reaction. Can you as-sume a complete reaction occurs? Yes. No CH4 appears in p, The system is the fuel cell (open, steady state). Because the process output is a composition will be mole fractions or moles, hence it more convenient to use moles rather than mass in this problem even though the quantities of CH4 are stated in kg. You can carry out the necessary preliminary conversions as fol lows: 300 kg A 1 mol A ----=---= -..;;;.......-- = ] 29.0 kg A kg mol A in 16.0 kg 1 kgmolC~ . 1 k CH = 1.OOkgmolCH4 ln 6.0 g 4 10.35 kg mol 0.21 ----'''---...=. = 2.17 kg mol O2 in 10.35 kg mol 0.79 kg mol N2 . 1 kg mol A = 8.18 kg mol N2 In Step S Since no particular basis is designated we will pick a convenient basis Basis: 16.0 CH.t F 1 mol CH4 Steps 6 amd 7 The degree-of-freedom analysis is (A has calculated): Variables: 8 FP P P P P A A , ,neol , nN1, no2, nH201 nOl' nN2, Equations: 8 Basis: F = 1 kg mol Element material balances: 4 (independent) C,H.O,N Specifications and calculated quantities: 2 n~l = 2.17. n~2 = 8.18 Implicit equation: 1 Inr = The degrees of freedom are zero, j
- 320. 5ec.,10.3 Material Balances Involving Combustion If you use species balances in solving the problem. you have to involve the re-action: + 202 ~ + 2H20: Then (A has been calculated) StepS Variables: 10 P ,A P P P P P t- , f I'IN2, I'ICH..' ncO:l' nN2' DOl' nHzO. '" Equations: 10 Basis: F = 1 Species balances: 5 Calculated quantities: 2 (as above) Specifications: 1 , ~ = 1 (because the reaction is complete) ImpHcirequation: 1 Inf = P After introduction of the specified and calculated quantities. the element material balances are (in moles): Out In nt02(1 ) = 1(1) H: n~20(2) = 1(4) 0: nt02(2) + 11.&(2) + nk20(1) = 17(2) 2N: p nN2 :::: 8.18 The species material balances are: Out In mol P ne"" = 1.0 J x 1 = 0 °2: n& = 17 - 2xl = 0.17 N2: p nN2 8.18 'Ox 1 :::: 8.18 = - CO2: p ncOz :::: 0 + 1 x 1 :::: LO H~O: p nH10 = 0 + 2xl :::: 2.0 Step 9 The solution of either set of equations gives nt~ :::: O. n& = 0.17, n~2 = 8.18, ntoz = to. nk~ = 2.0, P = 11 289 __
- 321. 290 Material Balances for Processes Involving Reaction Chap. 10 and the mole percentage composition of P is y~2 = 1.5%, YN2 = 72.1 %, YC02 = 8.8%, and YH20 = 17.6% Step 10 You can check the answer by determining the total mass of the exit gas and comparing ir to total mass entering (316 kg), but we will omit this step here to save space. Example 10.9 Combustion of Coal A local utility bums coal having the following composition on a dry basis. (Note that the coal analysis below is a convenient one for aUf calculations, but is not necessarily the only type of analysis that is reported for coal. Some analyses contain much less information about each element.) Component Percent C-~ 83.05 W 4.45 ° 3.36 N 1.08 S" 0.70 Ash 7.36 Total 100.0 The average Orsat analysis of the gas from the stack during a 24-hour test was Component Percent COl + S02 ) 5.4 CO 0.0 °2 4.0 N2 ~ Total 100.0 Moisture in the fuel was 3.90%, and the air on the average contained 0.0048 Ib H20llb dry air. The refuse showed 14.0% unburned coal, with the remainder being ash. You are asked to check the consistency of the data before they are stored in a database. Is the consistency satisfactory? What was the average percent excess air used? Solution This is an open, steady-state process with reaction. The system is the furnace. j
- 322. Sec. 10.3 !Material Balances Involving Combustion , An of the infonnation given in UUI."'U' statement has been placed on Fig-ure E1O.9, Because the gas analysis is on a dry we a flowstream W for the exit water to the process The compositions of F and R are moles. mass, those of P and A in .2!L C 83,05 H 4,45 0 3.36 N 1,08 S 0.70 Ash 7.36 100.0 Added H20 3.90 Ib Co .. { H: 0.433 Ib mol nhilmng 0: 0.217 Ib mol StepS a Coal F(lb} W (Ib mol) H20(g) 100% Fumace Air A (Ib Mol fr N2 0.79 °2 Q.6.1 1.00 Stack P(lb mol) Added H20 0.0048 IbIlb air C ......... {H: 0.0154 Ib mol/mol A Omalnmg 0: o.con Ib mol/mol A Figure EIO.9 ::::::: 100 Ib as convenient. 2L CO~ + S02' 15.4 CO 0.0 4.0 N2 eO.6 100.0 Refuse R (Ib) C-.;.H+O+N+S 14.0 Ash 100.0 must first calculate some extra information com-in the diagram (the information has already been added H20 in coal: 3.901b Ib 1 lb mol H20 2 Ib mol H 18 lb H 2 0 I lb mol H 2 0 = 0.433 Ib mol H (with O.2171h mol 0) 1 lb mol H20 Ib mol H20 ---- 18 lb H 0 = 0.0077 Ib mol (with 0.01541b mol HJlb mol A) (with 0.0077 Ib mol Ollb mol A) 2 291
- 323. Material Balances for Processes Involving Reaction Chap. 10 You might neglect the C, H, 0; Nt and S in the refuse, but we will include the amount to show what calculations would be necestw)' if the amounts of the elements were significant. To do we must make a preliminary mass balance for the ash ash balance (lb):7.36 = O.86(R) R;;; 8.561b The unburned coal in the refuse is 8.56(0.14) :: 1.20lb If we assume that combustibles in the refuse occur in the same proportions as they do in the coal (which may not be true), the quantities of combustibles in R on an ash-free basis are: Component mass % Ib Ib mol C 89.65 1.016 0.0897 4.80 0.058 0.0537 0 0.0436 0.0027 N 1.11 0.014 0.0010 S 0,76 Q..QQ2 g,0003 100.00 1.20 0.1474 Steps 6 and 7 can write only four element balances because S and C be combined inasmuch as these two elements are combined in the stack gas analysis. Presumably one of the four equations is redundant and can used to check the calculations. You can examine the composite matrix Appendix L) if you wish to find its rank. The degree-af-freedom analysis is Variables: 4 F,A, W.P Equations: 5 Basis: = 100 Element balances: 4 H. O. S + C use a degree-of-freedom analysis based on species would be quite complex. and win be omitted. Step 8 element balances in moles are
- 324. 10.3 Material Balances Involving Combustion In Out F A w p R C + S. 83.05 0.70 . 12,0 + + o "" 0 + P(O, 154) + 0.0897+0.0003 H: 0: N: Step 9 4.45 + 0.433 + 1.008 0.0154 A :: + o + 3.36 + 0.217 + 0.2IA(2)+O.OO7A ::: W + 2P{O.154+0.040) + 1 14.0 + 2(0.79A) = 0 + 2P(O.806) + 0.0537 0.0027 2(0.001) Solve the equations on a computer, or by hand in order: + S to P, N to getA. and H to W. values are (in moles) P = 44.5. A = 45.4, and W = 2.747. Step 10 Use the 0 balance to results 19.8 == 20.3 The difference is about 1 %. Inasmuch as the data provided are actual measurements, in view of the random and possibly biased errors in the data, the round-off error introduced in the calculations, and possible leaks the furnace. the data seem to be quite satisfactory. Try calculating W. a small number, from both H and 0 balances. What size error do you find? To calculate the excess air, because of oxygen the coal and the tence of unburned combustibles. we will calculate the total oxygen in and the reqU'ired oxygen: % excess air = 100 X entering - O2 required O2 required Assume that no oxygen is required by the ash. required 02 is Component C H o N S Reaction c + --+ CO2 Hz + 11202 --+ H20 -' Ib 83.05 4.45 0.70 Ibmol 0.210 0.022 Required O2 (lb mol) 6.921 1.104 (0.105) 7.942 293
- 325. 294 Material Balances for Processes Involving Reaction Chap. 10 and the oxygen entering in the air is (45.35)(0.21) :::; 9.5241b mol. . 9.524 - 7.942 % excess rur = 100 X 7.942 = 19.9% If you (incorrectly) calculated alone. you would get % excess air from the wet stack gas analysis 100 X 4.0 = 23.8% 15.4 + 2.746/2 From the viewpoint of the increases the CO2 concentration in the atmos-phere, would the CH4 in Example 10.8 or the coal in Example 10.9 contribute more CO2 kg? The HlC ratio moles in CH4 was 411, whereas in the coal it was 0.0537/0.0897 = 0.60. Figure 10.5 shows how the HlC ratio varies with different types of fuel. 0.5 1.0 Peat Residue Distillate Tar Heavy I Ught / Sand Oil 'Oil 'I Ught Products 1.5 Hie mole mHo Figure 10.5 Variation the HIe mole ratio selected fuels. SELF .. ASS SSMENT TEST Questions 1. Explain the difference between a flue gas analysis and an Orsat analysis; wet basis and dry basis. What does an SOr free basis mean? 3. Write down the equation relating percent excess air to required air and entering air. 4. Will the percent excess always be the same as the percent excess oxygen in combus-tion (by oxygen)? S. In a combustion process in which a specified percentage of excess air is used. and in which CO is one of the products of combustion. will the analysis of the resulting exit gases contain more or less oxygen than if aIJ the carbon had burned to CO2?,
- 326. Sec. 10.3 Material Balances Involving Combustion 295 6. Answer the following questions true or false. a. air for combustion calculated using the assumption of complete reaction whether or not a reaction takes place. b. For the typical combustion process the products are CO2 gas and H20 vapor. c. In combustion processes, since any oxygen in the coal or fuel oil inert, it can be nored in the combustion calculations. d. The concentration of N2 in a flue is usually obtained by direct measurement. Problems I. Pure carbon burned in oxygen. The flue gas analysis CO2 75 mol % CO 14 mol % 02 11 mol % What was the percent excess oxygen used? 2. Toluene, C7HSt is burned with 30% excess air. bad burner causes 15% of the carbon to fonn soot (pure C) deposited on the walls of the What is the Orsat analysis of the leaving the furnace? 3. A synthesis gas analyzing CO2: 6,4%,°2: 0.2%, CO: 40.0%, and H2: 50.8% (the balance is N2) is burned with excess dry The problem is to determine the composition of the flue How many degrees of freedom exist this problem, that is. how many addi-tional variables must specified? /4. A coal analyzing 65.4% C, 5.3% H. 0.6% 1.1% N, 18.5% 0. and 9.1 % ash is burned so that all combustible is burnt out of the ash. The flue gas analyzes 13'(X>% CO2, 0.76% CO, 6.17% ° 2, 0.87% and 79.20% N2. All of sulfur bums to S02' which is included in the CO2 in the gas analysis (Le., CO2 + S02 = 13.00%), Calculate: a. Pounds of coal fired per 100 Ib mol of dry flue gas as analyzed; b. Ratio of moles of total combustion to moles of dry air supplied; c. Total moles of water vapor in the stack gas per 100 Ib of coal if the entering air is dry~ d. Percent excess air. S. A hydrocarbon fuel is burnt with excess air. The Orsat analysis of flue gas shows 10.2% CO2, 1.0% CO, 8.4% O2, and 80.4% N2. What is the atomic ratio of to in the fuel? Thought Problems 1. In a small pbannaceutical plant, it had not possible for a period of two months to more than 80% of rated output from a boiler rated at 120,000 Ib of stearn per hour. boiler had complete flow metering and combustion control equipment, but the steam flow could not be brought to more than 100,000 lblhr. What would you recommend be done to fmd the cause(s) of the problem and alleviate it? 2. In connection with the concern about global warming, because of the increase in CO2 concentration in the atmosphere, would you recommend the use coal, ethanol, fuel oil. or natural gas as a fuel?
- 327. 296 Material Balances for Processes Involving Reaction Chap. 10 Discussion Problems 1. In-situ biorestoration of subsurface materials contaminated with organic compounds is being evaluated by the EPA and by industry as one technique for managing hazardous· wastes. The process usually involves stimulating the indigenous subsurface microflora to degrade the contaminants in place, although microorganisms with specialized metabolic capabilities have been added in some cases. The ultimate goal of biodegradation is to convert organic wastes into biomass and harmless byproducts of microbial metabolism such as CO2, CH4, and inorganic salts. Bioremediation of trichloroethylene, and cis· and trans-dichloroethylene was tigated in a test plot in a trial. The aquifer was not pretreated with methane and oxygen to stimulate growth of the methanotrophs. The biotransformation of transdichloroethylene, cis-dichloroethylene, and trichloroethylene added at 110. and 130 JL.g/L was 65, 45, and 25%, respectively, which suggests that the less-chlorinated compounds are more readily degraded than are the highly chlorinated compounds. What other influences might have affected the results obtained? 2. Suppose you are asked to serve as a consultant on the problem of how to produce oxygen on the moon as economically as possible. The raw material readily available is FeTi03, SiOll and/or FeO. The energy to carry out the reactions is presumably available from the sun or atomic energy that provides electricity or high pressure steam. Discuss some possible methods of O2 generation and draw a simple flowsheet for the process. Some very useful references pertaining to this problem are (a) L. A. Taylor, "Rocks and Minerals in the Regolith of the Moon: Resources for a Lunar Base," pp. 29-47 in Advanced Materials-Applications Mining and Metallurgical Processing Principles, ed. V. 1. Lakshmanan. Soc. Mining. Mineral, &arnpersand; Exploration, Littleton, CO (1988); (b) L. A. Taylor. "Resources for a Lunar Base: Rocks, Minerals and Soils of the Moon," in 2nd Symp. on Lunar Base and Space Activities of the 21st Century, ed. W. W. Mendell, Lunar &ersand; Planetary mst., Houston. TX (1993); (c) L. A. Taylor and D. W. Carner, "Oxygen Production Processes on the Moon: An Overview and Evaluation, n Resources in New-Earth Space, Univ. Arizona Press, Tucson, AZ (1993). looking Back In this chapter we applied Equation (10.1) and its analogs processes involving reaction. If you make element balances, the generation and consumption terms in Equation (10.1) are zero. If you make species balances, the accumulation and consumption terms are not zero, and you have to use the extent of reaction. Both open and closed systems with reaction thus admit no new principles. You simply apply the general material balance with reaction to these systems recognizing their characteristics (e.g.. closed systems have no flow into or out across the system boundary), I
- 328. Sec. 10.3 Material Balances Involving Combustion 297 GLOSSARY OF NEW WORDS Complete combustion The complete reaction of a reactant to produce C021 S02' and H20. Element balances Material balances involving chemical elements. Excess air (or oxygen) The amount of air (or oxygen) in excess that required for complete combustion. Flue or stack gas All of the gases resulting from combustion including the water vapoft also known as a wet basis. Minimal set of reactions The smallest set of chemical reaction equations th~t includes aU of the species involved in the reactions. Orsat analysis Also known as a dry basis. All of the gases resulting from combus-tion not including the water vapor. (Or8at refers to a of analysis.) Partial combustion Combustion that produces least some CO. Required air (or oxygen) The amount of air (or oxygen) required complete reaction to occur (see Theoretical air). Species balances Material balances involving chemical species. Theoretical air (or oxygen) The amount of air (or oxygen) required to be brought into a process to accomplish complete combustion. Also known as required air (or oxygen), SUPPLEMENTARY REFERENC S In addition to general references listed in the Frequently Asked Questions in the front material. the foHowing are Croce, ''The Application of the Concept of Extent of Reaction," 1. Chem. Educ.) 79, 506-509 (2002). Nevers, N. Physical and Chemical Equilibrium for Processes. WiJey-Interscience, New York (2002). Moulijin, lA.. M. Markkee. and A. Van Diepen. Chemical Process Technology. Wiley. New Yark (200 1 ). Web Sites http://voyager5.sdsu.edultestcenter/features/states/states.html http://www.ePin.ncsu.edulaptiJOC2000/mOdulel/materialmateriallhtm WWWjOhnZink.com
- 329. 298 Material Balances for Processes Involving Reaction 10 ROBl MS *10.1 Pu re I. n ph a se a reactor. of this is converted to B through the re-action A ~ 3 What the mole fraction of in the exit "'~ .. "." ..... What is extent of "' ... "'."' .... " "'10.2 One of the most common commercial methods for the production pure silicon that is to be for the manufacture of semiconductors is the Si~mens process ure PIO.2) of chemical vapor deposition (CVD). A chamber contains a heated silicon rod, and a mixture of high purity trichlorosilane mixed with high purity hydrogen that is passed over the Pure silicon (EGS-electronic grade silicon) deposits on the rod as a polycrystalline solid. (Single crystals of Si are made by subsequently' melt-ing the EGS and drawing a single from melt.) The reaction is: Hig) + SiHC]3(g) ~ SICs) + 3HCI(g). . The rod initially has a mass of 1,460g, and the mole fraction of Hi in exit" is 0.223. The mole fraction of in the to the reactor 0.580, and feed enters at the rate kg mollhr. What will the mass the rod at the of 20 ..... u .. ,..,'" Gu Gu SiHCI3 --~ ',...--....... Hel H2 L-'====--1 SiHCla H2 Figure PI0.l "10.3 A low-grade pyrite containing 32% S is mixed with 10 Ib of pure sulfur per 100 lb of pyrites so the mixture will bum readily with fonning a gas analyzes 13.4% S02' 2.7% °2, and 83.9% N2• No sulfur left in cinder. Calculate the percentage of the sulfur fired burned to S03' (The S03 is not detected by the analysis.) "'10.4 Examine the reactor in Figure PlO.4. Your boss something has gone wrong with the yield of CH20, and it is up to you to find out what the problem is. You start by making material balances (naruralIy!). Show all calculations. Is there some problem? Methanol (CH30H) ---IIII-i Reactor Air Figure PIOA Product 62.S N2 1 02 H20 4.6CH20 12.3CHsOH 1.2 HCOOH
- 330. Chap. 10 Problems 299 r ·10,5 A problem statement was: A dry sample of limestone is completely soluble in HCII and contains no Fe or AI. When a 1.000 g sample is ignited the loss in weight is found to be 0.450 Calculate the percent CaC03 and MgC03 in the limestone. The solution was: x (1.000 - x) 0.450 --+ =-- 84.3 100 44. 100x + 84.3 - 84.3 x = (0.450)(84.3)(100)/44 x ;::: 0.121 MgC03 = I 1 % CaCO) = 87.9% Answer the fonowing questions: 1. What information in addition to that in problem statement had to be ob-tained? 2. What would a diagram for the process look like? 3. What was the basis for the problem solution? 4. What were the known variables in the problem statement, their values. and their units? 5. What were the unknown variables the problem statement and their units? 6. What are the types of material balances that could be made for this problem? What type(s) of material balance was made for this problem? 8. What was the degree of freedom for this problem? Was the solution correct? '/""10.6. 1n order to neutralize the acid in a waste stream (composed of H2S04 and H20). ground limestone (composition 95% CaC03 and 5% inerts) is mixed in. The dried sludge collected from the process is only partly analyzed by firing it in a furnace which results in only CO2 being driven off. By weight the CO2 represents 10% of the dry sludge. What percent of the pure CaCO) in the limestone did not react in the neutralization? $$10.7 Capper as CuO can be obtained from an ore called Covellite which is comprised of CiS and gange (inert sulids). Only part of the <;:uS is oxidized with air to CuO. The gases the roasting process analyze: S02 (7.2%). 02 (8.1%). and N2 (84.7%), Unfortunately, the method of gas analysis could not detect S03 in the exit gas, but S03 is known to exist. Calculate the percent the sulfur in the of the CuS that reacts that forms S03' Hint: you can consider the un reacted CuS as a compound that comes and out of the process untouched, and thus is from the process and can be ignored. ·10.8 A is to remove Si02 from a wafer in semiconductor manufacturing by contacting the Si02 surface with HF. The reactions are: 6 HF(g) + SiOl(s) -; H2SiF6(l) + H20(l) H2SiF6(1) -; SiF4(g) + 2 HF(g)
- 331. 300 Material Balances for Processes Involving Reaction Chap. 10 Assume the reactor is loaded with wafers baving a silicon oxide surface, a flow of and 50% nitrogen is started, and the reactions proceed. he reaction 10% of the is consumed. What is the composition of the ex~ haust strea ? In the anae obic fermentation of grain, the yeast Saccharomyces digests glucose fro plants to form the products ethanol and propenoic acid by the following overall rea tions: actio 1: C6HI20 6 ~ 2C2HsOH + 2C02 Reaction 2: Hl20 6 ~ 2C2H3C02H + 2H20 In an open. flow reactor 3500 kg of~ 12% glucose/water solution flow in. During fermentation, 120 kg of carbon dioxide are produced together with 90 kg of unreacted glucose. What are the weight percenrs of ethyl alcohol and propenoic acid that exit in the broth? Assume that none of the glucose is assimilated into the bacteria. Semiconductor microchip processing often involves chemical vapor deposition (CVD) of thin layers. The material being deposited needs to have certain desirable properties. For instance, to overlay on aluminum or other bases. a phosphorus pentoxide- doped silicon dioxide coating is deposited as a passivation (protective) coating by the simultaneous reactions Reaction 1: SiH4 + 02 -7 Si02 + 2H2 Reaction 4PH3 + 502 -7 2P20 S + 6H2 Determine the relative masses of SiH4 and PH) required to deposit a film of 5% by weight of phosphorus oxide (P20S) in the protective coating. ··10.11 Printed circuit boards (PCBs) are used in the electronic industry to both connect and hold components in place. In production, 0.03 in. of copper foU is laminated to an insulating plastic board. A circuit pattern made of a chemically resistant polymer is then printed on the board. Next, the unwanted copper is chemically etched away by using reagents. If copper is treated with Cu(NH)4C12 (cupric ammonium chloride) and NH40H (ammonium hydroxide), the products are water and Cu(NH3)4C1 (cuprous ammonium chloride). Once the copper is dissolved, the polymer is removed by solvents leaving the printed circuit ready for further processing. If a single-sided board 4 in. by 8 in. is to 75% of the copper layer removed using the reagents above, how many grams of each reagent will be consumed? Data: The density of copper is 8.96 glcm3. ··10.12 The thermal destruction of hazardous wastes involves the controlled exposure of waste to high temperatures (usually 9000c or greater) in an oxidizing environment. Types thermal destruction equipment include high-temperature boners~ cement kilns. and industrial furnaces in which hazardous waste is burned as fueL In a properly designed system, primary fuel (100% combustible material) is mixed with waste to produce a feed for the boiler. (a) Sand containing 30% by weight of 4,4'-dichlorobiphenyl [an example of a poly· chorinated biphenyl (PCB)] is to be cleaned by combustion with excess hexane
- 332. Chap. 10 ·10.13 '10.1S "10.17 ·10.19 .3, ,0 : /'10.20 Problems 301/ to produce a feed that is 60% combustible by weight. To decontaminate 8 tons of such contaminated sand. how many pounds of hexane would be required? (b) Write the two reactions that would take place under ideal conditions if the mixture of hexane and the contaminated sand were fed to the thermal oxidation process to produce the most environmentally satisfactory products. How would you suggest treating the exhaust from the burner? Explain. (c) The incinerator is supplied with an oxygen-enriched airstream containing 40% 02 and 60% N2 to promote high-temperature operation. The exit gas is found to have a xCQ,: = 0.1654 and XQ,: = 0.1220. Use this information and the data about the feed composition above to find: (1) the complete exit gas concentrations and (2) the % excess 02 used in the °reaction. synthesis gas analyzing 6.4% CO2, 0.2% 2.40.0% CO, and 50.8% "2' (the balance is N2), is burned with 40% dry excess air. What the composition of the flue gas? Hydrogen-free carbon in the form of coke is burned: (a) With complete combustion using theoretical air (b) With complete combustion using 50% excess air (c) Using 50% excess air but with 10% of the carbon burning CO only. In each case calculate the gas ana.lysis that will be found by testing the flue gases on a dry basis. Thirty pounds of coal (ana.lysis 80% C and 20% H ignoring the ash) are bum~ with 600 Jb of air. yielding a gas having an Orsat analysis which the ratio of CO2 to CO is 3 to What is the percent excess air? gas containing only CH4 and N2 is burned with air yie1ding a flue gas that has an Orsat analysis of CO2: 8.7%, CO: 1.0%, 02: 3.8%, and N2: 86.5%. Calculate the percent excess air used in combustion and the composition of the CH4-N2 mixture. A natural gas consisting entirely of methane (CH4) is burned with an oxygen enriched air of composition 40% O2 and 60% N2. The Orsat analysis of the product gas as reported by the laboratory is CO2: 20.2%, 02: 4.1 %. and N2: 75.7%. Can the reported analysis be correct? Show all calculations. Dry coke composed of 4% inert solids (ash). 90% carbOD, and 6% hydrogen is burned in a furnace with dry air. The solid refuse left after combustion contains 10% carbon and 90% inert ash (and no hydrogen). The inert ash content does not enter into the reaction. The Orsat analysis of the flue-gas gives 13.9% CO2' 0.8% CO, 4.3% 02' and 81.0% N2. Calculate the percent of excess air based on complete combustion of the coke. A with the following composition is burned with 50% excess air in a furnace. What is the composition of the flue gas by percent? CH4:60%; C2H6:20%; CO:5%. °2:5%; N2: 10% A flare is used to convert unburned gases to innocuous products such as CO2 and H20. If a gas of the following composition (in percent) burned in the flare-CH4: 70%. <;3Ha= 5%, CO: 15%,°2; 5%, N2: 5%-and the flue gas contains 7.73% CO2, 12.35% H20, and the balance is O2 and N2, what was the percent excess air used?
- 333. 302 Material Balances for Processes Involving Reaction Chap. 10 *10.21 In underground coal combustion in the phase CO + 1/202 ~ CO2 reactions take place including + 1/2°2 ~ CH4 + 312°2 ~ CO + 2H20 where the CO, and CH4 come from coal pyrolysis. If a gas phase composed of CO: 13.54%, CO2: 15.01 %, CH4: 3.20%, and the balance N2 burned with 40% excess air, (a) how much is needed per 100 moles of gas, (b) what will the analysis of the product gas on a wet basis? *10.22 Ethanol (CH3CH20H) is dehydrogenated in the presence of air over a catalyst. and the following reactions take place CH3CH20H~ + 2 CH3CH20H + 302 ~ 4C02 + 6H2 2 CH3CH20H + 2H2 -4 4CH4 + 02 Separation of product, (acetaldehyde), as a liquid leaves an output gas with the following Orsat analysis: CO2:O.7%. °2:2.1 %, CO:2.3%. H2:7.1 %, CH4:2.6%, and N2:85.2% How many kg acetaldehyde are produced per of ethanol fed into the process? "10.23 So]vents emitted from industrial operations can become significant pollutants it: not disposed of properly. A chromatographic study of the waste exhaust gas from a synthetic fiber plant has the following analysis in mole percent: CS2 40% S02 10 50 It has been suggested that the gas be disposed of by burning with an excess of air. The gaseous combustion products are then emitted to the air through a smokestack. The local air pollution regulations say that no stack gas is to analyze more than 2 S02 by an Orsat analysis averaged over a 24-hr period, Calculate the minimum percent excess air that must be used to stay within this regulation. *10.24 The products and byproducts from coal combustion can create environmental prob-lems if the combustion process is not out properly. boss asks you to out an analysis of the combustion in boiler No.6. carry out the work as-signment using existing instrumentation, and obtain following Fuel analysis (coal): 74% 14% H, 12% ash Flue gas analysis on a dry 12.4% C02t 1.2% CO, 5.7% 02 and 80.7% N2 Wbat are you going to report to your boss? "'10.25 The Air Act requires automobile manufacturers to warrant their control sys-tems as satisfying the emission standards for 50,000 mL It requires owners to have
- 334. Chap. 10 Problems their engine control systems serviced exactly according to manufacturers' specifica~ tions and to always use the correct gasoline. In testing an exhaust having a known Orsat analysis of 16.2% CO2, 4.8% 02' and 79% N2 at the oudet, you find to your that at the end of the muffler the Orsat analysis is 13.] % CO2- this discrepancy caused by an air leak into the muffler? (Assume that the analyses are satisfactory,) If so, compute the moles of air leaking in per mole of exhaust gas leaving the engine, /*"'10.26 One of the products of sewage treatment is sludge. After microorganisms grow in the activated sludge process to remove nutrientc.; and organic material, a substantial amount of wet sludge is produced. sludge must dewatered. one of the most expensive parts of most treatment plant operations. How to dispose of the dewatered sludge is a major problem. Some organizalions seH dried sludge for fertilizer, some spread the sludge on farmland, and in some places it is burned. To bum a dried sludge. fuel oil is mixed with it, and the mixture is burned in a furnace with air. If you collect the foHowing analysis for the sludge and for the product gas Sludge(%) Product Gas (%) S 32 S02 1.52 C 40 CO2 10.14 H2 4 °2 4.65 °2 24 N2 81.67 CO 2.02 (a) Determine the weight percent of carbon and hydrogen in the fuel oil. (b) Determine the ratio of pounds of dry sludge to pounds of fuel oil in the mixture to the furnace. *10.27 Many industrial processes use acids to promote chemical reactions or produce acids from the chemical reactions occurring in the process. As a result, these acids many time end up in the wastewater stream from the process and must be neutralized as part of the wastewater treatment process before the water can be discharged from the process, Lime (CaO) is a cost effective neutralization agent for acid wastewater. Lime is dissolved in water by the following reaction: CaO + 1/202 ~ Ca(OHh which reacts directly with acid. for H2S04, H2S04 + Ca(OHh ~ CaS04 + 2H20 Consider an wastewater steam with a flow rate of lOOO gaVrnin with an acid concentration 2% H2S04, Determine the flow rate of lime Ib/rnin necessary to the acid in this stream if 20% excess lime is used. Calculate the production rate of CaS04 from this process . ~ons/yr, Assume that the specific gravity of the acidic wastewater stream is 1.05,
- 335. 304 Material Balances for Processes Involving Reaction Chap. 10 11110.28 Nitric acid (HNO) that is used industrially for a variety of reactions can be produced by the reaction of ammonia (NH3) with by the foHowing overall reaction: The product basis): NH3 + 202 ~ HNO) + H20 from such a reactor has the following composition (on a water free O.8%NH3 9.5%HN03 3.8% 02 85.9% N2 Determine the percent conversion of NH3 and the percent excess air used. ·10.29 Ethylene oxide (C2H40) is a high volume chemical intermediate that is used to produce glycol and polyethylene glycol. Ethylene oxide is produced by the partial oxidation of ethylene (~H4) using a solid catalyst in a fixed-bed reactor: 1 C2H4 + '2°2 -+ C2H40 In addition, a portion of the ethylene reacts completely to CO2 and H20: C2H4 + 302 ~ 2e02 + 2H20 The product gas leaving a fixed-bed ethylene oxide reactor has the fonowing water free composition: 20.5% C2H40; 72.7 N2~ 2.3 O2; and 4.5% CO2, Determine the cent excess air based in the desired reaction. and the Iblh of ethylene feed required to produce 100,000 tonJyr of ethylene oxide. ··10.30 Glucose (C6H120 6) and ammonia form a sterile solution (no live cells) fed continuously into a vesseL Assume complete reaction. One product formed from the reaction contains ethanol. cells (CHLsOo.sNo.2)' and water. produced is CO2, If the reaction occurs anerobically (without the presence of oxygen), what is the minimum amount in kg of feed (ammonia and glucose) required to produce 4.6 of ethanol? Only 60% of the moles of glucose are converted to ethanol. remainder is con· verted to cen mass, carbon dioxide. and water. ·10.31 Refer to Example 10.9. Suppose that during combustion a very small amount (0.24%) of the entering nitrogen reacts with oxygen to form nitrogen oxides (NOx)' Also. suppose that the CO produced is 0.18% and the S02 is 1.4% of the CO2 + 502 in the flue The emissions listed by the EPA in the load units (ELU)/kg of gas are: NOli. 0.22 CO 0.27 CO2 0.09 802 0.10 What is the total for the stack gas? Note: The are additive.
- 336. CHAPTER 11 MATERIAL BALANCE PROBLEMS INVOLVING MULTIPLE UNITS Your objectives in studying this chapter are to be able to: 1. Write a set of independent material balances for a process involving more than one unit. 2. Solve problems involving several serially connected units. If you have driven past an industrial plant, power station, or waste disposal facility, you must have noticed how complex the equipment is. Such plants involve a large number of interconnected processing units. Based on what you have learned in Chapters 6-10, are you now prepared to solve problems involving an entire plant? If not, study this chapter to find out what to do. Looking Ahead In this chapter we are going to discuss how to treat and solve material balance problems for systems of serially coupled units. You will be pleased to learn that principles employed in previous chapters still apply. All you have to do is apply them to individual subsystems andlor to the overall system. Main Concepts There are nine and sixty ways of constructing friba/lays, and every one of them is right. Kipling 305
- 337. 306 Material Balance Problems Involving Multiple Units Chap. 11 A process nowsheet (flowchart) is a graphical representation of a process. A flowsheet describes the actual process in sufficient detail that you can use it to formulate material (and energy) balances. Flowsheets are also used for troubleshooting, control of operating conditions, and optimization of process performance. You will find that flow sheets are also prepared to represent proposed processes that involve new techniques or modifications of existing processes. Figure 11.1 is a picture of a section of a plant. Figure 11.2a is a flow sheet of the process indicating the equipment sequence and the flow of materials. Figure 11.2b is a block diagram corresponding to Figure 11.2a. The units appear as simple boxes called subsystems rather than as the more elaborate portrayal in Figure 11.2a. You should note that the operations of mixing and splitting are clearly denoted by boxes in Figure lI.2b. whereas the same functions appear only as intersecting lines in Figure 11.2a. Figure 11.3a illustrates a serial combination of mixing and splitting stages. In a mixer, two or more entering streams of different compositions are combined. In a splitter, two or more streams exit, all of which have the same composition. In a separator, the exit streams can be of different compositions. Figure 11.1 Section of a large ammonia plant showing the equipment in place.
- 338. 'Materia' Balance Probtems Involving Muftiple Un~ 307 HIGH·PRESSURE RECYCLE PURGE Reactor . lOW-PRESSURE Ri:CYCLE ,. ' . . ; AMMONIA 8. Ammonia b. Figure 11.2 (8) Flowsheet of the ammonia plant that includes major pieces of equipment and the materials flow. (b) Block diagram of the infonnation flow corresponding to Figure 11.2a. Examine Figure I L3a. Which streams must have the same composition? Is the composition .of stream 5 the same as the composition inside the' unit represented by , " .. ;.the bo,,-? It'lwHl be the same if the contents of the unit are wen mixed, the usual as- ... " sqmption in this'text. If no reaction takes place in the unit, the output composition in stream 5 is the properly weighted average of the input compositions 3 and 4. Do
- 339. 308 Material Balance Problems Involving Multiple Units Chap. 11 4 1 3 2 5 1----.... 6 Figure 11.38 Serial mixing and 7 splitting in a without reaction. Streams 1 plus 2 mix to form Stream 3, and Stream 5 is split into 6 and 7. streams 5, 6, and 7 have the same composition? Yes, because streams 6 and 7 flow from a splitter fed by stream 5. How many material balances can you fonnulate for the three process units shown in Figure 11.3a? First, you can make overall material balances. namely balances on a system that includes an of the units within the boundary denoted by the dashed line shown as I in Figure 11.3b. In addition, you can make balances on each of the units that make up the overall process, 'as denoted by the boundaries fined by the dashed lines n. ill, and IV in Figure 11.3c. Finally, you can make balances about combinations of two or more units simultaneously, as indicated by the boundaries defined by the dashed lines V. Vt and vn in Figures 11.3d to 11 respectively. You can conclude for the three process units shown in Figure 11.3a that you can make material balances on seven different systems. The important question is: How many independent material balance equations can be written for the process? In general, you can write an independent material balance equation for each component present in each unit or subsystem except for splitters. For splitters, 4 :,'-- ------1----- ----, I I I 3 5 I I I I I I I I I 1 L I I " I ... .. I ------_ ....... _--- r" " I 2 7 6 Figure l1.3b The dashed line I designates the boundary for overall material balances made on the process in Figure 11.38.
- 340. Chap. 11 Material I 3 I - - - .. I I , I , I - I I -- I _J II 2 Problems Involving Multiple Units 309 I I 5 I I I I - - --......... -; I I III I -- "" ,I I } -"" IV 7 6 Figure ll.3c lines n, m, IV designate boundaries for material around each of the dividual units comprising the overall process. only one independent balance equation be written regardless of the number of components present in the streams. an example, lef s count the material balances that for the process shown in Figure 11 For the process shown in Figure 11 assume that 3 components are present and components present in the unit por-trayed the How independent balances can you write for the process with these assumptions? Did you get 81 is the count: 3, box: 4, splitter: I, for a total of 8. Next, how many different material balances can you write in total for the process including both independent and dependent You can make overall balances (one for each component and the total balance) plus a total of material balances single units and material baJances pairs of units. you can a total of 34 balances of which only 8 are in-dependent. For this example, wbich 8 of the 34 possible material balance equations should you select? Certainly you should select an independent of equations. As an ex-ample of what not to do, do not a component a 4 ,-....... i--." I I I 3 I I I 1 I " ... ,J .... _--------, v 2 I I I I I I J 5 7 6 Figure 11.3d The dashed line V designates the boundary for material balances around a system comprised of the mixing point plus the unit portrayed by the box.
- 341. 310 3 1 2 Material Balance Problems Involving Multiple Units Chap. 11 I I J I I I I I 4 ,,--l--------. 5 , ! I I I I I I , I ... '" .... --- -_ .... _-- VI 7 6 Figure 11.3e The dashed line VI designates the boundary for material balances about a system comprised of the unit portrayed by the plus nent for each of the individual units in the process plus an overall balance for the same component. This of equations would not independent because, as you know, the overall balance for each species is just the sum of the respective species balances for individual unit As another example, don't use material balances for the total flow either overall or for an individual unit together with al1 of the respective species balances for the system. What should you use to select the particular unit or subsystem with . which to start fonnulating your independent equations for a process comprised of a """'-II ........ ' .. "" .... of connected units? A good, but time-consuming, way to decide is to deter-the of freedom for various subsystems (single units or combinations of units) selected by inspection. A subsystem with zero degrees of freedom a good starting point. Frequently! the best way start is to make material balances for the overall process, ignoring information about the internal connections. If you ignore of the internal streams and variables within a set of connected subsystems, you can the overall system exactly as you treated a single system in Chapters 7 through 10 by drawing a boundary about the entire set of subsystems as in Figure 11.3b. 4 1 3 5 J ~------------------ -, ~ , '--~--------------- -- Figure 11.:11 The dashed V II VII designates the boundary for material balances about a system of 2 7 the plus the splitter.
- 342. Chap. 11 Material Balance Problems Involving Multiple Units EXAMPLE 11.1 Determination of the Number of Independent Material Balances in a Process with Multiple Units Examine Figure 1.1. No reaction takes place. The system is open and steady state. The arrows designate flows. The composition of each stream is as foHows: (1) Pure A B (3) A and B. concentrations known: wA = 0.800, iLlB = 0.200 (4) Pure C (5) A, B, and C. concentrations known: wA = 0.571, wB = 0.143, we = 0.286 (6) PureD (7) A and D. concentrations known: wA = 0.714, wn = 0.286 (8) Band C. concentrations known: wB = 0.333, we = 0.667 1---8 Figure Ell.1 What the maximum number of independent mass balances that can be generated to solve this problem? Write down the possible equations. Do they form a unique set? Solution Select each of the units as a system. With respect to the material bal· ances for the individual units. you can make 9 species equations as foHows for the three units (ignoring any total balances for the 3 units, plus overall species and total balance, plus balances for combinations of units): At unit I. two species are involved At unit II, three species are involved At unit III, four species are involved Total Total number of species balances 2 3 4 9 However, not all of the balances are independent. In the following list. all of the known concentrations been inserted, and F represents the stream flow designated by the superscript. 311
- 343. 312 Material Balance Problems Involving Multiple Units Subsystem 1 { A: FI (1.00) + FO) = F3(O.800) Balances B:Fl(O) + F2(1.00) = FO.20) Subsystem 11 { A: F3(O.800) + reO) = FS(O.571) Balances B: F3(O.200) + reO) = F5(0.143) c: F3(O) + F4( 1.00) = F5(O.286) Subsystem III Balances A: F5(0.571) + ~(O) = F7(O.714) + F8(O) B:F5(O.143) + P(O) = F7(O) + F8(O.333) c: FS(O.286) + P(O) = F7(O) + p8(O.667) D:F5(O) + F6(1.00) = F7(O.286) + F8(O) Chap_ 11 (a) (b) (c) (d) (e) (f) (g) (h) (i) If you take as a basis Fl = 100, seven values of P; are unknown, hence only seven independent equations need to be written. Can you recognize by inspection that among the entire set of 9 equations two are redundant, and hence a unique solution can be obtained using 7 independent equations? If you solved the 9 equations sequentially starting with Equation (a) and ending with Equation (i), along the way you would norice that Equation (d) is redundant with Equation (c) and Equation (h) is redundant with Equation (g). The redundancy of Equations (c) and (d) becomes apparent if you recal1 that the sum of the mass fractions in a stream is unity, hence an implicit relation exists between Equations (c) and (d) so that they are not independent. Why are Equations (g) and (h) not independent? If you inspect the set of Equations (a) through (0 from the viewpoint of solv· ing them sequentially, you will find that each one can be solved for one variable. Look at the following list: Equation Determines Equation Determines (a) F3 (e) F4 (b) F2 (f) F7 (c) Fs (g) F8 (d) Fs (h) Fs (i) F6 If you entered Equations (a) through (i) into a software program that solves equations, you would receive an error notice of some type because they are not an independent set of equations.
- 344. Chap. 11 Material Balance Problems Involving Multiple Units 313' 'Ilf the fresh facts which come to OUT knowledge all fit themselves into the " ........ ,..'" then our hypothesis may gradually become a solution. " Sherlock Holmes in Conan Doyle's The Complete Sherlock Holmes, '''The Adventure of Wisteria LIU.u: ..... you make one or more material balances around combination of subsystems I plus II, or or I plus HI in Example 11.1, or around the en· of three units, no additional material balances will generated. you substitute one of the mass balances an independent I-'''''''".A ..... ''' mass balance? Yes (as long as the precision of the about the same), unless it causes the equation to no longer independent In calculating the degrees of freedom for prohlems involving multiple you must he careful to involve only independent material balances and not miss any ,"VULUU unknowns. All the same principles apply to processes with mUltiple units that were discussed in 7 through 10. Table 1 L 1 is a simplified checklist to help keep in mind possible unknowns and equations for a process. Formal preparation of in such a list will help you to identify which variables and equations to in solving for the unknowns, and, of course, to ensure that the degrees of freedom are zero before starting solve the set of equa- TABLE 11.1 of Variables and Equations to Consider in a Degree-DC-Freedom Analysis Variables total flow) entering and leaving for each subsystem total flow) entering and the overall system any) in set for each subsystem Equations for subsystem or the overall .:>.al' ... ". Material balances (species or element): each species or element (or total) in each subsystem For each species or element (or their total) in the overall system Specifications (for each subsystem and overall) vv _ .. n_~ compositions flow rates .... 1-"" ..... , ...... flow ratios .... "' ........ u., .... conversions or extents of reaction ":-",liH_p restrictions equations (sum of mole or mass fractions)
- 345. 314 Material Balance Problems Involving Multiple Units Chap. 11 lions. Chapter 30 is a detailed discussion of determining the degrees of freedom for a complicated process. The solution goes on famously; but just as we have got rid of the other unknowns, beholdl V disappears as well, and we are left with the indisputable but irritating conclusion 0=0 This is a favorite device that mathematical equations resort to, when we propound stupid questions. Sir Arthur Eddington Frequently Asked Questions 1. In carrying out a degree-of-freedom analysis, do you have to include at the start of the analysis every one of the variables and equations that are involved in the -entire process? No. What you do is pick a system for analysis, and then you have to take into account only the unknowns and equations pertaining to the streams cut by the system boundary (plus those inside the system if it is an unsteady state system). For example, note in Example 11,1 that each subsystem was treated independently. If you had picked as the system the combination of all three of the units, then only the variables and equations pertaining to streams I I 2, 4, 6, 7, and 8 would be involved in the analysis. 2. Should you use element material balances or species material balances in solving problems that involve multiple units? For processes that do not involve reaction, use species balances. Element balances will involve redundancy and prove to be quite inefficient. For processes that do involve reaction, if you are given the reaction equations and information that will enable you to calculate the extent of reaction, species balances are easy to use. (If you are not specifically given the reaction equations, sometimes you can formui-ate them based on your experience, such as C burning with O2 to yield CO2,) OtheIWise use element balances. lust make sure that they are independent equations. . 3. Do you have to understand the details of the elements of equipment in a plant in order to make materi al balances? No. Collections of elements comprising a unit of equipment can be deemed to be the system. In fact, the balances do not have to represent any specific physical element of a unit For convenience in analysis, a hypothetical system may be designated that has no corresponding physical presence. For example, the reactor in Figure 11.2a might be represented as a connected sequence of hypothetical individual reaction systems even though the inside of the reactor forms a continuous bed of catalyst. d
- 346. Chap. 11 Material Balance Problems Involving Multiple Units We next look at some examples of making and solving material balances for systems composed of multiple units. Example 11.2 Material Balances for Multiple Units in Which No Reaction Occurs Acetone is used in the manufacture of many chemicals and also as a solvent. In its latter role, many restrictions are placed on the release of acetone vapor to the environment. You are asked to design an aceto~e recovery system ~ving the flowsheet U1ustrated in Figure E 11.2: All the concentrations shown in E 11.2 of both the gases and liquids are specified in weight percent in this special case to make the calculations simpler. Calculate, A. F. W. B, and D per hour. G = 1400 kglhr. Air Air 0.995 -I Water (100% : 1 W (kg A Water 0.005 (I (kg) 1.00 Absorber r-+- Distillation Column Column ® I) D(k9) Con denser ~ .. Distillate -_............ Acetone 0.99 Water 0.01 1.00 CD ® B (kg) G::: 1400 (kglhr) Entering Gas Air 0.95 Acetone 0.03 Water 0.02 1.00· Solution r Acetone 0.19 F (kg)· Water 0.81 1.00 Figure ~U.2 : Bottom A cetone 0.04 Water 0.96 1.00 This IS an open, steady-state process without reaction. Three subsystems exist. Steps 1, 2, 3, and 4 An the stream compositions are given. An of the unknown stream flows are designated by letter symbols in the figure. Step 5 Pick 1 hr as a basis so that G = 1400 kg.
- 347. 316 Material Balance Problems Involving Multiple Units Chap. 11 Steps 6and 7 could start the analysis of the of freedom with overall balances, but the subsystems are connected serially, we will start the analysis with the o:ln£!,n .. n,~T' column, Unit 1, and then to Unit and then to Unit Unit 1 (Absorber) Variables: 16 W, G, F, A (4 flow streams); SPtlcaes mass fractions in each stream = 3 so that 3 x 4 = 12 more variables Equations: 16 G Species material balances: 3 (one for each species) Specifications: 12 G G G Wt. WAC, Ww W~ = UJ~C = o. wtt: = 1.00 W~. t4 and w1c = 0 w~c. wW. and w~ = 0 Implicit equations (such as I. Wi = 1): all redundant given the specifications Degrees of freedom: 0 proceeding to calculate the of freedom for Unit 2 (the distilla-tion column). you should note the complete lack of information about the properties of the stream going from the distillation column to Unit 3 (the condenser). In general it is best that you avoid making material balances on systems include such streams. as they contain no information. Thus, the next system and Clelzre~e· of-freedom analysis we will be for the system composed of Units 2 aDd 3 combined. Units 2 and 3 (Distillation Column plus Condenser) Variables: 9 B (3 streams)~ species ma~s fractions in each stream::::: 2 so that 2 X 3 = 6 more vanables Equations: 9 oec:les ma.terial balances: 2 (one for each species) Specifications: 6 F D D B B WAC. WAC. Ww. WAC. Ww
- 348. Chap. 11 Material Balance Problems Involving Multiple Units Implicit equations: all redundant F is determined by first solving the equations for Unit 1 (the Absorber) Degrees of freedom: 0 if F is known~ 1 otherwise What would happen if a correct analysis of the degrees of freedom for a subsystem + I? Then you would hope that the value for one of the unknowns in the subsystem could be determined from another subsystem in the overall system. In fact, for this Example, if you started the analysis of the degrees of freedom with the combined Units 2 plus you would obtain a value of + 1 because the value of would not be known prior to solving the equations for Unit 1. StepS The mass balances for Unit 1 after introduction of the basis and other specifications are as follows: Acetone: In 1400 (0.95) 1400 (0.03) OUI = A(O.99S) = F(O.l9) Water: 1400 (0.02) + W(1.00) :::: F(O.SI) + A(O.OOS) (Check to make sure that the equations are independent.) Step 9 Solve Equations (a), (b). and (c) Polymath to get A = 1336.7 kgIhr B = 221.05 kglhr = 157.7 kglhr Step 10 (Check) Use the total balance. G+W~ A+F 1400 1336 157.7 221.05 1557.7 == 1557.1 StepS The mass balances for the combined Units 2 plus 3 are: (a) (b) (c) 317 /'
- 349. 3U) Step 9 MateriaJ Balance Problems Involving Multiple Units Acetone: Water: 221.05(0.19) = D(O.99) + 8(0.04) 1.05(0.81) = b(O.Ol) + B(O.96) , Solve Equations (d) and (e) simul~eo.u sly. to D, IF 34.90 kgIbr . B ::;;;.186.1 kglhr , Step 10 (Cbeek) Use the total balance F;:::: D + B or'221.05 == 34.90+ 186.1 :::: 221.0 Chap. 11 (d) (e) As a matter of interest. what other mass balances could be written for the system and substituted for anyone of the Equations (a) through (e)? Typical balances would be the overall balances In Out Air. G (0.95) ;:::: A(0.995) (f) , Acetone: G(0.03) ::: D(0.99) + B(O.04) . (g) Water: (0.02) + W ;:::: A(O,OO5) + D(O.Ot) + B(0.96) • ·(h) +w :::: A + D + e' , (i) Equations (f) through (i) do not add any extra infonnation to the problem; the grees of freedom are still zero. But any. of equations can be substituted for one of Equations (a) through (e) as long as you make sure that the resulting set of equa· tionsisindependent ' Example 11.3 Material Balances for a Process Involving Multiple Units and Reactions In the face higher fuel costs and the unc~rtainty of the supply a particu-lar fuel, many companies operate two furnaces, one fired with natural gas and the other with fuel oil. In the RAMAn'Corp., each'ftimace has own supply of oxy· gen, The furnace uses air while oil furnace uses an oxidation stream that ~alyzes: 02' 20%; N2, 76%; and CO2, 4%. The stack gases go up a common 'stack:. See Figure 1.3.
- 350. Chap. 11 Material Balance problemSlnvolving Multiple Units 6 Nt: 84.93% Oz: 4.13% S02' 0.10% CO2: 10.840/" 100.00% Air: A 0.21 P 6205 hr Air: A" 02: 0.20 Nz: 0.76 N2: 0.79 Gos Oil 0.04 1.00 Furnace Furnaee 'Nat~ CH .. : 0.96 mol rr ~H2: 0.02 Il't!I ir (F 0.02 mol fr 1.00 c: mOl fr Hi!: 0.47 mol ff 5: mol fr tOO Figure Ell,) (Note that two outputs are shown from the common stack to point out that the stack analysis is on a dry basis but water vapor also The fuel oil composition is given in mole fractions to save you the bother of converting mass fractions to rnole fractions.) During one blizzard, transportation to the RAMAD Corp. was cut off, and officials were worried about the dwindling reserves of fuel oil because the natural , " ,gas supply was being used at its maximum rate possible. At that timet the reserve of fuel oil was only. 560 bb1. How many hours could the company operate before shut· ting down if no"additional fuel oil was attainable? How many Ib mol/hr of natural gas were being consurned? The minimum heating load for the company when transtated into the stack gas output was 6205 lb mollhr of dry stack gas. Analysis of the fuels and stack gas at that time were: Fuel oil (API gravity:::: 24.0) Stack gas Natural gas (Mol %) (Onet CH4 96% C 50 N2 84.93% C2H2 2% H2 47 °2 4.13% CO2 2% S 3 CO2 10.84% S02 0.10% Also, calculate the percent incrtase ill toxic emissions of and mercury per hour caused by the combustion of fuel oil rather than natural gas. /" 319
- 351. 320 Data: Natural gas Material Balance Problems Involving Multiple Units ""'",L<JU factors Arsenic 2.30 X 10-4 1b/l ft3 3.96 X 10-4Ib/lQ3 gal 5.92 X lO-4tb/l03 gal Chap. 11 The molecular weight of the fuel oil was 7 .911bllb mol, and its density was 7.578 Ib/gal. Solution This is an open, steady-state process with reaction. Two subsystems exist. We want to calculate and G in Ib mollhr and then Fin bbllhr. Steps 1,2,3, and 4 We will use elements for the material balances. units of an the variables whose values are unknown will be pound moles. Rather than making balances for each furnace, we do not have any information about individual outlet of each furnace, we will make overall balances, and thus draw the system boundary around both furnaces. StepS 1 hr. so that P ::::: 6205 lb mol Steps 6 and 7 The simplified degree-of-freedom analysis is as fonows. You have five ments in problem and five streams whose values are unknown: A. G. F, A and W; hence, if the elemental mole balances are independent. you can obtain a unique solution for the problem. StepS The overall balances for the elements are (in pound moles) 2H: G(1.94) + 2N~ A(0.79) + 20: A(0.21) + F(0.03) c: G(0.96) + + In Out F(0,47) A "(0.76) A 4(0.20 + 0.04) + G(0.02) (2)(0.02) + 0.02 F(0.50) + O.04A" ::::: ::::: ::::: ::::: ::::: W(l) 6205(0.8493) 6205(0.0413 + 0.001 + 0.1084) +W('/2) 6205(0.0010) 6205(0.1084) The balances can be shown to be independent.
- 352. Chap. 11 Material Balance Problems Involving Multiple Units Step 9 Solve the S balance for F (inaccuracy in the S02 concentrations will cause some error in F, unfortunately); the sulfur is Ii tie component. Then solve for the other four balances simultaneously for G. The results are: F = 207 lb mollhr G = 499 Ib mollhr Finally, the fuel oil consumption is 207 lb mol 7.91 Ib bbl _ 5 1Ihr hr lb mol 7.5781b 42 gal - .14 bb . If the fuel oil reserves were only 560 bbl, they could last at the most 560 bbJ = 109 hr 5.14 ~l Basis: 1 hour The arsenic and the mercury produced are: Oil (5.14 bbllhr)(42 gallbbl) = 216 gallhr Arsenic: 216 &aI13.96 X 1O-41b = 8.55 X 1O-51b gal 216 15.92 X lO-4lb Mercury: . I oJ = 12.78 X 10-5 Ib 1 gaJ (4981b mollhr)(359 ft3nb mol) = 1.79 X lOS ft3/hr Ar . 1.79 X lOS ft3 2.30 X 10-4 Ib 0-5 1 senlC: = 4.11 X 1 b M 1.79 X 10-5 ft3 1.34 X 1O-41b .011 0-5 b ercury: = 2."fV X 1 1 The increase in levels of arsenic and mercury are: X 10-5 Arsenic: (100) = 108% X 10-5 Mercury: --------.........;---(100) = 433% 321 "
- 353. 322 Material Balance Problems InvoMng Multiple Units EXAMPLE 11.4 An.!lysis of a Sugar Recovery Process Involving Multiple Serial Units Figure E 11.4 shows the process and the known data. You are asked to calculate the compositions of every flow ~tream •. and the fraction of the sugar in the cane that is recovered. . F Cone 16% Sugar 25% Water 59% Pulp .--"-........ M Sugar, 10001b/l1r t . L--....- L Crystallizer '-----Water [ H Mill J Screen 1-----1001 Evaporab f--- Water 13% Sugar 1.....-""""11""'---' 14% Pulp 1.....-""""11""'---' D Boga5&e 80% Pulp Solution Steps 1, 2, 3, and 4 Solids G Conloin 95% Pulp Ftgure Ell." An of the known data have been placed on Figure E 11.4. If you examine the figure, two questions naturally arise~ What basis should you pick, and what system should you pick to start the analysis with? Some potential bases and systems lead to more equations to be solved simultaneously than others. You could, pick as a basis F = 100 lb, M::::: 1000 Ib (the same as I hour), or the value of any of the intermediate flow streams. You could pick an overall process as the system with which to start, or any of the individual units, or any consecutive combinations of units. StepS Basis: 1 hour ( M = 1000 Ib) Another important decision you must make is: what are the comppsitions of streams D, E. G. and H? Stream F has three components, and presumably stream K contains only sugar and water. Does stream H contain pu]p? Presumahly not, be· cause if you inspect the process flowsheet you will not find any pulp exiting anywhere downstream of the evaporator. Presumably streams D and G contain sugar and water because the problem implies that not all of the sugar in stream F is recovered. What happens if you assume streams D and G contain no water or no sugar?
