SOLUCION Manual de análisis, síntesis y diseño de procesos químicos , tercera edición

Solutions Manual for Analysis, Synthesis,
and Design of Chemical Processes
Third Edition
Richard Turton
Richard C. Bailie
Wallace B. Whiting
Joseph A. Shaeiwitz
Prepared by Jessica W. Castillo
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HALL
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ISBN-13: 978-0-13-702385-1
ISBN-IO: 0-13-702385-5
Text printed in the United States at OPM in Laflin, Pennsylvania.
First printing, January 2009

Chapter 1
1.1 Block Flow Diagram (BFD)
Process Flow Diagram (PFD)
Piping and Instrument Diagrams (P&ID)
(a) PFD
(b) BFD
(c) PFD or P&ID
(d) P&ID
(e) P&ID
1.2 P&ID
1.3 It is important for a process engineer to be able to review a 3-dimensional model prior to
the construction phase to check for clearance, accessibility, and layout of equipment,
piping, and instrumentation.
1.4 (1) Clearance for tube bundle removal on a heat exchanger.
(2) NPSH on a pump – affects the vertical separation of feed vessel and pump inlet.
(3) Accessibility of an instrument for an operator – must be able to read a PI or
change/move a valve.
(4) Separation between equipment for safety reasons – reactors and compressors.
(5) Crane access for removing equipment.
(6) Vertical positioning of equipment to allow for gravity flow of liquid.
(7) Hydrostatic head for thermosiphon reboiler – affects height of column skirt.
1.5 Plastic models are no longer made because they are too expensive and difficult to
change/revise. These models have been replaced with virtual/E-model using 3-D CAD.
Both types of model allow revision of critical equipment and instrument placement to
ensure access, operability, and safety.
1.6 Another reason to elevate the bottom of a tower is to provide enough hydrostatic head
driving force to operate a thermosiphon reboiler.
1-1

1.7 (a) PFD or P&ID
(b) PFD
(c) PFD
(d) P&ID
(e) BFD (or all PFDs)
1.8 A pipe rack provides a clear path for piping within and between processes. It keeps piping
off the ground to eliminate tripping hazards and elevates it above roads to allow vehicle
access.
1.9 A structure – mounted vertical plant layout is preferred when land is at a premium and the
process must have a small foot print. The disadvantage is that it is more costly because of
the additional structural steel.
1.10 (a) BFD – No change
PFD – Efficiency changed on fired heater, resize any heat exchanger used to extract
heat from the flue gas (economizer)
P&ID – Resize fuel and combustion air lines and instrumentation for utilities to fired
heater. Changes for design changed of economizer (if present)
(b) BFD – Change flow of waste stream in overall material balance
PFD – Change stream table
P&ID – Change pipe size and any instrumentation for this process line
(c) BFD – No change
PFD – Add a spare drive, e.g. D-301 → D-301 A/B
P&ID – Add parallel drive
(d) BFD – No change
PFD – No change
P&ID – Note changes of valves on diagram
1.11 (a) A new vessel number need not be used, but it would be good practice to add a letter to
donate a new vessel, e.g. V-203 → V-203N. This will enable an engineer to locate the
new process vessel sheet and vendor information.
(b) P&ID definitely
PFD change/add the identifying letter.
1-2

1.13 (a) (i) Open globe valve D
(ii) Shut off gate valves A and C
(iii)Open gate valve E and drain contents of isolated line to sewer
(iv)Perform necessary maintenance on control valve B
(v) Reconnect control valve B and close gate valve E
(vi)Open gate valves A and C
(vii) Close globe valve D
(b) Drain from valve E can go to regular or oily water sewer.
(c) Replacing valve D with a gate valve would not be a good idea because we loose
the ability to control the flow of process fluid during the maintenance operation.
(d) If valve D is eliminated then the process must be shut down every time
maintenance is required on the control valve.
1-4

1.16 (a) For a pump with a large NPSH – the vertical distance between the feed vessel and the
pump inlet must be large in order to provide the static head required to avoid cavitating the
pump.
b) Place the overhead condenser vertically above the reflux drum – the bottom shell
outlet on the condenser should feed directly into the vertical drum.
c) Pumps and control valves should always be placed either at ground level (always for
pumps) or near a platform (sometimes control valves) to allow access for maintenance.
d) Arrange shell and tube exchangers so that no other equipment or structural steel
impedes the removal of the bundle.
e) This is why we have pipe racks – never have pipe runs on the ground. Always elevate
pipes and place on rack.
f) Locate plant to the east of major communities.
1.17
1-6

1.17 HT area of 1 tube = πDL = π
1
12
⎛
⎝
⎜
⎞
⎠
⎟ 12 ft( )= 3.142 ft2
Number of tubes = (145 m2
)⋅
3.2808 ft
m
⎛
⎝
⎜
⎞
⎠
⎟
2
1
3.142 ft2
⎛
⎝
⎜
⎞
⎠
⎟ = 497 tubes
Use a 1 1/4 inch square pitch ⇒
Fractional area of the tubes =
π
4
1 m
1.25 in
⎛
⎝
⎜
⎞
⎠
⎟
2
= 0.5027
m
in
⎛
⎝
⎜
⎞
⎠
⎟
2
AVAP = 3 ALIQ ∴CSASHELL = 4 ALIQ
ALIQ =
497
0.5027
⎛
⎝
⎜
⎞
⎠
⎟
in
m
⎛
⎝
⎜
⎞
⎠
⎟
2
π
4
⎛
⎝
⎜
⎞
⎠
⎟ 1 m( )
2
= 777 in2
CSASHELL = 4( ) 777( )= 3108 in2
⇒
π
4
D2
SHELL = 3108 in2
DSHELL =
4( ) 3108 in2
( )
π
= 62.9 in =1.598 m
Length of Heat Exchanger = (2 +12 + 2) ft =16 ft = 4.877 m
Foot Print =1.598 × 4.877 m
1-7

1.18 From Table 1.11 towers and reactors should have a minimum separation of 15 feet or 4.6
meters. No other restrictions apply. See sketch for details.
1-8

1.21 (a) A temperature (sensing) element (TE) in the plant is connected via a capillary line to
a temperature transmitter (TT) also located in the plant. The TT sends an electrical
signal to a temperature indicator controller (TIC) located on the front of a panel in the
control room.
(b) A pressure switch (PS) located in the plant sends an electrical signal to …
(c) A pressure control valve (PCV) located in the plant is connected by a pneumatic (air)
line to the valve stem.
(d) A low pressure alarm (PAL) located on the front of a panel in the control room
receives an electrical signal from …
(e) A high level alarm (LAH) located on the front of a panel in the control room receives
a signal via a capillary line.
1-11

1.22
2” sch 40 CS
LE LT LIC
PAL LAH
LY
1
3
2
V-302
2” sch 40 CS
4” sch 40 CS
To wastewater treatment1
To chemical sewer2
Vent to flareP-402 3
P-401
2
LE LT LIC
LAL LAH
LY
3
2
2 2
P-401A P-401B
V-302
2” sch 40 CS
2” sch 40 CS
4” sch 40 CS
To wastewater treatment
To chemical sewer
Vent to flare
1
2
3
List of Errors
1. Pipe inlet always larger than pipe outlet due to NPSH
issues
2. Drains to chemical sewer and vent to flare
3. Double-block and bleed needed on control valve
4. Arrows must be consistent with flow of liquid through
pumps
5. Pumps in parallel have A and B designation
6. Pneumatic actuation of valve stem on cv is usual
7. Level alarm low not pressure alarm low
1-12
= Error
Corrected
P&ID

Chapter 2
2.1 The five elements of the Hierarchy of Process Design are:
a. Batch or continuous process
b. Input – output structure of process
c. Recycle structure of process
d. General separation structure of process
e. Heat-exchanger network/process energy recovery
2.2 a. Separate/purify unreacted feed and recycle – use when separation is feasible.
b. Recycle without separation but with purge – when separation of unused reactants is
infeasible/uneconomic. Purge is needed to stop build up of product or inerts.
c. Recycle without separation or purge – product/byproduct must react further through
equilibrium reaction.
2.3 Batch preferred over continuous when: small quantities required, batch-to-batch
accountabilities required, seasonal demand for product or feed stock availability, need to
produce multiple products using the same equipment, very slow reactions, and high
equipment fouling.
2.4 One example is the addition of steam to a catalytic reaction using hydrocarbon feeds.
Examples are given in Appendix B (styrene, acrylic acid.) In the styrene process,
superheated steam is added to provide energy for the desired endothermic reaction and to
force the equilibrium towards styrene product. In the acrylic acid example, steam is added
to the feed of propylene and air to act as thermal ballast (absorb the heat of reaction and
regulate the temperature), and it also serves as an anti-coking agent – preventing coking
reactions that deactivate the catalyst.
2-1

2.5 Reasons for purifying a feed material prior to feeding it to a process include:
a. If impurity foul or poison a catalyst used in the process.
e.g. Remove trace sulfur compounds in natural gas prior to sending to the steam
reforming reactor to produce hydrogen.
CH4 + H20 → CO + 3H2
b. If impurities react to form difficult-to-separate or hazardous products/byproducts.
e.g. Production of isocyanates using phosgene. Production of phosgene is
Remove trace sulfur
Platinum catalyst v. susceptible to
sulfur poisoning
CO + Cl2 → COCl2
The carbon monoxide is formed via steam reforming of CH4 to give CO + H2. H2 must
be removed from CO prior to reaction with Cl2 to form HCl, which is highly corrosive
and causes many problems in the downstream processes.
c. If the impurity is present in large quantities then it may be better to remove the impurity
rather than having to size all the down stream equipment to handle the large flow of
inert material.
e.g. One example is suing oxygen rather than air to fire a combustion or gasification
processes. Removing nitrogen reduces equipment size and makes the removal of CO2
and H2S much easier because these species are more concentrated.
2.6 IGCC H2O + CaHbScOd Ne + O2 → pCO2 + qH2 + rH2O + sCO + tNH3 + uH2S
Coal
In modern IGCC plants, coal is partially oxidized (gasified) to produce synthesis gas CO
+ H2 and other compounds. Prior to combusting the synthesis gas in a turbine, it must be
“cleaned” or H2S and CO2 (if carbon capture is to be employed.) Both H2S and CO2 are
acid gases that are removed by one of a variety of physical or chemical absorption
schemes. By removing nitrogen from the air, the raw synthesis gas stream is much smaller
making the acid gas removal much easier. In fact, when CO2 removal is required IGCC is
the preferred technology, i.e. the cheapest.
2-2

2.7 Ethylebenzene Process
a. Single pass conversion of benzene
Benzene in reactor feed (stream 3) = 226.51
kmol
h
Benzene in reactor effluent (stream 14) = 177.85
kmol
h
Xsp =1−
177.85
kmol
h
226.51
kmol
h
= 21.5%
b. Single pass conversion of ethylene
Ethylene in reactor feed (stream 2) = 93.0
kmol
h
Ethylene in reactor effluent (stream 14) = 0.54
kmol
h
Xsp =1−
0.54
kmol
h
93.0
kmol
h
= 99.4%
c. Overall conversion of benzene
Benzene entering process (stream 1) = 97.0
kmol
h
Benzene leaving process (stream 15 and 19) = 8.38 + 0.17
kmol
h
Xov =1−
8.55
kmol
h
97.0
kmol
h
= 91.2%
d. Overall conversion of ethylene
Ethylene entering process (stream 2) = 93.0
kmol
h
Ethylene leaving process (stream 15 and 19) = 0.54 + 0
kmol
h
Xov =1−
0.54
kmol
h
93.0
kmol
h
= 99.4%
2-3

2.8 Separation of G from reactor effluent may or may not be difficult. (a) If G reacts to form a
heavier (higher molecular weight) compound then separation may be relatively easy using
a flash absorber or distillation and recycle can be achieved easily. (b) If process is to be
viable then G must be separable from the product. If inerts enter with G or gaseous by-
products are formed then separation of G may not be possible but recycling with a purge
should be tried. In either case the statement is not true.
2.9 Pharmaceutical products are manufactured using batch process because:
a. they are usually required in small quantities
b. batch-to-batch accountability and tracking are required by the Food & Drug
Administration (FDA)
c. usually standardized equipment is used for many pharmaceutical products and
campaigns are run to produce each product – this lends itself to batch operation.
2-4

2.10 a. Single pass conversion of ethylbenzene
Ethylbenzene in reactor feed (stream 9) = 512.7
kmol
h
Ethylbenzene in reactor effluent (stream 12) = 336.36
kmol
h
Single pass conversion = 1−
336.36
kmol
h
512.7
kmol
h
= 34.4%
b. Overall conversion of ethylbenzene
Ethylbenzene entering process (stream 1) = 180
kmol
h
Ethylbenzene leaving process (stream 19, 26, 27 & 28) = 3.36 + 0.34 = 3.70
kmol
h
Overall conversion = 1−
3.70
kmol
h
180
kmol
h
= 97.9%
c. Yield of styrene
Moles of ethylbenzene required to produce styrene = 119.3
kmol
h
Moles of ethylbenzene fed to process (stream 1) = 180
kmol
h
Yield =
119.3
kmol
h
180
kmol
h
= 66.3%
Possible strategies to increase the yield of styrene are
(i) Increase steam content of reactor feed – this pushes the desired equilibrium
reaction to the right.
(ii) Increasing the temperature also pushes the equilibrium to right but increases
benzene and toluene production.
(iii) Remove hydrogen in effluent from each reactor – this will push the
equilibrium of the desired reaction to the right and reduce the production of
toluene from the third reaction – use a membrane separator, shown on
following page.
2-5

2.11 Route 1: 2A → S + R
Key features are that no light components (non-condensables) are formed and only one
reactant is used. Therefore, separation of A, R, and S can take place using distillation
columns.
Route 2: A + H2 → S + CH4
Unlike Route 1, this process route requires separation of the non-condensables from A and
S. If hydrogen is used in great excess (as with the toluene HDA process), then a recycle
and purge of the light gas stream will be required. Otherwise, if hydrogen conversion is
high, the unreacted hydrogen along with the methane may be vented directly to fuel gas.
Route 1 – PFD sketch
S
Route 2 – PFD sketch – gas recycle shown dotted since it is only needed if H2 is used in
(considerable) excess and must be recycled.
R
A
Recycled A
Tower 1
Reactor
Tower 2
S
H2 + CH4
A
Reactor
Recycled A
Tower
Gas Separator
Compressor
2-7

Route 1 is better since:
• Simpler PFD
• No gas recycle (no recycle compressor)
• No build up of inerts (CH4) so recycle stream is not as large
• All products are valuable – fuel gas in Route 2 has a low value
2-8

2.12 a. Good when product(s) and reactant(s) are easily separated and purified (most often by
distillation.) Any inerts in the feed or byproducts can be removed by some unit
operation and thus recycle does not require a purge.
b. When unused reactant(s) and product(s) are not easily separated (for example when
both are low boiling point gases) and single pass conversion of reactant is low.
c. This is only possible when no significant inerts are present and any byproducts formed
will react further or can reach equilibrium.
2-9

2.13 a.
b.
Alternative 1
Alternative 1 assumes butanol and acetic acid can be sold as a mixed product ⇒ very
unlikely so probably have to add another column to separate.
H
C2H5OH → C2H4O + H2
Acetaldehyde C2H4
C2H5O 2C2H5OH → C4H8O2 + 2H2
H2
Ethyl Acetate
2C2H5OH → C4H10O + H2O C4H8O
Butanol
C4H10
C2H5OH + H2O → C2H4O2 + 2H2
Acetic Acid C2H4O
Order of volatility is acetaldehyde, water, ethyl acetate, ethanol, isobutanol, acetic acid.
2-10

