![B. g to variation of the process variables with time
teady state
Accordin
S
sitions, flow
rates) do not change with time. For example, batch or semi batch processes.
nsteady state or transient
The values of all process variables (i.e. all temperatures, pressures, compo
U
bles change with time. Continuous processes may be either
eady-state or transient.
.3 The General Balance Equation
Figure 2.1: Process balance
omentum) in a system as show in figure 2.1 can be written in the following general way:
Input + generation - output - consumption = accumulation [2.1]
here,
Accumulation - builds up within system
xample 2.1
move out, 22000 are born, and
9000 die. Write a balance on the population of the city.
put + generation - output - consumption = accumulation
0000 + 22000 – 75000 – 19000 = - 22000
lation decreases by 22,000 people each year.
Any of the process varia
st
2
A balance on a conserved quantity (total mass, mass of a particular species, energy, and
m
w
Input - enters through system boundary
Generation - produced within system
Output - leaves through system boundary
Consumption - consumed within system
E
Each year 50000 people move into a city, 75000 people
1
In
5
Therefore, the city’s popu
19](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-19-320.jpg)
![Two types of balances:
. Differential balance
A
of the balanced quantity unit divided by the time unit (people/yr,
¾ This is the type of balance usually applied to a continuous process.
. Integral balance
¾ Balance that indicates what is happening in a system at an instant in time.
¾ Each term of the balance equation is a rate (rate of input, rate of generation, etc)
and has units
barrels/day).
B
ount of the balanced quantity and has the
nt after the input takes place and the moment before the
product is withdraw.
he following rules can be used to simplify the material balance equation:
is total mass, generation = 0 consumption = 0:
Input - output = accumulation
ce is non-reactive species, generation = consumption = 0:
Input - output = accumulation
If a system is at steady state, accumulation = 0:
Input + generation = output + consumption
.4 Balance on Continuous Steady-State Process
the general balance
quation, (Eq 2.1), equals zero and the equation simplifies as below:
Input + generation = output + consumption [2.2]
on and
onsumption terms equal zero and the equation reduces to the equation as below:
Output [2.3]
¾ Balance that describes what happens between two instants of time.
¾ Each term of the equation is an am
corresponding unit (people, barrels).
¾ This type of balance is usually applied to a batch process, with the two instants of
time being the mome
T
¾ If the balanced quantity
¾ If the balanced substan
¾
2
For continuous processes at steady state, the accumulation term in
e
If the equation is on a non-reactive species or on total mass, the generati
c
Input =
20](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-20-320.jpg)
![2.5 Integral Balance on Batch Process
pplied to any substanc
Integral Balance on Batch Process
pplied to any substanc
Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t=0 there
are n0 mol of NH3 in the reactor, and at later time tf the reaction terminates and the
contents of the reactor, which include nf mol of ammonia are withdraw. Between t0 and tf
no ammonia enters or leaves through the reactor boundaries, so general balance equation
(1) is simply generation = accumulation. Moreover, the quantity of ammonia that builds
up (accumulates) in the reactor between t0 and tf is simply nf – n0, the final amount minus
the initial amount. The same reasoning may be a
Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t=0 there
are n0 mol of NH3 in the reactor, and at later time tf the reaction terminates and the
contents of the reactor, which include nf mol of ammonia are withdraw. Between t0 and tf
no ammonia enters or leaves through the reactor boundaries, so general balance equation
(1) is simply generation = accumulation. Moreover, the quantity of ammonia that builds
up (accumulates) in the reactor between t0 and tf is simply nf – n0, the final amount minus
the initial amount. The same reasoning may be a e participating in a
the following equation:
nition)
= generation – consumption
herefore, equating these two expressions for the accumulation yields:
Initial input + generation = final output + consumption [2.4]
xample 2.3: Balance on a Batch Mixing Process
combined with 150g of the second, what are the mass and composition of the product?
he process can be depicted schematically as follows:
s of Eq (4) can be omitted so that all balance have the simple form
’.
e
g = m
= 350g
e participating in a
the following equation:
nition)
= generation – consumption
herefore, equating these two expressions for the accumulation yields:
Initial input + generation = final output + consumption [2.4]
xample 2.3: Balance on a Batch Mixing Process
combined with 150g of the second, what are the mass and composition of the product?
he process can be depicted schematically as follows:
s of Eq (4) can be omitted so that all balance have the simple form
’.
e
g = m
= 350g
batch process to obtain
batch process to obtain
Accumulation = final output – initial input (by defi
Accumulation = final output – initial input (by defi
T
T
E
E
Two methanol water mixtures are contained in separate flasks. The first mixture contains
40 wt% methanol, and the second contains 70 wt% methanol. If 200g of the first mixture
Two methanol water mixtures are contained in separate flasks. The first mixture contains
40 wt% methanol, and the second contains 70 wt% methanol. If 200g of the first mixture
is
is
T
T
200g
Observe that the input and output ‘streams’ shown on the chart denote the initial and final
states for this batch process. Since no reactions are involved, the generation and
consumption term
Observe that the input and output ‘streams’ shown on the chart denote the initial and final
states for this batch process. Since no reactions are involved, the generation and
consumption term
0
0.6g H2O/g
0
0.3g H O/g
(1 –x )(g H2O/g)
.4g CH3OH/g
150g
m (g)
x (g CH3OH/g)
.7g CH3OH/g
2
‘input = output
‘input = output
Total Mass Balanc
Total Mass Balanc
200g + 150
200g + 150
m
m
22](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-22-320.jpg)
![total number of moles of hexane evaporated occupied a liquid volume of 10 cubic meters
and the specific gravity of liquid hexane is 0.659, the accumulation term equals:
14
6
3
2
.
86 kg
m
The output term in th
3
45
.
76
1
659
.
0
10 H
C
kmol
kmol
kg
m
n −
=
×
×
−
=
e balance is the rate at which hexane is leaving the system [0.1n
mol C6H14/min)] times the total process time, tf (min). The balance (accumulation = -
0.1n tf
n = 0.111 kmol /min
s.
¾ If no stream amount or flow rate is specified in the problem statement, take as a
2) Draw the flowchart of the process, using boxes or other symbols to represent the
3) Fully label the chart when it is first drawn with values of known process variables
being written for each input and output stream.
5) If the problem is solvable, the starting balance should be an equation with minimum
6) After the one unknown in step 4 has been calculated, use that value to give an
7) As each unknown variable is determined, its value is filled so that the chart provides
ontinuous record of where the solution stands and what must still be done.
Δ
(k
output) is therefore;
- 76.45 kmol C6H14 = -
tf = 6887 min
2.7 General procedure for material balance calculations
1) Choose a basis of calculation an amount or flow rate of one of the process stream
basis an arbitrary amount or flow rate of a stream with a known composition.
process units, and lines with arrows to represent input and output streams.
and symbols for unknown variables
4) Do the degree-of-freedom analysis.
number of unknowns, preferably with only one unknown.
equation with one variable for another unknown.
a c
Notes:
¾ The maximum numbers of independent equations that can be derived by
¾
(Σ xi = 1.0).
writing balances on a non-reactive system equal the number of chemical
species in the input and output streams.
The additional equations can be written from the fact that the component mass or
mole fractions of a stream adds up to 1.0
(You can use these constrains to reduce the number of unknowns in the flow
chart (step 2 above) to a minimum.)
24](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-24-320.jpg)
![where;
Inlet Energy + Heat - Outlet Energy – Work = Accumulation
Inlet energy and outlet energy is summation/total of all energy such as potential,
kinetic and internal energy.
oving
relative to the surface of the earth is
/s) m ve
[3.2]
s into a 2 cm internal diameter (ID) pipe at a rate of 2.00 m³/h. Calculate
m
elocity (u) = Volumetric flowrate =
3.4 Kinetic Energy Equation (Ek)
` Kinetic energy, Ek (J) of an object of mass m (kg) m with velocity u (m/s)
[3.1]
2
2
1
mu
Ek =
` If the fluid enters a system with a mass flow rate m
& (kg and unifor locity u
(m/s), the rate at which kinetic energy K
E
& (J/s) is transported into the system is
Example 3.1
2
1
u
m
E
& =
k
E
&
2
k
&
Water flow
in J/s.
for this strea
Solution:
First, we calculate the velocity (u).
V
)
( 2
r
A
V
u
π
&
=
Pipe cross sectional area
Then, we calculate the mass flow rate of the water ( ).
m
&
s
kg
s
h
m
kg
h
m
V /
556
.
0
3600
1
1000
00
.
2
3
3
=
×
×
=
= ρ
&
m
&
s
m
cm
h
/
77
.
1
3600
1
)
1
( 2
2
2
2
=
π s
h
m
cm
m 1
100
1
00
.
2 2
2
3
×
×
×
=
2
2
u
m
K
&
& =
1
E
36](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-36-320.jpg)
![K
E
&
Finally, calculate
s
J
s
m
N
s
m
kg
N
s
u
m
EK
1
2 2
×
×
=
&
m
s
kg
/
870
.
0
/
.
870
.
0
/
.
1
)
77
.
1
(
/
556
.
0
2
1 2
2
2
=
=
=
` l potentia nergy, E bjec is giv below
` the fluid enters a system with a mass flow rate (kg/s) and an elevatio
relative to th
` e interested in the change of potential energy during energy
balance calculation;
[3.5]
at a rate of 15.0 kg/s from a point 220 meters below the
int 20 meters above the ground level. Calculate the attendant rate of
potential energy.
&
3.5 Potential Energy Equation (Ep)
Gravitationa l e n t en
p of a o
[3.3]
mgz
Ep
If m
& n
z
=
e potential energy reference plane.
[3.4]
Normally we ar
gz
m
E &
& =
p
)
( 1
2
1
2
z
z
g
m
E
E
E p
p
p −
=
−
=
Δ &
&
&
&
Example 3.2
Crude pump oil is pumped
earth’s surface to a po
increase of
Solution:
20 m
220 m
Ground level
[ ]
N s
J
s
m W
s
m
kg
s
s
35300
/
35300
/
.
35300
/
.
1 2
2
=
=
=
N
m
m
kg
z
z
g
m
E
E
E p
p
p
1
)
220
(
20
81
.
9
15
)
( 1
2
1
2
×
−
−
×
×
=
−
=
−
=
Δ &
&
&
&
37](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-37-320.jpg)
![3.6 Energy Balances on Closed System
` Closed system is defined as no mass is transferred across the system boundaries
hile the process is taking place.
` nergy balance for closed system;
here;
w
E
w
Initial energy system = )
(
)
( initial
P
initial
K
initial E
E
U +
+
)
(
Final energy system = U )
(
)
(
)
( final
P
final
K
final E
E +
+
Net energy transfer = W
Q −
changes, phase changes, or
Final System Energy – Initial System Energy = Net Energy Transferred to the System
W
Q
E
E
E
E
U
U initial
P
final
P
initial
K
fina
K
initial
final −
=
+ +
− −
l
− )
(
)
(
)
( )
(
)
(
)
(
(
)
(
)
( )
or
[3.6]
W
Q
E
E
U p
k −
=
Δ
+
Δ
+
Δ
` When applying energy balance equation to a given process, the following point
must be aware;
1. The internal energy of a system depends almost entirely on the chemical
composition, state of aggregation (solid, liquid, or gas), and temperature of
the system materials. If no temperature
chemical reactions occur in a closed system and if pressure changes are
less than a few atmospheres, then ∆ ≈ 0.
U
2. If a system is not accelerating, then ∆ K
E = 0. If a system is not rising or
falling, then ∆ P
E = 0.
3. If a system and its surroundings are at the same temperature or the system
is perfectly insulated, then Q = 0. The process is then termed adiabatic.
4. Work done on or by a closed system is accomplished by movement of the
system boundary against a resisting force or the passage of an electrical
current or radiation across the system boundary. If there no moving parts
or electrical current at the system boundary, then W = 0.
en System
is done
on the surrounding by mass that emerges from the systems.
` Both work terms must be include in the energy balance for open system.
3.7 Energy Balances on Op
` In open system, mass is transferred across the system boundaries while the
process is taking place.
` Therefore work must be done on open system to push mass in and work
38](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-38-320.jpg)
![` he net ork d y an open system;
process fluid on a moving part within
the
d at the system outlet minus rate of
ol is used to denote the specific property (property divided by mass or by
T w one b
where;
& - shaft work (rate of work done by the
S
W
system such as a pump rotor.)
& - flow work (rate of work done by the flui
fl
W
work done by the fluid at the system inlet.)
` ^ symb
mole) such as specific internal energy (U
)
kJ/kg), specific volume ( V
)
m3
/kg) and
so on.
ne important property for energy balance
` on open system is specific enthalpy
(
O
Ĥ kJ/kg).
` Sometimes, universal gas law constant can be used as a conversion factor to
cific enthalpy.
of helium at this temperature and pressure, and the rate at which enthalpy is
ansported by a stream of helium at 300 K and 1 atm with a molar flowrate of 250
Solution:
o convert L.atm/mol into J/mol, we need the factor J/(L.atm). From the gas constant
table,
ol.K)
y dividing these two,
in
in
out
out
in
out
fl
fl
s W
W
W &
&
& +
=
V
P
W &
& =
− V
P
W
W &
&
& −
=
[3.7]
V
P
U
H ˆ
ˆ
ˆ +
=
evaluate spe
Example 3.3
The specific internal energy of helium at 300 K and 1 atm is 3800 J/mol, and the specific
molar volume at the same temperature and pressure is 24.63 L/mol. Calculate the specific
enthalpy
tr
kmol/h.
mol
atm
L
mol
J
mol
L
atm
mol
J
V
P
U
H /
.
63
.
24
/
3800
)
/
63
.
24
)(
1
(
/
3800
ˆ
ˆ
ˆ +
=
+
=
+
=
T
0.08206 L.atm/(mol.K) = 8.314 J/(m
the desired factor will be obtained;
B
)
.
/(
3
.
101
)
.
/(
.
08206
.
0
)
.
/(
314
.
8
atm
L
J
K
mol
atm
L
K
mol
J
=
39](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-39-320.jpg)
![So;
mol
J
J
atm
L
mol
atm
L
mol
J /
6295
.
3
.
101
.
63
.
24
/
3800
ˆ =
⎥
⎦
⎤
⎢
⎣
⎡
×
+
=
If n = 250 kmol;
H
kW
s
J
s
h
kmol
mol
mol
J
h
kmol
H
n
H 437
/
437150
3600
1
1
1000
6295
250
ˆ =
=
×
×
×
=
=
&
` Energy balances equation for open system;
[3.8]
Answer:
where;
3.8 Reference States and State Properties
` It is not possible to know the absolute value of Û and Ĥ for a process material,
but we can determine the change in ΔÛ and change in Δ Ĥ corresponding to a
specific change of state (temperature, pressure, phase).
` A convenient way to tabulate ΔÛ and Δ Ĥ is to choose a temperature, pressure
and state of aggregation (i.e. phase) as a reference state.
` Since Ĥ cannot be known absolute, for convenience we may assign a value
0
ˆ =
O
H to be a reference state. Then ;
0
ˆ
ˆ −
=
Δ H
H ;
0
ˆ
ˆ −
=
Δ H
H and so on.
1
1 2
2
` Some enthalpy tables give the reference states on which the listed values of Ĥ are
based and others do not.
` However, we do not have to know the reference state to calculate Δ Ĥ for the
transition from one state to another state since the values are based on the same
reference in the table.
s
p
k W
Q
E
E &
&
&
&
& −
=
Δ
+
Δ
+
H
Δ
∑
∑
∑
∑
∑
∑
−
=
Δ
−
=
Δ
−
=
Δ
stream
input
j
j
output
j
j
P
stream
input
j
j
stream
output
j
j
K
stream
input
j
j
stream
output
j
j
gz
m
gz
m
E
u
m
u
m
E
H
m
H
m
H
&
&
&
2
2
ˆ
ˆ
2
2
stream
40](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-40-320.jpg)
![Column 6-8. The specific enthalpies; for liquid/fluid, for intermediate (liquid and
gas), for gas.
f
h fg
h
g
h
Column 9 – 11. The entropies. for liquid/fluid, for intermediate (liquid and gas),
for gas.
f
s fg
s
g
s
Page 2 (Appendix 1) list the same properties as page 1(Appendix 1), except the pressure
is the first column and temperature the second and the table covers a much broader range
of temperatures and pressures. Appendix 2 is known as superheated steam table. It is used
for superheated properties of steam.
Example 3.5
(a) Determine the pressure, specific internal energy and specific enthalpy of saturated
steam at 330.8 °C.
(b) Show that water at 400 °C and 10 bar is superheated steam and determine its specific
volume, specific internal energy and specific enthalpy.
Solution:
(a) Data in page 1 (Appendix 1) does not go to 330.8 °C, so we take a look at Page 2.
p = 130bar; = 2496kJ/kg; = 2662kJ/kg
g
u g
h
(b) The data in Appendix 1 does not cover temperature more than 374.15, therefore it is
superheated steam.
v = 0.3065; = 2957kJ/kg; = 3262kJ/kg
g
u g
h
3.9 Energy Balance Tips
` When labeling flowchart, write down together the temperature, pressure and state
of aggregation of the process material.
` Normally (depend on the process description) for chemical process unit; shaft
work, kinetic and potential energy change tend to be negligible compared to heat
flows, internal energy and enthalpy changes.
` Then simplified energy balance become;
For closed system
[3.9]
Q
For open system
[3.10]
U
Δ
=
H
Q &
& Δ
=
43](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-43-320.jpg)
![Problems
1. Liquid methanol is pumped from a large storage tank through a 1 inch internal
pipe (ID) at the rate of 3.00 gal/min. At what rate in ft.lbf/s and hp is kinetic
energy being transported by the in the pipe methanol? (Given: Density of
methanol = 49.5 lbm/ft3
).
2. Air at 300 °C and 130 kPa flows through a horizontal 7 cm ID pipe at velocity of
42.0 m/s. Calculate )
(W
EK
& , assuming ideal gas behavior.
3. If you pour 1 gallon of water on a yowling cat 10 ft below your bedroom window,
how much potential energy (ft.lbf) does the water lose?
4. Carbon monoxide (CO) at 120 K and 45 atm has a tabulated specific volume of
5.23 cm3
/g and specific internal energy of 1813 J/mol. Calculate the specific
enthalpy of CO in this state.
5. Oxygen at 150 K and 41.64 atm has a tabulated specific volume of 4.684 cm³/g
and a specific internal energy of 1706 J/mol. Calculate the specific enthalpy of O2
in this state.
6. Values of the specific internal energy of a fuel gas at two conditions are listed
below.
State
[Fasa]
T(K) P(bar) V
ˆ (L/mol) Û (kJ/mol)
Liquid
[Cecair]
320 0.505 0.0978 0.000
Vapor
[Wap]
320 0.550 97.78 35.4
i) What reference state was used to generate the listed specific internal
energies?
ii) Calculate (kJ/mol) for a process in which a fuel gas vapor at 320 K is
condensed at constant pressure. Then, calculate
Û
Δ
Ĥ
Δ (kJ/mol) for the same
process. Finally, calculate Ĥ
Δ (kJ) for 25 mol of the fuel gas that undergo
the process.
44](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-44-320.jpg)
![7. Complete the following table.
T(O
C) P(Bar) V
ˆ (m3
/kg) Û (kJ/kg) Phase
[Fasa]
88 ........... ................. ................ Saturated steam
.......... ........... ................. 1511 Water
.......... 32 ................. ................ Water
475 15 ................. ................ ........................................
.......... 197 ................. ................ Saturated steam
45](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-45-320.jpg)
![` Specific enthalpy change:
Ideal gas : exact
∫
=
Δ
1
)
(
ˆ
T
p dT
T
C
H
2
T
Nonideal gas : exact only if pressure (P) is
constant
Solid & liquid
∫
+
Δ
=
Δ
2
1
)
(
ˆ
ˆ T
T
p dT
T
C
P
V
H
4.6 Heat Capacities, Cp
` Estimation of heat capacities, Cp
o Kopp’s rule is the simple empirical method for estimating Cp of solid or
liquid at 20 °
C based on the summation of atomic heat capacities of the
molecular compound.
H
P
O
P
Ca
P
OH
Ca
P a
a
a
C
C
C
C )
(
2
)
(
2
)
(
)
( 2
)
( +
+
=
C
mol
J O
.
/
79
)
6
.
9
2
(
)
17
2
(
26 =
×
+
×
+
=
` Estimation for heat capacities of mixtures
= Cp for ith
component
i
P
C
= mass or moles fraction
i
y
∑
= )
(
)
(
)
( T
C
y
T
C pi
i
mix
p
Example 4.1
Calculate the heat required to raise 200 kg nitrous oxide (N2O) from 20 °
C to 150 °
C in a
constant volume vessel. The constant volume heat capacity of N2O in this temperature
range is given by this equation;
where T is °
C.
Solution:
T
C
kg
kJ
Cv
4
10
42
.
9
855
.