- 354. Chap. 11 Material Balance Problems Involving Multiple Units Then you would write a set of material balances that are not independent and/or are inconsistent (have no solution), Try Let S stand for sugar, P stand for pulp, and W stand for water. , Steps 6 and 7 Pick as the initial system the crystallizer. Why? Because (a) if you check the , degrees of freedom for the crystallizer, only a small set of unknowns are involved with zero degrees of freedom, and (b) the crystallizer is at one end of the process. The unknowns are L. and w~. You can make two species balances, S and W, and know that 0.40 + wt(. = I. Conseque'ntly, the degrees of freedom are zero, and the crystallizer seems to be a goo'd system with which to start. If you pick another basis, say F = 100 lb, and another system, say the mill. , you would have 5 unknowns: D, E, wf, w~, and w~. You could make 3 species balances and employ two implicit equations, 2.wf = 1 and 'ZwF = l, hence the degrees of freedom are zero. Steps 8 and 9 For the crystallizer the equations are (using w~ = 1 0.40 0.60) Sugar: K (0.40) :::: L (0) + 1000 Water: K (0.60) = L + 0 from which you get K 0= 2500 Ib and L = 1500 lb. Step 10 Check using the total flows 2500 = 1500 + 1000 = 2500 The next stage in the solution is to pick the evaporator as the system, and repear ~~e degree-~f-freedom analysis 3 variables: H, j, and w((, (w~ = 0) 3 equations: 2 independent species balances P balance is 0.::::: 0), and 'ZwfJ = 1 I The degrees of freedom are zero, You can solve the equations, then proceed upstream one unit, solve the equations for the screen, and lastly solve the equations for the mill. results for of the variables are: 323
- 355. 324 Material Balance Problems Involving Multiple Units Chap. 11 lb mass fraction D;;;;;. 16,755 wf = 0.174 £=7,819 lLI~ = 0.026 F;;;;;. 24,574 wij, = 0.73 G= 1,152 lJ)~ = 0.014 H= 6,667 w~ = 0.036 J = 4,167 w!(, = 0.85 K= 2,500 It)~ = 0.60 L= 1,500 M=lOOO The fraction of sugar recovered is [1000/(24.574)] (0.16) = 0.25. EXAMPLE 11.5 Production of a Hormone in Connected Reactors Figure E 11.5 shows two different reactor configurations to produce a honnone. Option 1 Option 2 F J F = 15 Uhr :.: I :z I F F ~ csln = 10 gil X1 x2 1 2 Figure E 11.5 The relation that gives the growth (generation in Equation (10.1» of the hormone in each vessel in the steady state is the Monod equation J-Lmu.Cs 1-'= Ks + C! where p, = the specific growth rate, llhr Ikrrw. ;: maximum specific growth rate (the state at which the concentration of nutrient becomes the limiting factor in the growth of cells). IIhr c of = nutrient concentration in the vessel, g nutrientIL substrate ~ = Monad constant, g nutrientIL substrate
- 356. Chap. 11 Material Balance Problems Involving Multiple Units substrate the liquid containing water, nutrients. and cells. A coefficient that relates the g of dry cells formed/g nutrient consumed is usually given the symbol Y xis' is called the biomass yield coefficient. Two related for the growth of a hormone have proposed. Op-tion I is to use a single well mixed vessel having a volume of 1 DOL. Option 2 is to use two well mixed vessels each of volume connected in series so that the output of vessell becomes the input to vessel 2. Which option win give the highest concentration (g cellslL of substrate)? Data: The substrate flow rate the vessel for Option 1, or into 1 Option is 15 Uhr. and contains no cells, but contains a nutrient with a concen-tration of 10 gIL. Y xis is equal to g cells1g nutrient consumed. :: 2 gIL. and J..imax :: O.4lhr. The consumption nutrient is ~out/Y)t/8 Solution Ell shows the stream information. Two material balances are involved in the solution to problem. One balance is the cell balance. The concentration of cells will be denoted by x with the units g ceHslL. The other balance is the nutrient balance with Cs being the nutrient concentration. Because the are weB mixed. the concentration of and nutrient~ respectively. effluent the same as concentration the vesseL Assume process is in the steady-state so that accumulation in each (the respective systems) zero. We will use Equation (10.1) for the bal-ances. The Monad equation applies to each vessel. Option 1 Cells Nutrient F o - - xout + J,LX OU1 - 0 = 0 V F V J,LX OU1 +0--- 0 Yx/s (a) (b) In literature of biotechnology the ratio FIV tenned the dilution D, e.g.. the number of culture volumes through the vessel per unit time. whereas in the chemical engineering literature the ratio VIF known as the resi· dence Equation (a): (; - I-' )XOu! = 0 c - c - F hence- = V From Equation (b): F (in out xout V s s Yx/s =0 hence xout = Y (cin xis $ (c) (d) 325
- 357. Materia! BaJance PrGblems Involving Multiple Units Chap. 11 Next, introduce the values for the parameters into the equations. FIV is 151100 per hour, and is equal to J.t so that . 15 L 0.4 c~ut -- Ihr looL 1 hr 2 g + c~Qt from which c~ut 1.2 g nutrientJL substrate. IntrodUce this value into equation (d) to get 0.2 cells . = ....:----------;..... :;;:: 1.76 g cells/4 .substrate 1 g nutrient consumed Option 2 The mat~rial balances are essential thesame as for Option eKcept that ,the vessel volume is 50 L. and the o~tput concentrations from vessel-l become the input concentration to vessel 2. Thus, FN;;;; (15 Uhr)50 ~ = 0.3/hr.: For vessell: 1 L O.4c~~· --- "'- M 50 L ;- 2 + "osu.1 t hence c~jt = 6 g nutrientiL substrate. From equation (d) xyut = 0.2 (10 - 6) = 0.8 g cellslL substrate For vessel 2: Equation (a) now is Foul F -XI V V Equation (b) now is J.tx°ut _cout __ COllt + 0 __ 2_ = 0 V s.1 V s.2 y xis (e) (f) . The solution of Equations (e), (0, plus the.Monad equation (they form a non· linear set of equations) gives C~.2t = 1.35 g nutrientiL substrate and = 1.73 g cellslL substrate. In view of the number significant figures in the values gi ven for the data, the respective answers are essentiaUy the same. Howcan you increase X OU1? I
- 358. 1 Chap. 11 Material Balance Problems Involving Multiple Units 327 " SELF·ASS S MENT T T Questions 1. Can a system be comprised of more than one unit or piece of equipment? 2. Can one piece of equipment treated as a set of several subsystems? 3. Does a flow sheet for a process have show one subsystem for each process unit that is connected to one or more other process units? 4. If you count the of freedom for each individual unit (subsystem) and add them up, can their total be different the degrees of freedom for the overall system? Problems 1. A separations unit is shown in SAT1I Pl. Given that the input stream FI is 1000 lb/hr. calculate the value of and composition of F1 .4 Toluene o o o 0.01 Toluene 0.99 Bet'llene PI I .4 Benzene .2 Xylene 12 0.95 TOJueflt 0.05 Benrene P2() r--"i"" 2 0.1 o Toluene 0.9 o Xylene P28 Figure SATllPl 2. A simplified process for the production S03 to used in the manufacture of sulfuric acid is illustrated in Figure SAT11P2. Sulfur is burned with 100% excess air in the burner, but the reaction S + 02 ~ S02' only 90% conversion of the S to S02 is in the burner. the the conversion of S02 to SOl is 95% complete. Calculate the kg of air required of sulfur burned, and the concentrations of the components in the exit from the burner and from the converter in mole fractions. Air s S (Unburnedl Figure SATllP2
- 359. 328 Material Problems Involving Multiple Units Chap. 11 3. In for the production of pure acetylene, C2H2 (see SATIIP3), pure methane (CH4), and pure oxygen are combined in the burner, .. lh ......... the following reactions occur: C~ + 202 --;. 2H20 + C02 C~ + I! --;.2H20 + CO 2C14 --;. + 3H2 (1) (2) (3) a. the ratio of £he moles of to moles of CH4 fed (0 the burner. b. On of 100 Ib rna] of gases leaving the condenser, calculate how many pounds of water are removed by the COI1tOellser c. What is overall percentage yield product (pure) C2H2• on the carbon in the natural entering the burner? L- .ca.>. ::I ttl tJ, CH4 Waste H2O Solvent CO2 and C2H2 Solvent and Figure SA TllP3 The gases from the burner are cooled in the condenser removes all of the analysis of the gases leaving the condenser is as follows: These gases are sent to an removed with the solvent. % 8.5 CO 58.3 CO2 3.7 4.0 Total 100.0 where 97% of the C2~ solvent from the ~h(e,nrhpr essentially all the CO2 are sent to the CO2 stripper, J
- 360. Chap. 11 Material Balance Problems Involving 329,- where all the CO2 is removed. The analysis of the stripper is as follows: stream Mol % 7.5 92.S 100.0 the top of the CO2 The solvent from the C2Hi as a pure product. is pumped to the C2H2 stripper, which removes all the Thought Problem 1. When choosing a in a process for which to start making material l.I<u ....... ,...,""' what criterion should you use in making the selection? Discussion Problems 1. repreSl~ntc~ as a small process. Material (and energy as well) flows in and out. ¥'re:oaJre of a household that includes the kitchen, laundry, lava· tory. toilets • .., ............... , .. , and air conditioners, and look up so that you can estimate all of material flows in and out of the system as weB as flows. 2. Look up a in an encyclopedia that includes a flow of the process .. "O ..... ,LLL reasons for your systems to be used in making material 1IJ ........ ,,~."'''''. Looking Back In chapter you have seen how systems composed of more than one subsys-tem can treated by the same principles that you used to systems. Whether you use combinations of material balances from each of subsystems aU the units into one system, all you have to do check to that the num-equations you prepare is adequate to solve for variables whose Flow sheets can help in the preparation the equation GLOSSARY OF NEW WORDS Block diagram A sequence of u..., ........... '" tional features of the process flowsheet Connections Streams flowing between and so on to represent opera-
- 361. 330 Material Balance Problems Involving Multiple Units Flowchart A graphical representation of the process layout. Mixer Apparatus to combine two or more flow streams. Overall process The entire system composed of subystems (units). Chap. 11 Process Oowsheet A graphical representation of the process. See Aowchart. Separator Apparatus that produces two or more streams of different composition from the fluid(s) entering the apparatus. ' Splitter Apparatus that divides the flow into tWQ or more streams of the same composition. Subsystem A designated part of the complete system. Well mixed Material within the system (equipment) is of unifonn composition, and the exit stream(s) is of the same composition as the material inside the system. SUPPLEMENTARY REFERENCES In addition to the general references listed in the Frequently Asked Questions in the front material, the following are pertinent. ' " Baldea, M. "Dynamics and Control of Process Networks with, Recycle and Purge," Paper 252 presented at AIChE meeting, Indianapolis, IN, November 4 (2002). Nagiev, M. The Theory of Recycle Processes in Chemical Engineering. MacMillan, New York (1964). , . ' PROBLEMS $11.1 For Figure Pll .. l how many independent equations 'are obtained from the overall balance around the entire system plus the overall balances on units A arid B? Assume that only one component exists in each stream. r--f2---------~-I: !1G! 3qIT51 ~I I 4 I L~-_--------- __ ~ Figure PI!.1 J
- 362. Chap. 11 Problems ·11.2 What is the maximum number of independent material balances that can be written for the process in Fig. PI1.2? 5 A Pure 8 5 } Mix1ure of A and 8 Sfeam flows (unknown) Figure Pll.2 ··11.3 What is the maximum number independent material balances that can be wrjtten for the process in PI t .31 The stream flows are unknown. Suppose you find out that A and B are always combined in each of the streams in the same ratio. How many indepe~dent equations could you write? .4 $Ieom composilioo (known) I A and B 2 Pure C 3 A,8,ond C 4 Dond r 5 Pure £ 6 Pure D 7 AIB.C; and D Figure Pll.3 "'11.4 The diagram Figure PI 1.4 represents a -typical but simplified distillation column. Streams 3 and 6 consist of steam-and water, "and do not come in contact with the fluids in the column that contains two components. Write the total and component matebalances for the three sections of the column. How many independent equations would balances represent? Assume that stream I contains n components. Column 1 Reboller 2 5 ........,~ __ 6 7 Condenser 3 4 Figure Pll.4 1 Feed 20verhtod '3 Cooling HzO 4 Product 5 Reflux 6 Steom 7 Bottoms e liquid flow to reboller 9 Vapor flow to column
- 363. 332 Material Balance Problems Involving Multiple Units Chap. 11 IIi·U.S A distillation process is shown in Figure PI} You are asked to solve for aU the values the stream flows and compositions. How many variables and unknowns are there in the system? How many independent material balance equations can you write? Explain each answer and show all details how you reached your decision. For each stream (except F, the only components that occur are labeled below the stream. 1000 kG/hI ? Cz o.S Ca 0.4 Cl 0.3 c) 'I (;. 1000 kg/hI 0.3 'I O.Z C, ? c. f Figure PUS 2 ? 4 0.14 ··"11.6 In Figure PI 1.6 you wilt see two successive liquid separations colunms operating in tandem in the steady state (and with no reaction taking place), The compositions of the feed and products are as shown in the figure. The amount of W 2 is 20% of the feed. Focus only on the material balances. Write down the names of the material bal· ances for each column treated as separate units (one set for each column). along with the balances themselves placed next to the names of the balances. Also. place an asterisk in front of the names of the balances that will comprise a set of independent equations for each column. Write down in symbols the unknowns for each column. Calculate the degrees of freedom for each column separately. °/0 A 60 B 20 PI C 20 100 kg A 15 B30 C .. W1 Figure Pll.6 A 10 ® B 85 C 5 W2 A 0.2 B ? C ?
- 364. Chap. 11 Problems Then determine the of independent equations overall system composed of the two columns together by listing the duplicate (by symbols) and specifications as welt as redundant equations. and appropriately adding or subtracting them from the found in the first paragraph Deter-mine the of freedom for the overall ",v",,,""rn Check you results obtained in paragraph 2 above by repeating the entire analy-sis of independent equations, unknowns, and degrees freedom for the overall tern. Do they agree? They should. Do not solve any the equations in this problem. "'11.7 Figure PII.7 provided through the courtsey of Professor Mike Cutlip. a. Calculate the molar flow rate of D I' D2> Bland B2. b. Reduce flow rate for each one of the compounds by 1 tum. Calcu-late the rates of D l' D2• B I and Do you notice something unusual? Explain your "' ......... ...". 15% Xylene Styrene 40% Toluene F =70 moUm!n D B FigurePU.' 70/0 Xyiene 4% Styrene 54% Toluene 35% Benzene 18% Xylene 24% Styrene I....-____ ~ 42% Toluene D2 82 16% Benzene 15% Xyiene 10% Styrene Toluene 21% Benzene 24% Xylene Styrene 10% Toluene 2% RAF"A"A *11.8 Figure PI 1.8 shows a schematic for making fresh water sea water by freezing. The pre-chilled sea water is a vacuum at a low The cooling re· quired to some of the feed sea water comes from evaporation of a fraction of
- 365. 334 Material Balance Problems Involving Multiple Units Chap. 11 the water entering the chamber. The concentration of the: brine stream, B. is 4.8% salt. The pure salt-free water water vapor is compressed and fed to a melter at a higher pressure where the heat of condensation of the vapor is removed through the heat of fusion of the ice which contains no salt. As a result, pure cold water and concentrated brine (6.9%) leave the process as products. (a) Determine the flow rates of streams W and D if the feed is 1000 kg per hour? (b) Detennine the flow rates of streams C, B and A per hour? Compressor 1000 kglhr Chilled sea water feed 3.45% NaCI pure water vapor, A Flash freezer Ice + brine B 4.8% NaCI Pure Ice C Figure P 11.8 Pure fresh chilled water, W -------- Chilled brine, D 6.9% NaCI ·11.9 Monoclonal antibodies are used to treat various diseases as well as in diagnostic tests. Figure P11.9 shows a typical process used to produce monoclonal antibodies. A stirred tank bioreactor grows the cells of the antibody of interest, namely immunoglobulin G (lgG). After fermentation in the reactor. a batch of 2200 L contains 220 grams of the product IgG. The batch is processed though a number of stages as shown in Figure Pl1.9 before the purified product is obtained. In the diafiltration stage, 95% of the IgG entering the filter is recovered. in the ultra filtration stage 95% of the entering IgG is recovered, and in the chromatography 90% is recovered. Feed Diluant Centrifugation Homogenization Diafiltratlon 1 Buffer-A Product Eluant Waste Chromatography Ultraflttratton 2 Figure P 11.9
- 366. Chap. 11 Problems 335 TABLEPll.9 Component T9tallnlet Tota) Outlet Proouct Ammonium '64.69 64.69 Biomass 0.00 0.87 Glycerol ' 1 I ,0.00 0.22 0.14 Growth Media -21.76 8.41 Na3Gitrate 0.80 0.80 Phosphoric Acid , t.~.96 1.040.96 Sodium Hydrophosphate 6.83 I Sodium Chloride 55.18 55.19 Sodium Hydroxide 6.83 6.81 Tris-HCl 0.69 0.69 Water 11.459.59 11,458.80 Injection water 18.269.54 Total ' 30,928.72 30!928.71. 0.14 . Table Pl1.9 lists the essential components entering and leaving the overall process in kg per batch. What is the fractional yield of product IgG of the 220 g produced in reactor? .... '11.10 Several are mixed as shown in Figure PI1.l0. Calculate the flows of stream in kg/so CI/O - 4.0 5.0 , 4.0 NaCI A Hel H2SO4 , H2O % B 9.0 'Inert solid 91.0 H20 C :% ,0 2.0, Hel 2.0 H~04 H20, F :;: 290 kg/min % NaCI 1.38 HCI 2.55 H2S0~ 2.21 H20 Inen Solid 1 Figure PH.to ··"'11.11 In 1988, the U.S. Chemical Manufacturers Association (CMA) embarked upon an ambitious and comprehensive environmental improvement effort-the Responsible Care initiative. Responsible Care commited all of the 1 members of the CMA to ensure continual improvement in the areas of health, safety, and environmental
- 367. 336 f2 Material Balance Problems Involving Multiple Units Chap. 11 ity, as wen as in eliciting and responding to public concerns about their products and operations, One of the best ways to reduce or eliminate hazardous waste is through source reduction. Generally, this means using different raw materials or redesigning the production process to eliminate generation hazardous byproducts. As an example, the following countercurrent extraction process (Figure PI1.Il) to recover xylene from a stream that contains 10% xylene and 90% solids by weight. The stream from which xylene is to be extracted enters Unit 2 at a flow rate 2000 kglhr. To provide a solvent for the extraction, pure benzene is fed to Unit 1 a flow rate of 1000 kgIhr. The mass' fractions of the xylene the solids stream (f) and the clear liquid stream (S) have the following relations: 10 cok1ylene = ~~Iene and p2 _ Sl 10 C:OXylene - Ol:Xylene' Determine benzene and xylene concentrations in aU of tl;le streams. What is the percent recovery of the xylene encering the process at Unit 10oa/e> Benzene 1000 kg/hr Waste stream Solids Unl11 Clear lIQuid (no solids) F1 0.9 Solids Unit 2 Figure P11.11 Product stream 1-----1 ... (no solids) SolidS} 0.1 Xylene 2000 k.gIhr "'11.12 Figure PI!. shows a three-stage separation process. The ratio of P31D3 is the ratio of P 102 is 1. and the ratio of A to B in stream is 4 to 1. Calculate the compo-sition and percent of each component in stream Hint:Although the problem comprises connected units; application of the standard strategy of problem solving will enable you to solve it without solving an exces~ number of equations simultaneously. , B 0.23 C 0.27 1.00 F -100 kg (2) Pl a.SOA 0.20 B O.30C E B 0.10 C 0.73 1.00 ® Pa Flpre Pll.ll D3:::: 10 kg (no C) 0 P3 A 0.70 a 0.30 1.00
- 368. Chap. 11 Problems 337 "11.13 In a tissue paper machine (Figure PII.I stream N contains 85% Find the un-known fiber values (all values in the figure are in kg) in kg each stream. p Fiber? N ~--------- Newpup ? Fiber 2.34 Water 7452 Water 3291 Stock chest Water 18 E Fiber? Water 4161 (reservoir) 1----L- -.,.. ... Figure PU.13 Fiber 1 03,26 Water 3309 ··11.14 Metallurgical-grade silicon is purified to electronic grade for use in the semiconductor industry by chemically separating it from its impurities. Si metal reacts in varying degrees with hydrogen chloride gas at 300I.'lC to form several polychlorinated sHanes. Trichlorosilane is liquid at room temperature and is easily separated by fractional distillation from the other gases. 100 kg silicon reacted as shown in 'Figure Pll.14, how much trichlorosilane is produced? Mole % 21.42 H2SiCIz 14.29 SiCI. HCllql 64.29 Ha A D Sils) ..!.... Reacior c DisIIaIioIl [ 100% HSiCI3 Figure PU.14 ..... 11.15 A furnace bums fuel gas of the following composition: 70% Methane (CH,J. 20% Hydrogen (HZ) and 10% Ethane (~H6) with excess air. An oxygen probe placed at the exit the furnace reads 2% oxygen in exit gases. The are passed then through a long duct to a heat exchanger. At entrance to the heat exchanger the Orsat analysis of the reads 6% 02' the discrepancy due to the fact that the first analysis is on a wet basis and the second analysis on a dry basis (no water condenses in the duct). or due to an leak in the duct? If the former. gi ve Orsat analysis of the exit gas from the furnace. If the latter. calculate the amount air that leaks into the duct per 100 mole fuel gas burned.
- 369. 338 Material Balance Problems Involving Multiple Units Chap. 11 **11.16 A power company operates one of its boilers on natural gas and another on oil. The analyses of the fuels show 96% CH4, 2% C2H2• and 2% CO2 for the natural and CnH/.8n for the oiL The flue from both groups enter the same stack, and an Orsat analysis of this combined flue gas shows 10.0% CO2, 0.63% CO, and 4.55% °2- What percentage of the total carbon burned comes from the oil? 110*11.17 Sodium hydroxide is usually produced from common salt by electrolysis. The essen-tial elements of the system are shown in Figure 1. (a) What is the percent conversion salt to sodium hydroxide? (b) How much chlorine gas is produced per pound of product? (c) Per pound of product, how much water must be evaporated in evaporator? FJgure PH.17 Producl 50% NaOH 7"10 NoCI 43%HP ···U.18 The flowsheet shown in Fig. P1l.IS represents the process for the production of titanium dioxide (Ti02) used by Canadian Titanium Pigments at Varennis, Quebec. Sorel slag of the following analysis: Ti02 Fe I nert silicates Wt% 70 8 22 is fed to a digester and reacted with H2S04, which enters as 67% by weight H2S04 in a water solution. The reactions in the digester are as follows: Ti02 + H2S04 ~ TiOS04 + H20 (I) Fe ~02 + H2S04 ~ FeS04 + H20 (2) Both reactions are complete. theoretically required amount of H2S04 for the Sorel slag is fed. Pure oxygen is fed in the theoretical amount for aU the in the Sorel-slag. Scrap iron ,(pure Fe) is added to the digester to reduce the formation of ferric sulfate to negligibJe amounts. Thirty-six pounds of scrap iron are added per pound of Sore) slag. The products of the digester are sent to the clarifier, where aU the inert silicates and unreacted Fe are removed. The solution of TiOS04 and FeS04 from the clarifier is cooled; crystallizing the FeS04' which is completely removed by a filter. The prod- J
- 370. Chap. 11 Problems 339 !.let TiOS04 solution from the filter is evaporated down to a slurry that is 82% by weight TiOS04. The slurry is sent to a dryer from which a product of pure hydrate, TiOS04- H20, is obtained. The hydrate crystals are sent to a direct-fired rotary kiln, where the pure Ti02 is produced according to the following reaction: Reaction (3) is complete. On the basis of 100 Ib of Sorel slag feed. calculate: (a) The pounds of water removed by the evaporator. (3) (b) The exit Ib of H20 per Ib dry air from the dryer if the air enters having 0.036 moles H20 per mole dry air and the air rate is 18 Ib mol of dry air per 100 Ib of Sorel slag. (c) The pounds of product Ti02 produced. Sorel Slag 67% by wtf----1 HzSO. Scrap Iron (Pure Fe) Inert S,lu::ale!'o Unreocled Fe Filler Figure PI1.IS Rotary I( iln Dryer *11.19 An enzyme is a protein that catalyzes a specific reaction, and its activity is reported in a quantity called "units". The specific activity is a measure of the purity of an enzyme. The fractional recovery of an enzyme can be calculated from the ratio of the specific activity (units per mg) after processing occurs to the initial specific activity. A three stage process for the purification of an enzyme involves 1. Breakup of the cells in a biomass to release the intercel1uar products. 2. Separation of the enzyme from the interceUuar product. 3. Further separation of the enzyme from the output of stage 2.
- 371. Material Balance Problems Involving Multiple Units Chap. 11 Based on the following data for one batch of biomass. calculate the percent recovery of the enzyme after each of the stages of the process. calculate the purifi-cation the enzyme that is defined as the ratio of the to the initial specific activity. Fill in the blank columns of the following table. Activity Prote1n Speclftc activity Pertent Stale No. (units) preseat (mg) (UDitslmg) recovery Purificatlon 1 6860 16,200 2 6800 2,200 3 5300 267 J
- 372. R CYCLE, BYPASS, PURGE, AND THE INDUSTRIAL APPLICATION OF MA RIAL BALANC S 12.1 Introduction 12.2 Recycle without Chemical Reaction 1 with Chemical Reaction 1 Bypass and Purge 1 The Industrial Application of Material Balances Your objectives in studying this chapter are to be able to: 1. Draw a flow diagram or sketch for problems involving recycle. bypass, and purge. 2. Apply the 10-step strategy to without chemical involving streams. 3. Solve problems involving a modest number of making appropriate balances. problems (with and bypass, and/or units by 4. the concepts of extent of reaction, overall r<r>f"""""r.,.,,.,. and single- (once-through) conversion in solving problems involving 5. Explain the purpose of a reC'/Cle stream, a bypass stream, and a purge stream, 6. Understand in a sense how material balances in industry. 342 347 355 365 373 341
- 373. 342 Recycle, Bypass, Purge, & the Industrial Application Material Balances Looking Ahead I think that my friends would agree How much recycle confuses me ft's something quite new Only clear to a few r d rather be watching IV DMH Chap. 12 In this chapter we discuss material balances involving recycle-instances in which material from down stream of the process, and the process again. with and without reaction will be discussed. Purge and bypass will also be explained along with the industrial uses of material balances. 12 .. 1 Introduction In Chapters 9 and lOwe restricted the discussion and examples to a single unit with stream inputs and outputs as illustrated in 1 la. In Chapter 11 you encountered multiple units but the stream flows still were an in a forward direction representing serial sequences as in Figure 1 b. this chapter we take up processes which material is recycled, that is fed back from a downstream unit to an upstream unit, as shown in Figure 12.lc. The stream containing the recycled material known as a recycle stream. Feed --........ 1---...... Products a. Feed __ .... 1---..... Products b. Feed __ • 1---... Products c. Figure 12.1 Figure 1 la shows a single unit with serial flows. Figure 12.b shows multiple units but still with serial flows. Figure 12.1c shows the addition ofrecycJe. I
- 374. Sec. 1 1 Introduction 343 What a recycle system? A recycle is a system streams. You can see Figure 12.1c that the recycle stream is mixed with the feed and the combination fed to Process 1. The products from Process 1 are in Process 2 (a) the products and (b) the stream. The recycle is returned Process 1 for further processing. 12.2 illustrates a more complex process involving several streams. Recycle of ~=!11'------- To water recycle Butyl ~ alcohol 1 2 Recycle steam from another L process tank reactors further processing Figure 12.2 A process involving multiple recycle streams comprised of a tanizer and distillation columns. Recycle streams from adjacent processes are feeds process, and some of the products process recy-cle streams fed to other pro1ces:ses Recycle systems can be found in everyday life, Used newspaper is collected from households, processed to remove the and used to new Clearly j more newspapers recycled, the trees that to be consumed to produce newspapers. Recycling of glass, aluminum cans, plastics, copper, and iron are also common. Recycle systems also occur in narure. example, consider the "water cycle" shown 12.3. If a of the is the system, recycle stream con-sists of evaporated water that falls to earth as precipitation, flow of III creeks and rivers brings the water back into system. of the relatively high cost of industrial feedstocks, when chemical re-actions are involved in a recycle of unused reactants the reactor can significant economic for high-volume processing systems. Heat recovery within a processing unit recycle) the overall consumption of the process. Process integration is terminology applied material en-ergy in process are some examples of the application of material recycling in the ""'l"r.I""",.r>c:o industries.
- 375. 344 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 - Clouds ~ - - - - - - -- -1 Evaporation ____ I ~, I ~ ~,t , ' Rain I ~~'=~. ~. : ~ Figure 12.3 A portion of the water cycle. 1. Increased reactant conversion. Recycling the reactants back to the feed to a reactor can significantly increase the overall conversion of the reactants. For certain systems recycle allows the reactor to be operated at low conversion levels, yielding improved selectivity, with recycling of the unreacted reactants making it possible to attain a high overall degree of conversion. 2. Continuous catalyst regeneration. Catalysts are used to increase the rate of chemical reactions, but their effectiveness can diminish with use (catalyst deactivation). Processes that use catalysts that deactivate at a relatively fast rate may require the onsite regeneration and recycling of the catalyst. For example. in a fluidized catalytic cracking (FCC) process (Figure 12.4). the cracking catalyst deactivates almost immediately upon contact with the gas oil feed at the reaction temperature because of the fonnation of coke on the surface of the catalyst. Therefore. the deactivated (spent) catalyst is transported to the catalyst regenerator where most of the coke is burned off the surface of the catalyst to restore the activity of the catalyst. Flue Gas Catalyst Regenerator Cracked Products Reactor GasOI! m~l1A rna~~ofaFCC process. !
- 376. l Sec. 12.1 Introduction Room Air a.. Expansion Valve Compressor a.. Atmosphere Figure U.5 Schematic for a closed refrigeration cycle. 3. Circulation of a working fluid. A number of processes use the closed circulation of a working fluid for heating or refrigeration. Refrigeration systems (Figure 12.5), including home air conditioning systems, circulate a refrigerant gas by a compressor so that the gas absorbs heat from the room air and discharges heat to the outside atmosphere. Frequently Asked Questions If you feed material to a stream continuously, as in Figure 12.lc. why does the amount of the material in the recycle stream not increase and continue to build up? What is done in this and subsequent chapters is to assume, often without so stating. that the entire process including aU units and an of the flows of material are in the steady state. the process starts up or shuts down, the flows in many of the streams change. but once the steady state is reached. "what goes in must come out" applies to the recycle stream as wen as the other streams in the process. You win find that systems involving recycle streams can be quantitatively analyzed without applying any new principles. The procedures developed in Chapters 9, 10, and 11 can be directly applied to systems with recycle. S L .. ASSESSM NT T ST Questions 1. the purpose of using recycle in a process. 2. Will a recycle stream always have the same composition as a product stream?
- 377. -346 Recycle, Bypass, Purge, & the Industrial Application Material, Balances Chap. 12 Problems 1. How many recycle streams occur in Figure SAT12.lP I? P1 - ! XI cR4~a -F-1 ---1:1 ~-------. P2 X2 Figure SAT12.1Pl 2. The Hooker Chemical Corporation operates a process' in Michigan for the purification of Hel. Figure SATl2.lP2 shows the flow sheet for the Hooker ,process. The streams from the bottoms of the five towers are liquid. The streams from ', the tops of the towers are gases. Hel is insoluble in the HCB (hexach1oro~utadiens), The various stream compositions are shown in Figure SAT12.1 P2. How many recycle streams are there in the Hooker process? wt.% Helga.7 t CI2 9.3 , (Feed) Pure wt.% eCI. CC1490.7 HCI 5.6 Cli" 18.2 wt.% CCISO.5 HC149.5 ~ II IV, 289~, .L.3.40atm+ .Q-..... Compressor . wt.'% HCS68.S CC1431.S v ~ wt.% " HCS88.S eCI. 11.5 ~ Pure HCB "'"---~~ HCI and CI2 recycled to chlorination units Figure SA T12.~ P2 ]
- 378. 1 Recycle without Chemical Reaction 347 12.2 Recycle without Chemical Reaction Recycle of material occurs in a variety of processes that do not involve chemical reaction, including distillation, crystallization. and heating and refrigeration sys-tems. an example of a recycle system, look at process of drying lumber shown in 12.6. If dry is used to dry the wood, the lumber will warp and crack. By recycling the moist air that exits from the drier and mixing it with dry air. the inlet air can be maintained at a safe water content prevent warping and crack-ing the lumber. Recycle Dry Air --'------+-1 ~---..J'--__ Moist Air Drier Wet Lumber Dry Lumber Figure 12.6 Lumber drying prClce~iS Another example. shown in Figure 12.7, is a distillation column used to separate two compounds. Note that a portion of the exit flow from the accumulator recycled back into the column as reflux while reboiler vaporizes part the liquid in bottom of the column to create the vapor flow up the column. The recycle of vapor from the reboiler and return of the liquid from the accumulator back into the column maintain good vapor and liquid contact on the trays inside the column. The contact aids in concentrating the more volatile components in the overhead vapor stream and concentrating the less volatile components in the liquid collected in the bottom of the column. You can formulate material balances for recycle systems without reaction exactly as you formulated material balances for processes without recycle~ as explained in Chapter 7 and subsequent chapters. Feed --11»-1 v t f-----I Accumulator Overhead t-<IIIIRi-e-fIU-x"-'" Product ~,.::s::7- Steam Reboller Bottoms L--_ .... Product Figure 12.1 Schematic of a twoproduct distillation column, "
- 379. 348 Recycle, Bypass. Purge, & the Industrial Application Material Balances Chap. 12 , The first step in problem solving to pick a good system(s) for analysis. amine Figure 12.8. You can write material balances for several different systems, four of which are shown by dashed lines in Figure 12.8, namely: 1. About the entire process including the recycle stream~ as indicated by the dashed lines identified by 1 in Figure 12.8. These balances contain no information about the recycle stream. Note that the fresh feed enters the overall tern and the overall or net product is removed. 2. About the junction point (mixing point) at which the fresh feed is combined with the recycle stream (identified by 2 in Figure 12.8) to produce the total, or gross, feed. These balances do contain information about the recycle stream. 3. About the basic process itself (identified by 3 in Figure 12.8). These balances do not contain any information about the recycle stream. Note that the total (gross) feed the process and gross product is removed. 4. About the junction point at which the gross product separated into recycle and overall (net) product (identified by 4 in 12.8). These balances do contain infonnation about the recycle stream. In addition, you can make balances (not shown in Figure 12.8) about combinations of subsystems, such as the process plus the separator (3 plus 4), or the mixing point plus the process plus 3). These balances added to the set of balances for the individual units would not be independent balances, but might be convenient to use, and serve as substitutes for some of the unit balances. Note that in Figure 12.8 the recycle stream associated both with the mixer, which is located at the beginning of the process, and with the separator, which located the end of the process. As a result, recycle problems lead to coupled equations that must be solved simultaneously. Therefore, sets of equations involving rew cycle typically require robust computer software to avoid trouble in solving the equations. You will find that overall material balances (l Figure 12.8), particu- -- ------ - - ------ Recycle R Figure 12.8 with recycle (the ----~------1--------- numbers designate possible system boundaries for the material balances; see the text).
- 380. Sec. 1 Recycle without Chemica! Reaction lady involving a tie component, are usually a good place to 'start when solving recycle problems. If you solve an overall material balance(s). calculate all or some of the unknowns, the rest the problem can usually solved by sequentially applying single unit material balances through the process. If you solving a recycle problem by writing material balances for individual units, and skip the overall balances, you will probably write an excessive number of equations that have be solved simultaneously. EXAMPLE 1.2.1 A Continuous Crystallizer Involving a Recycle Stream Figure E12.1a is a schematic of a process for the production of flake NaOH. which is used in households to plugged drains in the plumbing (e.g., Drano) , F 10,OOOl'Olhr wt% NaOH Evaporator R G 50 wfD/o NaOH Filtrate 45 wt% NaOH Figure Ell.la p NaOH Filter Crystallizer Cake and Filter 5% (8 45 solution of NaOH) The fresh feed to the process is 10,000 Iblhr of a 40% aqueous NaOH solution. The fresh feed is combined with the recycled filtrate from the crystallizer, and fed to the evaporator water removed to produce a 50% NaGH solution, which in tum is fed to the crystallizer. The crystallizer produces a filter cake that is 95% NaGH crystals and 5% solution that itself consists of 45% NaOH. The filtrate contains 45% NaOH. a. are asked to determine the flow rate of water removed by the evaporator. and the rate for this process. b. Assume that the same production rate of NaOH flakes occurs, but the filtrate is not recycled. What would be the total feed of 40% NaOH have to then? Assume that the product solution from the evaporator still contains 50% NaOH.
- 381. 350 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 Solution Open, steady-state process a. Steps 1, 2, 3, and 4 Figure E12.la contains the information needed to solve the problem. Step 5 Basis: 10,000 lb fresh feed (equivalent to 1 hour) Steps 6 and 7 The unknowns are W, G, P, and R. You can make two component balances about three systems: the mixing point A, the evaporator, and the crystallizer as well as two overall component balances. You can also make total balances for the same selection of systems. What balances should you choose to solve the problem? If you plan to put four equations in an equation solver, it does not make any difference as long as the equations are independent. But if you solve the problem by hand, you should count the number of unknown variables involved for each of the three subsystems and the overall system as follows: Component balances Mixing point Evaporator Crystallizer Overa11 Unknowns R plus feed (and compositions) to evaporator (not labeled) W, 0, and feed to evaporator 0, P, andR WandP You can see that by using just two overall component balances (you can substitute the overall total balance for one component balance) you can determine the values of Wand P. Consequently, you should start with overall balances. Steps 8 and 9 Overall NaOH balance (0.4)(10,000) = [0.95 + (0.45) (0.05)]P P = 41131b Overall H20 balance (0.6) (10,000) = W + [(0.55)(0.05)](4113) W= 58871b (or use the overall total balance 10,000 = 4113 + W) The total amount of NaOH exiting with P is [(0.95) + (0.45)(0.05)](4113) = 4000 Ib I .J.
- 382. Sec. 12.2 Recycle without Chemical ReacUon Are you surprised at this result? You shouldn't be. If you put 4000 lb of NaOH into the process, 4000 lb should come out. The amount of water in Pis 113 lb. As a check 1 i 3 + 5887 = 6(){)o Ib as expected. Steps 6 and 7 (repeated) Now that you know Wand 'P, the next step is to make balances on a system that involves the stream R. Choose either the mIxin'g point A or the crystallizer. Which one should you pick?1hc crystaHiz"~r involves three unknowns, and you now know the value of P, so that Oo,nly two unknowns are involved versus introducii'lg a considerable number of new unknowns if you chose mixing point A as the system. Steps 8 and 9 (repeated) NaOH balance on the crystallizer 0.5 G = 4000 + 0.45R H20 balance on the crystallizer 0.5 G ~ 113"+ 0.55 R (or use the total balance G = R + 41 1"3) R;;.; 38,870 Ib h. Now, suppose recycle from the crystallizer does not occur, but the production and composition of P remains the same. Then the output of the crystallizer is just P. as indicated in Figure E 12.1 b. How should you proceed? Do you recognize that the problem is analogous to the ones that you read about in Chapter II? w t G F th/h 50 wr'/o Crystallizer 40 wt% NaOH Evaporator NaOH and Filter , H 45 wi%. NaOH Filtrate Figure E12.1h Step 5 The basis is now P = 4U3 Ib (the same as I hour) p 95%N .sOH FI aka 5D/o (a 4 5 WfOIo ,solutlon of NaOH) 351
- 383. 352 Recycle, Bypass, Purge, & the Industria! Application Material Balances Chap. 12 Steps 6 and 7 The unknowns are now F. W. OJ and H. You can make two component balances on the evaporator and two on the crystallizer plus two overall balances. Only four are independent. The evaporator balances would involve W, and G. The crystallizer balances would involve and R while the overall balances would involve F, W, and H. Which balances are best to start with? If you put the equations in an equation solver. it makes no difference which four equations you use as long as they are independent The crystallizer balances are best to start with by hand because then you have to solve just two pertinent equations for G and Steps Sand 9 NaOH balance on the crystallizer 0.5G::: [(0.95) + (0.05)(0.45)](4113) + 0.45H H20 balance on the crystallizer 0.5G == [(0.05)(0.55)(4113) + 0.55H H== 38.870 lb An overall NaOH balance gives the new F Overall NaOH balance O.40F == 0.45(38.870) + 4000 F= 53,730 Ib Note that without recycle, the feed rate must be times larger than with recycle to produce the same amount of product, not to mention the fact that you would have to dispose of a large volume of filtrate. SELF .. ASS SSM NT T ST QuestIons Why have we not considered the buildup of material in recycle streams this chapter? 2. Under what circumstances might material be accumulated or depleted in a recycle stream? 3. Can you make material balances in both steady-state and unsteady-state flow processes that involve recycle? 4. Can you fonnulate sets of equations that are not independent if recycling occurs in a system? Problems 1. ball mill grinds plastic to make a very fine powder. Look at Figure SAT12.2Pl.
- 384. Sec. 12.2 Recycle without Chemical Reaction ;- ~L.. ___P_ a __ rtl_ c_l e_- rCo~l1ect_o_r_...J I I t I Uncollected powder 1-------.... to waste Ball Mill Product (10,000 kg) fine powder Figure SAT12.2Pl 353 At the present time 10,(X)() kg of powder are produced per day. You observe that the process (shown by the solid lines) is inefficient because 20% of the feed is not recovered as powder-it goes to waste. You make a proposal (designated by the dashed lines) to recycle the uncollected materia] back to the feed so that it can be remiUed. You plan to recycle 75% of the 200 kg of uncollected material back to the feed stream. If the feed costs $ L201kg, how much money would you save per day while producing 10,000 kg of fine powder? 2. Sea water is to be desalinized by reverse osmosis using the scheme indicated in Figure SAT12.2P2. Use the data given in the figure to detennine: (a) the rate of waste brine removal (B); (b) the rate of desalinized water (called potable water) production (P); (c) the fraction of the brine leaving the reverse osmosis cen (which acts in essence as a separator) that is recycled. 1000lblhr Sea water 3.1% salt Brine Recycle 4.0% salt Reverse osmosis cell P Desalinized Water 500 ppm salt Figure SATll.2P2 Brine waste (B) 5.25% salt 3. A material containing 75% water and 25% solid is fed to a granulator at a rate of 4000 kglhr. The feed is premixed in the granulator with recycled product from a dryer. which follows the granulator (to reduce the water concentration of the overall material fed into
- 385. 354 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 the granulator to 50% water, 50% solid). The product that leaves the dryer is 16.7% water. In the air is passed over the solid being dried. The air entering the dryer contains 3% water by weight (mass), and the air leaving the dryer contains 6% water by weight (mass). . a. What is the ratio of the recyc1e to the feed entering the granulator? b. What is the rate of air flow to the dryer on a dry basis? . 4. Benzene, and other aromatic compounds can be recovered by solvent extraction with sulfur dioxide (S02)' Figure SAT12.2P4 is the process schematic. As an example, a catalytic reformate stream containing 70% benzene and 30% nonbenzene material is passed through the countercurrent extractive recovery scheme shown in Figure SAT12.2P4. 1000 lb refonnate and 3000 Ib of S02 are fed to the system per hour. The benzene product stream contains 0.15 lb of S02 per lb of benzene. The raffinate stream contains all the initial1y charged nonbenzene material as well as Ib of benzene per Ib of non benzene material. The remaining component in the raffinate stream is S02' How many Ib of benzene are extracted in the product stream on an hourly basis? How many lb of raffinate are produced per hour? F S02 Feed 3,ooOlblhr p Thought Problem Product 2 1 R Raffinate Figure SAT12.2P4 1. Centrifugal pumps cannot run dry, and must have a minimum fluid flow to operate properly- to avoid cavitation, and subsequent mechanical damage to the pump. A storage tank is to be set up to provide liquid flow to a process, but sometimes the demand will drop below the minimum flow rate (10-15% the rated capacity of the pump). What equipment setup would you recommend be implemented so that the pump is not damaged by the low flows? Draw a picture of the layout so that the minimum flow can go through the pump no matter what the level of liquid is in the feed tank and no matter what the outlet pressure and demand may be. J
- 386. .12.3 Recycle with Chemical Reaction 355 Discussion Problems 1. of limitations in supply as well as economics, many industries reuse their water over and over again. example~ recirculation occurs in cooling towers, boilers, powdered coal transport, multistage evaporation~ humidifiers. and many devices to wash agri. cultural products. Write a report discussing one of these processes, and include in the report a descrip-tion of the process, a simplified flow problems with recycling, the extent of purge. and, if you can find the information, the made by recycling. A of with forward and reverse flow, and among is known as a cascade. The enrichment of natural uranium by of uranium hexafluoride is a well-known example of a process involving a cascade. Deve]op a flowsheet of a cascade. and indicate what material balances are for individual units at the beginning, middle, and end of the cascade, The Celanese company appealed to the Court the application an rule concerning benzene leakage to the atmosphere, Four of their plants used benzene in their processes. recycling the benzene over and over the'plants. The company believed that EPA rule on use applied to inventory (storage) of benzene. 'which was tess than the lOoo-metric-ton regulatory threshold "use," whereas the EPA interpretation of the "use' the benzene was it should counted each it circulated through a plant. The EPA argument was that every time the benzene circulated through a plant, it had the same chance of leaking into the as new benzene would. What is your opinion about this ar-gument? (The U.S. Court of Appeals with EPA.) 12.3 Recycle with Chemical Reaction The most common application of recycle for systems involving chemical reaction is the recycle of reactants, an application that is used to increase th~ overall con-in a reactor. Figure 12.9 shows a simple example for reaction A-tB From data in 12.9 you can see that the steady-state material balances for the mixer, reactor, and separator are satisfied. Also, that for the overall material balance if you calculate the extent of reaction you will find it 100 reacting g molls. Reactor 100 gmalls I--~ B 900 gmolls A Figure 12.9 A simple recycle system with chemical
- 387. 356 Recycle, Bypass, Purge. & the Industrial Application Material Balances Chap. 12 If you calculate based on B extent of reaction for the overall process in Figure 12.9 100 0 e-ovemll = 1 = 100 moles reacting If you use material balances to calculate the output P the reactor (on the basis of I second) you get A =900 g mol B = 100 g mol and the extent of reaction based on B for the by itself as the system is 100 - 0 ---= 100 moles reacting 1 ereactor In the extent of reaction is the regardless of whether an overall material balance is used or a material balance for the reactor Is used. This im· portant fact can be used in solving material balances for recycle systems with reactions. You win encounter two types of conversion when reactions occur: 1. Overall fraction conversion: mass of reactant in the fresh of reactant in mass (moles) of reactant in the 2. Single - pass ("once· through") fraction conversion: mass of reactant fed into the reactor - mass (moles) of reactant ~~.~ ..... the reactor mass (moles) of reactant fed into the reactor As the name indicates. the overall conversion depends only on what enters and leaves the overall process, while the single-pass conversion depends on what enters and leaves the reactor. For simple recycle reactor in Figure 12.9, the overall conversion is 100% 100 - 0 X 100 = 100% 100 and the single-pass conversion is 10% 1000 - 900 x 100 = 1000 When the fresh feed consists of more than one reactant, the conversion can be ex .. pressed for a single component, usnaUy the limiting reactant, or the most impor-
- 388. Sec. 12.3 Recycle with Chemical Reaction 357 tant (expensive) reactant. Remember Chapter 10 that in a reactor (the single-pass conversion) can be by chemical equilibrium chemical kinetics, but the overall conversion limited by the efficiency the separator in separating compounds to be recycled from the compounds that are not recycled. overall conversion if with a subscript OA) and the conversion SP) can be in terms of the extent of that was 10. From the definition of the overall conversion, note that the numerator equal to the product of the reaction times of the stoichiometric coefficient of the (A), and the denominator is the moles of of the reactant (A): Overall conversion of species A = faA = --- (12.1) single-pass conversion. numerator is the same as in Equation (12.1) while the denominator equal to the amount of the l,.<Cll'I..-UJ,JlU. fed to the reactor: .; conversion = f SP :::: nr;.ae t or f ee d (12.2) you solve Equations (12.1) and (12.2) for the extent of reaction, equate the ex-tents, and use a balance at the mixing point n~ctor feed = feed + nl~ycle, you can obtain the following relationship between overall and conversion: n~esh feed = --:---:-:---:-----:- (12.3) f OA nrsh feed + n~ccyclc you now apply Ua. .. 1Vl (12.3) to the simple recycle example in Figure 12.9, what value do you get for the the single-pass to overall conversion? Do you get 0.1, which agrees with previously? EXAMPLE 12.2 Recycle in a Process in Which a Reaction Occbrs Cyclohexane (C6H12) can be made by the with hydrogen according to the following reaction: C6H6 + 3H2 --i> C6H 1'2 For the process shown in Figure E12.2, determine of the recycle stream to " the fresh feed if overall conversion of U"' .. ,L. ..... ~.'" is 95%. and the single-pass conversion is 20%. that 20% excess is used in the fresh feed, and that the composition of the recycle stream is % benzene and 78.26 mol % hydrogen.
- 389. 358 Recycle, Purge, & the Industrial Application Material Balances Chap: 12 Reactor feed Product P F Fresh Feed nl Mixer f 20% excess H2 " V nF H2 Figure E12.2 O.A" .....'..".. ~. R Recycle 22.74'% Bz n.26% Separator of a recycle reactor. Also calculate the environmental impact of the product gas by environmental index based on the foHowing threshold limit values spective components: Solution Benzene Cyclohexane Hydrogen The process is open and steady state. Step S (ppm) 0.5 300 WOO an for the re- A convenient basis to choose would be 100 mol (g mol or Ib mol) of DentZeflC feed. althQu,gh you could the recycle to be 100 mol. Steps 1, 2, 3, and 4 Figure E12.2 contains all of the information available the flowstreams except the amount of H2> which is in 20% ex.cess (for complete reaction, remember) n~2 = 100(3)(1 0.20) 360 mol and the total fresh feed is 460 mol. From Equation (12.l) for (vHt. = -I) -( -l)~ 0.95 = 100 you can calculate that ~ = 95 reacting ·mole~.· Steps 6 and 7 unknowns are R, n' . , nk~' and ntlHrl' You can write three species ances for of the three systems, point, the reactor. and the
- 390. l 12.3 Recycle with Chemical Reaction plus overall (not all of which are independent, of Which systems should you to start with? The process, because then you can use the calculated for the extent of reaction. Steps 8 and 9 species overall are nptlt = n!n + I I Bz: n&z=lOO+(-1)(95) =5mol H2: nf'I2 = 360 + (. )(95) - mol C6H12 nt6H12 = 0 + (1)(95) = 95 mol P = 175 mol next step is to use the final piece information, the information about the conversion and Equation (12.2), R. The system is now the re-actor. The amount of the Bz to the reactor 100+ 0.2274R, and , = (the same a.s from the overall conversion). Thus. for benzene and Finally, the ratio -(-1)95 0.20 = 100 + 0.2274R R:= 1649 mol to fresh feed is R = 1649 mol = 3.58 F 460 mol The higher the TLV. the more exposw-e that can should use the of the TLV appropriately tolerated, hence an index __ "" .. ._. You can use concentrations or mole fractions as weights. Environmental index = 1;5 ( o~s) + -1- C~) + !7}S C~o o) = 0.059 Note that the benzene contributes 96% of the EXAMPLE 12.3 Recycle in a Process with a Reaction Occurring Immobilized isomerase is as a catalyst in producing frutose from glucose in a fixedMbed (water is the solvent), For the system shown in Figure El percent conversion of glucose results on one pass through the reactor of the exit to recycle stream mass units is equal to 8.33? II -+ C12H220U Glucose Fructose
- 391. 360 Recycle, Bypass, Purge. & the Industrial Application Material Balances Chap. 12 Recycle reed Fixed-Bed Product 40'" Glucose 4' .. F'ruclo.se Reactor In Water (0) Figure E12.3a Solution The process is an open. steady-state prqcess with a reaction occurring and a ',recycle. Steps 1, 2, 3, and 4 Figure E12.3b includes all the known and unknown values of the variables using appropriate notation (W stands for water, G for glucose. and F for fructose). Note that the recycle stream product stream have the same composition. and consequently the same mass symbols are used in the diagram for each stream. StepS R(kg) .------ =1 ~=? ~=? 1.00 (b) Figure E12.3b Pick as a basis S = 100 kg. given the data shown in Figure E12.3b. Step 6 We have not provided any notation for the reactor exit stream and composition because we will not be using these values in our balances. Let f be the fraction conversion for one pass through the reactor. The unknowns are R, P, T. w~. w~, w{t" wb. w~, and!. for a total of 9. Step' The balances are Lwf = 1. ~wr = 1, R = PIS.33, plus 3 species balances each on the mixing point 1, separator and the reactor as well as overall balances. J
- 392. Sec. 12.3 Recycle with Chemica! Reaction We will assume we can frnd 9 independent balances among the lot and proceed. We do not have to solve all of the equations simultaneously. The units are mass (kg). Steps I) and 9 We will start with overall balances as they are easy to form and are often decoupled for solution. OveroU balances Total: p::: S = 100 (How simple!) Consequently. 100 R;;;;;:: - = 12.0 8.33 Overall no water is generated or consumed, hence Water: 100(0.60) = P~ = l00~ ~ = 0.60 We now have 6 unknowns left for which to solve. We start somewhat arbitrarily with mixing point 1 to calculate some of the unknowns Miring point 1 No reaction occurs so that species balances can be used without involving the extent of reaction: or Total: 100 + 12 = T = 112 Glucose: Fr'!J.ctose: 100(0.40) + 12cu~ = 112wb o + 12w~ = 112(0.04) w~ = 0.373 Also, because w~ + w~ + w(t, = 1, w~ = 1 - 0.313 - 0.600 = 0.021 Next, from the glucose balance wb = 0.360 Next, rather than make separate balances on reactor and separator, we will combine two into one system (and thus avoid having to calculate values associ-ated with the reactor stream). 361
- 393. 362 Recycle. Purge. & the Industrial Application MateriaJ B~ances Reactor plus Seporator 2 Total: 12 + 100;:::; 112 (a redundant' equation) Step 10 Check wbr - (R + P)(£r)~) = (f)(wbr) (0.360)( 112) -:-( 112)(0:027) ~ f{0:360)(:112) 40.3 - 3.02 = f ( 40.32) , j= 0.93 Equation (12.2) and 3.02 - 40 = 37 j -1 '~ . extent of reaction -) )(37) , , ~---=0.93 40 EXAMPLE 12.4 A Bioreactor with R~cycle Chap. 12 Reactors that biological (bioreactors) use living organisms to "produce a .' products. Bioreactors ' for 'prodUcing 'ethanol, 'antibi-otics, and proteins dietary supplements and medkal diagnbSis'. E12.4 shows a recycle bioreactor in which the overall conversion of ~e proprietary com~ ponent in the fresh to product is 100%. The conversion of the 'Proprfenlry com-ponent to product per pass·inthe reactor is'_40%. the amount and the mass percent of component in the stream the product 90% product, and the to the reactor 3 wt % of the component. Fresh Meduim (F) '--_-' 10"'.k component 90% water Recycle Component water P 3% Component Bloreactor Figure E12.4 ProdU¢straam (P) , 10% water 90% ~roduct Live cell retum Cell Separator Waste stream (W) Water Oeadcells
- 394. 12.3 Recycle with Assume that the component and weight, and that the waste contains only the same molecular Solution Steps 2, 3, and 4 AU of the data have been in Figure Step 5 Pick a convenient of tOO kg of fresh feed (F), Steps 6 and 7 Pick the overall nrr.rD"'''''-ll$ the system. Variables (9): 3 stream flows Equations (9): 2 compositions in each stream The degrees of Steps 8 and 9 of mass fractions in 3 streams UV ....... .I..lJ"'al .. VJJli>. One composition in F (the other is redundant) One composition in P OJ .. L ...... """''' Component and water (or total) .......... ' .......... 1' overall conversion Total _"""' ... are zero. Total balance: 100 = P + W Component eu ..... ',",,,", 0.10 (l00) ;:: 0.90 P P = 11.1 kg W = 88.9 kg plus the product recovery unit as the system. Steps 6 and 7 I 3 1 1 2 9 .............. "', ... analysis can be omitted of proprietary component 40% single-pass conversion. component in the recycle stream. all that is needed to is a component balance of recycle and w be the 8and9 Apply Equation (10.1) using as units kg (the molecular of the compo-nent can be eliminated from each tenn by division). The consumption tenn represents conversion of 40% of the input term. 383' "
- 395. 364 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 Accumulation Input Output Generation Consumption o :: {l00 (0.10) + Rw] - Rw + 0 -DAD [100 (0.10) + Rw] Rw :: 15 kg of component in the recycle stream Next, pick the mixer as the system. Component balance: 100 (0.10) + 15 :: 0.03 FI Total balance: R= - 100 15 w = - = 0.0205 SELF-ASSESSMENT TEST Questions 1. If the components in the feed to a process appear in stoichiometric quantities and the subsequent separation process is complete so that all of the unreacted reactants are recycled, what is the ratio for reactants in the recycle stream? " . 2. Answer the following questions true or a. The general material balance applies for processes that involve recycle with reaction as it does for other processes. b. The key extra piece of information in material balances on processes with recycle in which a reaction takes place is the specification of the fraction conversion or extent of reaction. c. The degrees of freedom for a process with recycle that involves chemical reaction are the same as for a process without recycle. Cite two reasons using recycle in a process. Problems A catalytic dehydrogenation process shown in SATI2.3Pl, produces 1,3 butadiene (C4H6) from pure normal butane (C4HIO). The product stream contains 75 mollhr of H2 and 13 mollhr C4HIO as well as C4H6. The recycle stream is 30% (mol) C4HlO and 70% (mol) C4H6• and the flow is 24 mollhr. Pure ~---to~r--"" C4H10 F :::: ? moles/hr Recycle xC.Ho::::: 0.30 XC,He :: 0.70 Figure SA TI2.3Pl 1---100- nH~ =: moles/hr "C,H10 :: 13 moles "C4H6 ==?
- 396. 12.4 Bypass and Purge (a) What the feed rate, p, and the product flow rate of C4H6 1eaving the (b) What the single-pass conversion butane in the process? ,,- 365 2. (C3Hg) from EI Paso is dehydrogenated catalytically in a continuous process to obtain (C3H6). All of the hydrogen is separated from the exit gas with no hydrocarbon. The hydrocarbon is then fractionated to a product stream 88 mole % propylene and 12 mole % propane. The other stream. which is 70 % propane and 30 mole % propylene, is recycled. The one-pass conversion in the reactor 25%, and 1000 kg of fresh propane are hour. Find (a) .the kg of product stream per hour, and (b) the kg of recycle stream per ) Ethyl ether is made by the dehydration ethyl alcohol in the presence of sulfuric acid at 140°C: 2C;HsOH -t C;HSOC2Hs + ~O Figure SATI2.3P3 a simplified process diagram. H 87% conversion of the alcoho1 fed to the reactor occurs per in the reactor. calculate: (a) kilograms per hour of fresh and (b) kilograms recycle. 93% HzSO .. alcohol 5% weter Reactor Discussion Problem -"'Pure diethyl ether (1200 kg/hr) Ether separation Recycle 92°4 alcohol 8% weter Figure SAT12.3P3 Waste Sulfuric acid Alcohol and water 1. Numerous techniques have been proposed desulfurization of flue The techniques C8[leQ'(lin7j~ by the phase in which u"' .... , • ..., ..... that influence the choice 12.4 Bypass and Purge additional commonly 12.10 and 12.11. eaCOOllS occur: gas or solution. What are the process for flue gas desulfurization? types of process streams shown in
- 397. 366 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 Bypass 8 Feed ;..--.... Product Figure 12.10 process with a bypass stream. 3. A bypass stream-a stream that skips one or more stages of the process and goes directly to another downstream stage (Figure 1 10). A bypass stream can be used to control the composition of a final exit stream from a unit by mixing the bypass stream and unit exit stream in proportions to obtain the final composition. b. A purge stream bled off from the process to remove an accumula-tion of inerts or unwanted material that might otherwise build up the recycle stream (Figure 12.11). Recycle R 1---.... Purge Feed 1---.... Product Figure 12.11 A process with a recycle stream with purge. Many companies have had unfortunate experience that on startup of a new process, trace components not considered in the material balances used in the design of the process (because the amounts were so small) build up in one or more recycle loops. Look at Figure 12.12 for an example of a process involving recycle with a Note how in the steady state the argon concentration is different each sue· cessive recycle stream so that 1 % argon occurs in the feed stream while 25% argon occurs in the third recycle purge stream. For effective operation, the Ar concen~ tration cannot be allowed to increase further. Remember the process operates continuously in the steady state so that the AI concentration is constant in each individual recycle stream. Calculations for involving bypass and purge streams introduce no new principles or techniques beyond those presented so far. Two examples will make that clear.