Alternative 2
This alternative recycles C2H5OH and produces “pure” acetaldehyde – the remaining
streams are considered waste – incineration of organics or wastewater treatment are
possible ways to remove organics.
2-11

2.14
• A and R are both condensable and may be separated via distillation
• C may be separated by absorption into water
• R will be absorbed into water
• G and S cannot be separated except at very high pressure or low temperature
• After reaction, cool and condense A and R from other components.
• Separate A from R using distillation and recycle purified liquid A to the front end of
the process
• Treat remaining gas stream in a water absorber to remove product C
• Separate C and from water via distillation
• Recycle unused G containing S – since S does not react further – we must add a purge
to prevent accumulation of S in the system. This stream must be recycled as a gas
using a recycle gas compressor.
Reactor
Flash
Distillation
Absorber
G+S Purge
(to WT)
G+S Recycle
R
Water
(to WT)
C
A+R
C+G+S
Feed A
Feed G
Water
Distillation
If the value of G was very low, then consider not recycling G (and S.)
2-12

2.15
Malt Whiskey Process
Grain Whisky Process
2-13

Chapter 3
3.1. What is a flowshop plant?
A flowshop plant is a plant in which several batch products are produced using all or a sub-
set ofthe same equipment and in which the operations for each batch follow the same
sequence. Thus the flow ofany batch through equipment A, B, C, D ... is always
A-+B-+C-+D-+.... Omissions ofequipment are possible but no reversal in direction is
allowed.
3.2. What is ajobshop plant?
A flowshop plant is a plant in which several batch products are produced using all or a sub-
set ofthe same equipment but for which the operations ofat least one batch product do not
follow the same sequence, e.g., A-+C-+D-+B
3.3. What are the two main methods for sequencing multiproduct processes?
Either use multi-product campaigns or multiple single-product campaigns.
3.4. Give one advantage and one disadvantage ofusing single-product campaigns in a
multiproduct plant.
Advantage - sequencing of single-product campaigns is relatively simple and repetitive and
probably less prone to operator error since the batch recipe remains the same over the entire
campaign.
Disadvantage - significant final product storage will be required since all products will not
be made all the time and in order to even out supply some inventory ofproducts will have
to be maintained in storage. Single-product campaigns may be less efficient than multi-
product campaigns.
3.5. What is the difference between a zero-wait and a uis process?
A zero-wait process is one in which the batch is transferred immediately from the current
piece of equipment to the next piece ofequipment in the recipe sequence. This type of
process eliminates the need for intermediate storage (storage ofunfinished products or
intermediates).
A uis (unlimited intermediate storage) process is one in which any amount ofany
intermediate product may be stored. Such a process maximizes the use of the processing
equipment but obviously requires an unlimited amount of storage.
3-1

3.6 Number ofbatches ofA is twice that for B or C - repeat Example 3.3 with this restriction
using a SOO h cycle time.
Table E3.3: Equipment times needed to produce A, B, and C
Product Time in Mixer
Time in Time in Time in
Reactor Separator Packaging
A 1.S 1.S 2.S 2.S
B 1.0 2.S 4.S 1.S
C 1.0 4.S 3.S 2.0
Using Equation (3.6) with tcycle,A = 2.S, tcycle,B = 4.5, and tcycle,C = 4.S
If x is the number of batches of Products B and C, then 2x is the number of batches
ofProduct A
SOO
T =500 =2x(2.S) +x(4.S +4.5) => x == - =3S.7
14
Number of batches for each product are A = 70, B=3S, C=3SI
3.7 For Examples 3.3 and 3.4, determine the number of batches that can be produced in a
month (SOO h) usmg a multi-product campaign strategy with the sequence
ACBACBACB. Are there any other sequences for this problem other than the one used
in Example 3.4 and the one used here?
The multi-product cycle time = 2.S + 2.0 + 3.S + 4.S = 12.S h
Number of batches per month = (SOO)/(12.S) = 40 each of A, B, and C
The only sequences that can be used for multi-product campaigns are ABCABCABC
(Example 3.4) and ACBACBACB as used here.
3-2

3.8 Consider the multi-product batch plant described in Table P3.8
Table P3.8: Equipment Processing Times for Processes A, B, and C
Process Mixer Reactor Separator
A 2.0 h S.O h 4.0h
B 3.0h 4.0 h 3.S h
C 1.0h 3.0 h 4.S h
It is required to produce the same number ofbatches of each product. Determine the
number ofbatches that can be produced in a SOO h operating period using the following
strategies:
(a) using single-product campaigns for each product
Using Equation (3.6) with tcycle,A =S.O, tcycle,B = 4.0, and tcycle,C = 4.5
Ix= 37 batches I
SOO
T =SOO= x(S.0+4.0+4.S) => x=-= 37.0
13.S
(b) using a multi-product campaign using the sequence ABCABCABC...
- - .~
. . [cJ :-- :
o
Ai jA
1 1 1 1 1 1 1
A "-I:el :,A-~l-' :1.-.:JIIIII:C _1 1 1 1 1
1 1 1
A:~,-..-----I~---~-----I!
__ :I~CA I :-------1-
II 1 1 1 1
II 1 1 1 1
4 10.511 12.5 16.5 23 25 31.5
From this diagram we see that the cycle time for the multi-product campaign using the
sequence ABC is 12.S h.
Therefore, the number of batches, x, of each product that can be made during a SOO h
period is given by:
Ix = 40 batches
SOO
T =SOO=12.Sx=> x=-=40
12.S
3-3

(c) using a multi-product campaign using the sequence CBACBACBA...
tgJ~ : A :
I I :iC'1UII :A:I I I I I I I I
Li C 1....-I I I
A
III~
: Ll-__~__~ I: I
A
i ci_... I I I I I
iA i-~~-~-:~c;=~:__ A
I I I I I I
I I I I I I
o 1.5 6.5 8.5 12 13.5 15 17.5 20 22 25.5 31
From this diagram we see that the cycle time for the multi-product campaign using the
sequence ABC is 13.5 h.
Therefore, the number of batches, x, of product that can be made during a 500 h period is
given by:
Ix = 37 batches
500
T = 500= 13.5x => x =-=37.0
13.5
3-4

3.9 Consider the process given in Problem 3.8. Assuming that a single-product campaign
strategy is repeated every 500 h operating period and further assuming that the
production rate (for a year = 6,000 h) for products A, B, C are 18,000 kgly, 24,000 kgly,
and 30,000 kg/y, respectively, determine the minimum volume ofproduct storage
required. Assume that the product densities ofA, B, and C are 1100, 1200, and 1000
kglm3
, respectively
The tables below shows the results using data given from Problem 8
Rate Product A ProductB Product C
Volume (m';) ofproduct 18,00011211,100 24,000/1211,20 30,000/1211,000
reiluired per month =1.36 0 = 1.67 =2.5
Cycle time (h) 5.0 4.0 4.5
Production rate, rp (m3
/h)
(1.36)/(37)(5) (1.67)/(37)(4) (2.5)/(37)(4.5)
= 0.007371 = 0.01126 = 0.015015
Demand rate, rd (m3
/h)
(1.36)/(500)
0.003333 0.005
= 0.002727
Product Campaign time, rp-rd Minimum volume of
tcamp (m3
/h) product storage, Vs
(h) (m3
)
0.007371 - 0.00273 = (0.004644)(185) =
A (37)(5) = 185
0.004644 0.859
0.01126 - 0.003333 = (0.007928)(148) =
B (37)(4) = 148
0.007928 1.173
(37)(4.5) = 166.5
0.0.15015 - 0.005 = (0.010015)(166.5) =
C
0.010015 1.668
3-5

3.10
Table P3.lOA: Production rates for A, B, and C
Product
Yearly Production
production in 500 h
A 150,000 kg 12,500 kg
B 210,000 kg 17,500 kg
C 360,000 kg 30,000 kg
Table P3.1OB: Specific ReactorlMixer Volumes for Processes A, B, and C
Process A B C
Vreact (d/kg-product) 0.0073 0.0095 0.0047
tcycle (h) 6.0 9.5 18.5
Let the single-product campaign times for the three products be tA, tB, and te, respectively.
Applying Equation (3.6), the following relationship is obtained:
The number ofcampaigns per product is then given by tjtcycle,x and
b h · (k /b h) production ofxatc SIze g atc = =------
tx/tcycle,x
(3.9)
(3.10)
Furthermore, the volume ofa batch is found by multiplying Equation (3.10) by Vreac,x, and
equating batch volumes for the different products yields:
(production ofx)(vreacl x)
Volume of batch = '
tx / tcycle,x
(3.11)
(12,500)(.0073) (17,500)(.0095) (30,000)(.0047)
= = (3.12)
3-6

Solving Equations (3.9) and (3.12), yields:
fA =57.8 h
fs =166.8 h
fe =275.4h
Vreact.A = Vreact.B = Vreact,e = 9.47 m
3
#batches per campaign for product A = fA /6.0 = 9.6
#batches per campaign for product B = tB / 9.5 = 17.6
#batches per campaign for product C = te/ 18.5 = 14.9
Clearly the number ofbatches should be an integer value. Rounding these numbers yields
For product A
Number ofbatches = 10
fA = (10)(6.0) = 60 h
VA =(12,500)(0.0073)/(10) = 9.13 m3
For product B
Number ofbatches = 17
fB= (17)(9.5) =161.5 h
VB = (17,500)(0.0095)/(17) = 9.78 m3
For product C
Number ofbatches =15
fe = (15)(18.5) = 277.5 h
Vc =(30,000)(0.0047)/(15) = 9.40 m3
Total time for production cycle = 499 h ~ 500 h
Volume ofreactor = 9.78 m3
(limiting condition for Product B)
3-7

3.11
Table P3.11: Batch step times (in hours) for Reactor and Bacteria Filter for Project 8 in
Appendix B
Product Reactor* Precoating of Filtration Mass
Bacteria of produced
Filter Bacteria per batch,
L-aspartic Acid 40 25 5
L-phenylalanine 70 25 5
*includes 5 h for filling, cleaning, and heating plus 5 hours for emptymg
(a) let
tA = campaign time for L-aspartic acid
tp = campaign time for L-phenylalanine
Assuming equal recovery ratios for each amino acid we have
tA +tp =8000
Solving we get
tA = 1944 h
tp = 6056 h
(716)(tp) =1.25 (1020)(tA)
(70) (40)
kg
1020
716
yearly production ofL-aspartic acid = (1944)(1020)/(40) = 49,560 kg
yearly production ofL-phenylalanine = (6056)(716)/(70) = 61,950 kg
Number ofbatches per year for L-aspartic acid = (1944)/(40) = 48
Number of batches per year for L-phenylalanine = (6056)/(70) = 86
Ratio of
product,
s
1
1.25
(b) For each product calculate the average yearly demand and production rate in m3
/h and then
find the storage needed for each product
3-8

Rate L-aspartic acid L-phenylalanine
Volume (mJ) of product (49,560)(0.9)/(1,200) = (61,950)(0.9)/(1,200) =
required per year 37.17 46.46
Cycle time (h) 40 70
Campaign time (h) 1944 6056
Production rate, rp (mJ/h) (37.17)/(1944) =0.019125 (46.46)/(6056) =0.0076714
Demand rate, rd (m3
/h) (37.17)/(8,000) = 0.004646 (46.46)/(8000) =0.005808
rp-rd (m3
/h) 0.014479 0.001864
Storage Volume (m3
) (0.014479)(1944) =28.14 (0.001864)(6056) = 11.29
(c) Rework part (b) using a 1 month cycle time = 8,000112 = 666.67 h
tA+tp =666.67
(716)(tp) =1.25 (l020)(tA)
(70) (40)
Solving we get
tA = 162 hand tp = 504.7 h
monthly production ofL-aspartic acid = (4)(1020) =4,080 kg
monthly production ofL-phenylalanine = (7)(716) =5,012 kg
Number of batches per month for L-aspartic acid = (162)/(40) = 4
Number of batches per month for L-phenylalanine = (504.7)/(70) = 7
Note that these are rounded down so that integer numbers are given per month this gives rise
to a slightly lower production rate per year than before.
Rate L-aspartic acid L-phenylalanine
Volume (m3
) of product
(4,080)(0.9)/(1,200) =3.06 (5,012)(0.9)/(1,200) = 3.76
required per month
Cycle time (h) 40 70
Campaign time (h) 160 490
Production rate, rp (m3
/h) (3.06)/(160) = 0.019125 (3.76)/(490) =0.0076714
Demand rate, rd (m
3
Ih) (3.06)/(666.7) = 0.00459 (3.76)/(666.7) =0.005638
rv-rd (m3
/h) 0.014535 0.002033
Storage Volume (m3
) (0.014535)(160) =2.33 (0.002033)(490) = 1.00
These values are (not surprisingly) approximately 1112 ofthe previous results
3-9

3. 12
(a) Referring to Project B.8, Figures B.8.2 and B.8.3 and using batch reaction times for L-
aspartic acid and L-phenylalanine of25 and 55 h, respectively. We get the following
information:
Conversion ofL-aspartic acid = 42% (84% of equilibrium) (base case =45%)
Exit concentration ofL-phenylalanine = 18.5 kg/m3
(base case = 21%)
Product Reactor* Precoating Filtration Mass produced Ratio of
ofBacteria ofBacteria
Filter
L-aspartic Acid 35 25 5
L-phenylalanine 65 25 5
let
fA =campaign time for L-aspartic acid
fp = campaign time for L-phenylalanine
fA +fp =8000
Solving we get
fA = 1776 h
fp= 6224 h
(630.8)(fp) = 1.25 (952)(fA)
(65) (35)
per batch, kg
(42/45)(1020) =
952
(18.5/21)(716) =
630.8
yearly production of L-aspartic acid =(1776)(952)/(35) =48,316 kg
yearly production ofL-phenylalanine = (6224)(630.8)/(65) = 60,395 kg
Number of batches per year for L-aspartic acid = (1776)/(35) = 50 or 51
Nmnber ofbatches per year for L-phenylalanine = (6224)/(65) = 95 or 96
Therefore, the number of batches increases but the yearly production decreases
products
1
1.25

(b) Referring to Project B.8, Figures B.8.2 and B.8.3 and using batch reaction times for L-
aspartic acid and L-phenylalanine of35 and 65 h, respectively. We get the following
information:
Conversion ofL-aspartic acid = 47% (94% ofequilibrium) (base case = 45%)
Exit concentration ofL-phenylalanine =21.5 kglm3
(base case =21%)
Product Reactor* Precoating Filtration Mass produced Ratio of
ofBacteria ofBacteria per batch, kg
Filter
L-aspartic Acid 45 25 5 (47/45)(1020) =
1065
L-phenylalanine 75 25 5 (21.5/21)(716) =
let
fA = campaign time for L-aspartic acid
fp = campaign time for L-phenylalanine
fA +fp =8000
Solving we get
fA = 1986 h
fp=6014h
(733)«(p) =1.25 (1065)«(A)
(75) (45)
yearly production ofL-aspartic acid =(1986)(1065)/(45) = 47,002 kg
yearly production ofL-phenylalanine = (6014)(733)/(75) = 58,777 kg
Number of batches per year for L-aspartic acid = (1986)/(45) = 44
Number of batches per year for L-phenylalanine = (6014)/(75) = 80
733
products
1
1.25
Therefore, the number ofbatches decreases and the yearly production decreases - plot the results
from problems 11 and 12
3-11