0
)
.
/
( −
×
+
=
o
( )
∫
−
×
+
=
Δ
C
C
dT
T
kg
kJ
U
o
o
150
20
4
10
42
.
9
855
.
0
)
/
(
[ ]
kg
kJ
T
T
C
C
C
C
/
56
.
121
491
.
10
15
.
111
2
)
30
150
(
10
42
.
9
)]
20
150
(
855
.
0
[
2
10
42
.
9
855
.
0
2
2
4
150
20
2
4
150
20
=
+
=
⎥
⎦
⎤
⎢
⎣
⎡ −
×
+
−
=
⎥
⎦
⎤
⎢
⎣
⎡ ×
+
=
−
−
o
o
o
o
49](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-49-320.jpg)
![Example 4.2
15 kmol/min of air is cooled from 430 °
C to 100 °
C. Calculate the required heat removal
rate using 1) heat capacities formulas from Heat Capacities Table (Appendix 3) and 2)
Specific Enthalpies Table (Appendix 4).
Solution:
1. Write the energy balance for open system:
Δ
Q H
n
H
n
H
n
H
W
Q
E
E
H
in
air
air
out
air
air
s
p
K
ˆ
ˆ
ˆ
,
,
&
&
&
&
&
&
&
&
&
=
−
=
Δ
=
+
=
Δ
+
Δ
+
0
0 0
2. Using the heat capacities table (Appendix 3):
∫
=
Δ
C
C
p dT
T
C
mol
kJ
H
o
o
100
430
)
(
)
/
(
ˆ
[ ]
∫
−
−
−
−
×
−
×
+
×
+
×
=
Δ
C
C
dT
T
T
T
mol
kJ
H
o
o
100
430
3
12
2
8
5
3
10
965
.
1
10
3191
.
0
10
4147
.
0
10
94
.
28
)
(
ˆ
[ ]
/
C
C
C
C
C
C
C
C
T
T
T
T
mol
kJ
H
o
o
o
o
o
o
o
o
100
430
4
12
100
430
3
8
100
430
2
5
100
430
3
4
10
965
.
1
3
10
3191
.
0
2
10
4147
.
0
10
94
.
28
)
/
(
ˆ
⎥
⎦
⎤
⎢
⎣
⎡ ×
−
⎥
⎦
⎤
⎢
⎣
⎡ ×
+
⎥
⎦
⎤
⎢
⎣
⎡ ×
+
×
=
Δ
−
−
−
−
[ ]
⎥
⎦
⎤
−
×
−
⎥
⎦
⎤
⎢
⎣
⎡ −
×
+
⎥
⎦
⎤
⎢
⎡ −
×
+
−
×
=
Δ
−
−
−
−
4
)
430
100
(
10
965
.
1
3
)
430
100
(
10
3191
.
0
2
)
430
100
(
10
4147
.
0
)
430
100
(
10
94
.
28
)
/
(
ˆ
4
4
12
3
3
8
2
2
5
3
T
mol
kJ
H
⎣
⎢
⎣
⎡
mol
kJ
mol
kJ
H /
98
.
9
)
/
(
ˆ −
=
Δ
3. Using Tabulated Enthalpies (Appendix 4):
Read the value directly from the table according to the temperature desired:
mol
kJ
mol
kJ
H
mol
kJ
C
H
mol
kJ
C
H
/
98
.
9
/
)
179
.
12
19
.
2
(
ˆ
/
179
.
12
)
430
(
ˆ
/
19
.
2
)
100
(
ˆ
−
=
−
=
Δ
=
=
o
o
4. Calculate the heat removal rate.
kW
s
ol
s
ol
Q
kJ
kW
m
kJ
km
mol
kmol
H
n
H 2495
/
1
0
1
98
.
9
6
min
1
1
1000
min
15
=
×
ˆ −
×
×
×
Δ
= =
Δ
= &
&
50](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-50-320.jpg)
![Note: We can use Table in Appendix 4 if the gases are covered in the table. If not,
you have to calculate using the harder way that is heat capacities (Appendix 3).
Example 4.3
Calculate the heat required to bring 150 mol/h of a stream containing 50% C2H6 and 50%
C3H8 by volume from 0 O
C to 400 O
C. Determine a heat capacity for the mixture.
Solution:
From heat capacities table;
( )
3
12
2
8
5
3
10
28
.
7
10
816
.
5
10
92
.
13
10
37
.
49
6
2
T
T
T
C H
C
P
−
−
−
−
×
+
×
−
×
+
×
=
( )
3
12
2
8
5
3
10
71
.
31
10
11
.
13
10
59
.
22
10
032
.
68
8
3
T
T
T
C H
C
P
−
−
−
−
×
+
×
−
×
+
×
=
Try to integrate and calculate by yourself.
3
0
12
2
8
5
3
10
28
.
7
10
816
.
5
10
92
.
13
10
37
.
49
5
.
0
)]
.
/(
[
) T
T
T
C
mol
kJ
C
mix ∫
−
−
−
−
×
+
×
−
×
+
×
=
o
o
3
400
0
12
2
8
5
3
10
71
.
31
10
11
.
13
10
59
.
22
10
032
.
68
5
.
0 T
T
T
C
C
∫
−
−
−
−
×
+
×
−
×
+
×
+
o
o
mol
kJ
H /
19
.
36
ˆ =
400 C
o
(Cp
Δ
Example 4.4
A stream of gas containing 10 % CH4 and 90 % air by volume is to be heated from 20 °
C
to 300 °
C. Calculate the required rate of heat input in kW if the flow rate of the gas is
2.00 x 103
liters (STP)/min.
Solution:
1. Draw the flow chart.
2. Change the flow rate in STP to mol:
CH4= 0.1(89.3 mol) =8.93 mol; Air = (89.3-8.93)mol =80.4mol
kW
s
h
mol
kJ
h
mol
H
n
H 51
.
1
3600
1
19
.
36
150
ˆ =
×
×
=
Δ
=
Δ
= &
&
&
Q
mol
STP
L
mol
STP
L
3
.
89
)
(
4
.
22
1
min
)
(
2000
=
×
=
&
n
51](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-51-320.jpg)
![6. Calculate the heat.
[ ]
[ ]
kW
s
kJ
mol
kJ
mol
mol
kJ
mol
mol
k
mol
mol
kJ
mol
H
n
H
n
H
Q
out in
i
i
i
i
94
.
12
60
min
1
min
5
.
776
)
/
0
min)(
/
93
.
8
(
)
/
144
.
0
min)(
/
4
.
80
(
/
)
17
.
8
min)(
/
4
.
80
(
)
/
1
.
12
min)(
/
93
.
8
(
ˆ
ˆ
=
×
=
+
−
−
+
=
−
=
Δ
= ∑ ∑ &
&
&
&
4.7 Phase Change Operations
Phase change such as melting and evaporation are usually accompanied by large changes
in internal energy and enthalpy.
` Latent heat
o Specific enthalpy change associated with the phase at constant
temperature and pressure.
` Heat of fusion or heat of melting, ΔĤm (T,P)
o Specific enthalpy different between solid and liquid forms of
species at T & P.
o Heat of solidification (liquid to solid) is negative value of heat of
fusion.
` Heat of vaporization, ΔĤv (T,P)
o Specific enthalpy different between liquid and vapor forms of
species at T & P
o Heat of condensation (vapor to liquid) is negative value of heat of
vaporization.
The latent heat of phase change may vary considerably with the temperature at which the
changes occurs but hardly varies with the pressure at the transition point.
4.8 Estimation of Heat of Vaporization
1. Trouton’s rule – accuracy between 30%.
where;
b
T = Boiling point temperature
alcohol
MW
low
or
water
109
.
0
)
/
(
ˆ
liquid
nonpolar
088
.
0
)
/
(
ˆ
b
v
b
v
T
mol
kJ
H
T
mol
kJ
H
=
=
Δ
Δ
2. Chen’s equation – accuracy between 2%.
)
/
(
07
.
1
]
log
0297
.
0
0327
.
0
)
/
(
0331
.
0
[
)
/
(
ˆ 10
c
b
c
c
b
b
v
T
T
P
T
T
T
mol
kJ
H
−
+
53
−
=
Δ](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-53-320.jpg)
![STEP-by-STEP procedure of calculation:
y Material balance calculation of reactor.
y Choose reference states (usually 25 °C, 1 atm).
y Calculate extent of reaction.
y Prepare inlet-outlet enthalpy table.
y Calculate unknown component enthalpy.
y Calculate ΔH for the reactor.
evaluated
be
can
then
reactions)
(multiple
ˆ
ˆ
ˆ
reaction)
(single
ˆ
ˆ
ˆ
p
k
in
in
out
out
reaction
o
rj
j
in
in
out
out
o
r
E
E
H
W
Q
H
n
H
n
H
H
H
n
H
n
H
H
Δ
+
Δ
+
Δ
=
−
−
+
Δ
=
Δ
−
+
Δ
=
Δ
∑
∑
∑
∑
∑
ξ
ξ
Example 5.4
Normal heptane is dehydrocyclized to toluene in a continuous vapor-phase reaction:
C7 H16 ÎC6H5CH3 + 4H2
Pure heptane at 400 °C is fed to the reactor. The reactor operates isothermally at 400 °C
and the reaction goes to completion.
(a) Taking basis of 1 mol of heptane fed, draw and label a flowchart.
(b) Taking elemental species [C(s), H2(g)] at 25 °C as references, prepare and
fill in an inlet-outlet enthalpy table.
(c) Calculate the required heat transfer.
Given: C7H16(g)= 137.44x10-3
+ 40.85x10-5
T – 23.92x10-8
T2
+ 57.66x10-12
T3
.
Solution:
1) Draw the flowchart.
Basis: 1 mol heptane fed
62](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-62-320.jpg)
![3) Insert the H values calculated inside the inlet-outlet table.
References: C(s), H (g) at 25 °C
Substance Inlet Outlet
in
n
& in
Ĥ out
n
& out
Ĥ
C7H16 1 1
Ĥ = -
108.46
- -
C7H8 - - 1 2
Ĥ = 110.17
H2 - - 4 3
Ĥ = 10.89
5) Calculate the standard heat of reaction.
6) Calculate the overall enthalpy:
mol
kJ
mol
kJ
Hr
/
8
.
237
/
)]
8
.
187
(
00
.
50
)
0
(
4
[(
ˆ 0
=
−
−
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
−
Δ
=
Δ
= ∑
∑
∑ reactants
products
o
fi
i
o
fi
i
i
o
fi
i
o
r H
v
H
v
H
v
H
ˆ
ˆ
ˆ
Δ
∑
∑ −
+
Δ
=
Δ in
in
out
out
o
r H
n
H
n
H
H ξ
1
1
1
16
7
16
7 ,
=
=
=
H
C
r
H
C
v
n
ξ
&
mol
kJ
H
n
H
n
H
H in
in
out
out
o
r
/
2
.
262
]kJ/mol
(1)(79.34)
-
(4)(10.89)
)
[(1)(60.17
J/mol)
(1)(237.8k
ˆ
ˆ
ˆ
=
+
+
=
−
+
Δ
= ∑
ξ ∑
Δ
64](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-64-320.jpg)
![CHAPTER 6
BASIC CONCEPT OF HEAT TRANSFER
6.1 Introduction
o What is heat transfer?
Heat transfer is thermal energy in transit due to a temperature difference. In other
words, heat transfer is to predict the energy transfer between material bodies.
[In the simplest of terms, the discipline of heat transfer is concerned with only two
things: temperature, and the flow of heat. Temperature represents the amount of
thermal energy available, whereas heat flow represents the movement of thermal
energy from place to place.]
o How Science of heat transfer explain:
¾ How heat energy may be transfer?
¾ Predict the rate at which the exchange will take place?
¾ Different between thermodynamics and heat transfer ?
o Thermodynamics deal with system in equilibrium, it may used to predict of
energy required to change a system from one equilibrium state to another. But it
does not predict how fast a change will take place.
Example 6.1
Consider the cooling of a hot steel bar which is placed in a pail of water.
Thermodynamics may used to predict the final equilibrium temperature of steel bar-water
combination. However, thermodynamics will not tell:
¾ how long to reach the equilibrium or,
¾ what the temperature of steel bar after certain length of time?
67](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-67-320.jpg)
![Table 6.1: Comparison between heat transfer and thermodynamics
HEAT TRANSFER THERMODYNAMICS
• Heat transfer is a study which
predicts the energy transfer which
takes place between material
bodies.
• It is due to the temperature
difference
• Heat transfer explains how heat
energy may be transferred.
• It also predicts the rate at which the
exchange will take place under
certain condition.
• Thermodynamics deals with
systems in equilibrium.
• Thermodynamics may used to
predict the amount of energy
required to change a system from
one equilibrium state to another.
• Thermodynamics may not used to
predict how fast a change will take
place since the system is not in
equilibrium during the process.
Three modes of heat transfer are:
(1) Conduction
(2) Convection
(3) Radiation
6.2 Conduction Heat Transfer
o Energy transfer from high temperature region to low temperature region. We said
that the energy is transferred by conduction. And, the heat transfer rate per unit
area is proportional to the normal temperature gradient.
q/A ~ dT/dx
where,
q/A = heat transfer rate (W/m2
)
dT/dx = temperature gradient in the direction of the heat flow
When proportionality constant is inserted,
q = - kA dT/dx [6.1]
o The positive constant k = thermal conductivity of the material. The minus (-) sign
is inserted so that the second principle of thermodynamics will be satisfied i.e.
heat must flow downhill on the temperature scale as indicated in Figure 6.1.
68](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-68-320.jpg)
![Figure 6.3: Convection heat transfer from a plate
o The velocity is being reduced to zero at the plate as a result of viscous action.
Since the velocity of fluid layer at the wall will be zero, the heat must be
transferred only by conduction. Thus we might compute the heat transfer, using:
q = -kA (dT/dx)
but with little changes. We use thermal conductivity of fluid and the fluid
temperature gradient at the wall.The temperature gradient is dependent on the rate
at which the fluid carries the heat away. High velocity produces a large
temperature gradient. Thus, the temperature gradient at the wall depends on the
flow field. To express the overall effect of convection, we use Newton’s law of
cooling:
q = hA (Tw-T∞) [6.2]
where,
h = convection heat-transfer coefficient (heat-transfer coefficient = film
conductance)
h = watt per square meter per Celsius degree (W/m² .°C) when the heat flow is in
Watt.
o Heat-transfer rate is related to overall temperature difference between the wall
and fluid and the surface area.
6.3.2 Free and Forced Convection
o If heated plate were exposed to ambient room air without an external source of
motion, a movement of the air would be experienced as a result of the density
gradients near the plate.
o We call this as free convection. When the mass motion of the fluid is caused by
an external device like a pump, compressor, blower or fan, the process is called
forced convection.
Example 6.3
72](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-72-320.jpg)
![q emitted = σAT4 [6.3]
where;
σ is the proportionality constant and is called the Stefan-Boltzmann constant with
value of 5.669 x 10
-8
W/m
2
.K
4
.
o This equation is called Stefan-Boltzmann law of thermal radiation and it apply to
only blackbodies. This equation is valid only for thermal radiation; other type of
electromagnetic radiation may not be treated so simply.
q emitted = σAT4 govern only radiation emitted by a blackbody.
The net radian exchange between two surface will be proportional to the difference in
absolute temperature to the fourth power. i.e: q net exchange/ A (T1
4
– T2
4
)
q net exchange/ A = σ (T
1
4
– T2
4
)
6.4.1 Blackbody
o It is a perfect emitter of radiation. At a particular temperature the blackbody
would emit the maximum amount of energy possible for that temperature.
o This value is known as the blackbody radiation. It would emit at every
wavelength of light as it must be able to absorb every wavelength to be sure of
absorbing all incoming radiation.
o The maximum wavelength emitted by a blackbody radiator is infinite. It also
emits a definite amount of energy at each wavelength for a particular temperature,
so standard black body radiation curves can be drawn for each temperature,
showing the energy radiated at each wavelength (Fig. 6.4).
74](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-74-320.jpg)
![Figure 6.4: Theoretical black body curve for 5000K
o Again, blackbody is a body which radiates energy according to the T
4
law. Other
type of surface, such as glossy painted surface or a polished metal plate, do not
radiate as much energy as the blackbody.
o However, the total radiation emitted by this body still generally follows the T1
4
proportionality. To take account of the “gray” nature of such surfaces we
introduce another factor, called the emissivity, which relates the radiation of the
“gray” surface to that of an ideal black surface. In addition, we must take into
account the fact that not all the radiation leaving one surface will reach the other
surface since electromagnetic radiation travels in straight line and some will lost
to the surrounding.
o Therefore two new factors in (Eq 8.2)take into account both situation, so that
q= F
FG
σ A (T1
4
- T2
4
) [6.4]
F
= emissivity function
F
G
= geometric “view factor” function
A simple radiation problem is encounter when we have a heat-transfer surface at
temperature T
1
completely enclosed by much larger surface maintained at T
2. The net
radiant exchange in this case can be calculated with,
q =
1
Aσ
1
(T
4
1
- T
4
2
)
where,
1
= emissivity of material
Example 6.5
Two infinite black plates at 800
0
C and 300
0
C exchange heat by radiation. Calculate the
heat transfer per unit area. Given: σ = 5.669 x 10
-8
W/m
2
.K
4
Solution:
q/A = σ (T1
4
– T2
4
)
= 5.669 X 10
-8
)W/m
2
.K
4
(10734 - 5734)K
4
= 69.03 kW/m
2
75](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-75-320.jpg)
![7.2 Heat exchanger types
Heat exchangers are classified according to flow arrangement and type of construction.
Six (6) types of heat exchanger are:
(a) Concentric Tube Heat Exchangers
(b) Cross-Flow Heat Exchangers
(c) Cross-counter Flow (Coil) Heat Exchangers
(d) Shell-and-Tube Heat Exchangers
(e) Compact Heat Exchangers
(f) Plate-Type Heat Exchangers
(a) Concentric Tube Heat Exchanger
ƒ It also called double-pipe heat exchangers or co-axial flow heat exchangers
[Figure 7.1].
ƒ One fluid flows inside the tube and the other fluid flows inside the annulus.
ƒ In parallel-flow heat exchangers, the two fluids enter the exchanger at the same
end, and travel in parallel to exit at the other side. [Figure7.2a]
ƒ In counter-flow heat exchangers the fluids enter the exchanger from opposite ends.
Counter current design is more efficient, in that it can transfer more heat [Figure
7.2b].
Figure 7.1: Concentric tube heat exchangers
(a) Parallel flow (b) Counterflow
Figure 7.2: Concentric tube heat exchangers
77](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-77-320.jpg)
![(b) Cross Flow Heat Exchanger
ƒ In a cross-flow heat exchanger, the fluids travel roughly perpendicular to one
another through the exchanger.
ƒ In finned tubular heat exchangers, the fin-side fluid is unmixed because the fins
confine the flow to one direction. Example: automobile radiator.
ƒ In unfinned tubular heat exchangers, the fin-side fluid is mixed because the flow
in transverse direction is possible. [Figure 7.3]
ƒ The use of fins to improve the convection coefficient of fin-side fluid by
increasing the outside surface area.
Figure 7.3: Cross-flow heat exchangers
(c) Cross-counter Flow (Coil) Heat Exchangers
(
(a
a)
) F
Fi
in
nn
ne
ed
d w
wi
it
th
h b
bo
ot
th
h f
fl
lu
ui
id
ds
s
u
un
nm
mi
ix
xe
ed
d
(
(b
b)
) U
Un
nf
fi
in
nn
ne
ed
d w
wi
it
th
h o
on
ne
e f
fl
lu
ui
id
d m
mi
ix
xe
ed
d
a
an
nd
d t
th
he
e o
ot
th
he
er
r u
un
nm
mi
ix
xe
ed
d
78](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-78-320.jpg)
![ƒ Baffled heat exchanger with one shell pass and two tube passes and with two shell
passes and four tube passes are shown in Figures 7.6a and 7.6b, respectively.
Figure 7.6: Shell-and-tube heat exchanger. (a) One shell pass and two tube passes. (b)
Two shell passes and four tube passes
ƒ Shell-and-tube heat exchangers are classified according to the number of shell and
tube passes involved.
ƒ Heat exchangers in which all the tubes make one U-turn in the shell, for example,
are called one-shell-pass and two-tube-passes heat exchangers.
ƒ Heat exchanger that involves two passes in the shell and four passes in the tubes
is called a two-shell-passes and four-tube-passes heat exchanger [Figure 7.7]
80](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-80-320.jpg)
![(a) General layout (b) Detail of plate design
Figure 7.9: Plate-Type Heat Exchangers
The Overall Heat Transfer Coefficient
• The heat transfer between the two fluids across the solid wall involves convection
of fluid films adjacent to the wall and conduction across the wall.