- 398. Sec. 12.4 Bypass and Purge Dis ti lIation Reboiler Liquid NH3 Reactor 5% argon Condenser Liquid NH3 Reactor 11 .2% argon Distillation Liquid NH3 Reactor 25% argon Figure 12.12 A process to manufacture ammonia that involves three reactors and three distillation columns. Note the stepwise rise in the concentration of Ar in the recycle streams. EXAMPLE 12.5 Bypass Calculations In the feedstock preparation section of a plant manufacturing natural gasoline, isopentane is removed from butane-free gasoline. Assume for purposes of simplification that the process and components are as shown in Figure E12.5. What. fraction butonize( ,I Oe- I I I CD to o kg I Butene fre~ teed ,. /' / / - 0 IsopentGne side streo "..- i-CsH'2, 100°4 ' Iso- pentone ~ tower I 0 J Ollerall sy!.tem boundary n-C5Ht2 1000~M4 iX I To natural U!.oline , t~ ,/ ® plant r.tJ 90% n-C~12 to" i-CsH12 Figure EI2.S 367
- 399. 368 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 of the butane-free gasoline is passed through the isopentane tower? Detailed steps will not be listed in the analysis and solution of this problem. The process is in the steady state and no r,eaction occurs. Solution By examining the flow diagram you can see that part of the butane-free gasoline bypasses the isopentane tower and proceeds to the next stage in the natural gasoline plant. An the compositions (the streams are liquid) are known. Select a basis: 100 feed What kind of balances can you write for this process? You can write the following: 8. Overall balances (each stream is designated by the letter F, S, or P with the units being kg) Total material balance: In Out -=~- 100 S + p (a) Component balance for n-C5 (tie component) In Out ---=----- 100(0.80) S(O) + P(O.90) (b) Consequently. (0.80) P = 100 0.90 = 88~9 kg S= 100 - 88.9::: ILl kg The overall balances will not tell you the fraction of the feed going to the isopentane tower. For this calculation you need another balance. b. Balance around isopentane tower: Let x be the of butane-free gas going to the isopentane rower, and y be the kg of the n-CsH 12 stream leaving the isopentane tower. Total material balance: In Out -= xlI.I + y Component balance for n-Cs (a tie component): x(O.80) = y Consequently, combining (c) and (d) yields x 55.5 kg. or the desired fraction is 0.55. (c) (d) I
- 400. 12.4 Bypass and Purge Another approach to this problem is to make a balance at mixing points 1 and 2. Although there are no pieces of equipment at those points, you can see that streams enter and leave the junctions. c. Balance around mixing point . material into junction:;:: material out TotaL material: (100 - x) + y = 88.9 (e) Component (iso-Cs): (100 x)(O.20) + 0 :;:: 88.9(0.10) (f) Equation (0 avoids use of y. Solving yields x ::::: .5 kg as before EXAMPLE 12.6 Purge Considerable interest exists in the conversion of coal into more convenient liquid products for subsequent production of chemicals. Two of the main gases that can generated under suitable conditions from insitu (in the ground) coal combustion in the presence of steam (as occurs narurally in the presence of groundwater) are H2 CO. cleanup, these two gases can be combined to yield methanol according to the foHowing equation CO + 2H2 ~ CH30H E 12.6 illustrates a steady-state process for the production of methanoL AU of the compositions are in mole fractions or percent. stream flows are in moles. { 67.1 HI! Feed F 32.5 CO 02 CH4 ,rI --------- I I Mix l Reactor Seporotor I , I I Recycle R I I H 2 Split yCO r L _____________________ _ Figure EIl.6 ----., E Putge P I I I I I I I I I . I CH 10 Note in Figure 2.6 that some enters the process, but not partici-pate in the reaction. purge stream is used to maintain the CH4 concentration in the' from the separator at no more than 3,2 moJ%. and prevent hydrogen buildup as well. once-through conversion the CO in the reactor is 1 Compute the moles of recycle. eHlOH. and purge per mole of feed, and also compute the purge gas composition. 369
- 401. 370 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 Solution Steps 1, 2~ 31 and 4 All of the known information has been placed on the diagram. The process is in the steady state with reaction. The purge and recycle streams have the same composition (implied by the spliter in the figure). The mole fraction of the components in the purge stream have been designated as X, y, and 'z for H2, CO, and CH4, respectively. Step 5 Select a convenient basis: F = 100 mol Step 6 The variables whose values are unknown are x. y. z. E, p. and R. You can ignore the stream between the reactor and separator as no questions are asked about it. Step 7 Because the problem is presented in teImS of moles, making an overall mass balance is not convenient. Instead we will use element balances. You can make three independent element balances for the overall process: H, C, and 0 balances. If you make a CO species balance on the reactor plus separator, you can use the information about the percent conversion of CO to provide one additional balance. How can you obtain fifth and sixth balances so that the system of equations is determinate? One piece of information given in the problem statement that has not been used is the information about the upper Ilmit on the CH4 concentration in purge stream. This limit can be expressed as z S 0.032. Assume that the purge stream contains the maximum allowed CH4 so that you can get an equation, thus making z;;;;: 0.032 (a) Another piece of information is the implicit mole fraction balance in the recycle stream x+y+z=l Steps 8 and 9 The overall element balances are (in moles): 2H: 67.3 + 0.2(2):: E (2) + P (x + 2z) C: 32.5+0.2 =E(l)+P(y+z) 0: 32.5 =E(1)+P(y) (b) (c) (d) , (e) For a system composed of the reactor plus the separator (chosen to avoid calculating the unknown infonnation about the direct output of the reactor), the CO bal-ance is
- 402. 12.4 Bypass and Purge In Out Consumed co: [32.5 + Ry] - [y(R + P)J = (32.5 + Ry)(O.lS) (f) Equation (8) can be substituted into Equations (b) through (0. and the resultfive equations solved by successive substitution or by using a computer The resulting values obtained are (in moles) E CH]OH 31.25 p purge 6.25 R recycle 705 x H2 0.768 Y CO 0.200 Z CH4 0<032 Step 10 Check to see each of the balances (b)-(f) is satisfied. If you want use the. extent of reaction to make the calculations. you must first calculate gmax for CO H2• and then find that was the limiting reactant. You could reach the same conclusion by inspection. Then 0.18 E = 0 + (1)~ would equations to use. SELF .. ASS SSM NT T ST Questions 1. Explain what bypass means in words and also by a diagram. 2. Answer the following questions true or 371 -' a. Purge is used to maintain a concentration of a minor component of a process stream below some set point so that it not accumulate the b. Bypassing means that a process stream the process in advance of the to the process. c. A trace component in a or produced a reactor has negligible on the overall material balance when occurs. 3. the waste stream the same as a purge stream a process? Problems 1. In the famous process (Figure SAT12.4PI) to manufacture ammonia, me reaction is carried out at pressures of 800 to 1000 atm and at 500 to 600°C using a suitable catalyst Only a small fraction of the material entering the reactor reacts on one pass, so recy-is needed. Also, because nitrogen is obtained from air, it contains almost 1 % rare (chiefly that do not react. The rare gases would continue to build up in
- 403. 372 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 the recycle until their on the react.1Cm equilibrium would become adverse. There-fore, a small purge stream is + N2 --flo- 2NH:s Nz NH:s(Uquid) Ar (Gas) SAT12.4Pl composed of 16% H2• 24.57% N2• and 0.27% AI is mixed and enters the reactor with a composition of 79.52% H2. The separator contains 80.01 % H2 and no ammonia. The product ammoruacolllwllS no Per 100 moles of fresh feed: a. are and purged? b. conversion of hydrogen per pass? 2. a simplified process to make ethylene dichloride have been placed on the figure. Ninety percent conversion of the through the reactor. The overhead stream from the separator the separator, 92% of the entering C2H4• and 0.1 % of enter-percent of the overhead from the separator is purged. (a) the flow rate and (b) the composition of the purge stream. p urge Recycle ".---...... S e p a r Feed I Reactor a C2H4 + CI2 .......... C2H",CI.2 t 00 mollhr i o moUhr 0 n .. Product Figure SA T12.4P2
- 404. Sec. 12.5 The Industria! Application of Material Balances 373 12 .. 5 The Industrial Application of Material Balances Process simulators, which initially were used for material and energy balances, are now used by engineers for a number of important activities. including process design, process analysis, and process optimization. Process design involves selecting suitable processing units (e.g., reactors, mixers, and distillation columns) sizing them so that feed to the process can be efficiently converted into desired products. Process analysis involves comparing predictions of process variables using models of the process units with the measurements made the operating process. By comparing corresponding values of variables, you can determine if a particular process unit is functioning properly. If discrepancies exist, the predictions from model can provide insight into the root causes of problems. In addition, process models can used to carry out studies that evaluate alternate processing approaches and studies of debottlenecking, that is, methods designed to increase the production rate of overall process. Process optimization at detennin-the most profitable to operate the process. process optimization, models of the major processing units are used to detennine the operating conditions, such as product compositions and reactor temperatures, that yield maximum profit for the process (subject to appropriate constraints). For of three process applications, models the processing units are based on material balances. For simple equipment, just a few material balances for each component in the system are sufficient to modeJ the equipment. For more complex equipment such as distillation columns, will find the models involve mater- ial balance equations for component on each in a column, and• some indus-trial columns have over 200 trays. For process design and most of process analysis, each processing unit can be analyzed and solved separately. Modern computer codes make it possible to solve extensive sets simultaneous equations. For example, optimization model for an ethylene plant usually has over 150,000 equations with material balances comprising over 90% of the equations. 12.5-1 Issues In the Solution of Equations in Models The simultaneous solution of the large number of equations in process models presents a major challenge for commercial software vendors who develop and maintain process models used process design, process analysis, and process optimization. Computational efficiency and solution reliability (including stability and convergence of algorithms) are two important factors affecting the use of commercial simulators. If an excessive amount of computer time is required to solve the model equations, the utility of the simulators can be undennined, ticularly for process optimization applications, because they involve a large number
- 405. 374 Recycle, Bypass, Purge. & the Industrial Application Material Balances Chap. 12 of equations and naturally require considerable computer time for their solution. Also, optimization applications are applied continuously to many processes so that a long time to achieve a solution, or failure of the algorithm used to solve the equations, seriously degrades the performance of the software, and can make it impossible to obtain any expected benefits. You should be aware that the computational efficiency and reliability of software are affected by the way in which you formulate the process model equations and the order in which you enter them into the computer. In general, the more linear is a set of model equations, the faster the set can be solved, and the more reliable the solution. You may recall that in Chapter 7 it was pointed out that writing material balances using the mole fractions as unknowns results in a nonlinear set of equations when compared to using the component flow rates. By writing the model equations in open equation form, you can improve the computational and reliability achieved by equation solvers for large-scale problems. Consider system of equations in which aI and a2 are known values: Given the values of Y, and Y2' the first equation can be used to calcu1ate XI' Then the value of x2 can be calculated using the second equation. This procedure is called the sequential modular approach, that is, the equations are solved individually or in relatively small groups, and the results of one set are used to solve for other unknowns in other sets. The open equation fonn for the example is written as I, (XI' x2) = YI - alx1 h (XI' x2) = Y2 - a/xI - ar2 By specifying the values of Y, and Y2 and settingil=i2=O, a set of two equations and two unknowns results. The solution to this set of equations is the solution of the original problem. You solve the equations simultaneously for both XI and x2. This procedure is also called the simultaneous moduJar approach. Both methods yield identical solutions if the computer code is robust to variations in the character of the equation. Why is the open equation fonn used for industrial-scale model applications? Because it is more computationally efficient anti reliable for large-scale problems. The open equation form affords easier development of a standardized model formation so that engineers can combine models and software produced by different programmers without difficulty. In addition, with the open equation form it is easier to use the same model to calculate various parameters in the equations. For example, for the previous two - --------------------------------------
- 406. Sec. 1 The Industrial Application of Material Balances 375 equations, the values of y,! Y21 X" and the following equations can solved determine the a I and It (a l • ~) ::::: - al Xl 12 (al' ::::: - alx1 - Note that the equations have not changed, only the known and unknown. 12.5 .. 2 Material Balance Closure for Industrial Processes One important way in which individual ally to check that "in = out," that is, to rlAlr"'M"n balances are applied industri-how well material balances bal-ance using process in equations. You look for what is caned clo· sure, namely that the error between "in" and "out" acceptable. The flow rates and measured compositions for all the streams entering and exiting a process unit are substituted the appropriate material balance equations. Ideally, amount (mass) of each component entering the system should equal the amount that component leav-the system, Unfortunately, the amount of a component entering a process rarely equals the amount leaving the process when you make such calculations. The lack of closure for material balances on industrial occurs for reasons: 1. process is operating in steady state. Industrial processes are al-most always in a state of flux, and rarely reach precise steady-state behavior. 2. The flow and composition measurements have a variety errors associated with them. First, sensor readings noise (variations the measurement to more or random variations the readings do not correspond to changes in the process). The sensor readings can also be inaccurate for a wide variety of other reasons. example, a sensor may require recalibration be· cause it degrades, or it may be used for a measurement for which it was not 3. component of interest may be generated or consumed inside process by reactions that the process engineer has not considered. As a result, material balance closure to within 5% for balances for most industrial processes considered reasonable. (Here closure is defined as the calculated difference between the amount of a particular material entering and exit-the process by the amount entering multiplied by 100.) special atten-tion is paid to calibrating material closure of 2 to 3% can be tained. If special high accuracy sensors are used, smaner closure of the material balances can be attained, but if faulty sensor readings are used, much greater errors material balances are observed. In material can to deter-mine when faulty sensor readings exist.
- 407. 376 Recycle, Bypass! Purge, & the Industrial Application Material Balances Chap. 12 Looking Back Do the words of the jingle at the start of this chapter still apply to you? From the explanation and examples presented in this section you should have concluded that problems involving recycle, purge, and bypass are no different from the viewpoint of how they are analyzed than any of the problems solved in earlier chapters of this book. The one new factor brought out in this chapte"r is that recycle for a reactor usually involves information about the fraction conversion of a reactant or extent of reaction. GLOSSARY OF NEW WORDS By-pass stream A stream that skips one or more units of the process, and goes di-rectly to a downstream unit. Fresh feed overall feed to a system. Gross product The product stream that leaves a reactor. Once-through fraction conversion The conversion of a reactant based on the amount of material that enters and leaves the reactor. Open equation form Model equations in which the sum of the tenns of each equation a deviation from zero. Overall fraction conversion The conversion of a reactant in a process with recycle based on the fresh feed of the reactant and the overall products. Overall products The streams that exit a process. Process feed The feed stream that enters the reactor usually used in a process with a and recycle. Purge A stream bled off from the process to remove the accumulation of inerts of unwanted material that might otherwise build up in the recycle streams. Recycle Material (or energy) that leaves a process unit that is downstream and is returned to the same unit or an upstream unit for processing again. Recycle stream The stream that recycles material. Recycle system A system that includes one or more recycle streams. Sequential modular The sequential solution of model equations. Simultaneous modular Simultaneous solution of model equations. Single-pass fraction conversion conversion based on what enters and leave a reactor. See once-through conversion.
- 408. Chap. 12 Problems an SUPPLEMENTARY REFERENCES In addition to the general references listed in the Frequently Asked Questions in the front material, the following are pertinent. Cheremisinoff. P.N., and P. Cheremisinoff. Encyclopedia of Environmental Control Technology:Wastll Minimization and Recycling, Gulf Publishing, Houston. TX (1992), Lund. H.F., Ed. McGraw-Hill Recycling Handbook, 2nd ed., McGraw-Hill, New York (2000). Luyben. W.L., and WenzeL Chemical Process Analysis:Mass and Energy Balances. IntI. Ser. In Phys. & Chern. Engin. Sci, PrenticedHaU, Englewood Cliffs, N.J. (1988). Myers. A.L., and w.n. Seider. Introduction to Chemical Engineering and Computer Calcu-lations. Prentice-Han, Englewood NJ. (1976). NoH, K.E., N, Haas, C. Schmidt, and P. Kodukula. Recovery, Recycle. and Rewe of Industrial Wastes (Industrial Waste Management Series), Franklin-Book Co. (1985). Veslind. Unit Operations in Resource Recovery Engineering. Prenticea Ha11. Upper Saddle River, N.J. (1981). Web Sites http://www.capec.kt.dtu.dklmainl36445/simulators.pdf http://www.dur.ac.uk1a.k.hugheslkptlmooule2.html http://www.it.che. wusU.edu/josephJ477 Ihomeworksl http://www.nap.edulbooksl0309063779Ihtm1l28.html PROBl MS $12.1 How many recycle streams exist in each of the following processes? (a) 1----tIO- Product Figure Pll.la
- 409. 3'J.8 Recycle I Bypass. Purge, & Industrial Application MateriatBaiances (b) , ,; , (c) ~ ____ ~ UNIT ~ ____________________ -, 8 ~ UNIT 1--_2_~""1 1 4 ~ , UNIT 2 13 , 10 UNIT S Figure PU.lb Fipre P12.1c UNIT ':f -, ; -t·, .. -"8 -.' 11- ,- ~ 5 UNIT '< 4 , ~', ,- s ": . . I , UNIT 5 l: ! .. .! ., • ~f I I UNIT, " 12 7 . , ,"" Chap. 12 , ' 9 , ..... "
- 410. Chap. 12 (d) Figure P12.1d *12.2 Find the of recycle/kg feed if the amount of waste (W) is 60 kg A. (F) A20 8BO 'Yowt % wt R (100% A) G 40% A FigureP12.2 w = 60 kg 1000/0 A P A5%wt B 95%wt 379
- 411. 380 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 ·12.3 Find the kg RJlOO kg fresh feed. 40%KCf t 60 kg H2O 6O"'A. H2O Evaporator CD Product Fresh feed Recycle A FigureP12.3 ·12.4 In the process shown in the Figure P12.4 Unit I is a liquid-liquid solvent extractor and Unit II is the solvent recovery system. For the purposes of designing the size of the pipes for stream C and 0, the designer obtained from the given data values of C = 9,630 Ib/hr and 0::: 1,510 Iblhr. Are these values correct? Be sure to show all details of your calculations or explain if you do not use calculations. E Known Data: A B C o E Flow rate (lblhr) 5,000 10,000 A 1-----8 Figure P12.4 Butene 0.75 0.05 -I--D Composition Butadiene 0.25 1.00 0.95 0.01 Solvent 0.99 ·12.5 The ability to produce proteins through genetic engineering of microbial and mammalian cells and the need for high purity therapeutic proteins bas established a need for efficient large scale protein purification schemes. ---- ~------------
- 412. Chap. 12 Problems 381 The system of Continuous Affinity-Recycle E~traction (CARE) combines the advantages of well accepted separation methods, e.g. affinity chromatography, liquid extraction and membrane filtration, while avoiding the drawbacks inherent in batch and column operations. The technical feasibility of the system was studied using /3-galactosidase affmity purification as a test system. Figure P12.S shows the process. What is the recyflow rate in mLIhr in each Sb'eam? Assume that the concentrations of U are equivalent to concentrations of the {3-galactosidase in solution. and that steady state exists. Feed 600mLlh 1.37 U/mt. ABSORBING STAGE Waste Desorblng 60 mL1h Buffer 0 Ulml Figure Pl1.5 DESORBING STAGE &omUh 12.9 UlmL "12.6 Cereal is being dried in a vertical drier by air flowing countercurrent to cereal. To prevent breakage of the cereal flakes, exit air from the drier is recycled. For each 1000 kg/hr wet cereal to the drier. calculate the input of moist fresh air in kglhr and the recycle rate in kglhr. Data on stream compositions (note some are mass and others mol fractions): Fresh Wet cereal Exlt air Dried cereal Alr antl'llrinl'l 0.200 0.283 0.050 0.066 Wet cereal Recycle Dry cereal Fresh air
- 413. 382 Recycle, Bypass, Purge, & the Industrial Applic'ation Material Balances Chap. 12 ··12.7 Examine What the quantity of the recycle stream 4% water and 96% KN03 kglhr? In stream C the composition W 100°/. H2O 30011 F I E vopcrotcr I Feed R 10,000 kt/hr 20'*0 KN~ Solution Recy cle 100" F C ryalalli18r Silty ( 0.6 kg ) rated Solution --0 kg Hz KNOll Crystals plus HtO c Figure FIl.7 / ··12.8 Sea water is to be desalinized by reverse osmosis using the scheme indicated in Fig-ure P12.8. the data given the figure to determine: (a) The of waste brine removal (B) (b) The rate of desalinized water (called potable water) production (D) (c) fraction of the brine leaving the reverse osmosis cell (which essence as a separator) that is recycled. Brine Recycle Sea 3.1 Weter 4.0% Reverse 1000 % Salt Salt Osmosis Cell D Desalinlsed Ytblef 500 ppm Salt Brine Weste t8} % Solt *12.9 A plating plant has a waste stream containing zinc nickel in quantities in excess of that allowed to be discharged into the sewer. The proposed process to be used as a fa.rst step reducing the concentration of Zn Ni is shown in Figure P Each stream contains water. The concentrations of several of the streams are listed in the table. What is the flow (in Uhr) of the recycle stream if the feed is 1 Uhf? J
- 414. Chap. 12 Problems Stream F CD P, Concentration (gIL) Zn 100 190.1 3.50 4.35 o 0.10 0 Figure P12.9 P2 Ni 10.0 17.02 2.19 2.36 o 1.00 W (H,P 100%) 0) 383 -- 0 *·12.10 UltrafHtration is a method for cleaning up input and output streams from a number of industrial processes. The lure of the technology is its simplicity. merely putting a membrane across a stream to sieve oul physically undesirable oil, dirt, metal particles, polymers, and the like. The trick, of course, is coming up with the right membrane. The screening material has to meet a fonnidable set of conditions. It has to very thin (less than I micron), highly porous, yet strong enough to hold up month after month under severe stresses of liquid flow, pH, particle abrasion, temperature, and other plant operating characteristics. A commercial system consists of standard modules made up of bundles of porous carbon tubes coated on the inside with a series of proprietary inorganic compositions. A standard module is 6 inches in diameter and contains I tubes each 4 feet long with a total working area of sq. ft and daily production of 2,000 to 5,000 gallons of filtrate. Optimum tube diameter is about 0.25 inch. A system probably will last at least two to three years before the tubes need replacing from too much residue buildup over the membrane. A periodic automatic chemical clean out of the tube bundles is part of the system's nonnal operation~ On passing through the filter. the exit stream concentration of oB plus dirt is increased by a factor of 20 over the entering stream.
- 415. 384 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 Calculate the recycle rate in gallons per day (g.p.d.) for the set up shown in Figure P12.l0, and calculate the concentration of oil plus dirt in the stream that enters the filtration module. The circled values in Figure P12.10 are the known concentra· tion of oil plus dirt. Ultraflltratlon cleans water for re-use To process From process -~.~===~ 2910 g.p.d. Makeup water 90 g.p.d. t 6-1nch-o.d. module Water 2910 g.p.d. 011, dirt, and water 3000 g.p.d. Figure Pll.10 Oil-dirt concentrate discharge 90 p.d.-~~~ 8 J R Rearculating concentrate 100 p.i.1. 160" F j --12.11 To save energy, stack gas from a furnace is used to dry rice. The flow sheet and known data are shown in Figure P12.11. What is the amount of recycle gas (in lb mol) per 100 Ib of P if the concentration of water in the gas stream entering the dryer is 5.20%1 Stack gas S (Ib mol) 4.73% H2O Rice teed F (Ib) 75% Rice 25% Water ....... RecycleR ~ Dryer Figure P12.1l W Wet gas (Ib mol) 9.31% H20 Rice feed P (Ib) 95% Rice 5% Water
- 416. Chap. 12 Problems 385.- ·"'12.12 This problem is based on the data of Payne, "Bioseparations of Traditional Fer-mentation Products" in Chemical Engineering Problems in Biotechnology. ed. M.L. Schuler. American Institute of Chemical Engineers. New York: 1989. Examine me PI2.12. Three separation schemes are proposed to separate the desired fermentation products from the rest of the solution. Ten liters/min of a broth containing 100gIL of undesirable product is to be separated so that the concentration in the exit wastestream reduced to (not more than) 0.1 gIL. Which of the three flow sheets requires the least fresh pure organic solvent? Ignore any possible density changes in the solutions. Use equal values of the organic solvent in (b), Le., Fi + + Fi = FO. The relation between the concentration of the undesirable material in the aqueous phase and that in the organic phase in 10 to 1 that is, cAlco = 10 the outlet streams of each unit. FO Organic solvent 100% Aqueous phase FA = 10Umln c~ "" 100g1L FA == 10Umin ...-...... c~.= 0.1 gil F?Organlo 100% c? (a) F~Organic 100% II rJl (b) II (c) Figure Pll.ll FrOrganic 1000/13 III ~Organrc 100% III FA == 10L .............. C~=O.1
- 417. '-. f " , , " . " ~ 386 Recycle, BYRassJ P.tlrg~,··& the· lndustria~ Application Material Balances Chap. 12 .. , I r* j.~' . , '~." ,'1' " ···12.13 B~nzene. !o]uene -and~.~~r:aroma~c compounds can be recovered by solvent extraction witli~lfui:!dioxide~;As,an example. a catalytic refonnate stream containing 70% by ~ejg~t ~nzene and 30% non-benzene material is passed through the countercurrent ~~~tivS' recovery scheme shown in the diagram in Figure P12.13. One thot,sana kg-of. the reform ate stream and 3000 kg of sulfur dioxide are fed to the system' 'hour. The benzene product steam contains O.IS.kg of sulfur dioxide per of benzene. the waste steam contains all the initially charged non-benzene material as weB ~ O~5 kg of benzene per kg of the non-benzene material. The remaining component in'the waste steam is the sulfur dioxide. (a) How many kg of benzene are extracted per hour (are in the product stream)? (b) If 800 kg of benzene containing 0.25 kg of the non-benzene material kg of benzene are flowing per hour at paint A and 700 kg of benzene containing 0.07 of the non-benzene material per kg of benzene are flowing at point B. how many kg (exclusive the sulfur dioxide) are flowing at points C and D? O. Product benzene includes --.;;;......--.;;;;. kgBz B - Sulfur dioxide 3000 kglhr 3 Product stream - en Figure P12.13 *12.14 Examine Figure PI 14 (data for I hour), f 2 A "-"" ~ 00 1000 kglhr catalytic reformate 70% a. What the single pass conversion H2 in the reactor? b. What is the single pass conversion of CO? c. What the overall conversion of H2? d. What is the overall conversion of CO? 1 Waste 0.25 kg Bz kg the non-Bz
- 418. r Chap. 12 Problems 387 CO Recycle H2 1.979 mol co + 2H2 - CH30H CO 1 mol .....-_----, H2 1 mol H2 1.98 mol .----'-_-. CO 0.01 mol CH30H 1.146 mol CO Feed -----'------1.. ... , Reactor CH:,0H 0.99 mol Separator Prod mor Figure P12.14 11<12.15 Hydrogen. important for numerous processes, can produced the shift reaction: CO + -) CO2 + H2 the reactor system shown in conditions of conversion adjusted so that the content of the effluent from the reactor is 3 mole %. on the data in Figure 2.5: (a) Calculate composition of the feed. (b) Calculate the moles recycle mole of hydrogen produced. Recycle Reactor Figure % C02 48 1---""" H2 48 CO 4 ·12.16 Acetic acid (HAc) is to be generated by the addition of 10 percent excess sulfuric acid to calcium acetate (Ca(Ach). The reaction Ca(Ac)2 + H2S04 ~ CaS04 + 2HAc to 90 percent completion based on a single pass through the reactor. The unused Ca(Ach separated from the products of the reaction and recycled. The HAc HAc Ca(Ac~ Reactor Figure
- 419. 388 ... ,-""-..... Bypass, Purge. & the Industrial Application Material Balances Chap. 12 arated from the remaining products. Find the amount of recycle per hour based on 1000 kg of Ca(Ach feed per hOUT, also calculate the kg of HAc per hour. See Figure P12.16 that the process. (Ac = '12.17 The reaction of ethyl-tetrabromirle with dust proceeds as shown Figure P12.17. C:,!H:zBr" Gross ZnBr2 Reactor Product r a t R P12.17 The C2H2Br4 + 2Zn -:) on one pass through the re-the C2H2Br4 remainder recycled. On the basis of CZH2Br4 fed to the reactor per hour. calculate: much C2H2 is produced per hour (in kg) of recycle in kglhr (c) The rate necessary for Zn to be 20% in excess (d) The mole ratio of ZnBr2 to C2H2 in the final products '12.18 Examine the accompanying figure. NaCl solution react to fonn In the reactor the conversion of CaC03 is 76% complete. Unreacted CaC03 is ded. (a) the kg of N~C03 exiting per 1000 kg of feed, and (b) the kg recycled per 1000 of Nael 90°/., Feed 1000 kg hr Ns2C03 10°/0 + Reactor Figure P12.18 Separator '12.19 In the process ..:I ..... w_ ... ~ ..... below, Na2C03 is produced by + CaC03 ~ Na2C03 +
- 420. Chap. 12 Problems 389 The reaction is 90% complete on one pass through the reactor and the amount of CaC03 entering the reactor is 50% in excess of that needed. Calculate on the basis 1000 Iblhr of fresh feed: (a) the Ib of N~S recycled. and (b) the Ib of N~C03 solution formed hour. Fresh 1000 Ibmr Feed 40% Na2S H20 30%CaCOs Reactor 30% CaC03 soh .......... ----/ CaCOasoln. 60 Iblhr caCO:) Figure PIl.19 1......---1 80% Ns2CO 20% H~p soln. Toluene reacts with H2 to fonn benzene (B), but a side reaction occurs in which a byproduct diphenyl (D) is formed: --+ benzene + methane (a) C7HS + H2 Toluene hydrogen -+ C12HlO diphenyl 2C~ (b) The process shown Figure P12.20. Hydrogen is added to the gas recycle stream to make the ratio ofH2 to CH4 1 to 1 before the enters the mixer. The ratio of to toluene entering reactor at is 4H2 to 1 toluene. The conversion of toluene to benzene on one pass through the is 80%. and the conversion of toluene to the by-product diphenyl is 8% OD same pass. Calculate the moles of Ro and moles RL hour. Data: Compound: MW: 2 78 92
- 421. 390 Recycle, Bypass, Purge, & the Industrial Application Material Balances Makeup M 100% H2 Gas recycle P Purge Chap. 12 F Feed 1 OOt%~ toluene '---,.----' 3450 tblhr B Separator i-=-Se-n-z-en-e.. ....... Uquid recycle Rl 100% toluene Figure P12.20 100% D Diphenyl 100% /· ... 12.21 The process shown in Figure P12.21 is the dehydrogenation of propane (C3HS) to propylene (C3H6) according to the reaction C3Hg -7 C,3H6 + H2 The conversion of propane to propylene based on the total propane feed into the reactor at F2 is 40%. The product flow rate Fs is 50 kg rnollhr. (a) Calculate all the six flow rates FI to F6 in kg mollhr. (b) What is the percent conversion of propane in the reactor based on the fresh propane fed to the process (F!). Fl F2 Fresh C3Ha feed Catalytic Reactor 80% CaHs 20% C3H6 F3 Fa Recycle Figure P12.2I H2 Absorber & Distillation Tower CsHe
- 422. Chap. 12 ·"12.22 /"12.23 /' Problems 391 Natural gas (CHJ is burned a furnace using 15% excess air based on the complete combustion of CH4. One the concerns is that exit concentration of NO (from the combusrion of N2) is about 415 ppm. To lower the NO concentration in the stack gas to 50 ppm it suggested that system redesigned recycle a portion of the gas back through the furnace. You are asked to calculate the amount of recycle required. Will the scheme work? Ignore the effect of temperature on the conversion of N2 to that is, assume the conversion factor constant. Sulfur dioxide may converted to S03' which has many uses induding the pro-duction of H2S04 sulphonation of detergent. A gas stream having the composi· tion shown in P12.23 to be passed through a two-stage converter. The tion conversion of the to S03 (on one pass though) in the stage is and in the second stage 0.65. To boost overall conversion to 0.95, some of the exit gas from stage 2 is recycled back to the inlet of stage 2. How much must be recycled 100 moles of inlet (stream F)? Ignore the of temperature on the conversion. 802 10.0% F O2 9.0%--....... po..j 2 N2 a1.oof" R (kg mol) Figure P12.23 ***12.24 Nitroglycerine> a widely used high explosive, when mixed with wood flour is called "dynamite." It is by mixing high-purity glycerine (99.9 + percent pure) with nitration acid. which contains 50.00 percent 43.00 percent HN03• and 7.00 percent water by weight. The reaction is: C3HsO" + 3 HNO) + (H2S04) -? C)HS0 3(N02h + 3 H20 + (H2S04) The acid not take in the reaction, but is present to "catch" the water formed. Conversion of the glycerine in the nitrator is complete. and there are no reactions, so of the glycerine fed to nitrator nitroglycerine. The mixed entering the nitrator (Stream G) contains 20.00 percent excess HN03 to assure that all glycerine Figure P 12.24 is a flow diagram. nitration, the mixture nitroglycerine and acid (HN03• H2S04, water) goes to a separator settling tank). The nitroglycerine is insoluble in the acid, and its density is less, so it to the top. It is carefully drawn off as product stream P and sent to wash tanks for purification. The spent acid is withdrawn from the bottom of the separator and sent to an recovery tank. where the HNO) and H2S04 are The H2S04-H20 mixture is Stream tv, and is concentrated and sold for indusnial purposes. The recycle stream to nitrator is a 70.00% by
- 423. 392 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 G F-...... --....... ....------ Glycerine Nitrator Recycle, R Nitric acid, water Figure P12.24 weight solution of RNO) in water. In above diagram, product stream P is 96.50% nitroglycerine and 3.50% water by weight. To summarize: Stream ;;: 50.00 wt% H2S04, 43.00% HN03• 7.00% H20 Stream G contains 20.00 percent excess nitric acid Stream p:: 96.50 wt% nitroglycerine, 3.50 wt% water R = 70.00 wt% nitric acid, 30.00% water. (a) 1.000 X 103 kg of glycerine per hour are fed to the nitrator, how many kg hour of stream P result? (b) How many kg hour are in the recycle u~_,_u (c) How many kg fresh stream are fed per hour? (d) Stream W how many kg per hour? What is its analysis in weight percent? Molecular weights: glycerine::: 92.11, nitroglycerine::: 227.09, nitric acid ;;: 63.01, sulfuric acid = 98.08, and water::: 18.02. Caution: Do not try this process at home. $111"'12.25 A for methanol synthesis shown in Figure PI2.25. The pertinent chemical reactions involved are CH4 + 2H20 .....;. CO2 + 4H2 + H20 ~ CO + 3Hz 2CO+ ~2C02 CO2 + 3H2 .....;. CH30H + reformer reaction) (reformer side reaction) (CO converter reaction) (methanol synthesis reaction) (a) (b) (c) (d) Ten percent excess steam, based on reaction (a), is fed to the reformer, and con-version of methane is 100%, with a 90% yield of CO2, Conversion the methanol reactor is on one pass through the reactor. stoichiometric quantity of ox.ygen is to the CO converter, and the CO is completely converted to Additional makeup CO2 is then introduced to establish a 1 ratio of H2 to CO2 in the feed stream to the methanol reactor.
- 424. I Chap. 12 Problems 393 " 12 10 e CIlllOeftllf COl Moktl, 6 7 9 ''.,1I0II0I 11 RfI<)clOf Meitl<lllo! 5 StlllliOCI OIfttll 4 CO Coo¥tfl8f P MtllwIItt FmI 2 SUIIIII 1 RtllblMf Figure Pll.lS The methanol reactor effluent is cooled to condense all the methanol and water, with the nODcondensible gases recycled to the methanol reactor feed. The HiC02 ratio in the recycle stream is also i, Because the methane feed contains 1 % nitrogen as an impurity. a portion of the recycle stream must be purged as shown in Figure PI to prevent the accumula-tion of nitrogen in the system. The purge stream analyzes 5% nitrogen. On the basis of 100 mol of methane feed (including the N2). calculate: (a) How many moles of H2 are lost in the purge (b) How many moles of makeup CO2 are required (c) The recycle to purge ratio in moVmol (d) How much methanol solution (in kg) of what strength (weight percent) is produced. '12.26 Alkyl halides are used as an alkylating agent in various chemical transformations. The alkyl halide ethyl chloride can be prepared by the fonowing chemical reaction: 2C2H6 + Cl2 -; 2C2HsCl + H2 C2H. W CI2 elz H2 f2 Reoctor S p 60'. Separator ~H6 Conversion C2HC 100% C2H,CI ell! CIt (100,0 C2HIl CI excess C12 ) Ha R Figure Pll.l6
- 425. 394 Recycle, Bypass, Purge, & the Industrial Application Material Balances Chap. 12 j ''12.27 In the reaction process shown in Figure P 12.26, fresh ethane and ehlorine gas recycled ethane are combined and fed into the reactor. test shows that if 100% excess chlorine is mixed with ethane, a single-pass optimal conversion of 60% results, and of the ethane that reacts, all is converted to products and none into undesired products, You are asked to calculate: (a) The fresh feed concentrations required for operation (b) The moles of c;HsCl produced P per mole of C2H6 in the fresh feed Fl' What difficulties will you discover in the calculations? Many chemical processes generate emissions of volatile compounds that need to controlled. the process shown in Figure P 12.27 I the exhaust of CO is eli minated by its separation from the reactor effluent and recycling of 100% the CO generated in reactor together with some reactant back to the feed. Although the product is proprietary, information provided that feed SlTeam contains 40% reactant, 50% inert, and 10% CO, and that on reaction 2 moles of reactant yield 2.5 moles product. Conversion of reactant to product is only on one pass through the reactor, and 90% overall. You are to calculate the ratio moles recycle to moles of product. What do you discover is wrong with this problem? Feed stream (gas) 4100 me Vhr Recycle {CO 80% Reactant 20% CO Product Reactor Separator Inert Reactant Figure P12.27 Product Inert Reactant ·"PI2.28 The foHowing problem is condensed from Example 10.3-1 in the book by D. T. Allen and D. R. Shonnard. Green Engineering published by Prentice Hall, Englewood CUffs, NJ, 2002. Acrylonitrile (AN) can produced by the of propylene with ammonia in the phase C)H6 + NH3 + 1 02 --t C3H)N + 3 H20 Figure P12.28 is the flowsheet for the process with the data superimposed. only contaminate concern is the ammonia. Answer the foHowing questions. 2. Can of the waste streams that are collected and sent to treatment be used to replace some the boiler water feed? b. What streams might considered as candidates to replace some of the to scrubber?
- 426. Chap. 12 Problems 450"C, 2atm 2-pha5e stream always with 1 kg/s H20 but no H20 In the AN layer mass fraction of AN AI'IIIIfIW'A c. umn :s;10ppmNHa }-_......., II:Nl:lbl!lr .dllIU"" o ~IJ" H:zO 0.5 kg/$ H.tO () kOla AN U kg/$ AN 1......-,....--' 0 ppm ~ I.....-_....i 39 ppm NHe e.5klJ'l~ 4.5~1P't AN 14 ppm NH3 (YIQIUI't1) ~ I----Joo-I listllatJon 1.0 legis !-ItO ooUnn 1-+------. 4.2 kWft AN '--,....--' 10 ppm NH3 L.....,--' Figure Pll.28 ,,' 395 stream from the condenser distillation col-back to the scrubber to replace 0.7 used in the what changes in the flows and concentrations will occur in the process?
- 427. PART 3 GASES, VAPORS, LIQUIDS, AND SOLIDS CHAPTER PAGE 13 Ideal Gases 401 14 Real Gases (Compressibility) 435 15 Real Gases (Equations of State) 459 16 Single Component, Two Phase Systems (Vapor Pressure) 475 17 Two-Phase Gas-Liquid Systems (Saturation Condensation, and Vaporization) 509 18 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) 537 19 The Phase Rule and Vapor-Liquid Equilibria 560 20 Liquids and Gases In Equilibrium with Solids 590 We now take up the third prominent topic in this book, how to predict physical properties of pure components and mixtures. By property we mean any measurable characteristic of a substance, such as pressure, volume, or temperature, or a characteristic that can be calculated or deduced, such as internal energy. which discussed in Chapter 21. The state of a system is the condition of a system as specified by its properties. You can find values for properties of compounds and mixtures in many formats, including: 1. Experimental data 2. Tables 3. Graphs 4. Equations as indicated in Figure Part 3A. You have to become familiar with techniques of correlating and predicting physical properties because values of properties underlie the design, operation, and 396
- 428. Part 3 Gases, Vapors. Liquids, and Solid~ J ~9 i~ ! !l.6 -!. 0.5 1 0,4 J ~~ 0.1 • 0.0 397 ;;:: 0.0 0.1 Q.2 0.3 004 o..s 0.6 0.1 0.8 0.9 1.0 Xl. mole tw:d0ll melb~ chloride iii 6qWd 32 0.0886 0.1806 O.OHiOZ 34 0.0961 0.19S1 0.01602 36 0.1041 0.21:20 0.01602 38 0.1126 0.2292 0.01602 40 0.1211 0.2418 0.01602 42 0.1315 0..2.67?' 0.01fi02 44 0.1420 0.2891 0.01fi02 Figure Part 3A Sources of information used to retrieve physical properties. troubleshooting in aU processes. Companies and researchers make a significant investment in collecting data and assessing their validity before storing the data and equations that evolve from data in a data bank-a capital asset. Not everything lhat can be counted counts, and not everything that counts can be counted. Albert Einstein Clearly you cannot realistically expect to have reliable, detailed experimental data at hand for the properties of aU of the useful pure compounds and mixtures with which you win be involved. Consequently t in the absence of experimental infonna* tion, you have to estimate (predict) properties based on empirical correlations or graphs so that you can introduce appropriate parameters in material and energy
- 429. 398 o 0 0 o 0 ,00 o 0 0 o 0 gas unorganized 0 0 Gases, Vapors, Liquids, and Solids liquid some orgar:lizatlon Solid highly organized Part 3 Figure Part 3D Three phases of a compound showing the classification by deof organization. balances. The foundation the estimation methods ranges from quite theoretical to completely empirical. and the reliability of the methods ranges from excellent terrible. At any temperature and pressure, a pure compound can exist as a gas, liquid, or solid~ and at certain specific values of T and mixtures of phases exist, such as when boils or freezes, as indicated in Figure Part Thus, a compound (or a mixture of compounds) may consist one or more phases. A phase is defined as a completely homogeneous uniform state of mat- Liquid water would be a and would another phase. Two immiscible liquids in the same container, such as mercury and water, would represent two different phases because the liquids, although are homogeneous~ have different properties. In the chapters we first discuss ideal and real gas relationships. You win about methods of expressing the p- V-T properties of real by means of equations of state, and. alternatively, by compressibility factors, Next, we introduce the concepts of vaporization, condensation. and vapor pressure, and illustrate how material balances are made for saturated and partially saturated gases. Subsequently, we discuss vapor-liquid relations. and finally examine relations for gases liq-absorbed on solids. GLOSSARY OF NEW WORDS Phase homogeneous and uniform state of matter, not necessarily continuous. Property A measurable or calculable characteristic of material.
- 430. I j Gases, Vapors Liquids, and Solids 399 SUPPlEM NARY R FERENCES and F. Olti. Computer Aided Chemical Thermodynamics of Gases and Liquids: Models and Programs. Wiley, New York (1985), and P. M. Thermodynamics for Process " AIChE J, 48, 194-200 (2002). Daubert. T. E., and R. P. Danners, pounds, AIChE, New York, of Pure Com- Mathias, M., and H. Klotz. a Look at Models," Chemical Engineering Process, (June 1994). Poling, B. E., J. M. Prausnitz, and J. P. O'Connell. Properties of Gases and Liquids, McGraw- Hill, New York (2000). Van der Waals. J. R. On the Continuity of Gases and Liquid States li:llfl_t,.;;U. by 1. J. Robin-son). North Holland, (1988).
- 432. CHAPT R13 IDEAL GAS S 13.1 The Ideal Gas Law 13.2 Ideal Gas Mixtures and Partial Pressure 13.3 Material Balances Involving Ideat Gases Your objectives in studying this chapter are to be able to: 1. Write down the ideal gas law and define all its variables and parameters. 2. Calculate the values and units of the ideal law constant R in any of untts from the standard conditions. 3. Convert gas volumes to moles (and mass). and vice versa. 4. Calcu the value of one variable P, V. or n, from a given set values 01 the other variables. 5. Calculate the specific gravity of a gas aven if the reference condition is not clearly specified. 6. Calculate the density of a g.as given its specific gravity. Define and use partial pressure in gas calculations. 8. Show that under certain assumptions the volume fraction equals the mole fraction in a gas. 9. Solve material balances involving gases. 402 412 416 You have been exposed to the concept of the ideal gas in chemistry and physics. Why study ideal gases again? At least two reasons exist. First, the experimental and theoretical properties of ideal gases are far simpler than the co.rrespond~ ing properties of liquids and solids. Second, use of the ideal con~ept of considerable industrial importance. A review prior to solving material balances involving will refresh your memory. 401
- 433. 402 Ideal Gases Chap. 13 Looking Ahead In this chapter we explain how the ideal gas law can be used to calculate the pressure. temperature, volume, or number of moles in a quantity of gas, and define the partial pressure of a gas in a mixture of gases. We also discuss how to calculate the specific gravity and density of a gas. Then we apply the concepts to solving material balances. 13.1 The Ideal Gas Law In July 1984 an explosion killed a forklift operator in the Port of Houston while he was unloading canisters containing aluminum phosphide (a pesticide) from a Brazilian ship. The Coast Guard later discovered that the batch of pesticide, which was stored in a nitrogen.cooled van, had begun heating up and destabilizing, causing a risk of explosion. They decided to dispose of all of the cylinders by dumping them in the ocean. The chemical becomes poisonous phosphine gas when it contacts air or water. Initially the canisters containing the pellets simply were unplugged and tossed into the Gulf. After sinking a hundred feet or so, the chemical began reacting on contact with the sea water. and the gas trapped in each bottle forced them back to the surface, where the canister bobbed up and down on the waves. The Coast Guard issued fles to sharpshooters to blast each canister before it drifted dangerously away. nIt was a carnivaln said a local federal legislator in disgust. If you had been consulted about the disposal of the cylinders, could you have predicted the outcome? What might you have done? What about applying the ideal gas law to determine how much gas might be generated and act as a propellant to drive a cylinder through the water? Certainly the most famous and widely used equation that relates p, V, and T for a gas is th.e ideal gas law pV=nRT where p = absolute pressure of the gas V = total volume occupied by the gas n = number of moles of the gas R = ideal (universal) gas constant in appropriate units T = absolute temperature of the (13.1 ) You can find values of R in various units inside the front cover of this book. Sometimes the ideal gas law is written as A pV = RT (l3.1a) ,.. where in the equation V is the specific molar yolume (volume per mole) of the gas. When gas volumes are involved in a problem, V will be the volume per mole and not the volume per mass. Figure 13.1 illustrates the surface generated by Equation (13.1 a) in terms of the three properties p. V, and T. Look at the projections of the surface in Figure 13.1 onto the three two-parameter planes. The interpretation as follows:
- 434. Sec. 13.1 The Ideal Gas Law T i ~P3 I ~: p , 2 A V I P, I 403 Figure 13.1 Representation of the ideal gas law in three dimensions as a surface. 1. The projec}ion to the left onto the p - T plane shows straight lines for constant values of V. Why? Equation (13.1a) for constant specific volume is p == (constant) T, the equation of a straight line. A 2. The projection to the right onto the p - V plane shows curves for values of constant T.,., What kinds of curves are they? For constant T, Equation (l3.1a) becomes pV = constant, namely a hyperbola. A 3. The projection downward onto the T - ... V plane again shows straight lines . Why? Equation (13.1 a) for constant p is V = (constant) T. Equation (13.1) can be applied to a pure component or to a mixture. What are the conditions for a gas to behave as predicted by the ideal gas law? The major ones are 1. The molecules of an ideal gas do not occupy any space; they are infinitesimally small. 2. No attracti ve forces exist between the molecules so that the molecules move completely independently of each other. 3. The gas molecules move in random, straight-line motion" and the collisions between the molecules. and between the molecules and the walls of the container, are perfectly elastic.
- 435. Ideal Gases Chap. 13 Gases at low pressure and/or high temperature meet these conditions. Solids~ liquids. and gases at high density, that is, high pressure andlor low temperature. do not. From a practical viewpoint within reasonable erroflYou can treat air, oxygen, nitro-hydrogen, carbon dioxide, methane. and even water vapor under most of the ordinary conditions you encounter as ideal gases. Several arbitrarily specified standard states (usually known as standard condi-tions, or or S for standard temperature and pressure) of temperature and pressure have specified for gases by custom. Refer to Table 13.1 for the most TABLE 13.1 Common Standard Conditions for the Ideal Gas System T P V" SI 273.15K 101.325 kPa 22.415m3Jkg rno) Universal scientific D.O°C 760mmHg 22.415 literslg mol Natural industry 59.0oP 14.696 psia 379.4 ft311b mol (IS.O°C) (101.325 kPa) American engineering 491.67°R (32°F) 1 atm 359.05 ft31lb mol common ones. The fact that a cannot exist as a at ooe and 1 atm is im-material. Thus, as we see later, water vapor at ooe cannot exist at a pressure than its vapor of 0.61 kPa (0.i8 Hg) without condensation occurring. However, you can calculate the imaginary volume at standard conditions, and it is just as useful a quantity in the calculation of volume-mole relationships as though it could exist. In what follows, symbol V will stand for total volume and the sym- p. bol V for volume per mole. Because the SI, Universal Scientific, and standard conditions are identical, you can use the values in Table 1 1 with their units to change from one system of units to another. If you learn the standard conditions, you win find it to work with mixtures of units different systems. following example illustrates how you can use standard conditions to convert mass or moles to volume. After reading it. see if you can explain how to convert volume to moles or mass. EXAMPLE 13.1 Use of Standard Conditions to Calculate Volume from Mass Calculate volume, in cubic meters, occupied by 40 kg of at standard conditions assuming acts as an ideal gas.
- 436. Sec. 13.1 Ideal Law Solution 4OkgofCe2 40 kg CO2 1 kg mol CO2 m3 ----C-o..:::. = 20.4 m3 CO2 at S.C. kg CO2 1 kg mol 2 Notice in this problem how the information that .42, at S :;:: 1 kg mot is applied to transform a known number otmoJes. an equivalent number of cubic meters. An alternate way to, calcula'te the volume at standard conditions is to use Equation (13.1). 405 ? Incidentally. whenever you use volumetric measure, you must ~tabtish the conditions of temperature and pressure at which the volumetric measure exists, since the term um3" or "ft3 ttl standing alone. is really not any particular quantity of material. You can apply the ideal gas law, Equation (13.1), directly introducing val· ues three of four quantitiest n, P. T. and V. and solve for the fourth. To do so, you need to look up or calculate R in the proper units. Inside the front cover of this book you will find selected values of R for different combinations of units. Example illustrates how" to calculate the value of R in any set of units you want from the values of p, T. and V at standard conditions. EXAMPLE 13.2 Calculation of R Using the Standard Conditions , ' Find the value for universal constant R to match the following combi-nation of units: For 1 g mol of ideal gas when the pressure in atm. the volume is in em3 I and temperature is in K. Solution fonowing values are ones to use (along with their units). At standard conditions: p = 1 attn V = 15 cm3/g mol T = 273.15 K = 1 atm 22,415 crn3 (cm3)(atm) - -------= 82.06------ T 15 K 1 g mol (K)(g mol)
- 437. 406 Ideal Gases 13 In many processes going from an initial state to a fmal state, you will find it convenient to use the ratio of the ideal laws in the respective and eliminate R as fonows (the subscript 1 designates initial state, and the subscript 2 designates the final state) P VI - nlRTI P2V2 n2RT2 or (13 Note that Equation (13.2) involves ratios of the same variable. This result has the convenient feature that pressures may be expressed in any system of units you choose, such as kPa. in. Hgl mm Hg. atmt and so on, as long as the same units are used for both conditions of pressure (do not forget that the pressure must be absolute pressure in both cases), Similarly, the ratio of the absolute temperatures and of the volumes results in ratios that are dimensionless. Note how the ideal gas constant R is eliminated in taking the ratios. Let's see how you can apply ideal gas law both in the form of Equation (13.2) and Equation (13.1) to problems. EXAMPLE 13.3 Application of the Ideal Gas Law to Calculate a Volume Calculate the volume occupied by 88 lb of CO2 at 15°C and a pressure of ft water. Solution Examine Figure calculated as Ib To use Equation (I Stole I ttl ot Figure En.3 the initial volume has to shown in Example 13.1. Then the final volume can be calculated via Equation (13.2) in which Rand (n/n2) cancel out:
- 438. I Sec. 13.1 The Ideal Gas Law Assume that the given pressure is absolute pressure. At. S.c. (state I) At (state 2) p;;;;: 33.91 ft H20 p = 32.2 ft H20 T=273K T;;;;:273+15=288K Basis: 88 lb of CO2 88lb CO2 359 ft3 288 33.91 ----- --- - -- = 441b CO2 1 Ib mol 273 32.2 lIb mol CO2 Calculation of VI 798 ft3 CO2 at 32.2 ft H20 and 15°C You can mentally check your calculations by saying to yourself: The temperature goes up from O°C at S.C. to 15°C at the final state, hence the volume must increase from S.c., hence the temperature ratio must be greater than unity. Similarly. you can say: The pressure goes down from S.c. to the final state, so that the volume must increase from S.C., hence the pressure ratio must be greater than unity. The same result can be obtained by usin$ Equation (13.1). First obtain the value of R in t!le same units as the variables p, V, and T. Look it up or calculate the value from p, V, and T at S.C. At S.C., ... pV R=T P = 33.91 ft H20 V = 359 ~Ilb mol T = 273 K 33.91 359 (ft H20)(tt3) R = -- 273 = 44.59 (lb mol)(K) Now, using Equation (13.1), insert the given values, and perform the necessary calculations Basis: 88 lb of CO2 nRT 88 lb CO2 44.59 (ft H20){ft3) 288K V = -- = ---=- ------.,;...--..;.....:",.---.:.... ---- p 44lb CO2 (lb mol)(K) 32.2 ft H20 lb mol CO2 = 798 ft3 CO2 at 32.2 ft H20 and 15°C If you will inspect both solutions closely, you will observe that in both cases the same numbers appear, and that the results are identical. 401
- 439. 408 Ideal Gases Chap. 13 To calculate the volumetric flow rate of a gas through a pipe, you divide the volume of the gas passing through the in a time interval such as 1 second by the value of the time interval to get m3/s or ft3/s. To get the velocity, v, of the flow, you divide the volumetric flow rate by the area, A. of the pipe v = Av hence v == VIA (13.3) The density of a gas is defined as the mass per unit volume and can be expressed in kilograms per cubic meter, pounds per cubic foot, grams per liter, or other units. Inasmuch at;; the mass contained in a unit volume varies with the temperature and pressure, as we have previously mentioned, yop should always be careful to specify these two conditions in calculating density. If not otherwise specified, the densities are presumed to be at S.C. EXAMPLE 13.4 Calculation of Gas Density What is the density of N2 at 27°C and 100 kPa in 81 units? Solution Basis: 1 m3 of N2 at 27°C and 100 kPa 1 m3 273 K 100 kPa 1 kg mol 28 kg ~- 300K 101.3 kPa 22Am3 1 kg mol = 1.123 kg density = 1. kg/m3 of N2 at 27°C (300K) and 100 kPa The specific gravity of a gas is usually defined as the ratio of the density of the gas at a desired temperature and pressure to that of air (or any specified reference gas) at a certain temperature and pressure. The use of specific gravity occasionally may be confusing because of the sloppy manner in which the values of specific gravity are reported in the literature. You must be very careful in using literature values of the specific gravity to ascertain that the conditions of temperature and pressure are known both for the gas in question and for the reference gas. Thus, this question is not well posed: What is the specific gravity of methane? This question may have the same answer as the question: How many grapes are in a bunch? Unfortunately, occasionally one may see this question and the best possible answer is of methane SC. sp gr = ---'--------density of air at S.C. in which the temperature and pressure of the methane and reference air are clearly specified.
- 440. Sec. 13.1 The Ideal Gas Law EXAMPLE 13.5 Calculation of the Specific Gravity of a Gas What is the specific gravity of N2 at 800 P and 745 mm Hg compared to air at 80°F and 745 mm Hg? Solution One way to solve the problem is to calculate the densities of N2 and air at their respective conditions of temperature and pressure, and then calculate the specific gravity by taking a ratio of their densities. Example 13.4 covers the calculation of the density of a gas, and therefore, to save space, no units will appear in the intermediate calculation here: Basis: 1 ft3 of air at Soop and 745 mm Hg 1 492 745 29 3 - 540 760 359 - = 0.0721 Ih/ft at 800P and 745 mm Hg Basis: I ft3 of N2 at Soop and 745 mm Hg 1 492 745 28 - - - - - = O.06971b/ft3 at 80°F and 745 mm Hg 540 760 359 ) 0.0697 Ib N2/ft3 at BO°F, 745 mm Hg (sp gr N2 = = 0.967-----------.;;;- 0.0721 Ib air/ft3 air at 80°F. 745 mm Hg ,- 409 Did you note from Example 13.5 that for a gas and reference at the same temperature and pressure, the specific gravity is just the ratio of the respective molecular weights? You can gerive this result as foHows. Let A be one gas and B be another. Keep in mind that V for a gas is the volume per mole and not the volume per mass. Thus ... pV = RT or p= " 1 V = = [RTlp] p/MW (p)(MW) RT sp gr - :; - (:::.' ::.:)(~:)(;;) (13.3) and at the same temperature and pressure for A and B the specific gravity is just the ratio of the respective molecular weights.
- 441. 410 Chap. 13 SELF .. ASSESSMENT TEST Questions 1. What is the idea] gas law? Write it down, 2. What are the dimensions of T. p, V, n, and R? 3. List the standard conditions for a gas in the SI and AE systems of units. 4. How do you calculate the density of an ideal gas at S.c.? 5. Can you use the respective specific molar densities (mole/volume) of the gas and the reference gas to calculate the specific gravity of a gas? 6. Does the value of R depend on pressure, temperature, density, or number of moles of a gas? 7. If a gas meets these three conditions, will it behave like an ideal a. The volume occupied by molecules is so sman compared to total space that essentially the molecules occupy no b. The collisions are so few that it can be said that no energy is lost due to collisions. c. The average distance between the molecules is great enough to neglect the effect of the intermolecular forces. Problems 1. Calculate the volume in ft3 of 10 lb mol of an ideal gas at 68°F and 30 psia. 2. A steel cylinder of volume 2 m3 contains methane (CHJ at 50°C and 250 kPa solute. How many kilograms of methane are in the cylinder? 3. What is the value of the ideal gas constant R to use if the pressure is, to expressed in atm, the temperature in kelvin, the volume in cubic feet, and the quantity of material in pound mo]es? 4. Twenty-two kilograms per hour of CH4 are flowing in a gas pipeline at 30°C and 920 mm What is the volumetric flow rate of the CH4 in m3 per hour? 5. What is the density of a gas that has a molecular weight of 0.123 kglkg mol at 300 K and 1000 kPa? 6. What is the specific gravity of CH4 at 70°F and 2 atm compared to air at S.c. '] 7. An automobile tire is j nflated to a pressure of 35 psig at a temperature of OOE Calculate the maximum temperature to which the may be heated without the gauge pressure ex-ceeding 50 psia. (Assume that the volume of the tire does not change). Thought Problems 1. A candJe is placed vertically in a soup plate, and the soup plate filled with water. Then the candle is lit. An inverted water is carefully placed over the candle. The candle soon goes out, and the water rises inside the It is often said that this shows how much oxygen in the air has been used up. Is this conclusion correct?
- 442. Sec. 13.1 The Ideal Gas Law 411- - 2. A scientific supply house markets aerosol-type cans containing compressed helium for filling balloons, doing demonstrations, and the Like. On the la~1 there appears the notice: "Because the can contains helium, it quite naturally feels empty. It is actually lighter full than empty." Is this statement correct? If so, why? If not, why not? 3. Some reviewers of books and articles have suggested that the gas constant R in the ideal gas equation be forced to take the value of unity (1). What would this step require as far as using the ideal gas equation? 4. In an article by P. Hickman in Physics Teacher, vol. 25, p. 430 (1987) it was suggested that the density of air be determined using marshmallows as follows: 1. Put the marshmallows in a vacuum so they expand and release trapped air. Repressurize the vessel and compress the marshmaUows to a fraction of their original volume. " 2. Weigh the marshmallows on a balance before and after the vacuum treatment. The difference should be the mass of air trapped inside. 3. Measure their volume before and after the vacuum treatment. The difference should be the volume of the air trapped inside. Is this a sound way of getting the density of air? Discussion Problems 1. Three identical glasses are arranged as in Figure DP13.1(pl). Hollow stirrers are also needed. Glasses A and B are comp]etely filled with water (by submerging them jointly in a bucket or sink and joining the mouths before removing), and C is empty. Glasses A and B are carefully placed on a few hollow stirrers as shown in the figure. How can most of the water in glass A be transferred to glass C without ever touching or moving the glasses or their supporting stirrers? Glasses filled with water ==~=T==J4-- Supporting hollow straws C Empty glass Additional hollow straws ~~ Figure DP13.IPl
- 443. 412 !deal Gases Chap. 13 2. An employee was cleaning a cylindrical vessel that contained CS2 in which solid residues had built up on stirrer. had been pumped out and blanketed with nitrogen. The manhole cover was removed and solid residue removed from the with a C!,.. .. ~' ... "'r rod. The employee went to lunch. leaving the manhole cover off. and after returning to complete the job, started a flash fire with Iii spark from the stirrer with the scraper. What might be some of the causes of the accident? Sea breezes provide welcome the summer heat for residents who dose to the No matter what parr of the world-the coast of California, Australia where sea IhrPU~7r>1i: can be very even along the shores of the Lakes-the daily in liummerUme is the same. sea breeze, a wind blowing from sea to land, begins to develop l PI' 4 hOlU'1 tlfter iwuise and its peak intensity by mid-afternoon. It may penetrate inlapej ~ mucfias60 or 70 km. The sea breeze dies out in the evening and 3 or 4 hours after sunset may be ropl~ by a land blowing from land to the sea. The breeze, much woaker thaD .. sea breeze, reaches its intensity sunrise. Whal caUlI01 these '''1',,,,,''7,..'': 13.2 Ideal Ga, tJllxtures and Partial Pressure Frequently. as an you will want to make calculations for mixtures instead individual You can use the ideal gas law under the proper as-sumptions. of for a mixture of by p as the total absolute pressure the mixture, V as the volume occupied by the mixture, n as the total number moles of aU components in the mixture, and T as absolute temperature of the mixture. As the most obvious example. is composed of N2, 0 21 Ar, CO2, Ne, He, and trace gases, but we can treat air as a compound in applying ideal gas law. When the $150 miUion Biosphere project in Arizona in September 1991, it was as a sealed utopian planet in a bottle. where would be ded. Its inhabitants lived for two years in first large self-contained habitat for humans. But slowly the oxygen disappeared from air-four women and four men in the 3.15 acres of glass domes eventually were breathing air with an oxygen content similar to that found at an altitude of about 13.400 The "thin" left the group so and aching that they sometimes gasped for breath. Finally. the leaders of Biosphere 2 had to pump 21,000 lb of oxygen into the domes raise the oxygen level from 14.5 to 19.0%. Subsequent investigation the cause of the decrease in concluded that microorganisms in the soil that took up oxygen, a factor not accounted for in the of the biosphere, were the cause the problem. Engineers use a fictitious but usefuJ quantity caned the partial pressure in many of their calculations involving gases. The partial pressure of Dalton, Pi' namely the pressure that would be exertoo by a single component in a gaseous mix-
- 444. 13.2 Ideal Gas Mixtures and Partial Pressure 413 ture if it existed alone in the same volume as that occupied by the mixture and at the same temperature of the mixture, is defined by (13.4) where Pi the partial pressure of component i in the mixture. If you divide Equation (13.4) by Equation (13.1), you find that and Ptotal V total ntotal RT total ni Pi = Ptotanl-- = PtotalYi total (13.5) where Yi the mole fraction of component i. In air the percent of oxygen is 20.95, hence at the standard conditions of one atmosphere, the partial pressure of oxygen is P02 = 0.2095( 1) = 0.2095 atm. Can you show that Dalton's law of the summation of partial pressures is true using Equation (13.5)1 Pl + P2 + ... + Pn = Ptotal (13.6) Although you cannot measure the partial pressure of a gaseous component directly with an instrument, you can calculate value from Equations (13.5) and/or (13.6). To illustrate the significance of Equation (13.5) and the meaning of partial pressure, suppose that you carried out the following experiment with two nonreacting ideal gases. Examine Figure 13.2. Two tanks each of 1.50 m3 volume, one containing A at 300 kPa and the other B at 400 kPa (both gases being at the same temper-ature of 20°C), are connected to an empty third tank of similar volume. An the in tanks A and B is forced into tank C isothermally. Now you have a 1.50-m3 tank of A C 300 kPa A+B 20°C 20"C 1.50 m3 1.50 m3 700kPa B 400 kPa 1.50 mS Figure 13.2 lliustration of the meaning of partial pressure of the components of an ideal mixture.