65,000
-+- L-Aspartic Acid
___ L-Phenylalanine
>.
60,000
-0')
-<If+"
~
0:::
C 55,000
0
;:;
U
::I
"'C
010...
Il.. 50,000
45,000
-5 -2.5 o 2.5 5
Deviation from base batch-reaction time, hr
This shows that the base case conditions are close to optimal.
3-12

Chapter 5
5.1 For ethylbenzene process in Figure B.2.1
Feeds: benzene, ethylene
Products: ethylbenzene, fuel gas (by-product)
5.2 For styrene process in Figure B.3.1
Feeds: ethylbenzene, steam
Products: styrene, benzene/toluene (by-products), hydrogen (by-product), wastewater
(waste stream)
5.3 For drying oil process in Figure B.4.1
Feeds: acetylated castor oil
Products: acetic acid (by-product), drying oil, gum (waste stream)
5.4 For maleic anhydride process in Figure B.5.1
Feeds: benzene, air (note that dibutyl phthalate is not a feed stream)
Products: raw maleic anhydride (Stream 13), off gas (waste stream)
5.5 For ethylene oxide process in Figure B.6.1
Feeds: ethylene, air, process water
Products: fuel gas (by-product), light gases (waste stream), ethylene oxide, waste water
(waste stream)
5.6 For formalin process in Figure B.7.1
Feeds: methanol, air, deionized water
Products: off-gas (waste - must be purified to use as a fuel gas), formalin
5.7 The main recycle streams for the styrene process in Figure B.3.1 are:
ethylbenzene recycle (Stream 29) , reflux streams to T-401 and T-402
5.8 The main recycle streams for the drying oil process in Figure B.4.1 are:
acetylated castor oil (Stream 14) , reflux streams to T-501 and T-502
5-1

5.9 The main recycle streams for the maleic anhydride process in Figure B.5.1 are:
Dibutyl phthalate (Stream 14), circulating molten salt loop (Steam 15 and 16), and reflux
to T-601 and T-602
5.10 Process description for ethylbenzene process in Figure B.2.1
Raw benzene (Stream 1), containing approximately 2% toluene, is supplied to the
Benzene Feed Drum, V-301, from storage. Raw benzene and recycled benzene mix in
the feed drum and then are pumped by the benzene feed pump, P-301A/B, to the feed
heater, H-301, where the benzene is vaporized and heated to 400°C. The vaporized
benzene is mixed with feed ethylene (containing 7 mol% ethane) to produce a stream at
383°C that is fed to the first of three reactors in series, R-301. The effluent from this
reactor, depleted of ethylene, is mixed with additional feed ethylene and cooled in
Reactor Intercooler, E-301, that raises high pressure steam. The cooled stream at 380°C is
then fed to the second reactor, R-302, where further reaction takes place. The effluent
from this reactor is mixed again with fresh ethylene feed and cooled to 380°C in Reactor
Intercooler, E-302, where more high pressure steam is generated. The cooled stream,
Stream 11, is fed to the third reactor, R-303. The effluent from R-303, containing
significant amounts of unreacted benzene, Steam 12, is mixed with a recycle stream,
Stream 13, and then fed to three heat exchangers, E-303 – 305, where the stream is
cooled. The energy extracted from the stream is used to generate high- and low-pressure
steam in E-303 and E-304, respectively. The final heat exchanger, E-305, cools the
stream to 80°C using cooling water.
The cooled reactor effluent is then throttled down to a pressure of 110 kPa and sent to the
Liquid Vapor Separator, V-302, where the vapor product is taken off and sent to the fuel
gas header and the liquid stream is sent to column, T-301. The top product from T-301
consists of purified benzene that is recycled back to the benzene feed drum. The bottom
product containing the ethylbenzene product plus diethylbenzene formed in an unwanted
side reaction is fed to a second column, T-302. The top product from this column
contains the 99.8 mol% ethylbenzene product. The bottom stream contains
diethylbenzne and small amounts of ethylbenzene. This stream is recycled back through
the feed heater, H-301, and is mixed with a small amount of recycled benzene to produce
a stream at 500°C that is fed to a fourth reactor, R-304. This reactor converts the
diethylbenzene back into ethylbenzene. The effluent from this reactor, Stream 13, is
mixed with the effluent from reactor R-303.
5-2

5.11 Process description of drying oil process in Figure B.4.1
Acetylated castor oil (ACO) is fed to the Recycle Mixing Vessel, V-501, where it is
mixed with recycled ACO. This mixture is then pumped via P-501A/B to the Feed Fired
Heater, H-501, where the temperature is raised to 380°C. The hot liquid stream, Stream
4, leaving the heater is then fed to the Drying Oil Reactor, R-501, that contains inert
packing. The reactor provides residence time for the cracking reaction to take place. The
two-phase mixture leaving the reactor is cooled in the Reactor Effluent Cooler, E-501,
where low-pressure steam is generated. The liquid stream leaving the exchanger is at a
temperature of 175°C and is passed through one of two filter vessels, V-502A/B, that
removes any gum produced in the reactor. The filtered liquid, Stream 7, then flows to
the ACO Recycle Tower, T-501. The bottom product from this tower contains purified
ACO that is cooled in the Recycle Cooler, E-506, that raises low-pressure steam. This
stream is then pumped via P-504A/B back to V-501 where it is mixed with fresh ACO.
The overhead stream, Stream 9, from T-501 contains the drying oil and acetic acid
produced from the cracking of ACO. This stream is fed to the Drying Oil Tower, T-502,
where the ACO is taken as the bottom product and the acetic acid is taken as the top
product. Both the acetic acid, Stream 11, and the ACO, Stream 12, are cooled (not shown
in Figure B.4.1) and sent to storage.
5-3

5.12 Process description for ethylene oxide process in Figure B.6.1
Ethylene oxide (EO) is formed via the highly exothermic catalytic oxidation of ethylene
using air. Feed air is compressed to a pressure of approximately 27 atm using a three
stage centrifugal compressor, C-701-3, with intercoolers, E-701 and E-702. The
compressed air stream is mixed with ethylene feed and the resulting stream, Stream 10, is
further heated to a reaction temperature of 240°C in the Reactor Preheater, E-703. The
reactor feed stream is then fed to the first of two reactors, R-701. The feed passes
through a bank of catalyst filled tubes submerged in boiler feed water. The resulting
exothermic reaction causes the boiler feed water (bfw) to vaporize and the pressure is
maintained in the shell of the reactor to enable the production of medium pressure steam.
Combustion of the ethylene and ethylene oxide also occur in the reactor. The reactor
effluent is cooled in E-704 and is then recompressed to 30.15 bar in C-704 prior to being
sent to the EO Absorber, T-701. The EO in the feed stream to the absorber, Stream 14, is
scrubbed using water and the bottom product is sent to the EO column, T-703, for
purification. The overhead stream from the absorber is heated back to 240°C prior to
being fed to a second EO reactor, R-702 that performs the same function as R-701. The
effluent from this reactor is cooled and compressed and sent to a second EO absorber, T-
702, where the EO is scrubbed using water. The bottom product from this absorber is
combined with the bottom product from the first absorber and the combined stream,
Stream 29, is further cooled and throttled prior to being fed to the EO column, T-703.
The overhead product from the second absorber is split with a purge stream being sent to
fuel gas/incineration and the remainder being recycled to recover unused ethylene.
The EO column separates the EO as a top product with waste water as the bottom
product. The latter stream is sent off-site to water treatment while the EO product is sent
to product storage. A small amount of non-condensables are present as dissolved gases in
the feed and these accumulate in the overhead reflux drum, V-701, from where they are
vented as an off gas.
5-4

Chapter 6
6.1 Methods for setting pressure ofa distillation column
a. Set based on the pressure required to condense the overhead stream using cooling water
(minimum ofapprox. 45°C condenser temperature)
b. Set based on highest temperature ofbottom product that avoids decomposition or reaction
c. Set based on available highest hot utility for reboiler
6.2 Run a distillation column above ambient pressure because
a. The components to be distilled have very high vapor pressures (very "light" components)
and the temperatures at which they can be condensed at or below ambient pressure are
<Tcoolingwatero Thus, we run the column at a high pressure to avoid the use of refrigeration
in the condenser.
b. The size ofthe column (diameter) reduces as the pressure decreases. Although the
separation is usually more difficult at higher pressures, it may be more cost effective to run
the column at or above atmospheric pressure.
Run a distillation column below ambient pressure because
c. The components requiring separation have very low vapor pressures (very "heavy"
components) and if distilled at ambient pressure would require excessively high
temperatures to reboil.
d. For thermally sensitive materials, e.g., foods, pharmaceuticals, and some organics, the
bottom temperature in an atmospheric column might cause the material to decompose or
react. For such systems, the pressure is reduced below atmospheric to a point where the
bottom temperature does not cause thermal degradation ofthe product.
6.3 Run reactions at elevated temperature because
a. The rate ofreaction rate is faster. This results in a smaller reactor and/or higher single-pass
conversion
b. The reaction is endothermic and equilibrium limited and increasing the temperature shifts
the equilibrium to the right.
Run reactions at elevated pressures because
c. The reaction is gas phase and the concentration and hence rate is increased as the pressure
is increased. This results in a smaller reactor and/or higher conversion.
d. The reaction is equilibrium limited and there are fewer product moles than reactant moles.
An increase in pressure shifts the equilibrium to the right.
6-1

6.4 Running a process above 250°C is undesirable because
a. In order to heat the streams to that temperature the use ofa fired heater is required, which
. .IS expensIve.
b. At temperatures in excess of400°C process equipment may require more expensive
materials of construction.
Examples for doing this for a reactor are: to increase the reaction rate or improve
equilibrium for an endothermic reactor. An example for doing this for a distillation
column is: to provide a vapor-liquid system for a heavy (high boiling point) component
6.5 A "condition ofspecial concern" is a process condition that deviates from an "ideal" or
"low-cost" operating condition. There are many examples given in this chapter.
Operating at pressures outside the range of 1 - 10 atmospheres or temperatures outside of
the range 45° - 2500
/400°C are examples ofconditions ofspecial concern. Justification
for operating at high temperatures and pressure might be to increase the rate ofa desired
reaction.
6.6 Many ofthese products are thermally labile meaning that they degrade at quite low
temperatures. The use ofvacuum conditions allows vapor-liquid equilibrium and vapor-
solid equilibrium (freeze drying or lyophilization) to occur at temperatures below which
thermal degradation occurs.
6.7 Distillation ofa binary mixture, the effect ofan increase in column pressure on:
a. The tendency to flood at a given reflux ratio will decrease because the density ofvapor
will increase and hence the superficial velocity in the tower will decrease thus moving
away from flooding.
b. For a given top and bottom purity with a fixed number ofstages the tendency to flood
will increase with pressure. This is because as pressure increases, the separation becomes
more difficult and the equilibrium line moves closer to the xy line. The only way to
compensate is therefore to increase the reflux ratio that in tum increases the internal
flows in the column - hence the vapor flow and velocity will increase and move the
column towards flooding.
c. The number of stages will increase with pressure for the same reason given in (b) above.
With XD, XB, and R fixed as the separation gets more difficult, the number ofstages must
be increased.
d. As pressure increases, the condenser temperature will increase - this is consistent with
Antoine's equation that as temperature increases so does the vapor pressure
6-2

6.8 Required reboiler utility at 290°C or higher. Assume that the exit temperature of utility is
fixed at 290°C. Look at the T-Q diagram for the reboiler for both cases:
a. Using high-pressure (42 bar) steam superheated to 320°C. Since hps condenses at 254°C,
no condensation ofthe steam will occur and all heat transfer will be by cooling only. This
will lead to a very large heat exchanger (because U will be limited by the low steam':side
heat transfer coefficient) and a large flow ofsuperheated steam.
Duty,Q
b. Using saturated steam at 320°C requires a pressure of 112.7 bar. This is very high and
would cause the exchanger to be very expensive and possibly requiring special materials
ofconstruction. However, the overall heat transfer coefficient U would be high and the
exchanger area would be relatively small. Alternatively we could choose to throttle and
desuperheat the steam to a saturation temperature of290°C and a pressure of74.4 bar
which is still high but somewhat less costly.
uo
~ 320 r---------------------~ 320
~ or
S 290 290
~
Duty, Q
6-3

6.9 Ifthe column is designed to produce a saturated overhead product and reflux then a
change in cooling water will affect the column pressure. For example, ifthe column is
working at a pressure Pwinter in the winter when the cooling water is available at a
temperature of27°C then as the cw temperature increases, the temperature driving force
also drops and not as much vapor can be condensed. This is illustrated below. As less
vapor is condensed, vapor will accumulate in the top ofthe column (because it cannot be
condensed) and the pressure increases. As the pressure in the column increases, so does
the temperature at which the vapor can condense (dew point). A new equilibrium will be
reached when the temperature driving force in the condenser is restored to its original
value - thus allowing the correct amount ofvapor to be condensed. Thus the column is
to some extent selfregulating. This swing in pressure (higher in the summer and lower
in the winter) occurs quite slowly and would be noticed as a slow drift ofdays or weeks.
uo
f 37
!
Duty,Q
6-4
T-Q diagram for
overhead condenser
27

6.10 Benzene at 1 atm and 25°C to be vaporized and pressurized to 10 atm and 250°C.
a. Pumping a liquid requires less power than compressing a vapor so this suggests that
pumping and heating will be better than heating and compressing a vapor.
b. Use a basis of 1,000 kg/h ofbenzene, simulation results (Chemcad - SRK) are given
below
ElUip. No.
Name
output pressure aan
E:fficiency
Cal.cul.ated power klf
Cal.cul.ated Pout aan
Head JII.
Vo1. :flow rate m.3/h
Mass flow rate kq/h
ElUip. No.
Name
1st Stream. dp aan
1st Stream. TOut C
Cal.c Ht Duty MJ/h
LMTD Corr Factor
1st Stream. Pout aan
Name
Pressure out aan
Type o:f Compressor
E:ffieiency
Actual. power klf
Cp/Cv
Theoretical. power klf
Ideal. Cp/cv
Cal.e Pout atm
Cal.e. mass :flawrate
(kq/h)
Pomp Summary
1
10.3000
0.7500
0.4001
10.3000
110.0674
1.1455
1000.0000
Heat Exchanqer Summary
2 3
0.3000 0.3000
250.0000 140.0000
741.1040 579.4749
1.0000 1.0000
10.0000 0.7000
Compressor Summary
4
10.0000
1
0.7500
45.3997
1.0813
34.0498
1.0763
10.0000
1000
6-5
Adjust exchanger duties so
Streams 3 and 6 are both at
250°C
Pump and Heat
Cost = (0.4)(0.06) + (741.1)(.015)
= $11.14/b
Heat and Compress
Cost = (45.4)(0.06) + (579.5)(.015)
= $11.41/b
Heat and Compression is slightly
more expensive than pump and heat.
The fact that the two answers are so
close is in part due to the low cost of
electricity relative to the heating
utility used in this problem.