• The rate of heat transfer can be expressed by a single equation like Newton’s law
of cooling, with the overall heat transfer coefficient U incorporating convection
and conduction terms:
[7.1]
m
q UA T
= Δ
where,
ΔTm = mean temperature difference between
the two fluids along the exchanger length
• For the unfinned tubular heat exchanger, U can be calculated as follows:
1 1 1
i i o o
UA U A U A
= = [7.2]
ln( / )
1 1 1
2
fi fo
o i
i i i o o o
R R
D D
UA h A A kL A h A
π
= + + + + [7.3]
83](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-83-320.jpg)
![where , inside and outside heat transfer areas
, overall heat transfer coefficients based on
inside and outside surface areas
, inside (tube-side) convection coefficient
i o
i o
i o
A A
U U
h h
=
=
= and
outside (shell-side) convection coefficient
, fouling resistances at inside and outside surfaces
, inside and outside diameters of the tube
k thermal conductivity o
fi fo
i o
R R
D D
=
=
= f the tube wall
tube length of heat exchanger
L =
7.3 Heat Exchanger Analysis: Use of the Log Mean Temperature Difference
(LMTD)
To design or to predict the performance of a heat exchanger, it is essential to relate the
total heat transfer rate to quantities such as the inlet and outlet fluid temperatures, the
overall heat transfer coefficient, and the total surface area for heat transfer. A few steps
to design or predict the performance of a heat exchanger:
• Step 1
- Write down the overall energy balances between heat gain of cold fluid, heat
loss of hot fluid, and heat transfer across the wall separating the two fluids.
c c h h lm
q m H m H UA T
= Δ = Δ = Δ
) )
& & [7.4]
- If the fluids are not undergoing a phase change and constant specific heats are
assumed, the equation becomes:
[7.5]
, , , , , ,
( ) ( )
h p h h i h o c p c c o c i lm
q m C T T m C T T UA T
= − = − =
& & Δ
- Determine any unknown that can be directly calculated from the above
relations.
84](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-84-320.jpg)
![( ) (
ln[( ) ( )]
hi ci ho co
lm
hi ci ho co
T T T T
T
T T T T
)
− − −
Δ =
− −
Figure 8.11: Temperature scheme for parallel flow
(b) Counterflow
For counterflow with no phase change,
ΔHh = Cph(Thi-Tho), ΔHc = Cpc(Tco-Tci)
( ) ( )
ln[( ) ( )]
ho ci hi co
lm
ho ci hi co
T T T T
T
T T T T
− − −
Δ =
− −
Figure 8.12: Temperature scheme for counterflow
(c) Condensers
ΔHh = λh, ΔHc = Cpc(Tco-Tci)
ln[( ) ( )]
co ci
lm
h ci h co
T T
T
T T T T
−
Δ =
− −
86](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-86-320.jpg)
![Figure 8.13: Temperature scheme for condenser
(d) Evaporators
ΔHc = λc, ΔHh = Cph(Thi-Tho)
ln[( ) ( )]
hi ho
lm
hi c ho c
T T
T
T T T T
−
Δ =
− −
Figure 8.14: Temperature scheme for evaporator
where λ= latent heat of vaporization
- Miscellaneous
• Overall heat transfer coefficient (U), if not known, can be determined
Equation (10.3).
• Heat transfer area (A) is related to tube length as A = 2πrL.
• Step 3
- Determine any other unknowns from the overall energy balances (Eq. 8.4)
- Determine the HE effectiveness and number of transfer units.
87](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-87-320.jpg)
![• If the heat exchanger other than the double pipe is used, the heat transfer is
calculated by using a correction factor applied to the LMTD for a counter flow
double-pipe arrangement with the same hot and cold fluids temperatures.
• The heat-transfer equation becomes :
q = UAF ΔTm [7.6]
• Values of the correction factor F are plotted in Figures 7.14 to 7.17 for several
different types of heat exchangers.
• When phase changed is involved, as in condensation or boiling (evaporation), the
fluid normally remains at essentially constant temperature.
• For this condition, P and R becomes zero and F = 1.0 (for boiling and
condensation)
Figure 7.15: Correction factor plot for exchanger with one shell pass and two, four, or
any multiple of tube passes.
88](https://image.slidesharecdn.com/chapter1introductiontoengineeringcatabello-210216214003/85/Chapter-1-introduction_to_engineering_ca-ta-bello-88-320.jpg)
Chapter 1 introduction_to_engineering_ca ta bello
- 1. CHAPTER 1 INTRODUCTION TO ENGINEERING CALCULATIONS 1.1 Introduction What do chemical engineers do? Although their backgrounds and professional skills are similar, chemical engineers work in a wide variety of industries, in addition to chemicals and petroleum, such as: Biotechnology Lime and cement Consulting Man-made fibers Drugs and pharmaceuticals Metallurgical and metal products Fats and oils Paints, varnishes, and pigments Fertilizer and agricultural chemicals Pesticides and herbicides Environment Waste water treatment All the industries as mentioned above are involving numerous of chemical process unit. Therefore, chemical engineers play an important role on design, operation, control, troubleshooting, research and management in the chemical process. Chemical process is a combination of process equipment designed to efficiently convert raw materials into finished or intermediate products. Figure 1 shows the example of chemical processes converting the raw material into desired product. Figure 1.1: Chemical process 1.2 Units and Dimensions “What are units and dimensions and how do they differ?” Dimensions are basic concepts of measurement such as length (L), mass (M), time (t), temperature (T), amount of substance (n) and so on. Besides, units are the mean of expressing the dimensions as feet or centimeters for length, or hours or seconds for time. By attaching units to all numbers that are not fundamentally dimensionless, you are able to easy interrelating the physical meaning to the numbers use. Moreover, a logical 1
- 2. approach to the problem rather than remembering a formula and plugging numbers could also help the chemical engineers in engineering calculation. SI units are universally accepted for engineering calculation. Thus, American engineering system (based on British standards) is still used extensively in the U.S. Example 1.1 What are the dimensions of mass flux (mass flow rate per unit area perpendicular to the flow)? G A dm dt = 1 t L M 2 dimensions are The rules for handling units are essentially quite simple by addition, subtraction or equality. ¾ Values could be added if UNITS are the same. ¾ Values cannot be added if DIMENSIONS are different. Example 1.2 (i) 6 ft + 10 0 C =??? * Different dimensions: length, temperature -- cannot be added * Same dimension: length, different units -- can add 2
- 3. Table 1.1: SI Units Physical Quantity Name of Unit Symbol for Unit* Definition of Unit Basic SI Units Length metre, meter m Mass kilogramme, kilogram kg Time second s Temperature kelvin K Amount of substance mole mol Derived Sl Units Energy joule J kg.m2 .s-2 Force newton N kg.m.s-2 = J.m-1 Power watt W kg.m2 .s-3 = J.s-1 Density kilogram per cubic meter kg.m-3 Velocity meter per second rn.s-1 Acceleration meter per second squared rn.s-2 Pressure newton per square meter, pascal N.m-2 , Pa Heat Capacity joule per (kilogram - kelvin) J.kg-1 ,K-1 Alternative Units Time minute, hour, day, year min, h, d, y Temperature degree Celsius °C Mass tonne, ton (Mg), gram t, g Volume litre, liter (dm3 ) L Table 1.2: American Engineering System Units Physical Quantity Name of Unit Symbol Basic Units Length feet ft Mass pound (mass) lbm Force pound (force) lbf Time second, hour s, hr Temperature degree Rankine °R Derived Units Energy British thermal unit, foot pound (force) Btu, (ft)(lbf) Power horsepower hp Density pound (mass) per cubic foot lbm/ft3 Velocity feet per second ft/s Acceleration feet per second squared ft/s2 Pressure pound (force) per square inch lbf/in2 Heat capacity Btu per pound (mass) per degree F Btu/lbm.0 F 3
- 4. 1.3 Conversion of Units Conversion factors are statements of equivalent values of different units in the same system or between systems of units. The concept is to multiply any number and its associated units with dimensionless ratios termed conversion factors to arrive at desired answer and its associated units. The factors for conversion units are show in table 1.3. Table 1.3: Factors for unit conversions Example 1.3 Convert an acceleration of 1 cm/s2 to its equivalent in km/yr2 . 2 9 2 2 2 2 2 2 2 2 2 2 2 / 10 95 . 9 1 365 1 24 1 3600 1000 1 100 1 1 yr km yr day day hr hr s m km cm m s cm × = × × × × × 4
- 5. Do It Yourself: Convert 400 in3 /day to cm3 /min. (Answer: 4.56 cm /min) .4 Processes and Process Variables fore, several rocess variables are associated through input or output of a process stream. .3.1 Instructional Objectives he objectives in studying this section are to be able to: g a process, showing input and output streams, and essential process variables. nowledge of the third quantity for any species of known density and molar mass. 3. ds of measuring temperature and at least two for (c) The meaning of the terms absolute pressure and gauge pressure. luid to the equivalent pressure expressed as a force per unit area, and vice versa an open end manometer, a sealed end manometer and a differential manometer. 6. Convert among temperatures expressed in K, °C, °F and °R. 3 1 A process is any operation or series of operations by which a particular objective is accomplished. Those mentioned operations are involving a physical or chemical change in a substance or mixture of substances. The material that enters a process is referred to as the input or feed, and that which leaves is the output or product. There p 1 T 1. Draw a simple block flow diagram representin 2. Calculate the quantities of mass (or mass flow rate), volume (or volumetric flow rate), and moles (or molar flow rates) from a k Explain: (a) The meaning of gram-mole, lb-mole, mol and kmol; (b) At least two metho measuring fluid pressure; 4. Convert a pressure expressed as a head of a f 5. Convert a manometer reading into a pressure difference for 5
- 6. 1.3.2 Process input and output. Meanwhile, the details about ocess variables are stated in table 1.4. Figure 1.2: Process streams with various process variables. Table 1.4: Process variables A process is any operation or series of operations that cause a physical or chemical change in a substance or mixture of substances. Figure 1.2 shows a process stream with several examples of process variables of pr 6
- 7. 1.3.3: Mass and Volume as a conversion factor to relate the ass and the volume of a quantity of the substance. xample 1.4 y of carbon tetrachloride is 1.595 g/cm , the mass of 35 cm3 of CCl4 is erefore, 55.825 g nd the volume of 9.3 lbm of CCl4 is = 2647.47 cm3 ty ρ of the substance to the ensity ρref of a reference substance at a specific condition: (1.1) commonly used for solids and liquids is water at 4 oC, which has the = 62.43 lbm/ft sity units called egrees Baumé (°Bé), degrees API (°API) and degrees Twaddell (°Tw). 35 cm3 1.595 The density of a substance is the mass per unit volume of the substance (kg/m3 , g/m3 , lb/ft3 , etc). The specific volume of a substance is the volume occupied by a unit mass of the substance; it is the inverse of density. Densities of pure solids and liquids are essentially independent of pressure and vary relatively slightly with temperature. Densities of many pure compounds, solutions and mixtures can be found in standard references. The density of a substance can be used m E The densit 3 th = g cm3 a 9.3 lbm 3 454 g cm The specific gravity of a substance is the ratio of the densi 1 lbm 1.595 g d The reference most following density: ρH2O (4 °C) = 1 g/cm3 = 1000 kg/m3 3 If you are given the specific gravity of a substance, multiply it by the reference density in any units to get the density of a substance in the same units. Special den d SG ρ ρ = ref 7
- 8. Example 1.5 abulated specific gravity, and calculate iven that the specific gravity of mercury at 20 C as 13.546) Volume = = 0.56 ft3 o It Yourself: ) ) d v. What volume is occupied by18g? (Answer: 36 cm ) .3.4: Flow rate te (mass/time) or as a volumetric flow rate (volume/time) as show in figure 1.3 as below. Figure 1.3: Flow rate 215 kg 3 Calculate the density of mercury in lb/ft3 from a t the volume in ft3 occupied by 215 kg of mercury. o (G 3 3 / 43 . 62 546 . 13 . ft lb G S ref Hg × = × = / 67 . 845 ft lb = ρ ρ lbm ft 0.454 kg 845.67 lb D A liquid has a specific gravity of 0.5. i. What is its density in g/cm3 ? (Answer: 0.5 g/cm3 ii. What is its specific volume in cm3 /g? (Answer: 2 cm3 /g) 3 iii. What is its density in lb/ft3 ? (Answer: 0.032 lb/ft iv. What is the mass of 3 cm3 of this liqui ? (Answer: 1.5 g) 3 1 Most processes involve the movement of material from one point to another. The rate at which a material is transported through a process line is the flow rate of that material. The flow rate of a process stream can be expressed as a mass flow ra 8
- 9. Example 1.6 Given that the molecular weight of CCl4 = 153.838 and density of ρ CCl4 = 1.595 g/cm3 .) i. What is the mass flow rate? The volumetric flow rate of CCl4 in a 1.0-cm-diameter pipe is 100 cm3 /min. ( ρ V m • • = = 100 cm3 /min x 1.595 g/cm3 = 159.5 g/min ii. at is the molar flow rate? 159.5 g-CCl4/min x g mol-CCl4/153.838 g-CCl4 = 1.034 g mol-CCl /min iii. What is the linear velocity of CCl4? o It Yourself: he mass flow rate of n-hexane (ρ=0.659 g/cm3 ) in a pipe is 6.59 g/s. ) What is the linear velocity of hexane in the pipe with internal diameter of 5 cm? (Answer: (a) 10 cm /s; (b) 0.509 cm/s) .3.5: Chemical Composition f estimating physical properties of a mixture om the properties of the pure components. .3.5(a): Moles and Molecular Weight Wh = 4 min / 32 . cm 127 / 595 . 1 4 ) 1 ( min / 5 . 159 3 2 cm g cm g A m v = × = = π ρ D T (a) What is the volumetric flow rate of the hexane? (b 3 1 Most materials encountered in nature and in chemical process systems are mixtures of various species. The physical properties of a mixture depend strongly on the mixture composition. In this section we will review different ways to express mixture compositions and also outline the methods o fr 1 The atomic weight of an element is the mass of an atom on a scale that assigns. The atomic weights of all the elements in their naturally isotopic proportions are listed in the table given. The molecular weight of a compound is the sum of the atomic weights of the atoms that constitute a molecule of the compound: For example atomic oxygen (O) has an atomic weight of approximately 16 and therefore molecular oxygen (O2) has a molecular weight of approximately 32. A gram-mole (g-mole or mol in SI units) of a 9
- 10. species is the amount of that species whose mass in grams is numerically equal to its molecular weight. Hence, one g-mole of any species co Avogadro’s number) molecules of that species. ntains approximately 6.02x1023 example, and therefore there is 454 ol/lb-mole, regardless of the substance involved. xample 1.7 ow many of each of the following are contained in 100g of CO2 (M=44.01)? i. Mol CO2 2.273 mol CO2 ii. lb-moles CO2 = 5.011 x 10-3 lb-mole CO2 , each 6.02x1023 molecules of CO2 (1mol) contains 1 mol C, 1 mol O2, or 2 mol O. Thus, iii. mol C 2.273 mol C ( Moreover, there are other types of moles such as kg-moles, lb-moles, and ton-moles. For example: Carbon monoxide (CO) has a molecular weight of 28; 1 mol of CO therefore contains 28g, 1 lb-mole contains 28 lbm, 1 ton-mole contains 28 tons and so on. Besides, the same factors used to convert masses from a unit to another can also be used to convert the equivalent molar units: there is 454 g/lb for m 100 g CO2 1 mol CO2 44.01 g CO2 100 g CO2 1 mol CO2 44.01 g CO2 100 g CO2 1 mol CO2 44.01 g CO2 E H = 2.273 m l CO2 o 1 lb-mol Each molecules of CO2 contains one atom of C, one molecule of O2 or two atoms of O. Therefore 453.6 mol = 2.273 m l CO2 o 1 mol C 1 mol CO2 10
- 11. iv. mol O = 4.546 mol O v. mol O2 = 2.273 mol O2 vi. gram O = 72.7 g O vii. gram O2 = 72.7 g O2 viii. molecules of CO2 1.37 x 1024 molecules o It Yourself: anufacture of lime nd cement. Calculate the number of lb mol of calcium carbonate in: . ) 100 lb of CaCO3. (Answer: (a) 0.11 lb mol; (b) 3.3 lb mol; (c) 1 lb mol) .3.5 (b): Mass and Mole Fractions e used to define the composition of a mixture of substances, cluding a species A. 2.273 m l CO2 o 2 mol O 1 mol CO2 2.273 m l CO2 o 1 mol O2 1 mol CO2 4.546 mol O 16 g O 1 mol O 2.273 mol O2 32 g O2 1 mol O2 = 2.273 m l CO2 6.02 x 1 lecules o 023 mo 1 mol D Calcium carbonate is a naturally occurring white solid used in the m a (a) 50 g mol of CaCO3 (b) 150 kg of CaCO3. (c 1 Process streams occasionally contain one substance, but more often they consist of mixtures of liquids or gases, or solution s of one or more solutes in a liquid solvent. The following terms can b in 11
- 12. Example 1.8 solution contains 15% A by mass (xA = 0.15) and 20 mole % B (yB = 0.20) a) alculate the mass of A in 175 kg of the solution. 26.25 kg A ) Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53 lb/h. 7.95 lb A/ h e molar flow rate of B in a stream flowing at a rate of 1000 mol/min. 200 mol B/min total solution flow rate that corresponds to a molar flow rate of 28 kmol B/s. 140 kmol solution/s Calculate the mass of the solution that contains 300 lb of A. 2000 lb solution 175 kg solution A C = 0.15 kg A kg solution b = 53 lb 0.15 lb A h lb c) Calculate th = 10 l 0.2 B 00 mo mol min mol d) Calculate the = 28 k l B 1 n mo k mol solutio s 0.2 k mol B e) = 300 lb A 1 lb solution 0.15 lb A 12
- 13. 1.3.5 (c): Concentration is the mass of this a component is the number of moles of the component per nit volume of the mixture. alue of the molar concentration of the solute expressed g-moles solute / liter solution. xample 1.9 gravity of the solution is 1.03 and molecular weight of sulfuric acid 98.08. Calculate: The mass concentration of H2SO4 in kg/m = The mass concentration of a component of a mixture or solution omponent per unit volume of the mixture. c The molar concentration of u The molarity of a solution is the v in E A 0.5 molar aqueous solution of sulfuric acid flows into a process unit at a rate of 1.25 m3 /min. The specific is 3 a) 3 4 2 3 4 2 / 49 1 1000 1000 1 98 5 . 0 m SO H kg m L g kg mol g L SO H mol = × × × b) The mass flow rate of H2SO4 in kg/s = s SO H kg s m m SO H kg / 02 . 1 60 min 1 min 25 . 1 49 4 2 3 3 4 2 = × × c) The mass fraction of H2SO4 h l volumetric flow rate and the lution density. The mass fraction of H2SO4 equals the ratio of the mass flow rate of H2SO4 to the total mass flow rate, w ich can be calculated from the tota so 3 3 / 1030 ) 1000 ( 03 . 1 m kg m kg Solution = × = ρ 13
- 14. 14 s kg s solution m kg solution m s kg solution m / 46 . 21 60 min 1 1030 min 25 . 1 ) ( 3 3 = × × = solution kg SO H kg s solution kg s SO H kg m m x solution SO H SO H / 048 . 0 / 46 . 21 / 1 4 2 4 2 4 2 4 2 = = = 1.3.6: Pressure pr A essure is the ratio of a force to the area on which the force acts. Pressure units are 2 ), dynes/cm2 , and lbf/in2 or 0 = ρ g h is called the gauge he absolute pressure. P abs = P gauge + P atm force units divided by area units such as N/m or Pascal (Pa si. p Hydrostatic pressure = pressure at the base of a fluid column P = P 0 + ρ g h P-P 0 = ρ g h If P 0 is atmospheric pressure, then P-P pressure, and P is t Fluid Pressure Measurement ost common pressure measuring devices are stated in Figure 1.4. B M ourdon gauge anometers can show measurement nearly perfect vacuums to about 700 atm. eanwhile, manometers only can measure pressures below about 3 atm. m M
- 15. anometer principle is showing in figure 1.5. The fluid pressure must be the same at any o points at th ure 1.5: Manometer principle gd2 + ρfgh P1 – P2 = (ρf –ρ) gh, since ρ1 = ρ2 Figure 1.4: Pressure measurement device. M tw e same height in a continuous fluid. Fig General manometer equation: P1 + ρ1gd1 = P2 + ρ2 Differential manometer equation: 15