- 445. 414 Ideal Gases Chap. 13 A + B at 700 kPa and 20°C for this mixture. You could say that gas A exerts a partial pressure of 300 kPa and gas B exerts a partial pressure of 400 kPa in tank C. Of course you cannot put a pressure gauge on the tank and check this conclusion because the pressure gauge will read only the total pressure. These partial pressures are hypothetical pressures that the individual gases would exert if they were each put into separate but identical vo]umes at the same temperature. In tank C the partial pressures of A and B are according to Equation (13.5) PA = 7oo(~) = 300 kPa PB = 700( ~) = 400 kPa EXAl1PLE 13.6 Calculation of the Partial Pressures of the Components in a Gas from a Gas Analysis Few organisms are able to grow in solution using organic compounds that contain just one carbon atom such as methane or methanol. However, the bacterium methylococcus capsulates can grow under aerobic conditions (in the presence of air) on C-l carbon compounds. The resulting biomass is a good protein source that can be used directly as feed for domestic animals or fish. In one process the off-flue gas analyzes 14.0% CO2, 6.0% 02' and 80.0% N2. It is at 400°F and 765.0 mm Hg pressure. Calculate the partial pressure of each component. Solution Use Equation (13.5): Pi ;;; P'Oio./ Yi Basis: 1.00 kg (or 1b) mol flue gas Component kg (or Ib) mol p(mmHg) CO2 0.140 107.1 °2 0.060 45.9 N2 0.800 612.0 Total 1.000 765.0 On the basis of 1.00 mole of flue gas, the mole fraction Y of each componenr, when multiplied by the total pressure, gives the partial pressure of that component. If you find that the temperature measurement of the flue gas was actually 437°F but the pressure measurement was correct, would the partial pressures change? Hint: Is the temperature involved in Equation 13.5?
- 446. l ,- Sec. 13.2 Ideal Gas Mixtures and Partial Pressure 415 SELF-ASSESSMENT TEST Questions 1. A partial pressure of oxygen in the lungs of 100 mm Hg is adequate to maintain oxygen saturation of the blood in a human. 1s this value higher or lower than the partial pressure of oxygen in the air at sea level? 2. An exposure to a partial pressure of N2 of 1200 mm Hg in air has been found by experience not to cause the symptoms of N2 intoxication to appear. Will a diver at 60 meters be affected by the N2 in the air being breathed? 3. If 1.72 m3 of a gas mixture of 30% CO2 and 70% N2 at 20°C and 105 kPa in a balloon is heated to 70°C, will the partial pressure of the CO2 in the balloon increase or decrease? 4. If you add neon isothermally to helium in a fixed volume vessel. will the partial pressure of the helium change? Problems I. A gas has the following composition at 1200 P and 13.8 psia. Component a. What is the partial pressure of each component? Mol % 2 79 19 b. What is the volume fraction of each component? 2. a. If the C2H6 were removed from the gas in Problem 1, what would be the subsequent pressure in the vesse1? b. What would be the subsequent partial pressure of the N2? Discussion Problem 1. A distillation column reboiler in a room, as shown in Figure DP13.2P 1, had been cleaned, but the manhole cover was not securely fastened on startup again. As a result, benzene vapor escaped from the manhole and one operator died by asphyxjation. How could this accident have occurred? Figure DP13.2Pl
- 447. 416 Ideel Gases Chap. 13 13.3 Material Balances Involving Ideal Gases Now that you have had a chance to practice applying the ideal gas law to simple problems. lef s apply the ideal law in material balances. The only difference between the subject matter of Chapters 6 through 12 and this chapter is that here the amount of material can be specified in terms of p, V. and T rather than solely mass or moles. For example. the basis for a problem, or the quantity to be solved for, might be a volume of gas at a given temperature and pressure rather than a mass of gas. The next two examples illustrate material balances for problems similar to those you have encountered before, but now involve gases. EXAMPLE 13.7 Material Balances for a Process Involving Combustion evaluate the use of renewable resources, an experiment was carried out to pyrolize rice hulls. The product analyzed 6.4 % CO2, 0.1 % O2, 39% CO, 5 t .8% 0.6% CH4• and 1% Nz. It entered the combustion chamber at 90°F and a pressure 35.0 in. Hg. and was burned with 40% excess air (dry), which was at 70°F and an atmospheric pressure of 29.4 in. Hg; 10% of the CO remained unburned. How many cubic feet of air were supplied per cubic foot of entering gas? How many cubic feet of product gas were produced per cubic foot of entering gas if the exit was at 29,4 in. Hg and 400°F? Solution This is an open, steady-state system with reaction. tion chamber. Steps If 2, 3, and 4 system is the combus- Figure illustrates the process and notation. With 40% excess air, cer-tainly all of the CO. H2• and CH4 should burn to CO2 and H20; apparently. for some unknown reason, not a11 the CO burns to CO2, The components of the product gas are shown in the figure. Comp. - CO2 O2 CO H2 CH4 NI!. .....-__ -, 400° F alld 29.4 in. H9 90° F and 35.0 ill. HW Gas 100 Ib mol 1-------- Product '0 ------..., CombustiO{1 c:: mol 02 reqd. P Ib mDI 6.4 Air A ... ? (lb moll COt ? 0.1 (0.1) Or 0.21 H20 ? 39.0 19.5 N2 0.79 CO ? 51.8 25.9 -- O2 ? 0.6 L~ 1.00 NI!. ? 2. , 40'.. Xl 100.0 46.5 10~ f and 29.4 in. Hg Figure E13.7 _.
- 448. 1 Material Balances Involving Idea! _a.i;J'W'.;;:I StepS You could take 1 at 90°F and 35.0 in. Hg as the basis, and convert the vol-ume to moles, but it is just as easy to take 100 Ib mol as a basis because then % = Ib moL At the end of problem you can convert lb mol to ft3. : 100 Ib m01 pyrolysis Step 4 (continued) The entering can calculated from the specified 40% excess air; the reac-tions for complete combustion are 1 CO + 2 1 H2 + 202-H20 moles of oxygen required are listed in Figure Excess 02: 0.4(46.5) = 18.6 Total = 46.5 + 18.6 = 65.1 in is 65.1 G~) = 244.9 Total moles of air in are 244.9 + 65.1 = 310.0 Ib moL Steps 6 and 7 The excess oxygen is Let's make a reduced degree-of-freedom analysis. Five unknowns exist, five products. You can make four element balances and you know fraction of the entering CO that exists in P so that you can calculate ~ag = 0.10(39)' = 3.9. Hence the problem has zero degrees of freedom. Steps 8 and 9 Make element balances moles thies, substitute the value of 3.9 In 2N: 2.1 + 244.9 C: 6.4 + 39.0 + 0.6 51.8 + 0.6(2) 20; 6.4 + 0.1 + 0.5(39) + 65.1 calculate the values of the unknown quanthe number of moles of CO exiting. Out = nNz = + 3 = nH20 ;;:: + nco) 411
- 449. 418 Ideal Gases Chap. 13 The solution of these equations is 7/N2 = 247 n~ = 20.55 and nco = 3.9 The total moles exiting' sum to 366.6 Ib moL If you prefer to make species balances, let l1io ut be the moles of species i exiting the process. The degree-of-freedom analysis is (the basis is 100 g mol entering): Equations and specifications (11). Material balances CO2: nf:~ -39 = -€. 02: ~t -1 - 65.1 = -0.5€ I - 0.5t'2 - 2t'3 CO: n~~ -6.4 = €! + €2 H2: n~~l -51.8 ;: 6 CH4: n~~ -0.6 = -6 H20: n~~b -0 == € 1 + 2€3 N2: n~u; -2.1 - 244.9 = 0 (no reaction) Specifications: nltt :::::;; 0 nCol.Rlt t -- 0 ~6 = 0.1(39) = 3.9 Implicit relations: ~n'oUl = P I The degrees of freedom are zero. Solution of these equations gives the same results as the element balances, as expected. FinaUy, you can convert the lb mol of air and products that were calculated on the basis of 100 Ib mol of pyrolysis gas to the volumes of gases at the states requested: Tg as -- 90 + 460 -- 5500R Tair = 70 + 460;; 5300R T product = 400 + 460 = 8600R
- 450. Sec. 13.3 Material Balances Involving Ideal Gases 100 1b mol entering ttl at SC 5500R -29-.9-2- -=- - 1 1b mol 492°R 35.0 in. Hg 3 i 0 lb mol air 359 at SC 5300R 29.92 in. I Ib mol 492QR 29.4 Hg 1220 X 102 3 366.61b mol P 359 ft3 at SC 8600R 29.92 in. Hg 2 ft of product: 1 ib mol 4290R 35.0 in. Hg = 2255 X 10 The answers to questions are -] 2-2-0- X- :1:0-2 = ft3 air at and 29.4 in. Hg 3.56----------"- 343 X 1 ft3 at 5500R and .0 2255 X 102 = 6.57_=--__at_ 8_6_00_R_a_n_d_2_9_.4_in_,_H-'-g x and 35.0 in. Hg EXAMPLE 13.8 Material Balance without Reaction at and ]05 kPa is flowing through an irregular duct. To determine the rate of flow of the gas, CO2 from a tank is passed into the gas stream. The gas analyzes 1.2% CO2 by volume before 3.4% CO2 by volume after the addition. the CO2 that wa.~ injected left the tank, it was passed through a rotameter, and found to flow at the rate of 0.0917 m3 fmin at and 13] kPa. What was the rate of flow of the entering in the duct cubic meters minute? Solution This is an open, steady-state system without reaction. The system is the ·duct. Figure 3,8 is a sketch of process. Steps 1, 2, 3, and 4 The data are presented in Figure 3.8. F .. tSGC and 105 kPa % C(h 1.2 Other 98.8 100.0 Figure E13.8 15°C and 105 kPa % CO2 3.4 Other 96.6 100.0 419
- 451. 420 Ideal Gases Chap. 13 Both F and P are at the same temperature and pressure. StepS Should you take as a basis 1 min == 0.0917 m3 of CO2 at 7°C and 131 kPa? The gas analysis is in volume percent. which is the same as mole percent. We could convert the 0.0917 m3 to moles and solve the problem in terms of moles, but there is no need to do so because we just as easily convert the known flow rate of addition of CO2 to a volume at I and 105 kPa, and solve the problem using m3 for each stream since all the streams are at the same conditions. We could similarly convert all of the data to 7°C and 131 kPa, but more calculations would be required to get the answer than for 15°C and 105 kPa. Thus, we start with the basis of 3 273 + 15 131 = 0.1177 m3 at 150C and 105 kPa 0.0917 m 273 + 7 105 Steps 6 and 7 We do not know F and P, but can make two independent component bal· ances, and "other," hence the problem has zero degrees of freedom. Steps 7, 8, and 9 "Other" balance (in m3 at 15°C and 105 kPa): F(0.98B) ::::; P(0.966) (a) CO2 balance (in m3 at 15°C and 105 kPa): F(O.012) + 0.1177 ::::: P(O.034) (b) Total balance (in m3 at 15°C and 105 kPa): F + 0.1177 == P (c) Note that the "other" is a component Select Equations (a) and (c) to solve. The solution of Equations (a) and (c) gives F::::: 5. m3 /min at 15°C and 105 kPa Step 10 (Check) Use the redundant equation: By Equation (b): 17 (0.012) + 0.1177 == 0.180 J. 5.17(0.988/0.966)(0.034) = 0.180 equation checks out to a satisfactory degree of precision. SELF-ASSESSMENT T ST Questions 1. How does the introduction of ideal balances? affect the principles involved in making material
- 452. Sec. 13.3 Material Balances Involving Ideal Gases 421 2. Does including the ideal gas law together with material balances add an additional equation to the analysis of the degrees of freedom? Problems 1. A furnace is fired with 1000 ft3 per hour at 60°F and 1 atm of a natural gas having the following volumetric analysis: 80%, C2H6: 1 02: 2%. CO2: 1%, and N2: 1 %. The exit flue gas temperature 800°F and the pressure is 760 mm Hg absolute; 15% excess air is used and combustion is complete. Calculate the (a) volume CO2 produced per hour; (b) volume of H20 vapor produced per hour; (c) volume of N2 produced per hour~ and (d) total volume of flue gas produced per hour. 2. A flue gas contains 60% N2 and mixed with air to cool it. If the resulting mixture flows at a rate of 250,000 ft3 Ihr and contains 70% Nz, what is the flow rate of the flue gas? State all your assumptions concerning the temperatures and pressures of the streams. 3. Two tanks containing N2 at the following conditions sit next to each other TankA TankB Volume (m3) 1 5 Temperature (ae) 25 40 Pressure (kPa) 300 ? AmouDr of (g mol) '] ? After the two tanks are connected and reach equilibrium, the conditions in the combined tanks are 700 kPa and 35°C. What was the pressure in Tank B? Thought Problems 1. In a test of the flow of gases through a pipe, pure hydrogen was found to flow at a volumetric flow rate 22 times that of carbon dioxide. When the hydrogen was diluted with carbon dioxide entering the pipe midway from the ends, the exit flow was less than that of pure hydrogen. Explain the observed differences. 2. A pair of identical balloons are inflated with air to the same pressure, and tied to a stick that is held in the center by a balloons are each same distance from the center of the stick so that the stick remains horizontal to the ground. When the lefthand baUoon is carefully punctured, will the stick rotate down from the left, up from the left, or remain horizontal? Discussion Problem 1. In a demonstration. a 30 em diameter balloon was filled to two-thirds of its maximum pressure with SF6, a gas. Students measured the balloon's diameter for 10 days. at which time the balloon burst. No one ever touched it. Explain how this could happen. (Note: The baUoon was not defective.)
- 453. 422 Ideal Gases Chap. 13 Looking Back We reviewed the ideal gas law as applied to pure components and gas mixtures, and explained about density and specific gravity for gases. In making balances, process measurements are frequently made as volumetric flows rather than molar or mass flows, and the molar flows can be calculated from the volumetric flows as wen as the reverse. GLOSSARY OF NEW WORDS Dalton's Law summation of each of the partial pressures of the components in a system equals the total pressure. The other related law (of partial pressures) is that the total pressure times the mole fraction of a component in a system is the partial pressure of the component. Density of gas Mass per unit volume expressed in kglm3, Ib/ft3, gIL, or equivalent units. Ideal gas constant constant in the ideal gas law (and other equations) denoted by the symbol R. Ideal gas law Equation relating p. V. n, and T that applies to many gases at low density (high temperature and/or low pressure). Partial pressure The pressure that would be exerted by a single component in a mixture if it existed alone in the same volume as occupied by the mixture and at the temperature of the mixture. Specific gravity Ratio of the density of a gas at a temperature and pressure to the density of a reference gas at some temperature and pressure. Standard conditions (S.C.) Arbitrarily specified standard states of temperature and pressure established by custom. SUPPLEMENTARY R F R NC S In addition to the general references listed in the Frequently Asked Questions in the front material, the following are pertinent. Black, W.Z., and J.G. Hartley. Thennodynamics, Harper & Row, New York (1985). Howell. I.R.. and RD. Buckius. Fundamentals of Engineering Thennodynamics, McGrawHill, New York (1987).
- 454. Chap. 13 Problems 423" Masavetas, K.A. "The Mere an Ideal Gas," Math. Comput. Modelling, 12, 1-657 (1989). Van Wylen. GJ .• and R.E. Sonntag. Fundamentals of Classical Themwdynomics. 3rd ed., John Wiley, New York (1985). Wark, Thermodynamics, 5th ed., McGraw-Hill, New York (1988). Web Sites hnp:/lantoineJrostburg.edulchemlseneseilO I/matter/resources .shtml hUp:/lchemistry .ohiostate.edulbetbalrealgaslaw/fr22.html http://eng.sdsu.edultestcenterlTest/solv ... tlidealgasideaIgas/idealgasidealgas.html http://mccoy.lib.siu.edulprojectslchem200/audioHnks.html http://voyagerS.sdsu.edultestcenterlhome.html PROBl MS "'13.1 How many pounds H20 are in 100 of vapor at mrn Hg and 23°C? *13.2 One liter of a gas is under a pressure of 780 mm Hg. What will be its volume at standard pressure, the temperature remaining constant? *13.3 A gas occupying a volume 1 m3 under standard pressure is expanded to 1.200 m3, the temperature remaining constant. What is the new pressure? *13.4 Determine the mass specific volume molal specific volume air at 78°F and psia. An oxygen cylinder used as standby source of oxygen contains 1.000 ft3 of O2 at 70°F and 200 psig. What will be the volume this 02 in a dry-gas holder at 900P and 4.00 in. H20 above atmospheric? barometer reads 29.92 in. *13.6 You have 10 Ib of CO2 in a 20-ft3 fire extinguisher tank at 30°C. Assuming that the ideal law holds. what will the pressure gauge on the tank in a test to see if extinguisher is full? *13.7 work as far as 500 ft below the water surface. Assume the water tempera-ture is What is the molar specific volume (ft3llb mol) for an ideal under these conditions? *13.8 A 2S-L glass vessel is to contain 1.1 g of nitrogen. The can withstand a pressure of only 20 kPa above atmospheric pressure (taking into account a suitable safety factor). What is the maximum temperature to which the N2 can be raised the vessel? *13.9 An cylinder as a standby source of oxygen contains 02 at 70°F. To cali-brate the gauge on the 02 cylinder which has volume 1.01 fi3, all the oxygen. initially at 70°F, is released into an evacuated tank of known volume (15.0 ft3), At
- 455. 424 Ideal Gases Chap. 13 equilibrium, the gas pressure was measured as 4 in. H20 gauge and the gas tempera-ture in both cylinders was 75°F. See Fig. P13.9. The barometer 29.99 in. Hg .. What did the pressure gauge on the oxygen tank initially read in psig if it was a Bourdon 1.01 tt3 V:::. 16.0 75"F 4 in. H20 Gauge Figure Pt3.9 "13.10 The U-tube manometer depicted in Figure P13.l0 has a left leg 20 inches high and a right leg 40 inches high. The manometer initially contains mercury to a depth of 12 inches in each leg. Then the left leg is closed with a cork, and mercury poured in the right leg until the mercury in the left (closed) leg reaches a height of 14 inches. How deep is the mercury in the right leg from the bottom of the manometer? Cork Hg Figure P13.10 ··13.11 0 ne 0 f th e expen.m ents 1. 0 t he fuel-testing laboratory has been giving some trouble because a particular barometer gives erroneous readings owing to presence of a small amount of air above the mercury column. At a true atmospheric pressure of755 mm Hg the barometer reads 748 nun Hg, and at a true 740 mm the reading is 736 mm Hg. What will the barometer read when the actual pressure is 760 mm Hg?
- 456. Chap. 13 Problems 425 "'13.12 An average person's lungs contain about 5 L of under normal conditions. If a diver makes a free dive (no breathing apparatus); the volume of the lungs is compressed when the pressure equalizes throughout the body. If compression occurs below irreversible lung damage win occur. Calculate the maximum safe depth for a free dive in sea water (assume the density is the same as fresh water). 1113.13 An automobile tire when cold (at 75°F) reads 30 psig on a tire After driving on the freeway, the temperature in the becomes 140°F. Will the pressure in the tire exceed the pressure limit of psi manufacturer stamps on the tire? "'13.14 You are making measurements on an air conditioning duct to test its load capacity. The warm flowing through the circular duct has a density of 0.0796 pounds per cubic foot. Careful measurements of the velocity of the air in the duct disclose that the average air velocity is 11.3 per second. The inside radius of the duct is 18.0 inches. What is (a) the volumetric flow rate the in ft3lhr, and (b) what is the mass flow rate of the air in Ib/day? "'13.15 Flue at a temperature of 18000P is introduced to a scrubber through a pipe which has an inside diameter of 4.0 ft. The inlet velocity to and the outlet velocity from the scrubber are 25 ftls and 20 respectively. The scrubber cools the flue gas to 550°F, Determine the duct size required at the outlet the unit. ·13.16 Calculate the number of cubic meters of hydrogen sulphide, measured at a tempera-ture of 30°C and a pressure of 15.71 em of Hg, which may be produced from 10 of iron sulphide (FeS). *13.17 One pound mole flue gas has the foHowing composition. it as an ideal How many ft3 will the gas occupy at 100°F and 1.54 atm? "13.18 Monitoring of hexachlorobenzene (HCB) in a flue gas from an incinerator burning 500 lblhr of hazardous wastes is to be conducted. Assume that all of the RCB is removed from a sample of the flue and concentrated 25 of solvent. The analytical detection limit for HCB is 10 micrograms per milliliter in the solvent Determine the minimum volume of flue gas that has to be sampled to detect the existence of HCB in the flue gas. Also, calculate the time needed to collect a gas sample if you can collect LO L minute. The flue gas flow rate is 427,000 ft3lhr measured at standard conditions. *13.19 Ventilation is an extremely important method of reducing the level of toxic airborne contaminants in the workplace. Since it is impossible to eliminate absolutely all leak· from a process into the workplace, some method is al ways needed to remove toxic materials from the air in closed rooms when such materials are present in process streams. Occupational Safety and Health Administration (OSHA) has set the permissible exposure limit (PEL) of vinyl chloride (VC, MW = 78) at 1.0 ppm as a maximum time-weighted average (TWA) for an eight-hour workday, because VC is believed to be a human carcinogen. If VC escaped into the air. its concentration must be maintained at or below the If dilution ventilation were to be used. you can estimate the required air flow rate by assuming complete in workplace air,
- 457. 426 Ideal Gases Chap. 13 and then assume that the volume of air flow through the room will carry VC out with it at the concentration of 1.0 ppm. If a process loses 10 g/min of VC into the room air, what volumetric flow rate of air will be necessary to maintain the PEL of 1.0 ppm by dilution ventilation? (In practice we must also correct for the fact that complete mixing will not be realized in a room so that' you must multiply the calculated air flow rate by a safety factor, say a factor of 10.) If the safety analysis or economics of ventilation do not demonstrate a safe concentration of VC exists, the process might have to be moved into a hood so that no VC enters the room. If the process is carried out in a hood with an opening of 30 in. wide by 25 in. high, and the "face velocity" (average air velocity through the hood opening) is 100 ft/s, what is the volumetric air flow rate at SC? Which method of treating the pollution problem seems to be the best to you? Explain why dilution ventilation is not recommended for maintaining air quality. What might be a problem with the use of a hood? The problem is adapted with permission from the publication Safety, Health, and Loss Prevention in Chemical Processes published by The American Institute of Chemical Engineers, New York (1990). -13.20 Ventilation is an extremely important method of reducing the level of toxic airborne contaminants in the workplace. Trichloroethylene (TeE) is an excellent solvent for a number of applications, and is especially useful in degreasing. Unfortunately, TeE can lead to a number of harmful health effects, and ventilation is essential. TCE has been shown to be carcinogenic in animal tests. (Carcinogenic means that exposure to the agent might increase the likelihood of the subject getting cancer at some time in the future.) It is also an irritant to the eyes and respiratory tract. Acute exposure causes depression of the central nervous system, producing symptoms of dizziness, tremors, and irregular heartbeat, plus others. Since the molecular weight of TCE is approximately 131.5, it is much more dense than air. As a first thought, you would not expect to find a high concentration of this material above an open tank because you might assume that the vapor would ,sink to the floor. If this were so, then we would place the inlet of a local exhaust hood for such a tank near the floor. However, toxic concentrations of many materials are not much more dense than the air itself, so where there can be mixing with the air we may not assume that all the vapors will go to the floor. For the case of trichloroethylene OSHA has established a time-weighted average 8 hr permissible exposure limit (PEL) of 100 ppm. What is the fraction increase in the density of a mixture ofTCE in air over that of air if the TCE is at a concentration of 100 ppm and at 25°C? This problem has been adapted from Safety, Health. and Loss Prevention in Chemical Processes, New York: American Institute of Chemical Engineers, (1990): 3. -13.21 Benzene can cause chronic adverse blood effects such as anemia and possibly leukemia with chronic exposure. Benzene has a PEL (pennissible exposure limit) for an 8-hr exposure of 1.0 ppm. If liquid benzene is evaporating into the air at a rate of 2.5 cm3 of liquid/min. what must the ventilation rate be in volume per minute to keep the concentration below the PEL? The ambient temperature is 68°F and the pressure is 740 mm Hg. This problem has been adapted from Safety, Health, and Loss Preven- -----_.. ----_._ ---_ ._- . .
- 458. Chap. 13 Problems in Chemical Processest New York: American Institute Chemical Engineers, (1990): 6. *13.22 A recent report states: for fuel gas measure volume of gas usage on a standard temperature, usually 60 degrees. contracts when it's cold expands when warm. Ohio Gas Co. figures that in chilly Cleveland, the homeowner with an outdoor meter more gas than the he does, so that's built into company's gas guy who loses is the one an indoor meter: If his stays at 60 degrees or over, he'll pay for more than gets. (Several make temperature-compensating meters, but cost more and aren't widely Not surprisingly, they are sold mainly to utilities in the North.)" Suppose that the outside temperature drops from 60°F to 10°F. What is the in the mass of passed by a noncompensated outdoor meter that at constant pressure? that the gas is CH4• $13.23 Soft ice cream is a commercial/ice cream whipped usually with CO2 (as the O2 in the air causes deterioration.) You are working in the Pig-in-a-Poke Drive-In and to make a full (4 gal) of soft ice cream. The unwhipped mix a 0.95 and the local forbids you to cream of less than of 0,85. Your CO2 tank a No~ 1 cylinder (9 in, by in. high) of com- (99.5% min. carbon dioxide. In the pressure gauge you note it 68 psig. Do you have to order another cylinder of Be sure to specificaIJy assumptions you for this problem. Additional Data: Atmospheric pressure 752 mm Hg cream::::: 0.84 milk == 0.92 content of soft ice cream is less than 14% *13.24 Prom known standard conditions, calculate the value of law constant R the foHowing sets of units: (a) caV(g mol)(K) (d) J/(g mol)(K) (b) BtuJ(1b (e) (cm3)(atm)/(g mol)(K) (c) (psia)(ft3)/(1b mol)(OR) (0 (ft3)(atm)/Ob mol)(OR) $13.25 What is density of 02 at lOOoP and 740 mm Hg in (a) lb/ft3 (b) gIL? *13.26 What is density of propane (C3HS) in kg per meter at 200 kPa and What is the specific gravity propane? *13.27 is the specific gravity of propane (C3Hg) at l000P 800 mm Hg relative to air at and 760 mm Hg? What is the mass of 1 of H2 at 5°C and 110 kPa? What is the gravity of this H2 compared to at SoC and 110 kPa? natural gas has following composition: CH4 (methane) 87% (ethane) 12% (propane) 1 % I
- 459. 428 (a) What is the composition in weight percent? (b) What is the composition in volume percent? Ideal Gases Chap. 13 (c) How many m3 will be occupied by 80.0 kg of the gas at 9°C and 600 kPa? (d) What is the density of the gas in kg/m3 at SC? (e) What is the specific gravity of this gas at 9°e and 600 kPa referred to air at SC? "13.30 A mixture of bromine vapor in air contains 1 % bromine by volume. (a) What weight percent bromine is present? (b) What is the average molecular weight of the mixture? (c) What is its specific gravity? (d) What is its specific gravity compared to bromine? (e) What is its specific gravity at lOOoP and 100 psig compared to air at 60DF and 30 in. Hg? *1331 A gas used to extinguish fires is composed of 80% CO2 and 20% N2. It is stored in a 2 m3 tank at 200 kPa and 25°C. What is the partial pressure of the CO2 in the tank in kPa? *13.32 A natural gas has the foHowing composition by volume: CH4 94.1% N2 3.0 1.9 1.0 100.0% This gas is piped from the well at a temperature of 20°C and a pressuro of 30 psig. It may be assumed that the ideal gas law is applicable. Calculate the partial pressure pf the oxygen. *13.33 A Bter of oxygen at 760 mm Hg is forced into a vessel containing a Hter of Qitrogen at 760 mm Hg. What will be the resulting pressure? What assumptions are necessary for your answer? "13.34 The contents of a gas cylinder are found to contain 20 percent CO2, 60 percent °2, and 20 percent N2 at a pressure of 740 mm Hg and at 20°C. What are the partial pressures of each of the components? If the temperature is raised to 40°C. will the partial pressures change? If so, what will they be? *13.35 Methane is completely burned with 20% excess air, with 30% of the carbon going to CO. What is the partial pressure of the CO in the stack gas if the barometer reads 740 mm Hg, the temperature of the stack gas is 300"F. and the gas leaves the stack at 250 fa above the ground level? *13.36 Answer the following questions true or false: a. The volume of an ideal-gas mixture is equal to the sum of the volumes of each individual gas in the mixture. b. The temperature of an ideal-gas mixture is equal to the sum of the temperatures of each individual gas in the mixture. c. The pressure of an ideal-gas mixture is equal to the sum of the partial pressures of each individual gas in the mixture.
- 460. ;. Chap. 13 Problems 429 -13.37 A 0.5-m3 rigid tank containing hydrogen at 20°C and 600 kPa is connected by a valve to another 0.5-m3 rigid tank that holds hydrogen at 30°C and 150 kPa. Now the valve is opened and the system is allowed to reach thermal equilibrium with the surroundings, which are at 15°C. Detennine the final pressure in the tank. ··13.38 A 400 ft3 tank of compressed H2 is at a pressure of 55 psig. It is connected to a smaller tank with a valve and short line. The small tank has a volume of 50 ft3 and contains H2 at 1 atmosphere absolute and the same temperature. If the interconnecting valve is opened and no temperature change occurs, what is the final pressure in the system? ··13.39 A tank of N2 has a volume of 100 ft.3 and an initial temperature of 80°F. One pound of ![2 is removed from the tank, and the pressure drops to 100 psig while the temperature of the gas in the tank drops to 60oP. Assuming N2 acts as an ideal gas, calculate the initial pressure reading on the pressure gage. -13.40 Measurement of flue gas flow rates is difficult by traditional techniques for various reasons. Tracer gas flow measurements using sulfur hexafloride (SF6) have proved to be more accurate. The accompanying figure shows the stack arrangement and the injection and sampling points for the SF6. Here is the data fOf .one experiment: Volume of SF6 injected (converted to SC): Concentration of SF6 at the flue gas sample point: Relative humidity correction: Calculate the volume of the exit flue gas per minute. Sample point (Elevation above base = 70 ft) Diameter == 15 ft Temperature = saoe 12 ft plastic flue t #1 Fan Tracer-gas injection I point, 15 ft x 8 ft 60 ft #2 Fan 28.8 m3/min 4.15 ppm none T.mp.ratu65~ft 1 + / A-I Gas I 1~,I " "- ----. flow Gas flow = ~ _... _ ~ Baghouse outlet flue, 15f1x8ft Figure P13.40
- 461. 430 Ideal Gases Chap. 13 "'13.41 An ideal gas at 60°F and 31.2 in. Hg (absolute) is flowing through an irregular duct. determine the flow rate of the gas, CO2 is passed into the gas stream. The ana-lyzes 1.2 mole % CO2 before and 3.4 mole % addition. 'The tank is placed on a scale and found to lose 15 lb in 30 minutes. What is the flow rate of the entering gas in cubic per minute? ··13.42 One important source of emissions from gasoline-powered automobile engines that causes smog is the nitrogen oxides NO and N02. They are formed whether combustion is complete or not as follows. At the high temperatures that occur in an internal combustion engine during the burning process, oxygen and nitrogen combine to fonn nitric oxide (NO). higher the peak temperatures and the more oxygen available, the more NO is formed. There is insufficient time for the NO to decompose back to 02 and N2 because the burned gases cool too rapidly during the expansion and exhaust cycles in the engine. Although both NO and nitrogen dioxide (N02) are significant air pollutants (together termed NOx)' the N02 is formed in the atmosphere as NO oxidized. Suppose that you coHect a sample of a NO-N02 mixture (after having r<>,. .. ,,"""""'''' the other combustion gas products including N2• and H20 by various separations procedures) in a l00-cm3 standard cell at 30°C. Certainly some the NO will have been oxidized to N02 2NO+ 42N02 during the collection. and processing of the combustion gases, so that measurement of NO alone will be misleading. If the standard cell contains 0.291 g of N02 plus NO and pressure measured in the cell is 170 kPa, what percent of the NO + N02 is in the form of NO? *13.43 In the manufacture of dry a fuel is burned to a flue gas which contains 16.2 per-cent CO2, 4.8 percent 02' and the remainder N2. This flue gas passes through a heat exchanger and then goes to an absorber. data show that the analysis of the flue entering the absorber is 13.1 percent CO2 with the remainder 02 and N2 . Apparently something has happened: To check your initial assumption that an air leak developed in the heat exchanger, you collect the following data on a dry on the heal exchanger: Entering flue in a 2-rnin period 47.800 at 6000P and 740 mm of Hg flue gas in a 2·min period 30,000 ft3 at 600P and 720 mm of Hg your assumption about an air leak a good one. or was perhaps the analysis of the error? Or both? ."13.'.". .A.I A mmom. a at 10 0° C and 150 kPa is burned with 20% excess 02 The r~action is 80% complete. The NO is separated from the NH3 and water, and the 1S recycled as shown in Figure P13.44.
- 462. Chap. 13 Problems 431 O2--..... 1---..... NO NH3 100°C 150 kPa 1---..... O2 1....----,_----1 NH3 onfy 150"'C 150 kPa Calculate the m3 of NH3 recycled at 1 and 150 kPa . m3 of NH3 fed at 1 QOoC .... "'r' .......... (C6H~ is converted to 12) by direct reaction with H2. The fresh feed to the process is 260 Umin of plus 950 Umin of H2 at 100°C and 1 kPa. The single pass conversion of in the reactor is 48% while the overall of H2 the process is stream contains 90% Hl and the benzene (no cyc1ohexane). P13,45. (a) the molar flow rates of H2• and C6H 12 in the exiting product (b) Determine the volumetric flow rates of the components in the product stream if it at 100 kPa and 200°C. ( c) the molar flow rate of the if the recycle stream is at 100°C and 100 kPa. Recycle R 90%H2• Reactor FJgure P13.4S and the volumetric flow rate P (gas) Ce He H2 C6 H12 Pure (C2H4) and oxygen are fed to a process for the manufacture of ethylene oxide (C2H40): Figure P13.46 is the flow diagram for the orocess. reactor oper-ates at 300°C and 1 atm. At these conditions, on the re-actor show 50% of the ethylene entering reactor is per pass, and of this, 70% is to ethylene oxide. The rl".T'nl:ll·nn .. '1" of the ethylene reacts to form CO2 and water.
- 463. 432 Ideal Gases Chap. 13 C2H4 + 302 -7 2eOl + 2H20 a daily production of 10,000 kg of ethylene oxide: (a) Calculate the m3lhr of total gas entering the reactor at SC if the ratio of the 02 (g) fed to fresh C2H4 (g) is 3 to 2. (b) Calculate the recycle ratio, m3 at 10°C and 100 kPa of C2H4 recycled per m3 at SC of fresh C2H4 fed; (c) Calculate the m3 of the mixture of 02' CO2 and H20 leaving the separator per day at Boac and 100 kPa. F,e~h Ctl'!. (g} Catalytic Pure O2 to} Reoc:lof ReoelCI Product Olilput Stpototor Figure P13.46 CO2 tgl H20 (9' O2 (g) ··13.47 Three thousand cubic meters per day of a gas mixture containing methane and n-butane at 21°C enters an absorber tower. The partial pressures at these conditions are 103 kPa for methane and 586 kPa for n-butane. In the absorber. 80% of the butane is removed and the remaining gas leaves the tower at 38°C and a total pressure of 550 kPa. What is the volumetric flow rate of gas at the exit? How many moles per day of butane are removed from the gas in this process? Assume ideal behavior. *"'13.48 A heater bums normal butane (n-C4HlO) using 40.0 percent excess air. Combustion is complete. The flue gas leaves the stack at a pressure of 100 kPa and a temperature of 260°C. (a) Calculate the complete flue gas analysis. (b) What is the volume of the flue gas in cubic meter per kg mol of n-butane? ··13.49 The majority of semiconductor chips used in the microelectronics industry are made of silicon doped with trace amounts of materials to enhance conductivity. The silicon initially must contain less than 20 parts per million (ppm) of impurities. Silicon rods are grown by the foHowing chemical deposition reaction of trichlorosilane with hydrogen: HSiC13 + H2 --..,.. 3HCl + Si Assuming that the ideal gas law applies, what volume of hydrogen at 1000°C and I atm must be reacted to increase the dihmet~!-Gf a rod 1 m long from 1 cm to 10 cm? The density of solid silicon is 2.3_Wcm>:/ ·"'13.50 An incinerator produces a dry exit gas of the foHowing Orsat composition measured at 60°F and 30 inches of Hg absolute: 4.0% CO2, 26.0% CO, 2.0% CH4, 16.0% H2 and 52.0% N2. A dry natural of the foHowing (Orsat) composition: 80.5% CH4• 17.8%, ~H6' and 1.7% N2 is used at the rate of 1200 fts/min at 60°F and 30 inches
- 464. Chap. 13 Problems of Hg absolute to burn the incineration off gas with air. The final products of combustion analyze on a dry basis: 12.2% CO2, 0.7% CO, 2.4% O2, and 84.7% N2· Calculate (a) the rate of flow in ft3/min of the incinerator exit gas at 60°F and 30 inches of Hg absolute on a dry basis, and (b) the rate of air flow in f(!Jmin. dry. at gO°F and 29.6 of Hg absolute. A mixrure consisting of 50 mol % hydrogen and 50 mol % acetaldehyde (C 2H40) is initially contained in a rigid vessel at a total pressure of 760 mm Hg abs. The formation of ethanol (C2H60) occurs according to After a time it was noted that the total pressure in the rigid vessel had dropped to 700 mm Hg abs. Calculate the of completion of the reaction at that time using the following assumptions: (1) All reactants and products are in the gaseous state; and (2) the vessel and its contents were at the same temperature when the two pressures were measured. ··13.52 Biomass (CH1.80o.sNo . s) can be converted to glycerol by anerobic (in the absence of air) reaction with ammonia and glucose. In one batch of reactants, L CO2 measured at 300 K and 95 kPa was obtained per mole of glucose in the reactor. The molar stoichiometric ratio of nitrogen produced to ammonia reacted in the reaction equation is one to one, and the mol CO2/mol C6HJ20 S = 2. How much (in g mol) (a) glycerol was produced and (b) how much biomass reacted to produce the 52.4 L of CO2 ? "'13.53 The oxygen and carbon dioxide concentrations in the gas phase of a 10 L bioreactor operating in steady-state control the dissolved oxygen and pH in the liquid phase where the biomass exists. a. If the rate oxygen uptake by the liquid is 2.5 X 10-7 g moJJ(1000 cells)(hr). and if the culture in the liquid phase contains 2.9 X 106 cellS/mL. what is the rate of oxygen uptake in miHimoUhr? If the gas supplied to the phase is 45lJhr containing 40% oxygen at 110 kPa and 25°C, what is the rate of oxygen supplied to the bioreactor in miHimollhr? c. Will the oxygen concentration in the gas phase increase or decrease by the end of one hour compared to the initial oxygen concentration? ·P13.54 When natural gas (mainly CH4) is burned with 10 percent excess air, in addition to the main gaseous products of CO2 and H20. other gaseous products result in minor quantities. The Environmental Protection Agency (EPA) lists the following data: Emission Factors (kg/loti m3 at SC) S02 N02 CO COl utility boiler. uncontrolled 9.6 3040 1344 1.9 X 1()6 utility boiler, controlled gas recirculation 9.6 1600 1344 1.9 X l()6 Residential furnace 9.6 1500 640 1.9 X 1()6
- 465. Ideal The data are based on 106 m3 at SC of methane burned. What is the approximate fraction of S02' N02, and CO (on a for each class of combustion equipment? Estimate the emissions of each compound produced in m3 measured at ton 0000 kg) of No. 6 fuel oil burned in an oil-frred burner with no given the following data: "s sulfur Particulate matter 1.5 Chap. 13 6 fuel oil contains 0.84% sulfur. and composition from Perry of No. 6 a specific gravity of 0.86 at 1 mass Compute C H o N S Ash with a specific gravity of 0.86 is in 10.49 0.84 0.04 each component per 103 L of and compare the resulting emis-listed in the EPA analysis in Problem P13
- 466. REAL GAS S: COMPRESSIBILITY Your objectives In studying this chapter are to be able to: 1. Explain what Define the 1'",',,"1""'" state. states means. Calculate the reduced temperature, reduced and ideal volume, and use any two of these three parameters the compressibility factor, from the compressibility charts. 4. Use compressibility factors and appropriate to predict V-Tbehavior of a or, given required data, find a compressibility factor. Calculate compressibility z using the acentric 6. Use Kay's method of pseudocritical values to calculate the reduced values, and predict P. V, T, and n via the compressibility factor. Looking Ahead Predicting gas features has appeal Bya law that appears 10 real somehow the law Seems to have flaws Because gases are rarely DMH In this we how the critical properties of can be employed to facilitate calculation of a factor, a factor that transforms the ideal gas law into a relation that can be used to for V, n, T for singleand mUlti-component real gases.
- 467. 436 Real Gases: Compressibility Chap. 14 Main Concepts 14.1 Corresponding States When gases do not confonn to the assumptions underlying ideality that were explained in Chapter 13. we call them real gases. For example, if you compress argon to 1000 atm in a chamber with a viewport, the argon will have about the same density as water (but it is still a gas), and you can float a material in the argon that floats on water. You cannot apply the idea] law under such c9nditions, Figure 14.1 shows the error involved in using the ideal gas law to predict p-V-T properties for steam. In the attempt to devise some truly urn versal gas law that predicts well at low and high pressures, the idea of corresponding states was developed. Early experimenters found that at the critical point all substances are in approximately the same state of molecular dispersion. Consequently. it was felt that their thermodynamic and physical properties should be similar. The law of corresponding states expresses idea that in the critical state all substances should behave alike. What does the critical state (point) mean? For a pure component it means the maximum temperature and corresponding pressure at which liquid and vapor can coexist. You can fmd many definitions, but the one most suitable for general use with pure component systems as well as with mixtures of gases is the followmg: The critical state for the gas-liquid transition is the set of physical conditions at which the density and other properties of the liquid and vapor become identical. 1 O'----.L----""------l o 40 80 120 Pressure (AtM) s. 2....---------........, 1 o ~ ______ -4 ___ ~~ 250 350 450 550 Temperature (K) b. Figure 14.1 Graph of the measured specific volume V of stearn as It function of (a) the pressure and (b) the temperature contrasted with predictions by the ideal gas law, Note that rhe deviations at high pressure and low temperature are relatively quite large.
- 468. Sec. 14.2 Critical State 437 14 .. 2 Critical State One way to determine the critical is by acoustic because ,>1..1 ... ,""' .... of sound in a fluid dIOpS to a minimum wh.en the fluid in the criticru. fact. in 1822 acoustic effects were exploited unwittingly by Cagniard de 1a Tour when heated alcohol in a sealed gun barrel and to the musket ban roIling Figure 14.2 shows measurements by Andrews 1863 of the pressure versus the volume for CO2, Note the point C at 31 the highest temperature at which liquid and gaseous CO2 can coexist in equilibrium. Above 31°C only fluid exists, so that the critical temperature for is 31°C (304K). The corresponding pressure 72.9 atm (7385 Also note at higher temperatures, such as 50°C. the data can be rep-by the ideal law because p V = constant, a hyperbola. You can experimental values of the temperature (Tc) and the, cal pressure (Pc) for various compounds in Appendix D on the CD that accompanies book. If you cannot find a desired critical value this text or a hand-book, you can always consult Poling a]. (refer to references at end for this chapter), which and evaluates methods estimating constants various compounds. p(atm) 80 72.9 Liquid .. •• - •• I 4O"C ',-----31"C ~------- 20"C 40 -------:---......-....- --10°C ~ _ _L __ ~ ____________ ~ ________________ ~ __ • 100 94 75 50 Specific volume. V (crrt3/g mol) Figure 14.2 Experimental measurements of carbon dioxide by Andrews. The solid lines smoothed is the highest temperature which any liquid exists, At the solid dots Hquid and vapor coexist.
- 469. Real Gases: Compressibility Chap. 14 A supercritical fluid is a compound in a state above critical point. Super-critical fluids are used to replace solvents such as trichloroethylene and methylene chloride, the emissions from which, and the contact with, have severely limited. For example. coffee decaffeination, the remova] of cholesterol from egg yolk with CO2! the production of vanilla extract, and the destruction of undesirable organic compounds all can take place using supercritical water. Oxidation in supercritical water has been shown to destroy 99.99999% of all of the major types of toxic agents present in chemical weapons, which are being eliminated. 14 .. 3 Reduced Variables Other terms with which you should become familiar are the reduced variables. These are corrected, or normalized, conditions of temperature. pressure, or volume, normalized (divided) by their respective critical conditions. as follows: T Tr =- p p, = Pc A V V, = -;;:- Vc In theory, the law of corresponding states indicates that any compound should have the same reduced volume at the same reduced temperature and reduced pressure so that a universal gas law might be (14.1) Unfortunately Equation (14.1) does not make accurate predictions universally. You can check this conclusion by selecting a compound such as water, applying Equation (14.1) at some low temperature and ~igh pressure to calculate V, and comparing your results with the value obtained for V for the corresponding conditions from the steam tables that are the folder in the back of this book. 14.4 Compressibili How can the ideas presented above be used? One common way to modify the ideal gas law by inserting an adjustable coefficient z, the compressibility factor, a factor that compensates for the nonideality of the gas, and can be looked at as a !
- 470. Sec. 14.4 Compressibility 439"· measure nonideality. Thus, the ali zed equation of state. gas law turned into a gas law. a gener .. pV=znR 04.2) or " pV = zRT (14.2a) One way to look at z is to consider it to be a factor makes Equation (14.2) an equality. Although we treat only in this chapter, the same idea has applied to liquids. If you plan on using Equation (14.2), where can you find the values of z to use in it? Equations exist in the literature specific compounds and classes of compounds, such as those found in petroleum refming. Refer to the references at the end of this chapter. But you will find graphs or tables of z to be convenient sources for 1.4,...------------, Temperature" 100· C PJeS5ure., atmospheres (0) Reduced pressure. P, ( b) Figure 14.3 (a) Compressibility factor at 100DC for several gases as a function of pressure~ (b) compressibility factor for several as a function of reduced temperature and reduced pressure.
- 471. 440 Real Gases: Compressibility Chap. 14 values of z. If the compressibility factor is plotted for a given temperature against the pressure for different like Figure 14.3a result. However, if the compressibility is plotted against the reduced pressure as a function the reduced temperature~ then for most the compressibility values at the same reduced temperature and reduced pressure faU at about the same point. as illustrated in 14.3b, 14.5 Compressibility Charts Because of the generaHzation feature shown in 14.3b~ you can use what is called the generalized compressibility factor z for your gas calculations. Figures 14.4a and 14.4b shows two examples of the generalized compressibility factor charts, or z-factor charts. prepared by Nelson and Obert.'" These charts are based on data for 30 gases. 14.4(a) represents Z for 26 gases (excluding H2, He, NH3• and H20) with a maximum deviation of 1 %, and H2 and H20 within a deviation of 1.5%. Figure 14.4(b) for nine gases and errors can be as high as 5%. Note that the vertical axis is not z but zTr in Figure 14.4(b). To use the charts for H2 and He (only). make corrections to the aetna1 constants pseudocritical constants 0.801----+----1-- 0.70 I---I--+-+---+--I---I--+-+-~....q 0.601---I--+-+--+--1---I--+-+-- 0.50 t--t---+-t--+----1f--+-+-+--+--+-+--1-----i-+----I--.J:->...:.-I--+~ Reduced pressure, P, Figure 14.4a Generalized compressibility chart for lower function of p,., T,., and it '/ showing z .as a "Ne!son. and 14.4(b) include and E.P. Obert. Chern. Eng .• v. 61. No.7, pp. 203-208 (1954). Figures 14.4(a) reported by P.E. Liley. Chern. Eng .• p. 123 (July 20. 1987). !
- 472. Sec. 14.5 Compressibility Charts 441 0.0 1.0 1.5 2..0 7.0 7.5 Figure 14.4b Generalized compressibility chart higher values of T' c = T c 8 K p' c = Pc + 8 atm and then you can use Figures 14.4a and 14.4b for these two critical con~ts as replacements for their true values. You will fwd these two charts and a itional charts for other ranges of p y and Ty on the CD this book in fonnat that can be expanded to get better accuracy. Instead/ of the specific volume, a third parameter shown on the charts is the dimensionless volume defined by where V" ej ume), .and is A Both V c. and V r. are I I estimated for a compound. A V V -- y. - A I Ve. I volume (not the experimental value of the critical vol- A RTc V =- Cj Pc ( 14.3) Tc and Pc are presumed known or can development of the generalized compressibility
- 473. 442 Real Gases: Compressibility Chap. 14 charts is of considerable practiool as well as pedagogical value because their existence enables you to make engineering calculations with considerable ease~ and also pennits the development of thermodynamic functions for gases for which no experimental data are available. Frequently Asked Questions " 1. What is in the blank region in Figure 14.4a below the curves for Tr and V 7,? The blank re-gion corresponds to a different phase-a liquid. A 2. Win pV = zRT work for a liquid phase? Yes, but relations to calculate z accurately are more complex than those for the g~s phase. Also, liquids are not very compressible so that for our purposes we can ignore p-V-T relations for liquids. A 3. Why should I use pV = zRT when I can look up the da,!a needed in a handbook or on the Web? Although considerable data exists, you can use pV = zRT to evaluate the accuracy of the data and" interpolate within data points. If you do not have data in the range you want. use of pV = zRT is the best method of extrapolation. Finally, you may not have any data for the gas of interest. EXAMPLE 14.1 Use of the Compressibility Factor to Calculate a Specific Volume In spreading liquid ammonia fertilizer. the charges for thaemEt of NH3 used are based on the time involved plus the pounds of NH3 inject nto the soil. After the liquid has been spread, there is still some ammonia left in e source tank (volume:::: 120 ft3), but the fonn of a Suppose that your weight tally, which is obtained by difference, shows a net weight of ] 25 Ib of NH3 left'in the tank at 292 psig. Because the tank is sitting in the sun, the temperature in the tank is 125°E Your boss complains that his calculations show that the specific volume of the NH3 is 1.20 ft3 lIb, and hence that there are only 100 Ib of NH3 in the tank. Could he be correct? See Pi gure E 14.1 Figure E14.1 Solution Basis: 1 Ib of NH)
- 474. Sec. 14.5 Compressibility Charts You can apply p V = wRT to calculate n and check the amount of NH3 in the Apparently, your boss used ideal law in getting his figure of l. ft 3ltb NH3 (psia) (ft3) R = 10.73 (lb mol)C'R) p = 292 + 14.7 = 306.7 psia T= + 460 = 585°R lib n=---- ii1bJ1b mol 1 7( 10.73)( ) nRT t t V = = ~---- = 1.20 ft3 per Ib p 306.7 However, he should have included the compressibility factor in gas law because NH3 does not behave as an ideal gas under the observed conditions temperature and pressure. Let us again compute the mass of in the tank, this time using =mRT You know all of the values of the variables in the equation except z. addi· tiona! infonnation needed (taken from Appendix or the CD) is Tc = 405.5K ~ 729.9°R = 111.3 atm ~ 1636 psia z is a function of T, and p,... T, T ---- 0.801 729.9" R 306.7 p - -~--=O.l87 r - Pc - 1636 psia From the Nelson and Obert chart. Figure 14.4(a), you can read z ~ 0.855. (The value may be somewhat in error because ammonia was not one of the gases included in preparation the figure.) Now V can be calculated the ratio of pV real::: Zreal nRTto pVideal = Zidea1 nRT, the net result of which is V"'real lreal ideal ,=~- On the of lIb NH3 tireal = 1.20 ~t: ideal O.~55 = L03 ft3 IbNH3
- 475. On the basis of Real Gases: Compressibility ft3 in the tank 1 Ib NH3 120 ft3 = 1171b NH3 1.03 Chap. 14 Certainly 117 Ib a more realistic figure than 1001b, it is easily possible to be in error by 8 Ib if the residual weight of NH3 in the tank determined by dif~ ference. As a matter of interest. as an alternative to miling the calculations, you could look the specific v21ume of NH3 at the conditions in the tank in a handbook. You would find that V = 0.973 ft3l1b, and hence the compressibility factor calculation yielded a volume an error of about 5.9% versus an error of 23% using the ideal EXAl1PLE 14.2 Use of the Compressibility Factor to Calculate a Pressure Liquid oxygen is used the steel industry, in the chemical industry, in hospitals, as a rocket fuel oxidant, and for wastewater treatment as well as many other applications. A hospital tank 0.0284 volume is filled with 3.500 kg of liquid O2 that will vaporize at -25°C. Will the pressure in the tank exceed the safety limit of the tank specified as 1 Q4 kPa? Solution Basis: 3.500 kg O2 You can find from Appendix D or CD that for oxygen Tc': 154.4 K Pc 49.7 atm e 5,035 However, you cannot proceed to solve this problem in exactly the same way as you did the preceding problem because you do not know the pressure of the O2 in the tank to begin with. Thus, you to use the pseudoparameter, V'I' that available as a parameter on the Nelson and Obert charts. First, calculate A , 0.0284 m3 V(speclfic molal volume) = 3.500 kg 1 kg mol;::: 0.260 m3lkg mol Note that the specific molar volume must be used in calculating V'j since V C; is a volume per mole. 154.4 K - -----'-........,;----'--.;... 5,035 kPa = 0.255 Pc
- 476. Sec. 14.6 Calculating the Compressibility Using the Pitzer Factors & 445 Then A - ~ = 0.260 = 1 02 V,; -" 0.255 . Vel you know the values of two parameters, V'j and = 248 K = 1.61 T, 154.4 K From the Nelson and Obert chart (Figure 14.4(b» you can read p,::::: 1 Then p:::: PrPc :::; 1.43 (5,035) :::; 7200 kPa The pressure of 1 ()4 kPa will not be exceeded. Even at room temperature the pressure wiU be less than 1 Q4 kPa. 14.6 Calculating the Compressibility Factor Using the Pitzer Factors zfJ and Z1 Several methods have appeared in the literarure and in computer codes to calculate z via an equation in order to obtain more accurate values of z than can be obtained from charts. Equation (14.4) employs the Pitzer acentric factor, w z: = zO + Z1 W (14.4) where zO and z1 are listed tables in Appendix C as a function of and PI" and w is unique for each compound, and can be found in the CD that accompanies this book. Table 14.1 is an abbreviated table of the acentric factors from Pitzer. The acentric factor w indicates the degree of acentricity or nonsphericity of a molecule. For helium and argon, w is equal to zero. For higher molecular weight hydrocarbons and for molecules with increased polarity, the value of w increases. Table 14.2 lists the value of Z obtained for ethylene (C2 H4) at two conditions by three different methods: (a) by Equation (14.4), (b) from the generalized compressibility charts, and (c) from the ideal gas law. The three values are compared with the experimental value (from Perry's Handbook. 7th edition). If you redo Example 14.1 using Equation (l4.4) with the Pitzer acentric factor for Pr = 0.187 and = O.801~ by linear interpolation of data in the tables in Appendix C, you will find that ZO = 0.864 and ,1 = -0.103 so that z = 0.838, and the Ib of NH3 = 119. If you want to redo Example 14.2 using Equation (14.4), you win have
- 477. 446 Real Gases: Compressibility Chap. 14 TABLE 14.1 Values of the Pitzer· Acentric Factor '. Compound Acentric Factor Compound Acentric Factor Acetone 0.309 Hydrogen sulfide 0.100 Benzene 0.212 Methane 0.008 Ammonia 0.250 Methanol 0.559 Argon 0.000 n-butane 0.193 Carbon dioxide 0.225 n-pentane 0.251 Carbon monoxide 0.049 Nitric oxide 0.607 Chlorine 0.073 Nitrogen 0.040 Ethane 0.098 Oxygen 0.021 Ethanol 0.635 Propane 0.152 Ethylene 0.089 Propylene 0.148 Freon-12 0.176 Sulfur dioxide 0.251 Hydrogen -0.220 Water vapor 0.344 ·Pitzer, K.S., J. Am. Chem. Soc. 77 (1955):3427. to use a trial-and-error solution because you do not know p at the start. What you do is to assume a sequence of values of p, calculate the related sequen£e of values of p,.. and next calculate the associated sequence of values of z and then V. When you find the value of the specific volume that matches the value specified in the problem, you have identified the correctp,.. and can then calculate p = PcPr TABLE 14.2 A Comparison of Values of the Compressibility Factor z for Ethyleoe Determined via Three Different Methods with the Experimental Values Specified CondJtioos* At 350K and S atm At 300 K and 30 attn z % deviation Z % deviation Experimental 0.983 0.812 Compressibility chart 0.982 0.0 0.815 0.0 Ideal gas law 1 1.8 1 23.1 Equation (14.4) 0.983 0.0 0.812 0.0 zO = 0.983 zO = 0.812 zl = 0.00476 zl =0.0131 ·w = 0.089, Tc = 282.8K, Pc = 50.5 atm, zO and ,I are from the tables in Appendix C. ·Kay. W.B. "Density of Hydrocarbon Gases and Vapors at High Temperature and Pressure," Ind. Eng. Chem. 28, 1014--1019 (1936).
- 478. Sec. 14.7 Real Mixtures 447 14.7 Real Gas Mixtures How can you apply the concept of compressibility to problems involving gas mixtures? Each component in the mixture will have different critical properties. Numerous ways have been proposed to properly weigh the critical properties so that an appropriate reduced temperature and pressure can be used to obtain z. Refer to the references at the end of this chapter for some examples. One simple way that is reasonablyaccurate, least for our purposes, is Kayts method.· In Kay's method, pseudocritical values for mixtures of gases are calculated on the assumption that component in the mixture contributes to the pseudocritical value in the same proportion as the mol fraction of that component in the gas. Thust the pseudocritical values are computed as foHows: I _ Pc - PCj,YA PcBYs (14.5) T'c = TCj,YA TcyYB (14.6) where Yi is the mole fraction, piC is the pseudocriticaI pressure and T~ is the pseudocritical temperature. You can see that these are linearly weighted mole pseudocritical properties. As an example of how the averaging works, look at Figure 1 which compares the true critical values of a gaseous mixture of and SOl with the pseudocritical values. respective pseudoreduced variables are p'r = I Pc 100~----------------------------------~ 90 60 -E 70 I:) t{ 60 50 40 Locus of actual cri ticol points for mixtUres of C02 and 502 --7-- Locus of pseudocriticol poin1s for mixtures of C02 and 502 compufed by Koy1s rule Vapor pressure curves " "C For 60°/" 502 - 40"/" CO2 mixture 200 Figure 14.5 Critical and pseudocritical points for mixtures of COl and SOl-
- 479. 448 Real T T' r =c Compressibility Chap. 14 Kay's method is known as a two-parameter rule only Pc and Tc for each component are involved ill the calculation of z. If a third parameter such as or the Pitzer acentric factor, or V ci is included in the detennination of the compressibility factor, then you would have a three-parameter rule. Other pseudocritical methods with additional parameters provide better accuracy in predicting p-V-T properties than Kay's methodt but Kay's method can suffice for our workt and it is easy to use. EXAMPLE 14.3 Calculation of p .. v-T Properties for a Real Gas Mixture gaseous mixture has the following composition (in Methane, CH4 20 Ethylene, C2H4 30 Nitrogen, Nz 50 percent): at 90 atm and 100°C. Compare the volume per mole as computed. by the methods of: (8) the ideal gas law and (b) pseudoreduced technique (Kay's method). Solution Basis: 1 g mol of mixture Additional data needed are: Component 191 283 1 Pc (atm) 45.8 50.5 .5 (cm3)(atm) R = 82.06 (g mol)(K) 8. Ideal gas A Kl' 1(82.06)(373) V = P = 90 == 340 cm3/g mol at 90 atm and 373 K b. According to Kay's method, you for the mixture calculate the pseudocritical values pIC == PC,J,YA + PcsYB + = 41.1 atm = (45.8)(0.2) + (50.5)(0.3) + (33 (0.5)
- 480. Sec. 14.7 Real Gas Mixtures T'c Tc,.,YA + TcnYB + TccYc = (191 )(0.2) + (283)(0.3) + (126)(0.5) = 186K Then you calculate pseudoreduced values for the mixture p 90 P' = - = = 219 r p' c 41.2 ., T T'r = T' = 186 = 2.01, c With the aid of these two parameters you can find from Figure 14.4b tha.t zT~ = 1.91 and thus z = 0.95. Then zRT 0.95( 1 )(82.06)(373) - = = 323 cmJ/g mol at 90 atm and 373 K p 90 ... V If you decided to use Equation (14.4) to calculate z fOf the mixture. how might you average /J, . and w? " 449 In instances in which the temperature or pressure of a gas mixture is unknown, to a trial-and-error solution using the generalized compressibility charts. you can compute pseudocritical ideal volume and a pseudoreduced ideal volume Vr., thus j " RT' C VI c· -- -,- I Pc and A 1 _ V V ,. - -::::- , V Cj V;. can I used in lieu of P / r or r the compressibility charts. SELF .. ASSE SMEN T ST Questions 1. What is pseudocritical volume? What is the advantage of using V c.? I 2. Answer the following questions true or a. Two fluids, which have the same values of reduced temperature and pressure and the same reduced volume, are said be in corresponding states. b. It is expected that aU gases will have the same z at a specified and Pr Thus a COtTelation of l in tenns of Tr and P r would apply to all c. The Law of Corresponding States states that at critical state (Tc' Pc) all substances should behave alike. d. The critical state of a substance is the set of physical conditions at which the density and other properties of the liquid and vapor become identical. e. Any substance (in theory) by the Law of Corresponding States should have the same reduced volume at the same reduced T,. and p,.