6.11 Production ofhigh purity oxygen via cryogenic distillation.
a. Normal Boiling Point of02 = 90.2 K, NBP ofN2 = 77.7 K
b. Assuming nearly pure compositions ofproducts, nitrogen at the top (~78 K) and oxygen
at the bottom (~90 K)
c. Critical temperatures of02 and N2 are 155 K and 126 K, respectively. Thus neither
component can be liquefied at 40°C
d. Obviously, distillation at ambient temperature is impossible because a 2-phase mixture
cannot be produced at 40°C. Ifdistillation is used then it must occur at cryogenic
temperatures. Typically these units are run at about 5-6 atm pressure which gives top and
bottom temperatures about 20 K above those in part (a).
6.12 Since the synthesis reaction to produce ammonia is highly exothermic, a high temperature
tends to push the reaction to the left (undesirable). The reason for the high temperature
must be to increase the kinetics rather than improve the equilibrium conversion (the iron
catalyst is only effective above temperatures ofabout 400°C). Since there are fewer
product moles than reactant moles, a high pressure pushes the equilibrium to the right
(desirable) and also increases the concentration ofall species, which in tum increases the
kinetic rates.
6.13 The conversion is limited by eqUilibrium and so it would be increased by using lower
temperature and higher pressure. As pointed out in Problem 12, the lower temperature
slows the reactions and would lead to much larger and expensive reactors (that are
already expensive because ofthe high pressure and relatively high temperature). Higher
pressures could be used but again this will lead to increased costs. Another alternative is
to remove ammonia during or after each reactor (using several reactor stages in series)
this will significantly increase the single pass conversion per reactor but require a
significant amount ofadditional, cooling, reheating, and separation equipment.
6-6

6.14 Drying Oil Process
Reactors and Separators Other Equipment
Tables 6.1 - 6.3 Table 6.4
Equipment High Low High Low
Non-
Stoich. Compr Exch Heater Valve Mix
Temp Temp Pres Pres
Feed
E-501
E-502
E-503
E-504
E-505
E-506
H-501
P-501
P-502
P-503
P-504
R-501 X
T-501 X
T-502
a. PCM is shown above
b. High temperature in R-501 - need high temperature to initiate cracking reactions. High
temperature in T-501 - heavy components (ACO and drying oil) need high temperature
to form a two-phase mixture.
c. Remedy for high temperature in R-501 - possibly find a catalyst that would promote the
cracking reaction at a lower temperature. Note that pressure is not a variable since the
cracking reaction occurs in the liquid phase.
Remedy for the high temperature in T-501 - this may be a problem since reaction (and
gum formation) may be occurring in reboiler and would cause plugging and fouling of
exchanger surfaces and trays/packing. One remedy would be to operate the tower at
vacuum to lower the bottom temperature. For example, at 30 kPa the bottom temperature
would drop to about 300°C.
6-7

6.15 Styrene process
a. PCM
Reactors and Separators Other Equipment
Tables 6.1 - 6.3 Table 6.4
Equipment
High Low High Low
Non-
Stoich Compr Exch Heater Valve Mix
Temp Temp Pres Pres
. Feed
C-401 X
E-401
E-402
E-403 X
E-404 X
E-405
E-406
E-407
E-408
E-409
H-401
P-401
P-402
P-403
P-404
P-405
P-406
R-401 X X
R-402 X X
T-401
T-402
V-401
V-402
V-403
b. High temperatures in R-401 and R-402 - the desired reaction is slightly endothermic and
may be equilibrium limited. Therefore, the high temperature may be required to push the
equilibrium to the right and/or increase the reaction rate.
Non-stoichiometric feed to R-401 and R-402 - high temperature steam is added to the
reactor feed. Steam is not required as a reactant; its purpose is to push the equilibrium to
the right by diluting the reaction mixture.
6-8

The pressure ratio for C-401 is slightly greater than 3 - since the pressure ratio is so close
to 3, it is probably not worth the cost ofadding a second stage with intercooling.
The I1Tlmfor E-403 and E-404 are both greater than 100D
C - shows low heat integration
but represents a simple, low-cost configuration.
c. A possible remedy to using such high temperatures in the reactors is to use a lower
pressure but since the pressure is already quite low (170 kPa) this would lead to larger
reactors and possibly vacuum operations.
The use ofsteam as a diluent in the reactors, improves equilibrium conversion. If steam
is not added, then the reaction would be pushed back to the left - not any viable
alternatives to this (higher T and lower P -because MOC or vacuum problems, using say
nitrogen as the diluent would mean producing a contaminated hydrogen stream.)
The excessive compression ratio in C-401 could be investigated by looking at a 2nd
stage
with intercooling.
The high I1Tlm for E-403 and E-404 could be eliminated with better heat integration.
6-9

Chapter 7
7.1 (i) Capacity or size (for heat exchanger this would be heat exchanger area)
(ii) Operating (or more correctly the design) pressure
(iii) Materials of construction
7.2 CEPCI is used to adjust purchased costs ofequipment for different times. It is a measure
for the inflation of costs associated with the manufacture ofchemical process equipment.
7.3 Total module cost represents the all costs associated with the purchase and installation of
new equipment for an existing chemical facility. The grass roots cost includes the total
module cost plus costs associated with the off sites and utilities needed for a completely
new "grass roots" or "green field" facility.
7.4 Use a cost exponent or 0.6 to estimate the change in cost associated with a modest change
in capacity for a whole chemical process.
7.5 The economy ofscale refers to the fact that the cost exponent for chemical plant equipment
is (usually) less than one. Therefore, as a chemical plant's capacity increases, the unit cost
ofequipment ($/unit ofproduction) decreases.
7.6 A Lang factor is a constant (between approximately 3 and 5) that when mUltiplied by the
purchase cost ofthe equipment gives an estimate ofthe total installed cost (capital
investment) for a chemical process.
7.7 Most ofthe cost ofa heat exchanger involves machining and tube costs. The relative
change in these costs for an increase in pressure is much smaller than for a process vessel
whose purchase price is directly affected by wall thickness and hence operating pressure.
7-1

7.8 Actual Cost = $540 million
For a Class 1 estimate the expected range of accuracy is +6% to -4%. Thus the range of
expected cost estimates would be
540 to 540 => $509.4 to $562.5 million
(1 + 0.06) (1- 0.04)
For a Class 3 estimate, the range ofaccuracy is 2 to 6 times that ofa Class 1 estimate. Use
a mid.point of4 times the accuracy. Thus the range of expected cost estimates would be
540 to 540 => $435.5 to $642.9 million
(1+(0.06)(4)) (1-(0.04)(4))
Similarly, for a Class 5 estimate use a mid·point value of (
4
+ 20) = 12. Range of
2
estimates
540 to 540 => $314.0 to $1,038 million
(1 + (0.06)(12)) (1- (0.04)(12))
7-2

7.9 The figures in Appendix A are plotted with the y-axis as the purchased cost per unit of
capacity. For a cost exponent of<1 the slope ofthe curves should be negative. Searching
are the figures in Appendix A we find only two instances ofpositive slopes. These are
7.10
Figure A.I _ Rotary compressor} These are the only equipment that does
Figure A.9 - Auto batch separator not exhibit the "economy of scale" over
the whole range ofequipment sizes.
Year
1993
1998
2007
Capacity (mJ) Cost (£ CEPCI
75 7,800 359
155 13,800 390
120 ? 500
C =C(12 ). Ca=(Aa)1l2 I 1 ' C AI b h
3 (500)For 75 m : C2007 = - (£7,800) = £10,864
359
(
500)For 155 m3
: C2007 = - (£13,800) = £17,692
390
1 j 10,864)
n = o~17,692 =0.6719
75 m3
lool---=-
'
01
155 m3
/
Exchange Rate
$1.40/£
$1.65/£
$2.00/£
C::' =(1~~:::T71'(£10,864) = £14,897 = ($2.00/£)(£14,897) =I$29,795 I
7-3

7.11
Year
1978
1988
1998
Capacity (lOJ e;al Purchased Cost ($)
? 35,400
105 45,300
85 45,500
[
219)For 105 gal: C1978 = - ($45,300) = $28,923
343
For 85 gal: C1978
= ( 219)($45,500) = $25,5S0
. ~390
$3S,400
$28,923
1 J$28,923)
n = 0$25,550 =0.587
lo~ 105 gal
" 85 gal
(
X )00587 => X =10S($35,400)Yo0587
lOS $28,923
IX= 148.2 gal
7-4
CEPCI
219
343
390

7.12
Year Capacity (kW) Purchased Cost (103
$) CEPCI
? 1,000 645.93 ?
2000 500 500.00 394
2007 775 811.68 500
C =C(I2 ). Co =(AlI)ll2 1 I 'c A1 b h
In 1986 CEPCI was 318 - use this information to estimate year.
(
318)For 500 kW: C1986 = - ($500)= $403.55
394
(
318)For 775 kW: C1986 = - ($811.68) = $516.23
500
{
I 000 kW)OoS619
For 1,000 kW: Cl986 = ($403.55 ' =$595.73
500kW
CEPC! = (318)($645.93) =344.8
year $595.73
CEPCI of 344.8 corresponds to about 1988. The compressor was not purchased in 1976
when the CEPCI was approximately 192.
7-5

7.13 and 7.14
t = PD + CA = (150bar)(3.2m) +0.00635
2SE -1.2P 2(0.9)S-1.2(150bar)
Iluignt (1-1) '" 1S rn
Dlmm::tc, (D) '" 3.2 rn
p", 150 bur
[: '" 0.9
CA =0.0015:15 rn
l'" PD,'(2SE-1.2P)-CP..
Wu:ghl {W.i =:rDHlp
dUII:;i1)!, f1 ~ &::IOU ~glm'
1,0 I
C'"i:).§: 0.8 I~ i
.... 0.61~cO.·: 1-,_ _- - " . . , . . . .........;....--~
] 0.2 r· ~ ... ~ _. - SS
.... I0.0 ...__.._~ ____~ __v _ _ _•_ _ _ _ _ ~J
YJO ::.~O ~'"O ·150 530
Tl>lnpo:,alu,o;,. C
ASME SA5t5 - Grade 55 Carbon St430l ASME SA·2·m· Grado 316 Stainies:; Stool
I crn::II,lriJll1fC S 'I','cight RcliJ[,vu t,;osl Tornpolal~ro S v'olyhl r~olalNu t;Ost
·:C om rn kg S "C bur rn Io;g ~
300 !J5C 0.3201 3130.130 :3BO.1:JO 30U B.30 0.3716 44El.344 1.::145.032
32D 95G U.3201 3130.130 3B5.13U 32U l!15 0.3793 457.5B!J 1.::I72.75!J
340 !J5G 0.3201 3136.130 366.13U 340 BOO 0.3873 467.231 1.401.692
J60 D2G U.3:J'5 399.976 389.975 36G 790 0.392E! 47:.1.691 1.421.673
3!.iO 1J6C (J.3572 4:.1G.94D 43C.94!J JBG 7B5 0.3056 477.2S'4 iA31.B!.i2
40n !JOC 0.357:.1 467.231 467.2::11 40(; '7130 0.39B5 460.:'41 1.442.242
42D 720 O.4::J65 525.530 526.5JU 42(: 776 G.4UOE! 4B::I.547 1.450.640
44,;) MG O.5(J02 60::l.4GO ti03,40U 44G 7n 0.4G32 4B5.37!J 1,459.1313
4&0 560 O.5E!61 107.0U6 107.005 46C 76El 0.4U56 469.240 i,467.7::!!)
4&0 4!.iS 0.699(; 1l43.24:J U4::1.243 4BG 764 0.4060 492.147 i,4"f6,441
50;) :J9D U.9259 i.1Hi.9(:W i.1i6.90U SOD 760 0.4104 495.GB:! 1,4B5.25;)
For this temperature range carbon steel is always the best (cheapest) choice ofmaterial of
construction (MOC).
7-6

7.15 - 17.19
Add EQLipment IUnit Number 100
Edit EQLipment 500
Problem 15 ~E·l0l Floating Head 5 30 Stainless Sleell Carbon Steel 160 40,100 $ 191.000
Problem 16 ~E·l02 Floaling Head 30 5 Slainiess Sleell Slainiess Sleel 160 40,100 $ 270.000
Problem l7a ~ E·l03 Floaling Head 20 20 Carbon Sleell Carbon Sleel 400 126.000 $ 435.000
Problem l7b ~ E·l04 Floaling Hoad 20 LV Carbon Sloell Carbon Sleel 400 74,900 $ 256.000
f"'
20 Slainiess Sleel Vatv.. 14 2 Carbon Steel 157,000 $ ::El.ooo
Used for
Trays
problems T·l02 20 Stainless Steel Vatv.. 14 2 Stainless Steel 157.000 $ 543,000
18 and 19
Trays
T·l03
20 Stainless Steel Vatv.. 14 2 Titanium 157.000 $ 1.020.000
-
Problem 7.15 and 7.16
Heuristics are consistent with Table 11.11, heuristic 3 - namely the tube side is for corrosive,
fouling, scaling, and high pressure fluids. Based on this, it is best to keep the high pressure
corrosive fluid in the tubes as given in Problem 15.
This saves approximately $270k - $191k =$79,000
Problem 7.17
(c) The results clearly indicate the principles of the economy of scale. Much cheaper to use a
single shell than multiple shells ($256k for single shell vs. $435k for 4 shells) - of course, this is
not always possible!
Problem 7.18
For Carbon Steel, CBM =$381,000
Problem 7.19
(a) 10% nitric acid solution - From Table 7.9 - use stainless steel CBM =$543,000
(b) 50% sodium hydroxide solution- From Table 7.9 - use carbon steel CBM = $381,000
(c) 10% sulfuric acid solution- From Table 7.9 - use Titanium CBM = $1,020,000
(d) 98% sulfuric acid solution- use carbon steel because there is not enough water for ionization
CBM =$543,000 when concentration drops to below about 90% need to use graphite or glass
lining
7-7

7.20 - Toluene Hydrodealkylation Process - Unit 100
Add EQuipment
~dit Equipmert 500
E·l02 Floating Hoad 23 Stainloss Stoell Stainless St",,1 763 $ 134.000 878.000
r
E·l03 Multipl~ Pip~ Carbon St••11 Carbon St••1 11 $ 5.110 $ 18.800
E·l04 Floating H~ad 2 2 Carbon Ste.11 Carbon Steel 35 $ 23.800 $ 78.300
E·l05 MultiploPipo Carbon Steoll Carbon St.ol 12 $ 5.530 $ 18.200
E·106 Floating Head 10 Carbon Ste.11 Carbon Ste.1 80 $ 29.200 $ 91.000
29 1.5 CarbonSt••1 2 $ 119.000 $ 324.000
T·l02
~
10 meters of 304 Stainless 14.2 2.3 Stainless Steel 24 $ 152.000 ;$ 2.030.000
Reactor - R - 101
Vertioal 3.5 1.1 CarbonSte.1 23 $ 7.280 :$ 5e.l01)
Vortioal 3.5 1.1 CarbonSt••1 1.5 $ 7.280 :$ 29.500
Horizontal 3.9 1.3 CarbonSto.1 1.5 $ 9.380 :$ 28.200
Total8are Module Cost $ 7,380,700
7-8