- 16. 1.3.7: Temperature Temperature is a measurement of the average kinetic energy possessed by the substance ust be determined indirectly by measuring some temperature-dependent al resistance of a conductor) similar metals) c) Pyrometer (by spectra of emitted radiation) e o a fixed mass of fluid) : .15 ) = 1.8 T (K); T(0 F) = 1.8 T(0 C) + 32 (a) 60 mi/hr to m/s (b) 30 N/m2 to lbf/ft2 . Chan its: (a) 235 g to pounds. (b) 610 L to cubic feet. (c) 30 g/L to pounds/cubic feet. (d) 14.7 lb/in2 to kg/cm2 . Con antities to the ones designated: (b) 25 psig to psia. . The of a fuel oil is 0.82. (a) What is the specific gravity 20°C/4°C of this material? s a white crystalline salt, used in marking inks, medicine ny kilograms of silver nitrate (AgNO3) are there in: (a) 13.0 lb mol AgNO3. (b) 55.0 g mol AgNO3 molecules. It m physical properties of another substance. The temperature measuring devices are: a) Resistance thermometer (by means of electric b) Thermocouple (by voltage at the junction of two dis d) Thermometer (by volum f The temperature conversions are T(K) = T(0 C) + 273 ; T(0 R) = T(0 F) + 459.67; T(0 R Problems: 1. Convert the following to the desired units: (c) 16.3 J to Btu (d) 4.21 kW to J/s 2 ge the following to the desired un 3 vert the following qu (a) 42 ft2 /hr to cm2 /s. (c) 100 Btu to hp-hr. 4 specific gravity (a) What is the density of oil in lb/ft3 ? 5. The density of a liquid is 1500 kg/m3 at 20 °C. (b) What volume (ft3 ) does 140 lbm of this material occupy at 20°C? 6. Silver nitrate (lunar caustic) i and chemical analysis. How ma 16
- 17. 7. Complete the table below with the proper equivalent temperat res. ° ° u C F K ° R - 40.0 77.0 698 69.8 17
- 18. CHAPTER 2 FUNDAMENTALS OF MATERIAL BALANCES .1 Introduction put will only give one ton of total output, i.e. total mass of input = total mass f output. rates of different process streams entering or leaving chemical or physical rocesses. .2 Process Classification A. ased on manner of delivering process streams atch process 2 Material balances are important first step when designing a new process or analyzing an existing one. They are almost always prerequisite to all other calculations in the solution of process engineering problems. Material balances are nothing more than the application of the law of conservation of mass, which states that mass can neither be created nor destroyed. Thus, as an example, you cannot specify an input to a reactor of one ton of naphtha and an output of two tons of gasoline or gases or anything else. One ton of total material in o A material balance is an accounting for material. Thus, material balances are often compared to the balancing of current accounts. They are used in industry to calculate mass flow p 2 B B the boundary between the time the feed is charged and the time the product is removed. ts and unconsumed actants sometime later when the system has come to equilibrium. ontinuous process The feed is charged into a vessel at the beginning of the process and the vessel contents are removed some time later. No mass transfer across Example: Rapidly add reactants to a tank and remove the produc re C Inputs and outputs flow continuously through the duration of the process. nt rate and eadily withdraw product streams from the top and bottom of the column. emi batch process Example: Pump a mixture of liquids into a distillation column at consta st S ny process that is neither batch nor continuous. hdrawn. A Example: Allow the contents of pressurized gas container to escape to the atmosphere; slowly blend several liquids in a tank from which nothing is being wit 18
- 19. B. g to variation of the process variables with time teady state Accordin S sitions, flow rates) do not change with time. For example, batch or semi batch processes. nsteady state or transient The values of all process variables (i.e. all temperatures, pressures, compo U bles change with time. Continuous processes may be either eady-state or transient. .3 The General Balance Equation Figure 2.1: Process balance omentum) in a system as show in figure 2.1 can be written in the following general way: Input + generation - output - consumption = accumulation [2.1] here, Accumulation - builds up within system xample 2.1 move out, 22000 are born, and 9000 die. Write a balance on the population of the city. put + generation - output - consumption = accumulation 0000 + 22000 – 75000 – 19000 = - 22000 lation decreases by 22,000 people each year. Any of the process varia st 2 A balance on a conserved quantity (total mass, mass of a particular species, energy, and m w Input - enters through system boundary Generation - produced within system Output - leaves through system boundary Consumption - consumed within system E Each year 50000 people move into a city, 75000 people 1 In 5 Therefore, the city’s popu 19
- 20. Two types of balances: . Differential balance A of the balanced quantity unit divided by the time unit (people/yr, ¾ This is the type of balance usually applied to a continuous process. . Integral balance ¾ Balance that indicates what is happening in a system at an instant in time. ¾ Each term of the balance equation is a rate (rate of input, rate of generation, etc) and has units barrels/day). B ount of the balanced quantity and has the nt after the input takes place and the moment before the product is withdraw. he following rules can be used to simplify the material balance equation: is total mass, generation = 0 consumption = 0: Input - output = accumulation ce is non-reactive species, generation = consumption = 0: Input - output = accumulation If a system is at steady state, accumulation = 0: Input + generation = output + consumption .4 Balance on Continuous Steady-State Process the general balance quation, (Eq 2.1), equals zero and the equation simplifies as below: Input + generation = output + consumption [2.2] on and onsumption terms equal zero and the equation reduces to the equation as below: Output [2.3] ¾ Balance that describes what happens between two instants of time. ¾ Each term of the equation is an am corresponding unit (people, barrels). ¾ This type of balance is usually applied to a batch process, with the two instants of time being the mome T ¾ If the balanced quantity ¾ If the balanced substan ¾ 2 For continuous processes at steady state, the accumulation term in e If the equation is on a non-reactive species or on total mass, the generati c Input = 20
- 21. E 500 kg B/h 500 kg T/h m1 (kg T/h) 475 kg T/h 450 kg B/h m2 (kg B/h) xample 2.2 ene and toluene to alculate the unknown component flow rates in the output streams. he process can be depicted schematically as follows: o nonzero generation or consumption terms. For all Input = Output 0 kg B/h + m2 m2 = 50 kg B/h + 475 kg T/h m1 = 25 kg T/h ence, 1kg/h + m2 kg/h + 475 kg/h 000 kg/h = 1000 kg/h One thousand kilograms per hour of mixture of benzene (B) and toluene (T) containing 50% benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. The operation is at steady state. Write balance on benz c T Since the process is at steady state there can be no buildup of anything in the system, so the accumulation term equals zero in all material balances. In addition, since no chemical reactions occur, there can be n balances, Eq 2.3 are apply here: For Benzene balance: 500 kg B/h = 45 For Toluene balance: 500 kg T/h = m2 H Total Mass Balance: 1000 kg/h = 450 kg/h + m 1 21
- 22. 2.5 Integral Balance on Batch Process pplied to any substanc Integral Balance on Batch Process pplied to any substanc Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t=0 there are n0 mol of NH3 in the reactor, and at later time tf the reaction terminates and the contents of the reactor, which include nf mol of ammonia are withdraw. Between t0 and tf no ammonia enters or leaves through the reactor boundaries, so general balance equation (1) is simply generation = accumulation. Moreover, the quantity of ammonia that builds up (accumulates) in the reactor between t0 and tf is simply nf – n0, the final amount minus the initial amount. The same reasoning may be a Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t=0 there are n0 mol of NH3 in the reactor, and at later time tf the reaction terminates and the contents of the reactor, which include nf mol of ammonia are withdraw. Between t0 and tf no ammonia enters or leaves through the reactor boundaries, so general balance equation (1) is simply generation = accumulation. Moreover, the quantity of ammonia that builds up (accumulates) in the reactor between t0 and tf is simply nf – n0, the final amount minus the initial amount. The same reasoning may be a e participating in a the following equation: nition) = generation – consumption herefore, equating these two expressions for the accumulation yields: Initial input + generation = final output + consumption [2.4] xample 2.3: Balance on a Batch Mixing Process combined with 150g of the second, what are the mass and composition of the product? he process can be depicted schematically as follows: s of Eq (4) can be omitted so that all balance have the simple form ’. e g = m = 350g e participating in a the following equation: nition) = generation – consumption herefore, equating these two expressions for the accumulation yields: Initial input + generation = final output + consumption [2.4] xample 2.3: Balance on a Batch Mixing Process combined with 150g of the second, what are the mass and composition of the product? he process can be depicted schematically as follows: s of Eq (4) can be omitted so that all balance have the simple form ’. e g = m = 350g batch process to obtain batch process to obtain Accumulation = final output – initial input (by defi Accumulation = final output – initial input (by defi T T E E Two methanol water mixtures are contained in separate flasks. The first mixture contains 40 wt% methanol, and the second contains 70 wt% methanol. If 200g of the first mixture Two methanol water mixtures are contained in separate flasks. The first mixture contains 40 wt% methanol, and the second contains 70 wt% methanol. If 200g of the first mixture is is T T 200g Observe that the input and output ‘streams’ shown on the chart denote the initial and final states for this batch process. Since no reactions are involved, the generation and consumption term Observe that the input and output ‘streams’ shown on the chart denote the initial and final states for this batch process. Since no reactions are involved, the generation and consumption term 0 0.6g H2O/g 0 0.3g H O/g (1 –x )(g H2O/g) .4g CH3OH/g 150g m (g) x (g CH3OH/g) .7g CH3OH/g 2 ‘input = output ‘input = output Total Mass Balanc Total Mass Balanc 200g + 150 200g + 150 m m 22
- 23. Methanol Balance: ) ) ( ) ( ) ( ( ) 7 . 0 150 ( ) 4 . 0 200 ( 3 3 g x g m g OH gCH g g OH gCH g × = × + × x = 0.529 g CH3OH/g .6 Integral Balance on Semi-batch Process 3OH gCH en so far. However, some problems of this type are relatively straight forward lutions. xample 2.4 e. Use an integral balance to estimate the time required to vaporize 10m3 of e liquid. he process can be depicted schematically as follows: id nor reacts with hexane in the process unit, the balance reduces to input = utput: 2 Integral balances can also be written for semi-batch and continuous processes. The procedure is to write a differential balance on the system and then to integrate it between two instants of time. In most cases the required calculations are more complex than those we have se so E Air is bubbles through a drum of liquid hexane at a rate of 0.1 kmol/min. The gas stream leaving the drum contains 10 mole% hexane vapors. Air may be considered insoluble in liquid hexan th T 0.1 kmol /min n 0 l 0.9 kmol air/kmol .1 kmol C6H14/kmo We begin with a differential balance on air. Since we assume that air neither dissolves in the liqu o min / 111 . 0 min 9 . 0 min 1 . 0 kmol kmol n kmol air kmol air kmol = × = n We next write an integral hexane balance, proceeding from time t=0 to tf (min), the time we seek to calculate. The balance has the form accumulation = - output. The accumulation term, which is the total change in the moles of liquid hexane in the system during time tf, must be negative since hexane is being lost from the system. Since the 23
- 24. total number of moles of hexane evaporated occupied a liquid volume of 10 cubic meters and the specific gravity of liquid hexane is 0.659, the accumulation term equals: 14 6 3 2 . 86 kg m The output term in th 3 45 . 76 1 659 . 0 10 H C kmol kmol kg m n − = × × − = e balance is the rate at which hexane is leaving the system [0.1n mol C6H14/min)] times the total process time, tf (min). The balance (accumulation = - 0.1n tf n = 0.111 kmol /min s. ¾ If no stream amount or flow rate is specified in the problem statement, take as a 2) Draw the flowchart of the process, using boxes or other symbols to represent the 3) Fully label the chart when it is first drawn with values of known process variables being written for each input and output stream. 5) If the problem is solvable, the starting balance should be an equation with minimum 6) After the one unknown in step 4 has been calculated, use that value to give an 7) As each unknown variable is determined, its value is filled so that the chart provides ontinuous record of where the solution stands and what must still be done. Δ (k output) is therefore; - 76.45 kmol C6H14 = - tf = 6887 min 2.7 General procedure for material balance calculations 1) Choose a basis of calculation an amount or flow rate of one of the process stream basis an arbitrary amount or flow rate of a stream with a known composition. process units, and lines with arrows to represent input and output streams. and symbols for unknown variables 4) Do the degree-of-freedom analysis. number of unknowns, preferably with only one unknown. equation with one variable for another unknown. a c Notes: ¾ The maximum numbers of independent equations that can be derived by ¾ (Σ xi = 1.0). writing balances on a non-reactive system equal the number of chemical species in the input and output streams. The additional equations can be written from the fact that the component mass or mole fractions of a stream adds up to 1.0 (You can use these constrains to reduce the number of unknowns in the flow chart (step 2 above) to a minimum.) 24
- 25. ¾ If you are given that the mass of stream 1 is half that of stream 2, label the masses of these streams m and 2m rather than m1 and m2; if you are not normally written on volumetric quantities. rate is given, it is generally useful to calculate the mass or molar f . know that there is three If a volumetric flow low rate for the equires an environment of humid ir enriched in oxygen. Three input streams are fed into an evaporation chamber to mposition. : Air (21 mole% O2, the balance N2) he output gas is analyzed and is found to contain 1.5 mole% of water. Draw and label a am variables he process can be depicted schematically as follows: e variable name (n1) is chosen for the air flow rate, the given information ion of H2O in the outlet stream is known to be 0.015, once the mole times as much nitrogen (by mass) in a stream as oxygen, label the mass fractions of O2 and N2 y and 3y rather than y1 and y2. (This can also reduce the number of unknowns in step 2 above, or can provide additional equation needed for solution.) ¾ Balances balance Example 2.5 An experiment on the growth rate of certain organisms r a produce an output stream with the desired co A: Liquid water, fed at a rate of 20 cm3 /min B C: Pure oxygen, with a molar flow rate one fifth of the molar flow rate of stream B. T flowchart of the process, and calculate all unknown stre T 0.21g mol O2/mol 0.79 mol N2/mol 0.2 n (mol 1 n3 0.015 mol H O/mol n1 (mol air/min) 2 y mol O2/mol (0.985 – y) (mol N2/mol) 3 20 cm H2O/min n2 (mol H2O/min) Notes on the labeling: 1. Since the one known flow rate (20 cm3 H2O/min) is given on a per minute basis, it is most convenient to label all stream flow rates on this basis. 2. Once th about the ratio of the air and O2 flow rates may be used to label the O2 flow rate 0.2n1. 3. The mole fractions of the components of any stream must add up to 1. Since the mole fract 25
- 26. fraction of O2 is labeled y, that of N2 must be 1 – (y + 0.015) = (0.985 – y) (mol N2/mol). The quantity n2 can be calculated from the given volumetric flow rate and the density of liquid water: min / 11 . 1 02 . 18 1 1 min 20 2 3 2 3 2 O H mol g mol cm g O H cm n = × × = The three remaining unknowns (n1; n2 and y) can be determined from balance, all of ut = output for this non-reactive steady state process. The art. Water balance: in) = n3 (0.015 mol H2O/min) 1 + n1 + n2 = n3 n1 = 60.74 mol/min Nitrogen Balance: which have the simple form inp balances are easily written by referring to the flow ch n2 (mol H2O/m n3 = 74 mol/min Total Mole Balance: 0.2n mol O mol y y mol N mol y mol n mol N mol mol n / 337 . 0 ) 985 ) 985 . 0 ( min 79 . 0 min 2 2 3 2 1 = − − n n . 0 ( 79 . 0 3 1 = ∴ × = × or the unknown values of m. Kelly says that 4 values of m are missing, that you an write 3 component material balances, and that you can use 3 relations for ∑mi =1, ne for each stream, a total of 6 equations, so that a unique solution is not possible. Who right? Do It Yourself: In the steady state flow process shown in the figure below, a number of values of m (mass fraction) are not given. Mary says that nevertheless the problem has a unique solution f c o is 26
- 27. (Answer: Stream F, m = 0.90; stream A, m = 0.5; stream P, m = 0.75, m3 = 0.075) .7 Balancing a process onsid umbe umbe iF iD iW ion constraint of each stream sults in the overall balance. Thus the number of independent equations = 7. ince number of unknowns = number of independent equations, the problem is solvable. hen how will you solve the problem in the easiest way? F 10 = kg 2 2 2 2 C er the following process flowchart for a separation unit at steady state: N r of unknowns = 7 N r of mass balance equations: 4 component balances: F x = D x + W x for i = A, B, C, and D 1 overall balance: F = D + W 3 mass fraction constraints of streams: Σ x iF = 1.0, Σ x iD = 1.0, Σ x iW = 1.0 Total number of equations = 8. They are not all independent, because the summing of the four component balances and then using the mass fract re S T A = 6kg P = 16kg m1 = 0.01 m2 =? m1 = 0.30 m3 = 0.20 m2 =? m3 =? m1 = 0.175 m2 =? 27
- 28. Example 2.6 An aqueous solution of sodium hydroxide contains 20% NaOH by mass. It is desired to produce an 8.0% NaOH solu f pure water. Calculate the tion by diluting a stream of the 20% solution with a stream ratios (liters H 2 O needed / kg feed solution) and (kg product g feed solution). 0 k tep 2: teps 1 and 2): 1 m 1 , and m 2 ) Number of independent equations = 2 2-2 = 0. Thus the problem is solvable. uation that contains only one unknown is NaOH balance since it is not NaOH mass balance: 100 x 0.2 = 0.08 m2 he remaining two balances (H2O and overall) contain 2 unknowns, and simplest of these overa verall ass b lance: So, m1 = m2 – 100 o solution / k Solution: Basis: 10 g of 20% feed solution S Step 3 (in conjunction with S Degree of freedom analysis: umber of unknowns needed to be solved = 2 (V in terms of N Since it is a nonreactive process and involves two species, So, degree of freedom = Step 4 (solution procedure): Number of mass balance equations involved: 2 The balance eq present in pure water stream. So choose it. m2 = 250 kg T is ll balance, so choose it. O m a 100 + m1 = m2 = 250 – 100 = 150 kg 28