- 481. 450 Real Gases: Compressibility Chap. 14 f. The equation pV = znRT cannot be used for ideal gases. By definition a fluid becomes supercritical when its temperature and pressure exceed the critical point. h. Phase boundaries do not exist under supercritical conditions. I. For some under normal conditions, and for most gases under conditions of high pressure. values the gas properties that might be ob~ained using the ideal Jaw would be at variance with the experimental evidence. 3. Explain meaning the following equation for the compressibility factor z = f(T,., p,) 4. What is the value of z at p = O? Problems l 1. Calculate the compJeSSibiHty factor z. and determine whether or not the following gases can be treated as iJeal, the listed temperature and pressure: (a) water at 1,OOO°C and } 2.000 kPa; (b) oxygen at and 1,500 kPa~ and (c) methane at 10°C and 1.000 kPa. 2. In a proposed low pollution vehicle burning H2 with °2, the respective gases are to be stored in two separate tanks 2000 psia. The vehicle has to operate from -40 to 130oP. a. Is ideal gas Jaw a sufficiently good approximation to use in the design of these tanks? b. A practical operating range that 3 Ibm of hydrogen be stored. How large must the hydrogen tank be if the is not to exceed 2000 psia? c. The Hi02 ratio is 2 on a molar basis in the proposed fuel. How must the oxygen tank be? 3. A carbon dioxide frre extinguisher has a volume of 40 L and is to be charged to a pressure of 20 atm at a storage temperature of 20°C. Detennine the mass in kilograms CO2 at 1 atm. 4. Calculate the pressure of 4.00 g mol CO2 contained in a 6.25 X 10-3 m3 fire extinguisher at 25°C. 5. One pound mole of a mixture containing 0.400 Ib mol N2 and 0.600 Ib mol C2H4 at 50°C occupies a volume of 1.44 ft3. What is the pressure in the vessel? Compute your an· swer by Kay's method. Thought Problems 1. At a swimming pool the pressure regulator attached to the chlorine cylinder did not seem to function properly. the valve on the cylinder was closed. After standing for 8 months someone disconnected the regulator to replace it, and chlorine gas spurted into his face. He and other people had to hospitalized. What happened to cause the dent? Pressure vessels and rigid piping have to be protected against overpressure by using safety devices. For example, when liquid is trapped within a rigid piping system, it ex"
- 482. 14.7 Real Gas Mixtures 451 pands. and a small expansion by a temperature increase will produce a large pres~ sure rise in the in the vapor above the liquid. What happens in a space containing 400 L of liquid and a bubble of 4.0 L with an initial pressure of 1 atm when the liquid expands 1 % so that the volume is compressed? Will the piping fail? 3. The sum of the mass fractions for an ideal gas mixture equal to 1. Is the sum equal to 1 for a real mixture? 4. Municipal is being to sterile and biodegradable liquid efflu-ent in a mile-deep well at Longmont, Colorado. The reduces chemical demand, that is, the oxygen to oxidize organic compounds the sludge, by up to 68%, and destroys living organisms. Tubes of appropriate are in the to create an annual space two-phase down and up the wel1. Oxidation takes place at the bottom of the well with air. Why is it advantageous to oxidize the sludge at the bottom of the well than in a pond at the Discussion Problems 1. A letter to (he editor was headed" Not Sold on Hydrogen." In part it Your innovative story "Fuel Cells: A Lot Hot Air?" concerned me. drogen under pressure is difficult to contain. Leaks are difficult to detect. and you need to virtual zero leakage. Hydrogen is explosive and flammable, and bums with an invisible flame. Finally. your economics do not address the entire process from beginning to end, and come to Jess than realistic conclusions. Comment on points that author of the makes. Do they damage the potential of hydrogen-based fuel cells? In what aspects is he correct, and in what wrong? 2. In the 1970s it was believed that geopressured brines might appreciable po-tential for electric power production because the thermal and kinetic energy of the waters approximately 50% of the total energy in the fluids. However, field tests have indicated that temperatures at 15.000 feet are usually in the of 275 to 300°F, mak-ing electricity production However, the brines are saturated with natural Ten test wells have been tested in coastal Texas and Louisiana at an average cost $2 million each. Test results have mixed. Brine salinities have ranged from 13,000 to 191,000 ppm. Most tests indicate at or near saturation levels. Temperatures ranged from 237 to 307°F and forma-tion pressures from 11.050 to 13,700 psia. Flow rates have from 13,000 to 29,000 barrels day of water. Because natural gas content is proportional to temperature and pressure but inversely proportional to salinity. the natural gas in design wells "_",~_" from 19 to 50 standard feet per barrel. All of the wells have indicated some carbon dioxide (C02) in the produced. The concentration seems to be correlative with temperature, ranging 4 to 9% of total content. The CO2 has potential to cause scaling problems, requiring use of inhibitors. Reservoir limits are very large. Transient pressure testing on two wells indicates no barriers (0 an outer limit of about 4
- 483. 452 Real Gases: Compressibility Chap. 14 Is recovery natural gas from such weBs a viable source of energy? report that includes economic, engineering. and environmental considerations pro-posed gas production. Looking Back In this chapter we described the law of corresponding states and reduced variables of temperatures and These led to the idea com-pressibility factor that could be in the ideal gas law to generalize it applica-tion to real Both equations were used to estimate z. In addition~ we described how to apply Kay's method (the pseudocritical method) to predict p. V, n, and Tin reat mixtures. GLOS ARY OF N W WORDS Acentric factor A parameter that indicates the degree of nonsphericity of a molecule. Compressibility charts of the compressibility factor as a function of re-duced temperature. reduced pressure. and ideal reduced volume. Compressibility factor factor that is introduced into the ideal ~Vll"'(,l~"'" for the nonideality of a gas. Corrected Normalized. Corresponding states Any gas should reduced temperature and reduced pressure, same reduced law to com-at the same Critical state The set of physical conditions which the density and other of liquid and become identical. Generalized compressibility chart See Compressibility charts. Generalized equation of state The ideal law converted to a real gas law by inserting a compressibility factor. Ideal critical volume '" i = RTr ! p c " A Ideal reduced volume V r, i = V N c, i Kay's method for calculating the compressibility factor for a mixture of Law of corresponding states See Corresponding states. Pitzer acentric factor Acentric
- 484. Sec. 14.7 Real Gas Mixtures 453 Pseudocritical Temperatures. pressures, and lor specific volumes adjusted to be used with charts or equations used to calculate the compressibility factor. Real whose behavior not conform the assumptions underlying ideality. Reduced variables Corrected or normalized conditions of temperature, pressure, and volume, normalized by their respective critical conditions. Supercritical fluid Material in a state above its critical point SUPPLEMENTARY REFERENCE In addition to the references listed in the Frequently Asked Questions in the front material, the following are pertinent. Castillo, C. A. HAn Alternative Method for the Estimation of Critical Temperatures of Mix-tures H • J" 33, 1025 (1987), Compressed Gas Association~ Inc. Handbook of Compressed Gases. Van Nostrand Reinhold, New York, 1990. Daubert, T. E., and R. P. Danners, eds. Data Compilation Tables of Properties of Pure Compounds. New York. AlChE. 1985 and supplements. Dean, 1. A., . Lange's Handbook of Chemistry. 14th ed .. McGraw-HilI. New York, 1992. Department of Labor. Occupational Safety, and Health Administration. Process Safety Management of Highly Hazardous Chemicals Compliance Guidelines and Enforcement Procedures. OSHA Publishing. Washington, 1992. Edmister, Wayne C. Applied Hydrocarbon Thennodynamics, 2nd ed. Vol. i, 1984; VoL Gulf Publishing, Houston. 1988. EI-Gassier. M. M. "Fortran Program Computes Gas Compression," Oil Gas 1987). 88 (July Gomes, J. "Program Calculates Critical Properties," Hydrocarbon Processing, 67, 110 (September i 988). Lyderson. A. R. A. Greenkom, and O. Hougen. Generalized Thermodynamic Proper-ties of Pure Fluids. Univ. Wisconsin Expt. Station. Madison (1955). Polling, B. E., J. M. Prausnitz. and J. P. O'Connell. Properties of Gases and Liquids. 5th ed., McGraw-Hm. New York (2001). Selover, T. B. t ed. National Standard Reference Data Service of the U.S.S.R. Hemisphere Publishing. New York, 1987. Simmrock, et at Critical Data of Pure Substances, 2 parts, v. Dechema, Frankfurt (1986). Sterbacek, B. Biskup, and Calculation of Properties Using Corresponding State Methods. Elsevier Scientific, New York (1979).
- 485. 454 Real Gases: Compressibility Chap. 14 Teja, A.S., and P. Rice. "A Multifluid Corresponding States Principle for Thermody-namic Properties Fluid Mixtures," Chem. Science. ~, 1-6 (1981). Yaws. c.L., Chen. H.C. Yang, Tan, and D. Nico: "Critical Properties of Chemicals," "14.6 *14.7 Hydrocarbon Processing, 68, 6J (July 1989). PROBLEMS Seven pounds 0' N2 are in a cylinder 0.75 ft3 volume at 120cF. Calculate the pressure in the cylinder in atmospheres: a, Assuming N2 to be an ideal gas. b. Assuming N2 is a real gas and using compressibility factors. Two gram moles of ethylene (C2H4) occupy 418 em3 at 95°C. Calculate the pressure. (Under these conditions ethylene is a nonideal gas.) Data: = 283.1K, Pc = 50.5 atm, Ie = 124cm3/gmol. = 0.270. The critical temperature of a real is known to be 500 K, but its critical pressure is unknown. Given that 3 Ib mol of the at occupy 50 ft3 at a pressure of 463 psia. estimate the critical pressure. The volume occupied by 1 Ib of n-octane at atrn is 0.20 ft3. Calculate the ture of the n-octane. A block of ice weighing 50 Ib was dropped into an empty steel tank. the volume of which is 5.0 ft3. The tank was heated until the pressure gage read 1600 psi. What was the temperature of the gas? Assume an of the CO2 became gas. cyUnder containing 10 of CH4 exploded. It had a bursting pressure J 4.000 kPa gauge and a safe operating pressure of 7,000 kPa gauge. The cylinder had an terna! volume of 0.0250 m3. Calculate the temperature when the cylinder exploded. A cylinder has a volume of 1.0 ft3 and contains dry methane at gO°F and 200 psig. What weight of methane (CH4) is in the cylinder? The barometric pressure is 29.0 m. Hg. -14.8 How many kilograms can put into a 25 liter cylinder at room temperature (25DC) and 200 kPa absolute pressure? "'14.9 A natural gas composed of 100% methane is to be stored in an underground reservoir 1000 psia and 120oP. What volume of reservoir is required for 1,000,000 cubic feel of measured at 60°F and psia? *14.10 Calculate the specific volume of propane at a pressure of 6000 kPa and a temperature of 230°C. "'14.11 State whether or not foHowing gases can be treated as ideal gases in calculations. (a) Nitrogen at 100 kPa and 25°C. ____ (b Nitrogen at 101000 kPa and (c) Propane at 200 kPa and 25°C. (d) Propane 2000 kPa and II
- 486. Chap 14 Problems 455" (e) Water at 100 kPa 25°C. (0 Water at 1000 kPa and (g) Carbon dioxide at 1000 kPa ~OC. (h) Propane at 400 kPa and DoC. -14.12 One g of chlorobenzene (C6HsCl) just fills a tank at 230 kPa and 380 K. What is the volume of the tank? ··14.13 A size 1 cylinder of ethylene ::: 9.7°C) costs $45.92, F.O.B., New Jersey. The outside cylinder dimensions are: 9 diameter, 52 high. The gas 99.5% (mini-mum) C2H4! and the cylinder charge is $44.00. Cylinder pressure is 1500 psig, and the invoice says it contains 165 cu-ft. An identical cylinder of CP grade methane at a pressure of 2000 psig is 99.0% (minimum) CH4, and costs $96.00 Illinois. CH4 cylinder contains 240 cu. ft. of The ethylene cylinder is supposed to have a gross weight (including cylinder) of 163 Ib while the CH4 cylin-der a weight of 145 lb. Answer the foHowing questions: (a) What does the "165 cu,ft." and "240 cu.ft." of gas probably mean? Explain with calculations. (b) Why does CH4 cylinder have a gross weight less than the C2H4 cylinder when it seems to contain more Assume the cylinders are at 80oP. (c) How many pounds of are actually in each cylinder? ...... 14.14 You have been asked to settle an argument. argument concerns the maximum al-lowable working pressure (MWAP) permitted in an Al gas cylinder. of your coworkers says that calculating pressure in a tank via ideal law is best because it gives a conservative (higher) value the pressure than can actually occur in the tank. The other coworker says that everyone knows the ideal gas law should not be to calculate gas as it gives a lower value than the true Which coworker is correct? u14.15 practices in modern laboratories call for placing gas cylinders hoods or in util~ ity corridors. In case leaks, a toxic gas can properly taken care of. A cylinder of CO, that has a volume of 1 ft3 is received the distributor of on Friday with a gauge reading 2000 psig and placed in the utility corridor. On Monday when you are ready to use the you the reads 1910 The tempera-ture has remained constant at 76°F as the corridor is air conditioned so you conclude that the tank has CO (which not smell). (a) What has been the leak rate from the tank? (b) [f tank was placed in a utility corridor whose volume is 1600 ft3• what would be the minimum time that it would for CO concentration in the hallway to reach the UCeiling Threshold Li mit Value" (TL V -C) of 100 ppm set by the state Pollution Control Commission the air conditioning did not operate on weekend? (c) In the worst case, what would be the concentration of CO the corridor if the leak continued from Friday, 3 PM to Monday, 9 (d) Why would either case (b) or (c) occur in practice?
- 487. 456 Real UCl,;)C;:) Compressibtlity Chap. 14 *14.16 Levitating solid materials during processing is the way known to ensure their purity. High-purity materials, which are in great demand. in electronics, other areas, usually are produced by melting a solid. Unfortunately, the containers used to hold the material tend to contaminate it. heterogeneous nucleation occurs at the container walls when molten material is cooled. Levitation avoids these problems because the material processed is not in contact with the container. Electromagnetic levitation requires that sample be electrically conductive, but with a levitation method based on buoyancy, the density of the is the only limiting factor. , ...... ,....,,, that a such as argon is to be compressed at room temperature so that silicon (sp gr 2.0) floats in the What must the the argon If wanted to use a lower pressure, what different gas might be selected? Is there a limit to the processing temperature for this manufacturing strategy? "·14.17 determine the temperature that occurred in a fITe in a warehouse, the arson investi-gator noticed that the val ve on a methane tank had popped open at 3000 psig. the rated value. Before the fire started, the tank was presumably at ambient con-ditions, about and the read 1950 If the volume of the tank was 240 , estimate the temperature the flre. List any assumptions you *14.18 When a scuba diver goes to the dive shop to have her scuba tanks fined with air the tank is connected to a compressor and filled to about 2100 psia while immersed a tank of water. (Why immerse the tank in water?-So that the compres~ion of air into the tank will appro~imately isothermaL) Suppose that the tank filled without inserting it into a water bath, air at 27°C is compressed rapidly from 1 atm absolute to the same final pressure. final temperature would about Compute fractional or r'lpr'fi"~:~"" in the final quantity of air put in the tank relative to the isothennal case assuming that air behaves as a pure component real gas with Pc = 37 and Tc = 1 Use 0,036 for the Pitzer ascentric factor. *14.19 A gas is flowing at a rate cubic feet/per hour). What is actual volumetric flow rate if pr alm and temperature is 6oo0R? The critical temperature is 40.0° and critical pressure is 14.3 atm. ·14.20 A steel cylinder contains ethylene (C2HJ at 200 psig. The cylinder and weigh 222 lb. The supplier refills the cylinder with ethylene until the pressure reaches 1000 psig, at which time the cylinder and weigh 250 lb. The temperature is constant at 25°C. Calculate the to be made for the ethylene the ethylene is sold at I pound, and what the weight of the cylinder is for use in' billing the freight charges. Also find the volume the empty cylinder in cubic feet. "'14.21 A gas following composition: 10% 40% 50%
- 488. Chap 14 Problems It is desired to distribute 33.6 Ib of this gas per cylinder. Cylinders are to be designed so that the maximum pressure will not exceed 2400 psig when the temperature is 180°F. Calculate the volume of the cylinder required by Kay's method. $14.22 A gas composed of 20% ethanol and 80% carbon dioxide is at 500 K. What is its pressure if the volume per g mol is 180 cmJ/g mol? ·14.23 A sample of nawral gas taken at 3500 kPa absolute and 120°C is separated chro-matography at standard conditions. It was found by calculation that the grams of each component in the gas were: Component (g) Methane (CH4) [00 Ethane (C2H6) 240 Propane (C3Hg) 150 Nitrogen (N2) 50 Total 540 What was the density of the original gas sample? '*14.24 A gaseous mixture has the following composition (in mol %): 57 40 3 at 120 pressure and 25°C. Compare the experimental volume of 0.14 Ug mol with that computed by Kay's method. ~2S You are in charge of a pilot plant using an inert atmosphere composed of 60% ethyl· ~e (C2H4) and 40% argon (A). How big a cylinder (or how many) must be purchased if you are to use 300 ftl of gas measured at the pilot plant conditions of 100 atm and 300oP. Buy the cheapest array. Cylinder type lA 2 3 cost $52.30 42.40 33.20 pressure (pslg) 2000 1500 1500 Ib gas 62 47 3S State any additional assumptions, You can buy only one type of cylinder. A feed for a reactor has to be prepared comprised of 50% ethylene and 50% nitrogen. One source of gas is a cylinder containing a large amount of gas with the composition 20% ethylene and 80% nitrogen. Another cylinder that contains pure ethylene 1450 psig and 700P has an internal volume of 2640 cubic inches. If aU the ethylene in the latter cylinder is used up in making the mixture, how much reactor feed was prepared and how much of the 20% ethylene mixture was used?
- 489. Real Compressibility Chap. 14 In a high pressure separations process, a gas having the mass composition of 50% benzene, 30% toluene, and 20% xylene is fed into the process at the rate of 483 m3lhr at 607 K and 26.8 atm. One stream a vapor containing 91.2% benzene, 7.2% toluene, and 1.6% xylene. A second exit stream is a liquid containing 6.0% benzene, 9.0% toluene, and 85.0% xylene. What is the composition of the third stream if it is liquid flowing at the rate 9,800 kglhr, and the ratio of the benzene to the xylene in the stream is 3 kg benzene to 2 kg xylene?
- 490. CHAPTER 15 EAL GASES: EQUATIONS OF STATE Your objectives in studying this chapter are to be able to: 1. Cite two reasons for using equations of state predict p- V-T properties of gases. 2. Solve an equation of state the of the coefficients in equation and values of three ot the four variables in the p, V, n, and T. Convert the coefficients in an equation of from one set of units to another. 4. the values of the coefficients for equations of in the literature on the Internet. In making calculations with the p- V-T properties a gas you would like to have accuracy. reliability, and computational efficiency. In Chapter 14 you learned that compressibility factor z can be introduced into the ideal equation so that the latter can be used to predict properties of real gases. Is there another method to predict physical properties? Read on. looking Ahead In chapter we describe how equations derived from theory or experiment can be fit from experimental data and used to sol ve for V. n, or T for real JiioI"'~''''''U' Main Concepts Although this may seem a paradox, all exact science dominated by the idea of approximation. Bertrand Russell 459
- 491. 460 Real .... u. ............. Equations of State Chap. 15 We now describe another way of predicting p, V. n, and T for gases (either pure components or mixtures), one that differs from the method discussed in Chapter 14, namely by using equations of state. The simplest example of an equation of state the gas law itself. Equations of can just empirical relations se-lected to fit a set, or they can be based on theory. or a combination of the two. Even if you have experimental data availablet in of their complications, equations of are important for several reasons. They permit a concise summary of a large mass of experimental data and also pennit accurate interpolation between experimental data points. They provide a continuous function to facilitate calculation of physical properties based on differentiation and integration of p-V-T relation- Finally, they provide a point of departure for treatment of the properties of mixtures. Some of the advantages of using an equation of state versus other prediction methods are: 1. of p-V-T can be predicted with reasonable error in regions where no data exist. 2. Only a values of coefficients are needed in the equation to be able diet gas properties versus collecting large amounts data by experiment. 3. The equations can manipulated on a computer whereas graphics methods cannot. Some disadvantages are: 1. fonn of an equation is hard to change to fit new data. Inconsistencies may between equations for p-V-T equations for other physical properties. 3. Usually the equation is quite complicated and may not be easy to solve for p. VI or T because of its nonlinearity T I consider that I understand an equation when I can predict the properties of its solutions, without actually solving it. Paul Dirac If you plan to use a equation of state such as one of those listed in Table 15.1, you have numerous choices, no one of which win consistently the best results. Examine Mathias and Klotz* and the references therein, consult the erences at the end of this chapter, or the Internet you to find a more comprehensive of equations. You can calculate the values of the coefficients in -Mathias P.M., and H.C. Klotz. "Take a Closer Look at Thermodynamic Property Models," Chern. Progress, 67-75 (June (994)
- 492. Chap. 15 Real Gases: Equations of State 461 TABLE 15.1 Examples of Equations of State (for 1 g molet van der Waals: (p ;2)(Y - b) '" RT a = (21) R2T~ . 64 Pc b = (~)RTc 8 Pc Peng-Robinson (PH equation): RT aa P = A A A ~ V - b V(V + h) + h(Y - b) a '" 0.45724( R2r~ ) b '" O.07780( :c) a = [1 K(l - T;12)f K = 0.37464 + L54226a> - 0.26992w2 Benedict-Webb-Rubin (BWR equation): A U pY = . + A "'2 V Y CO f3 RT Bo - Ao - r 2 w "5 V u = bRT - a + ..£. exp( - ! ) T2 y2 '1 = cyexp( - Z2) w=aa A Soave-Redlich-Kwong (SRK equation): RT a'A P=A -"A V - b V(Y + b) 0.42748 a'=---- 0.08664 RTc b=---...;:. Pc A = [l + K(l - T~I2)J2 . K = (0.480 + 1.514w - 0.17w) Redlich-Kwong (RK equation): RT a p = A -- A A (V - b) T1J2V(V b) R2y~.5 a == 0.42148-- Pc RTc b = 0.08664- Pc Kammerlingh .. Onnes (a viriaJ equation): pY '" RT( I + ~ + y2 + ... ) Holborn (8 vinal equation): A pV = RT(l B'p C1p2 ... ) *Tc and Pc are explained in Chapter 14; V is the specific volume; w the acentric factor.
- 493. 462 ReaJ Gases: Equations of State Chap. 15 several of the equations of state from the critical properties plus the acentric factor of a compound as indicated in Table 15 .1. You should note that no adjustable parameters are needed in the van der Waals, Peng-Robinson (PR), Soave-RedlichKwong (SRK), and Redlich-Kwong (RK) equations as they are for the virial equations. Because of its accuracy, the data bases in many commercial process simulators make extensive use of the SRK equation of state. These equations in general will not predict p-V-T values across a phase change from gas to liquid very well. Keep in mind that under conditions such that the gas starts to liquefy, the gas laws apply to the vapor phase portion of the system. How accurate are equations of state? Cubic equations of state such as RedlichKwong, Soave-Redlich-Kwong, and Peng-Robinson listed in Table 15.1 can have an accuracy of 1-2% over a large range of conditions for many compounds. Figure 15.1 compares the percent deviation from experimental data of the predicted z using the SRK equation for (a) a polar compound, namely steam, and (b) a light hydrocarbon (C2H~, the latter being the gas for which the SRK equation should yield better predictions as it does. Other classical equations of state are formulated as a power series (called the " A virial fonn) with p being a function of IN or V being a function of p with 3 to 6 terms. Computer databases offer several choices of equations for most compounds. Such relations may be (but may not be) more accurate than cubic equations of state. Equations of state in databases may have as many as 30 or 40 coefficients to achieve high accuracy (see, for example, the AIChE DIPPR reports that can be located on the AIChE web site). Keep in mind you must know the region of validity of any equation of state and not extrapolate outside that region, particularly not into the liquid region! E ~ 6r--------.........., ~ ~ Tr = 1.0 ..... CIS 00 c - 4 .+o= ~c <tI Q) '> E ~ '':: 2 -c ~)( ~ w 0 L--........... ;::;;;;;..--==:::::....L_~..:.i.J ~ 0 0.4 0.8 1.2 Reduced Pressure (p/pJ a. E 2 10 ..... IIS Pr N_ =0 .391 ..... <tI 8 00 c-. o. ~c 6 ,~ CI,) ~ ,~ 4 oGi -cc .)(. 2 ea> w C2H. 0 &. 0.5 1 1,5 Reduced Temperature (Trrc) b. Figure 15.1 Comparison of the percent deviation in the value of l calculated using the SRK equation for a polar compound (srrearn) and a nonpolar compound (C2H4) from the respective experimental values,
- 494. l Chap. 15 Equations of State 8. -2 b. 4 x 463 Figure IS.2 Gra.ph (a) shows the existence of one positive real root whereas graph (b) shows the of two rea] roots. One feature of the cubic equations of state listed in Table 15.1 com-ment. In solving for n or VI solve a cubic equation that more than one real root, as indicated in 15.2. For example. the van der Waals can easily solved explicitly for p as foHows. p= However, if you cubic in V (or n): V (or n). you can see that the equation becomes f(V) = V3 - (nb and can have multiple roots (see Figure 15.2), You want a positive real root in the gas phase. In Figure 15.2b, the smaller positive root figuratively pertains to the pressure of the liquid. See Pratt'll for a good about the meaning of the multiple roots of an equation of state. On the disk that accompanies this book you will find a computer program caned Polymath that can nonlinear equations, and can be used to solve for V (or n) if you have a reasonable initial guess for V (or n). say from "Pratt, R.M. "Beware of Bogus Roots with 278-281 (Fall 1999). ,. Chern Eng. Educ ..
- 495. f(V) Solution f(V}:: 0 Real Gases: Equations of State Chap. 15 Calculated values of f(V) Figure 15.3 Plot of f( V) versus V showing points above and below f(V) ::: O. The solid line is used in graphical interpolation. the ideal gas law. Refer to Appendix L2 for the methods of solving nonlinear equations. An alternative graphical technique for finding the roots of a function of a sin-gle variable such asftV) is to substitute into the function a series of values of V. and find the value of V that causes the value of I(v) to cross the horizontal axis f(V) = 0 as V changes. Look at Figure 1 You can interpolate for V after bracketingf(V) = o with values above and below O. EXAMPLE 15.1 Application of van der Waals's Equation to Calculate a Temperature A cylinder 0.150 m3 in volume containing 22.7 kg propane C3Hg stands in the hot sun. A pressure shows that the pressure is 4790 kPa gauge. What is the temperature the propane in the cylinder? Use Van der Waals's equation. Solution Basis: 22.7 kg of propane You can obtain the van der Waals constants from any suitable handbook; or by using the critical properties shown in Table 15.1. They are 3 )2 a = 9.24 X 106 atm( cm .gmot em3 b = 90.7-- grool p + (n:~) (V - nb) ~ nRT AU the additional infonnation you need is as follows:
- 496. J Chap. 15 Real Gases: Equations of State p = (4790 + 101.3) kPa IO:.~~a = 48.3 atm , . 82.06(cm3)(atm) R the proper Units = (g mol)(K) 22.7 n = = mol propane 44 kglkg mol [ (0.516 X loJ)2(9.24'X 106)] 6 nRT = 48.3 + [0.150 X 10 , (0.150 X 1 - (0.516 X 103)(90.7)] = (0.516 X 103)(82.06)(T K) T=384 K Suppose you want to apply the PR or SRK equation in solving problem posed in Example 15.1. It may not be immediately obviou$, but both of these equa-tions a value for T in to obtain values for the coefficients in respec-tive and you do not know T. Look at Table I 1. Can you locate where T is involved in calculating the values of the coefficients? You might augment the PR or equation with the equation for A. and solve pair for T to avoid a trial-am: l-eJror solution. Because A does not change with temperature, one or two iterations using assumed values of A should suffice. EXAMPLE 15.2 Solution of van der Waals's Equation for V Given the 'UlU' ........ " a vessel of p = 679.7 psia a = 3.49 X 10· n = L1361b mol b=L for the volume [he vessel using Van der Waals's equation. Solution Wrice van V. Waals's equation as a cubic equation in one unknown f(V) = V3 _ -,-pn_h_+_n_RT_ p / (a)
- 497. 466 Real Gases: Equations of State Chap. 15 Let's apply Newton's (refer to Appendix L2) to obtain the desired root: (b) wherer(Vk) is the derivative ofj{V) with to V evaluated at Vkt and the sub-script k designates the stage of the calculation 2 2(pnb + nRT) n2a r(V}.) = 3Vk - Vk + (c) p p In many cases you can obtain a reasonably <l...,.n' .. nv ....... """n ... to V (or n) from the ideal gas law, useful at least for first trial nRT 1.1361b mol 10.73 (psia)(ft3) Vo = p = (Ibmol)(OR) -- = 12.26 ft3 at 679.7 psia and 683°R The second and subsequent estimates of V will be calculated f(Vo) VI = Vo - !'(Vo) ) 3 (679.7)(1.137)(1.45) + 1.1 f(Vo = (12.26) - 679.7 (b). ( (L 137)2(3.49 X 104) (1.1 + 679.7 ( 12.26) - ...:...------:.--:.....----=-=----=- = 738.3 2 2[( 679.7) l. i' (V 0) = 3 ( 12.26) - ----'----:..--~~-...:...-.-----:....::..---=-..::..-...:...;;. ( 1 (1.137 +...;.-~--.:---~-=- = 216.7 738.3 - 216.7 = 8.85 On the next iteration I(VI ) V2 = VI - !'(Vd and so on until change in Vk from one iteration to the next is sufficiently smalL The Polymath program on the disk in the back of the book will execute this teenfor you (and save you considerable time), The final solution is 5.0 ft3 at 679.7 and
- 498. Chap. 15 Real Gases: Equations of State 467 The group contribution method has been successful in estimating p-V-T properties of pure components (as well as other thermodynamic properties). As indicated by the name, the idea is that compounds can be constructed from combinations of functional groups, the contribution of each group to a property can be tabulated, and the group contributions can be correlated andlor summed to give the desired property of the compound, The assumption is that a group as -CH3, or -OH, behaves identically no matter what the molecule may be in which it appears. This assumption is not quite true, so that any group contribution method yields approximate values for gas properties. Probably the most widely used group contribution method is UNIFAC, * which forms a part of many computer databases. UNlQUAC is a variant of UNIFACt and is widely used in the chemical industry in the modeling of nonideal systems (systems with strong interaction between the molecules). To this point. we have discussed predicting p- properties for pure compo-nents. How should you treat mixtures of gases? An enormous literature describing proposals for mixing rules, that is, rules ~o weight the coefficients for each pure component so that the weighted coefficient can used in the same equation of state as used for a pure component. Even if there is only one possible unified theory it is just a set of rules and equations Stephen Hawking For example, one of mixing rules for the Soave-Redlich-Kwong equation suggests to proceed as follows. Let the weighted coefficient (the overlay bar denotes average) be calculated as n b = LYibi i=1 where n is the number of components in the mixture, i and j are the indices indicating the pairs of components, Yj is the mole fraction of component i, and (al A)ij is the product of af and A for each the possible pairs of components. Other simple methods include: Fredenslund, 1. Grnehling. and Rasmussen. Vapor-Liquid Equlibria Using UNIFAG. sterdam: Elsevier, 1977; D. Tiegs, J. GrnehHng. P. Rasmussen, and A. Fredenslund. Ind. Eng. Chem. Res,. 26, 159 (1987):
- 499. 468 Real Gases: Equations of State Chap. 15 1. A verage the coefficients in the equation by the mole the compo-nent gases. the results of the of the equations for by the mole fractions of the component 3. z for each component from the respective equations, and average 4. coefficients obtained by experiments with mixtures, and interpolate and extrapolate over the variables. For more infonnation on end of the chapter. topic of mixing references at the S LF-ASSESSMENT T T Questions 1. Explain why the van for p and hard to solve and Peng-Robinson ........ "' ..... (of state) are easy to solve v. A Explain some of the in obtaining an equation of f(p, V, T) = 0 for a that is valid for any set of p. V. and T. 3. Under what conditions an equation of state be the accurate? 4. Will equations of state with many coefficients be more accurate than equations state that involve only two or coefficients? How can you get for the coefficients in equations of state given the experimental for p, V. and 6. How can you of the equations of state that are explicit in p to a form in z? 7. What are the units of a and b SI system for Redlich·Kwong equation? Problems 1. Convert the virial (power series) equations of Kammerlingh.:.Onnes and Hotborn (in Table 15.1) to a form that yields an expression for z. 2. Repeat for the Waals and Peng-Robinson equations. A 3. For the Peng-Robinson equation of state, plot p (in MPa) on an arithmetic versus V (in m3/kg) on a loglO showing data and the corresponding predicted values p V for a compound. = 1. C Tr ::.: 0.85, and 0.73 as the temperature isotherms for the predictions and A 4. Predict V at 310K and 7500 (a) ideal gas law. (b) {f""T" .... ~ charts, (c) equation (B' ::.: -0.00851 I and C' = 2.40 X
- 500. Chap. 15 Real Gases: Equations of State 469 and (d) yan der Waals equation. The Pitzer relatjon gives z = 0.514. The experimental value is V = 3.90 cm3/g. 5. You measure that 0.00220 tb mol of a certain gas occupies a volume of 0.95 ft3 at 1 attn and 32°F. If the equation of state for this gas is p V ;:: nRT( I + bp), where b is a constant, find the volume at 2 atm and 71°F. 6. Calculate the temperature of 2 g mol of a gas using van der Waals's equation with a = 1.35 X 10-6 (m6) (atm)(g mol-2), b = 0.0322 X 10-3 (m3)(g mo)-l) if the pressure is 100 kPa and the volume is 0.0515 m3. 7. 'Calculate the pressure of to kg mol of ethane in a 4.86 m3 vessel at 300 K using two equations of state: (a) ideal gas and (b) Soave-Redlich-Kwong. Compare your answer with the observed value of 34.0 atm. S. The van der Waals constants for a gas are a = 2.31 X 106 (atm)(cm3)/(g mol)2 and b ;:: 44.9 cm3Jg mol. Find the volume per kilogram mole if the gas is at 90 atm and 373 K. Thought Problems 1. Data pertaining to the atmosphere on Venus (which has a gravitational field only 0.81 of that of the Earth) shows that the temperature of the atmosphere at the surface is 474 ± 20°C and the pressure is 90 ± 15 atm. What do you think is the reason(s) for the difference between these figures and those at the Earth's surface? 2. A method of making protein nanopartic1es (0.5 to 5.0 j.tm in size) has been patented by the Aphio Corp. Anti-cancer drugs that small can be used in novel drug delivery systems. In the process described in the patent, protein is mixed with a gas such as carbon dioxide or nitrogen at ambient temperature and 20,000 kPa pressure When the pressure is released. the proteins break up into fine particles. What are some of the advantages of such a process versus making powders by standard methods? Discussion Problems 1. Fossil fuels provide most of our power, and the carbon dioxide produced is usually discharged to the atmosphere. The Norwegian company Statoi! separates carbon dioxide from its North Sea gas production. and, since 1996, has been pumping it at the rate of 1 million tons per year into a layer of sandstone 1 km below the seabed. The sandstone traps the gas in a gigantic bubble that in 2001 contained 4 million tons of carbon dioxide. Will the bubble of carbon dioxide remain in place? What problems exist with regard to the continuous addition of carbon dioxide in future years? Is the carbon dioxide actually a gas? 2. What are some of the tests you might apply to determine if an equation of state fits p- V-T data? 3. What factors in a real gas cause the gas to behave in a non ideal manner? Are these factors taken into account in the equations of state listed in Table 15.1?
- 501. 410 Real Gases: Equations of State Chap. 15 Looking Back We provided a condensed view of equations of state for pure components in this chapter, citing their advantages and disadvantages, expected accuracy, and examples of application. We also mentioned briefly one way to treat gas mixtures. GLOSSARY OF NEW WORDS Benedict-Webb-Rubin An eight-parameter equation of state that relates the physical properties p. V, T. and n for a gas. Group contribution method A technique of estimating physical properties of compounds by using properties of molecular groups of elements in the compound. Holborn A multiple parameter equation of state expanded in p. KammerHngh.Onnes A multiple parameter equation of state expanded in Peng .. Robinson A three-parameter equation of state. Soave-Redlich-Kwong A three-parameter equation of state. UNIFAC A group contribution method of estimating physical properties. UNIQUAC An of the UNIFAC method of estimating pbysical proper-ties. Van der Waals A two-parameter equation of state. Vinal equation of state An equation of state expanded in successive terms of one of the physical properties. UPPLEMENTARY REF R NCES In addition to the genera] references listed in the Frequently Asked Questions in the front material, the following are pertinent. Barrufet. M. A., and Eubank. "Generalized Saturation Properties of Pure Fluids via Cubic Equations of State," Chem. £.ng. Educ .. 168-1 (Summer 1989). Benedict, P .• and Olti. Computer Aided Chemical Thermodynamics of Gases and Liquids, Wiley-interscience, New Yark (1985). Chaa~ K. C., and R. L Robinson. Equations of State in Engineering and Research, American Chemical Society, Washington, D.C. (1979). Copeman. T. W., and P. M. Mathias. "Recent Mixing Rules for Equations of State:' ACS Symposium Series 300~ 352-369. American Chemical Society. Washington, D.C. (1986).
- 502. Chap. 15 Problems 411 Eliezer, S. et al. An Introduction to Equations of State: Theory and Applications, Cambridge University Press, Cambridge, U.K. (1986). Elliott. J. R., and T. Daubert. "Evaluation of an Equation State Method for Calculating Critical Properties of Mixtures,U Ind. Chem. Res. 26, 1689 (1987). Gibbons, R. M. uindustriai Use Equations of State:' Chemical Thermodynamics In-dustry, ed. T. I. Barry. Blackwell Scientific. Oxford, (1985). LawaI, A. "A Consistent Rule for "" ..... '''' ...... ,1''. Roots in Cubic Equations of State," Inti Eng. Chern. 26, 857-859 (1987). Manavis, M. Volotopoulos, M. Stamatoudis. "Comparison of Fifteen Generalized Equations of State to Predict Enthalpy," Chern. Eng. Commu.) 130, 1-9 (1994). Mathias, P. M., and M. S. Benson. "Computational Aspects of Equations of State," AlChE 32, 2087 (1986). Mathias, M., and H. C. Klotz. "Take a Closer Look at Thermodynamic Property Models," Chern. Eng. Progress, (June 1994). Orbey, H .• S.I . Sander, and D. S. Wong. "Accurate Equation of State Predictions at High Tempeatures and Pressures using the Existing UNIF AC Model," Fluid Phase Equil., 8S, 41-54 (1993). Polling, B. ,1. M. Prausnitz, and D. P. O'Connell. The Properties of 5th ed. McGraw-Hill, New York (2001). and Liquids, Sandler, S. 1., H. Orbey, and B. I. "Equations of State.H In Modeling for Thermody-namic and Phase Equilibrium Calculations, Chapter 2, S. I. Sander, Ed. Marcel Dekker, New York (1994). Span, Muitiparameter Equations of State. Springer, New York (20Cl0). Web Sites http://che201.vu.msu.edul701/topic07/1essonllllllp05.htm http://www .che. ufl.eduJcouTsesJECH3023Ilectures/acomplexiacomplex.html http://flory.engr.utk.edu/che3301L6-1 http://forums.mit.edulacad?13@104.jaYQFqdl''20@.ee6cfl4 www.virtualmaterials.comlcep200 1.html PROBLEM ·15.1 You want to obtain an answer immediately as to the specific volume of ethane at 700 and List in descending order the teChniques you would use with the most preferable one at the top of the list: (a) Ideal gas law (b) Compressibility charts (c) An equation state
- 503. 472 Real Gases: Equations of State (d) Look up the value on the web (e) Look up the value in a handbook Explain your choices. Chap. 15 *15.2 Which procedure would you recommend to calculate the density of carbon dioxide at 1200P and 1500 psia? Explain your choice. (a) Ideal gas law (b) Redlich-Kwong equation of state (c) Compressibility charts (d) Look up value on the web (e) Look up value in a handbook *15.3 Finish the following sentence: Equations of state are preferred in PVT calculations because *15.4 Use the Kammerlingh-Onnes virial equation with four terms to answer the following questions for CH4 at 273 K: "15.5 *15.6 *15.7 ""'15.9 (a) to what pressure is one term (the ideal gas law) a good approximation? (b) Up to what pressure is equation truncated to two terms a good approximation? (c) What is the error in using (a) and using (b) for CH4? Data: At the values of the virial coefficients are: B == A cm3/mo! C == 2,620 cm6/mol-2 D == 5000 cm9/mol-3 You are asked to design a steel tank in which CO2 will be stored at 290K. The tank is IDA m3 in volume and you want to store 460 kg of CO2 in it. What pressure will the CO2 exert? Use the Redlich-Kwong equation to calculate the pressure in the tank. Re· peat using the SRK equation. Is there a significant difference in the predictions of pressure between the equations? The Peng-Robinson equation is listed in Table 15.1. What are the units of a, b, and a the equation if p is in V" is in Ug mol, and Tis K? The pressure gauge on an 02 cylinder stored outside at oop in the winter reads 1375 By weighing the cylinder (whose volume is 6.70 ft3) you find the net weight, that is, the 02' is 63.9 lb. the reading on the pressure gauge correct? Use an equa-tion of state to make your calculations. What would be developed if 100 tt3 of ammonia at 20 ann and 4000P were compressed into a volume of 5.0 ft3 at 3500P? Use the Peng-Robinson equation to get your answer. An interesting patent (U.S. 3,718,236) explains how to use CO2 as the driving gas for aerosol sprays in a can. A plastic pouch is filled with small compartments containing sodium bicarbonate tablets. Citric acid solution is placed in the bottom of the pouch, and a small amount of carbon dioxide is charged under pressure into pouch as a starter propellant. As the product is CO2 dispensed, the carbon dioxide expands, rupturing the lowest compartment membrane. thus dropping bicarb tablets into the cirtic
- 504. f Chap. 15 """$15.10 "'*15.12 Problems acid. That generates more carbon dioxide, giving more pressure in the pouch, which expands and helps push out more product. (The CO2 does not escape from the can, just the product.) How many grams of NaHC03 are needed to generate a residual pressure of 81.0 psig in the can to deliver the very last em3 of product if the cy Hndrical can is 8.10 cm in diameter and 17.0 em high? Assume the temperature is 25°C. Use the Peng-Robinson equation. Pi rst commercialized in the 1970s as extractants in "natural" decaffeination processes, SCFs (supercritical fluids)-panicularly carbon dioxide and water-are finding new applications, as better, less-expensive equipment lowers processing costs, and regulations drive the chemical industries away from organic vents. extraction capabilities are now being exploited in a range of new pharmaceutical and environmental applications. while supercritical extraction, oxidation and precipitation are being applied to waste cleanup challenges. A compressor for the carbon dioxide compresses 2,000 m3/min at 20°C and 500 kPa to llOoC and 4800 kPa. How many m3/min are produced at the high pressure. Use Vander Waals' equation. A tank of H2 is left out overnight in Antarctica. You are asked to determine how many g moles of H2 are in tank. The pressure gauge reads 39 atm gauge and the temperature is - 50°C. How many g moles of H2 are in the tank? Use the van der Waals and Redlich-Kwong equations state to solve this problem. (Hint: The nonlinear-equation-solving program on the CD in the pocket at the back of this book win make execution of the calculations quite easy.) Find the molar volume (in cm3/g mol) of propane at K 21 atm. Use the Redlich-Kwong and Peng-Robinson equations, and solve for the molar volume using the nonlinear equation solver on the CD in pocket at the back of this book. The acentric factor for propane to use in the Peng-Robinson equation is 0.1487. 4.00 g mol of is contained in a 6250-cm3 vessel at 298.15 K and 14.5 atm. Use the nonlinear equation solver on the in the back of the book to solve the RedlichKwong equation for molar volume. Compare the calculated molar volume of the CO2 in the vessel with the experimental value. ··15.14 The tank cited in problem 15.5 is constructed and tested, and your boss informs you that you forgot to add a safety factor in the design of the tank. It tests out satisfacto- t rily to 3500 kPa, but you should have added a safety factor of 3 to the design. that the tank pressure should not exceed (3500/3) 1167 kPa. say 1200 kPa. How many kg of CO2 can be stored in the tank if the safety factor is applied? Use the RedlichKwong equation. Hint: 'Polymath will solve the equation for you. .. I J ··15.15 A graduate student wants to use van der Waa]s· equation to express the pressurevolume- temperature relations for a gas. Her project required a reasonable degree of precision in the p- V-T calculations. Therefore, made the following experimental measurements with her setup to get an idea how easy the experiment would be:
- 505. 414 Temperature, K 273.1 273.1 Real Gases: Equations of State Pressure, Atm 200 1000 Volume, ft31lb mol 1.860 0.741 Chap. 15 Determine values for the values of constants a and b to be used in van der Waals' equation that best fit the experimental data. ""'15.16 An 80-tb-block of ice is put into a lO-ft3 container. and heated to 900K. What is the final pressure in the container. Do (his problem two ways: (1) use the compressibility factor method, and (2) use the Redlich-Kwong equation. Compare your results. "15.11 What weight of ethane is contained in a gas cylinder that is I.0-ft3 in volume if the gas is at lOO°F and 2000 psig? Do this problem two ways: (1) use Van def Waals' equa.tion, and (2) use the compressibility factor method. The ex.perimental value is 21.4 lb. "'15.18 Answer the foHowing questions: a. wm the constant a in van der Waals' equation be higher or lower for methane than for propane? Repeat for the other Van der Waals constant b. b. Will the constant al in the SRK equation be higher or lower for methane than for propane? Repeat for the other van der Waals constant h. "'15.19 Oxygen at -18°C is take from a tank at a rate of 0.190 m3/hr, and mixed with ethane [0 be fed to an engine along with 100% excess air. The exhaust from [he engine is 2593 m3 mea.c:ured at standard conditions per hour, and has the foHowing composition: CompOllent Percent 9.89 74.41 0.63 9.42 What is the pressure in the tank from which the oxygen is being withdrawn? "15.20 A feed for a reactor has to be prepared comprised of 50% ethylene and 50% nitrogen. One source of gas is a cylinder containing a large amount of gas with the composition 20% ethylene and 80% nitrogen. Another cylinder that contains pure ethylene at 1450 psig and 70°F has an internal volume of 2640 cubic inches. If aU the ethylene in the latter cylinder is used up in making the mixture, how much reactor feed was prepared and how much of the 20% ethylene mixture was used? Use van der Waals' equation and compare your answers with those obtained Problem 14.26.
- 506. CHAPTER16 SI NG LE-COM PON ENT TWO-PHASE SYSTEMS (VAPOR PRESSURE) 16.1 Phase Diagrams 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature Your objectives In studying this chapter Bre to be able to: 1. Define and explain several important terms related to the properties of water that are listed in Section 16.1. 2. Sketch the vapor pressure of a pure compound as a function of temperature at constant volume and a function of volume at constant temperature_ 3. Sketch the processes of vaporization, condensation, compression, and expansion of a pure compound on p' versus T and p. versus V charts. 4. Explain what occurs in terms of p', V, and Tduring a phase transition. 5. Calculate the vapor pressure of a substance from an equation mat relates the vapor pressure to the temperature (such as the Antoine equation) given values for the coefficients in the equation. 6. Look up the vapor pressure in reference books. Calculate the temperature of a substance from a vapor-pressure equation given the values for the coefficients in the equation and the vapor pressure. S. Estimate the vapor pressure of a compound via a Cox chart. 9. Use the steam tables to retrieve data for the vapor pressure of water. 476 485 47S
- 507. 476 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 When you read subsequent chapters you will need to quite clear to how to obtain vapor pressure and use it your calculations. Now your opportunity to familiarize yourself with the pertinent graphs, equations, and tables so that you will avoid confusion later on. Looking Ahead In this chapter we want add a number new terms to your vocabulary re-lated liquids and vapors. In addition, we describe some useful types of diagrams. Then we proceed to the prediction of vapor pressure as a function of temperature, and introduce you to the steam tables in which you can look the vapor pressure of water. 16.1 Phase Diagrams You can conveniently display the properties of compounds via phase dia-grams. A pure substance can simultaneously in many phases. of which, as you know, solid, liquid, and are the most common. Phase diagrams enable you to view the properties of two or more phases as functions of temperature, pressure, specific volume, concentration, and other abIes. Tables and equations may yield greater accuracy, but "a picture worth a thousand wordsH or in our case, a thousand bits of We are going to discuss phase diagrams in terms of water because presumably you are familiar with the three phases of water, namely ice, liquid water, and water vapor (stream), but the discussion applies to all other pure substances. The terms vapor and gas are used very loosely. A that exists below its critical temperature is usually caned a vapor because it can condense. We will reserve the word vapor to describe a gas below its critical point in a process in which the phase change of primary interest, while the word gas or nODcondensable gas will be used to describe a gas above the critical point or a in a process at conditions under which it cannot condense. Suppose you carry out some experiments with the apparatus shown in Figure 1 1. Place a lump of ice in the chamber below the piston, and evacuate the chamber to remove all of the (you want to only water in the chamber). Fix volume of the chamber by fixing the position of the piston, and start slowly (so that the phases of that will be in equilibrium) heating the you plot the measured pressures as a function of temperature. you will get 16.2-a phase diagram-in which all of the measurements you make have been fitted by a continu-ous smooth curve for clarity. .
- 508. Sec. 16.1 Phase Diagrams Temperature Pressure Figure 16.1 ~pparatus used to explore the p, V. and T properties of water. The initial conditions of p and T in the chamber are at 0 in Figure 16.2 with the solid in equilibrium with the·vapor. As you raise the temperature the ice starts to melt at point A, the triple point. the one p-T-V" combination at which solid, liquid, and vapor can be in equilibrium. Further increases in the temperature cause the pressure to rise along the curve AB that represents liquid and vapor in equilibrium. B is the critical point at which vapor and liquid properties become the same. At point A, if you had kept the temperature constant and raised the pressure on the ice, ice would still exist and be in equilibrium with liquid water along the line AC (C is not the tennination of the line, just a marker for convenience in descrip-c Uquid T, °c Supercritical fluid B Vapor F1pre 16.1 Results of an experiment in heating a fixed amount of a pure compound in a closed vessel (8 constant volume) as shown OD a phase diagram of p versus T at constant V.
- 509. 478 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 tion) , line AC is so vertical that you can use the saturated liquid properties for the compressed liquid. For example, saturated water at 400K and p* = 245.6 kPa has a density of 937.35 kglm3. If the pressure on the water is increased at 400K to 1000 kPa, the density is 937.79 kglm3, an increase of just 0.047%. skating possible because the high pressure exerted by the thin blade on ice fOnTIS a liquid layer with low friction on the blade. usually do not exert enough pressure to melt snow, in fact skis are frequently waxed to reduce friction, If the vapor and liquid of a pure component are in equilibrium, then the equilibrium pressure is caUed the vapor pressure, which we"will denote by p., At a given temperature there is only one pressure at which the liquid and vapor phases of a pure substance may exist in equilibrium. Either phase alone may exist, of course, over a wide of conditions. We next take up some terminology associated with processes that are conveniently represented on a p* phase diagram such as Figure 16.3 (the letters in parenthesis refer to a process shown in 16.3). You will note that some of the states or processes, for historical reasons, have duplicate names. 'boiling bubble point The change of phase from liquid to vapor (e.g .. D to A to C, or M to 0). The temperature at which a liquid just starts to vaporize (N, H, and are examples). G 760~--r-----------------~ ~ E .E r::i. 526 Triple point: SOlid. liquid and vapor in equilibrium Liquid .(subcooled) "'"r--------+--+---'---'-;......;..;;..;;;..;.;..=-----! 20.0 5b!:i:==~~----.L---.l-L.-___ --~O.67 9Oioo Figure Solid and vapor in equilibrium 60 T, Various common processes as represented on a p" -T diagram.
- 510. Sec. 16.1 condensation dew point evaporation freezing (solidifying) melting (fusion) melting curve normal boiUng point normal melting point saturated liquid/saturated vapor subcooled liquid sublimation sublimation curve sublimation pressure supercriticai region superheated vapor '" 419 The change of phase from vapor to liquid , F to D, C to A. or 0 to M). The temperature at which the vapor just gins to condense at a specified namely temperature values along the vapor curve (N. H, and E are examples). change of phase from liquid to vapor D to F. A to C, or M to 0). phase from liquid to solid (N to L). change In phase from solid to liquid M). equilibrium curve starting and continuing vertical1y M. which the vapor pressure kPa)-point B for water. at which the solid melts at 1 atm(101 kPa). Values liquid and vapor equilibrium curve (vapor pressure curve). Liquid melting curve and the vapor curve (D is an example). The change In from solid to vapor (1 to K). The solid-vapor equilibrium curve from J (and lower) to the tripie point The pressure along the curve function of temperature) p-T values for liquid and above the critical point. Vapor at temperatures and pressures ing those at saturation is an example). degrees of superheat are the (0 minus N or F minus E) in temperature be- . tween the actual T and the saturated T at the given pressure. For example, steam at 500°F and 100 psia (the saturation temperature for 100 psia is 327.8°F) has (500 - 327.8) = t 72.2°F of superheat.
- 511. 480 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 two-phase region vaporization Conditions of T and p in which two phases can coexist at equilibrium (the region com-prising vapor and liquid compressed into curve N-H-E-B in Figure 16.3) The change of phase from liquid to vapor, that boiling (D to F). In Figure 16.3 process of evaporation and condensation of water at 1 atm is represented by the ABC with the phase transformation occurring at 100°C. Sup-pose that you went to the top of Pikes and repeated the of evaporation and condensation in the open What would happen then? The process would be the same (points D-E-F) with the exception of the temperature at which the water would begin to boil, or condense. Since the pressure of the at the top of Pikes Peak lower than 101.3 kPa, the water would start to displace the or boil, at a lower temperature. Some unfortunate consequences might result if you expected to kill certain of disease-causing bacteria by boiling! To conclude, you can see that (a) at any given temperature water exerts its vapor pressure (at equiHbrium)~ (b) as the temperature increases, vapor pressure increases as wen; and (c) it makes no difference whether water vaporizes into into a cylinder closed by a piston, or into an evacuated cylinder-at any temperature it still exerts same vapor pressure as long as the liquid water is in equilibrium with its vapor. A pure compound can change at constant volume from a liquid to a vapor. or reverse~ via a constant temperature process as well as a constant pressure process. A process of vaporization or condensation at constant temperature is illustrated by the lines G-H-/ or JaH-G, respective]y, Figure 16.3. Water would vaporize or condense at constant temperature as the pressure-reaches point H on the vapor-pressure curve. TQe change that occurs H the increase or in the fraction vapor or 1iquid~ respectively. at the fixed temperature. The sure does not change until all of the vapor, or liquid, has completed the phase transition. Now let's go back to th~ experimental apparatus and collect data to prepare a graph of values of p and V as a function of temperature. This time you want to hold the temperature in the chamber constant and adjust the volume while measuring the pressure. Start with compressed liquid water (subcooled water) rather than ice. and raise the piston so that water eventually vaporizes as the pressure drops. Figure 16.4 illustrates by dashed lines the measurements two different temperatures, T) and T2. the pressure is reduced at constant Ttt V increases very slightly (liquids are not very compressible) until the liquid pressure reaches p., the vapor pressure, at point A.
- 512. Sec. 16.1 Phase Diagrams heat addition Super Crfttcal Region Critical heat addltion Saturated mlXlUfe liquid and wpor heat addItIOn r- 1 aim - I pISton B Saturated V8pOf p I • I I I I I I ' I mpreued! Uquld Region A • Solid • • point Superheated vapor Region Saturated Satld • Vapor Region •• Figure 16.4 Experiments to obtain a p-V phase diagram. The dasbed lines are measurements made at constant temperatures T and T2. The dots represent the points at which vaporization or condensation. respectively, of the saturated liquid. or vapor, occurs, and form an envelope about the two-phase region. 481 Then, as the piston continues to rise, both the pressure and temperature remain constant as the liquid vaporizes until all of the liquid is vaporized (point B) on the saturated vapor line. A Subsequently, as the pressure decreases, the value of V can be calculated via an ideal or real gas equation. Compression at constant T2 is just a reversal of the process at T1• The dots in Figure 16.4 represent just the measurements.made when saturation of liquid and vapor occurs, and are deemed to fonn the envelope for the two-phase region that from a different perspective appears in Figures 16.2 and 16.3 as the vapor pressure curve. The two-phase region (e.g., A to B or D to C in Figure 16.4) represents the conditions under which liquid and vapor can exist at equilibrium. Note from Figure 16.4 that a cbange in the specific volume occurs in going from a liquid to a solid at the triple point. In other words, water expands when it freezes and this is why ships trapped in the polar ice can be crushed by the force of the expanding ice. A new term, quality, the fraction or percent of the total vapor and liquid mixture that is vapor (wet vapor), applies to the two-phase region. Examine Figure 16.5. You can calculate the volume of the liquid-vapor mixrure at B in Figure 16.5 by adding the volume fraction of material that is saturated liquid to the volume fraction that is saturated vapor
- 513. 482 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 p Figure .1.. 6.5 Representation of quality 011 a p-V phase diagram. A is saturated liquid and C is saturated vapor. The compound at B is part liquid and part V sat. V vapor. and the vapor is cal.Jed vapor the quality. A A A V (1 - x) V saturated liquid + X V saturated 'Vapor (16.1) where x is the fractional qUality. Solving for x yields A A V Vsat. - liquid x = -:A----.-A-- Vsat. VstU• vapor liquid of physical properties frequently display Hnes of constant quality in the tworegion. Examine Figure J2 for CO2 in Appendix Figures 16.3 and 16.4 £an be reconciled by looking the three-dimensional surface that the p-V -T functionality for water, Figure 16.6. You can see that the vapor pressure curve shown in Figures 16.2 and 16.3 corresponds to a two-dimensional projection of a three-dimensional onto the p-T plane. The view you see directly across the surface representing the liquid-vapor region. A slight of the projection in the lower left shows more tail. each temperature you can read the corresponding press}lre at which vapor and liquid exist in equilibrium. Similarly. the p-V projection corresponds to a view you would have looking across the three-dimensional surface standing" in front of the specific volume axis. Frequently Asked Questions 1. saturation really refer both to the liquid and to the vapor? both can be termed saturated. 2. Does the dew point mean the same thing as saturated? Yes, for saturated vapor. 3. Does the bubble point mean same thing as saturated? for saturated liquid.
- 514. Sec. 16.1 Phase Diagrams 483 ,, .. . ·, · · , ·, , 1 ICE , . . . . , .. ,, . , -- ' A Figure 16.6 The p-V -T surface for water (a compound that expands on freezing) showing th_e two-dimensional projections for pairs of the three variables. 4. Is the saturation pressure really the same thing as the vapor pressure? Yes. 5. Why can you use some of the properties of saturated liquid for liquid at higher pressures than saturation? Inspection of a p-V diagram shows that the phase boundary (line) between the liquid and vapor is essentially a vertical line. Consequently, the specific volume, e.g., the density, of the liquid remains essentially unchanged as the pressure increases. SELF-ASSESSMENT TEST Questions 1. If CuS04 (a solid) is heated to 653°C it begins to decompose to CuO, S02' and 02' An article shows a plot of the partial pressure of S02 versus temperature with the caption stat- ing "Vapor pressure of copper sulfate as a function of temperature." Do you think the caption is correCt? 2. You wish to, produce solid nitrous oxide using the same equipment and process as used for solid carbon dioxide. Where would you locate your plant-at low or high altitude?