7.21 - Ethylbenzene Process - Unit 300 - Project B.2
E·303
E·304
E·305
E·306
E·307
E·30a
E·30S
P·302
P·303
P·304
P·305
T·302
V·302
V·303
V·304
Add EQuipment I
&dit EqUipme~
Floating Head
Floating Head
Fi.ed, Sheet, or U·Tube
Floating Head
Floating Hoad
Floating Head
Centrifugal
Centrifugal
Positio. Displacement
Trays
76 Stainless St.el Sieoe
Trays
Vertical
Horizontal
Horizontal
'~.~
Vertical
V·306 Vertical
V·307 Vertical
V·30S Vertical
Reactors R- 301-304
41
41
5
21
2
2
1
1.4
2.7
21.5
29
4.86
4.45
4.14
11
12
12
5
lJojrNllmber
21
21
21
21
21
,.
41
1.1
1.5
1.62
1.4S
1.38
1.12
1.85
1.97
0.95
300
500
Stainless Steel IStainless Steel
StainlessSteel IStainless Steel
Carbon Steel ICarbon Steel
Carbon Steel I Carbon Steel
Carbon Stool I Carbon Steel
Carbon SteelI Carbon Steol
Carbon Steoll Carbon Steel
Carbon Steel
CarbonSt.el
Carbon Steel
CarbonSt.el
Carbon Steel
Carbon Ste.1
Carbon Steel
CarbonSt.el
Carbon Steel
Stainles Steel
StainlesSteel
StainlesSteel
Stainles Steel
7-9
19
19
80.1 $
546 $
1570 $
348 $
57.8 $
54.6 $
22.6 $
11.5 $
3 $
3
1.5 $
2 $
2 $
21 $
21 $
21 $
21 $
Tolal Bare Module Cost
29,200
97,600 $
166,000 $
67,100 $
25,000 $
26,000 $
20,600 $
23,200 $
6,170 $
6,170 $
6,330 $
9,190 $
147,000 $
177,000 $
$
11,500 $
10,300 $
27,600 $
33,000 $
36,400 $
1,520 $:
203,000
679,000
55000
224,000
a5,SOa
85,400
69,500
16,300
24,600
24,600
2S,OOO
31,600
408,000
435,000
58.000
34,BOO
31.000
102.000
889.000
1,030.000
123.000
9,050,600

7.22 - Styrene Process - Unit 400 - Project B.3
Add Equipment I lJnjtNnmher 400
~dit Equipmert 500
E·402 Floating Head 4.65 O.S Stainless Steel I Stainless Steel 456 $ $ 514,000
E·403 Floating Head 41 0.35 Stainless Steel I Stainless Steel 2010 $ 354,000 $ 2,460,000
E·404 Floating Head 5 1.2 Carbon Steel I Carbon Steel 2130 $ 374,000 $ 1,230,000
E·405 Floating Head 5 1.05 Carbon Steel I Carbon Steel 2900 $ 510,000 $ 1,680,000
E·406 Floating Head Carbon Steel I Carbon Steel 76.7 $ 28,800 $ 9UOO
E·407 Floating Head Carbon Steel I Carbon Steel 127 $ 35,600 $ 117,000
E·408 Floating Head 0 Carbon Steel I Carbon Steel 902 $ 158,000 $ 521,000
E·409 Floating Head 0 Carbon Steel I Carbon Steel 680 $ 120,000 $ 393,000
P·402 Centrifugal Carbon Steel 0.4 $ 6,170 $ 24,600
~
,.
P·403 Centrifugal 2 Carbon Steel 1.1 $ 6,610 $ 26,300
P·404 Centrifugal 0.38 Carbon Steel 1 $ 6,170 $ 24,600
P·405 Centrifugal 0.75 Carbon Steel $ 6,170 $ 24,600
P·40S Centrifugal 2.65 Carbon Steel 1.1 $ 6,930 $ 27,600
28 3.15 Carbon Steel O.S
T·402 34.5 meters of Ceramic 34.5 4.1 Carbon Steel O.S 357,0'00 $ 2,080,000
7-10

Trays
O.S 467,000 $ 1,240,000
T·402 34.5 meters of Ceramic 34.5 4.1 Carbon Steel 0.6 357,000 $ 2,080,000
V·402 Horizontal 3.86 1.29 Carbon Steel 9,220 27.100
Horizontal 3.86 1.29 CarbonSteol 0.6 $ 9,220 :$: 27,700
Vortical 20 3.6 Stainlos Stool 0.7 $ 160,000 :$: 1.270.000
Vortical 20 3.6 Stainlo. Stool 0.7 $ 160,000 $ 1,270.000
Vortical 20 3.6 Stainlo. Stool 0.7 $ 160,000 $ 1.270,000
Vortical 20 3.6 Stainlo. Stool 0.7 $ 160,000 :$: 1.270.000
Vortical 20 3.6 Stainlo. Stool 0.7 $ 160,000 :$: 1.270,000
Vortical 20 3 Stainlo. Stool 0.7 $ 114,000 1: 905,000
Vortical 20 3 Stainlo. Stool 0.7 $ 114,000 :I: 905.000
Vortical 20 3 Stainlo. Stool 0.7 $ 114,000 :$: 905,000
Vortical 20 3 Stainlo. Stool 0.7 $ 114,000 $ 905,000
Vortical 20 3 Stainlo. Stoel 0.7 $: 114,000 $ 905,000
103 Reactors R-401 a-e and R-402 a-e Total Baro Modulo Co.t $32,968,400
7-11

7.23 Drying Oil Process - Unit 500 - Project BA
Add EquipmentJ Unit Number 500
Edit Equipmert 500
E·502 Floating Head 2 0.38 Stainless Steel/Stainless Steel 57.5 $ 26.300 $ 162.000
E-503 Floating Head 0.25 Stainless Steel / Stainloss Steol 2.95 $ 25,000 $ 154,000
r r ..E-504 Floating Head 0.25 41 Stainless Steel/StainlessSteel 64.S $ 27,200 $ 174,000
E-505 Floating Head 0.05 Stainless Steel/StainlessSteel 0.58 $ 25,000 $ 154,000
Floating Head 5 0.25 StainlessSteel / Stainle.s St.ol 919 $ lSI,OOO $ 994,000
,. ".
V-502 Vertical 4.6 0.9 Stainlos Stoel 0.48 6,760 53,500
V-503 Horizontal 3 Stainlo. Steel 0.25 $ 6,450
* 40,100
V-504 Horizontal 1.5 0.5 Stainle. St.ol 0.05 :$ 3,030 t 18,SOO
V~05 Vortical 5.3 0.53 StainlosSteel $ 4,250
* 33.600
Reactor R-501 Total Bare Module Cost $ 4.374.200
7-12

7.24 Maleic Anhydride - Unit 600 - Project B.5
92 !
E·S02
E·S03
E·S04
E·S05
E·SOS
E·S07
P·S02
P-S03
P·604
P·605
P·SOS
T·S02
V·S02
V·S03
V'S,
Floating Head
Floating Head
Floating Head
Floating Head
Floating Head
Centrifugal
Reciprocating
Centrifugal
Centrifugal
Centrifugal
Trays
42 Stainless SteelSieve
Trays
Horizontal
Horizontal
Vertical
Reactor R-601
2
2
... r
5
2
0
40
3.8
0.1
6.75
0.7
2.4
18
13.2
3.9
41 Stainless Steel/Stainless Steel S1.S $
41 Stainless Steel/Stainless Steel 17S0 $
Stainless Steel/StainlessSteel 1090 $
41 Stainless Steel/Stainless Steel 131 $
Stainless Steel/Stainless Steel 11.7 $
40 Stainless Steel/Stainless Steel 192 $
StainlessSteel $
Stainless Steel $
StainlessSt.el $
StainlessSteel 3 $
Stainless Steel 0.5 $
4.2 StainlessSteel 1.1
1.05 StainlessSteel 1.1
4.4 StainlesSteel 0.1 $
1.3 StainlesSteel 0.1 $
3.8 Stainles Steel $
TotalSare Module Cost
7-13
26,800 $ 171,000
309,000 $ 1,970,000
132,000 $ 813,000
3S,100 $ 230,000
24,300 $ 150,000
44,SOO $ 309,000
7,470 $
11,900 $ 60,400
8,720 $ 43,300
S,170 $ 30,700
6,810 $ 33,800
1,700,000
68,500 $ 243,000
119.000 $ 740.000
9,380 $ 58.300
68,700 $ 764,000
S 12.745,300

7.25 Ethylene Oxide - Unit 700 - Project B.6
fdil Equipmert
C-702 CentriFugal 23000
C-703 Centrifugal 21500
C-704 Centrifugal 5600
C-705 Contrifugal 5500
0-702 Electrio - EHplosion Proof 23000
0-703 Electric - Ellplosion Proof 21500
0-704 Eleotrio - ExplosionProof 5500
0-706 Electric- EKPlosion Proof 5500
E-703 Floating Hoad
E-704 FloatingH.ad
E-705 Floating Head 41
E-70S FloaUng H.ad 5
E-707 Floating He-ad
E-708 Floating Head 10.6
E-709 Floiling Ho.d 10
E-710 ~FloatingH.ad 10
E-711 FIo.ting Hud 10
Reactors R-701 and R-702
T-702
V-702
V-703
Tr.ays
20 Stainless SteelSieve
Trays
70 StalnlfoSs Steifl Sieve
Ve-rtioal
Vertical
12.2
43
10
10
IInitNnmb.r
2S.S
25.S
30
25.S
30
41
30
30
5.6
5.07
5.07
700
500
CarbonSt..1
Carbon Steel
Carbon Steel
CarbonStnl
Corbon Stool ICorbon Stool
Carbon Stool ICorbonSteel
Carbon Steel I Carbon St~1
Carbon Steel I Carbon 8tH'
CarbonSto.11 CarbonSt..1
Carbon Stull carbon StH'
StainlessSteel I Stainless Steel
Stainless Steel IStainless Sto.1
Corbon St••11 Carbon St..1
Carbon Steel
Stainless SteE'l
c.rbon Steel
CarbonSto.1
12100
14100
14100
13900
1480
566
154
22100
19000
30
10
30
30
"$
$
$
$
$
$
$
$
$
$
$
$
$:
$
$
$
$:
$
$
TotalBare Module Cost
7-14
5,870,000 $ IS,I00,OOO
5,S20,ooO $ 16,400,000
1,430,000 $ 3,910,000
1,430,000 $ 3,910,000
3,SSO.oOO $ 5,490,000
3,610,000 $ 5,420,000
1,130,000 $: 1,700,000
1,130,000 $ 1,700,000
tl00,ooO $ 3,S3O,OOO
2,130,000 $: 7,110,000
2,480,000 $ 8,290,000
2,480,000 $ 8,870,000
2,450_000 $: 8,190,000
259,000 $ 870,000
10tOOO $ 701,000
39,300 $: 245,000
3,890,000 $: 13,600,000
3,350,000 $ 11,700,000
t460,OOO $: 8,150,000
29,700,000 $ 99,400,000
lSS,OOO 3,810,000
169,000 $ 3,810,000
$ 262,033,500

7.26 Formalin Process - Unit 800 - Project B.7
Add Equipment Unit Number 800
gdit Equipmert 500
E·802 Floating Head 2.5 41 Carbon Steel/CarbonSteel 4.62 $ 25,000 $ 84,400
E·803 Floating Head 2.5 Carbon Steel/CarbonSte.1 28.2 $ 23,300 $ 76,600,.
E·804 Ketti. Reboiler 1.5 10 StainlessSt••1/ StainlessSt.el 37,3 $ 51,600 $ 320,000
E·805 Floating Head 1.5 StainlessSt..1/ StainlessSteel 269 $ 55,600 $ 342,000
E·806 Floating Head 5 Stainless Steel/Stainless Steel 41 $ 24,400 $ 150,000
E·a07 Floating He.d 10 2.5 Carbon Ste.1 / Carbon Steel 140 $ 37,400 $ 124,000
'I:.
Reactor R-801 costed as a heat exchanger
I T·aOl 10 meters of Ceramic 10 0.86 Carbon Steel 2 $ 18,300 $ 49,200
75 I
! 31 Stainless Steel Sie"eI T·802 19 2.5 StainlessSte.1 2 $ 200,000 $ 937,000I Trays76 !
77
Total Bare Module Cost $3,069,100
7-15

Chapter 8
8.1 The term O.18FCl has nothing to do with the interest on capital investment. From Table
8.2 we see that the following costs are based on the FCl:
• Maintenance and repairs
• Operating supplies
• Local insurance and taxes
• Plant overhead costs
• Administration costs
8.2 The number ofoperators per shift is multiplied by approximately 4.5 to get the total
number ofoperators required by the plant because it is assumed that a single operator
works 5 shifts a week and 49 weeks per year.
(
(dayXc XShift/ci) JTotal number ofoperators ~ (# operators per shift) ( year J .day
weeks/ (Shifts/)
/ operator '- 7week
(365 days/yr)(3 shifts/day) 45 /
~ . operators year
(49 weeks/operator)(5 shifts/week)
8 3 S F (SF)
# Days plant operated per year
. tream actor =
365
(Equation 8.5)
8.4 COMd = fiFCl, operating labor, raw material costs, utility costs, waste treatment costs)
8.5 The simple answer is that the cooling water cost does not vary very much with return
temperature because the main costs are associated with pumping (electricity), fans (power),
cost of water make up and chemical addition.
As Tew, return t, mew"!" .'. pump power.,!.. but WBD and WMake up t -so these compensate each
other.
Must not exceed TCWRetum = 45°C because of excessive fouling ofheat exchanger surfaces
above this temperature
8-1

8.6 No - this is not the reason for the factor of 1.23 multiplying CRM• The reason is that in the
derivation ofEquation (8.2) from the data in Table 8.2 some ofthe costs are based on the
COMD. Thus when all terms are added, the factor mUltiplying COMDis O.19COMD.
COMD =(CRM +CWT +CuT)+2.215CoL +O.146FCI+O.19COMD
O.81COMD =(CRM + CWf + CUT)+ 2.215COL + 0.146FCI
8.7 Yes, the 2.73 multiplier does include costs for supervisory and clerical labor but also
includes:
• Laboratory charges
• Plant overhead charges
• Administration costs
8.8 From Table 8.1 we have
• Direct Costs Costs that vary with the rate ofproduction
e.g. raw materials, utility costs.
• Indirect Costs Costs (in the plant) that do not vary with the
rate ofproduction, e.g. depreciation, taxes,
insurance
• General Expenses Costs associated with management level and
administrative functions not directly related
to the manufacturing process, e.g.
administration, research and development,
distribution, selling.
8-2

8.9
(a) Options (i) and (ii) are not reliable, and option (iii) is the best. If you do not have
time to dig up information on the new plant then (i) is probably better than (ii) unless
the dye and resin industries are known to have similar costs.
(b) Use Equation 8.2 when you have no/little information on a particular process or you
want to estimate manufacturing costs for a broad range ofindustries/processes.
8.10 (a) 10,000 kg/h, 58.6 barg - SH = 165°C
(i) Discharge P = 4" Hg absolute
From Table 8.5, theoretical steam required = 3.2 kg steam/kWh
Power from Turbine ~ 10,000 (k;;{) (0,80) ~2500kW ~ 2.5 MW
3.20 (kg/ )
7kWh
(ii) Discharge P = 4.82 barg
From Table 8.5, theoretical steam rate = 6.45 kg steam/kWh
. 10000 (ky{)Power from Turbme = ' ~X (0.80) = 1240 kW = 1.24 MW
6.45 g
kWh
(b) 10,000 kg/h saturated at 10 barg = 4.82 barg
From Table 8.5, theoretical steam rate = 28.80 kg steam/kWh
Power from Turbine ~ 10,000 (~) (0.80) ~277.8 kW
28.80 (k%Wh)
(c) a-(i) => Turbine #1 in Figure 8.6 (HP "'""'+ surface condenser)
a-(ii) => Turbine #2 (HP "'""'+ LP)
b => Turbine #4 (MP "'""'+ LP)
8-3