- 29. Volum Thus, e stream at the rate tream w chart of the process. (b) Calculate the composition (in %) of the three compounds in the unknown eam and its flow rate in kg/hr. nswer: (a) e = 92.5% rised of many interconnected units. However, umber of independent equations equals the number of variables whose values are Thus, e stream at the rate tream w chart of the process. (b) Calculate the composition (in %) of the three compounds in the unknown eam and its flow rate in kg/hr. nswer: (a) e = 92.5% rised of many interconnected units. However, umber of independent equations equals the number of variables whose values are e of diluents water, V 1 = m 1 /ρ w = 150 kg x 1.0 liter/kg = 150 liter = 150 kg x 1.0 liter/kg = 150 liter V1/100 kg = 150 liter/100 kg = 1.5 liters H2O/kg feed solution V1/100 kg = 150 liter/100 kg = 1.5 liters H2O/kg feed solution m2/100 kg m2/100 kg = 250 kg/100 kg = 2.5 kg product solution/kg feed solution on = 250 kg/100 kg = 2.5 kg product solution/kg feed solution on Do It Yourself: Do It Yourself: A solution composed of 50% ethanol (EtOH), 10% methanol (MeOH), and 40% water H2O) i A solution composed of 50% ethanol (EtOH), 10% methanol (MeOH), and 40% water H2O) i ( s fed at the rate of 100 kg/hr into a separator that produces of 60 kg/hr with the composition of 80% EtOH, 15% MeOH, and 5 ( s fed at the rate of 100 kg/hr into a separator that produces of 60 kg/hr with the composition of 80% EtOH, 15% MeOH, and 5% H2O, and a second s of unknown composition. (a) Sketch and completely label a flo % H2O, and a second s of unknown composition. (a) Sketch and completely label a flo str str A A 1 hr 10 % MeOH 40 % H2O y % MeOH H 15 % MeOH 5 % H2O 00 kg/hr 60 kg/hr 50 % EtOH 80 % EtO A kg/ x % EtOH z % H2O A Answ r: (b) A = 40 kg/hr; x = 5%; y = 2.5%; z nsw r: (b) A = 40 kg/hr; x = 5%; y = 2.5%; z 2.8 Balances on Multiple Unit Processes Plants in the process industries are comp 2.8 Balances on Multiple Unit Processes Plants in the process industries are comp you can easily apply the same techniques discussed in the previous sections to solve material balance problems in such plants. List and count up the number of variables whose values are unknown, making sure you do not count the same variable more than once, and then list and count up the n you can easily apply the same techniques discussed in the previous sections to solve material balance problems in such plants. List and count up the number of variables whose values are unknown, making sure you do not count the same variable more than once, and then list and count up the n independent balances that you can make, making sure that balances for one unit do not render formerly independent balances for another unit into dependent balances. If the number of independent balances that you can make, making sure that balances for one unit do not render formerly independent balances for another unit into dependent balances. If the number of u unknown, at least for a set of linear equations, you can generally solve the equations for a unique answer. nknown, at least for a set of linear equations, you can generally solve the equations for a unique answer. 29
- 30. If you ignore all the internal streams and variables within a set of connected subsystem, as as inputs Feed Streams 1, 2 and 3 and Product Stream 1, 2 and 3.Balances n this system are referred to as overall balances. The stream that connects Unit 1 and 2 flowing to Unit 1 is an output. Boundary C encloses Unit 1 (one put stream and two output streams). Boundary D encloses a stream splitting point (one nd two output streams). Boundary E encloses Unit 2 (two input streams and ne output stream). labeled flow chart of a continuous steady state process is shown below. Each stream ontains of two components; A and B, in different proportions. Three streams whose flow tes and compositions are not known are labeled 1, 2 and 3. you can treat the overall system exactly as you treated a single system by drawing a oundary about the entire set of subsystem. flow chart for a two un rocess is show as below. b A it p Feed 1 Feed 2 Five boundaries drawn above portions of the process define systems on which balances can be written. Boundary A encloses the entire process; the system defined by this boundary h o is internal system are referred to this system and so would not either into overall system balances. Boundary B encloses a feed stream mixing point. Feed Streams 1 and 2 are inputs to this system and the stream in input stream a o Example 2.7 A c ra Product Product Feed 3 Product Unit 1 Unit 2 A B C D E 30
- 31. 100 kg/h 30 kg/h 30 kg/h 40 kg/h 0.5 kg A/kg 0.5 kg B/kg 0.9 kg A/kg 0.1 kg B/kg 0.6 kg A/kg 0.4 kg B/kg x1 kg A/kg 1 - x1 kg B/kg x2 kg A/kg 1 – x2 kg B/kg m1 kg/h m2 kg/h m3 kg/h x3 kg A/kg 1 – x3 kg B/kg 0.3 kg A/kg 0.7 kg B/kg Solution: Overall Mass Balance: + m3 m3 = 60 kg/h : (kg A/h) ) = (0.9)(40) + (0.6)(30) + x3 (60) 33 kg A/kg 1: m1 = 60 kg /h g A/h) x3 = 0.233 kg A/kg eam Mixing Point: m2 = 90 kg/h (100 + 30) kg/h = (40 + 30) kg/h Overall Mass Balance on A (0.5)(100) + (0.3)(30 x3 = 0.08 Mass Balance on Unit 100 = 40 + m1 Balance on A in Unit 1: (k (0.5)(100) = (0.9)(40) + x1 (60) Mass Balance on Str m1 + 30 = m2; m1 = 60 kg/h 31
- 32. Balance on A in Stream Mixing point: (kg A/h) x1 m1 + (0.3)(30) = x2 m2; x1 = 0.233 kg A/kg m1= 60 kg /h m2= 90 kg /h kg A/kg tate double effect evaporator is shown below. he system of evaporator is designed to reduce water from an incoming brine (NaCl + a t r e eiving 12000 lb/hr of NaCl (along with 60 wt % H2O), determine: ) in lb/hr. r; V2= 2727.27 kg/hr) x2 = 0.255 Do It Yourself: A labeled flow chart of a continuous steady s T H2O) stream from 60 wt % to 20 wt %. If the ev pora o unit is r c (a) The product rate (P2) of brine in lb/hr. (Answer: P2 = 6000 kg/hr) (b) The water removed from 1 V2 (Answe .73 kg/h the brine in each evaporator (V r: V1= 3272 P1 P2 H2O 1.00 I II V1 V2 H Brine 12000 lb/hr NaCl 0.40 H2O 0.60 NaCl 0.55 H2O 0.45 NaCl 0.80 H2O 0.20 2O 1.00 32
- 33. Problems: . atch, (2) flow, (3) neither, or (4) both on time s (c) Catalytic converter on an automobile 2. many independent material balance equations can be formulated for this problem? How many variables whose values are unknown exist in the problem 3. ) pure solvent. Sketch the flow charts for the scaled processes and calculate the weight of each of the three stocks that must be blended together to ch are settled and removed from the remaining liquor. contains 50% NaOH, 2% NaCl, a b) The kilograms of salt precipitated per hour 1 Classify the following processes as (1) b a cale of one day: (a) Oil storage tank at a refinery (b) Flush tank on a toilet (d) Gas furnace in a home A continuous still is to be used to separate acetic acid, water, and benzene from each other. On a trial run, the calculated data were as shown in the figure. Data recording the benzene composition of the feed were not taken because of an instrument defect. The problem is to calculate the benzene flow in the feed per hour. How ? A liquid adhesive, which is used to make laminated boards, consists of a polymer dissolved in a solvent. The amount of polymer in the solution has to be carefully controlled for this application. When the supplier of the adhesive receives an order for 3000 kg of an adhesive solution containing 13 wt % polymer, all it has on hand is (A) 500 kg of a 90 wt % solvent, (B) a very large quantity of a 20 wt % solvent, and (C fill the order. 4. An evaporator is fed continuously with 25,000 kg/h of a solution containing 10% NaOH, 10% NaCl, and 80% H2O. During evaporation, water is boiled off, and salt precipitates as crystals, whi The concentrated liquor leaving the evaporator nd 48% H2O. Calculate: a) The kilograms of water evaporated per hour c) The kilograms of concentrated liquor produced per hour. 33
- 34. 5. % of pure nitric acid have to be added through the waste contains 40% H2SO4 (a) (b) the quantity of waste solution of acid and pure acid that’s needed to be added together to produce 1500 kg of mixed solution of 6. an incoming 3 wt %. If the evaporator unit is to produc (a) The feed rate (F) of brine in lb/hr. (b) The water removed from the brine in each evaporator (V1 V2 V3) in lb/hr. A waste solution of acid from titration process contains 33% of sulfuric acid (H2SO4), 36% of nitric acid (HNO3) and 31% of water (H2O) by mass. 98% of pure sulfuric acid and 78 solution of acid to produce final mixed solution of acid which and 43% HNO3. Draw and completely label a flow chart of the process. Calculate acid. A labeled flow chart of a continuous steady state triple effect evaporator is shown below. The system of evaporator is designed to reduce water from brine (NaCl + H2O) stream from 75 wt % to e 14,670 lb/hr of NaCl (along with 3 wt % H2O), determine: I II III Brine F lb/hr P1 P2 P3 14670 lb/hr V1 V2 V3 NaCl 0.25 H2O 0.75 NaCl 0.33 H2O 0.67 NaCl 0.50 H2O 0.50 NaCl 0.97 H2O 0.03 2O 1.00 2O 1.00 2O 1.00 H H H 34
- 35. CHAPTER 3 GY AND ENERGY BALANCES ` ional motion of the system as a whole relative to ` ergy (U) due to translation, rotation, vibration & electromagnetic olecules, atom and subatomic particle within the across the system boundaries while ocess is taking place), energy may be transferred between such a system and f flow is always from a higher temperature to a low one. Heat is defined as positive when its transferred to the system from the as a force, a torque or a voltage. e when it is done by the system on the at energy can neither be created nor estroyed. ` eneral form of first law of thermodynamics ENER 3.1 Forms of Energy Three component of total energy of a system: Kinetic energy (Ek) energy due to the translat ◦ some frame of reference (usually the earth’s surface) or to rotation of the system about some axis. ial energy (E ) ` Potent p ◦ energy due to the position of the system in a potential field (such as a gravitational or electromagnetic field). Internal en ◦ all energy possessed by a system other than kinetic and potential energy; or ◦ Energy interactions of the m system. 3.2 Transfer of Energy ` In closed system (i.e. no mass is transferred the pr its surroundings in two ways as heat or work. Heat ◦ Energy that flows as a result of temperature difference between a system and its surroundings. The direction o ◦ ◦ surroundings. Work ◦ Energy that flows in response to any driving force other a temperature difference, such ◦ ork is defined as positiv surroundings. W 3.3 First Law of Thermodynamics ` Law of conservation of energy, which state th d G 35
- 36. where; Inlet Energy + Heat - Outlet Energy – Work = Accumulation Inlet energy and outlet energy is summation/total of all energy such as potential, kinetic and internal energy. oving relative to the surface of the earth is /s) m ve [3.2] s into a 2 cm internal diameter (ID) pipe at a rate of 2.00 m³/h. Calculate m elocity (u) = Volumetric flowrate = 3.4 Kinetic Energy Equation (Ek) ` Kinetic energy, Ek (J) of an object of mass m (kg) m with velocity u (m/s) [3.1] 2 2 1 mu Ek = ` If the fluid enters a system with a mass flow rate m & (kg and unifor locity u (m/s), the rate at which kinetic energy K E & (J/s) is transported into the system is Example 3.1 2 1 u m E & = k E & 2 k & Water flow in J/s. for this strea Solution: First, we calculate the velocity (u). V ) ( 2 r A V u π & = Pipe cross sectional area Then, we calculate the mass flow rate of the water ( ). m & s kg s h m kg h m V / 556 . 0 3600 1 1000 00 . 2 3 3 = × × = = ρ & m & s m cm h / 77 . 1 3600 1 ) 1 ( 2 2 2 2 = π s h m cm m 1 100 1 00 . 2 2 2 3 × × × = 2 2 u m K & & = 1 E 36
- 37. K E & Finally, calculate s J s m N s m kg N s u m EK 1 2 2 × × = & m s kg / 870 . 0 / . 870 . 0 / . 1 ) 77 . 1 ( / 556 . 0 2 1 2 2 2 = = = ` l potentia nergy, E bjec is giv below ` the fluid enters a system with a mass flow rate (kg/s) and an elevatio relative to th ` e interested in the change of potential energy during energy balance calculation; [3.5] at a rate of 15.0 kg/s from a point 220 meters below the int 20 meters above the ground level. Calculate the attendant rate of potential energy. & 3.5 Potential Energy Equation (Ep) Gravitationa l e n t en p of a o [3.3] mgz Ep If m & n z = e potential energy reference plane. [3.4] Normally we ar gz m E & & = p ) ( 1 2 1 2 z z g m E E E p p p − = − = Δ & & & & Example 3.2 Crude pump oil is pumped earth’s surface to a po increase of Solution: 20 m 220 m Ground level [ ] N s J s m W s m kg s s 35300 / 35300 / . 35300 / . 1 2 2 = = = N m m kg z z g m E E E p p p 1 ) 220 ( 20 81 . 9 15 ) ( 1 2 1 2 × − − × × = − = − = Δ & & & & 37
- 38. 3.6 Energy Balances on Closed System ` Closed system is defined as no mass is transferred across the system boundaries hile the process is taking place. ` nergy balance for closed system; here; w E w Initial energy system = ) ( ) ( initial P initial K initial E E U + + ) ( Final energy system = U ) ( ) ( ) ( final P final K final E E + + Net energy transfer = W Q − changes, phase changes, or Final System Energy – Initial System Energy = Net Energy Transferred to the System W Q E E E E U U initial P final P initial K fina K initial final − = + + − − l − ) ( ) ( ) ( ) ( ) ( ) ( ( ) ( ) ( ) or [3.6] W Q E E U p k − = Δ + Δ + Δ ` When applying energy balance equation to a given process, the following point must be aware; 1. The internal energy of a system depends almost entirely on the chemical composition, state of aggregation (solid, liquid, or gas), and temperature of the system materials. If no temperature chemical reactions occur in a closed system and if pressure changes are less than a few atmospheres, then ∆ ≈ 0. U 2. If a system is not accelerating, then ∆ K E = 0. If a system is not rising or falling, then ∆ P E = 0. 3. If a system and its surroundings are at the same temperature or the system is perfectly insulated, then Q = 0. The process is then termed adiabatic. 4. Work done on or by a closed system is accomplished by movement of the system boundary against a resisting force or the passage of an electrical current or radiation across the system boundary. If there no moving parts or electrical current at the system boundary, then W = 0. en System is done on the surrounding by mass that emerges from the systems. ` Both work terms must be include in the energy balance for open system. 3.7 Energy Balances on Op ` In open system, mass is transferred across the system boundaries while the process is taking place. ` Therefore work must be done on open system to push mass in and work 38
- 39. ` he net ork d y an open system; process fluid on a moving part within the d at the system outlet minus rate of ol is used to denote the specific property (property divided by mass or by T w one b where; & - shaft work (rate of work done by the S W system such as a pump rotor.) & - flow work (rate of work done by the flui fl W work done by the fluid at the system inlet.) ` ^ symb mole) such as specific internal energy (U ) kJ/kg), specific volume ( V ) m3 /kg) and so on. ne important property for energy balance ` on open system is specific enthalpy ( O Ĥ kJ/kg). ` Sometimes, universal gas law constant can be used as a conversion factor to cific enthalpy. of helium at this temperature and pressure, and the rate at which enthalpy is ansported by a stream of helium at 300 K and 1 atm with a molar flowrate of 250 Solution: o convert L.atm/mol into J/mol, we need the factor J/(L.atm). From the gas constant table, ol.K) y dividing these two, in in out out in out fl fl s W W W & & & + = V P W & & = − V P W W & & & − = [3.7] V P U H ˆ ˆ ˆ + = evaluate spe Example 3.3 The specific internal energy of helium at 300 K and 1 atm is 3800 J/mol, and the specific molar volume at the same temperature and pressure is 24.63 L/mol. Calculate the specific enthalpy tr kmol/h. mol atm L mol J mol L atm mol J V P U H / . 63 . 24 / 3800 ) / 63 . 24 )( 1 ( / 3800 ˆ ˆ ˆ + = + = + = T 0.08206 L.atm/(mol.K) = 8.314 J/(m the desired factor will be obtained; B ) . /( 3 . 101 ) . /( . 08206 . 0 ) . /( 314 . 8 atm L J K mol atm L K mol J = 39
- 40. So; mol J J atm L mol atm L mol J / 6295 . 3 . 101 . 63 . 24 / 3800 ˆ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × + = If n = 250 kmol; H kW s J s h kmol mol mol J h kmol H n H 437 / 437150 3600 1 1 1000 6295 250 ˆ = = × × × = = & ` Energy balances equation for open system; [3.8] Answer: where; 3.8 Reference States and State Properties ` It is not possible to know the absolute value of Û and Ĥ for a process material, but we can determine the change in ΔÛ and change in Δ Ĥ corresponding to a specific change of state (temperature, pressure, phase). ` A convenient way to tabulate ΔÛ and Δ Ĥ is to choose a temperature, pressure and state of aggregation (i.e. phase) as a reference state. ` Since Ĥ cannot be known absolute, for convenience we may assign a value 0 ˆ = O H to be a reference state. Then ; 0 ˆ ˆ − = Δ H H ; 0 ˆ ˆ − = Δ H H and so on. 1 1 2 2 ` Some enthalpy tables give the reference states on which the listed values of Ĥ are based and others do not. ` However, we do not have to know the reference state to calculate Δ Ĥ for the transition from one state to another state since the values are based on the same reference in the table. s p k W Q E E & & & & & − = Δ + Δ + H Δ ∑ ∑ ∑ ∑ ∑ ∑ − = Δ − = Δ − = Δ stream input j j output j j P stream input j j stream output j j K stream input j j stream output j j gz m gz m E u m u m E H m H m H & & & 2 2 ˆ ˆ 2 2 stream 40
- 41. ` Ĥ and Û can be said as state properties (property whose change of value in any process depend only on it initial and final states and do not depend on the path take to reach the state. Example 3.4 The following data are for methyl chloride: State T(°F) P(psia) ) / ( ˆ 3 m lb ft V ) / ( ˆ m lb Btu H Liquid -40 6.878 0.01553 0.0 Vapor 0 18.9 4.969 196.23 Vapor 50 51.99 1.920 202.28 (a) What reference state was used to generate the given enthalpies? (b) Calculate Ĥ Δ and U Δ̂ for the transition of saturated methyl chloride vapor from 50°F to 0°F. Solution: (a) Liquid at -40 °F and 6.878 psia (the state at which 0 ˆ = H ). (b) m m initial final lb Btu lb Btu H H H / 05 . 6 / ) 28 . 202 23 . 196 ( ˆ ˆ ˆ − = − = − = Δ From equation 3.7, ( ) m m m lb Btu U psia ft Btu lb ft psia lb Btu PV H U V P U H / 96 . 4 . 73 . 10 987 . 1 / . ) 821 . 99 914 . 93 ( / 05 . 6 ) ˆ ˆ ˆ 3 3 − = Δ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × − − − = Δ − Δ = Δ ∴ Δ + Δ = Δ 3.9 Steam Table The simplest phase diagrams are pressure-temperature diagrams of a single simple substance, such as water. Figure 3.1 shows the water phase diagram. The axes correspond to the pressure and temperature. The phase diagram shows, in pressure-temperature space, 41
- 42. the lines of equilibrium or phase boundaries between the three phases of solid, liquid, and gas. The phase boundary between liquid and gas does not continue indefinitely. Instead, it terminates at a point on the phase diagram called the critical point. This reflects the fact that, at extremely high temperatures and pressures, the liquid and gaseous phases become indistinguishable, in what is known as a supercritical fluid. In water, the critical point occurs at around Tc=647.096 K (1,164.773 °R), pc=22.064 MPa (3,200.1 psi) and ρc=356 kg/m³. Figure 3.1: Water Phase Diagram Appendix 1 lists properties of saturated liquid water and saturated steam at temperature from 0.01 °C to 100 °C. The following properties can be determined for each tabulated temperature: Column 2. The pressure Column 3. The specific volumes, (m³/kg), of gas or saturated steam. g g v v / ˆ Column 4 – 5. The specific internal energies; for liquid/fluid, for gas. f u g u 42