- 515. 484 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 How would you ship your product to a distant point (say, Houston) where the solid N20 is to be liquified and bottled? How would you liquify the solid? 3. Does iodine have a vapor pressure? Does it melt at room temperature? 4. Why does dry ice sublime at room temperature and pressure? S. Is it true that foods cook more quickly in a pressure cooker because the temperature inside the cooker is higher than the temperature obtained in an open pot? Problems 1. A tank contains 1000 kg of acetone (C3HuO), half of which is liquid and the other half is the vapor phase. Acetone vapor is withdrawn slowly from the tank, and a heater in each phase maintains the temperature of each of the two phases at 50°C. Determine the pressure in the tank after 100 kg of vapor has been withdrawn. 2. Draw a p-T phase diagram for water. Label the following clearly: vapor-pressure curve, dewpoint curve, saturated region, superheated region, subcooled region, and triple point. Show the processes of evaporation, condensation, and sublimation by arrows. Thought Problems 1. A cylinder containing butadiene exploded in a research laboratory, killing one employee. The cylinder had been used to supply butadiene to a pilot plant. When butadiene gas was required, heat was supplied to the cyHnder to raise the pressure of the butadiene in the tank. The maximum temperature that could be achieved in the tank on subsequent tests with a like tank was 160°C. At 152°C, the critical temperature for butadiene, the pressure is 628 Ibr'in.2, less than one-half of the pressure required to rupture'the tank by hydraulic test. Why did the tank explode? 2. The advertisement reads "Solid dry ice blocks in 60 seconds right in your own lab! Now you can have dry ice available to you at any time, day or night, with this small, safe, efficient machine and readily available CO2 cylinders. No batteries or electrical energy are required." How is it possible to make dry ice in 60 seconds without a compressor? 3. An inventor is trying to seU a machine that transforms water vapor into liquid water without ever condensing the water vapor. You are asked to explain if such a process is technically possible. What is your answer? Discussion Problems 1. The following description of a waste disposal system appeared in Chemical Engineering, June 1993, p. 23. The first commercial application of VerTech Deep Shaft, a wet oxidation process that'takes place in a reactor suspended in a borehole about 1,250 m deep. has been commissioned to treat 25,000 m.t.lyr of sewage sludge at a municipal wastewater treatment plant in Apeldoorn, The Nether1ands. The reactor consists of three steel tubes suspended in a 1,250-m drilled shaft of 95 cm dia. A 5% sludge solution is pumped down the inner 19.5-cm-dia. open-ended rube, whHe the oxidized sludge
- 516. Sec. 1 Modeling and Predicting Vapor Pressure as a Function of Temperature 485 and dissolved gases come up through the annular space between the first tube and the closed second tube (34 cm dia.), About 400 m down. pure oxygen is sparged into the sludge to promote oxidation of the suspended organic matter. The static head of the j ,200-m column prevents boiling at about 280°C and 100 bars. The alternati yes: incineration, composting. drying, or conventional wet oxidation have per m.t. costs DGL 790. DGL 790, DGL 700, and DGL 600, respectively. versus DGL 580 for the VerTech process. Check the consistency of the data. and offer an opinion as to the problems that might occur if the process operated on a 20% sludge solution in water. 2. The term (boiJing·liquid expanding-vapor explosion) refers to an explosion that occurs when the pressure on a liquid is suddenly reduced substantiaHy below its vapor pressure. Explain how such an explosion can occur. Might it occur if a tank car ruptures in an accident even if no occurs? Give one or two other examples of vessel failures that might cause a BLEVE. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature How can youfind the answer if you don't know the question? M.C. Sonby. Shell Development Co. In this section we are going to explain how to determine values for the vapor pressure given the temperature, or the temperature the vapor pressure. Severa] aids exist for you to use: (a) Equations of as a function of T; (b) charts of versus (c) tables of p" versus T; Dr. Sonby reported at one of the A.I.ChE. meetings that "In one of my first projects at Shen. I was trying to determine the rate of emission of sulfur vapor from a pit full of molten sulfur. I was able to locate four different references on the vapor pressure of molten sulfur. Unfortunately, all four references different vapor pressures. with the highest and lowest being an order of magnitude apart! As it turned out, the experimental methods used to determine the vapor pressures varied greatly, as did the purity of the sulfur in the experiments. " Thus, you need to be able to appraise the accuracy of the predictions values of the vapor pressure of even old standbys. 16.2-1 Prediction via Equations You can see from Figure 16.2 that the function of p* versus T is not a linear function (except as an approximation over a very small temperature range). Many functional forms have been proposed to predict p* from T (refer to Ruzicka, K., and
- 517. 486 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 V. Majer, "Simple and Controlled Extrapolation of Vapor Pressures toward the Triple Point," A.I.ChEJ., 42, 1723-1740 [J996] for an evaluation of numerous equations) and the reverse, some with numerous coefficients. We wiH use the Antoine equation in this book-it has sufficient accuracy for our needs, and coeffi- "' ......... ., for equation for over 5,000 compounds in the literature: B lnp '" = A - --T-where A. B, = constants for each substance = temperature, K ( 16.2) Refer to Appendix for values of A, B, and C for a small set of compounds. The CD in back this book~ based on data provided by Yaws,'" will enable you to retrieve vapor pressures for over 700 compounds. The National Institute of Standards and Technology a program (www.nist.gov/srdinist87.htm) that calculates the vapor pressures and displays tables for approximately 6,000 pure compounds. You can estimate the values of A, B. and C in Equation (16.2) from experimental data by using a regression program such as the one in Polymath on the CD that accompanies this book:. With just experimental values the vapor pres-sure versus temperature you can Equation (16.2), EXAMPLE 16.1 Vaporization of Metals for Thin Film Deposition Three methods of providing vaporized metals for thin film deposition are evaporation from a boat or from a filament transfer via an electronic beam. Fig-ure 1 illustrates evaporation from a boat placed in a vacuum chamber. Vacuum chamber To pump Electrode Electrode Figure E16.1 ilYaws. C. and Yang. "To ..... "'''' ............ Vapor Pressure Easily," Hydrocarbon ProceSSing, p. 65 (October 1989).
- 518. 1 Modeling and Predicting Vapor Pressure as a Function of Temperature 481 The boat made of tungsten a negligible vapor pressure 972oC. the ating temperature for vaporization of aluminum (which melts at 660°C and fiBs the boar). The approximate rale of evaporation m is given g/(cm2)(s) by p*(MW)112 m = 0.437---......:.--- T where p. is the vapor pressure in kPa and T is the temperature in What is the vaporization rate for Al at 972°C in g/(cm2)(s)? Solution You have to calculate p" for at 972°C. The Antoine equation is suitable if data are known for the pressure of AI. Considerable variation exists in the data for A 1 at high temperatures, but for Equation (16.2) we will use A :::; 8.779, B = 1.615 X lQ4, and C = 0 withp" in mm Hg and Tin K. '" 1.615 X 104 In P972°C = 8.799 - 972 + 273 = -4.17288 P;72"C 0.0154 mm Hg (0.00205 kPa) (0.00205 )(26.98) Jf2 m = 0.437 112 = 1.3 X 10-4 gI(cm2)(s) (972 + 273) 16.2 .. 2 Retrieving Vapor Pressures from the Tables You can find the vapor pressures of substances listed in tables in handbooks, physical property books, and Internet sites. We will use water as an example. Tabulations of the properties of water and steam (water vapor) are commonly called the steam tables. Furthermore, when you retrieve the properties a CD, such as the American Society of Mechanical Engineers' Properties of Steam, they probably were generated by an equation. We will often to the software on the CD in the back of this book that gives the properties of water and steam as the "steam tables." In book you will also find a foldout in back pocket that contains abbreviated steam tables in both and 51 units. CD in the back of this book will prove to be a saver because you can obtain for the properties of water in mixed units that are continuous over the permitted range of values. and avoid tedious single or double interpolation in tables. Several types of tables exist in the foldout. The following examples are from the tables in S1 and AE units:
- 519. 488 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 1. A table for saturated water and vapor listing p"" versus T as well as other properties such as the specific volume of the liquid (Vt) and vapor (V,), Properties of Satw'ated Water In SI Units Press. T Volwne~ m3lkg kPa K VI 0.80 276.92 0.001000 159.7 1.0 280.13 0.001000 129.2 1.2 282.81 0.001000 108.1 1.4 285.13 0.001001 93.92 1.6 287.17 0.001001 82.76 1.8 288.99 0.001001 .74.03 2.0 290.65 0.001002 61.00 2.5 294.23 0.001002 54.25 3.0 297.23 0.001003 45.61 4.0 302.12 0.001004 34.80 2. A table of for saturated water and vapor listing versus P"'1 as well as other properties such as V for the liquid and vapor. Properties of Saturated Water 10 81 Urub T Press. A Volume, mJJkg K kPa V, 213.16 0.6113 OJXHOOO 206.1 0.6980 0.001000 181.1 280 0.9912 0.001000 1 285 1.388 0.001001 94.61 290 1.919 0.001001 69.67 295 2.620 0.001002 51.90 300 3.536 0.001004 39.10 305 4.718 0.001005 . 29.18 310 6.230 0.001007 22.91 315 8.143 0.001009 17.80 3. A table listing superheated vapor (steam) properties as a fmiction of and p, including the specific volume of the vapor v in Ib/ft3. Prf'perties of Superheated Steam In AE HaUs AM. Press. IbIiD.1 Sat. Sat. (Sat. Temp.) Water Steam ~ 4:zooF 44O"F Sh 29.23 49.23 69.23 175 v 0.0182 2.601 2.130 2.814 2.897 (310.77) h 343.61 1196.7 1215.6 1227.6 ' 1239.9
- 520. f Sec. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature 489 Superheated Steam (continued) Abs. Press. Iblin.1 Sat. Sat. (Sat. Temp.) Water Steam 400~ 4200F 44O"F Sh 26.92 46.92 66.92 180 v 0.0183 2.532 2.648 2.731 2.812 (373.08) h 346.07 1197.2 1214.6 1226.8 1239.2 Sh 24.66 44.66 64.66 185 I 0.0183 2.466 2.570 2.651 2.731 (375.34) h 348.47 1197.6 1213.7 1226.0 1238.4 4. A table of subcooled (liquid) water properties as a function of p and T; p is the density; u and h are discussed in Chapter 21. Properties of LlquJd Water in SI Units T(K) 400 425 450 Puc. kPa x 10-3 0.2456 0.4999 0.9315 Sat. p, kglmJ 937.35 915.08 890.25 h, kJ/kg 532.69 639.71 748.98 U, kJ/kg 532.43 639.17 747.93 500 P. kg/rn3 937.51 915.08 h, kJ/kg 532.82 639.71 u, kJ/kg 532.29 639.17 700 p, kglm3 937.62 9J5.22 i4 kl/kg 532.94 639.84 u, kl/kg 532.19 639.07 Take the foldout from the pocket in the back of this book, locate each table, and use it to follow the explanations below. How can you tell which table to use to get the properties you want? One way is to locate the state of interest on a phase diagram for water. The other is to plow through the steam tables. . For example, do the conditions of 25°C and 4 atm refer to liquid water, a saturated liquid-vapor mixture, or water vapor? Examine Figure 16.7. The state occurs to the left of the saturated liquid curve in the tiny space between the saturated liquid line and the vertical axis on the 298K isotherm (just below the 300K isotherm). Note that the 25°C (298K) isothenn in the liquid region goes so straight up the chart that it is essentially a vertical line extending from the intersection of 298K and the saturated liquid curve. You can use the values in the steam tables plus what you know about phase to reach a decision about the state of the water. In the SI steam tables of T versus p. for
- 521. 490 Single-Component Two-Phase Systems (Vapor Pressure) 1000 100 10 3.536 Saturated liquid boundary {compressed) 2 Pha.se region of vapor and liquid Saturated vapor / boundary 300 K equilibrium line ---------------------------- Chap. 16 1.0 0.1 1 10 100 1Qoo Figure 16.7 Portion of the p-V phase diagram water. (Note that the axes are logarithmic scales). saturated water, Tis just less than 300K at which p. = 3.536 kPa, hence p. will be a bit less than 3.536 kPa. Because the given pressure is about 400 kPa, much higher than the saturation pressure at 298K, clearly the water sullcooled (compressed liquid), Can you locate the point p. = 250 kPa and V = 1.00 m3lkg? Do you find the water is in the superheated region? The specified volume is larger than the saturated vapor volume of 0.7187 m3lkg at 250 kPa. What is the state corresponding to T = 300K and V ;:;:; 0.505 m3lkg? Water at that state is a mixture of saturated liquid and vapor. You can calculate the quality of the water-water vapor mixture using Equation (16.1) as follows. From the steam tables the specific volumes of the saturated liquid and vapor are Ve = 0.001004 m3lkg Vg = 39.1Um3/kg Basis: 1 kg of wet steam mixture Let x:::: mass fraction vapor. Then 0.001004 m3 -(1 -- ---- 1 kg liquid 39.10 rn3 x kg vapor 5 3 -----~.---;=--- = O. 05 m 1 kg vapor x = 0.0129 (the fractional quality)
- 522. Sec. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature 491 If you are given a specific mass of saturated water plus steam at a specified temperature or pressure so that you know the state of the water is in the two-phase region, you can use the steam tables for various calculations. For example, suppose a 10.0 m3 vessel contains 2000 kg of water plus steam at 10 atm, and you are asked to calculate the volume of each phase. Let the volume of :vater be V {. and the volume of steam be Vg; then the masses of each phase are VeN e and V gIV g, respectively. From your knowledge of the total volume and total mass: Ve + Vg = 10 and A A VeNe + VglVg = 2000 A You can find from )he steam tables that the specific volumes are Ve = 0.0011274 m3/kg and V g = 0.19430 m3/kg. Solving these simultaneous equations for Ve and V gives the volume of the liquid as 2.21 m3, and the volume of the steam as 7. 79 m~. The mass of the liquid is 1960 kg, and the mass of the steam is 40 kg. Because the values in the steam tables are tabulated in discrete increments, for intermediate values you will have to interpolate to retrieve values between the discrete values. (If interpolation does not appeal to you, use the steam tables on the CD in the back of this book.) The next example shows how to carry out interpolations in tables. EXAl1PLE 16.2 Interpolating in the Steam Tables (a) What is the saturation pressure of water at 312K? (b) Steam is cooled from 6400P and 92 psia to 480°F and 52 psia. What is ~ V in ft3llb? Solution (a) To solve this problem you have to carry out a single interpolation. Look in the steam tables under the properties of saturated water to get p. so as to bracket 312K: T(K) 310 315 6.230 8.143 Figure E16.2 shows the concept of a linear interpolation between 310K and 315K. Find the change of p. per unit change in T. Il.p. (8.143 - 6.230) kPa 1.91 ~T - (315 - 310) K = -5- = 0.383 _ _. __ _ ._ _ ._ J
- 523. 492 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 18 p. (kPa) 8,143 ------- ..... ----- 7 6.230 ..f£"----+---_4_-.... 310 312 315 T(K} F.lgure E16.2 Multiply the fractional change and the number of degrees increase from 310K to get the change p", and add the result to the value of p" at 31 OK of 6.230 kPa: p·31 = /' 3 10K + AT 12 - 1 310) :::: 6.230 + 0.383 (2) :::;: 7.00 kPa (b) problem requires double interpolation in the steam tables. Set the double interpolation as shown below. Th~ volumes in the table below are in ft3llb. The step is to get data for V that bracket both 92 psia and 64Qop ¥ listed in columns 1, 3, and 4. The second step is to interpolate to get V at the intermediate pressure of 92 psia (see column 2) for 600°F (6.768) and then 700°F (7.437) as listed in the fifth and sixth columns, respectively. p (psia) 90 95 p (psia) 50 55 TeT> 600 700 6.916 7.599 .. 92 -------iIIIIo- 6.547 7.195 450 500 10.69 11.30 ... 52------~~ 9.703 10.26 reT> 600 700 640 6.768 7.437 7.036 450 500 480 10.295 10.88 10.646 The fin41 step is to interpolate at p :::;: 92 psia between 6QO°F and 700°F to get V = 7.036 ft31lb at 640° and 92 psia. Repeat the procedure for 52 psia to get if = 10.646 ft311b at 480oP. Fmally, calculate AV = 10.646 - 7.036 = 3.61 ttlllb Could you have interpolated first in temperature and then in pressure? Of course.
- 524. Sec. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature 493 " 16.2 .. 3 Predicting Vapor Pressures from Reference Substance Plots Because of the curvature of the vapor pressure versus temperature data (see 16.2), no simple equation with two or three coefficients will fit the data accurately from the triple point to the critical point. Othmer proposed in numerous articles (see. for example. Othmer, D.F., Ind. Eng. Chern. 32, 841(1940) and Perry, J.H., and E.R. Smith, Ind. Eng. Chem., 25, 195 (1933» that reference substance plots (the name will become clear in a moment) could convert the vapor pressure versus temperature curve into a straight line. One well-known example is the Cox chart (Cox, E.R., Ind. Eng. Chern .• 15, 592[ 1923]). You can use the Cox chart to retrieve vapor pressure values to test reliability of experimental data, to interpolate, and to extrapolate. Figure 16.8 is a Cox chart. Here is how you can make a Cox chart. 1. Mark on the horizontal values of Jog p* so as to cover the desired range of p* for the compound of interest. 2. Next draw a straight line on the plot at a suitable angle. say 45° I that covers the range of T that is to marked on the vertical axis. Vapor pressure, kPo /log 10 scale) tOO 1000 47 200 150 _ 2501---1--- .- 100 .!! 0 "~" 11:1 "8 2001--+---+- VI fOO LI... '" 75 50 25 ~ IJ- "- 50 3 .a. 25 0 ... c.. .e.. I- -25 -25 1.0 10 100 1000 10,000 Vapor pressure psio lIoQlo scale) Figure 16.8 Cox chart. The vapor pressure of compounds other than water is observed to fall on straight lines when plotted on the special developed for the Cox chart by using a reference substance water).
- 525. 494 Single-Component Two~Phase Systems (Vapor Pressure) Chap. 16 3. To calibrate the vertical axis in common integers such as 25, 50, 100, 200 degrees, and so on, you use a reference substance namely water. For the index tic mark for the first integer, say lOOoP, you look up the vapor pressure water at 100°F in the steam tables, or calculate it from the Antoine equation, to 0.9487 psia. 0.9487 psis on the horizontal axis, and proceed verti-cally until you hit the 45° straight line. Then proceed horizontally left until you hit the vertical axis. Mark the scale at the intersection as 100°F. 4. Pick the next temperature, say 20QoP, for which the vapor pressure of water is 11 psia. From 11 psi a on the horizontal axis proceed vertically to the 45° straight line, and then horizontally to the vertical axis. Mark the scale at the intersection with the vertical axis as 200oP. Continue as in 3 and 4 until the vertical scale is established over for the temperature. desired A plot of the temperature versus vapor pressure for other compounds will yield straight lines, as shown in Figure 16.8. What proves useful about the Cox chart is that the vapor pressures of other substances plotted on this specially prepared set of coordinates will yield straight lines over extensive temperature ranges, and thus facilitate the extrapolation and interpolation of vapor-pressure data. It has been found that lines so constructed closely related compounds, such as hydrocarbons, aU at a common point. Since straight lines can be obtained in a Cox chart, only two points of vapor-pressure data are needed to provide complete information about vapor pressure of a substance over a con<siderable temperature range. Lets look at an example of preparing and using a Cox chart. EXAl1PLE 16.3 Extrapolation of Vapor .. Pressure Data control of solvents was ftrst described in the Federal Register, v. no. 158, August 14, 1971> under Title 42, Chapter 4, Appendix 4.0, Control Organic Compound Emissions. Chlorinated solvents and many other solvents used in industrial finishing and processing, dry-cleaning plants, metal degreasing, printing operations, and so forth, can be recycled and reused by the introduction of carbon adsorption equipment. To predict the size of the ad sorber. you frrst need to know the vapor pressure of the compound being adsorbed at the process conditions. The vapor pressure chlorobenzene is 400 mm Hg at HO°C and 5 attn at 205°C. Estimate the vapor pressure at 245°C and also at the critical point (359°C). Solution vapor pressures will estimated by use of a Cox chart. You construct the temperature scale (vertical) and vapor pressure scale (horizontal), as described in connection with Figure 16.8, Prepare the horizontal axis on the chart by marking
- 526. 16.2 Modeling and Predicting as a Function of Temperature a log (or 1n) scale from 0.1 to 104. pressures of water from 3.72 to 3094 psia corresponding to 1500P to and mark the respective integer temper-atures marked on the vertical as shown in Figure E16.3. 100,-... 6001--- 15 ~ 5001--- I ~I---I---r-.~.~-+~~ ~ ~OOI---I---c""-r---+I-T-+-+-!~ .u. ... I 200 150 ! 100 OJ to 100 1000 10,000 Figure E16.3 convert the two given vapor pressures of 400 mm Hg 14.7 psis psia 760 mrn Hg = 7.74 psi a 230"P pressures. 5 aIm ----"'- -~- = 73.5 psia 1 atm Hg ~ 401 two points on the graph paper. Examine the line between the encircled points and extrapolate At these two temperatures, you can read Estimated: 471 OF (245°C) 150 psia 678°F (359°C) 700 psia Experimental: 147 psia 666 psia for comparison. dots. Finally. 471°F (245°C) estimated vapor Experimental You could chart using Sf units if you wanted to save conversions. EXAMPLE 16.4 Solvent Selection Based on OSHA PEL Limits and Potential Hazard The limitations of toxicity of chemicals established by the Occupa-tional Safety and Health Administration (OSHA) as permissible exposure limits (PEL) can be used to estimate the relative hazard of solvents. Also, the relative
- 527. 496 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 vapor pressures of solvents can be used as a measure of occupat.ional exposure. Based on the following data for the OSHA PEL Ethyl acetate Methyl ethyl ketone n-butyl acetate OSHA PEL (in air), ppm by volume 400 200 1.3 estimate the relative hazards of the three solvents taking into account both toxicity and exposure. Solution A combined hazard criterion would be to take the inverse of the OSH PEL as a potential hazard due to toxicity and multiply by the pertinent vapor pressure, a measure of exposure. The vapor pressures of the respective compounds at 25°C are: Ethyl acetate Methyl ethyl ketone n-butyl acetate -From the CD "From Perry p. (mm Hg) The combined criteria in increasing order of hazard are: Ethyl acetate: Methyl ethyl ketone n-butyl acetate 96.9/400 = 0.24 94.81200 = 0.47 20/1.3 = 15.4 Clearly n-butyl acetate would be the poorest choice from the viewpoint of its hazardous nature. Although this chapter treats the vapor pressure of a pure component, we should mention that the tenn vapor pressure has been applied to solutions of multiple components as well. For example. to meet emission standards, refiners formulate gasoline and diesel fuel differently in the summer than in the winter. The rules on emissions are related to the vapor pressure of a fuel, which is specified in tenns of the Reid Vapor Pressure (RVP), a value that is detennined at 100°F in a bomb that permits partial vaporization. For a pure component the RVP is the true vapor pressure, but for a mixture (as are most fuels) the RVP is lower than the true vapor pressure of the mixture (by roughly 10% for gasoline). Refer to Vazquez-Esparragoza, 1. J., G. A. Iglesias-Silva, M. W. Hlavinka, and J. Bulin. "How to Estimate RVP of Blends," Hydrocarbon Processing, 135 (August 1992), for specific details about estimating the RVP.
- 528. Sec. 16.2 Modeling and Predicting Vapor Pressure a Function of Temperature 497 S LF .. ASS SSM NT TES Questions 1. Metallic lead can be recovered by smelting of the battery plates lead-acid batteries. If the plates are composed of 50% lead sulfate and 50% metallic lead. how would you gest eliminating most lead emissions from the process? 2. A solid can vaporize into a without going through a liquid phase. What form of an equation would you recommend to represent pressure of the gas when it is at equiHb-with the solid? 3. Do the steam tables. and similar tables for other compounds, provide more accurate values of the vapor pressure than use of the Antoine equation. or modifications of it? 4. [s it possible to prepare a Cox chart for water? Problems 1. Describe the state and values of the pressure of water initially at 200P as the temperature is increased to 2500P in a fixed volume. 2. Look in Appendix J the diagram CO2" a. At what pressure is solid in equilibrium with both CO2 liquid and vapor? b. What happens if the solid is placed in the atmosphere? ' 3. Use the Antoine equation to calculate the vapor pressure of ethanol at 50°C. and compare the result with the experimental value. 4. Determine the nonnal boiling point of benzene from the Antoine equation. 5. Prepare a Cox chart from which the vapor pressure of to1uene can be predicted over the temperature range - 20 to 140°C. 6. Make a large copy of Figure SAT16.2P6, a substance that contracts on freezing like most Figure SA T16.1P6
- 529. 498 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 substances (except water), and on the copy label all of the boundaries that represent equilibria between the phases. Also label the surface to show where the solid, liquid. and vapor regions themselves occur. 7. In Figure SATI6.2P7 label the ax.es and the contours on the special phase chart for water with the appropriate symbols for the variables involved. Figure SA T16.2P7 Thought Problems 1. In the start-up of a process, Dowtherm, an organic liquid with a very low vapor pressure, was being heated from room temperature to 335°F. The operator suddenly noticed that the gauge pressure wa' not the expected 15 psig but instead was 125 psig. Fortunately, a relief valve in the ex.it line ruptured into a vent (expansion) tank so that a serious accident was avoided. Why was the pressure in {he exit line so high? 2. A cylinder containing butadiene exploded in a research laboratory, killing one employee. The cylinder had been u~ed to supply butadiene to a pilot plant. When butadiene gas was required, heat was supplied to the cylinder to raise (he pressure of the butadiene in the tank. The maximum temperature that could be achieved in the lank on subsequent tests with a like tank was 160°C. At 152°C, the critical temperature for butadiene. the pressure in 628 Ibf lin. 2, les~ than one-half of the pressure required to rupture the tank by hydmulic test. Why did the tank explode? 3. "Careless Campers Contaminate Mountain Water" was a recent headline in the newspaper. The article went on: J
- 530. 16.2 Modeling and Predicting Vapor Pressure as a Function of Temperature 499 Beware! There are little monsters loose in those seemingly clean. pristine mountain streams. Their name: Giardia, and of specific interest to humans and pigs, Giardia lamblia. Giardia is a beet~shaped organism with no less than eight geila. It is of concern to anyone who happens to slurp any down because it attac. hes itself by means of a sucking organ1sm to the intestinal mucous membranes. The result is severe diarrhea, bordering on dysentery. The incidence of Giardia in the wilderness areas of New Mexico Colorado has vastly increased over the past five years. The disease it causes, giardiasis, is contracted by drinking water containing the organism. Unfortunately for all backpackers, horse packers, and day hikers. many of the lakes and streams are already tainted. No problem, you say-just drop in a chemical purification tablet and let it do the job? While chemical purification such as Halazone, iodine, or chlorine may kill many bacteria, the hale and hearty giardia goes unscathed. What steps would you take to avoid the problem other than carrying a potable water sup-ply with you? You need at least 210°F to kill the organisms boiling water. Discussion Problems 1. The following has been taken from the Science Essays of Dr. Ronald Delorenzo with permission. WHY HOT-WATER PIPES WATER Houses without cellars or basements usually have a crawl between the ground and the first floor for servicing plumbing. People living houses with crawl spaces may find their water pipes frozen after a very cold winter night. To their surprise. it is usuaUy the hot-water pipe that is frozen and not the cold~wa.ter pipe. Hot water pipes freeze before cold water pipes. Let's see if we can explain why this occurs. Can you dissolve more gas in hot water or in cold water? More gas can be dissolved cold water than in hot water. Most students answering this question for the first time give the wrong answers, probably cause they know that hot water will dissolve more solid solutes (like salt and sugar) than cold water. However, colder solvents dissolve more gas than warmer solvents. One reason soda pop is refrigerated is to keep it from losing its carbonation and going flat. Warm soda pop goes flat faster than cold soda pop. You've probably observed a related phenomenon when heating water to the boiling point. Before boiling occurs, many litde gas bubbles you see are dis-solved coming out of solution. When water is in a hot-water heater, the water is degassed. Dis-solved air escapes from cold water as it is heated in a hot-water Water in a hot-water pipe (supplied by the hot-water heater) contains dissolved gas than water in a cold-water pipe. Now can you explain why hot-water pipe before the cold-water pipe freezes? (For the answer refer to www.educationcenter.orgllorenzo.html.)
- 531. 500 Single-Component Two-Phase Systems (Vapor Pressure) Chap,16 2. Many distillation columns are designed to withstand a pressure of or SO psig. The re-boiler at the bottom of the column is where the used to vaporize the fluid in the column is introduced. What would you recommend as to the type of heat source among these three: (1) steam (heat exchanger), (2) fired heater (analogous to a boiler), and (3) hot oil (heat exchanger)? GLOSSARY OF N W WORDS Antoine equation Equation that relates vapor pressure to absolute temperature. Boiling Change from liquid to vapor. Bubble point The temperature at which liquid changes to vapor (at some pressure). Condensation The change of phase from vapor to liquid. Degrees of superheat The difference in temperature between the actual T and the saturated a pressure. Dew point The temperature at which the vapor just begins to condense at a specified pressure, that the values of the temperature along the vapor curve. Equilibrium A of the system in which there no tendency to spontaneously change. Evaporation The of phase of a substance from liquid to vapor. Freezing The change of phase of a substance from liquid to solid. Fusion Melting. Melting The change of phase of a substance from solid to liquid. Noncondensable gas A gas at conditions under which it cannot condense to a liq-uid or a solid. Norma.l boiling point The temperature at which the vapor pressure of a substance (P") is 1 atrn (101 kPa). Normal melting point The temperature at which a solid melts at 1 atm (101.3 kPa). Phase diagram Representation of the different phases of a compound on a two( or three-) dimensional graph. Quality Fraction or percent of a liquid-vapor mixture that is vapor (wet vapor). Reference substance The substance used as the reference in a plot substance Reference substance plot A plot of a property of one substance versus the same property of another (reference) substance that results in an approximate straight line. Saturated liquid Liquid that is in equilibrium with its vapor.
- 532. 16.2 Modeling and Predicting Vapor Pressure a Function of Temperature 501 Saturated vapor Vapor that is in equi1ibrium with its liquid. Sublimation Change of phase of a solid directly to a vapor .. Sublimation pressure The pressure given by the melting curve (a function of temperature) . Superheated vapor Vapor at values of temperatures and pressure exceeding those that exist at saturation. Steam tables Tabulations of properties of water and (water vapor), Subcooled liquid Liquid at values temperature and pressure less than those that at saturation. Supercritical region portion of a physical properties plot in which the sub-stance at combined p-T values above critical point. Triple point The one p-T-V combination at which solid, liquid, and vapor are all in equilibrium. Two-phase (region) Region on a plot physical properties in which two phases exist simultaneously. Vapor A gas below critical point in a system in which the vapor can condense. Vaporization The change from liquid to vapor of a substance. SUPPLEM NTARV REF RENeE In addition to references listed the Frequently Asked Questions in the front material, the following are pertinent. American National Standards Inc. ASTM D323-79 Vapor Pressure of Petroleum Products (Reid Method), Philadelphia (1979). Boublik, V. Fried. and The Vapor Pressure of Pure Substances, 2nd ed .• Else-vier, New York (1984). Chapey. N. P., and G. Hicks, eds. Handbook of Chemical Engineering Calculations. McGraw-Hill. New York (2003). De Nevers. N. Physical and Chemical Equilibrium for Chemical Engineers, Wiley- Interscience, New (2002). EOA Scientific Systems. Water, Vapor. Liquid, Solid. EOA Scientific Systems (2001). Pallady. P. "A Simple Way to Calculate Water Vapor " Chem. Eng., 133 mary 1993). Saul. A., and W. Wagner. uA Fundamental Equation for Water Covering the Range from the I Melting Line to 1273K at Pressures up to 25,000 MPa," J. Phys. Chern. Ref Data, 18, 1537-1 (1989). Yaws, C. Handbook of Vapor Pressure (4 volumes), Gulf Publishing, Houston, ( 1993-1995).
- 533. 502 Single-Component Two-Phase Systems (Vapor Pressure) Web Sites http://antoine.fsu.umd.edu⪙liquids/faq/antoine-vapor-pressure.shtml http://chemengineer.abouLcomllibrary/weekl y laa082 800a.h tm http://w ww .questcons u It.comJ -j rmJdew bu b.html http://www.taftan.comJsteam.htrn PROBLEMS -16.1 Select the correct answer(s) in the following statements: Chap. 16 ] . In a container of 1.00 L of toluene, the vapor pressure of the toluene is J 03 mm Hg. The same vapor pressure will be observed in a container of (1) 2.00 L of toluene at the same temperature~ (2) 1.00 L of toluene at one-half the absolute temperature~ (3) 1.00 L of alcohol at the same temperature~ (4) 2.00 L of alcohol at the same temperature. 2. The temperature at which a compound melts is the same temperature at which it (1) sublimes; (2) freezes; (3) condenses; (4) evaporates. 3. At what pressure would a liquid boil first? (1) 1 atm; (2) 2 atm; (3) 200 mrn Hg; (4) 101.3 kPa. 4. When the vapor pressure of a liquid reaches the pressure of the atmosphere surrounding it, it will (I) freeze; (2) condense; (3) melt; (4) boil. 5. Sublimation is the phase change from (1) the solid phase to the liquid phase; (2) the liquid phase to the solid phase; (3) the solid phase to the gas phase; (4) the gas phase to the solid phase. 6. A liquid that evaporates rapidly at ambient conditions is more like than not to have a (1) high vapor pressure; (2) a low vapor pressure; (3) a high boiling point; (4) strong attraction among the molecules. -16.2 Based on the following phase diagrams, answer the questions below, and explain your answers. (a) What is the approximate normal melting point for compound A? (b) What is the approximate nonnal boiling point for compound A? Compound A Compound B 1.5 10 E..- -E a ~ 1.0 ~ a. Q. 6 4 2 o 40 80'20160200 20 40 60 ao 1 00 120 T (0C) T (DC)
- 534. Chap. 16 Problems (c) What the approximate triple point temperature for compound B? (d) Which compounds sublime at atmospheric pressure? $16.3 Figure P 16.3 is a phase diagram for a pure compound. On a copy of the diagram place the name in the fonowing list next to the associated letter in the diagram: (a) the saturated curve (b) the saturated liquid curve (c) the saturated vapor curve (d) the liquid phase (e) the vapor phase c speciflc volume Figure P16.3 (0 the liquid-vapor two phase region (g) the critical point (h) a constant temperature Hne (i) a constant pressure line 11.'16.4 Draw a p-T diagram for a pure component Label the curves and points that are listed in P16.3 on it. "16.S One fonn of cooking is to place the food in a pressure cooker (8 sealed pot). Pressure cookers decrease the time require to cook the food. Some explanations of how a pressure cooker works are as fonows. Whlch of the explanations are correct? (a) We know PI Tl = P2T2' So if the pressure doubled, the temperature should be doubled and resulting to quicker cooking. (b) Pressure cookers are based on the principle that p oc: T, i.e .• pressure is directly proportional to temperature. With the volume kept constant, if you increase the pressure, the temperature also increases, and it takes less time to cook. (c) Food cooks faster because the pressure is high. This means that there are more impacts of molecules per surface area, which in tum increases the temperature of the food.
- 535. 504 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 (d) If we increase the pressure under which food is cooked, we have more collisions of hot vapor with the food, cooking it faster. On an open stove, vapor escapes into the surroundings, without the food more than once. (e) As the pressure inside the sealed cooker builds, as a result of the vaporization of water, the boiling point of water is thereby increasing temperature at which the food cooks-hotter temperature. time. ·16.6 Answer the following questions true or false: (a) The vapor curve separates the liquid phase from the vapor phase in a p-T diagram. (b) The vapor pressure curve the liquid phase from vapor phase in a p-V diagram. (c) The freezing curve separates the liquid phase from the solid phase in a p-T wagram. (d) The freezing curve separates the liquid phase from the solid phase in a p-V diagram. (e) At equilibrium the triple point, liquid and solid coexist. (f) At equilibrium at the triple point, solid and vapor coexist. ·16.7 Explain how the for a component changes (higher. lower, no change) for the following scenarios: (a) A system containing saturated liquid is compressed at constant temperature. (b) A system containing liquid is expanded at constant temperature. (c) A system containing saturated liquid is heated at constant volume. (d) A system containing saturated liquid is cooled at constant volume. (e) A system containing saturated vapor is compressed at constant temperature. (f) A system containing saturated vapor expanded at constant temperature. A system containing saturated vapor is heated at constant volume. (h) system containing saturated vapor cooled at constant volume. (I) A system containing vapor liquid in equilibrium is heated at constant vol-ume. G) A system containing vapor liquid in equilibrium is cooled constant vol-ume. (k) A system containing a superheated gas is expanded at constant temperature. (1) A system containing superheated gas is compressed at constant temperature. "'16.8 Ice skates function because a lubricating mm liquid forms immediately below the small contact area of the skate blade. Explain by means of diagrams and words why this liquid film appears on ice at "*16.9 Methanol been proposed as an alternate fuel for automobile engines. Proponents point out that methanol can be made from many feedstocks such as natural gas, coal, biomass, and and that it emits 45% less ozone precursor than does Critics say that methanol combustion emits toxic formaldehyde and that methanol rapidly corrodes automotive Moreover. engines using methanol are hard to start at £emperatures below 4Q0 F. Why are engines hard to start? What wouJd you recommend to ameliorate the situation?
- 536. Chap. 16 Problems 11116.10 Calculate the vapor pressure of each compound listed below at the designated temperature using the Antoine equation and the coefficients in' Appendix G, Compare your results with the corresponding values of the vapor pressures obtained from the Antoine equation found in the physical properties package on the CD accompanying this book. (a) Acetone at O°C (b) Benzene at Soop j. (c) Carbon tetrachloride at 300 K *16.11 Estimate the vapor pressure of ethyl ether at 40°C using the Antoine equation based on the experimental values as follows: l'(lcPa): T(OC): 2.53 -40.0 15.0 -10.0 58.9 20.0 ·16.11 At the triple point. the vapor pressures of liquid and solid ammonia are respectively given by In p* == 15.16 - 30631T and In p* = 18,70 - 37541T where p is in atmospheres and T is in kelvin. What is the temperature at the triple point? 11116.13 In a handbook the vapor pressure of solid decaborane (B IOH14) is given as .. 2642 loglO P = 8.3647 - T • 3392 Jog 10 P = 10.3822 - T The handbook also shows the melting point of B lOH 14 is 89.8°C. Can this be correct? 11116.14 Calculate the normal boiling point of benzene and of toluene using the Antoine equation. Compare your results with listed data in a handbook or data base . • 11116.15 Numerous methods are employed to evaporate metals in thin film deposition. The rate of evaporation is p Mil2 W = 5.83 X 10-2 v 112 gI(cml)(s) (Pv in torr, T in K. M = molecular weight) T Since Vy is also temperature-dependent. it is necessary to define further the vapor pressure-temperature relationship for this rate equation. The vapor pressure model is where Tis in K. B log 10 Pv = A - T Calculate the temperature needed for an aluminium evaporation rate of 10-4 g/(cm2)(s). Data: A = 8.79, B = 1.594 X 104.
- 537. 1 506 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 16 *16.16 Take 10 data points from the steam for the vapor pressure of water as a func-tion of temperature from the freezing point to 500 K. and fit file following function: p* = exp [a+b T + c (in Til. + d (1n n3) j where p is in kPa and T is in K. *16.17 For each the conditions of temperature and pressure listed below for water, state whether the water is a solid phase. liquid 'phase1 superheated, or is a saturated mix-ture. and if the latter, indicate how you would ,calculate the quality. Use the steam ta- f bles (inside the back to assist the calculations. i i . State p (kPa) T(K) if (mJlkg) 1 2000 475 2 1000 500 0.2206 3 101.3 200 4 245.6 400 0.7308 5 ]000 453.06 0.001127 6 200 393.38 0.8857 *16.18 Repeat Problem 16.17 for the fonowing conditions: State p (psia) T(CF) if (ttlJ1b) 1 0.3388 927.0 2 1661.6 610 0.0241 3 308.82 420 0.4012 4 180.0 440 2.812 ·16.{9 Calculate the specific volume for water that exists at the following conditions: (a) T= 100°C, p:;; 101,4 kPa. x= 0.5 (in m3/kg) (b) T::; 406.70 K, p == 300.0 kPa. x ::; 0.5 (in m3/kg) (c) T= loo.OoP, p:::.: 0.9487 psis, x:: 0.3 (in ff1llb) Cd) T= 860.97"R. p ::; psia, x;;;:; 0.7 (in ft3l1b) "'16.20 Answer the following questions true or false: (a) A pot full of boiling water is tightly closed by a heavy lid. The water will stop boiling. (b) Steam quality is the same thing as steam purity. (c) Liquid water that is in equilibrium with its vapor is saturated. (d) Water can exist in more than three different phases. (e) Superheated steam at 300°C means steam at 300 degrees above boiling point. (f) Watercan be made to boil without heating it. ' t 1
- 538. Chap. 16 Probjems ) I, , I 501 In a vessel with a volume of 3.00 m3 you put 0.030 m3 of liquid water and 2.97 of water vapor so that the pressure is 101.33 kPa. Then you he~t the system until all of the liquid water just evaporates. What the temperature and pressure in the vessel at that time? - In a vessel with a volume of 10,0 ft3 you put a mixture of 2.01 lb of liquid water and water vapor. When equilibrium is reached the pressure in the vessel is measured as 80 Calculate the quality of the water vapor in the and the reslpecltlve masses and volume liquid and vapor at 80 psis. "16.23 A vessel that has a volume of 0.35 m3 contains 2 kg of a mixture of liquid water and water vapor at equilibrium with a pressure of 450 kPa. What is the quality of the water vapor? ·16.24 A with an unknown volume is fiUed with 10 kg of water at 90°C. Ins}pection of , the vessel at equilibrium shows that 8 kg of the water is in the liquid state, What is the pressure in the vessel, and what is the volume of the vessel? ·16.25 What is the velocity in ftls wben 25 ,000 lblhr of superheated at 800 psia and 9000P flows through a pipe of inner diameter 2.9 in.? "'16.26 Maintenance of a heater was out to remove water that had condensed in the bottom of the heater. By accident hot oil 150°C was released into the heater when the maintenance man opened the wrong valve. The resulting explosion caused serious damage both to the maintenance man and to the equipment he was working on. Explain what happened during the incident when you write up the accident report. ··16.27 Prepare a Cox chart for: (a) Acetone vapor (b) Heptane (c) Ammonia (d). Ethane from O°C to the critical point (for each substance). Compare the vapor pressure the critical point with the critical pressure. "16.28 'Estimate the vapor of benzene 1 from the vapor pressure data by preparing a Cox **16.29 Estimate the sure data. p. (atro) T(°F): p' (psia): 102.6 3.36 212 25.5 pressure of aniline at 350°C based on the following vapor pres- 184.4 21 254.8 292.7 1.00 2.00 5.00 10.00
- 539. 508 Single-Component Two-Phase Systems (Vapor Pressure) Chap. 15 ·16.30 Exposure in the industrial workplace to a chemical can come about by inhalation and skin adsorption. Because skin is a protective barrier for many chemicals, exposure by inhalation is of primary concern. The vapor pressure of a compound is one commonly used measure of exposure in the worJcplace. Compare the relative vapor pressures of three compounds added to gasoline: methanol, ethanol, and MTBE (methyl tertiary butyl ether), with their respective OSHA pennissible exposure limits (PEL) that are specified in ppm (by volume): Methanol 200 Ethanol 1000 MTBE 100
- 540. CHAPTER 17 TWO-PHASE GAS-LIQUID SYSTEMS (SATURATION, CONDENSATION, AND VAPORIZA ION) 17.1 Saturation 17.2 Condensation 17.3 Vaporization Your objectives in studying this chapter ate to be able to: 1. Define what saturated gas means. Explain how saturation of a vapor can occur in a noncondensable gas. 3. Calculate the partial pressure of the components of a saturated ideal given combinations of the temperature, pressure, volume, and/or number of moles present. 4. Calculate the number of moles of a vapor in a saturated given the pressure and the temperature. S. Determine the condensation temperature (dew point) of the vapor in a saturated given the pressure, volume, and/or number of moles. 510 514 Why does it rain or snow? How can you predict the conditions of a mixture of a pure vapor (which can condense) and a noncondensable gas equilibrium? Based on the discussion in Chapter 16 about phase diagrams and pressure, you are now ready to consider the principles that will help you answer such questions. 509
- 541. 510 Two-Phase Gas-Liquid Systems Chap. 17 Looking Ahead In this chapter we explain what saturation meanst why the concept useful, and how to detennine the composition and dew point of a vapor in a noncondensable gas. We also discuss how condensation and vaporization relate to saturation. 17.1 Saturation When any noncondensable gas (or a gaseous mixture) comes in contact with a liquid, the gas will acquire molecules from the liquid. If contact is maintained for a sufficient period of time, vaporization continues until equilibrium is attained, at which time the partial pressure of the vapor in the gas will equal the vapor pressure of the liquid the temperature of the system. At equilibrium. the rate of vaporization is equal to the rate of condensation; therefore, the amount of liquid and the amount of vapor remain constant As a result, regardless of the duration of contact between the liquid and after equilibrium reached no more net liquid will vaporize into tM phase. The gas is then said to be saturated with the particular vapor at the given temperature. We also say that the gas mixture is at dew point. The dew point for the mixture of pure vapor and noncondensable means the temperature at which the vapor just starts to condense. At the dew point the partial pressure of the vapor is the vapor What is an example? You are familiar with air and water. At a given temperature, what is the partial pressure of the water in air when it is saturated? It is the vapor pressure (p*) of water at that temperature. Suppose the partial pressure the water vapor less than p. at some Can the gas be called saturated? No. Suppose the partial pressure of the water vapor is greater than p* -r, the dewpoint. Will the be saturated? This a nonsense question because at equilibirum the partial pressure of the water vapor cannot be greater than p* at the dewpoint. Consider a gas partially saturated with water vapor at p and T. If the partial pressure of the water vapor is increased by increasing the total pressure on the system, eventually the partial pressure the water vapor will equal p'" at of the system. Because partial pressure of water cannot p* at that temperature, a further attempt to increase the pressure on the system will result in water vapor condensing at constant T and p. Thus, p" represents the maximum partial pressure that water can attain at a r. Do you have to have both liquid and vapor present for saturation to occur? Really, no; only a minute drop of liquid at equilibrium with its vapor, or a minute amount of vapor in eqUilibrium with liquid. will suffice. What use can you make of the information or specification that a noncondensable gas is saturated? Once you know that a gas is saturated, you can determine the composition of the vapor-gas mixture from knowledge of the vapor pressure of the vapor (or equivalently temperature of the saturated mixture) to use in material
- 542. Sec. 17.1 Saturation 511 balances. From Chapter 13 you should that conditions the ideal gas law applies to both air and water vapor with excellent precision. Thus, we can say that the following relations hold saturation: nH20RT - PairV nairRT or ,. PH20 P H2O nH20 Protal - Pair --= - - Pair Pair because V and Tare the same for the air and water vapor. Also PH20 Ptotal - PH20 PH20 ---=--- = 1 - Yair Prur PH20 (17.1) (1 (1 Asa 51°C. and the pressure of water vapor is find that pit: = suppose you have a saturated gas) say water in air ....... "u .. ...., on the system is 750 mm Hg absolute. What is the partial Furtbennore. saturated, you know that the partial oreSSUlre You can look in a handbooks or use the steam Then Pair = 750 - 98 = 652 mm Hg vapor air mixture has the following composition PH20 98 YH20 = -- = - = 0.13 Ptotal 750 Pair 652 Yrur = -- = 750 = 0.87 Ptotal You can use these compositions when applying material balances. Calculation of the Dew Point of the Products of Combustion burned at atmospheric so that of the carbon burns to CO2, the dew So]ution Examine Figure E17. L To get the dewpoint, a mine P~20' To get P~20 you have to calculate P~20 = with 248% excess air product gas. you have to deter. The solu-
- 543. 512 Two-Phase Gas-Liquid Systems Chap. 17 tion of the problem involves preliminary calculations foHowed by material balances: Figure E17.1 1. Calculate the combustion products via material balances. 2. Calculate the mole fraction of the water vapor in the combustion products. 3. Calculate PH20 in the combustion products. 4. Condensation (at constant total pressure) will occur when P~20 equals the calculated PSIO 5. Look up the temperature corresponding to P~lO in the steam tables. Basis: 1 mol H2C20 4 Chemical reaction equations: H2~04 + 0.5 O2 ~ 2 CO2 + H20 H2C20 4 ~ 2CO + H20 + 0.502 1 mol H2C204 0.5 mol ----~~--x --------~ 1 1 mol H2C204 Mol 02 entering: (1 + 2.48)(0.5 mol 02) = 1.74 mol O2 Element material balances in moles: Component 2H C o 2N Molin 2 2 4 + 2(1.74) 1.74(0.79/0.21 )(2) Mol out 2 nH20 nco + nCOl nH10 + nco + 2nco1 + 2no~ With ncOz = (0.65)(2) = 1.30, the problem has zero degrees of freedom, and the solution of the material balances is 1
- 544. Sec. 17.1 Saturation 513 Component Mol nH20 1.00 neo2 1.30 nco 0.70 nO'l 1.59 nN2 6.55 Total IT.14 )'rhO = 1 mol H20/! L 14 mol total = 0.0898 partial pressure the water in the product gas (at an assumed atmospheric pressure) determines the dew point of the stack gas: P~20 = (Protal) = 0.0898 (101.3 kPa) = 9.09 (1.319 psia) From steam tables, the dewpoint temperature is: T;;;; 316.5 K (43.4°C or 110°F). SELF .. AS SSMENT T ST Questions 1. What the term "saturated gas" mean? 2. If a container with a volumetric ratio of air to liquid water 5 is heated to 60°C and equi-librium is will there still be liquid water present? What about at 125°C? 3. If a gas is saturated with water vapor, describe the state of the water vapor and the air if it (a) heated at conftant pressure; (b) cooled at constant pressure; (c) expanded at constant temperature; and (d) compressed at constant temperature. 4. How can you lower the dewpoint of a poHutant gas before analysis? 5. In a gas-vapor mixture, when is the vapor pressure the same as the partial pressure of the vapor in the mixture? 6. Will a noncondensable gas influence the vapor pressure of a vapor as total pressure changes if the temperature remains constant? Problems The dew point of water in atmospheric is 82°P. What is the mole fraction of water vapor in the air if the barometric pressure is 750 mm Hg7 2. Ten pounds KCt03 is completely decomposed and the oxygen evolved is collected over water at 80°F. The barometer reads 29.7 in. Hg. What weight of saturated oxygen is obtained? 3. Calculate the composition in mol fmction of air that saturated with water vapor at a total pressure of 100 kPa and 21°C.
- 545. 514 Two-Phase Gas-Liquid Systems Chap. 17 4. A benzene-air mixture with the composition between 1.4% and 8.0% can explode if niled. Will a saturated benzene air mixture at 1 attn be a potentia) ~xplosive? 5. An 8.00-liter cylinder contains a gas saturated with water vapor at 25,0°C and a pressure of 770 mm What is the volume of gas when dry at standard conditions? Thought Problem Why is it important to know the concentration of water in au a boiler? 17.2 Condensation From Tom and Ray Magliozzi (Click and Clack Talk Cars on PBS): Question: I have a 1994 Buick LeSabre with 19,000 miles. The car runs fecdy, except after it is parked our Occasionally, in the morning. I find a water puddle under the exhaust pipe about 6 inches in diameter. There seems to be a black carbon subst~nce on It is not greasy and does not seem to oil. What is this stuff?-Sidney. Ray: It's good old H20, Sidney. Water is one of the byproducts of combustion, so it's produced whenever you run the engine. Tom: And when you use the car for short trips, the exhaust system never really gets hot enough to evaporate the water so some of it condenses and drips out end of the tailpipe, Ray: And since carbon (or "soot") is also a byproduct of (incomplete) combustion, aU exhaust systems some carbon in them. So water takes a little bit of with it, and that's what you see on the puddle, Tom: It's perfectly nonnal. Sidney. You might even advantage of it by the car with the tailpipe hanging over your perennial bed. That'lI save you from watering it a couple of times a week. Examine the setup for combustion gas analysis shown in Figure 17. t. What error has been made in the setup? If you do not heat the sample of gas collected by the probe and/or put an intermediate condenser before the pump, the analyzer win fill with liquid as the gas sample cools. and will function. Condensation is the change of vapor (in a noncondensable gas) to liquid. Some typical ways of condensing a vapor in a are: 1. Cool it at constant system total pressure (the volume changes, of course). 2. Cool it at constant total system volume (the pressure changes), Compress it isothermally (the volume changes).
- 546. Sec. 17.2 Condensation Pump Analyzer 502 NOx CO2 I-C:::::J O 2 Calibration gas 515 Figure 17.1 Instrumentation for stack gas analysis. Combinations of the three as well as other processes are possible. of course. As an example condensation let's look at cooling a system constant total pressure for a mixture of air and 10% water vapor. Pick the air-water vapor mixture as system, If the mixture is cooled at constant total pressure from .sloe and 750 mm Hg absolute (point A for the water vapor in Figure 17.2), how low can the temperature before condensation starts (at point B, the same as point in Figure 17.2a, but a different point in Figure 17 .2b)1 You can cool the mixture until temperature reaches the dew point associated with the partial pressure of water of B. b. Region of partial saturation Vapor pressure I curve "- , 75 Total pressure Air pressure Saturation reached 51 750 75 46"'C isotherm / Vapor-liquid region Region of partial saturation ~ Water for water (region of saturation) c 48"C Figure 17.2 Cooling of an air water mixture at constant. total pressure. lines and curves for ,the water are distorted for the purpose of'illustration-the scales are not arithmetic units.)
- 547. 516 Two-Phase Gas-Liquid Systems Chap. 17 From the steam tabJes you can find that the corresponding temperature is = 46°C (points Band C Figure 17.2a on the vapor pressure curve}. After reaching p* = 75 mm Hg at point B, if the condensation process continued, it would continue at constant pressure (75 mm Hg) and constant temperature (46°C) until all of the water vapor had been condensed to Hquid (point C in Figure 17 .2b). Further cooling would reduce the temperature of the liquid water below 46°C. If the same air-water mixture starts at 60°C and 750 rnm Hg. and is cooled at constant pressure, at what temperature will condensation occur for the same process? Has the dewpoint changed? It same because the mole fraction of the water vapor in the air is the same, and PH20 0.10 (750) = 75 nun Hg still. The volume of both the air and the water vapor can be calculated from p V = nRT untH condensation starts, at which point the ideal gas law applies only the residual water vapor, not the liquid. The number of moles of H20 in the system does not change from the initial number of moles until condensation occurs, at which stage the number of moles water in the phase starts to decrease. The number of moles of air in the remains constant throughout the process. Condensation can also occur when the pressure on a vapor-gas mixture is increased. If a pound of saturated at 75°F is isothermally compressed (with a reduction in volume, of course), liquid water win be deposited out of the just like water being squeezed out of a wet sponge Figure 17.3), For example, if a pound of saturated air at 75°P and 1 atm (the vapor pressure of water is 0.43 psi a 75°F) compressed isothennally to 4 atm (58.8 psia), almost three-fourths of the original content of water vapor now wiH be in the fann of liquid, and the air sliB has a dew point of Remove the liquid water, expand the air isothennally back to 1 atm, and you will find that the dew point has been lowered to about 36°P. Here how to make the calculations. Let 1 = state at 1 atm and 4 = state at 4 atm with z = 1.00 for both components. PH,o ~ Pair ;:: 14.3 P'otal :: 4 4.7 Saturated Air F, I O1m psio ~iO :: 0.43 Pair :: 58.4 Ptotol : 5R9 Saturated Air 75~ F. 4 otm psio ~o = 0.11 "bir = I 4.6 . Ptotol :: I 4.1 Figure 17.3 Effect of an increase of pressure on saturated air, removal of condensed water, and a return to the initial pressure at constant temperature.
- 548. Sec. 17.2 Condensation 517 Pick as a basis 0.43 mol of H20. For saturated air at 75°F and 4 atm: * ( P H,lO) = 0.43 PaJr 4 58.4 For the same air saturated at 75°F and 1 atm: (nH20) (P~20) 0.43 nair I = Pair 1 = 14.3 Because the moles of air in state 1 and in state 4 are the same, the material balances simplify to 0.43 (n4) = 58.4 = 14.3 = 0.245 n, H 2 0 0.43 58.4 14.3 that is, 24.5% of the original water will remain as vapor after compression. After the air-water vapor mixture is returned to a total pressure of I atm, to get the partial pressure of the water vapor the following two equations apply at 75°F: PH20 + Pair = 14.7 PH20 nH20 (0.245)(0.43) - = - = = 0.00717 Pair nair 14.7 From these two relations you can find that PH20 = 0.105 psia Pair = 14.6 PtotaJ = 14.7 psia The pressure of the water vapor represents a dew point of about 36°F. N ow let's look at some examples of condensation from a gas-vapor mix ture. EXAMPLE 17.2 Condensation of Benzene from a Vapor Recovery Unit Emission of volatile organic compounds from processes is closely regulated. Both the Environmental Protection Agency (EPA) and the Occupational Safety and Health Administration (OSHA) have established regulations and standards covering emissions and frequency of exposure. This problem concerns the first step of removal of benzene vapor from an exhaust stream using the process shown in Figure E17.2a. The process has been designed to recover 95% of the benzene from air by compression. What is the exit pressure from the compressor?
- 549. 518 Alr 26"C. 1 atm II Mot fro Benzene 0.018 Air 0.982 ~.OOO Solution Compressor Two-Phase Gas-Liquid Systems Chap. 17 Air + Benzene vapor Vapor + Uquid Separator liquid Benzene Figure E17.2a Figure E17.2b illustrates on a p versus V" chart for pure benzene what occurs to the benzene vapor during the process. The process is an isothermal compression at 26°C. The partial pressure of the benzene if the entering air at 26°C is less than the saturation pressure. As the total pressure increases, the partial pressure of the benzene reaches the saturation pressure at point A in Figure E17.2b. Subsequently, any increase in total pressure causes the benzene to condense. If you pick the com- ' pressor as the system, the compression is isothermal and yields a saturated gas. You can look up the vapor pressure of benzene at 26°C in a 'handbook or get it from the CD in the back of this book. It is p" = 99.7 mm Hg. P 99.7 mmHg Va po rlLiquid Region ~- - -- V Vapor Figure E17.2b Initial state -- / ---e Next you have to carry out a short material balance to determine the outlet concentrations from the compressor: Basis: 1 g mol entering gas at 26°C and 1 atm Entering components to the compressor: mol of benzene ;:: 0.018 (1) = 0.018 g mol mol of air = 0.982 (1) = 0.982 g mol total gas = 1.000 g mol
- 550. Sec. 17.2 Condensation Exiting components the gas phase from the compressor: mol of benzene = 0.018 (0.05) = 0.90 x 10-3 g mol mol of air gmol total gas 0.983 g mol 3 = 0.916 X 10-3 = ~~ 0.90 X 10- YBenzene exiting = 0.983 PTotal N ow the partial pressure of the benzene is 99.7 nun Hg so that 99.7 mm Hg ",, PTotal = = 109 x hr nun Hg (143 atm) 0.916 X . Could you the pressure at the exit of the compressor above 143 atm? Only if all of the benzene vapor condenses to liquid. Imagine that the dashed line in Figure 17.2b is extended to the left until it reaches the saturated liquid line (bubble point line). Subsequently, the pressure can be increased on the liquid (it would folIowa vertical line as liquid benzene is not very compressible). EXAMPLE 17.3 Smokestack Emissions and Pollution A local pollution-solutions group has reported the Simtron Co. boiler plant as being an air polluter, and has provided as proof photographs of heavy smokestack emissions on 20 different days. As the chief engineer for the Simtron Co .• you know that your plant is not a source of pollution because you bum natural gas (essentially methane), and your boiler plant operating correctly. Your boss believes the pollu~ tion-solutions group has made an error in identifying the stack-it must belong to the company next door that bums coal. Is he correct? Is the poUurlon·solutions group correct? See Figure E17.3a. Figure E17.3a Solution Methane (CHJ contains 2 kg mol H2 per kilogram mol of if you look up composition of coal in a handbook or on the Lltemel. yuu wiH finel that a typical ,- 519
- 551. 520 Two-Phase Gas-Liquid Systems Chap. 17 coal contains 71 kg of C per 5.6 kg of in 100 of coal. The coal analysis is equivalent to 71 molC -12-k-g-C- = 5.92 kg mol C 5.6 or a ratio of 2.78/5.92 = 0.47 kg mol of Hikg mol C. Suppose that each fuel bums with 40% excess and that combustion is complete. You can compute the mole fraction of water vapor in each stack gas. and thus get the respective partial pressures of water vapor in the respective flue gases. Steps I, 2, 3, and 4 The process is shown in Figure StepS Natural Gas Step 4 Required 02: Excess 02: N2: Steps 6 and 7 Known Fuel (Dala Given --1IPi in Tobles) Knowl'! Air O2 0.21 Nz 079 -1.00 Products fo--.... (Oaio Given ill Tablti) Figure El'.3b Basis: 1 kg mol C 2 2(0.40) = 0.80 kg mol (2.80)(79121) ;;; 10.5 kg mol combustion problem is a standard type problem having zero degrees of freedom in which both the fuel and air flows are given, and you can calculate the product flows directly. as shown next. Steps', 8, and 9 Tables will make the analysis and calculations compact. '-
- 552. Sec. 17.2 Condensation Components Total kg mol 1.0 2.0 Composition of combustion product gases (kg LO 2.0 1.0 2.0 0.80 10.5 10.5 total kilogram moles of produced are 14.3 and mole fraction is 2.0 14.3 = 0.14 Coal is Excess Components C 2H Air Total 1 + 0.47(112) = 1 kg mol ( 1.24 )(0.40) = 0,49 mol 1.40(79/21)[1 + 0.47(1/2)] = 6.50 kg mol kg mol 1 0.47 CompMition of combustion product 1 1 0.47 0.47 0.49 0.49 (kg mol) 6.5 6.5 The total kilogram moles of produced are 8.46 and the mole fraction H20 0.47 8.46 = 0.056 the barometric pressure is, 100 kPa. and if the stack became satu-rated so that water vapor would start to condense at PH20. condensed vapor could be photographed. The respective partial pressures of the water vapor in combustion gases are: Partial "' ...... ,,', .. Equivalent temperature: Natural gas 100(0.14) = 14 kPa 52.SoC Coal 100(0.056) = 5.6 kPa 35QC
- 553. 522 Two-Phase Gas-Liquid Systems Chap. 17 Thus. the stack emit condensed water vapor at higher ambient temperatures for a boiler natural than for one burning coal. The public. unfortunately. sometimes concludes all the emissions they perceive are pollution. Natural could appear to the public to a greater pollutant than oil or coal, whereas. fact, the emissions are just water vapor;The sulfur content coal and oil can be released as sulfur dioxide to atmosphere. and the capacities of mercury heavy metals in coal and oil are much greater than natural gas when all three are being burned properly. The sulfur contents as to the consumers are as follows: natural gas, 4 X )0-4 mol (as added mercaptans to provide smell for safety); number 6 fuel oil, up to 2.6%; and coal. from 0.5 to 5%. In addition. coal release particulate matter into the stack plume. By mixing the stack gas with air, and by convective above the stack. the water vapor can reduced. and hence the condensation temperature can be reduced. However, equivalent dilution, the coal·burning plant will always a iewer condensation temperature. With the information calculated above, how would you resolve the questions that were originally posed? EXAMPLE 17.4 Material Balance Involving Condensation Contamination from accidental discharges of compounds used by industries, airports. businesses, and homeowners a challenging problem for amelioration. contaminated with poly aromatic hydrocarbons can be with hot air and steam drive out the contaminants. 30.0 m3 of air at 100°C and 98.6 kPa with a point of 30°C is introduced into the contaminated soil. and in the soil the gas cools 14°C at a pressure of 109.1 what fraction of the vapor in the gas at 100°C wi]] condense out in the soil if the gas does not esc:ape too rapidly from the soil? Solution Assume the system at 14°C is at equilibrium. and select a fixed volume of initial gas as the SVS:tCt1I1. Step S Basis: F;::;; 30 wet air at lOOOC and 98.6 kPa. Steps 1,2,3, '" Some of the data been placed in Figure 7.4
- 554. Sec. 17.2 Condensation P1nj :: S8.6 kPo T jtO 100" C dew pcirrl == 30" C F Dry Air Figure E17.4 p Dry Air = l09kPa W==? In addition. you need to get data for the vapor pressure of water at 30°C and 14°e: at 300e p. = 4.24 kPa at 14°C p. = 1.60 kPa By subtracting the vapor pressures from the respective total pressures you can cula[e the respective partial pressures of the and find the compositions of the entering and gas streams. Keep in mind that during the transition from 1 DOoe to 300e the water vapor does not condense. The needed compositions are listed below in kPa than mole fractions: Component Air: H20: Total: Steps 6 and 7 F 98.6 - 4.24 = 94.36 4.24 p 109.1 - 1.60 = 107.5 1.60 We have two unknowns. and W, and two independent balances can be made, air and water, thus degrees of freedom are zero. Steps 8 and 9 Calculate F (in moles) using the ideal law: n = -------::----= 0.954 kg mol 8.314 (kg mol)(K) material balances to calculate P and W are Air: 0.954(94.36) = p( 107.50') 98.6 . 109.1 ( 4.24) ( 1.60 ) H20: 0.954 98.6 = P 109.1 + W P = 0.926 mol W = 0.0274 mol
- 555. 524 Two-Phase Gas-Liquid Systems The initial amount of water was (4.24) 0.954 98.6 = 0.0410 kg mol and the fraction condensed was Step 10 0.0274 = 0.668 0.0410 A check can be made using a balance on Ihe total moles: F = P + W or 0.954 == 0.926 + 0.0274 = 0.953 SELF-ASSESSMENT TEST Questions Chap. 17 1. Can a vapor condense in a at temperature T if the partial pressure of the vapor remains less than the vapor pressure of vapor at T? 2. Why is substantial excess air used in coal combustion? 3. Is the dew point of a vapor-gas mixture the same variable as the vapor pressure? 4. What and how can variables be changed in a vapor-gas mixture to cause the vapor to condense? 5. Can a gas containing a superheated vapor be made to condense? Problems 1. A mixture of air and benzene contains 10 mole benzene at 43°C and 105 kPa pressure. At what temperature can frrst liquid form? What is the liquid? 2. Two hundred Ib of water out 1000 lb is electrolytically decomposed into hydrogen and oxygen at 25°C and 740 mm Hg absolute. The hydrogen and oxygen are separated at 740 mm and stored in two different cylinders each at 25°C and a pressure of 5.0 atm ab-solute. How much water condenses from the in each cylinder? A 3. Draw a p. versus T and a p. versus V diagram illustrating the condensation of a super-heated vapor from an air-vapor mixture when cooled at constant volume in a vessel until all of the vapor condenses. 4. Repeat Problem #3 for cooling at constant pressure and variable volume.