8.11 Costs for cooling water are:
• Pumping costs (electric)
• Fan costs (electric)
• Process water make up
• Injection ofchemicals for water treatment
(517.2XO.067)
(a) Make Up Water (%) = _--...:('-1,0_0_0.<....)-·100 = 9.8%
(0.354)
(b) Power Cost (%) ~ (O.06~2.36;1.61) .100~67.3%
0.354
If power cost doubles:
Cost of Cooling Water = 0.354 + (0.06X2.36+ 1.61)= $0.59}{000 kg (167%)
(c) N C =0354 (517.3XO.067)=$0.389/ (10980 /)
ew ost . + (1,000) /1,000 kg . 1'0
8.12 (a) Using Chemcad and SRK package for k-values and enthalpy calculations, the
following data are found: at 45°C vapor pressure ofpropane = 1556 kPa
(b) At -50°C vapor pressure of propane = 69.47 kPa
(c) For case when there are no LlP in exchangers and condenser and evaporator pressures
are set to (a) and (b).
(d) Chemcad simulation gives:
Qevap = 168.7 MJ/h
mproppne =20 kmollh = 881.9 kg/h
. 103[GJ/MJ]
Propane reqUIred for 1 GJ = 881.9[kg/h] =5227.6 kg = 118.6 kmol
168.7[MJ/h]
Circulating 5227.6 kg/h ofpropane ::::::> 1 GJ/h in condenser
(e) Using 5 kPa M in exchangers and a 75% compressor efficiency we get:
For Qevap = 1 GJIh
Wcomp = 288.7 kW
Qcond = -2.040 GJ/h
Cost to produce 1 GJ ofrefrigeration at -50°C
Using propane = (288.7 kWXO.06)+ (2.04X0.354)
ICost = $18.04/GJ I
8-4

8.13 (a) Repeat problem 8.12 using propylene as refrigerant
Vapor pressure (45°C) = 1,876 kPa
Vapor pressure (-50°C) = 90.6 kPa
mpropylene = 4,909.3 kg/h (116.6 kmol/h)
Qcond= -2.015 GJ/h
Qevap = 1.00 GJ/h
Wcomp = 282.0 kW
Cost = (282XO.06)+ (2.015X0.354)= $17.6%J
(b) Repeat problem 8.12 using ethane as refrigerant
Vapor pressure (4S0C) = No result Tcriticai ~ 32°C
Vapor pressure (-SO°C) =554 kPa
We cannot use ethane in a single stage refrigeration cycle because we cannot
condense it at a temperature of 45°C - need to go to a 2-stage system to use ethane.
It is probably not worth using a 2-stage system for this temperature since other
systems (single-stage) can be used.
(c) Repeat problem 8.12 using ammonia as refrigerant
Vapor pressure (45°C) = 1,797 kPa
Vapor pressure (-50°C) = 38.7 kPa
mammonia = 1,012.4 kg/h
Qcond =-1.896 GJ/h
Qevap =1 GJ/h
Wcomp = 248.9 kW
Cost =(248.9 kWXO.06)+ (1.896 GJ/hXO.354) =$15.6YuJ
Note: For problems 12 and 13 only single stage compressors have been used.
Because the pressure ratios are all much greater than the recommended maximum of
3~4, multiple stage compressors with inter-cooling should be used. This will reduce
the above costs. For example, for ammonia the pressure ratio = 1,802 =46.6.
38.7
Therefore, using 3 stages with inter-cooling with each stage having aPR =3.6 we get:
Qcond = -1,490 - 24 - 323 = -1,837 MJ/h
Wcomp = 61.5 + 86.6 + 84.S =232.6 kW
Qevap = 1 GJ/h
Cost =(232.6 kWXO.06)+ (1.837 GJ/hXO.354) =$14.6Ya.J
8-5

8.14 Operating Costs for the Ethyl Benzene - Project B.2
Name Total Module Cost Grass Roots Cost UtilitiUsed Efficien£y Actual Usage Annual Utiliti Cost
E-301 $ 221,000 $ 266,000 High-Pressure Steam -1967 MJIh $ (289,738)
E-302 $ 240,000 $ 288,000 High-Pressure Steam -2592 MJ/h $ (381,800)
E-303 $ 801,000 $ 960,000 High-Pressure Steam -10080 MJ/h $ (1,484,778)
E-304 $ 650,000 $ 920,000 Low-Pressure Steam -12367.001 MJIh $ (1,366,753)
E-305 $ 264,000 $ 375,000 Cooling Water 4940 MJ/h $ 14,600
E-306 $ 101,000 $ 142,000 Low-Pressure Steam 9110 MJ/h $ 1,006,700
E-307 $ 101,000 $ 144,000 COOling Water 7280 MJIh $ 21,400
E-308 $ 82,000 $ 116,000 High-Pressure Steam 5280 MJ/h $ 777,900
E-309 $ 90,000 $ 128,000 Cooling Water 5260 MJ/h $ 15,500
H-301 $ 3,490,000 $ 4,610,000 Natural Gas 0.9 24900 MJlh $ 2,067,000
P-301 $ 83,000 $ 111,000 Electricity 0.75 20 kilowatts $ 10,070
P-302 $ 29,000 $ 39,000 Electricity 0.75 1.33 kilowatts $ 671
P-305 $ 44,400 $ 59,300 Electricity 0.75 3.6 kilowatts $ 1,810
T-301 $ 481,000 $ 640,000 NA
T-302 $ 513,000 $ 675,000 NA
V-301 $ 38,900 $ 55,400 NA
V-302 $ 68,400 $ 97,000 NA
V-303 $ 40,900 $ 58,200 NA
V-304 $ 36,500 $ 52,000 NA
V-305 $ 830,000 $ 880,000 NA
V-306 $ 1,050,000 $ 1,120,000 NA
V-307 $ 1,220,000 $ 1,290,000 NA
V-308 $ 145,000 $ 160,000 NA
CUT
~FCI= CTM
Totals $ $ 390,000
Note that full credit is given for the steam generated in E-301 - 304 (In CAPCOST, this credit is
activated by using a negative duty for the exchanger)
NOL =[6.29+31.7p2 +0.23Nllp J'S =[6.29+0.23(l6)t
S
=3,16 op:1:::rs
(Nnp = 9 exchangers + 1 heater + 2 towers + 4 reactors = 16)
Noperators = (4.5)(3.16) = 14.2 round up to 15
COL = (15)($52,900) = $794,000/yr
Number ofhr/yr = 80,000,00019538.6 = 8387 h/yr (based on 80,000 tonne/yr ofEB)
CRM, benzene = (7761.3)(0.657)(8387) =$42,770,000/yr
CRM, ethylene = (2819.5)(1.202)(8387) =$28,420,OOO/yr
FCI (from Problem 7.21 CAPCOST Output for CTM) = $10,700,000
CUT = $390,000/yr
COMd= O,180FCI + 2.73COL + 1.23(CUT + CWT + CRM) = 0.180(10.7) + 2.73(0.794) +
l.23(0.390 + °+ 42.77+28.42) = $92,140,000/yr
COMd = $92,140,000/yr
COMd= [$92,140,000/yr]/[80,000,000 kg/yr] = $1. 152/kg EB (slightly higher than in Table 8.4)
8-6

8.15 Operating Costs for the Styrene Process - Project B.3
Name Total Module Cost Grass Roots Cost Utili~Used Efficienc:: Actual Usa~e Annual Utlli~ Cost
C-401 $ 610,000 $ 869,000 NA
D-401 $ 375,000 $ 534,000 Electricity 0.9 422 kilowatts $ 213,000
E-401 $ 384,871 $ 544,000 High-Pressure Sleam 29700 MJlh $ 4,374,000
E-402 $ 607,000 $ 744,000 Low·Pressure Steam -198460.008 MJlh $ (21,933,039)
E-403 $ 2,901,000 $ 3,480,000 High-Pressure Steam -162800.028 MJlh $ (23,980,345)
E-404 $ 1,450,000 $ 2,070,000 Low..Pressure Steam -35160.001 MJlh $ (3,885,748)
E-405 $ 1,980,000 $ 2,820,000 Cooling Valer 396000 MJlh $ 1,170,000
E-406 $ 110,000 $ 159,000 Low..Pressure Steam 331000 MJlh $ 36,600,000
E-407 $ 138,000 $ 197,000 Cooling Valer 281000 MJlh $ 830,000
E-408 $ 615,000 $ 870,000 Low..Pressure Steam 341000 MJlh $ 37,700,000
E-409 $ 464,000 $ 661,000 Cooling Valer 342000 MJlh $ 1,010,000
H-401 $ 12,200,000 $ 17,400,000 Nalural Gas 0.9 353000 MJlh $ 32,610,000
P-401 $ 40,300 $ 54,200 Electricity 0.8 8 kilowatts $ 4,030
P-403 '"$ 31,100 '"$ 41,800 Electricity 0.8 2.5 kilowatts $ 1,260
T-401 $ 1,460,000 $ 1,960,000 NA
T-402 $ 2,450,000 $ 3,180,000 NA
V-401 $ 101,000 $ 143,000 NA
V-402 $ 32,700 $ 46,600 NA
V-403 $ 32,700 $ 46,600 NA
V-404 $ 1,500,000 $ 1,820,000 NA
V-405 $ 1,500,000 $ 1,820,000 NA
V-406 $ 1,500,000 $ 1,820,000 NA
V-407 $ 1,500,000 $ 1,820,000 NA
V-40B $ 1,500,000 $ 1,820,000 NA
V-409 $ 1,070,000 $ 1,300,000 NA
V-41 0 $ 1,070,000 $ 1,300,000 NA
V-411 $ 1,070,000 $ 1,300,000 NA
V-412 $ 1,070,000 $ 1,300,000 NA
V-413 $ 1,070,000 $ 1,300,000 A
FCI=CTM
Totals $ $ 64,700,000
Note that full credit for the HPS and LPS in Exchangers E-403 and E-404 has been taken and that credit
for the high temperature steam from E-402 has been taken as LPS (which is low).
[ 2 J"S [ ]0.5 operatorsNOL = 6.29+31.7P +0.23N/lp = 6.29+0.23(24) =3.44 .
ShIft
Nnp =(1 compressor + 1 drive + 9 exchangers + 1 heater + 2 towers + 10 reactors) =24
Noperators =(4.5)(3.44) = 15.5 round up to 16
COL =(16)($52,900) =(16)($52,900) =$846,000/yr
Assume 1yr = 8200 h
CRM, ethylbellzelle =(1.069)(19417)(8200) =$170,21O,OOO/yr
FC! (from Problem 7.22 CAPCOST Output for Cm) =$39,000,000
CUT =$64,700,000/yr
COM" = O. l80FCI + 2.73COL + 1.23(CuT + CWT + CRtvl) = 0.180(39.0) + 2.73(0.846) + 1.23(64.7 + 0 +
170.21) = $298,300,000/yr
Icou,= $298.}00,000/yr
COM" =[$298,300,OOO/yr]/[(8200 h/yr)(l2,432 kg/h)] =$2.93/kg-styrene
Note that the COMdfor styrene is $1.40 greater than the selling prices from Table 8.4. We must
greatly improve integration of Styrene and EB processes to make styrene production viable.
8-7

8.16 Operating Costs for the Drying Oil Process - Project B.4
Name Total Module Cost Grass Roots Cost Utili!yUsed Efficien£i! Actual Usage Annual Utili!y Cost
E-501 $ 168,511 $ 207,000 -6329.002 MJJh $ (699,457)
E-502 $ 191,000 $ 235,000
E-503 $ 182,000 $ 223,000 Cooling Water 1030 MJJh $ 3,030
E-504 '$ 205,000 '$ 250,000 High.Pressure Steam 719 MJIh $ 105,910
E-505 $ 182,000 $ 223,000 coon", Water 230 MJJh $ 680
E-506 $ 1,173,000 $ 1,440,000 Low.Pressure Steam -4961.999 MJJh $ (548,381)
H-501 $ 1,900,000 $ 2.700,000 Natuml Gas 0.9 14700 MJ/h $ 1,221,000
P-503 $ 36,200 $ 46,200 Electricity 0.8 1 kilowatts $ 503
T-501 $ 683,000 $ 871,000 NA
T-502 $ 144,000 $ 176,000 NA
V-501 $ 22,900 $ 32,600 NA
V-502 $ 63,100 $ 76,900 NA
V-503 $ 47,300 $ 57,000 NA
V-504 $ 22,200 $ 26,800 NA
V-505 $ 39,700 $ 48,300 NA
r--,CUT
Totals $ 5,160.000 $ 6,740,000 $ 84,670
Note that the utility cost for the Dowtherm exchanger (E-402) is accounted for in the natural gas
for the fired heater and Ips credit is given for E-S01 and E-S06.
[
2 " JO.5 [ ]0.5 operators
NOL = 6.29+31.7P +0.2.:>N'1fJ = 6.29+0.23(10) =2.93 .
ShIft
Nnp =(6 exchangers + 1 heater + 2 towers + 1 reactors) = 10
Noperators =(4.5)(2.93) = 13.2 round up to 14
COL =(14)($52,900) = $741,000/yr
Assume 1 yr = 8200 h
CRM. ACO =($1.764lkg)(1628.7 kg/h)(8200 h/yr) = $23,S60,000/yr
Use value of $1.764/kg ($0.80/Ib) from http://www.icis.com/StaticPages/a-e.htrn#top for acetylated
castor oil
FC] (from Problem 7.23 CAPCOST Output for CTM) = $5,160,000
CUT= $84,700/yr
COMd= 0.180FCI + 2.73COL + 1.23(CUT + CWT + CRM) = 0.180(5.16) + 2.73(0.741) + 1.23(.085
+ 0 +23.56) = $32,030,000/yr
COMd = $32,030,000/yr
COMd= [$32,030,000/yr]/[(8200 h/yr)(12S0.04)] = $3.l3/kg-DO
Note that acetic acid is also produced and so revenue from this byproduct reduces the net COM
for Drying Oil (DO).
8-8

8.17 Operating Costs for the Maleic Anhydride Process - Project B.5
Name Total Module Cost Grass Roots Cost Utili!'iUsed Efficien9! Actual Usage Annual Utili!'i Cost
C-601 $ 3,130,000 $ 4,460,000 NA
D-601 $ 469,000 $ 668,000 Electricity 0.95 3270 kilowatts $
E-601 $ 171,273 $ 210,000 Low-Pressure Steam 1750 MJIh $
E-602 $ 202,000 $ 246,000 High.Presstire Steam -16700 MJIh $
E-603 $ 2,323,000 $ 2,830,000 Hlgh.Pressure Steam -31400.003 MJIh $
E-604 $ 960,000 $ 1,180,000 Cooling Water 86900 MJIh $
E-605 $ 272,000 $ 331,000 High.Pressure Steam 19200 MJ/h $
E-606 $ 177,000 $ 217,000 3050 MJIh $
E-607 $ 365,000 $ 438,000
H-601 $ 2,540,000 $ 3,620,000 Natural Gas 0.9 29800 MJ/h $
P-601 $ 29,000 $ 39,000 Electricity 0.8 0.375 kilowatts $
P-606 $ 39,900 $ 50,900 Electricity 0.8 3 kilowatts $
T-601 $ 2,010,000 $ 2,440,000 NA
T-602 $ 286,000 $ 350,000 NA
V-601 $ 28,500 $ 40,500 NA
V-602 $ 870,000 $ 1,050,000 NA
V-603 $ 68,800 $ 83,000 NA
V-604 $ 900,000 $ 1,040,000 NA
~
FC1=CTM CUT
~Totals $ 15,000,000 $
Note that credit is taken for hps in E-602 and E-603
[
~ JO.5 [ ]0.5 operators
NOL = 6.29 +3 l.7P- +0.23Nnp = 6.29+0.23(13) = 3.05 shift
Nnp = (1 compressor + 1 drive + 7 exchangers + 1 heater +2 towers + 1 reactor) = 13
Noperators = (4.5)(3.05):::;; 13.7 round up to 14
COL =(14)($52,900) = $741,000/yr
CRM, benzene = (3304)(0.657)(8200) = $17,800,000/yr
FC1 (from Problem 7.24 CAPCOST Output for CTM) = $15,000,000
CUT =$610,000/yr
1,650,000
193,400
(2,459,900)
(4,625,201)
256,000
2,821,000
9,000
2,753,000
189
2,390
63
4,250
440
1,510
610,000
COMd= 0.180FCI + 2.73CoL + 1.23(CUT + CWT + CRM) = 0.180(15.0) +2.73(0.741) + 1.23(0.61
+ 0 + 17.80) = $27,370,000/yr
COMd = $27,370,000/yr
COMd = [$27,370,000/yr]/[(8200 hlyr)(2597 kg/h)]=$1.29/kg-MA
This is about $0.30/kg less than value in Table 8.4
8-9