- 43. Column 6-8. The specific enthalpies; for liquid/fluid, for intermediate (liquid and gas), for gas. f h fg h g h Column 9 – 11. The entropies. for liquid/fluid, for intermediate (liquid and gas), for gas. f s fg s g s Page 2 (Appendix 1) list the same properties as page 1(Appendix 1), except the pressure is the first column and temperature the second and the table covers a much broader range of temperatures and pressures. Appendix 2 is known as superheated steam table. It is used for superheated properties of steam. Example 3.5 (a) Determine the pressure, specific internal energy and specific enthalpy of saturated steam at 330.8 °C. (b) Show that water at 400 °C and 10 bar is superheated steam and determine its specific volume, specific internal energy and specific enthalpy. Solution: (a) Data in page 1 (Appendix 1) does not go to 330.8 °C, so we take a look at Page 2. p = 130bar; = 2496kJ/kg; = 2662kJ/kg g u g h (b) The data in Appendix 1 does not cover temperature more than 374.15, therefore it is superheated steam. v = 0.3065; = 2957kJ/kg; = 3262kJ/kg g u g h 3.9 Energy Balance Tips ` When labeling flowchart, write down together the temperature, pressure and state of aggregation of the process material. ` Normally (depend on the process description) for chemical process unit; shaft work, kinetic and potential energy change tend to be negligible compared to heat flows, internal energy and enthalpy changes. ` Then simplified energy balance become; For closed system [3.9] Q For open system [3.10] U Δ = H Q & & Δ = 43
- 44. Problems 1. Liquid methanol is pumped from a large storage tank through a 1 inch internal pipe (ID) at the rate of 3.00 gal/min. At what rate in ft.lbf/s and hp is kinetic energy being transported by the in the pipe methanol? (Given: Density of methanol = 49.5 lbm/ft3 ). 2. Air at 300 °C and 130 kPa flows through a horizontal 7 cm ID pipe at velocity of 42.0 m/s. Calculate ) (W EK & , assuming ideal gas behavior. 3. If you pour 1 gallon of water on a yowling cat 10 ft below your bedroom window, how much potential energy (ft.lbf) does the water lose? 4. Carbon monoxide (CO) at 120 K and 45 atm has a tabulated specific volume of 5.23 cm3 /g and specific internal energy of 1813 J/mol. Calculate the specific enthalpy of CO in this state. 5. Oxygen at 150 K and 41.64 atm has a tabulated specific volume of 4.684 cm³/g and a specific internal energy of 1706 J/mol. Calculate the specific enthalpy of O2 in this state. 6. Values of the specific internal energy of a fuel gas at two conditions are listed below. State [Fasa] T(K) P(bar) V ˆ (L/mol) Û (kJ/mol) Liquid [Cecair] 320 0.505 0.0978 0.000 Vapor [Wap] 320 0.550 97.78 35.4 i) What reference state was used to generate the listed specific internal energies? ii) Calculate (kJ/mol) for a process in which a fuel gas vapor at 320 K is condensed at constant pressure. Then, calculate Û Δ Ĥ Δ (kJ/mol) for the same process. Finally, calculate Ĥ Δ (kJ) for 25 mol of the fuel gas that undergo the process. 44
- 45. 7. Complete the following table. T(O C) P(Bar) V ˆ (m3 /kg) Û (kJ/kg) Phase [Fasa] 88 ........... ................. ................ Saturated steam .......... ........... ................. 1511 Water .......... 32 ................. ................ Water 475 15 ................. ................ ........................................ .......... 197 ................. ................ Saturated steam 45
- 46. CHAPTER 4 ENERGY BALANCE ON NONREACTIVE SPECIES 4.1 Introduction ` Normally in chemical process unit, S W =0; Δ P E =0; Δ K E =0; Then energy balance equation become: Close System Open System U Q Δ = H Q Δ = ` For this chapter, we will learn the procedure for evaluating ΔU and ΔH when table Ĥ and Û are not available for all process species. ` Method to calculate Δ Ĥ and ΔÛ associated with certain process such as: 1. Change in P, at constant T & constant state of aggregation. 2. Change in T, at constant T & constant state of aggregation. 3. Phase changes at constant T & constant P. 4. Mixing at constant T & constant P. 5. Chemical reaction at constant T & constant P. 4.2 Hypothetical Process Path ` State properties o Properties that depend on the state of the species (primarily on its temperature and state of aggregation, and to lesser extent on its pressure). o Specific enthalpy ( Ĥ ) and specific internal energy ( Û ) are state properties species. o When a species passes from one state to another state, both Δ Ĥ and ΔÛ for the process are independent of the path taken from the first state to the second state. ` We can construct a hypothetical process path which can consist of several step based on our convenience, as long as we reach to the final state starting from their initial state. ` For example, the enthalpy change (Δ Ĥ ) of solid phenol at 25 o C and 1 atm to phenol vapor at 300 o C and 3 atm. phenol Ĥ Δ = (vapor, 300˚C, 3 atm) – (solid, 25˚C, 1 atm) o Cannot determine directly from enthalpy table – must use hypothetical process path consist of several step. o Check Table of Physical Properties Data Appendix 5: P= 1 atm; Tm= 42.5°C and Tb= 181.4°C 46
- 47. 6 5 4 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ H H H H H H H Δ + Δ + Δ + Δ + Δ + Δ = Δ 4.3 Procedure Energy Balance Calculations 1. Perform all required material balance calculations. 2. Write the appropriate form of the energy balance (closed or open system) and delete any of the terms that are either zero or negligible for the given process system. 3. Choose a reference state – phase, temperature, and pressure – for each species involved in the process. 4. Construct inlet-outlet table for specific internal energy (close system) or specific enthalpy (close system) { For closed system, construct a Table with columns for initial and final amounts of each species (mi or ni) and specific internal energies (Û ) relative to the chosen reference states. { For an open system, construct a table with columns for inlet and outlet stream component flow rates (mi or ni) and specific enthalpies ( Ĥ ) relative to the chosen references states. 5. Calculate all required values of Ĥ or Û and insert the values in the appropriate places in the table. Then calculate Δ Ĥ or ΔÛ for the system. 6. Calculate any work, kinetic energy, or potential energy terms that you have not dropped from the energy balance. 7. Solve the energy balance for whichever variable is unknown (often Q). 47
- 48. Example of Inlet-Outlet Enthalpy Table: References: Ac (liquid, 20˚C, 5atm); N2 (gas, 25˚C, 1atm) Substance Inlet Outlet in n & in Ĥ out n & out Ĥ Ac (vapor) 66.9 3.35 Ac (l) - - 63.55 0 N2 33.1 33.1 4.4 Change in Pressure (P) at Constant Temperature (T) & Constant Phase ` Solid & Liquid o Nearly independent of pressure. P V H U Δ = Δ = Δ ˆ ˆ 0 ˆ ` Ideal Gases o Independent of pressure (unless undergo very large pressure changes). 0 ˆ 0 ˆ = Δ = Δ H U 4.5 Change in Temperature (T) at Constant Pressure (P) & Constant Phase ` Sensible heat is the heat that must be transferred to RAISE or LOWER the temperature of substance or mixture of substance. o P C - Heat capacity at constant pressure (given in Appendix 3 the form of polynomial and the equation is the function of temperature). o V C - Heat capacity at constant volume. Gas Ideal R C C Solid Liquid C C v p v p + = = & ` Specific internal energy change: Ideal gas : exact Solid or liquid : good approximation Nonideal gas : valid only if volume (V) is constant dT T C U T T v ) ( ˆ 2 1 ∫ = Δ 48
- 49. ` Specific enthalpy change: Ideal gas : exact ∫ = Δ 1 ) ( ˆ T p dT T C H 2 T Nonideal gas : exact only if pressure (P) is constant Solid & liquid ∫ + Δ = Δ 2 1 ) ( ˆ ˆ T T p dT T C P V H 4.6 Heat Capacities, Cp ` Estimation of heat capacities, Cp o Kopp’s rule is the simple empirical method for estimating Cp of solid or liquid at 20 ° C based on the summation of atomic heat capacities of the molecular compound. H P O P Ca P OH Ca P a a a C C C C ) ( 2 ) ( 2 ) ( ) ( 2 ) ( + + = C mol J O . / 79 ) 6 . 9 2 ( ) 17 2 ( 26 = × + × + = ` Estimation for heat capacities of mixtures = Cp for ith component i P C = mass or moles fraction i y ∑ = ) ( ) ( ) ( T C y T C pi i mix p Example 4.1 Calculate the heat required to raise 200 kg nitrous oxide (N2O) from 20 ° C to 150 ° C in a constant volume vessel. The constant volume heat capacity of N2O in this temperature range is given by this equation; where T is ° C. Solution: T C kg kJ Cv 4 10 42 . 9 855 . 0 ) . / ( − × + = o ( ) ∫ − × + = Δ C C dT T kg kJ U o o 150 20 4 10 42 . 9 855 . 0 ) / ( [ ] kg kJ T T C C C C / 56 . 121 491 . 10 15 . 111 2 ) 30 150 ( 10 42 . 9 )] 20 150 ( 855 . 0 [ 2 10 42 . 9 855 . 0 2 2 4 150 20 2 4 150 20 = + = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − × + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × + = − − o o o o 49
- 50. Example 4.2 15 kmol/min of air is cooled from 430 ° C to 100 ° C. Calculate the required heat removal rate using 1) heat capacities formulas from Heat Capacities Table (Appendix 3) and 2) Specific Enthalpies Table (Appendix 4). Solution: 1. Write the energy balance for open system: Δ Q H n H n H n H W Q E E H in air air out air air s p K ˆ ˆ ˆ , , & & & & & & & & & = − = Δ = + = Δ + Δ + 0 0 0 2. Using the heat capacities table (Appendix 3): ∫ = Δ C C p dT T C mol kJ H o o 100 430 ) ( ) / ( ˆ [ ] ∫ − − − − × − × + × + × = Δ C C dT T T T mol kJ H o o 100 430 3 12 2 8 5 3 10 965 . 1 10 3191 . 0 10 4147 . 0 10 94 . 28 ) ( ˆ [ ] / C C C C C C C C T T T T mol kJ H o o o o o o o o 100 430 4 12 100 430 3 8 100 430 2 5 100 430 3 4 10 965 . 1 3 10 3191 . 0 2 10 4147 . 0 10 94 . 28 ) / ( ˆ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × + × = Δ − − − − [ ] ⎥ ⎦ ⎤ − × − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − × + ⎥ ⎦ ⎤ ⎢ ⎡ − × + − × = Δ − − − − 4 ) 430 100 ( 10 965 . 1 3 ) 430 100 ( 10 3191 . 0 2 ) 430 100 ( 10 4147 . 0 ) 430 100 ( 10 94 . 28 ) / ( ˆ 4 4 12 3 3 8 2 2 5 3 T mol kJ H ⎣ ⎢ ⎣ ⎡ mol kJ mol kJ H / 98 . 9 ) / ( ˆ − = Δ 3. Using Tabulated Enthalpies (Appendix 4): Read the value directly from the table according to the temperature desired: mol kJ mol kJ H mol kJ C H mol kJ C H / 98 . 9 / ) 179 . 12 19 . 2 ( ˆ / 179 . 12 ) 430 ( ˆ / 19 . 2 ) 100 ( ˆ − = − = Δ = = o o 4. Calculate the heat removal rate. kW s ol s ol Q kJ kW m kJ km mol kmol H n H 2495 / 1 0 1 98 . 9 6 min 1 1 1000 min 15 = × ˆ − × × × Δ = = Δ = & & 50
- 51. Note: We can use Table in Appendix 4 if the gases are covered in the table. If not, you have to calculate using the harder way that is heat capacities (Appendix 3). Example 4.3 Calculate the heat required to bring 150 mol/h of a stream containing 50% C2H6 and 50% C3H8 by volume from 0 O C to 400 O C. Determine a heat capacity for the mixture. Solution: From heat capacities table; ( ) 3 12 2 8 5 3 10 28 . 7 10 816 . 5 10 92 . 13 10 37 . 49 6 2 T T T C H C P − − − − × + × − × + × = ( ) 3 12 2 8 5 3 10 71 . 31 10 11 . 13 10 59 . 22 10 032 . 68 8 3 T T T C H C P − − − − × + × − × + × = Try to integrate and calculate by yourself. 3 0 12 2 8 5 3 10 28 . 7 10 816 . 5 10 92 . 13 10 37 . 49 5 . 0 )] . /( [ ) T T T C mol kJ C mix ∫ − − − − × + × − × + × = o o 3 400 0 12 2 8 5 3 10 71 . 31 10 11 . 13 10 59 . 22 10 032 . 68 5 . 0 T T T C C ∫ − − − − × + × − × + × + o o mol kJ H / 19 . 36 ˆ = 400 C o (Cp Δ Example 4.4 A stream of gas containing 10 % CH4 and 90 % air by volume is to be heated from 20 ° C to 300 ° C. Calculate the required rate of heat input in kW if the flow rate of the gas is 2.00 x 103 liters (STP)/min. Solution: 1. Draw the flow chart. 2. Change the flow rate in STP to mol: CH4= 0.1(89.3 mol) =8.93 mol; Air = (89.3-8.93)mol =80.4mol kW s h mol kJ h mol H n H 51 . 1 3600 1 19 . 36 150 ˆ = × × = Δ = Δ = & & & Q mol STP L mol STP L 3 . 89 ) ( 4 . 22 1 min ) ( 2000 = × = & n 51
- 52. 3. Write the energy balance; H Q W Q E E H s p K & & & & & & Δ = + = Δ + Δ + Δ 0 0 0 4. Construct inlet-outlet table. References: CH4 (gas, 20˚C, 1atm); air (gas, 25˚C, 1atm) Substance Inlet Outlet in n & in Ĥ out n & out Ĥ CH4 8.93 0 8.93 1 Ĥ air 80.4 2 Ĥ 80.4 3 Ĥ 5. Calculate the enthalpies. Why we choose air at 25°C and 1 atm as our reference? This is because it is the reference state for gases in Table B.8, so we can still calculate the enthalpy at 20°C. mo ∫ = C C CH p dT C H o o 300 20 1 4 ) ( ˆ dT T T T C C kJ l / 1 . 12 ) 10 0 . 11 10 3661 . 0 10 469 . 5 10 31 . 34 ( 3 12 2 8 20 5 3 = × − × + × + × = − − − − ∫o 300o To calculate H2 and H3, use Table in Appendix 4: Why we calculate Ĥ Δ of air using Table B.8 instead of Table B.2? It is because the air is included in Table B.8. We can still integrate and calculate using the long way. You will get the same answer. mol kJ H mol kJ H mol kJ H C air C air / ) 0 144 . 0 ( ˆ / 144 . 0 ˆ / 0 . 0 ˆ 2 ) 20 , ( ) 25 , ( − − = Δ − = = o o mol kJ H mol kJ H mol kJ H C air C air / ) 0 17 . 8 ( ˆ / 17 . 8 ˆ / 0 . 0 ˆ 2 ) 300 , ( ) 25 , ( − = Δ = = o o Insert the value of enthalpies inside the inlet-outlet table. References: CH4 (gas, 20˚C, 1atm); air (gas, 25˚C, 1atm) Substance Inlet Outlet in n & in Ĥ out n & out Ĥ CH4 8.93 0 8.93 12.1 air 80.4 -0.144 80.4 8.17 52
- 53. 6. Calculate the heat. [ ] [ ] kW s kJ mol kJ mol mol kJ mol mol k mol mol kJ mol H n H n H Q out in i i i i 94 . 12 60 min 1 min 5 . 776 ) / 0 min)( / 93 . 8 ( ) / 144 . 0 min)( / 4 . 80 ( / ) 17 . 8 min)( / 4 . 80 ( ) / 1 . 12 min)( / 93 . 8 ( ˆ ˆ = × = + − − + = − = Δ = ∑ ∑ & & & & 4.7 Phase Change Operations Phase change such as melting and evaporation are usually accompanied by large changes in internal energy and enthalpy. ` Latent heat o Specific enthalpy change associated with the phase at constant temperature and pressure. ` Heat of fusion or heat of melting, ΔĤm (T,P) o Specific enthalpy different between solid and liquid forms of species at T & P. o Heat of solidification (liquid to solid) is negative value of heat of fusion. ` Heat of vaporization, ΔĤv (T,P) o Specific enthalpy different between liquid and vapor forms of species at T & P o Heat of condensation (vapor to liquid) is negative value of heat of vaporization. The latent heat of phase change may vary considerably with the temperature at which the changes occurs but hardly varies with the pressure at the transition point. 4.8 Estimation of Heat of Vaporization 1. Trouton’s rule – accuracy between 30%. where; b T = Boiling point temperature alcohol MW low or water 109 . 0 ) / ( ˆ liquid nonpolar 088 . 0 ) / ( ˆ b v b v T mol kJ H T mol kJ H = = Δ Δ 2. Chen’s equation – accuracy between 2%. ) / ( 07 . 1 ] log 0297 . 0 0327 . 0 ) / ( 0331 . 0 [ ) / ( ˆ 10 c b c c b b v T T P T T T mol kJ H − + 53 − = Δ
- 54. where; b T = c T c P = Normal boiling point temperature Critical temperature = Critical Pressure 3. Clausius-Clapeyron equation - plot ln p* versus 1/T. B RT H p In v + Δ − = ˆ * 4. Chaperon equation R H T d p In d v ˆ ) / 1 ( ) ( * Δ − = 5. Watson correlation – estimate ΔĤv at T2 from known ΔĤv at T1. 38 . 0 1 2 1 2 ) ( ˆ ) ( ˆ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − Δ = Δ T T T T T H T H c c v v where; = c T Critical temperature 4.8 Estimation of Heat of Fusion ΔĤm (kJ/mol) = 0.0092 Tm (K) metallic elements = 0.0025 Tm (K) inorganic compound = 0.050 Tm (K) organic compound Example 4.5 The normal boiling point of methanol is 337.9 K, and the critical temperature of this substance is 513.2 K. Estimate the heat of vaporization of methanol at 300 °C . Solution: First, we calculate the standard heat of vaporization using Trouton’s rule: Then, we calculate the latent heat using Watson’s correlation: mol kJ K T mol kJ H b V / 83 . 36 )) 9 . 337 ( 109 . 0 ( ( 109 . 0 ) / ( ˆ ≈ ) Δ ≈ ≈ mol kJ T T T T T H T H c c V V / 0 . 21 9 . 337 2 . 513 473 2 . 513 83 . 36 ) ( ˆ ) ( ˆ 38 . 0 38 . 0 1 2 1 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − Δ = Δ 54
- 55. Problems: 1. Estimate the specific enthalpy of steam (kJ/kg) at 350 °C and 100 bar relative to steam at 100 °C and 1 atm using: (a) The steam tables (b) Heat capacities table 2. Chlorine gas is to be heated from 100 °C and 1 atm to 200 °C. (a) Calculate the heat input (kW) required to heat a stream of the gas flowing at 5.0 kmol/s at constant pressure. (b) Calculate the hat input (kJ) required to raise the temperature of 5.0 kmol of chlorine in a closed rigid vessel from 100 °C and 1 atm to 200 °C. 3. A stream of water vapor flowing at a rate of 250 mol/h is brought from 600 °C and 10 bar to 100 °C and 1 atm. (a) Estimate the required cooling rate (kW) three ways: (i) from the steam tables, (ii) using heat capacity data and (iii) using specific enthalpy data. (b) Which of the answers in part (a) is most accurate and why? 4. A fuel gas containing 95 mole% methane and the balance ethane is burned completely with 25% excess air. The stack gas leaves the furnace at 900 °C and is cooled to 450 °C in a waste heat boiler, a heat exchanger in which heat lost by cooling gases is used to produce steam from liquid water for heating , power generation, or process applications. (a) Taking as abasis of calculation 100 mol of the fuel gas fed to the furnace, calculate the amount of heat (kJ) that must be transferred from the gas in the waste heat boilerto accomplish the indicated cooling. (b) How much saturated steam at 50 bar can be produced from boiler feedwater at 40 °C for the same basis of calculation? 55
- 56. CHAPTER 5 ENERGY BALANCE OF REACTIVE SYSTEMS 5.1 Types of reaction ` Exothermic reaction: the product molecules have lower internal energies than the reactants at the same T and P. ΔH = NEGATIVE, reaction produces heat. ` Endothermic reaction: the product molecules have higher internal energies than the reactants. ΔH = POSITIVE, reaction consumes heat. 5.2 Heat of reaction ` ΔH depends on how the stoichiometric equation is written. CH4 (g) + 2O2 (g) Î CO2(g) + 2H2O(l) ΔHr1 (25O C) = -890.3 kJ/mol Î for 1 CH4 2CH4 (g) + 4O2 (g) Î 2CO2 (g) + 4H2O (l) ΔHr2 (25O C) = -1780.6 kJ/mol Î for 2 CH4 ` ΔH depends on the states of aggregation (gas, liquid, or solid) CH4 (g) + 2O2 (g) Î CO2 (g) + 2H2O (l) ΔHr1 (25°C)= -890.3 kJ/mol CH4 (g) + 2O2 (g) Î CO2 (g) + 2H2O (g) ΔHr2 (25° C)= -802.3 kJ/mol 5.3 Standard heat of reaction (ΔHr o ) ` heat of reaction when both reactants and products are at reference conditions (usually 25 °C and 1 atm). C4H10 (g) + 13/2O2 (g) Î 4CO2 (g) + 5H2O(l) ΔHr1 (25 °C)= -2878 kJ/mol Example: For 2400 mol/s CO2 produced; s kJ mol kJ s mol Hr / 10 73 . 1 2878 / 4 2400 6 2 × − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = Δ 5.4 Reaction in a REACTOR (constant V) ΔUr(T)=Uproducts – Ureactants ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − − Δ = Δ ∑ ∑ reactants gaseous products gaseous ) ( ) ( i i r r v v RT T H T U 56
- 57. 5.5 Hess’s Law ` Look at this reaction. C (s) + ½ O2 (g) Î CO (g): ΔHr = ? (1) C (s) + O2 (g) Î CO2 (g): ΔHr1 = -393.51 kJ/mol (2) CO (g) + ½ O2 (g) Î CO2 (g): ΔHr2 = -282.99 kJ/mol ΔHr = ΔHr1 + (- ΔHr2) = (-393.51 + (-282.99)) = -110.52 kJ/mol Example 5.1 The standard heats of the following combustion reactions have been determined experimentally: 1. C2H6 + 7/2 O2 Î 2 CO +3 H2O: 2. C + O2 Î CO2: 3. H2 + ½ O2 Î H2O: mol kJ H mol kJ H mol kJ H r r r / 8 . 285 ˆ / 5 . 393 ˆ / 8 . 1559 ˆ 0 3 0 2 0 1 − = Δ − = Δ − = Δ Use Hess’s law and the given heats of reaction to determine the standard heat of reaction 4. 2 C + 3 H2 Î C2H6 ? ˆ 0 4 = Δ r H Solution: 1) Rearrange the given equations: mol kJ mol kJ Hr / 787 / ) 2 5 . 393 ( ˆ 0 2 − = × − = Δ mol kJ mol kJ Hr / 4 . 857 / ) 3 8 . 285 ( ˆ 0 3 − = × − = Δ mol kJ Hr / 8 . 1559 ˆ 0 3 − = Δ (Eq. 2 x 2) 2C + 2O2 Î 2CO2 (Eq. 3 x 3) 3H2 + 3/2 O2 Î 3H2O (Eq. 1) C2H6 + 7/2 O2 Î 2 CO +3 H2O 2) Substrate or add these equations to form the desired equation: 3H2 + 3/2 O2 Î 3H2O + 2C + 2O2 Î2CO2 3H2 + 7/2 O2 + 2C Î 3H2O + 2CO2 - C2H6 + 7/2 O2 Î 2 CO2 +3 H2O mol kJ Hr / 4 . 857 ˆ 0 3 − = Δ mol kJ Hr / 787 ˆ 0 2 − = Δ mol kJ Hr / 4 . 1644 ˆ 0 ) 2 3 ( − = Δ − mol kJ Hr / 8 . 1559 ˆ 0 1 − = Δ mol kJ Hr / 6 . 84 ˆ 0 4 − = Δ 3H2 + 2C ÎC2H6 57