- 556. Sec. 17.3 Vaporization 525 Thought Problem 1. Water was drained from the bottom of a gasoline tank into a sewer and shortly thereafter a flash fire occurred in the sewer. The operator took special care to make sure that none of the gasoline entered the sewer. What caused the sewer fire? Discussion Problems 1. Whenever fossil fuels containing sulfur are burned heaters or boilers, sulfur dioxide, carbon dioxide, and water are fonned analogously, when municipal solid wastes are incinerated Hel and HBr form as well as sulfur dioxide. These acid gases form quite corrosive solutions if the water vapor in the flue gas condenses. How would you go about estimating the dew point of a flue gas such as one that contained 8% CO2, 12% H20, 0.73% N2i 0.02% SOl. 0.015% HCl. 6% 02' and 0.01% HBr? What protective measures might used to avoid corrosion the heater or stack surfaces? Will the point alone adequate information to alleviate corrosion? Hint: Compounds such as ferric chloride are very hydro scopic even at high temperatures. 2. Expired breath at body temperature. that 37°C. and essentially saturated.. Condensa-tion of moisture in respiratory equipment from the expired breath occurs as the breath Such a high humidity level has several implications in developing space suits, diving equipment, oxygen masks in hospitals, and so on. What might they be? 11.3 Vaporization In June 1992 an explosion rear Brenham, Texas, kiHed three people, injured 19, and caused over $10 million of damage. Liquid petroleum gas was being pumped. into a dome for storage under pressure. As the brine (which lay below the liquid petroleum in the salt dome) was forced out by the incoming liquid through a pipe that led [0 a surface storage pond, apparently too much liquid was forced into the liquid petroleum storage space in the salt dome. The liquid pushed out an of the brine and continued up through the brine discharge Hne to the surface pond where the Hquid vaporized. Apparently a check valve intended to stop such flow failed, and, in addition, someone had turned off the sensor to detect gas at the surface of the pond. After some time the cloud reached a source of ignition. Vaporization is the reverse of condensation, namely the transfonnation of a ' liquid into vapor (in a noncondensable gas). You can vaporize a liquid into a nonCODdensable gas, and raise the partial pressure of the vapor in the until the saturation pressure (vapor pressure) is reached at equilibrium. Figure 17.4 shows how the partial pressure of water and air change with time as water evaporates into initially dry
- 557. 526 ... ::z: " Two-Phase Gas-Liquid Systems Chap. 17 .l.... E. 760 .... _---..;,.;.;....~- 101.3 f!" 1'--"""';;';;"'-- 101.3 i .! 573 76.4 .=. ~ Figure 17.4 Change of partial and total pressure during the vaporization of water into initially dry air: (a) at constant temperature and total pressure (variable volume); (b) at constant temperature and volume (variable pressure). : ~ ... .t 187 ~---24.9 ~_--24.9 011<.....-------'-- Timl- COnttllnl T,mplfGlwt DItd Talal PrHlllr. (l/oliGt!le Volume) (a) Timl- Con,'an' Tlmptfolure Gnd 1/00IIIIIe lVariable PFlUllr.) (b) air. On ap-Tdiagram, such as Figure 17.2, a liquid would vaporize at the saturation temperature C (the bubble point temperature, which is equal to the dew point temperature) until the air became satu,.r ated. Look at Figure 17.2b, a p-V diagram. Evaporation of the liquid would occur from C to B at constant temperature and pressure until the air was saturated. At constant temperature and total pressure, as shown in Figure 17.5, the volume of the air would remain constant, but the volume of water vapor would increase so that the total volume of the mixture would increase. You might ask: is it possible to have the water evaporate continuously into air and saturate the air, and yet maintain a constant temperature, pressure, and volume, in the cylinder? (Hint: What would happen if you let some of the gas-vapor mixture escape from the system?) You can use Equations (17.1) through (17.3) in solving material balances involved in vaporization problems. For example. if sufficient liquid water is placed in a dry gas that is at 15°C and 754 nun Hg, if the temperature and volume remain constant during the vaporization. what is the final pressure in the system? The partial pressure of the dry gas remains constant because n, V, and T for the dry gas are constant. The water vapor reaches its vapor pressure of 12.8 mm Hg at 15°C. Thus, the total pressure becomes Prot = PH20 + Pair = 12.8 + 754 ::: 766.8 rom Hg 1 aIm 101m (101.3 kPa) 1101.3 kPo) 1 otM (tou kPo) Figure 17.5 Evaporation of water at constant pressure and a temperature of 65°C.
- 558. Sec. 1 Vaporization EXAMPLE 17.5 Vaporization to Saturate Dry Air What is the minimum number of cubic meters of dry air at 20°C and 100 kPa that are necessary to evaporate 6.0 kg of liquid ethyl alcohol if the total pressure remains constant at 100 and the temperature remains Assume that the is blown through the alcohol to evaporate it in such a way that the exit pressure of the air-alcohol mixture is at 100 kPa. Solution are Look at Figure E17.5. The process is isothermal. additional data needed alcohol at 20°C = 5.93 kPa Mol. wt. ethyl alcohol = 46.07 20"C 100 kPa Air Figure E17.S 100 kPo SoIuroted oir-alcohol miJI.tur8 minimum volume of air means that the resulting mixture is saturated; any condition than saturated would require more air. Basis: 6.0 kg of alcohol, The ratio of moles of ethyl alcohol to moles of in the final gaseous mixture is the same as the of the partial pressures of two substances. Since we the moles of alcohol, we can find the number of moles of needed the vaporization. .. Palcohol nalcohol = Pair nair Once you calculate number of moles air, you can apply the ideal gas law. Since P:lcohol = 5.93 Pair = PlOta! - P~COhol = (100 - )kPa = 94.07 !cPa 6.0 alcohol 1 alcohol 94.07 mol air 46.07 kg alcohol k I al h I = 2.07 kg mol air g mo co 0 8.314 (lcPa)(mJ) 293 K Vai.r=----- (kg mol) (K) 100 kPa 2.07 50.4 m3 at 200e and 100 kPa 527
- 559. 528 Two-Phase Gas-Liquid Systems Chap. 17 Another way to view this problem is to say that the final volume contains V of alcohol at 5.93 kPa and 20Ge V m3 of air at 94.07 kPa and 20ce V m3 of air plus alcohol at 100 kPa and 20"C Thus. the volume could be calculated from the information about ilie alcohol using ideal gas law _ (~) 8.314 293 _ 1 , Valcohol - -- 5.93 - 53.5 m at 20 C and 5.93 kPa The volume of the air is the same but the is at 94.07 kPa 20oe. You can adjust the volume of the alcohol of m3 to obtain the volume of the air: 3 94.07 m Vair = 53.5 m 3 air := 50.3 of dry air at 100 kPa and 200C 100 total EXAMPLE 17.6 Vaporization of a Hazardous Component or an onSHck determine the of components from a thin layer of oil spread on water, a 3 nun layer of oil containing various alkanes was allowed to evaporate. It was determined that a simple formula gave good predictions x = xoexp(-Kt) where x ;;;;;;; mole fraction of a component at time t Xo initial mole fraction of a component K = a constant determined by experiment, min-( It turned out that K could be predicted reasonably well by the relation 10gl0 (p*) = I 10gIO (K) + 0.160 where p. is in atm. Calculate the half-time (the time required to reduce the concentration of a compound by one-half) of n-heptane in an oil layer 3 mm thick at 22.3°C. Solution The vapor pressure of n-heptane at 22.3°C from Peny is 40 mm Hg.
- 560. Sec. 17.3 Vaporization Then IOglO( 7:~) = 1.251og10 (K) + 0.160 from which K = 0.0706 min-I. In(:J = In(On = -Kt = -O.0706tln -0.693 98' t - = . mm 112 - -0.0706 Sublimation into a noncondensable gas can occur as well as vaporization. The Chinook: A Wind That Eats Snow Each year the area around the Bow River VaHey in Southwestern Canada experiences temperatures that go down to -40oP. And almost every year, when the wind called the Chinook blows, the temperature climbs as high as 60oP. In just a matter of a few hours, this Canadian area experiences a temperature increase of about lOO°F. How does this happen? Air over the Pacific Ocean is always moist due to the continual evaporation of the ocean. This moist air travels from the Pacific Ocean to the foot of the Rocky Mountains because air masses tend to move from west to east. As this moist air mass climbs up the western slopes of the Rocky Mountains. it encounters cooler temperatures, and the water vapor condenses out of the air. Rain falls, and the air mass becomes drier as it loses water in the form of rain. As the air becomes drier, it becomes heavier. The heavier dry air falls down the eastern slope of the Rockies, and the atmospheric pressure on the falling gas increases as the gas approaches the ground . . . . the air gets warmer as you increase the pressure. As the air falls down the side of the mountain, its temperature goes up 5.5°F for every thousand-foot drop. So we have warm, heavy, dry air descending upon the Bow River Valley at the base of the Rockies. The Chinook is this warm air (wind) moving down the Rocky Mountains at 50 mph. Remember, it is now winter in the Bow River Valley, the ground is covered with snow, and it is quite cold (-40°F). Since the wind is warm, the temperature of the Bow River Valley rises very rapidly. Because the wind is very dry, it absorbs water from the melting snow. The word Chinook is an Indian word meaning "snow eater. It The Chinook can eat a foot of snow off the ground overnight. The 529
- 561. 530 Two-Phase Gas-Liquid Systems Chap. 17 weather warmer and the snow is cleared away. It's the sort of thing they have fantasies about in Buffalo, New York. Reprinted with permission. Adapted from an article originally appearing in Problem Solving in General Chemistry. 2nd ed. By Ronald DeLorenzo, 1993, 13()-L32, Wm. C. Brown Publ. SELF .. ASS SSMEN TEST Questions 1. If a dry gas is isothermally mixed with a liquid in a fixed volume. will the pressure remain constant with time? 2. dry is placed in contact with a liquid phase under conditions of constant pressure, and allowed to come to equilibrium, will a. the total pressure with time? b. volume of gas plus liquid plus vapor increase with time? c. the temperature increase with time? 3. If a beaker of water is in a bell and the water is maintained at constant temper-ature while the bell jar is evacuated, what will happen as time goes on? Problems 1. Carbon disulfide (CS2) at 20°C a vapor pressure of 352 nun Hg. Dry air is bubbled through the CS2 at 20°C until 4.45 Ib of CS2 are evaporated. What was the volume of the dry air required to evaporate this CS2 (assuming that the air becomes saturated) if the air was initiaHy at 20°C 10 atm and the final pressure on the air-CS2 vapor mixture is 750 mm Hg? In an acetone recovery system, the acetone is evaporated into dry N2. The mixture of ace~ tone vapor and nitrogen flows through a 2 ft diameter duct at 10 fusec. At a point the pressure is 850 mm and temperature lOO°F. The dew point is 80°F. Calculate the pounds of acetonelhr passing through the duct 3. Toluene is used as a diluent in lacquer formulae. Its vapor pressure at 30°C is 36.7 mm Hg absolute. If the barometer falls from 780 mm to 740 mm Hg. will there any change in the volume of dry air required to evaporate 10 kg of toluene? 4. What is the minimum number of m3 of dry air at 21 and 101 kPa. required to evaporate 10 kg of water at 21 Thought Problems 1. To reduce problems of condensation associated with continuous monitoring of stack gases, a special probe and flow controller were developed to dilute the flue gas with outside in a controlled ratio (such Wi 10 to 1). Does this seem like a sound idea? What II
- 562. Sec. 17.3 Vaporization 531 problems might occur with continuous operation of the probe? 2. A large fermentation tank fitted with a 2-in. open vent was sterilized for 30 minutes by blowing in Bve steam at 35 After the steam supply was shut off. a cold liquid sub· strate was quickly added to the at which point the tank collapsed inward. What hap. pened to cause the tank to collapse? Looking Back In this chapter we explained what saturation means. and showed how the vapor pressure of a compound mixed with a noncondensable gas reaches saturated conditions, which are a function of the temperature and total pressure on the system. Vaporization and condensation problems involving saturation were illustrated via phase diagrams and calculations. GLOSSARY OF N W WORDS Condensation The change of a vapor in a noncondensable gas to liquid. Evaporation See Vaporization. Saturated vapor A condensable vapor at its dew point in a noncondensable gas. Vaporization The transformation of a liquid into a vapor in a non condensable SUPPL M NTARY REFERENCES In addition to the references in the Frequently Asked Questions in the front material. the following are pertinent. American Institute of Chemical Engineers. SatuTation and Material Balances (Moldular In· structional Series Vol. 2), A.I.Ch.E .• New York (1981). Carey. V. P. Vapor·Liquid Phase·Change Phenomena: An Introduction to the Thermo-physics of VaporilQtion and Condensation Processes Heat Transftr Equipment, Hemisphere Publishing, New York (1992). Kandilkar, S. S. Mashiro. and V. K. Dhir~ eds. Handbook of Phase Change: Boiling and Condensation, Hemisphere Publishing. New York (1999). Patten. J. M. Liquid to Gas and Rourke. Vero Beach. FL (1995). Trechset. H. R.. ed. Moisture Analysis and Condensation Control in Building Envelopes " (ASTM Manual Series, Mn140)J American Society for Testing Materials. West Con~ shohocken. PA (2001).
- 563. 532 Two-Phase Gas-Liquid Systems Chap. 17 Web Sites http://www.doc.mmu.ac. uklairc/eaelW eather/Older/Condensation. html http://eng.sdsu.edultestcenter/I'esusolv ... temslclosediprocess/generic.html http://www.netguide.co.nz/condensation http://www.usatoday.comlweather/whumdef.htm PRO lEMS ''17.1 A large chamber contains dry N2 at and 101.3 kPa. Water is injected into *17.3 "17.6 chamber. After saturation of N2 with water vapor, the temperature the chamber is (a) What is the pressure inside the chamber saturation? (b) How many moles H20 per mole of N2 are present in the saturated mixture? The vapor pressure of hexane (C6HIJ at -20°C is 14.1 mm Hg absolute. Dry air at this temperature is saturated with the vapor under a total pressure of 760 mm Hg. What the percent excess for combustion? Suppose that you place in a volume of dry gas that is in a flexible container a quantity of liquid, and allow system to come to equilibrium at constant temperature and total pressure. Will the volume of container increase, decrease or stay the same from the initial conditions? Suppose that the container is of a fixed instead of flexible volume, and the temperature is held constant as the liquid vaporizes. Will the pres~ sure increase, decrease or remain the same in the container? In a search for new fumigants, chloropicrin (CC1 3NOz) has been proposed. To be fective. the concentration of chloropicrin vapor must be 2.0% in air. The easiest way to get this concentration is to saturate with chloropicrin from a container of liquid. Assume that the pressure on the container is 100 What temperature should be used to the 2.0% concentration? From a handbook. the vapor pressure data are (T, vapor mm Hg): 0, 5.7; to, 10.4; 15. 13.8; 20, lS.3; 23.8; 30. 31.1. At this temperature and pressure, how many saturate 100m3 of air? of chloropicrin are needed to What is the dewpoint of a mixture of air and water vapor at 60°C and 1 atm in which the mole fraction of the air 12%? The total on the mixture is constant Hazards can arise you do not calculate the pressure in a vessel correctly. One galIon of a hazardous liquid that has a vapor pressure of 13 psia at SO°F is transferred to a tank containing 10 ft3 of air at 10 psig and 80°F. The pressure seal on the tank containing air will rupture at 30 psis. When the transfer takes place. will you have to worry about the seal rupturing? A room contains 12,000 ft3 air at and 29.7 in Hg absolute. The air has a dew point of 600 P. How many pounds of water vapor are in the air?
- 564. Chap. 17 Problems 533 '*11.8 One ganon of benzene (C6H6) vaporizes in a room that is 20 ft by 20 ft by 9 ft in at a constant barometric pressure of 750 mm Hg abs01ute and 70°F. The lower explolimit for benzene in air 1S 1.4%. Has this value been exceeded? "'*]7.9 mixture of acetylene (C2H2) with an excess of oxygen measured ft3 at and 745 mm Hg absolute pressure. After explosion the volume of the dry gaseous product was 300 ft3 at 60°C. and the same pressure. Calculate the volume of actylene and of oxygen in the original mixture. The final was saturated. Assume that all of the water resulting from the reaction was in gas phase after the reaction . . ··17.10 In a science question and answer column, the foHowing question was On a trip to see the elephant seals in California, we noticed that when the male elephant seals were benowing, you could see their breath. But we couldn't see our own breath. How come? "'17.11 One way that safety enters into specifications is to specify the composition of a vapor in air that could burn if ignited. If the range of concentration of benzene in in which ignition could take place is 1.4 to 8.0 what would the correspond-ing temperatures for saturated with benzene of a storage tank? The lOtal pressure in the vapor is 100 kPa. ··17.12 In a dry cleaning establishment warm dry is blown through a revolving drum in which clothes are tumbled until all of the Stoddard solvent is removed. The solvent may be assumed to be n-octane (CgH 1s) and have a vapor of 2.36 Hg at 120°F. J f the air at 120°F becomes saturated with octane, calculate the: (a) Pounds of air required to evaporate one pound octane; (b) Percent octane by volume in the the drum~ (c) ft3 of inlet air required per lb of octane. The barometer reads 29.66 in. Hg. "'17.13 When people are exposed to chemicals at relatively low but toxic con centra-tion~, the toxic effects are only experienced prolonged exposures. Mercury is such a chemical. Chronic exposure to low concentrations of mercury can cause permanent menta! deterioration, anorexia. instability, insomnia, pain and numbness in the hands and feet. and several other symptoms. The level of mercury that can cause these symptoms can be present in the atmosphere without a being aware of it because Jow concentrations of mercury in the air cannot be seen or smelled. Federal standards based on the toxicity of various chemicals have been sel for the "Pennissible Exposure Limit," or PEL. These limits are set by the Occupational Safety Health Administration (OSHA). The PEL is [he maximum level of exposure permitted in the workplace based on a time weighted average (TWA) exposure. TW A is the average concentration permitted for exposure day after day without causing adverse effects. It is based on exposure for 8 hours day for the worker's lifetime. The present Federal standard (OSHNPEL) for exposure to mercury in air is 0.] mg/m3 as a ceiling value. Workers must be protected from concentrations greater than 0.1 mglm3 if they are worldng in areas where mercury is being used. Mercury manometers are filled and cali braled in a small store room that has no ventilation. Mercury has been spil1ed in the storeroom and is not completely cleaned
- 565. 534 Two-Phase Gas-Liquid Systems Chap. 17 up because the mercury runs into cracks and cracks in the floor covering. What is the maximum mercury concentration that can be reached in the storeroom if the temperature is 20°C? You may assume that the room has no ventilation and that the equilibrium concentration will reached. Is this level acceptable for worker exposure? Data: P~g == 1.729 X to-4 kPa; the barometer reads 99.S kPa. This problem has been adapted from the problems in the publication Safety, Health, and Loss Prevention in Chemical Processes published by the American Institute of Chemical Engineers, New York (1990) with permission. *'17.14 Figure P 17.14 shows a typical n-butane loading facility. To prevent explosions either (a) additional butane must be added to the intake lines (a case not shown) to raise the concentration of butane above the upper explosive limit (UEL) of 8.5% butane in air, or (b) air must be added shown in the figure) to keep the butane concentration below the lower explosive limit (LEL) of 1.9%. The n-butane leaving the water seal is at a concentration of 1.5%, and the exit gas is saturated with water (at 20°C). The pressure of the leaving the water seal is 120.0 kPa. How many m3 of air per minute at 20.0°C and 100.0 kPa must be drawn through the 'system by the burner if the joint leakage from a single tank car and two trucks is 300 cm3/niin at 20.DoC and 100.0 kPa? Air ........ __ _ intake Water seal II Figure P17.14 ·17.15 When you fill your gas tank or any closed vessel. the air in the tank rapidly becomes saturated with the vapor of the liquid entering the tank. Consequently, as air leaves the tank and is replaced by liquid, you can often smell the fumes of the liquid around the filling vent such as with gasoline. Suppose that you are filling a closed five-gallon can with benzene at After the air is saturated. what will be the moles of benzene per mole of air expelled
- 566. Chap, 17 Problems 535 from the can? Will this value exceed the OSHA limit for benzene in air (currently 0,1 mg/cm3)? Should you fiU a can in your garage with the door shut in the winter? *17.16 All of the water is to removed from moist (a process called dehydration) by passing it through silica SO ft3/min of air at 29.92 in. Hg absolute with a dew point of SO°F is dehydrated~ calculate the pounds of water removed per hour. ··17.17 Sludge containing mercury is burned in an incinerator. The mercury concentration in the sludge is 0.023%. resulting (MW = 32) 40,000 Iblhr, at Sooop, and is quenched with water to bring it to a temperature of 1 The resulting stream is filtered to remove all particulates. What happens to the mercury? Assume the process pressure is 14.7 psia. (The vapor pressure of Hg at 150°F is 0.005 psia.) ·*17.18 To prevent excessive ice formation on the cooling coils in a refrigerator room, moist air is partially dehydrated and cOQled before passing it through the refrigerator room. The moist air from the cooler is passed into the refrigerator room at the rate 20,000 ft3124 hr measured at entrance temperature and pressure. At the end of 30 days the refrigerator room must be allowed to warm in order to remove the ice from the coils. How many pounds of water are removed from the refrigerator room when the ice on the coils in it melts? *17.21 Moist Air 37°C COOLER 7°e Dew point 22"C 800mm 805 mm Hg libs. 000001 Hg abs. I Liquid H:P REFRIGERATOR Wate•r fr om melted ice -ieee Air at 2SQC and 100 kPa has a dew point of 16QC. you want to remove 50% of the initial moisture in the air (at a constant pressure of 100 kPa). to what temperature should you cool the air? One thousand of air saturated with water vapor at 30°C and 99.0 kPa cooled to 14 cC and compressed to 133 kPa. How many kg of HzO condense OlJt? Ethane (C2H6) is burned wjth 20 percent excess in a furnace operating at a pressure of 100 kPa. Assume complete combustion occurs. Determine the dew· point temperature of the flue Coal as fired contains 2.5% moisture. On the dry basis the coal analysis C: 80%, H: 6%, 0: 8%, ash: 6%. The flue gas analyses CO2: 14.0%, CO: 0.4%, O2: 5.6%, N2: 80.0%. The air used has a dew point 50QF. The barometer is 29.90 in. Hg. Calculate the dew point of the stack gas. A synthesis gas of the fonowing composition: 4.5% CO2, 26.0% CO, 13.0% H2• 0.5% CH4, and 56.0% N2 is burned with 10% excess air. The barometer reads 98
- 567. 536 U17.24 *·17.25 Two-Phase Gas-liquid Systems Chap. 17 kPa. Calculate the dewpoint of the stack gas. To prevent condensation and conse-quent corrosion, stack must be kept wen above their dewpoint. CH4 is completely burned with air. The outlet from the burner, which contain no oxygen, are passed through an absorber where some of the water is removed by condensation. The gases leaving the absorber have a nitrogen mole fraction of 0.8335. If the exit gases from the absorber are at 130°F and 20 psia. calculate: (a) what temperature must this gas be cooled at constant pressure in order to start condensing more water? (b) To what pressure must this gas be compressed at constant temperature before more condensation will occur? 3M removes benzene from synthetic resin base sandpaper by passing it through a drier where the benzene is evaporated into hot air. The air comes out saturated with benzene at 40°C (1 04°F). p* of benzene at 40°C::: 181 mm Hg; the barometer::: 742 mm Hg. They recover the benzene by cooling to 10°C (P* = 45.4 nun Hg) and compressing to 25 psig. What fraction of the benzene do they recover? The pressure is then reduced to 2 and the air is recycled in the drier. What is the partial pressure of the benzene in the recycled air? Wet solids containing 40% moisture by weight are dried to 10% moisture content by weight by passing moist air over them at 200°F, 800 mm Hg pressure. The partial pressure of water vapor in the entering air is 10 rom Hg. The exit air has a dew point of 140°F. How many cubic feet of moist air at 200°F and 800 rom Hg must be used per 100 lb of wet solids entering. "~'l7.27 Aerobic growth (growth in the presence of air) of a biomass involves the uptake of oxygen and the generation of carbon dioxide. The ratio of the moles of carbon dioxide produced per mole of oxygen consumed is caned the respiratory quotient (RQ). Calculate the RQ for yeast cells suspended in the liquid in a wen-mixed steady state bioreactor based on the following data: 41 Volume occupied by the liquid: 600 m3 • Air (dry) flow rate into the gas (head) space: 600 m3lhr at 120 kPa and 300 K • Composition of the entering 21.0% 02 and 0.055% CO2 .. Pressure inside the bioreactor: 120 kPa " Temperature inside the bioreactor: 300 K .. Exit Saturated with water vapor and contains 8.04% 02 and 12.5% CO2 .. Exit pressure: 110 kPa • Exit gas temperature: 300 K
- 568. d CHAPTER 18 TWO-PHASE GAS-LIQUID SYSTEMS (PARTIAL SATURATION AND HUMIDITY) 18.1 Terminology Involved for Partial Saturation 18.2 Material Balance Problems Involving Partial Saturation Your objectives in studying this chapter are to be sbJe to: 1. Define relative saturation (relative humidity), molal saturation (molal humidity). and humidity by formulas involving the partial pressures of the vapor and the gas. 2. Given the value of the partial saturation in one form, calculate the corresponding values in the other three forms as well as the dew pOint. 3. Solve material balances involving partial saturation including vaporization and condensation. 538 544 Chapter 17 explained partial saturation, but we did not discuss the concept using the terminology commonly encountered by engineers. You need to become acquainted with that terminology and its applications. looking Ahead In this chapter you wiU again dealing with partially saturated gases. but this time using some familiar, and some not so familiar, definitions. You will be pleased 537
- 569. 538 Two-Phase Gas-liquid Systems (Partial Saturation and Humidity) Chap. 18 to learn that the material balances we treat in processes such as partial saturation. vaporization, condensation. drying, and so on involve no new principles. 18.1 Terminology Involved for Partial Saturation We often solve old problems by abandoning them. Then we go on to create new generations of problems. Philip J. David A while ago partial saturation was involved in an extremely important lawsuit. The lawsuit centered on the way the energy content of natural gas is measured. The at which producers sell to pipelines and indirectly to consumers is determined by the energy content of the gas, according to the 1978 Natural Gas Policy Act. Before Congress passed the 1978 law, natural was measured as though it were partially saturated with water vapor. The water reduced the energy content of the natural gas per unit volume, and a fixed quantity of water vapor (1.75%) was arbitrarily included in the volumetric measurement regardless of how much water was actually present. Because the actual water vapor content of the gas was rarely as much as 1.75%, and the gas was priced per presumed million Btu in the gas rather than the actual (higher) Btu, the producers had to provide more energy per unit volume than they deemed to be fair. To correct this matter, in 1980 the Federal Energy Regulatory Commission (FERC) decided to change the measurement of natural gas volume so that the water content was no longer assumed to be at 1 % of the volume of gas, but was instead to be the actual water content. The new ruling enabled the gas producers to charge more for the same volume of natural gas sent to a pipeline. In 1983 a federal appeals court voided the new rule, and furthennore required that gas producers refund an estimated $1 billion in gas overcharges! Several ways exist to express the concentration of a vapor in a mixture with a noncondensable gas besides volume percent. In ordinary 'conversation you will find that the terminology used frequently gets mixed up with the correct terminology. What does the weather person mean when he or she cites the humidity as "60%H? The next time you listen to a weather report check the images and the comments with the following definitions involving partial saturation. When the vapor is water vapor and the gas is air, the special tenn humidity applies. For other or vapors, the term saturation is used. Here are the terms we are going to consider: (a) Relative saturation (relative humidity) (b), Molal saturation (molal humidity) (c) Humidity
- 570. · 1 1 Terminology Involved for Partial Saturation 539 18.1 .. 1 Relative Saturation (Relative Humidity) Relative saturation defined as the partia1 vapor pressure of the vapor at the temperature of the of the vapor divided by the 'RS ::= --. - = relative saturation p (18.1 ) where Pvapor = partial pressure of the vapor in the mixture p II< = partial pressure of the vapor in the mixture if the gas were saturated at the given temperature of the mixture (i.e., the vapor pressure of the vapor component) Then, for brevity, if the subscript 1 denotes the vapor ns ::= !?L = p)/PtO[ = V1/VtOl __ -'--- __ m_a_s_s,,-- p ~ P; / Plot VI, satdlVtot n 1. satd mass 1. satd (18.2) You can see that relative saturation, in effect, represents the fractional approach to the tota1 saturation. If you listen to the radio or TV and hear the announcer say that the temperature is 25°C (77°F) and the relative humidity is 60%, he or she implies that P.H20 (100) = %'R1i = 60 P H20 with both the PH20 and the P * H20 being measured at 25°C. Zero percent relative saturation means no vapor in the gas. What does 100% relative saturation mean? It means that the partial pressure of the vapor the IS same as the vapor pressure of the substance. EXAMPLE 18.1 Application of Relative Humidity to Calculate the Dewpoint The weather report on the radio this morning was that the temperature this afternoon would reach 94C1F, the relative humidity would be 43%, the barometer was 29.67 in. Hg, part]y cloudy to clear. with the wind from SSE at 8 milhr. How many pounds of water vapor would be in 1 mil of afternoon air? What would be the dew point of this air? Solution The vapor pressure of water at 94°P is 1.61 in. You can ca1culate the tia} pressure of the water vapor in the air from the given percent relative humidity.
- 571. 540 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 PH10 = (i .61 in Hg)(0.43) = 0.692 in. Hg Now dew point the temperature at which the water vapor the air wilt first condense on cooling at constant and the gas is cooled you can see from Equation (18.1) that the relative humidity increases the partial pressure of the water is constant while the vapor pressure water decreases with temperature, When percent relative humidity reaches 100% PH20 100 t = 100% or PH,O = p"H20 = 0,692 in. Hg P H20 water vapor start to condense. Prom the steam tables you can see that this corresponds to a lemperarure of about 68-69QP. 18.1 .. 2 Moliaf Saturation Another way moles of vapor to vapor concentration in a moles of vapor-free gas: Ilvapor ----= molal saturation nvapor-free gas to use the of the ( If subscripts 1 and 2 represent the vapor and the dry nary systemt respectively t for a bi- PI + P2 = Ptat :;;;.eL=.!J. n2 P2 V2 --nl -- -...:::....:.:.--- - --"--- ntot - nl Vtot - V t (18.4) 18.1 .. 3 Humidity (Specific Humidity) The special term humidity (11.) refers to the mass of water vapor per mass of bone-dry air, and is frequently used in connection with the humidity charts that will described in Chapter 29. By multiplying Equation (18 by the appropriate mole-cular weights, you can find the mass of vapor per mass of dry gas:
- 572. l Sec. 18.1 Terminology Involved for Partial Saturation 541 ... (nvapor) (mol. wt'vapor) ~ =--~~------~--~- mass vapor (ndry gas) (mol. wt.dry gas) massdry gas (18.5) Sometimes you see the word "saturation" used when a vapor other than water is involved. Table 18.1 summarizes the three relations we have discussed. TABLE 18.1 Summary of the Relationships Used for Vapor-Inert Gas Mixtures Humidity (normally applied only to water vapor) massWBle1Vapot' (nvapor) X (MWvapor) (Pvapor) X (MWvapor) 1-l= = =---~--------=--- maSSdry gas (ndry gas) X (MW dry gas) (Prow - Pvapor) X (MW vapor) Molal saturation (molal humidity when applied TO water vapor) n vapor nvapor Pvapot Pvapor molal saturation = -----=-- = ---'--- = --'---- = ----'--- nvapor free gas Iltotal - nvapor Pvapor free gas Prow - Pvapor Relative saturation (relative humidity when applied to water vapor) Pvapor Pvapor Ptotal nvapor R8 (or 'R1-l) = -.- = -.- = --""---- Pvapor Pvapor nvapor at saturation Now let's look at an example that uses all of the above terms in a problem. EXAMPLE 18.2 Calculations Involving Various Partial Saturation Terms The humidity of air at 30°C (86°F) and a total pressure of 750 nun "Hg absolute (100 kPa) is 0.0055. Calculate (a) the percent relative humidity, (b) the molal humidity. and (c) the partial pressure of the water vapor in the air. What is the dew point of the air-vapor mixture? Solution Data from the steam tables are p" H20 at 30°C = 1.253 in. Hg = 31.8 rnm Hg = 4.242 kPa (c) To get the partial pressure of the water vapor in the air, stan with the ! value of the humidity of 0.0055 Jb waterllb dry air. I (MWH20)(nH,o) (18)(PH,O) 18(PH2o) 1-{ = = - = 0.0055 (MWair)(nair) (29)(Pair) 29(ptow "- PH2o)
- 573. 542 Two&Phase Gas-Liquid Systems (Partial Saturation and Humidity) from which you can get = 6.71 mm Hg. (8) To get the relative humidity. PH 0 2 .=: 6.71 =021 21~ ;t 31 8 . or () PH20 J. (b) The molal humidity is 6~1 -3 - 750 - 6.71 = 9.03 x 10 Plomi - PHzO Chap. 18 The dew point the temperature at which the water vapor in the would first commence to condense when cooled at constant rotal pressure, that when the gas becomes completely saturated. This would be at the vapor pressure of 6.7 mm Hg, or about 1°C (41°F). S LF .. ASSESSMENT TEST Questions 1. Write down an equation that expresses the molal saturation in terms of the relative satura-tion. 1. Write down an equation iliat expresses the humidity in terms of the relative humidity. 3. Can the value of the relative humidity and ilie molal saturation ever be equal? 4. At the dew point of an air-water vapor mixture. what are (a) the relative humidity, and (b) the humidity? Problems 1. A mixture of air and benzene is found to have a 50% relative saturation at 2T'C and an absolute pressure of 110 kPa. What the mole fraction of benzene in the air? 2. A TV announcer says that the dew point is If you compress the air to 110°F and 2 psig, what is the percent relative humidity? 3. Nine hundred forty-seven cubic feet of wet air at 70°F and 29.2 in. are dehydrated. If O.94tb of H20 are removed, what was relative humidity the wet air? 4. The in your bedroom is at 20°C and 75% relative humidity. What window temperature will cause water to start condensing on the glass? 5. Toluene is mix.ed with at 21°C in such proportions that the partial pressure of the vapor is 10 mm Hg. total pressure is 745 mm Hg. Calculate the following: s. The relative saturation h. The moles of toluene per mole of vapor-free gas Co The weight of toluene per unit weight of vapor-free gas d. The percentage of toluene by volume
- 574. Sec. 18.1 Terminology Involved for Partial Saturation 543' Thought Problems 1. A stirred tank containing liquid CS2 solvent had to be cleaned because of solid residue that had accumulated on the stirrer. To avoid a possible fire and explosion, the CS2 was pumped out and the tank blanketed with nitrogen. Then the manhole cover was removed from the top of the tank, and a worker started to remove the solid from the stirrer rod with a scraper. At this point the maintenance worker left for lunch. When he returned to complete the job, a spark caused by the scraper striking the stirrer rod started a flash fire. How could the fire have occurred despite the preventive measure of using a N2 -blanket? 2. From: To: Subj.: Marine Board of Investigation Commandant (MVI) SS V. A. FOGG, O. N. 244971; sinking with loss of life in Gulf of Mexico on 1 February 1972 Findings of Fact At 1240 on February I, 1972, the tankship V. A. FOGG departed Freeport. Texas, en route to the Gulf of Mexico to clean cargo tanks that carried benzene residue. The vessel was due to arrive in Galveston. Texas. at 0200. on February 2. At approximately 1545, February 1. the V. A. FOGG suffered multiple explosions and sank. All 39 persons aboard died as a result of this casualty. Three bodies were recovered; two of the bodies were identified and one remained unidentified. The other persons were missing and presumed dead. You are asked your opinion of the most probable cause of the incident. What is your explanation? Discussion Problem 1. Toward the end of football season, the weather often turns out to be cold, snowy, and rainy. One report from Pittsburgh said that 20 groundskeepers worked around the clock in 18-hour shifts, fighting to prevent snow. gusty winds, and temperatures that dropped below zero from wrecking the delicate artificial playing surface at Three Rivers Stadium. Although the average footbal.l fan considers artificial turf fairly immune from the effects of the elements. at least compared to grass fields. nothing could be further from the truth. Artificial turf can become a carpet of swamp. To combat the elements, the grounds crew brought a 3-million BTU heating unit and a large propane tank onto the field and situated it beneath the 18-gauge, herculite tarpaulins that cover the field. A fter anchoring the tarps with weights and ropes and poles. the crew switched on the heater, blowing the tarps up into a tent of warm air that stood about 15 feet high in the middle of the field and raised the surface temperature to 30 or 35°F. The crew then began a constant vigil to prevent winds from ripping the tarps of the field. What would the relative humidity be beneath the tarp? Hint: What happens to the combustion products? Would providing a vent to the tarp be better than sealing it to the field from the viewpoint of water removal from the field?
- 575. 544 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 18.2 Material Balance Problems Involving Partial Saturation Sophisticated smugglers are trying to outwit customs officials who want to prevent the illegal trade in ozone-destroying CFCs. Illegal CFCs, worth an estimated $300 million a year, are flooding the market, a London seminar on environmental crime was told. Duncan Brack of the Royal Institute of International Affairs in London outlined some of the techniques smugglers use to pass off illegal CFCs as legal substitutes. One technique, he says, is to hide cylinders of CFCs inside larger cylinders of another gas-the gaseous equivalent of the false-bottomed · suitcase. But in many countries, there is no need for smugglers to go to such lengths. For instance, addjng nitrogen to cylinders containing illegal CFC-12 raises the pressure in the cylinder to match that used for HCFC-22, a substitute gas that is still legal. A pressure reading is often the only check that is made. How could you detect the phony CFC? A ga~ analysis would reveal a vapor in a noncondensable gas, N2. Would cooling the contents of a tank reveal an unusual p-T relationship? Most likely. What other suggestions could you make as to detect smugglers of CFC? In the examples that follow we treat dehydration, hl!midification, and drying, all processes that involve partial saturation. Keep in mind that if you know the dew point. the relati ve humidity, the humidity, or any other measure of partial saturation, you can calculate the composition of the air-vapor mixture to use in a material balance. EXAMPLE 18.3 Dehydration of Moist Air To avoid deterioration of drugs in a container. you proceed to remove all (0.93 kg) of the H20 from the moist air in the container that is at 15°C and 98.6 kPa by adsorption in siHca gel. The residual dry air measures 1000 m3 at 20°C and 108.0 kPa. What was the relative humidity of the moist air originally in the container? Solution Steps 1, 2, 3, and 4 Figure E18.3 contains the known data. 15° C and F='? 98.6 kPo 0 Vapor ---I H20 W'" 0.93 kg H 2 1000 m3 Oru, Air - Dry Aif A=? of 20Q C ond 108.0 kPo Figure E18.3
- 576. Sec. 18.2 Material Balance Problems Involving Partial Saturation StepS Either the Wor the A stream can serve as basis. Let's pick Basis: 1000 bone-dry air (BOA) at 200e and 108.0 kPa Steps 3 and 4 You first need to calculate the amounts (in kg mol) of water vapor and dry air in the original air. W = 0.93 kg or 0.93 --=..........;;;...... ------= 0.05 kg mol H20 18 kg H20 As for the bone dry air (BDA) 1000 BDA 273 K 1108.0 kPa 11 kg mol 293 K 101.3 kPa 22,4 m3 = 44.35 kg mol BOA Steps 6, 1, 8 and 9 The materia] balances are trivial. All of the water and all of the air in the orig-inal moist are accounted for in the above calculations. The problem has zero. de-grees of freedom. To get the relative humidity the moist air, you first have to calculate the partial pressure of the water vapor in the moist PH20 nH20 0.0517 -3 ·PCOI = ntot = 0.0517 + 44.35 = 1.164 X 10 PHzO = (1.164 X W-3}(108.0) = 0.126 kPa Next you look up the vapor pressure at 15°C for water, namely 1.70 kPa. Conse-quently, fractional relative humidity of the original air was 0.126 = 0.074 1.70 EXAMPLE 18.4 Humidification of Air To condition the in an office building in the winter, 1000 m3 of moist air hour at 101 kPa 22°C with a dew point 11°C enter the system. The air leaves the system at 98 kPa with a dew point of 58°C. How many kilograms water vapor are added to each kilogram of wet entering the process? Solution Steps 1, 2, 3, and 4 The known data appear in Figure E18A . . " 545 ..
- 577. 546 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) F = 1000 mS ot 101 kPo ond 22° C Air Entering ----~ ~O Vapor DItW Point 11° C Figure ElS.4 The additional vapor pressure data needed are System Boundary 1---Ai-r --E~tting P =? ~ Vapor Oew Point 58° C Dew polot temp. (OC) p. HID (mm Hg) 11 9.S4 58 136.1 1.31 18.14 t These values give the panial pressures of the water vapor in the initial and final gas mixtures. Chap. 18 Do you understand that the partial pressures of the water vapor are the pressures at the dew point in each case, and that the dry air has a partial pressure, which is the difference between the total pressure and the partial pressure of the water vapor? Let the subscript W stand for the water vapor and BDA stand for the dry air: In: PBDA = Ptat - Pw = 10 1 - 1.31 = 99.69 kPa Oue: PBDA = 98 - 18.14 = 79.86 kPa Step 5 The basis is 1000 m3 at 101 kPa and 22°C. Other bases could be selected such as 10 I kg mol of moist entering air or 98 kg mol of moist exit air. Step 6 and 7 You have two unknowns. Wand P, and can make both an air and a water balance, hence the problem has zero degrees of freedom. Steps 7 t 8, and 9 1000 m31101 kPa 1273 Kil kg mol _ . 10 3 295 K 3 - 41.19 kg mol wet atr 1. kPa 22.4 rn The material balances are in kg mol: BDA balance: 41.19(99.69) = p(79.86) 101 98 P = 49.89 kg mol Total Balance: F + W = P W= 49.89 - 41.19 = 8.70 kg mol H20 J. t l
- 578. Sec. 18.2 Materia! Batance Problems Involving Partial Saturation Step to Check using the water balance. Waler balance: .1 w= 8.68 mol ok Step 9 (Continued) To the of wet air Component Dry air 41. .19(_1 I) 101 Total 41.19 (8.70)( 18) 1188.6 + W = 49.89( 1:~1) Mol. wi. 18 kg water kg wet air in EXAMPLE Condensation of Water from Air kg 1179.0 9.6 il88.6 It's a typical1ate summer Houston day. The temperature 100oP, humidity is 90%, and the barometer is 29.76 in. Hg. The comer for its "free air" supply. and compresses the but the work of compression increases the air to 120o P. Does water condense out of the air how many Ib of water per ton of dry air that enters? E 18.5 shows the process and the data. PrOi = 29.76 In. Hg at l00"F '" , .93 in. Hg w.,.? 1()()4/o Figure Prot::: 64.7 (131.1 in. Hg) lit 12O"F "" 3.45 in. Hg 547
- 579. 548 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 Steps 3 and 4 The problem involves making material balances on air and water. To do so, you need to get the compositions of the inlet and outlet streams by first calculating the respective partial pressures of the air and the water vapor. Inlet Stream F P~O . • = f!k'H. so that PHlo. = 1.93(0.90) = 1.74 m. Hg P H20 Pajr = Plot - PH20 = 29.76 - 1.74 = 28.02 in Hg You can initially assume that condensation takes place, and calculate the partial pressures of the air and water vapor, respectively. You can subsequently check to see if this assumption is false by detennining if W is positive or negative. The outlet gas P is saturated when condensation occurs. Outlet Stream P Pair = Ptot - PH20 = 131.7 - 3.45 = 128.3 in. Hg StepS Several bases could be selected: 1 ton of dry air entering, 1 Ib of dry air entering. 1 lb mol dry air entering, etc. We will take a somewhat unusual but convenient basis, narnel y Basis: F = 29.761b mol Steps 6 and 7 The number of unknowns is two: P and W. The number of independent balances is two: air and water, hence the degrees of freedom are zero. Steps 8 and 9 The balances to be used are (note we didn't bother to calculate the mole fraction compositions directly): 1.74 ( ( 3.45 ) H20: 29.76 29.76) = W + p 131.7 (a) ,- " ,~ Air: ~::~~ (29.76) = 0 + p( ~~~:~) (b) J
- 580. 1 Material Ba[ance Problems Invotving Partial Saturation The solution to Equations (a) and (b) is P = 28.76 !b 3 = 0.99 lb mol Since W is positive, water condenses during the compression process. On the basis of 1 ton of dry air Ib mol out 18 lb waterllb mol water 12000 Ib = Ib mol dry air Ib airllb mol I ton Ib mol condensed ton of dry air. Thus (43.9)(18.016);::::; 7861b water. SELF .. ASSES MENT TEST Problems 549 /1. a synthetic gas plant analyzes (on a dry basis) 4.8% CO2, °2- 24.4% 12.2% H2• 3.6% CH4• and 54.4% N2. coal used 70.0% C. 6.5% H, 0, and 12.5% ash. entering air for combustion has a partial of water equal to kPa. The barometer 101 kPa. show that 465 steam is supplied to the combustion vessel per ton of coal Calculate the dew point of the exit gas. 2. A liquid solution of pharmaceutical material to dried is sprayed into a stream of hot water evaporated from the solution with the exit gases. The solid is recov-means of cyclone separators. Operating data are: Inlet 100,000 ft3lhr. 600°F, 780 mm humidity of 0.00505 lb H20llb dry Inlet solution: 300 lblhr, t solids, 70°F Outlet dew Outlet Calculate the composition of the outlet solid-it is not entirely dry. 3. A gas leaves a solvent recovery system saturated with benzene at and 750 mm Hg. The gas analyzes. on a benzene-free basis, t 5% CO, 02' and the The is compressed to 3 atm and is subsequently cooled to percent condensed in process. What is the relative saturation the final gas? Thought Problems 1. A company advertises for sale a super efficient two-compartment refrigerator that uses only $15.00 of electricity per also claim that food keeps it because 100 percent humidity is maintained in the refrigerator. it to 100% hu-midity in a refrigerator? 2. A science question answer in response to the question "What produces visible vapor trails behind jets at high altitude" wrote as follows:
- 581. Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Vapor or condensation trails are clouds. When an airplane engine water is released into the atmosphere as a byproduct of the combustion. If high altitude of is cold and very moist, the water from the and condense to form a airplane. Chap. 18 fuel, at the wiU cool In order to produce a trail behind a jet engine, combustion of the jet fuel must take in an with very high relative humidity. The upper level moisture is provided by an incoming pressure that pushes moisture ahead of it in an upper level .... """ ...... The presence of condensation trails indicates upper level a sys-tem to the west or southwest which could result in precipitation a or two. The first precursors to an incoming system are condensation trails aloft and a "halo" around the moon. The halo effect is caused by light passing through the upper atmospheric ice crystals, while vapor trails are caused by the condensation of water in an atmosphere that is very cold and of high humidity. Is the explanation provided by the columnist satisfactory? 3. On two different and barometric are the same. On day 1 the the air the most dense? ideal gas law. Why does humidity is high, on 2 the humidity is low. On which Justify your answer with using Dalton's law the air feel "heavier" on day 1 than on day 2? Discussion Problem 1. a method for detennining loss from biological products. storage of potatoes, for depends on several factors such as species, maturity, harvest conditions, degree and storage environment The paper pertained to a method of measuring from potatoes based upon measuring the of moisture content of the air in a system containing toes. whether or not the analysis of the method is and what confounding might cause it to in serious error. input data for the calculations were: potato weight in air. jar volume, barometric pressure, specific gravity, dew-point temperature, time age of the tubers from harvest date. Table 1 lists the van· ables, which were following calculations. TABLE 1 Symbols and Units of Variables Used the Moisture Loss Symbol Variable Symbol Variable Vp volume (em3) T temperature Wp Potato weight in air (g) Pv Vapor pressure (psi a) SG gravity (-) Td temperature (oF) VA volume (ft3) W Humidity ratio (lhH2d1bdry air) VJ volume (m3) WI Total weight of vapor in jar (lb) P Barometric pressure (in. Hg) Wtj Initial total weight of vapor at zero time (Ib) PI pressure (psia) AWa Difference between initial vapor weight and WA (lb) vapor weight at any particular (Ib) 6Wp Weight loss per lb of tuber J
- 582. 1 Material Balance Problems Involving Partial Saturation 551 " volume of the tubers In by the specific was caIulated by dividing the potato weight water (1.0 glem3): Wp Vp=- (1) Then, the air volume in eu ft3 in volume: was ....... """ from the jar ... a ........... and tuber (VJ - V ----...::..-A - (2.54 X 12)3 The barometric pressure was converted from the readings in inches of mercury to psia ( 14.59) PI = P 29.92 (144) (3) The mass of the min was calculated from the ideal gas law equation: PI x VA WA. = ---'----- + 4.60) (4) The amount of moisture in the was determined from the dew-point tenrlpe:rature at a particular time and mass of in jar. First, the vapor was ealeu· lated from the dew-point temperature, using an equation suggested Pv - 54.6329 12301.688 - e - Td + 460 - 5.16923 in (Td + 460) The mass of the water vapor in the jar was determined from mass the in The lU.J'J.3LIIoU water vapor in Wt at any time W = 0.6219 Pv 14.696 - Pv ~ = W X WA loss from the tubers was determined jar by subtracting the initial water than zero 6.Wa = WI - Wti Then. the weight loss per pount of tuber was determined 6.Wa 6.Wp = W 1454 p (5) humidity ratio and the (6) (7) in mass of the the water mass (8) particular time. (9)
- 583. 552 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 Looking Back In this chapter we explained a number of new tenns, all of which are used to identify partially saturated conditions in a noncondensable gas. We also solved material ba]ance problems in which these tenns were used to get the stream compositions. GLOSSARY OF NEW WORDS Humidity The mass of water vapor per mass bone-dry Molal saturation Moles of vapor divided by the moles vapor-free gas. Partial saturation A noncondensable containing a vapor whose partial sure less than the saturation pressure. Relative humidity The partial pressure of water vapor in a non condensable gas vided by the vapor pressure of the water at the temperature of the Relative saturation The partial pressure of a vapor in a noncondensable gas vided by the saturation pressure the vapor at the temperature of gas. SUPPL M NTARY R FERENCES American Institute of Chemical Saturation and Material Balances. Molular In-struction Series AIChE. New York (1981). r/--", Instrument Society ( 1985). America, Moisture and HumiditY. Instrument Society of America Shallcross, C. Handbook of Psychrometric Charts-Humidity Diagrams for Chemical gineers, Kluwer Academic, New York (1997). Web Sites http://members.aol.coml_hCalRogluthierlhumidity.html http://w3.one.netJ-jwclyrner/wet.html http://www.agsci.kvl.dkJ-beklrelhum.htrn http://www.crh.noaa.gov/mkfsoo/docuJhumidity.htm ROBLEMS at 60.0°C and 1 ° 1.6 abs. has a molal humldity of 0,030, detennine: (a) the relative humidity (b) the dewpoint of the (in °C) J
- 584. Chap. 18 Problems 553 "'18.2 What is the relative humidity of 28.0 m3 of wet at 27.0°C that is found to contain -18.3 ·18.4 0.636 kg of water vapor? Air at 80°F and I atm has a dew point of 40°F. What the relative humidity of this air? If the air is compressed to 2 atm and what the relative humidity of the resulting air? If a gas at 140°F and 30 in. Hg abs. has a molal humidity of 0.03 mole of H20 per mole of dry air. calculate: (a) The relative humidity (%) (b) The dew point of the gas (OF) The Weather Bureau reports a temperature of 90°F, a relative humidity of 85%, and a barometric pressure of ] 4.696 psia. (a) What is the molal humidity? (b) What is the humidity (weight basis)? (c) What is the saturation temperature or the dew point? (d) What is the number of degrees superheat of the water vapor? (e) Determine the molal humidity and dew point if the air is heated to 105°P. the pressure remaining steady. (f) Determine the molal humidity and dew point if the air is cooled to 60°F, the pres-sure remaining steady. (g) What fraction of the original water is condensed at 60°F? The Envirorunental Protection Agency has promulgated a national ambient air quality standard for hydrocarbons: 160 J.Lglrn3 is the maximum 3-hr concentration not to be exceeded more than once a year. It was arrived at by considering the role of hydrocarbons in the formation of photochemical smog. Suppose that in an exhaust gas benzene vapor is mixed with air at 25°C such that the partial pressure of the benzene vapor is 2.20 mrp Hg. The total pressure is 800 mm Hg. Calculate: (8) The moles ofoenzene vapor per mole of gas (total) (b) The moles of benzene per mole of benzene free gas (c) The weight of benzene per unit weight of benzene-free gas (d) The relative saturation The micrograms of benzene per cubic meter (f) The grams of benzene per cubic foot Does the exhaust gas concentration the national quality standard? A constant volume bomb contains air at 66°F and psia. One pound of liquid water is introduced into the bomb. The bomb is then heated to a constant temperature of 180°F. After equilibrium is reached. the pressure in the bomb is 33.0 The vapor pressure of water at 1800P 7.51 psia. (a) Did all of the water evaporate? (b) Compute the volume of the bomb in cubic (c) Compute the humidity of the air in the bomb at the final conditions in pounds of water per pound of air. ·18.8 A mixture of ethyl acetate vapor and air has a relative saturation of 50% at 30°C and a lotal pressure of 740 mm Hg. Calculate (a) the analysis of the vapor and (b) the molal saturation.