8.18 Operating Costs for the Ethylene Oxide Process - Project B.6
Name Total Module Cost Grass Roots Cost Utili~Used Efficienc~ Actual Usage Annual Utili~ Cos
C-701 $ 16,000,000 $ 22,800,000 NA
C-702 $ 19,000,000 $ 27,000,000 NA
C-703 $ 18,200,000 $ 25,900,000 NA
C-704 $ 4,610,000 $ 6,570,000 NA
C-705 $ 4,610,000 $ 6,570,000 NA
0-701 $ 5,680,000 $ 8,080,000 Electricity 0.9 21100 kilowatts $ 10,630,000
E-701 $ 3,784,770 $ 5,390,000 Cooling Water 58500 MJIh $ 172,000
E-703 $ 8,390,000 $ 11,900,000 Hlgh.Pressur~ St~am 148000 MJIh $ 21,740,000
E-705 $ 10,500,000 $ 14,500,000 High.Pressure Steam 230000 MJIh $ 33,860,000
E-708 $ 830,000 $ 990,000 High.Pressure Steam 43800 MJIh $ 6,458,000
E-709 $ 290,000 $ 354,000 Cooling Wate, 14200 MJIh $ 42,000
E-710 $ 16,043,000 $ 22,400,000 Medium·Pressu,e Steam -33101 MJIh $ (3,908,870)
E-711 $ 13,800,000 $ 19,300,000 Medium·Pressure Steam -26179 MJIh $ (3,091,457)
T-701 $ 9,600,000 .$ 10,700,000 NA
T-702 $ 9,600,000 $ 10,700,000 NA
T-703 $ 117,300,000 $ 134,800,000 NA
V-701 $ 224,000 $ 247,000 NA
V-702 $ 4,490,000 $ 4,840,000 NA
V-703 $ 4,490,000 $ 4,840,000~ NA
FCI
I CUTTotals $ I 309,200,000 r $ 397,100,000 $ 98,500,000
Note that credit is taken for hps in R-701 and R-702 that correspond to E-710 and E711. This process
looses enormous amounts ofuseful energy in E-704 and E-706. Heat integration can significantly reduce
the utility burden. A turbine in Stream 29 could also reduce the electrical utilities. See note at bottom
[ .., 2 r5
[ t5 operatorsNOL = 6.29+.)1.7P +0.23Nnp = 6.29+0.23(24) =3.44 .
ShIft
N np = (5 compressors + 5 drives + 9 exchangers + 3 towers + 2 reactors) = 24
COL = (16)($52,900) = $846,000/yr
Assume 1 yr =8200 h
CRM, ethylene = (20,000)(1.202)(8200) = $197,130,000/yr
FC! (from Problem 7.25 CAPCOST Output for ClM) = $309,200,000
Cur =$98,500,000/yr
COM, = O. JSOFCI + 2.73COL + 1.23(CuT + CWT + CRlvD = 0.lS0(309.2) + 2.73(0.846) + J.23(9S.5 + 0 +
COM, = $421,600,000/yr
197.13)=$421,600,000/yr I
~--------------------~
COM, = [$421 ,600,OOO/yr ]I[(8200h/yr)(20,000 kg/h)] =$2.57/kg
This is about $O.SO/kg more than the selling price - thus we need to improve the energy integration
significantly - see note below utility table.
8-10

8.19 Operating Costs for the Formalin Process - Project B.7
Name Total Module Cost Grass Roots Cost Utili!)! Used Efficienc~ Actual Usage Annual Utili!)! Cos
C-801 $ 610,000 $ 869,000 NA
D-801 $ 126,000 $ 179,000 Electricity 0.65 282 kilowatts $ 141,700
E-S01 $ 295,000 $ 420,000 Medlum-Pressure Steam 4110 MJIh $ 485,500
E-802 $ 100,000 $ 141,000 Hlgh.Pressure Steam 76.8 MJ/h $ 11,305
E-803 $ 90,000 $ 129,000 Cooling Water 983 MJIh $ 2,900
E-804 $ 378,000 $ 463,000 Medium~Pressllre Steam 37800 MJ/h $ 4,458,000
E-805 $ 404,000 $ 495,000 Cooling Water 32500 MJIh $ 96,000
E-806 $ 177,000 $ 21S,OOO Cooling Water 1170 MJIh $ 3,450
E-807 $ 146,000 $ 208,000 Medium-Pressure Steam -8928 MJIh $ (1,054,300)
T-S01 $ 58,000 $ 7S,500 NA
T-802 $ 1,110,000 $ 1,330,000 NA
V-801 $ 30,000 $ 42,800 NA
Totals $ 3,630,000 $ 4,710,000 $ 4,150,000
Note that credit for the mps generated in the reactor is taken account ofin E-807, which
represents the reactor exchanger.
[
2 JO.5 ]0.5 operators
NOL = 6.29 +31.7P +0.23Nnp = [6.29 +0.23(10) = 2.93 .
ShIft
Nnp = (1 compressor + 1 drive + 5 exchangers + 2 towers + 1 reactor) = 10
COL = (14)($52,900) = $741,000/yr
CRM, methanol = (2464.75)(0.294)(8200) = $5,940,000/yr
FC] (from Problem 7.25 CAPCOST Output for CTM) = $3,630,000
CUT =$4,150,000/yr
COMeI = 0.180FCI + 2.73CoL + 1.23(CUT + CWT + CRM) = 0.180(3.63) + 2.73(0.741) + 1.23(4.15
+ 0 + 5.94) = $15,090,000/yr
COMeI= $15,090,000/yr
COMeI = [$ 15,090,000/yr]/[(8200 h/yr)( 3897.06)]=$0.472/kg-formalin
This is about the same price as the formalin with 6% methanol inhibitor given in Table 8.4
8-11

Chapter 9
Chapter 9 (Short Answers)
9.1. Simple interest is calculated such that the interest is based on the original principal.
Compound interest is calculated such that the interest is based on the accrued principal
including previously paid interest, so that there is interest on interest.
9.2 The nominal interest rate is a number based on interest payments once per year; however,
ifinterest is compounded multiple times per year, the interest rate for the period is the
nominal rate divided by the number of compounding periods per year. The effective
annual interest rate involves using the result ofcompounding ofinterest several times per
year to calculate an interest rate as ifthere was one interest payment at the end ofthe year.
These are equal only when there is one compounding period per year.
9.3
iiiSONDJFMAMay
In Jl A +++ ++++
9.4 An annuity is a unifonn (constant) series oftransactions at the same interval. Examples
include a loan payment, a fixed monthly deduction from a paycheck into savings, etc.
9.5 525/485 = 1.0825 => so the inflation rate is 8.25%
9.6 Interest is the return on an investment or the charge for borrowing money. Inflation is the
increase in cost over time ofgoods, commodities, and services, which is equivalent to the
decrease in purchasing power of money. The time value of money includes the effect of
interest in an investment or cost. The time value ofmoney is identical to interest, but it has
nothing to do with inflation.
9.7 This term is depreciation
9.1

9.8 Depreciation is an accounting procedure that allows a company to reduce their taxable
income based on capital expenditures (such as new construction, improvement projects).
In a new chemical plant, depreciation reduces the taxable income at the beginning ofthe
project life, so the profitability ofthe new plant is increased.
9.9 The time value ofmoney suggests that "a dollar today is worth more than a dollar
tomorrow." Therefore, the sooner depreciation occurs, the higher the net cash flow is to
the company.
9.10 A company can only use depreciation to offset revenue. Depreciation can never exceed
profit. Therefore, a small company, with only one project, may have to defer depreciation
if it exceeds their profit from the project. A large company usually has so many profitable
projects, that they can use the depreciation in one project that may exceed that project's
profits to offset profits elsewhere in the company.
9.11 After-tax profit is the net profit from the project taking into account all expenses including
depreciation. After-tax cash flow is the net cash flow generated by the project after taxes.
These two quantities are the same when no depreciation is taken, i.e., in years when the
book value ofthe assets is zero.
9.12 P =$1,000
F = $1,000(1 + iejf)
(a) F = $1,000 +$5(52 weeks) = $1,260
$1,260 =$1,000(1 + iefJ )
Ii<!fr =0.26 =26% I
(
.)12
$1,260 =$1,000 1+ 1~
Ii =0.23 =23% I
(c) The interest paid to Paulie is $5/wk or $21.67/month (average of4.33 wks/month)
The interest paid to the bank is $1,000(0.07112) = $5.83/month. So no matter how
long it takes to repay the loan, the interest paid to Paulie will always be greater than
that paid to the bank.
9.2

9.13
(a)
F ( i )nm-=: 1+-
P m
P =: $1,000
(
0095)365
Daily: F = $1,000 1+ -'- =: $1,099.65
365
(
010)12Monthly: F =: $1,000 1+i2 =$1,104.71
(
0.105)4
Quarterly: F =: $1,000 1+-4- = $1,109.21
10.5% p.a. compounded quarterly is the most profitable scheme.
(b) F = P(l + ieff)
$1,109.21 =$1,000(1 + ieff )
Iieff = .109 = 10.9% I
(c)
Iiejf =0.111=11.1% I
9.3

9.14 F=P(l+i)"
F=2P
2P = P(l + i)"
ln2
n= ( )=>P=l
InP l+i
~--~------~--~----~
i n =72/i ieff % error
0.05 14.2 14.4 1.34%
0.075 9.58 9.58 0.21%
0.10 7.27 7.27 0.96%
The accuracy ofthe "rule of 72" improves as i decreases. For today's investment it is a very
accurate approximation.
i ief(
0.04 0.0408
0.05 0.0513
0.06 0.0618
0.07 0.0725
0.08 0.0833
F = $5,000(1 +0.0408Y(1 +0.0513)2(1 +0.0618X1 +0.0725X1 + 0.0833)
1F =$7,686.031
F ( . )n1n
9.16 P = 1+ ~ ;Fcont =Pe
il1
Fyearly =$15,000(1 + 0.09ys =$54,637.24
(
0.09)(4)(15)
Fquarterly = $15,000 1+ -4- = $57,002.02
(
0.09)(12)(15)
F;/lOl1thly = $15,000 1+12 = $57,570.65
(
0.09)(365XI5)
Fdaily =$15,000 1+- = $57,851.75
365
Feont
, =e(O,09)(15)($15,000) = $57,861.38
9.4

9.17 -= 1+-
F ( i )/1111
9.18
P m
IfP = $1
Bank 1: F = $1 1+ 0.04 = $1.0408
( )
365
365
Bank 2: F = $1(1 + 0.041)= $1.041
A
(a)
t_ n _ _ n n _ n n _ n _ _ t 19 20 21 22
o 1 18 IIIl75 75 75 75
(b) For years 1-18: F =(1+iY-1::::>F=A(1+0.08YS-1
A i 0.08
For ears 19-22: P = (1+iY -1 ::::>P=$75,000 (1+0.08)4 -1
y A i(1+iY 0.08(1+0.08t
.A (1.08ys -1 = $75 000 1.08
4
-1
0.08 ' 0.08(1.08t
A =$6,633.11 yr
(c) $5,000 (1 + iYS -1 = $75,000 (1 + it -1
i i(1+it
lr-i-=-0-.1-0-3-=-10-.3-<X--'oI
9.S

9.19
(a)
10
5 5 5
t-------------- ----------------------- t
o 1 8 25
50
(b) Fg =$10,000(1 +0.08Y+$5,000 (1 +~~8l-1
IFg =$71,692; Yes, you will have enough for the down paymentj
(c) ; = (1 +i)" ;~ = (1 +ir-1
F =$10 000(1 +0.08)25 +$5 000 (1 +0.08y5 -I _ $50 000(1 +0.08)17
25' , 0.08 '
IF25 =$249,000 I
9.6

9.20
(a)
A A A
Jan
1000 1000 1000 1000
1500 1500 1500
1000 1000 1000
8000 8000
(b) F =(1 +i)n. F =(1 +it -1
p , A i
(c)
F~(A-$1'500{(1+O;~r +(t+O;~4r +(1+O;~4)']
(
0.04)9
F=$7000[(1+ 0.04)8 +(1+ 0.04)4]+$1000 1+12 -1
, 12 12 ' 0.04
(A -$1,500X1.0373 + 1.0338 + 1.0304) = $7,000(1.02698 + 1.0033)+ $1,000(9.1209)
1 A =$9,046 1
(1+O;~4)9 -1
($4,000-$1,500X1.0373+1.0338+1.030~+ =$7,00d).02698-t-1.003~+$1,00c(9.120~
0.04
12
I A = $1,716 1
(d) Finances while in college
9.7

9.21
(a) F =P(l+ ~Jm
F =$1,500(1 + 0.045ys +$1,000(1 + 0.045t + $750(1 + 0.045Y3 +
$1,000[(1 + 0.04S)12 +(I + 0.04St + (1 + 0.045Yo + (1 + 0.045Y + (1 + 0.045Y +(1 + 0.045Y]-
$2,000(1 + 0.04SY +$700[(1 + 0.045r + (1 + 0.045t]- $500(1 +0.04SY + $1,000(1 +1.04SY +
$300(1 + 0.04S)- $2,000
1 F =$10,504 1
(b) F25 = $10,504(1 +0.04SYo
1 F25 =$16,312 1
9.22 F = (l+it -1
A i
(a) $1,000,000 =$6,000 (1 + it -1
i
1 i =0.0629 =6.29% 1
(b) $2,000,000 =$6,000 (1 + it -1
i
1 i =0.0895 =8.95% 1
(c) F =$6000 (1 + 0.07tO -1
, 0.07
1 F =$1,197,811 1
9.8

9.23
(a) F =$4,000(: ,0.09,25)(; ,0.09,16)+$5,000(: ,0.09,16)
F = $4 000((1 + 0.09ys -lJ(l + 0.09)16 + $5000((1 + 0.09r -lJ
' 0.09 ' 0.09
IF =$1,510,171 I
(b) $2,000,000 ~ $4,000 (I +i~41 -IJ+$I,OO{(I +i~16 -IJ
Ii = 0.1002 = 10.02% I
9.24 A = i(1 +it
P (1+it-1
(0;~7)(1+ 0;~7rX'1
A = $25,000 (12)(3)
(
1+ 0.07) -1
12
IA =$772 I
9.9

9.25
A i(l+it
P - (l+i)"-l
(
• )( • ) (12X2)
$500 =$20 000 iz 1+iz
1+- -1
12
, ( i )(12)(2)
'---j=-0-.0-9-24-=-9-.2-4-"-%--'1
9.26 A = i(I + it
P (l+i)"-1
(O.~~}+ O.~~5rX30)
(a) A = $200,000 (12X30)
r - ,A-=-$-1,2-6-4.-06--',
(1+0.~:5) -1
(O.~~5)(1+ O.~~5rXlS)
(b) A =$200,000 (12)(15)
.--,A-=-$I-,7-42-.24--",
(1+0.~:5) -1
(c) 30 years: (360X$1,264.14) =$455,090.40
15 years: (180X$1,742.22) = $313,599.60
I$141,490.80 I
9.10