- 58. 5.6 Heat of combustion (ΔHc) ` The heat of the combustion of a substance with oxygen to yield specific products. C2H5OH (l) + 3 O2 (g) Î 2 CO2 (g) + 3 H2O (l): ΔHc (25 o C, 1 atm) = -1366.9 kJ/mol ` The value of Hc for the substance is indicated in Table B.1. Example 5.2: Calculate the heat of reaction for the dehydrogenation of ethane: C2H6 Î C2H4 + H2 Then, use Hess’s law to find the standard heat of reaction for the above equation, using these reactions: 1. C2H6 + 7/2O2 Î2CO2 + 3H2O 2. C2H4 + 3 O2 Î2CO2 + 2H2O 3. H2 + ½ O2 ÎH2O Compare the answer. Solution: 1) Find standard heat of combustion for every substance. From Appendix 5; 2) Use the formula to calculate the standard heat of reaction: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ − = ∑ ∑ ∑ products reactants ) ( ) ( ) ( i o c i i o c i i i o c i o r H v H v H v H Δ mol kJ H mol kJ H mol kJ H H c H C c H C c / 84 . 285 ) ( / 99 . 1410 ) ( / 9 . 1559 ) ( 2 4 2 6 2 0 0 0 − = Δ − = Δ − = Δ mol kJ H mol kJ H v H v H v H o r i o c i i o c i i i o c i o r / 9 . 136 ) / ) 84 . 285 99 . 1410 ( ) 9 . 1559 [( ) ( ) ( ) ( products reactants − = Δ − − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ − = Δ ∑ ∑ ∑ 58
- 59. 3) Find the standard heat of reaction for each equation using standard heat of combustion: mol kJ H mol kJ H mol Eq r Eq r / 84 . 285 / 99 . 1410 / ) 3 . ( 0 ) 2 . ( 0 − = Δ − = Δ kJ H Eq r 9 . 1559 ) 1 . ( 0 − = Δ 4) Substrate or add these equations to form the desired equation: (1) – (2) – (3) C2H6 + 7/2O2 Î2CO2 + 3H2O mol kJ H mol kJ H mol kJ H mol kJ H mol kJ H r Eq r Eq r Eq r Eq r / 93 . 136 / 84 . 285 / 91 . 148 / 99 . 1410 / 9 . 1559 0 ) 3 . ( 0 ) 2 1 . ( 0 ) 2 . ( 0 ) 1 . ( 0 − = Δ − = Δ − = Δ − = Δ − = Δ − - C2H4 + 3O2Î2 CO2 + 2 H2O C2H6 + ½ O2ÎC2H4 + H2O - H2 + ½ O2 ÎH2O C2H6 ÎC2H4 + H2 5.7 Heat of formation (ΔHf) ` Enthalpy change associated with the formation of 1 mole of a compound from its elemental constituents (in nature) at a reference T and P. N2 (g) + 2 H2 (g) + 3/2 O2 (g) Î NH4NO3 (c) ΔHr o = -365.14 kJ/mol 6 C(s) + 3 H2 (g) Î C6H6 (l) ΔHr o = 48.66 kJ/mol Note: standard heat of formation of an elemental species is ZERO. ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ = Δ ∑ ∑ ∑ reactants products o fi i o fi i i o fi i o r H v H v H v H Example 5.3 Determine the standard heat of reaction for the combustion of liquid n-pentane, assuming H2O(l) is a combustion product. C5H12 (l) + 8O2 (g) Î5CO2 (g) + 6H2O (l) Then, use Hess’s law to find the standard heat of reaction for the above equation, using these reactions: 1. 5C (s) + 6H2 ÎC5H12 2. C (s) + O2 (g) ÎCO2 3. H2 (g) + ½O2 (g) ÎH2O (l) Compare the answer. 59
- 60. Solution: 1) Find standard heat of formation for every substance. From table B.1; mol kJ H mol kJ H mol kJ H O H f CO f H C f / 84 . 285 ) ( / 5 . 393 ) ( / 0 . 173 ) ( 2 2 12 5 0 0 0 − = Δ − = Δ − = Δ 2) Use the formula to calculate the standard heat of reaction: 3) Find the standard heat of reaction for each equation using standard heat of formation and use Hess Law to calculate the desired equation: 1. 5C (s) + 6H2 ÎC5H12 2. C (s) + O2 (g) ÎCO2 3. H2 (g) + ½O2 (g) ÎH2O (l) The desired reaction can be obtained by; ) 1 ( ) 3 ( 6 ) 2 ( 5 − × + × , and the can be obtained from Hess Law. o r H Δ 5.8 Energy Balances (General Procedures) ` Heat of reaction method: mol kJ mol kJ H v H v H v o i o f i i o f i i i o f i o r / 3509 ) / ) 0 . 173 ( ) 84 . 285 6 ( ) 5 . 393 5 [( ) ( ) ( ) ( reactants products − = Δ − − − × + − × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ − = Δ ∑ ∑ ∑ H Hr 60
- 61. ` Heat of formation method: Some notes (extent of reaction) ` For a single reaction, extent of reaction (ξ) can be calculated: known product or reactant any is A where A in A out A v n n − − − = ξ Some notes (inlet-outlet enthalpy table) Components nin Hin nout Hout A B C Some notes ` Latent heat Î heat transferred without change of T. It could be as heat of vaporization (or condensation). There is a phase change. H H ˆ ˆ Δ = Δ ` Sensible heat Î heat transferred due to the T difference. There is a change of T, but no phase change. v Since Cp = f (T), then don’t forget to integrate it. See Appendix 3. ∫ = Δ 2 ˆ T T pdT C H 1 61
- 62. STEP-by-STEP procedure of calculation: y Material balance calculation of reactor. y Choose reference states (usually 25 °C, 1 atm). y Calculate extent of reaction. y Prepare inlet-outlet enthalpy table. y Calculate unknown component enthalpy. y Calculate ΔH for the reactor. evaluated be can then reactions) (multiple ˆ ˆ ˆ reaction) (single ˆ ˆ ˆ p k in in out out reaction o rj j in in out out o r E E H W Q H n H n H H H n H n H H Δ + Δ + Δ = − − + Δ = Δ − + Δ = Δ ∑ ∑ ∑ ∑ ∑ ξ ξ Example 5.4 Normal heptane is dehydrocyclized to toluene in a continuous vapor-phase reaction: C7 H16 ÎC6H5CH3 + 4H2 Pure heptane at 400 °C is fed to the reactor. The reactor operates isothermally at 400 °C and the reaction goes to completion. (a) Taking basis of 1 mol of heptane fed, draw and label a flowchart. (b) Taking elemental species [C(s), H2(g)] at 25 °C as references, prepare and fill in an inlet-outlet enthalpy table. (c) Calculate the required heat transfer. Given: C7H16(g)= 137.44x10-3 + 40.85x10-5 T – 23.92x10-8 T2 + 57.66x10-12 T3 . Solution: 1) Draw the flowchart. Basis: 1 mol heptane fed 62
- 63. 2) Prepare inlet-outlet Table References: C(s), H (g) at 25 °C Substance Inlet Outlet in n & in Ĥ out n & out Ĥ C7H16 1 1 Ĥ - - C7H8 - - 1 2 Ĥ H2 - - 4 3 Ĥ 3) Find all the enthalpies mol kJ mol kJ dT T T T mol kJ dT C H H C C C C p H C f / 46 . 108 / ) 34 . 79 8 . 187 ( 10 66 . 57 10 92 . 23 10 85 . 40 10 44 . 137 / 8 . 187 ˆ ˆ 400 25 3 12 2 8 5 3 400 25 ) ( 0 1 16 7 − = + − = × + × − × + × + − = + = ∫ ∫ − − − − o o o o 63 H3 can be obtained directly from Appendix 4: H mol kJ mol kJ dT T T T mol kJ dT C H H C C C C p H C f / 17 . 110 / ) 17 . 60 00 . 50 ( 10 33 . 80 10 86 . 27 10 00 . 38 10 18 . 94 / 00 . 50 ˆ ˆ 400 25 3 12 2 8 5 3 400 25 ) ( 0 2 8 7 = + + = × + × − × + × + + = + = ∫ ∫ − − − − o o o o mol kJ mol kJ H H C H C H / 89 . 10 / ) 0 89 . 10 ( ˆ ˆ ˆ ) 25 , ( ) 400 , ( 3 2 2 = − = − = o o
- 64. 3) Insert the H values calculated inside the inlet-outlet table. References: C(s), H (g) at 25 °C Substance Inlet Outlet in n & in Ĥ out n & out Ĥ C7H16 1 1 Ĥ = - 108.46 - - C7H8 - - 1 2 Ĥ = 110.17 H2 - - 4 3 Ĥ = 10.89 5) Calculate the standard heat of reaction. 6) Calculate the overall enthalpy: mol kJ mol kJ Hr / 8 . 237 / )] 8 . 187 ( 00 . 50 ) 0 ( 4 [( ˆ 0 = − − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ − Δ = Δ = ∑ ∑ ∑ reactants products o fi i o fi i i o fi i o r H v H v H v H ˆ ˆ ˆ Δ ∑ ∑ − + Δ = Δ in in out out o r H n H n H H ξ 1 1 1 16 7 16 7 , = = = H C r H C v n ξ & mol kJ H n H n H H in in out out o r / 2 . 262 ]kJ/mol (1)(79.34) - (4)(10.89) ) [(1)(60.17 J/mol) (1)(237.8k ˆ ˆ ˆ = + + = − + Δ = ∑ ξ ∑ Δ 64
- 65. Problems: 1. Use Hess’s law to calculate the standard heat of the water-gas shift reaction; CO(g) + H2O(v) ÎCO2(g) + H2(g) from each of the two sets of data given here. lbmole Btu Hr / 1226 ˆ 0 + = Δ (a) CO(g) + H2O(l) ÎCO2(g) + H2(g): H2O(l) ÎH2O(v): lbmole Btu Hr / 935 , 18 ˆ 0 + = Δ (b) CO(g) + ½ o2(g) ÎCO2(g): H2(g) + ½O2(g) Î H2O(v): 2. Formaldehyde maybe produced in the reaction between methanol and oxygen: 2CH3OH(l) + O2(g) Î 2HCHO(g) + 2H2O(l): The standard heat of combustion of hydrogen is H2(g) + ½ O2(g) Î H2O(l): Use these heats of reaction and Hess’s Law to determine the standard heat of the direct decomposition of methanol to form formaldehyde: CH3OH(l) Î HCHO(g) + H2(g) 3. Trichloroethylene, a widely used degreasing solvent for machine parts, is produced in a 2 steps reaction sequence. Ethylene is first chlorinated to yield tetrachloroethane, which is dehydrochlorinated to form trichloroethylene. C2H4(g) + 2Cl2(g) Î C2H2Cl4(l) + H2(g) C2H2Cl4(l) Î C2HCl3(l) + HCl(g) The standard heat of formation of liquid trichloethylene is -276.2 kJ/mol. lbmole Btu Hr / 740 , 121 ˆ 0 − = Δ mol kJ Hr / 76 . 385 ˆ 0 − = Δ lbmole Btu Hr / 040 , 104 ˆ 0 − = Δ (a) Use the given data and tabulated standard heats of formation of ethylene and hydrogen chloride to calculate the standard heat of formation of tetrachloroethane and the standard heat of the second reaction. (b) Use Hess’s Law to calculate the standard heat of the rection C2H4(g) + 2Cl2(g) Î C2HCl3(l) + H2(g) +HCl(g) 65
- 66. 4. The standard heat of reaction for the oxidation of ammonia isgiven below: 4NH3(g) + 5O2(g) Î 4NO(g) + 6H2O(v) 500 mol NH3/s and 800 mol O2 at 25 °C are fed into a reactor in which the ammonia is completely consumed. The product gas emerges at 300 °C. Calculate the rate at which heat must be transferred to or from the reactor, assuming operation at approximately 1 atm. 66
- 67. CHAPTER 6 BASIC CONCEPT OF HEAT TRANSFER 6.1 Introduction o What is heat transfer? Heat transfer is thermal energy in transit due to a temperature difference. In other words, heat transfer is to predict the energy transfer between material bodies. [In the simplest of terms, the discipline of heat transfer is concerned with only two things: temperature, and the flow of heat. Temperature represents the amount of thermal energy available, whereas heat flow represents the movement of thermal energy from place to place.] o How Science of heat transfer explain: ¾ How heat energy may be transfer? ¾ Predict the rate at which the exchange will take place? ¾ Different between thermodynamics and heat transfer ? o Thermodynamics deal with system in equilibrium, it may used to predict of energy required to change a system from one equilibrium state to another. But it does not predict how fast a change will take place. Example 6.1 Consider the cooling of a hot steel bar which is placed in a pail of water. Thermodynamics may used to predict the final equilibrium temperature of steel bar-water combination. However, thermodynamics will not tell: ¾ how long to reach the equilibrium or, ¾ what the temperature of steel bar after certain length of time? 67
- 68. Table 6.1: Comparison between heat transfer and thermodynamics HEAT TRANSFER THERMODYNAMICS • Heat transfer is a study which predicts the energy transfer which takes place between material bodies. • It is due to the temperature difference • Heat transfer explains how heat energy may be transferred. • It also predicts the rate at which the exchange will take place under certain condition. • Thermodynamics deals with systems in equilibrium. • Thermodynamics may used to predict the amount of energy required to change a system from one equilibrium state to another. • Thermodynamics may not used to predict how fast a change will take place since the system is not in equilibrium during the process. Three modes of heat transfer are: (1) Conduction (2) Convection (3) Radiation 6.2 Conduction Heat Transfer o Energy transfer from high temperature region to low temperature region. We said that the energy is transferred by conduction. And, the heat transfer rate per unit area is proportional to the normal temperature gradient. q/A ~ dT/dx where, q/A = heat transfer rate (W/m2 ) dT/dx = temperature gradient in the direction of the heat flow When proportionality constant is inserted, q = - kA dT/dx [6.1] o The positive constant k = thermal conductivity of the material. The minus (-) sign is inserted so that the second principle of thermodynamics will be satisfied i.e. heat must flow downhill on the temperature scale as indicated in Figure 6.1. 68
- 69. Equation (8.0) is called Fourier’s Law of heat conduction. Above is defining equation for the thermal conductivity and k has the unit of Watts per meter per Celsius degree (W/m.0 C), which the heat flow is expressed in watts. Figure 6.1: Sketch showing direction of heat flow Table 6.2: Lists typical values of the thermal conductivities of some metal (thermal conductivity at 300 K(W/m K)). Metal k Copper, pure 396 Aluminium 238 Carbon steel, 1% C 42 Plastics 0.2 - 0.3 Air 0.026 69
- 70. Figure 6.2: One –dimensional plane wall o Consider the one-dimensional plane wall shown in Figure above, if the system in a steady state, i.e., if the temperature does not change with time, then only integrate Equation (8.0) and substitute the appropriate values. 1 0 2 x T T q dx kdT A = ∫ ∫ o Where temperature at the left face (x=0) is uniform at T1 and the temperature at right face is uniform at T2. If k is independent of T, we obtain after integration : 2 1 T T q k A L − = − o Under the steady state conditions, where the distribution is linear, the temperature gradient may be expressed as: 2 1 T T dT dx L − = o And the heat transfer rate: 2 1 T T q k A L − = − or; 1 2 T T q T k k A L L − Δ = = 70
- 71. o Since dT/dx = - q/k for the same q, if k is low (i.e: for an insulator), dT/dx will be large .i.e. there will be a large temperature difference across the wall, and if k is high (i.e. for a conductor), dT/dx will be small, or there will be a small temperature difference across the wall. Example 6.2 One face of a copper plate 3 cm thick is maintained at 400 °C, and the other face maintained at 100 °C. How much heat is transferred through the plate? Given: k = 370 W/m.K. From Fourier’s Law, q d k A d = − T x Integrating gives ( ) 2 370 / . (100 400) 3 10 W m K K q T k A x m − − − Δ = − = Δ × q/A = 3.7 MW/m 2 6.3 Convection Heat Transfer o It is well known that hot plate of metal will cool faster when placed in front of a fan than when exposed to still air. We say that, heat is convected away, and we call the process, convection heat transfer. The velocity at which the air blows over the hot plate obviously influence the heat transfer rate. 6.3.1 Mechanism of convection o Consider the heated plate shown on Figure 8.3. Temperature of plate is Tw and temperature of the fluid is T∞.The velocity of the flow will appear as shown in the figure. 71
- 72. Figure 6.3: Convection heat transfer from a plate o The velocity is being reduced to zero at the plate as a result of viscous action. Since the velocity of fluid layer at the wall will be zero, the heat must be transferred only by conduction. Thus we might compute the heat transfer, using: q = -kA (dT/dx) but with little changes. We use thermal conductivity of fluid and the fluid temperature gradient at the wall.The temperature gradient is dependent on the rate at which the fluid carries the heat away. High velocity produces a large temperature gradient. Thus, the temperature gradient at the wall depends on the flow field. To express the overall effect of convection, we use Newton’s law of cooling: q = hA (Tw-T∞) [6.2] where, h = convection heat-transfer coefficient (heat-transfer coefficient = film conductance) h = watt per square meter per Celsius degree (W/m² .°C) when the heat flow is in Watt. o Heat-transfer rate is related to overall temperature difference between the wall and fluid and the surface area. 6.3.2 Free and Forced Convection o If heated plate were exposed to ambient room air without an external source of motion, a movement of the air would be experienced as a result of the density gradients near the plate. o We call this as free convection. When the mass motion of the fluid is caused by an external device like a pump, compressor, blower or fan, the process is called forced convection. Example 6.3 72
- 73. Air at 20 °C blows over a hot plate 50 by 75 cm maintained at 250 °C. The convection heat transfer coefficient is 25 W/m².°C. Calculate the heat transfer? Solution: From Newton’s Law of cooling; ( ) w q hA T T∞ = − q = (25)W/m 2 .K (0.5x0.75) m 2 (250-20)K q = 2.156 kW Example 6.4 An electric current is passed through a wire 1 mm in diameter and 10 cm long. The wire is submerged in liquid water at atmospheric pressure, and the current is increased until the water boils. For this situation h = 5000 W/m².°C, and the water temperature will be 100 °C. How much electric power must be supplied to the wire to maintain the wire surface at 114 °C. Solution: Total convection loss is ( ) w q hA T T∞ = − The surface area of the wire A = πdL = π (1x10 -3 )(10x10 -2 ) = 3.142 x 10 -4 m 2 Therefore, the heat transfer is q = (5000w/m².°C)(3.142 x 10 -4 m²)(114-100) = 21.99 W and this is equal to the electric power which must be applied. 6.4 Radiation Heat Transfer o Heat may also transfer through regions where perfect vacuum exist. The mechanism in this case is electromagnetic radiation which is propagated as a result of a temperature difference. This is called thermal radiation. o This is confirmed by our experienced everyday experience of being warmed by the sun, which separated from the earth by approximately 1.5 x 1011 m of empty space. o Thermal radiation can of course transport through any ‘transparent’ medium such as air. Ideal thermal radiator, or black body, will emit energy at a rate 73
- 74. q emitted = σAT4 [6.3] where; σ is the proportionality constant and is called the Stefan-Boltzmann constant with value of 5.669 x 10 -8 W/m 2 .K 4 . o This equation is called Stefan-Boltzmann law of thermal radiation and it apply to only blackbodies. This equation is valid only for thermal radiation; other type of electromagnetic radiation may not be treated so simply. q emitted = σAT4 govern only radiation emitted by a blackbody. The net radian exchange between two surface will be proportional to the difference in absolute temperature to the fourth power. i.e: q net exchange/ A (T1 4 – T2 4 ) q net exchange/ A = σ (T 1 4 – T2 4 ) 6.4.1 Blackbody o It is a perfect emitter of radiation. At a particular temperature the blackbody would emit the maximum amount of energy possible for that temperature. o This value is known as the blackbody radiation. It would emit at every wavelength of light as it must be able to absorb every wavelength to be sure of absorbing all incoming radiation. o The maximum wavelength emitted by a blackbody radiator is infinite. It also emits a definite amount of energy at each wavelength for a particular temperature, so standard black body radiation curves can be drawn for each temperature, showing the energy radiated at each wavelength (Fig. 6.4). 74
- 75. Figure 6.4: Theoretical black body curve for 5000K o Again, blackbody is a body which radiates energy according to the T 4 law. Other type of surface, such as glossy painted surface or a polished metal plate, do not radiate as much energy as the blackbody. o However, the total radiation emitted by this body still generally follows the T1 4 proportionality. To take account of the “gray” nature of such surfaces we introduce another factor, called the emissivity, which relates the radiation of the “gray” surface to that of an ideal black surface. In addition, we must take into account the fact that not all the radiation leaving one surface will reach the other surface since electromagnetic radiation travels in straight line and some will lost to the surrounding. o Therefore two new factors in (Eq 8.2)take into account both situation, so that q= F FG σ A (T1 4 - T2 4 ) [6.4] F = emissivity function F G = geometric “view factor” function A simple radiation problem is encounter when we have a heat-transfer surface at temperature T 1 completely enclosed by much larger surface maintained at T 2. The net radiant exchange in this case can be calculated with, q = 1 Aσ 1 (T 4 1 - T 4 2 ) where, 1 = emissivity of material Example 6.5 Two infinite black plates at 800 0 C and 300 0 C exchange heat by radiation. Calculate the heat transfer per unit area. Given: σ = 5.669 x 10 -8 W/m 2 .K 4 Solution: q/A = σ (T1 4 – T2 4 ) = 5.669 X 10 -8 )W/m 2 .K 4 (10734 - 5734)K 4 = 69.03 kW/m 2 75