- 585. 554 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 a gas mixture there are 0.0083 Ib mol of water per Ib mol of dry at a temperature of 80°F and a total of 2 atm. (a) Calculate the relative saturation of this mixture. (b) Calculate the temperature to which the mixture must be heated in order that relative becomes 20%. A drier must remove 200 kg of H20 per hour from a certain material. Air at 22°C and 50% relative humidity enters the and leaves 72°C and 80% relative humidity. What is the weight (in kg) of dry air used per hour? The barometer reads 103.0 kPa. ··18.11 thousand (1 metric ton) of a slurry containing 10% by weight of CaCOJ are to filtered in a rotary vacuum The filter from filter contains 60% water. This cake is then placed into a drier and to a moisture content 9.09 kg H20/1OOkg CaC03• the humidity of the entering the is 0.005 of water of dry and the humidity of the leaving the is 0.015 kg of dry air, calculate: (a) the kg of water removed by the filter (b) the wet air entering the drier "18.12 In the solid electronic chips, concern over contamination pre~ dominates concern over cost In specification of nitrogen, control hydrocar~ bon content is essential. Cryogenic separation is a preliminary in the purifica-tion process. If nitrogen containing a hydrocarbon at a relative"{aturation of 3.8% at 750mm Hg absolute and 300K is to be cooled at 750mm Hg to remove 90% of the hydrocarbon as liquid, what must temperature of the outlet gas be? The vapor sure of the hydrocarbon the range of interest is given by 1 ,.. = 159008 _ 2788.5] n p , T - 52.36 where p* is in mm absolute and T is in K. "18.13 wet gas at 30°C and 100.0 kPa with a relative humidity of 75.0% was compressed to 275 kPa, then cooled to 20°C. How many m3 the original was com~ pressed if 0.341 kg of condensate (water) was removed from separator that was connected to the *18.14 An absorber receives a mixture air containing 12 percent carbon (CS2). absorbing solution is and gas exits from the absorber with a content of 3 percent and a benzene content of 3 percent (because some of the ..,"' .. "' ........ evaporates). What fraction of the was recovered? ***18.15 If a liquid with a fairly high vapor pressure at room conditions is stored in a fixed size tank that breathes, that is, has a vent to the atmosphere. because of ambient tempera-changes, how much Joss per day occurs in g mol octane m3 of free space under the following conditions, name]y the material stored is n-octane at the 50°C during day and at at night. The space the octane consists of and octane vapor that ex.pands and contracts. Ignore changes the liquid density. J
- 586. Chap. 18 Problems 555 " ··18.16 Thermal pollution is the introduction of waste heat into the environment in such a way as to adversely affect environmental quality. Most thermal pollution results from the discharge cooling water into the surroundings. It has suggested that power plant..;; use cooling towers and recyc1e water rather than dump water into the streams and In a proposed cooling tower. air enters and passes through baffles over which warm water from the heat exchanger falls. The enters at a temperature of SO°F and leaves at a temperature of 70oP. The partial pressure of the water vapor in the entering is 5 mm and the partial pressure the water vapor in the air leaving the tower is 18 nun Hg. The total is 740 nun Hg. Calculate: (8) The relative humidity of the air-water vapor mixture entering and of the mixture leaving tower (b) The percentage composition by volume of the moist air entering and of that leaving (c) The percentage composition by weight of the moist air entering and of that leaving Cd) percent humidity of the moist air entering and leaving (e) The pounds of water vapor 1000 ft3 mixture both entering and leaving (f) The pounds of water vapor 1000 ft3 of vapor-free air both entering and leaving The weight of water evaporated if 800,000 ft3 air (at 740 mm and 80°F) enters the cooling per ',,/ ·18.17 A must evaporate 200 Iblhr H20. Air at 700P and 50% relative humidity en· ters drier, leaving at 140°F 80% relative humidity. What volume dry air is necessary hour? "'18.18 1000 ft3 of air, saturated with H20, at 30°C and 740 mm are cooled to a lower temperature and one-half of the H20 is condensed out. Calculate: (a) How many pounds H20 are condensed out (b) The volume of dry at 30°C and 740 mm Hg .... 18.19 Moist.air at 25°C and 100 kPa with a dew point of 19.5°C is to be dehydrated so that during its passage through a large cold room used for food storage excess ice forma-tion can be avoided on the chilling coils in room. Two suggestions have of· fered: (I) Cool the moist to below the saturation temperature at 100 or (2) compress the moist air above the saturation pressure 25°C. Calculate the saturation temperature (I) and the total pressure at saturation for (2). (a) If 60% of the initial water in the entering moist has to removed before air enters the cold room, to what temperature should the air in process (1) be cooled? (b) What pressure should moist in process (2) reach for the same perset removal? (c) Which appears to be the most satisfactory? Explain. $18.20 A hydrocarbon fuel is burned with bone-dry air in a furnace. The stack is at 116 kPa and has a dew point of 47°C. The analysis of the gas shows 10 mol % car~ bon dioxide; the balance of oxygen and nitrogen. What is tbe ratio of hydrogen to carbon in the hydrocarbon "18.21 Hot that is used to dry pharmaceuticals recycled in a closed loop to prevent the contamination of the moist material from atmospberic impurities. In the first condi· tioning for the 5000 kg mollhr at 105 kPa and 42°C with a 90% relative humidity are fed to a condenser to remove some of the water picked up previously in
- 587. 556 Two·Phase Gas-Liquid Systems (Partial Saturation and Humidity) Chap. 18 the process. The air exits the condenser at 17°C and 100 kPa containing 91 kg mollhr of water vapor. Next. the air is heated in a heat exchanger to 90°C, and then goes to the dryer. By the time the air enters the dryer, the pressure ot the stream has dropped to 95 kPa and the temperature is 82°C. (a) How many moles of water hour enter the condenser? (b) What is the flow rate of the condensate water in kglhr? (c) What is the dew point of the air in the stream exiting the condenser? (d) What is the dew point of the air in the stream entering the dryer? "18.22 A certain gas contains moisture, and you have to remove the moisture by compression and cooling so that the gas will finally contain not more than I % moisture (by volume). You decide to cool the final gas down to 21°C. ·18.24 (a) Determine the minimum final pressure needed. (b) If the cost of the compression equipment is cost in $ = (pressure in psia)L40 and the cost of the cooling equipment is cost in $ :;: (350 - temp. K) 1.9 is 21°C the best temperature to use? flue gas from a furnace leaves at 31 and has an Orsat analysis 16.7% 4.1 % 02' and 79.2% N1. It is cooled in a spray cooler and passes under slight suction through a duct to an absorption system at 32.0°C to remove CO2 for the manufacture of dry ice. The gas at the entrance to the absorber analyzes 14.6% C021 6.2% °2, and 79.2% due to leaking into the system. Ca1culate the cubic meters of air leaked in to spray cooler per cubic meter of gas to the absorber, both measured at the same temperature and A wet sewage sludge contains 50% by weight of water. A centrifuging step removes water at a rate of 100 lblhr. The sludge is dried further by air. Use the data in Figure P18.24 to detennine how much moist air (in cubic feet per hour) is required for the process shown. Air 7d' F 50% Reiotve_-"'IiJ!o-I ttunldlty 760mm rig Water 100 IbIhr _---t Heater .---J.._....., Air 100°F Dried SUdge 94· F Dewpoint 750mm rio 104>/., Water by Weight Figure P18.24
- 588. Chap. 18 J.·U~.2S Problems 557-- To insure a slow rate of drying and thereby prevent checking of the dried productJ an inlet relative humidity of 70% at is specified for the moist air entering dryer. Outside is mixed with recycled from the dryer The leaving the dryer has a relative humidity of 95% at If the outside air has a dewpoint of 40°F, what fraction of the air entering the dryer must come from the outside and what fraction must come fron the recycled air from the exit to provide the desired moisture content in the air to dryer? See P18.2S. Recycle Moist Air -"'-----1-Il0>-lII 11-----'---- Moist Air Dried Products """---11 ....-__- --1f 4--Wet Figure P18.15 !-18.26 A drier must take up 200 Iblhr of H20. fresh at 70°F and SO% 'Rfi is mixed with recycle (R) and enters the drier. leavjng at 140°F and 80% 1lfi (P). What is the volume of R (al conditions) per volume of P (at its conditions)? The barometer reads 14.62 psia. and P and R are at O.IS psia gauge pressure. The dew point the after the heater is lOO°F. Figure P18.26. F ~o 200 Iblhr R Figure P18.16 P • .... 1~27 A process being designed to crystalize a phannaceutical soax (S) from benzene (Bz) solution, The process is shown in PlS.27. A solution of S in Bz is to the top a packed column. Dry air is fed to the bottom of the column. As the liquid j.la..:l."' ..... down, evaporates into the air stream. By the time the liquid leaves the boltom of the column, enough Bz ha~ been removed so that some of the S crystallizes out of solution. Wet crystals of S are then removed in a rotary vacuum filter. The liquor from the filter is mixed with the feed solution and sent to top of the column. Calculations indicate that the relative saturation of the benzene in the exit air will 0.70 at 65.0°C. Answer the following questions: (a) Calculate the required feed rate of air, in ft3lhr at 770 mmHg~ 7S oC, (b) Before developing the final process design, it necessary to account for the fact that the feed solution really contains 180 ppm of an undesirable toxic material, A, which has the same solubility in Bz as does S. If the present design is put into operation. what will happen to the A? Suggest one (or more) simple modification{s) to the process which win yield crystals of S which contain
- 589. 558 Two-Phase Gas-Liquid Systems (Partial Saturation and Humidity) Exit Air PI = 765mm Hg ReI. Sal. ;::: 0.70 Column Dry Air P t ::: mm Hg abs. ____ ....J 75°C Figure P18.27 wet crystals 0.25 Ibs Bzllb S Chap. 18 Feed Solution 781 [bslhr 31 S ag% much less than 180 ppm of A. part b of this problem. do not attempt to perfonn a material balance on the modified process or make any calculations. Just explain qualitatively in a very few sentences what happens to the A in the present and how the modification will reduce the concentration of A in the product crystals. *·'*18.28 Refer to the process flow diagram (Figure P18.28) for a process that produces maleic anhydride by the partial oxidation of benzene. The moles of 02 fed to the reactor per mole of pure benzene fed to the reactor is 18.0. The benzene leaves the still 180°F Mois' air I Qtm tempefCIture :: 150°F 50% relative Pure humidity tOry basis) 0.76 mole % C4H40", 80..0 mole 0/ .. N2 Reactor Steam still preuure "23.2 psia Contaminated benzene '--JI"-.... Woste benzene Condenso1e (liquid H2O) Steom 8 contaminant Figure P18.28 lotm 142°F saturated with H20 Water scrubber 12.0 mole "/ .. C4H.04 88.0 mole H20 ....---!IJ>- Pure H20 vopor Dehvdrator
- 590. Chap. 18 i: l Problems 559' before entering the reactor. All of the maleic acid produced the reactor is removed with water in the bottom stream from the water scrubber. All of the C6H6• 02' CO2, and N2 leaving the reactor leave in the stream from the top of the water scrubber, saturated with H20. Originally. the benzene contains trace amounts of a nonvolatile contaminant that would inhibit the reaction. This contaminant is removed by steam distillation in the steam still. The still contains liquid phases of both benzene and water (benzene is completely insoluble in water). The benzene phase is 80% by weight, and the water phase is 20% by weight of the total of the two liquid phases in the stilL Other process conditions are given in the flow sheet. Use the foHowing vapor-pressure data: Temperature (oF) 110 120 130 140 150 160 170 180 190 200 The reactions are Benzene (psla) 4.045 5.028 6.195 7.570 9.178 11.047 13.205 15.681 18.508 21.715 o CH-C( OH c<O OH Water (psis) 1.275 1.692 2.223 2.889 3.718 4.741 5.992 7.510 9.339 1] .526 + 2~ + H20 ~~ + 7!Ch ---+ 6CO:z + 3H20 Calculate: (1) (2) (a) The moles of "' .... &li .... "" •.• '" undergoing reaction (2) per mole of benzene feed to the reactor (b) The pounds of H20 removed in the top stream from the dehydrator per pound mole of benzene feed to the reactor (c The composition (mote percent, wet basis) of the gases leaving the top of the water scrubber (d) The pounds pure liquid H20 added to the top of the water scrubber per pound mole of benzene feed to the reactor
- 591. CHAPTER 19 THE PHASE RULE AND VAPOR-LIQUID EQUILIBRIA 19.1 The Gibbs Phase Rule 19.2 Vapor-liquid Equlibrla in Binary Systems Your objectives In studying this chapter are to be able to: 1. Understand and apply the phase rule to systems composed of one or more and one or more components. 2. Use Raoulfs law and Henryfs law to predict the partial pressure of a solvent and a solute, respectively, in the vapor phase. Use the relationship Kj =y/xj to calculate anyone of the variables the other two. 4. Calculate the composition of binary systems at equilibrium for the liquid and vapor 561 565 Environmental awareness and associated regulations will require you in the ture to the conflicts that pit producers of fuel. consumers of governmental agencies, politiciansi and environmental activists against each other. you are going to make wise technical and economic decisions, you have to understand contents of this chapter. Looking Ahead In this chapter we first discuss the Gibbs phase rule that explains physical degrees of freedom for systems at equilibrium. Then we proceed to take up various relations that you can use to predict the values of p. V, T, and the composition in binary systems at equilibrium. 560 J
- 592. Sec. 1 1 Gibbs Phase Rule 561 19.1 The Gibbs Phase Rule Let us help one another to see things better. Claude Monet The phase rule pertains only to systems at equilibrium. Equilibrium to .. a state of absolute rest .. no tendency to change state .. no processes operating (physical equilibrium) .. no fluxes of energy, mass, or momentum .. no temperature, pressure, or concentration gradients .. no reactions occurring (chemical equilibrium) Thus, phase equilibrium means that the phases present a system are invari-ant, as are the phase properties. By phase we mean a part of a system that is chemically and physically uniform throughout. This definition does not necessarily imply that the phase is continuous. For example. air bubbles in water represent a system that consists of two phases even though the bubbles are a discontinuous phase. The important concept of phase for you to retain is that a gas and liquid at equilibrium can each be treated as having a uniform domain. You can assume that most vapors and liquids at eqUilibrium are uniform. The phase rule concerned only with the intensive properties of the system. By this statement we mean properties that do not depend on the quantity of material present. If you think about the properties we have employed so far in this book, do you get the impression that pressure and temperature are independent of the amount of material present? Concentration is an intensive variable, but what about volume? The total volume of a system is called an extensive variable because it does depend on how much material you have. The specific volume and the density, on the other hand, as in cubic meters per kilogram, are intensive properties because they are independent of the amount of material present. You should remember that the specific (per unit mass or mole) values are intensive properties; the total quantities are extensive properties. Furthermore, the state of a system is specified by the intensive variables, not the extensive ones. You will find the Gibb's phase rule a useful guide in establishing how many intensive properties, such as and temperature. have to be specified to definitely fix all the remaining intensive properties and number of phases that can coexist for any physical system. The rule can be applied only to systems in equilib-rium. It that F=2 P C (19.1)
- 593. 562 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 where = number of degrees of freedom (Le" the number of independent properties that have to specified to an of the intensive properties of each of the system of interest)-not to be confused with the degrees of freedom calculated in solving material balances that can involve both intensive and variables. P = number of phases that can exist in the system; a IS a .. '-'.uv""','"' neous quantity of material such as a gas, a liquid, a solution, or a homogeneous solid. C = number of (independent chemical species) in the sys- Without reaction the system, the number of components (species) is equal to With reaction in the system, the value for C may be reduced to a value than the number of species, as explained in Appendix In particular cases additional constraints must be included Equation (19.1) that reduce the number of degrees of freedom (such as equilibrium and e]ectroneutrality among ions solution), but we do not have space to discuss them. Refer to the references at the of this chapter. such as Rao (1985), for "- Look at Figure 19.1, the three-dimensional representation of the p-V-T proper-of water that you previously saw in 16.6 in Chapter 16, Consider the vapor phase. Three-dimensional of the properties of water as a surface with dimensions p. V, and T. J
- 594. Sec. 1 1 The Gibbs Phase Rule 563 You will remember a pure gas that we had to specify three the four vari-ables in the ideal gas equation pV = nRT in order to be able to detennine the remaining one unknown. You might conclude that F = 3. If we apply the phase rule, for a single phase P = 1. and for a pure gas so that C = 1 =2 P C=2-1 1 = 2 variables to specified How can we reconcile this apparent paradox with our previous statement? Eas- , ily! Since the phase is concerned with intensive properties only, the following are the phase-rule variables to included in the ideal gas law: ~ (specific molar volume) } 3 intensi ve properties Then you can write the g~s law as A pV = RT (19.2) and in this form you can see that by specifying the values of two intensive variables so that = 2, third can calculated. Thus, in the superheated region in the steam tables. you can fix all of the properties of the water vapor by specifying two intensive variables. An invariant system one in which no variation of conditions is possible without one phase disappearing. In Figure 19.1 the ice-water-water vapor system at only one temperature (O.OlOe) and pressure (0.611 kPa) along the triple point line (a point in a p.T diagram), and represents one of the invariant states in the water system =2-P C=2-3+1 0 With all three phases present, none of the physical conditions can be without the loss of one phase. As a corollary, if the phases are present, the tem-perature. the volume, and so on, must always fixed at the same values. This phenomenon is useful in calibrating thermometers and other instruments. Now let's look some examples of application of the phase EXAMPLE 19.1 Applications of the Phase Rule to Systems Without Reaction ......... J ..... U'LC;, ....... the number of degrees of freedom (how many additional intensive variables must be specified to the system) from the phase rule the foHowing materials at equilibrium:
- 595. 564 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 (a) liquid benzene (b) of ice and water only (c) of liquid benzene vapor, and helium (d) A mixture of salt and to achieve a vapor pressure. What might be specified in Solution 1. hence = 2 - 1 + 1 == The temperature in the range in which benzene remains a (b) N = I, P = 2, and C = 1. :F = 2 - 2 + 1 = 1. Once either the temperature or the is specified, intensive variables are fixed. (c) N = 2, P = temperature, and C = 2, hence :F = 2 - 2 + 2 = 2. A pair from or mole fraction can be specified. (d) N = 2, P = 2, Iar pressure is to the temperature of = hence :F = 2 - 2 + 2 - 2. Since a achieved, you would adjust the salt concentration and solution. in (a) and (b) it would be likely that a phase would exist in tice, increasing P by 1 and by one . EXAMPLE 19.2 in .................. of the Phase Rule to Systems Reactions Can Occur Calculate the number of degrees of freedom lowing systems at equilibrium the phase rule (8) You have a composed of CO, H20, and CH4. the fol~ (b) Sulfur can removed from high temperature gas streams a bed of zinc oxide ..................... The results of the decomposition of zinc oxide with carbon show presence of the following compounds. ZnO(s) C(s) CO(g) Zn(s) Solution (a) P = I, N and you need to detennine C. From the information given in Appendix Ll, you can detennine that the number independent species is to the number of present among compounds, namely 3. Consequently, C = 3. :F=2-1+3=4 - -- --- ------------------------------------""""""'"
- 596. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems 565 You might specify T, p, and two mole fractions. (b) :;:: 4 different soUd phases plus 1 phase). By use of the method explained in Appendix Lt, you can calculate that C::: 3. =2-4+3=1 You can fix Tor p. S LF-ASS SSM NT TEST Questions 1. List two intensive and two extensive properties. Answer the following questions true or false: (a) A phase is an agglomeration of matter having distinctly identifiable properties such as a distinct refractive index, viscosity, density, pattern, etc. (b) A solution containing two or more compounds comprises a phase. (c) A mixture real comprises a single phase. Fill the following table for water Number of Phases 2 3 Example Is the critical point a Problems Degrees of Freedom Number of Variables That Can be Adjusted at EquiUbrium 1. Determine the number of degrees freedom from the phase rule for the foHowing terns at eqUilibrium: a. Liquid water, water vapor, and nitrogen. b. Liquid water with dissolved acetone in eqUilibrium with their vapors. c. H20(g), H20(f), CuS04 .. 5H20(s), CuS04 in solution. 2. Can the following system exist at equilibrium: H20(s). "2°0), H20(g), decane(s). cane( 1), decane(g)? (Hint: Decane is insoluble in water.) 19 .. 2 Vapor ..... Llquld Equilibria In Binary Systems In Chapter 16 we discussed vapor-liquid equilibria for a pure component. In Chapters 17 18 we covered equilibria of a pure component in the presence of a noncondensable gas. In this section we consider aspects of amore general
- 597. 566 The Phase Rule and Vapor-Liquid Chap. 19 of circumstances. namely cases in which equilibrium exists for a binary vapor and liquid system, that is, the vapor and liquid phases each contain both components. Distilling moonshine a fennented grain mixture an example of binary vaporliquid equilibrium in which water and ethanol are the primary components the system. The end result of vapor-liquid equilibrium in distillation the more volatile component (the component with the higher vapor pressure a given temperature) tends to accumulate in vapor phase while the less volatile component to accumulate in the liquid phase. Distillation columns, which are used to separate a mixture into its components, are based on principle. Distillation columns are the most commonly used separation technique in the chemical process industry. A distillation column is comprised of trays that provide contact between liquid and vapor streams inside column. At tray, the concentration of the more volatile component is increased in the vapor stream leaving the tray and the concenttation the less volatile component is increased in the liquid leaving the tray. In this manner, applying a number of trays in series, the more volatile component becomes quite concentrated in the overhead from the column while the less volatile component becomes concentrated in bottom product In order to . and analyze distillation columns, you must be able to quantitatively describe vapor-liquid equilibrium. When you analyze binary systems at equilibrium, the number of variables "- volved increases beyond p. T, V. Compositions must specified, possibly in each phase. It hard to visualize phase phenomena that should reany be portrayed in three or more dimensions. You have to interpret projections of three-dimensional surfaces in two dimensions, such as 19.2 and 1 below. 19.2 .. 1 Ideal Solution Relationships An ideal solution a mixture whose properties such as vapor pressure, cific volume, and so on can be calculated from the knowledge only of the corresponding properties of the pure components the composition of the solution. For a solution to behave as an ideal solution: • All of the all have roughly the same size. • All of the molecules have essentially the same intennolecular interactions. Most solutions are not but some solutions are nearly ideal. a.. Raoultts Law The best known relation to make predictions for ideal dons is Raoult's Law:
- 598. Sec. 19.2 Vapor-Liquid Equilibria in Binary Syste'ms 567,.. (19.3) where Pi partial pressure of component i in the vapor phase xi = mole fraction of component i in liquid .............. p7(T) = pressure of componenti at T 19.2 how the vapor pressure of the two components an ideal binary solution sum to the total pressure at SO°C. Compare Figure 19.2, which is based on an solution, with Figure 19.3 which is based on a nonideal solution. Raoult's law used primarily for a component whose mole fraction ap-proaches unity (look at methyal in Figure 19.3 as xcs2 -i> 0), or for solutions of com-ponents quite similar chemical such as straight chain hydrocarbons. b. Henry's Law Henry's law is primarily for a component whose mole fraction approaches such as a dilute gas dissolved in a liquid: P·=H. t l (19.4) where Pi is the partial pressure in the gas phase of dilute component at equilibrium at some temperature, and the Henry'$ law constant. Note that limit nIh."""''''' Xi == 0, ::::: O. Values of Hi can be found in several handbooks and on the Internet. If the solute dissociates into two species such as ions when in solution, Equa£ tion (19.4) does not apply. T::: eo"c B Mol fraction benzene Figure Plot of partial pressures and total pressure as a function of composition calculated using Raoult's for an solution of and toluene (like species). The vapor pressures the pure components are shown by points A B atO 1.0 mole fraction benzene.
- 599. 568 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 T:::: Pto!.al 1,0 0.80 E ~ ~ 0.60 :::) ~ G) 0.40 .... Figure 19.3 Plot of the partial Il. 0.20 pressures and total pressure (solid lines) exerted by a solution of carbon " disulfide (CS2}-methylal CH2(OCHJ)2 0 as a function of composition. The 0 0.25 0.50 0.15 1.0 dashed lines represent the pressures that would exist if the solution were Mol fraction carbon disulfide ideal, Henry's law is quite simple to apply when you want to calculate the partial pressure of a in the gas phase that is in eqUilibrium with the gas dissolved in the liquid phase. Take, for example, CO2 dissolved in water at 40°C for which the value of H is 69,600 atm/mol fraction. (The large value of H shows that CO2(g) is only sparing soluble in water.) If .teo2 = 4.2 X 10-6, the partial pressure of the CO2 in the gas phase is From the viewpoint of an aquatic environment, a compound discharged into it will have a low risk potential if it has a large Henry's law constant, and vice-versa. From the viewpoint of an atmospheric environment, a gas with a global warming potential might have less effect if it had a small Henry's law constant. Even though CO2 is only slightly soluble, in 1986 the CO2 buildup in the deep water of Lake Monoun (the "killer lake") in the Cameroon exploded to the surface, causing a 260-foot tidal wave that flattened everything along the shoreline. Also, since CO2 is denser than air, it blanketed the ground so that 1.750 people and thousands of livestock perished from suffocation or drowning. Most people would say that when you shake a bottle of soda and see the bubbles form, the pressure in the bottle increases. Certainly you uncap the bottle after shaking, the liquid foams out the top. However, if the bottle stays at a fixed temperature, the CO2 reach an equilibrium state with the liquid, and shaking just displaces some CO2 from gas phase into the liquid phase. and it is bubbles of this (not the dissolved gas) that you see bubble out of the bottle.
- 600. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems i ! ~ Q. 19.2-2 Vapor-Liquid Equilibria Phase Diagrams Adventurers are easier of entrance than exit; and it is but common prudence 10 see our way out before we venture in. Aeson The concept of a phase diagram. which was discussed in Chapter 16. for a pure component can be extended to cover binary mixtures. In this section we treat liquid-vapor equilibria, but you can find information concerning solid-solid. solid-liquid. and solid-vapor equilibria in the references at the end of this chapter. Experimental data usually are presented as pressure as a function of composition at constant temperature, or temperature as a function of composition at a constant pressure. For a pure component, vapor-liquid eqUilibrium occurs with only one degree of freedom: F=2-P+C=2-2+ 1 =1 At one atmosphere pressure, vapor-liquid equilibria will occur at only one temperature- the nonnal boiling point. However, if you have a bjnary solution, you have two degrees of freedom: .1"=2-2+2=2 For a system at a fixed pressure, both the phase compositions and the temperature can be varied over a finite range. 1.0 0.80 0.62- O~sO 0.40 0 Vapor phase composttlon Uquld phase composition ~ I #<" : .. ~I : Gapov I U I I _ Figure 19.4 Phase diagram for a mixture of benzene and toluene at 80°C. At 0 mol fraction of benzene (point A) the pressure is the vapor pressure of to1uene at sOGe. At a mol fraction of benzene of 1 (point B) the I I pressure is the vapor pressure of .-Inttlal8tate 0.25 0.38 0.5 0.75 Mol fraction benzene benzene at sOGe. The tie line shows the 1.0 liquid and vapor compositions that are in equilibrium at a pressure of 0.62 atm (and SO°C).
- 601. 570 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 p::::; 0.50 atm liquid composition 70 60 50 o A 0.25 I I I I I I 0.5 Mol fraction benzene 0.75 1.0 Figure 19.5 diagram for a mixture of benzene and toluene at 0.50 atm. At 0 mole fraction of benzene (point A), the temperature corresponds to the vapor pressure of toluene of 0.50 atm. At a mol fraction of benzene of 1 (point B). the temperature corresponds to vapor pressure of benzene of 0.50 atm. The tie line shows the respective liquid and vapor compositions that are in equilibrium at a temperature of 80°C. (and 0.50 atm) an initial mix~ of 50% oenlZeJle and 50% toluene. Figures 19.4 and 1 show the vapor-liquid envelope for a binary mixture of benzene and toluene, which is essentiaHy ideaL You can interpret the information on the phase diagrams as follows. Suppose you start in Figure 19.4 at a 50-50 mixture of benzene-toluene at 80°C and 0.30 atm in the vapor phase. Then you increase the pressure on the system until you reach the dew point about 0.47 atm, at which point the vapor starts to condense. At 0.62 atm the mol fraction of benzene in the vapor phase will be about 0.75 and the mol fraction in the liquid phase will be about 0.38, as indicated by the tie hne. As you in-. crease the pressure from 0.62 to about 0.70 atm, aU of the vapor will be condensed to liquid. What will the composition of the liquid be? 0.50 benzene. of course! Can you carry out an analogous conversion of vapor to liquid using Figure 19.5, the temperature-composition diagram? Phase diagrams for nonideal solutions abound. 19.6 shows the tempera-ture- composition diagram for isopropanol in water at 1 atm. Note the minimum boiling point at a mole fraction of isopropanol of about 0.68, a point called an· azeotrope (a point for which the dew point curve and the bubble point curve coincide) that makes separation by distillation different. 19.2 .. 3 K .. Value (Vapor-Liquid Equilibrium Ratio) For nonideal as wen as ideal mixtures that comprise two (or more) phases, it proves to be convenient to express the ratio of the mole fraction in one phase to the mole fraction of the same component in another phase in terms of a distribution co .. efficient or equilibrium ratio K, usually called a K~value. For example.
- 602. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems 571 o o ....." ~ ~ "- CD a. E ~ p- 1.00atm 95 90 85 80 75~~~~~~~~--~~~~~~~ o 0.2 0.4 0.6 Mol fraction of Isopropanol 0.8 Figure 19.6 Phase diagram for a nonideal mixrure of isopropanol and water at 1 atm. Yi Vapor-liquid: - = K · Xi J Xli Liquid -liquid: - = K Li Xu (19.5a) 19.5b) and so on. Refer to Chapter 20 for soi1 sorption coefficients. If the ideal gas law (Dalton's law), Pi = Yi Ptota}, applies to the gas phase, and the ideal RaouJt's law ap-plies to the liquid phase, Pi = x;P; (T), then for the ideal system _ Yi _ p·i(T) Ki - - - (l9.5c) Xi Protal Equation (19.5c) gives reasonable estimates of Ki values at low pressures for components well below their critical temperatures, but yields values too large for components above their critical temperatures, at high pressures, andlor for polar compounds. For nonidea1 mixtures Equation (19.5a) can be employed if Ki is made a function of temperature, pressure, and composition so that relations for K; can be fit by experimental data and used directly, or in the fonn of charts, for design calculations, as explained in some of the references at the end of this chapter. Figure 19.7 shows how K varies for the non ideal mixture of acetone and water at 1 atm. K can be greater or less than one but never negative. For ideal solutions you can calculate values of K using Equation (l9.5c). For nonideal solutions you can get approximate K values from
- 603. 512 The Phase Rule and Vapor-liquid Equilibria Chap. 19 10 Total pressure == 1 atm K Kwaler 0.1 L-i--L--I----1-....L-..L.......I'--1..---L--'-....I..--'---lI....-L--'--'-.....L-.J.....",;L........I o 0.2 0.4 0.6 0.8 Mol fraction acetone in the liquid phase. 1. Empirical equations such as· Figure 19.7 Change of K of water with composition at p == 1 atm. K . = (pc,i)exp [7.224 - 7.534rrr.i - 2.598 In Tr,i] If Tc,/f > 1.2: I Ptotal 2. Databases (refer to the supplementary references at the end of the chapter). 3. Charts such as Figure 19.8. 4. Thennodynamic relations (refer to the references at the end of the chapter). Typical problems you should be able to solve that involve the use of the equilibrium coefficient Kj together with material balances are: 1. Calculate the bubble point temperature of a liquid mixture given the total pressure and liquid composition. 2. Calculate the dew point temperature of a vapor mixture given the total pressure and vapor composition. 3. Calculate the compos~tion of the vapor and liquid streams, and their respective quantities, when a liquid of given composition is partially vaporized (Dashed) at a given temperature and pressure (the temperature must He between the bubble and dew point temperatures of the feed). 4. Calculate the re]ated equilibrium vapor-liquid compositions over the range of mole fractions from 0 to 1 as a function of temperature given the total pressure, Analogous problems occur with respect to calculating the total pressure given a fixed temperature. Let's next outline the procedure to solve the first three problems cited above. ·SJ. Sandler, in Foundations o/Computer Aided Design, VoL 2, R.H.S. Mah and W.O. Seider, eds .• p. 83, American Institute of Chemical Engineers, New York (1981).
- 604. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems ISO BUTANE VAPORIZATION EQUILIBRIUM CONSTANTS 573 0.01 ___- Figure 19.8 K-values for isobutane as a function of temperature and pressure. From Nanna! Gasoline Association of America Technical Manual (4th 00) (1941) with permission (based on data provided by George Granger Brown). o 200 300 400 TEMPERATURE, of 1. To calculate the bubble point temperature (given the total pressure and liquid composition), you can write Equation (19.5a) as Yj = K; x,'" Also, you know that IYi = I in the vapor phase. Thus for a binary (19.6) in which the K;'s are functions of only the temperature. Because each of the K/s increases with temperature, Equation (19.6) has only one positive root. You can employ Newton' s method (or Polymath on the CD) to get the root (see Appendix L2) if you can express each K; as an explicit function of temperature. If not, you have to assume a series of temperatures, look up or calculate Ki, and then calculate each tenn in Equation (19.6). After you bracket the value of 1, you can interpolate to get T. which satisfies Equation (19.6).
- 605. 574 The Phase and Vapor-Liquid Equilibria Chap. 19 an ideal solution, Equation (19.6) becomes (19.7) LYi ~ ---'-- + --.....::.-.. = 1 Ptotal Ptota! (19.7a) No. te that this equation is nonlinear and requires an iterative solution to deter- mme You might use Antoine's equation for to formulate an equation ex-plicit in T. Solve it as described above in connection with Equation (19.6). Once the bubble point temperature is determined, you can calculate the vapor composition from .. X· I Yi = , Ptotal A degree-of-freedom analysis for the bubble point temperature for a bi-nary mixture shows the degrees of freedom are zero: Total variables = 2 2 + 2: Xl' X2; Yl' Y2; Ptotal; T Prespecified values of variables = 2 + 1: xl' Ptotal Independent equations::.: 2 + I: Y1 = K1x., Y2 = KzX2; Y1 + Y2 = 1 2. To calculate the dew point temperature (given the total pressure and vapor composition), you can write Equation (19.5a) as Xi = y/Ki• and you know k = 1 in the liquid phase. Consequently, you want to solve the equation: 1 = lL + Kl (19.8) in which s are a function of as explained for bubble point calculation. For an ideal binary solution, Equation (19.8) becomes Equation (19.8a) To calculate 1 = YI (19.8a) Y2 ] P~(Tdp) nonlinear requiring an iterative numerical solution. composition of the liquid phase, you use The degree-of-freedom analysis is similar to that for the bubble point temperacalculation. J
- 606. l Sec. 19.2 Vapor-liquid Equilibria in Binary Systems Composition Koown x,/ ". Composition'" ? Composilion = ? rl Figure 19.9 Flash vaporization with vapor and liquid in equilibrium. 3. calculate the amount of the respective vapor and liquid phases that evolve at equilibrium when a liquid of known composition XFi flashes (flash vapor .. ization) at a known temperature and pressure, you use Equation (19.5a) gether with a material balance. Figure J 9.9 illustrates the open, steady-state process. a binary mixture, a mole balance for component i gives = LXj VYi (19.9) where F the motes of liquid to be flashed. L is the moles of liquid equilibrium, and V the moles of vapor at equilibrium. Introduction of Yi = Ki Xi into Equation (19.9) gives so that Yr =------~--- - ------~----- L (F _ L) 1 _ L I __1 ) Ki . F Ki (19.10) where L/F the liquid fraction resulting from vaporization of the liquid feed. Consequently, for a binary mixture, since l: Yi = 1, you want to solve the foHowing equation 1 - ----'---- l_L(l-~J (19.11) for LIF (which must be positive), Numerous computer programs exist to solve the flash vaporization problem. Based on the above equations, you can prepare figures such as Figures 19. 19.6 for binary mixtures. Tab1e 19.1 summarizes the phase equilibrium calculations.
- 607. 576 and Vapor-Uquid Equilibria of the Information Involved Equilibrium Calculations Chap. 19 Known* Information Variables to be Calculated Equation to Use Bubble point temperature Dew point temperature Bubble point Dew point ........ '''~n'~ ... Flash PUll.aI' Xi PtOlaI' Yi T. Xi T. Yi Plolal, .. Ki is 8.,",sumed 10 be a known function of T and P10!.aI' T. Yi T. Xi PIOial' Yi PlOtlll' Xi L - Yi> Xi 1 19.6 19.8 11 We should remark that in this chapter we have assumed that any liquids are miscible. If they are completely immiscible. then each would exert an vapor pressure Pi be ~ Pj' • P i = YiPtotal, and the total on the would The next two examples illustrate calculations. of vapor-liquid equilibrium EXAMPLE 19.3 Bubble-Point Calculation Suppose that a liquid mixture of 4.0 mol % Il-hexane in n-octane is vaporized. What is the composition of the first vapor fonned if tOlal pressure is 1.00 atm? Solution Refer back to to view the type relation T versus. x at constant p that is involved in this The mixture can be as an ideal mixture be& cause the components are similar. As an intermediate step, you can ca1culate the bubble point temperature using Equation (19.7) or (19.7a), Let's use the Antoine equation to relate p" to substitute the result into Equation (19.7), and solve the resulting equation for to look up the coefficients of the Antoine equation to obtain the vapor pressures of the two components: .. B In(p ) = A - C + T where p'" is in mm Hg and is in K: n-hexane (Cli): n-octane (Cg): A 15.8366 15.9798 B 2697.55 3127.60 c
- 608. Sec. 19.2 Vapor-Liquid Equilibria in Systems Basis: 1 mol of liquid You want to solve the following equation to the bubble point tenape:ratlJre usmg a nonlinear equation solver: ( 2697.55 760 == exp 15.8366 - -48.784 + + exp( 15.9798 - ---polymath is T for which the vapor pressure of hexane is mm the vapor Dlc:ssmrc octane is 661 rom Hg. respective mole fractions in the vapor phase are 31 == 760 (0.04) = 0.164 Yea = 1 0.164 = 0.836 EXAMPLE 19.4 Flash Calculation for a Binary Liquid Mixture Calculate the fraction of liquid that will remain at equilibrium when a mixture of 68.6% hexane and 31.4% toluene is vaporized at 80°C and I atm. Solution You can treat the hexane toluene mixture as an solution, and use (t 9.11) in solving problem. 3and4 Remember given in the problem statement are in mass. so concentrations to fractions. Component Hexane Toluene The next Basis: 100 g Grams MoL wt. 68.6 86.17 31.4 92.13 100.0 data at 80°C are: p'" (mm Hg) ]020 to calculate the values of K = 1020 = I hexane = 760 Ptotal gmol Mol fr. 0.796 0.70 0.341 0.30 TTI7 r:oo Toluene 290 1 K ;;::;; 0.745 hexane '" 5n
- 609. 518 . 'The Rule and Vapor-liquid Equilibria Chap. 19 ,290 .' KI.O'ue~ ;" 760 = 0,382; 1 --= 2.621 KtollleJ1e Steps 8 and 9 Introduce the above values .Equati(,)n (19.11) to get 0.70 0.30 1 -= -------'---- + ------- L L. 1 - F(l - 0.745) 1 - -(1 - 2.621) The solution is L -::;:;;; F (The equality of UP and lIKhexane is purely coincidental.) EXAMPLE 19.5 Separation of a Virus from a t.:U,ltull'e Particles such as 8 virus in.8 <iUAiULI.l'L/lll rore by partitioning between two phases. culture solution is added and mixed with the can from the cul-two exist. The virus is found both solution immiscible with the After equilibrium is reached. culture solution and the polymer so-lution. The data for one experiment is as follows: 8. volume of the solution containing the culture and ViruS was L b. volume of the polymer solution added to the vessel was Vp 1 c. partition coefficient Kp/C for the virus between the two is Cp Kp/c = - = 100 .. Cc . where = concentration of the virus in the polymer phase. = concentration of particles in the culture phase. After reached. what is the fraction of the initial virus in ture phase that is in the polymer phase? Solution You want to CeVe
- 610. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems The virus particle balance is VOCo =Vpcp + Vccc Assume Vo = Vc = 3.0 L after equilibrium is reached. Then 3.0 L I Co virus particles = 0.1 L I Cp virus particles + 3.0 L Cc virus particles L L L Divide both sides of the equation by Co and introduce the partition coefficient for Cc to get O.1ep 3.0 Cp 3.0 = -- + ---'- Co 100 Co The solution is Cp ep 3.0 = O.13-sothat- = 23.1 Co Co and Cp X Vp = 0.77. Co Va SELF .. ASSESSMENT TEST Questions 579 1. The heading of a recent article in the Journal of Chemical Education, 72, 204-205 (1995) was: Raoult's Law is a Deception What did the author mean by such a provocative title? 2. When should you use Henry's Jaw and when should you use Raoult's law? 3. As you know, the higher the fuel volatility, the higher the emissions from the fuel. Refiners adjust the butane content of gasoline because it is a high octane, relatively cheap hydrocarbon that helps cars to start and wann up easily. Similarly, blending ethanol (more expensive than butane) with gasoline raises the volatility of the blend above that of straight gasoline, which makes the emissions problem worse. ff you ignore costs. wilI adding two mole % ethanol or butane to octane yield a product that has a higher pressure over the liquid so1ution? 4. Can you make a plot of the partial pressures, and total pressure of a mixture of heptane and octane given solely that the vapor pressure of heptane is 92 mm Hg and that of octane is 31 mm Hg at a given temperature?
- 611. 580 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 S. Why is it necessary to remove much of the water vapor that in a natural gas supply before sending the through a pipeline? Problems 14 Calculate the boiling point temperature of 1 of a solution of 70% ethylene glycol (an-tifreeze) in at I atm. the solution is ideal. a system comprised a liquid mixture of benzene and toluene at 1 atm. answer the following questions: a. normal boiling point of is SO. DOC. What is the mole fraction of benzene in the liquid and vapor in the cited system? b. The normal boiling point toluene is llO.4°C. What the mole fraction of benzene in liquid and vapor phases for cited system? c. At 1 what is the mole of the in the liquid and vapor phases? 3. When a mixture of 50 mol % benzene and 50 mol % toluene was flashed at equilibrium at 1 atm, 90% of the feed was vaporized. What were the mole fractions of the benzene in the liquid and vapor phases. and what was the temperature of the system? Thought Problems 1. The fluid in a large tank caught on fire 40 minutes after the start of a blending operation in which one of was being added to The fire was soon put out and the naphtha was moved to another tank. The next day blending was resumed in the second 40 minutes another started. Can you explain reason for this sequence of What be done to prevent such accidents? 1. CO2 can be used to optical or semiconductor surfaces and remove particles or or-ganic contaminants. A bottle of CO2 at 4000 kPa is attached to a jet that sprays onto the optical surface. Two precautions must be taken with this technique. The surface must be heated to about to minimize moisture condensation, and you must employ a CO2 source with no residual heavy hydrocarbons Oubricants) to minimize recontamination in critical cleaning applications. Describe physical conditions of as it hits optical Is it liquid. or solid? How does decontamination take place? 3. The advertisement "Solid dry blocks in 60 seconds right in your own lab! Now you can have dry available to you at any day or night, with this small, safe, effi-cient machine and readily available CO2 cylinders. No batteries or electrical energy are required." How is it possible to make dry in 60 seconds without a compressor? 4. An inventor is trying to a machine transfonns water vapor liquid water with-out ever condensing the water vat>or. You are to explain such a process is techni-cally possible. What is your Examine the statements a. 'The vapor pressure gasoline is about 97 kPa at 54°C." b. 'The vapor ............ c"' .. the system. water-furfural diacetate, is lOl kPa at 99.96°C."
- 612. Sec. 19.2 Vapor-Liquid Equilibria in Binary Systems 581 Are the statements correct? If not, correct them. Assume the numerical values are correct. 6. To maintain safe loading of hydrocarbon fluids, one of the many objectives of the Coast Guard is to prevent underpressuring of the tank(s) of the vessel being loaded. Overpressuring is easy to understand-to much fluid is pumped into a tank. How can underpressuring occur? Discussion Problems 1. Gasoline tanks that have leaked have posed a problem in cleaning up the soil at the leak site. To avoid .digging up the soil around the tank, which is located 5-10 m deep. it has been suggested that high pressure steam be injected underneath the gasoline site via wells to drive the trapped gasoline into a central extraction well which, under vacuum, would extract the gasoline. How might you design an experiment to test the concept of removal? What kinds of soils might be hard to treat? Why do you think steam was used for injection rather than water? 2. How to meet increasingly severe federal and state regulations for gasoline, oxygenated fuels, and low-sulfur diesel fuel represents a real challenge. The table below shows some typical values for gasoline components prior to the implementation of the regulations in the State of California. and the limits afterwards. Fuel Parameter Sulfur (ppmw) Benzene (vol %) Olefins (vol %) Oxygen (wt %) Boiling point for 90% of the gasoline (oF) Former (Typical GasoHne) 150 2 9.9 o 330 Current (Limit for Reftneries) 40 1 6 2.2 300 Read some of the chemical engineering literature, and prepare a brief report on some of the feasible and economic ways that have been proposed or used to meet the new standards. Will enforcing emission standards on old automobiles (or junking tbem) be an effective technique of reducing emissions relative to modifying the gasoline? What about control of evaporative emissions from the fuel tank. What about degradation or malfunction of emission controls? Etc. 1. The EPA negotiated an agreement on refonnuJated gasoJine that included a waiver permitting the use of higher vapor pressure gasoline with added ethanol than gasoline without ethanol. Considerable argument occurred because the ethanol-gasoline fuel leads to more volatile organic compounds finding their way into the atmosphere. Supposedly blending 10% ethanol into gasoline increases the vapor pressure of the mixture over ethanol free gasoline by 1 psi measured as Reid vapor pressure-RVP). Tests show that the evaporation of hydrocarbons increases by 50%. What would be the vapor pressure of a 10% ethanol-gasoline mixture versus the vapor pressure of gasoline alone at 25°C. and indicate whether the reported 50% increase in vaporization of hydrocarbons from the fuel
- 613. 582 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 seems reasonable. (Note the aromatics in the gasoline are not more than 25% and the zene not more than 1 % by volume.) What other must be take into account in b1ending gasoline? Looking Back In this chapter we showed how the phase rule applies to systems in equilib-rium. Then we described how to calculate partial pressure of the components in a vapor-liquid mixture at equilibrium Henry's law. Raoult's law, and the equi-librium coefficient Kj • Finally, we explained how to determine T. p, x, y, at equilibrium and the fraction of liquid vaporized for ideal and real solutions. GLOSSARY OF NEW T RMS Azeotrope Point which dew point and the bubble point curves coincide. Bubble point Temperature at which vapor first forms from a liquid at given pressure. Dew point Temperature at which a liquid first forms from a vapor at constant pressure. Distribution coefficient K-value. Equllibrium ratio See K -value. Flash Partial vaporization a liquid of a given composition at a fixed temperature and pressure. Gibb's phase rule A relation that gives the degrees freedom for intensive vari-ables in a system in terms the number of phases and number components. Henry's law A relation between the partial pressure of the gas the gas phase and the mole fraction of the gas the liquid phase equilibrium. Ideal solution system whose properties, as vapor pressure~ specific vol-ume, and so 001 can be calculated from the knowledge only of the corresponding properties of the pure components and the composition of the solution. Invariant system in which no variation of conditions is possible without one phase disappearing. K .. vaJue A parameter (distribution coefficient) used express the ratio of the mole fraction in one phase to the mole fraction of the same component in another phase.
- 614. Sec. 1 Vapor-liquid Equilibria in Binary Systems Phase A part of a system that is chemically and physical!y uniform throughout. This definition does not necessarily imply that the phase is continuous. Phase equilibrium The phases present in a system are invariant as are the phase properties. Raoult's Law relation that relates the partial pressure of one component in the vapor phase to the mole fraction of the same component in the liquid phase. Vapor-liquid equilibria Graphs showing the concentration of a component a vapor-liquid system as a function temperature and/or pressure. SUPPL MENTARY REF RENCES In addition to the "p'rf·nf',,.c listed the Frequently Asked Questions the front ma-terial, the following are pertinent. Phase Rule Alper, J. S. 'The Gibbs Phase Rule Revisited," 1. Chern. Educ .• 76, 1567-1569 (1999). Jensen. W. "Generalizing the Phase Rule," 1. &Juc.. 1369-1370 (2001). Perry. R. H. et Perry's Chemical Engineers' Handbook, McGraw-Hill, New York (2000). Rao, Y. "Extended Form of the Gibbs Phase " Chem. Educ., 40-49 (Winter 1985). Zhao, M.. Wang. and Xiao, "Detennining the Number of Independent Components by Brinkley's Method," 1. Chern. Educ., 69.539-542 (1992). Vapor-Liquid Equilibria Carroll, 1. 1. "What is Henry's Law?"Chem. Progr., 87, 48 (1 1). Coulson, J, M. et a1. "Coulson &ersand; Richardson's Chemical Engineering: Fluid Flow, Heat Transfer and Mass Transfer," 6th ed. Pergamon Oxford (1999). De Nevers, N. Physical and Chemical Equilibrium for Chemical Engineers, WileyInterscience, New York (2002). Gmehling, J. et a1. Vapor-Liquid Equilibria Data Collection, 13 parts, Dechema, Frankfurt, Gennany (1991). Han. and G.P. Rangaiah. HA Method Multiphase Equilibrium Calculations," Compo Chern. Eng .. 22,897-911 (1998). Henley, 1., and 1. D. Seader, Equilibrium· Stage Separation Operations in Chemical gineering. Wiley, New York (1981). Hines, A. and R. N. Maddox. Mass Transfer, Fundamentals and Applications, Prentice- Hall. Englewood Cliffs. N.J. (1985) King. C. 1. Separation Processes, 2nd ed., McGraw-Hill, New York (1980).
- 615. 584 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 McCabe. W. L., J. Smith and P. Harriott, Unit Operations of Chemical Engineering, McGraw-Hill, NY (999). Orbey, H. and I. Sandler, Modeling Vapor-Liquid Equilibria: Cubic Equations of State and their Mixing Rules, Cambridge University Press, New York (1998). Treybal. R. E. Mass Transfer Operations, 3rd ed., McGraw-Hili, NY (1980). Web Sites hrrp:/Ichemineer.miningco.com/sitesearch.htm http://eng.sdsu.edultestcenlerfTestlsolv ... tlidealgasidealgaslideaIgasidealgas.html http://www. v lecak.org http://www.mpch-mrunz.mpg.del-sander/reslhenry.html www.mnsi.netl-paslbrochure.htm www.net-link.netl-wdkovats www.public.iastate.eduJ-joUs www.owlnet.rice.edul-wgchap PROBl MS *19.1 A vessel contains liquid ethanol. ethanol vapor, and N2 gas at eqUilibrium. How many phases. components, and degrees of freedom are there according to the phase rule. *19.2 What is the number of degrees of freedom according to the phase rule for each of the following systems: (a) Solid iodine in equilibrium with its vapor (b) A mixture of liquid water and liquid octane (which is immiscible in water) both in equilibrium with their vapors *19.3 A mixture of water. acetic acid, and ethyl alcohol is placed in a sealed container at 40°C at equilibrium. How many degrees of freedom exist according to the phase rule for this system? List a specific variable for each degree of freedom. "19.4 (a) A system contains 2 components at equilibrium. What is the maximum number of phases possible with this system? Give your reasons for your answer. (b) A two-phase system is specified by fixing the temperature, the pressure. and the amount of one component. How many components are there in the system at equilibrium? "'19.5 Liquid water in equilibrium with water vapor is a system with how many degrees of freedom? $19.6 Liquid water in eqUilibrium with moist air is a system with how many degrees of freedom?
- 616. Chap. 19 Problems 11019.7 You have a closed vessel that contains NH4CI(s). NH3(g), and HCI(g) in equilibrium. How many degrees freedom exist in the system? 1119.8 In the decomposition of CaC03 in a sealed container from which the air was initiaHy pumped out, you generate COz and CaD. If not aU of the CaC03 decomposes at equi· librium. how many degrees of freedom exist for the system according to the Gibbs phase rule? 11'19.9 Answer the following questions true or false: (a) critical temperature and pressure are the highest temperature and pressure at which a binary mixture of vapor and liquid can exist at equilibrium. (b) Raoult's law is best used for a solute in dilute solutions. (c) Henry's law is best used for a solute in concentrated solutions. (d) A mixture of liquid butane and pentane can be treated as an ideal solution. (e) The liquid phase region is found above the vapor phase region on p - x - y chart. (t) The liquid phase region is found below the vapor phase region on a T - x - y chart. 11019.10 Detennine if Henry's Law applies to HzS in HzO based on the foHowing measurements at 30°C; Liquid mole fraction X 101 0,0003599 0.0004498 0.0005397 0.0008273 0.0008992 0.001348 0.001528 0,003194 0.004712 0.007858 0.01095 0.01376 0.01507 Pressure (kPa) 20 30 40 50 60 90 100 200 300 500 700 900 1000 Determine the equilibrium concentration in mg/L chloroform in water at 20°C and 1 attn assuming that gas and liquid phases are ideal, and the mole fraction of the chloroform in the gas phase is 0.024. The Henry constant for chloroform is H = 170 attn/mol fraction. Water in an enclosed vessel at 17°C contains a concentration of dissolved oxygen of 6.0 mgIL. At eqUilibrium. determine the concentration of oxygen in the air space above the water in mg/L Henry's law constant is 4.02 X 1()6 kPalmol fraction and the pressure in the vessel is I atm. You are to remove 90% the sulfur dioxide in a gas stream of air and sulfur dioxide that flows at the rate of cubic meters per minute and contains 3% sulfur
- 617. 586 The Phase Rule and Vapor-Liquid Equilibria Chap. 19 dioxide. The sulfur dioxide is to be removed by a stream of water. The entering water contains no sulfur dioxide. The temperature 290 and the pressure on the process is one atmosphere. Find the kilograms water per minute need to remove the sulfur dioxide assuming that the exit water is in equilibrium with the entering gas, and (b) the ratio of the water stream to the gas stream. The Henry's Law constant for sulfur dioxide at 290 K 43 atmlmol fraction. *19.14 A tank contains a liquid composed of 60 mole percent toluene 40 mole percent benzene in equilibrium with vapor phase and air at 1 atro and 60oP. What is the concentration of hydrocarbons in the vapor phase? If the lower flarrunabity limit for toluene in air is 1.27% and benzene is 1.4%, is the vapor phase flammable? ~ "'19.15 Fuel tanks for barbeques contain propane and n-butane. At 120oP, if an essentially full tank liquid that contains liquid and vapor in eqUilibrium and exhibits a pressure of 100 psia, what is the overall (vapor plus liquid) mole fraction of butane in the tank. "'19.16 Based on the following vapor pressure data, construct the temperature-composition diagram at 1 atm for the system benzene-toluene, assuming ideal solution behavior. Vapor mmHg Temperature °c Benzene Toluene 80 760 300 1078 432 100 1344 1 lOA 1748 760 ·19.17 Sketch a T-x-y diagram that shows an azeotrope locate and label the bubble and dew lines and the point. ·19.18 What is (a) the pressure in the vapor phase, and (b) the composition of the vapor phase in equilibrium with a liqUid mixture of 20% pentane and 80% heptane at 50°F? the mixture is an ideal one at equilibrium. *19.19 Methanol has a flash point at at which temperature its vapor pressure is 62 mm Hg. What is the flash point (temperature) of a mixture 75% methanol and 25% water. Hint: The water does not burn. 11119.20 Two kiJograms of a mixture of 50-50 benzene and toluene is at 60°C. As the total pressure on the system is reduced. at what pressure will boiling commence? What will be the composition of the first bubble liquid? *"'19.21 The normal boiling point of propane is -42.1°C and the normal boiling point of nbutane is -O.soC. (a) Calculate the mole fraction of the propane in a liquid mixture that boils at - 31.2°C and ] atm. Calculate the corresponding mole fraction of the propane in the vapor at - 31 (c) Plot temperature vs. propane mole fraction for the system of propane and butane. - *19.22 In the system n-heptane, n-octane at 200oP, determine the partial pressure of each component in the vapor phase at liquid mole fractions of n-heptane of 0, 0.2~ OA. 0.6, 0.8 and] Also calculate the total pressure above each solution.
- 618. Chap. 19 Problems 587 1 ; / ·"'19.25 "·19.27 Plot your results on a P-x diagram with mole fractions of C7 to the right and mole fractions of Cg increasing to the The ordinate should pressure 10 Read from the plotted graph the following information: The total and partial pressure of each at a mole frac-tion of C, = 0.47 in the liquid. Calculate the bubble point of a liquid mixture of 80 mole% n-hexane and 20 mole% n-pentane at 200 Calculate the dew point a vapor mixture of 80 mole% n-hexane and 20 mole% n-pentane at 100 A mixture of 50 mole% benzene and 50 mole% toluene is contained in a cylinder at 39.36 in. absolute. Calculate the temperature range in which a two phase system can eXIst. A liquid mixture of n-pentane and containing 40 mol per cent n-pentane is fed continuously to a flash separator operating at 250°F and 80 psia. Determine: (a) quantity of and liquid obtained the separator mol of feed. (b) The composition of both the vapor the liquid leaving the separator. hundred moles per minute of a binary mixture of A and B are separated in a two stage (serial) process. In the first the liquid vapor flow rates exiting from the are each 50 moles per minute. liquid then into a sec-ond separator that operates at same temperature as the first and the respec-tive exit streams of liquid and vapor from the second stage are each moles per minute. temperature is same for each stage. and at that temperature, the vapor pressure of A is 10 kPa while vapor pressure B is 100 kPa. Treat the liquids and vapors as ideal. Calculate the compositions of an the streams in the process, and calculate the pressure in each stage. ·"'19.28 Most combustible reactions occur in the phase. any flammable material to burn both fuel oxidizer must be present., and a minimum concentration of the flammable gas or vapor in the phase must also exist. mlnlmum concentration at which ignition will occur called the lower flammab1e limit (LFL). liquid temperature at which the vapor concentration reaches the LFL can be found experimentally. is usually measured a standard method called a "dosed cup flash point" The Hflash of a liquid fuel is thus the liquid temperature at which the concentration of fuel vapor in air is large enough for a flame to flash across the surface of the fue] if an ignition source is present The point and the LFL concentration are dosely related through the vapor pressure of the liquid. Thus, if the flash point is known, the concentration can be estimated, and if the concentration is known, the flash point can be estimated. Estimate the point (the temperature) of liquid n-decane that contains 5.0 mole percent pentane. The for pentane is 1.8% and that for n-decane is 0.8%. Assume the propane-n-decane mixture is an liquid. Assume the ambient pressure is 100 kPa. This problem has adapted from Safety, Health, and Loss Prevention in
- 619. 588 Chemical Processes, cal Engineers. New The Phase Rule and Vapor-Liquid Equilibria Chap. 19 lR. Welker and C. Springer, American Institute of Chemi( 1990), with permission. "'19.29 You are asked to maximum pressure at which steam distillation naphtha can be out at ] (the maximum allowable temperature). IS injected into the liquid naphtha to vaporize it If (1) the distillation is out at 160°F, (2) the liquid naphtha contains 7.8% (by weight) nonvolatile impurities. and (3) if the initial charge to the distillation equipment is 1000 Ib water and 5000 Ib of naphtha. how much water wiI1 left in the still when last drop of naphtha vaporized? Data: naphtha the MW is about 107, and p" (180°F) = 460 mm Hg, (l60°F) = 318 mm . "19.30 Late in evening of2l August 1986 a large volume toxic gas was released from beneath and within Lake Nyos in the Northwest Province of Cameroon. Ari aerosol of water mixed with toxic swept down the valleys to the north of Lake Nyos, leaving more than 1.700 and dying people in its wake. The lake had a surface area of 1.48 km2 and a depth of 200-250 m. It took 4 days to refill the lake, hence it was estimated to have lost about 200,000 tons of water during the gas emission. To the south of tbe lake and in the small cove immediately to the east of spillway a wave rose to a height about 25 m. conclusion of investigators studying this incident was that the waters of Lake Nyos were saturated with CO2 of volcanic origin. in the evening of 21 August a pulse of vo1canic gas-mainly CO2 but containing some H2S-was released above a volcanic vent in the northeast corner of the lake. The stream of bubbles rising to surface brought up more bottom waters highly charged with CO2 that gushed out increasing the gas flow and hence the flow of water to surface much as a warm soda overflows on release of pressure. At the surface, the of transformed the accompanying water into a mist and sent a wave of water ....... "''''"- ing across the lake. The aerosol water and CO2 mixed with a trace of H2S down valleys to the north of lake leaving a terrible toll of injury and death in its wake. If the solution at the bottom of the lake obeyed Henry' slaw, how much was ....... ,...." ........ with the 200.000 tons of water, and what would the volume of the at cubic meters? At 25°C Henry's law constant is I X lQ3 atmlmol fro "19.31 Examine the statements below: (a) "The vapor pressure of gasoline is about ]4 psia at 130oP." (b) "The vapor pressure of system, water-furfural diacetate, 760 mm Hg at 99.96°C." Are the statements correct? If not, correct them. Assume the numerical va]ues are correct. "'19.32 If pressure in the head space (gas space) in a bioreactor is ItO kPa and and the oxygen concentration in the head space is enriched to 39.7%. what is the mole fraction of the dissolved oxygen in liquid phase? What is the percent excess oxydissolved in the liquid phase compared with the saturation value that could obtained from air alone dissolved the liquid?
- 620. Chap. 19 Problems 589 "19.33 separate waste discharge streams a plant into a river contain the following respective chemicals in water Methyl ethyllretone (MEK) Phenol Concentration 5.5 1.1 2.1 K are from Aspen at 20°C. K 1.20 X 10-7 3.065 0.00485 Estimate the concentration of the compounds in the gas phase discharge stream at 20°C. Will volatilization from the discharge stream cant?
- 621. CHAP E 20 IQUIDS AND GASES IN EQUILIBRIUM WITH SOLIDS Your objectives In studying this chapter are to be able to: 1. Predict adsorption of gases and liquids on solids 2. Determine the values of coefficients in adsorption equilibrium relations from physical measurements. far in this book you have read about equilibria between fluids and solids are also important-they Looking Ahead and liquids. cannot be In this chapter we discuss the adsorption of gases and liquids on solids when the at equilibrium. You will learn what some of the are are used to the amounts of absorption, and what kinds of data are to the values the coefficients in the relations. Main Concepts 590 the adsorption of gases or liquids on solids. of petroleum fractions. from municipal water supplies. , ,
- 622. Chap. 20 Liquids and Gases in Equilibrium with Solids .. Decolorizing of vegetable and animal oils, and of crude sugar syrups . .. Clarification of beverages and pharmaceutical preparations. .. Purification of process effluents and gases for pollution control. 591 .. Solvent recovery from air such as in removing evaporated dry cleaning sol-vents . .. Dehydration of gases . .. Odor and toxic gas removal from air or vent gases. .. Separation of rare gases at low temperatures. .. Impurity removal from air prior to low-temperature fractionation . .. Storage of hydrogen. What goes on in adsorption? Adsorption is a physical phenomenon that occurs when gas or liquid molecules are brought into contact with a solid surface, the adsorbent. Some of the molecules may condense (the adsorbate) on the exterior surface and in the cracks and pores of the solid. If interaction between the solid and condensed molecules is relatively weak, the process is called physical adsorption; if the interaction is strong (similar to a chemical reaction), it is called chemisorption, or activated adsorption. We focus on equilibria in physical adsorption in this chapter. Equilibrium adsorption is analogous to the gas-liquid and liquid-liquid equilibria described in previous chapters. Picture a small section of adsorbent surface. As soon as molecules come near the surface. some condense on the surface. A typical molecule will reside on the adsorbent for some finite time before it acquires sufficient energy to leave. Given sufficient time, an equilibrium state will be reached: the number of molecules leaving the surface will just equal the number arriving. The number of molecules on the surface at equilibrium is a function of the (1) nature of the solid adsorbent, (2) nature of the molecule being adsorbed (the adsorbate), (3) temperature of the system, and (4) concentration of the adsorbate over the adsorbent surface. Numerous theories have been proposed to relate the amount of fluid adsorbed to the amount of adsorbent. The various theories lead to equations that represent the equilibrium state of the adsorption system. Theoretical models hypothesize various physical conditions such as (l) a solid surface on which there is a monomolecular layer of molecules, (2) a ffiultimolecular layer, or (3) capillary condensation. Figure 20.1 illustrates typical simple equilibrium adsorption isotherms for the adsorption of water vapor on Type SA molecular s