- 76. Problems 1. If 3 kW is conducted through a section of insulating material 0.6 m2 in cross- section and 2.5 cm thick and the thermal conductivity may be taken as 0.2 W/m0 C, compute the temperature difference across the material. 2. A temperature difference of 850 C is impressed across a fiberglass layer of 13 cm thickness. The thermal conductivity of the fiberglass is 0.035 W/m.0 C. Compute the heat transferred through the material per hour per unit area. 3. Two perfectly black surfaces are constructed so that all the radiant energy leaving a surface at 8000 C reaches the other surface. The temperature of the other surface is maintained at 2500 C. Calculate the heat transfer between the surfaces per hour and per unit area of the surface maintained at 8000 C. 4. One side of a plane wall is maintained at 100 C, while the other side is exposed to a convection having T = 100 C and h=10 W/m2.0 C. The wall has k = 1.6 W/m.0 C and is 40 cm thick. Calculate the heat transfer rate through the wall. 0 5. A vertical square plate, 30cm on a side, is maintained at 500 C and exposed to room air at 200 C. The surface emissivity is 0.8. Calculate the total heat lost by both sides of the plate. CHAPTER 7 HEAT EXCHANGERS 7.1 Introduction A heat exchanger is a device that facilitates exchange of heat between two fluids that are at different temperatures and separated by a solid wall. In other words, heat exchangers are devices built for efficient heat transfer from one fluid to another and are widely used in engineering processes. The specific applications of heat exchanger are space heating and air conditioning, power production, waste heat recovery, food and chemical processing, oil refining, and in vehicles. 76
- 77. 7.2 Heat exchanger types Heat exchangers are classified according to flow arrangement and type of construction. Six (6) types of heat exchanger are: (a) Concentric Tube Heat Exchangers (b) Cross-Flow Heat Exchangers (c) Cross-counter Flow (Coil) Heat Exchangers (d) Shell-and-Tube Heat Exchangers (e) Compact Heat Exchangers (f) Plate-Type Heat Exchangers (a) Concentric Tube Heat Exchanger ƒ It also called double-pipe heat exchangers or co-axial flow heat exchangers [Figure 7.1]. ƒ One fluid flows inside the tube and the other fluid flows inside the annulus. ƒ In parallel-flow heat exchangers, the two fluids enter the exchanger at the same end, and travel in parallel to exit at the other side. [Figure7.2a] ƒ In counter-flow heat exchangers the fluids enter the exchanger from opposite ends. Counter current design is more efficient, in that it can transfer more heat [Figure 7.2b]. Figure 7.1: Concentric tube heat exchangers (a) Parallel flow (b) Counterflow Figure 7.2: Concentric tube heat exchangers 77
- 78. (b) Cross Flow Heat Exchanger ƒ In a cross-flow heat exchanger, the fluids travel roughly perpendicular to one another through the exchanger. ƒ In finned tubular heat exchangers, the fin-side fluid is unmixed because the fins confine the flow to one direction. Example: automobile radiator. ƒ In unfinned tubular heat exchangers, the fin-side fluid is mixed because the flow in transverse direction is possible. [Figure 7.3] ƒ The use of fins to improve the convection coefficient of fin-side fluid by increasing the outside surface area. Figure 7.3: Cross-flow heat exchangers (c) Cross-counter Flow (Coil) Heat Exchangers ( (a a) ) F Fi in nn ne ed d w wi it th h b bo ot th h f fl lu ui id ds s u un nm mi ix xe ed d ( (b b) ) U Un nf fi in nn ne ed d w wi it th h o on ne e f fl lu ui id d m mi ix xe ed d a an nd d t th he e o ot th he er r u un nm mi ix xe ed d 78
- 79. Figure 7.4: Cross-counter Flow (Coil) Heat Exchangers (d) Shell-and-Tube Heat Exchangers ƒ Tubular heat exchangers consist of a tube bank enclosed by a shell. One fluid flows inside the tubes and the other flows inside the shell. ƒ Figure 8.5, show the simplest form of shell and tube heat exchanger which involves single tube and shell passes. ƒ Figure 7.5: Shell and tube heat exchanger with one-shell pass one-tube pass (1-1) (a) Two-pass (b) Four-pass (cross-counterflow mode of operation) ƒ Baffles are usually installed to increase the convection coefficient of the shell-side fluid by inducing turbulence and a cross-flow velocity component. 79
- 80. ƒ Baffled heat exchanger with one shell pass and two tube passes and with two shell passes and four tube passes are shown in Figures 7.6a and 7.6b, respectively. Figure 7.6: Shell-and-tube heat exchanger. (a) One shell pass and two tube passes. (b) Two shell passes and four tube passes ƒ Shell-and-tube heat exchangers are classified according to the number of shell and tube passes involved. ƒ Heat exchangers in which all the tubes make one U-turn in the shell, for example, are called one-shell-pass and two-tube-passes heat exchangers. ƒ Heat exchanger that involves two passes in the shell and four passes in the tubes is called a two-shell-passes and four-tube-passes heat exchanger [Figure 7.7] 80
- 81. (a) One-shell pass and two-tube passes (b) Two-shell passes and four-tube passes Figure 7.7: Multipass flow arrangement in shell-and-tube heat exchangers (e) Compact Heat Exchanger ƒ Used to achieve very large heat transfer area per unit volume. ƒ Have dense arrays of finned tubes or plates. ƒ Typically used when at least one of the fluids is a gas, characterized by small convection coefficient. ƒ The tubes may be flat or circular, and the fins may be plate or circular (Figure 7.8a, 7.8b and 7.8c). ƒ Parallel plates may be finned or corrugated and may be used in single-pass or multi-pass mode (Figure 7.8d and 7.8e). 81
- 82. Figure 7.8: Compact heat exchanger cores. (a) Fin-tube (flat tubes, plate fins; (b) Fin-tube (circular tubes, plate fins); (c) Fin-tube (circular tubes, circular fins); (d) Plate-fin (single pass); (e) Plate-fin (multi-pass) (f) Plate-Type Heat Exchangers • Gasketed plate exchanger, alternative to shell-and-tube exchangers for applications at moderate temperature and pressure. • Consists of many corrugated stainless steel sheets separated by polymer gaskets and clamped in a steel frame. • Inlet portals in the gaskets direct the hot and cold fluid to alternate spaces between plates. • Adjoining plates have different pattern or angle of corrugation. Corrugations induce turbulence for improved heat transfer. • Widely used in dairy and food processing 82
- 83. (a) General layout (b) Detail of plate design Figure 7.9: Plate-Type Heat Exchangers The Overall Heat Transfer Coefficient • The heat transfer between the two fluids across the solid wall involves convection of fluid films adjacent to the wall and conduction across the wall. • The rate of heat transfer can be expressed by a single equation like Newton’s law of cooling, with the overall heat transfer coefficient U incorporating convection and conduction terms: [7.1] m q UA T = Δ where, ΔTm = mean temperature difference between the two fluids along the exchanger length • For the unfinned tubular heat exchanger, U can be calculated as follows: 1 1 1 i i o o UA U A U A = = [7.2] ln( / ) 1 1 1 2 fi fo o i i i i o o o R R D D UA h A A kL A h A π = + + + + [7.3] 83
- 84. where , inside and outside heat transfer areas , overall heat transfer coefficients based on inside and outside surface areas , inside (tube-side) convection coefficient i o i o i o A A U U h h = = = and outside (shell-side) convection coefficient , fouling resistances at inside and outside surfaces , inside and outside diameters of the tube k thermal conductivity o fi fo i o R R D D = = = f the tube wall tube length of heat exchanger L = 7.3 Heat Exchanger Analysis: Use of the Log Mean Temperature Difference (LMTD) To design or to predict the performance of a heat exchanger, it is essential to relate the total heat transfer rate to quantities such as the inlet and outlet fluid temperatures, the overall heat transfer coefficient, and the total surface area for heat transfer. A few steps to design or predict the performance of a heat exchanger: • Step 1 - Write down the overall energy balances between heat gain of cold fluid, heat loss of hot fluid, and heat transfer across the wall separating the two fluids. c c h h lm q m H m H UA T = Δ = Δ = Δ ) ) & & [7.4] - If the fluids are not undergoing a phase change and constant specific heats are assumed, the equation becomes: [7.5] , , , , , , ( ) ( ) h p h h i h o c p c c o c i lm q m C T T m C T T UA T = − = − = & & Δ - Determine any unknown that can be directly calculated from the above relations. 84
- 85. rate of heat transfer , mass flow rate of cold fluid and hot fluid , , inlet, outlet, and mean temperature of cold fluid , , inlet, outlet, and mean temperature of hot fluid , c h ci co c hi ho h c h q m m T T T T T T H H = = = = = & & ) ) ( ) 1 2 1 2 1 2 specific enthalpies of cold fluid and hot fluid overall heat ransfer coefficient heat transfer surface area log mean temperature difference (LMTD) Δ Δ , where Δ and Δ are ln Δ Δ tempera lm U A T T T T T T T = = Δ = − = ture differences at the two ends of HE. ΔT1 ΔT2 c c h h lm q m H m H UA T = Δ = Δ = Δ ) ) & & Enthalpy change of cold fluid Enthalpy change of hot fluid Figure 7.10: Overall Energy Balances of the Hot and Cold Fluids of a Two-fluid Heat Exchanger • Step 2 - Enthalpies and LMTD depends on fluid temperature behavior: (a) Parallel flow For parallel flow with no phase change, ΔHh = Cph(Thi-Tho), ΔHc = Cpc(Tco-Tci) 85
- 86. ( ) ( ln[( ) ( )] hi ci ho co lm hi ci ho co T T T T T T T T T ) − − − Δ = − − Figure 8.11: Temperature scheme for parallel flow (b) Counterflow For counterflow with no phase change, ΔHh = Cph(Thi-Tho), ΔHc = Cpc(Tco-Tci) ( ) ( ) ln[( ) ( )] ho ci hi co lm ho ci hi co T T T T T T T T T − − − Δ = − − Figure 8.12: Temperature scheme for counterflow (c) Condensers ΔHh = λh, ΔHc = Cpc(Tco-Tci) ln[( ) ( )] co ci lm h ci h co T T T T T T T − Δ = − − 86
- 87. Figure 8.13: Temperature scheme for condenser (d) Evaporators ΔHc = λc, ΔHh = Cph(Thi-Tho) ln[( ) ( )] hi ho lm hi c ho c T T T T T T T − Δ = − − Figure 8.14: Temperature scheme for evaporator where λ= latent heat of vaporization - Miscellaneous • Overall heat transfer coefficient (U), if not known, can be determined Equation (10.3). • Heat transfer area (A) is related to tube length as A = 2πrL. • Step 3 - Determine any other unknowns from the overall energy balances (Eq. 8.4) - Determine the HE effectiveness and number of transfer units. 87
- 88. • If the heat exchanger other than the double pipe is used, the heat transfer is calculated by using a correction factor applied to the LMTD for a counter flow double-pipe arrangement with the same hot and cold fluids temperatures. • The heat-transfer equation becomes : q = UAF ΔTm [7.6] • Values of the correction factor F are plotted in Figures 7.14 to 7.17 for several different types of heat exchangers. • When phase changed is involved, as in condensation or boiling (evaporation), the fluid normally remains at essentially constant temperature. • For this condition, P and R becomes zero and F = 1.0 (for boiling and condensation) Figure 7.15: Correction factor plot for exchanger with one shell pass and two, four, or any multiple of tube passes. 88
- 89. Figure 7.16: Correction factor plot for exchanger with two shell pass and four, eight, or any multiple of tube passes. 89
- 90. Figure 7.17: Correction-factor plot for single-pass cross-flow exchanger, both fluids unmixed. 90
- 91. Figure 7.18: Correction-factor plot for single-pass cross-flow exchanger, one fluid mixed, the other unmixed. 91
- 92. Example 7.1 Water at the rate 68 kg/min is heated from 35 to 750 C by an oil having specific heat of 1.9 kJ/kg.0 C. The fluids are used in a counterflow double pipe heat exchanger and the oil enter the exchanger at 1100 C and leaves at 750 C. The overall heat-transfer coefficient is 320 W/m2 .0 C. Calculate the heat exchanger area. Solution: The total heat transfer is determined from the energy adsorbed by the water: . w w w q m C T = Δ (68)(4180)(75 35) 11.37 / min 189.5 MJ kW = − = = Since all the fluid temperature are known, the LMTD can be calculated by using the temperature scheme in Figure 10.12, 0 (110 75) (75 35) 37.44 ln(110 75) /(75 35) Tm C − − − Δ = = − − Then, since q = UAΔTm 5 2 1.895 10 15.82 (320)(37.44) x A m = = Example 7.2 Instead of the double-pipe heat exchanger of Example 1 above, it is desired to use a shell- and-tube exchanger with the making one shell pass and the oil making two tube passes. Calculate the area required for this exchanger, assuming that the overall heat-transfer coefficient remains at 320 W/m2 .0 C. Solution: To solve this problem we determine a correction factor from Figure x to be used with the LMTF calculated on the basis of a counterflow exchanger. The parameters according to the nomenclature of Figure 8.15 are: T1=35 0 C T2=75 0 C t1=110 0 C t2=75 0 C 92
- 93. 2 1 1 1 1 2 2 1 5 2 75 110 0.467 35 110 35 75 1.143 75 110 1.895 10 19.53 (320)(0.81)(37.44) t t P T t T T R t t x A m − − = = = − − − − = = = − − = = 2 1 1 1 1 2 2 1 75 110 0.467 35 110 35 75 1.143 75 110 t t P T t T T R t t − − = = = − − − − = = = − − So the correction factor is F = 0.81 And the heat transfer is q = UAFΔTm so that 5 2 1.895 10 19.53 (320)(0.81)(37.44) x A m = = Example 7.3 A cross flow heat exchanger is used to heat an oil in the tubes, c = 1.9 kJ/kg.0 C from 150 C to 850 C. Blowing across the outside of the tubes is steam which enters at 1300 C and leaves at 1100 C with mass flow of 5.2kg/sec. The overall heat transfer coefficient is 275 W/m2 .0 C and c for steam is 1.86 kJ/kg.0 C. Calculate the surface area of the heat exchanger. Solution: The total heat transfer may be obtained from an energy balance on the steam: . (5.2)(1.86)(130 110) 193 s s s q m c T kW = Δ = − = We can solve for the area from Eq.(8.6). The value of ΔTm calculated is as if the exchanger were counterflow double pipe. Thus, 0 (130 85) (110 15) 66.9 130 85 ln 110 15 m T C − − − Δ = = − ⎛ ⎞ ⎜ ⎟ − ⎝ ⎠ From Figure 8.18, t1 and t2 will represent the unmixed fluid (oil) and T1 and T2 will represent the mixed fluid (the steam) so that T1 = 1300 C T2 = 1100 C t1 = 15 0 C t2 = 850 C 93
- 94. and we calculate 130 110 0.286 85 15 R − = = − 85 15 0.609 130 15 P − = = − so the area is calculated from m q A UF T = Δ 7.4 Heat Exchanger Analysis: Use of the Effectiveness-NTU method 7.4.1 Introduction The driving temperature across the heat transfer surface varies with position, but an appropriate mean temperature can be defined. In most simple systems this is the log mean temperature difference (LMTD). Sometimes direct knowledge of the LMTD is not available and the NTU method is used. • The LMTD approach to heat-exchanger analysis is useful when inlet and outlet temperature are known or are easily determined. • However, when the inlet or exit temperatures are to be evaluated for a given heat exchanger, the analysis involves an interactive procedure because of the logarithmic function in the LMTD. • In these cases the analysis is performed more easily by utilizing a method based on the effectiveness of the heat exchanger in transferring a given amount of heat. • The heat exchanger effectiveness can be define as: Actual rate of heat transfer , Maximum possible rate of heat transfer Effectiveness ε = • The actual rate heat transfer, q may be computed by calculating either the energy lost by the hot fluid or the energy gained by the cold fluid. • For the parallel-flow exchanger: . . 1 2 2 1 ( ) ( h c h h h c c c q m c T T m c T T = − = − ) • For the counter-flow exchanger : 94
- 95. . . 1 2 1 2 ( ) ( h c h h h c c c q m c T T m c T T = − = − ) ) • The maximum possible heat transfer, qmax is the rate of heat transfer that a heat exchanger of infinite area would transfer with given inlet temperatures, flow rates, and specific heat. • qmax occurs when the fluid with minimum product of flow rate and specific heat changes temperature to the entering temperature of the other fluid. • Maximum possible heat transfer is expressed as . max min ( ) ( hi ci q mc T T = − • The minimun fluid may be either hot or cold fluid depending on mass-flowrates, m and specific heats, c. • For the parallel exchanger : . . . 1 2 1 2 . 1 1 1 1 ( ) ( ) h h h h h h h c h h h c m c T T T T T T m c T T ε − − = = − − . . . 2 1 2 1 . 1 1 1 1 ( ) ( ) c c c c c c h c c c h c m c T T T T T T m c T T ε − − = = − − Figure 7.19 : Temperature profile for parallel exchanger 95
- 96. • For the counterflow exchanger : . . . 1 2 1 2 . 1 2 1 2 ( ) ( ) h h h h h h h c h h h c m c T T T T T T m c T T ε − − = = − − . . . 1 2 1 2 . 1 2 1 2 ( ) ( ) c c c c c c h c c c h c m c T T T T T T m c T T ε − − = = − − Figure 7.20: Temperature profile for parallel exchanger • In the general way the effectiveness is expressed as : T ( minimum fluid) Maximum temperature difference in heat exchanger ε Δ = • The effectiveness is usually written for parallel flow double pipe HE : . . . . . 1 exp[( / )(1 / ) 1 / c c h c c c h c h UA m c m c m c m c m c ε − − + = + h min min max min max 1 exp[( / )(1 / ) 1 / UA C C C C C ε − − + = + where = capacity rate . C mc = • The number of transfer units (NTU) is indicative of the size of the heat exchanger. min / NTU UA C = • Figure 7.21 to 7-26 presented effectiveness ratios for various heat exchanger arrangements. • Table 7.1 and Table 7.2 summarizes the effectiveness relations 96
- 97. Figure 7.21: Effectiveness for parallelflow exchanger performance Figure 7.22: Effectiveness for counterflow exchanger performance 97
- 98. Figure 7.23 : Effectiveness for crossflow exchanger with one fluid mixed Figure 7.24: Effectiveness for crossflow exchanger with one fluid unmixed Figure 7.25 : Effectiveness for 1-2 parallel counterflow exchanger performance Figure 7.26 : Effectiveness for 2-4 multipass counterflow exchanger performance 98
- 99. Table 7.1: Heat exchanger effectiveness relations 99
- 100. Table 7.2: NTU relations for heat exchangers Problems 1. Water flowing at a rate of 0.723 kg/s enters the inside of a countercurrent, double- pipe heat exchanger at 300 K and is heated by an oil stream that enters at 385 K at a rate of 3.2 kg/s. The heat capacity of the oil is 1.89 kJ/kg K, and the average heat capacity of water over the temperature range of interest is 4.192 kJ/kg K. The overall heat-transfer coefficient of the exchanger is 300 W/m2 .K, and the area for heat transfer is 15.4 m2 . What is the total amount of heat transferred? 2. A shell and tube exchanger operates with two passes and four tube passes. The shell fluid is ethylene glycol (c= 2.742 kJ/kg.0 C), which enters at 140 C and leaves at 80 0 C with flowrate of 45000 kg/h. Water flows in tubes (4.174 kJ/kg.0 C), entering at 35 0 C and leaving at 85 0 C. The overall heat transfer coefficient for this arrangement is 850 W/m2 .0 C. Calculate the flow rate of water required and the area of the heat exchanger. 0 3. A double pipe heat exchanger is used to heat an oil with c = 2.2 kJ/kg.0 C from 50 0 C to 1000 C. The other fluid having c = 4.2 kJ/kg.0 C enters the exchanger at 160 0 C and leaves at 900 C. The overall heat transfer coefficient is 300 W/m2 .0 C. Calculate the area and effectiveness of the heat exchanger for a total heat transfer rate of 600 kW. 100
- 101. 101 REFERENCES 1. Felder, R.M., Rousseau, R.W., Elementary Principles of Chemical Processes, Third Edition, John Wiley & Sons, 2000. 2. Frank P. Incropera and David P. De Witt, Introduction to Heat Transfer, Fourth Edition, John Wiley & Sons, 2002. 3. Frank P. Incropera, David P. De Witt, Theodore, L. Bergman, Adrienne S. Lavine, Fundamentals of Heat and Mass Transfer, Fourth Edition, John Wiley & Sons, 2007. 4. J.P.Holman, Heat Transfer, Ninth Edition, Mc Graw Hill,2002. 5. Yunus A. Cengel, Heat Transfer: A Practical Approach, Second Edition, Mc Graw Hill, 2003.