9.27 A _ i(1 + i)"
P (l+i)"-1
(
i )( i )(12)(25)
$1,600 =$225,000 12 1+ 12
(
i )(12)(25)
1+- -1
r------.': 12
Ii =0.0707 =7.07% I
ieff =(l+i)"-1
i =(1 0.0707 )12eff + -1
12
Ii~tr =0.073 =7.3% I
9.28 A t(1 + i)"
P (1 +i)" -1
(0.06)( 0.06)n
$1,612 =$250,000 12 1+ 12
(
0.06)n
1 n =25 years 1
9.29 A = i(1 +it
P (l+i)"-1
1+- -1
12
(
0.07)(1 + 0.07 )(12XIO)
A = $50,000 12 12
(
0.07)(12XlO)
1+- -1
12
A =$580.54+ (0.0005X$50,000)
IA = $605.541
9.11

9.30
$75,000
$
25
tOO
r
$500,000
F=P(1+i)"
$250,000
$125,000
$100,000 1
F = $25;000(1 + i)4 + $75,000(1 + iY + $100,000(1 + if + $125,000(1 + i)+
$250,000 = $500,000(1 + iy
Ii=0.037=3.7%1
9.12

9.31 (a)
$60,000 $60,000
$50,000.........................................$50,000
i i i i i i
$250,000
(b) Cumulative, discounted cash flow diagram
Year Cash Flow
o -$250,000.00
1 $60,000.00
2 $60,000.00
3 $50,000.00
4 $50,000.00
5 $50,000.00
6 $50,000.00
7 $50,000.00
8 $50,000.00
(c) F=P(l+iY
F =$44,332.07(1 + 0.09Y
I F =$88,334 I
Discounted
Cash Flow
-$250,000.00
$55,045.87
$50,500.80
$38,609.17
$35,421.26
$32A96.57
$29,813.37
$27,351.71
$25,093.31
Cumulative, Discounted
Cash Flow
-$250,000.00
-$194,954.13
-$144A53.33
-$105,844.l5
-$70,422.89
-$37,926.32
-$8,112.96
$19,238.75
$44,332.07
(d) F =$250,000(1 +iy-$250,000(1 +0.09Y =$88,334
i =0.1125 =11.25%
9.13

9.32 After Tax Cash Flow = (R - d)(l- t)+ d
($60,000 $25iOOO)
R = $250000 = $83,522.73
(1-0.45)+ ~
($50,000- $25iOOO)
R= $250000 = $65,340.91
(1-0.45)+ ~
Before Tax Profit =Revenue - Depreciation
After Tax Revenue = (Before Tax Profit· 0.55)+ Depreciation
Note: All values in millions ofdollars.
Before Tax
Year MACRS Depreciation Revenue Profit
1 0.2 $50,000.00 $83,522.73 $33,522.73
2 0.32 $80,000.00 $83,522.73 $3,522.73
3 0.192 $48,000.00 $65,340.91 $17,340.91
4 0.1152 $28,800.00 $65,340.91 $36,540.91
5 0.1152 $28,800.00 $65,340.91 $36,540.91
6 0.0576 $14,400.00 $65,340.91 $50,940.91
7 $65,340.91 $65,340.91
8 $65,340.91 $65,340.91
9.14
AfterTax
Revenue
$68,437.50
$81,937.50
$57,537.50
$48,897.50
$48,897.50
$42,417.50
$35,937.50
$35,937.50

9.33
Discounted Cumulative, Discounted
Year Cash Flow Cash Flow Cash Flow
0 -$10.00 -$10.00 -$10.00
1 -$20.00 -$17.39 -$27.39
2 -$30.00 -$22.68 -$50.08
3 -$30.00 -$19.73 -$69.80
4 $25.00 $14.29 -$55.51
5 $25.00 $12.43 -$43.08
6 $25.00 $10.81 -$32.27
7 $25.00 $9.40 -$22.87
8 $25.00 $8.17 -$14.70
9 $25.00 $7.11 -$7.59
10 $25.00 $6.18 -$1.41
11 $25.00 $5.37 $3.96
12 $25.00 $4.67 $8.63
13 $45.00 $7.31 $15.95
(a)
9.15

(b)
-$69.80
() NPV--$10- $20 $30 _ $30 +
c - (1+0.15) o+0.15f (1+0.15Y
$25«1+0.15Yo -1) $20
----~---=IO--~~3+ I'
(0.15X1+0.15) (1+0.15) (1+0.15f
INPV =$15.9 million I
(d) F=$15.9(1+0.15)13
IF =$97.8 million I
(e) NPV=0=-$10- $20 _ $30 _ $30 +
(1 + i) (1 + if (1 + iy
$25«1 + iyo-1) $20
(0.15X1+iYO(1+iY + (1+ii
3
Ii =0.192 =19.2%I
(f)Proceed
9.16
$15.95

9.34
d SL _ undepreciated capital
k - remaining time for depreciation
dk
DDB =~[100-Id~DB]5 j=O
Four Year
Year DDB SL MACRS
1 25.00 25.00
2 37.50 21.43 37.50
3 18.75 15.00 18.75
4 9.38 12.50 12.50
5 2.34 6.25 6.25
Six Year
Year DDB SL MACRS
1 16.67 16.67
2 27.78 15.15 27.78
3 18.52 12.35 18.52
4 12.35 10.58 12.35
5 8.23 9.88 9.88
6 2.74 9.88 9.88
7 2.29 4.94 4.94
Nine Year
Year DDB SL MACRS
1 11.11 11.11
2 19.75 10.46 19.75
3 15.36 9.22 15.36
4 11.95 8.27 11.95
5 9.29 7.60 9.29
6 7.23 7.23 7.23
7 5.62 7.23 7.23
8 4.37 7.23 7.23
9 3.40 7.23 7.23
10 1.32 3.61 3.61
9.17

9.35
(a)
$75
45$ $45 $45 45$ $45 5$4 $45 $45 $45
o 1 2 3 4 5 6 7 8 9 1o
-$190
9.18

.0
I-'
.0
Amc·unts arc in mUbons of dollilirs.
Land 10
Fe[ l65
'NC 15
Sal"'agc 15
R
CDr.·l
0.14
0.4
70
25
(b)
'fear
Cash Flo~...'. _ - .
hn'C'sttllem _ .. H~ discounted COMo ;lAC[.tS Ii
nondiscounted
0 ·190.00 ,190..00
1 45.00 61.40 21.93 U.2 33.0U
2 45.00 53.86 I Q '''.d•••.1;. • lU2 52.g0
]- 45.00 .d·7 '1~• I .~_ 16..8'7 0.192 3L6!:l
4 45.00 41.45 14.80 O. n52 ]9.tH
5 45.00 36.36 12.98 lU J52 19.0]
6 45.00 31.89 11.39 0.0576 9.50
...,
45.00 "'7 Q'7 9.99,, "' •• ';'" I
8 45.00 'M ~.d.L ••_ • 8.76
9 45.00 21.53 7.69
10 75.0<.1 26.97 <i '7.d., .
( c ) (d) & (c) (0
Aftm' '[ax itfter' '("ax Cllmu(ath'e~
Prom Cash FIOl~' dtseGo ntC'd cash flow
0.00 190.00 ~ 190.00
3JH! 36.gf:{ "153.12
·10.90 41.90 ll1.22
~(1.71~~ 30.9'0 ·80.32
4.5!:l 23.59 ··56.73
2.62 21.63 ···35.11
(~.61) 16.111 ·19.01
10.79 10.79 8 ....."ole • ".l.L
9.47 9,47 1.25
g.30 !:l.30 9.55
12.14 12.14 21.69

(e)
o
-$190
(f)
o
!
1-$190.00
I
$41.90
$36.88 $30.90 $23.59 $21.63 $16 10 $ $12.14
. 10.79 $9.47 $8.30
1 2 3 4 5 6 7 8 9 10
$21.69
2 3 4 8 9 10
(g) $21.69 million - after-tax cash flow or cumulative cash flow at the end ofyear 10.
9.20

Chapter 10
Chapter 10 (short answers)
10.1 NPV = 0 at DCFROR. So, ifuse hurdle rate to calculate NPV, ifNPV > 0, rate ofretum
exceeds hurdle rate.
10.2 Choose the project with the greatest NPV.
10.3 If improvements are proposed for a process, an incremental economic analysis uses the
capital cost ofthe improvements and the changes in the cost ofmanufacture (operating
cost, revenue, etc.) to evaluate profitability. The entire process is not evaluated.
10.4 Yes. There would be a difference in the operating costs, which would affect the
profitability.
10.5 EAOC or Common Denominator or Capitalized Cost methods but not NPV because
equipment lives are different.
10.6 Risk cannot be eliminated. Monte Carlo simulation allows the risk to be quantified and
analyzed.
10.7 Advantages include quantifying risk and thereby obtaining a more comprehensive picture
ofthe effect ofdifferent factors on the profitability. Disadvantages include obtaining
predictions ofthe variability ofdifferent parameters from historic data (difficult to predict
the future).
10-1

10.8 The spreadsheet on the next page shows the cash flow diagrams.
(a) Non-Discounted Cash Flow Diagram shown on next page.
(b) (i) Cumulative cash position (CCP) = $132 million
Cumulative cash ratio (CCR) = 1.81
CCP = $132 million
CCR= 1.81
(ii) Payback period (PBP) = 4.0 years
I PBP = 4.0 years
(iii) Rate ofreturn on investment (ROROI) = 11.00%
I ROROI = 11.00%
(c) Discounted Cash Flow Diagram shown on next page.
(d) (i) Net present value (NPV) = $1.71 million
Present value ratio (PVR) = 1.01
NPV = $1.71 million
. PVR= 1.01
(ii) Discounted payback period (DPBP) = 6.7 years
I DPBP = 6.7 years
(iii) Discounted cash flow rate of return (DCFROR) = 9.73%
DCFROR=9.73%
10-2

~
oI
W
20.0
~ 0.0
"0 -20.0...
'0 -40.0
"'c
~ -60.0
§. -80.0
..i -100.0
>
'g -120.0
"£ -140.0
-160.0
-1 0
QenBr.JleCFD
Year Investment
0 0.00
0 750
1 60.00
2 3960
3 20.40
3 35.00
4
5
6
7
8
9
10
11
12
13
13
Discounted Cash Flow Diagram
150.0
e 100.0
J!!
"0...
~
~
5
50.0
0.0
§. -50.0..
'"~ -100.0
i
-200.0
2 3 4 5 6 7 8 9 10 11 12 13 14 -1
project Life (Years)
Discounted Profitibility Criterion Non-Discounted Profitibility Criteria
Net Present Va!ue (millions) 1.71 Cumularve Gash Position (millions) 132.00
Discounted cash Row Rate of Return 9.73% Ra:e of Retum on Investment 1100%
Discounted Payback Period (ysS'"s) 6.7 Payback Period (years) 4.0
cash FlOW (Non. CBshl=IOW CUmtJlative Cash I=low
d" !=OI.:Sd)( R COfv, (R.COMrd")"(1.t).d,, discounted) (discounted) (discounted)
120.00 0.00 0.00 0.00
120.00 (7.50) (7.50) (7.50)
120.00 (60.00) (54.79) (62.29)
120 00 (39.60) (33.03) (95.32)
120.00 (20AO) (1554) (110.86)
120.00 (35.00) (26.66) (137.52)
24.00 96.00 52.00 18.00 30.00 30.00 20.67 (116.65)
38AO 5780 52.00 18.00 35.76 35.76 22.72 (93.93)
2304 34.55 52.00 18.00 29.62 29.62 17.18 (76.75)
13.80 20.76 52.00 18.00 25.92 25.92 13.73 (63.02)
13.80 6.96 52.00 18.00 25.92 25.92 12.54 (50.48)
6.96 52.00 18.00 23.18 23.18 10.24 (40.24)
52.00 18.00 20.40 20AO 8.23 (32.01)
5200 18.00 20.40 20AO 752 (24.49)
52.00 18.00 20.40 20.40 G.87 (17.62)
52.00 18.00 20.40 20AO 6.27 (11.35)
42.50 13.06 1.71
Non-Dlscounted cash Flow Diagram
o 2 3 4 5 6 7 8 9 10 11 12 13 14
Project Ufe (Years)
CumulatIVe Cash FtM
(Non·discounted)
0.00
(7.50)
(6750)
(107.10)
(127.50)
(162.50)
(132.50)
(96.74)
(6712)
(4120)
(15.28)
7.90
28.30
4870
6910
89.50
132.00

10.9 The spreadsheet on the next page shows the cash flow diagrams.
(a) Non-Discounted Cash Flow Diagram shown on next page,
(b) (i) Cumulative cash position (CCP) = $132 million
Cumulative cash ratio (CCR) = 1.81
CCP = $132 million
CCR = 1.81
(iii) Payback period (PBP) = 4.4 years
I PBP = 4.4 years
(iii) Rate ofreturn on investment (ROROI) = 11.00%
I ROROI = 11.00%
(c) Discounted Cash Flow Diagram shown on next page.
(d) (i) Net present value (NPV) = -$1.05 million
Present value ratio (PVR) = 0.99
NPV = -$1.05 million
PVR=0.99
(iv) Discounted payback period (DPBP) = 7.1 years
I DPBP = 7.1 years
(v) Discounted cash flow rate ofretum (DCFROR) = 9.37%
I DCFROR=9.37%
The MACRS depreciation method improves profitability when compared to SL
depreciation.
10-4

Discounted Cash Flow Diagram Non-Dlscounted Cash Flow Diagram
0.0 lSO.0
~ -20.0 ~ 100.0
~
'"'0 -40.0
'is." ."
SO.O
'0 '0
'" -60.0 VI
<:
~ 0.0.E
!. -80.0 g -SO.O
~ -100.0
II>
::>
.. ~ -100.0>
tl -120.0
II>
0' £-1SO.0Ii: -140.0
-160.0 -200.0
-1 0 2 3 4 5 6 7 8 9 10 11 12 13 14 -1 0 2 3 4 5 6 7 8 9 10 11 12 13 14
Project Life (Years) Project Life (Years)
!1enora1e CFO
>-'
0
Discounted Profitibility Criterion Non-Discounted Profitibility CriteriaI
Vl
Net Present Value (millions) (1.05) Cumulative cash posinon (millions) 132.00
Discounted Cash Flcw Rate of Return 9.37% Rete of Return on Investment 11.00%
Discounted Payback Partod (years) 7.1 Payback Period (years) 4.4
Cash Flow Cash Flow Cumulatrve Cash Row OJmulative Cash Flow
'fea Investment cI, FCiL-ScI, R COM" (R.COM,,·<I,)"(1·t)+cI, (Non·dlscounted) (discounted) (discounted) (Non·discounted)
0 0.00 120.00 0.00 0.00 0.00 0.00
0 7.50 120.00 (7.50) (7.50) (7.50) (7.50)
1 60.00 120.00 (60.00) (54.79) (62.29) (67.SO)
2 39.60 120.00 (39.60) (33.03) (95.32) (107.10)
3 20.40 120.00 (20.40) (15.54) (110.86) (12750)
3 35.00 120.00 (35.00) (26.66) (137.52) (162 50)
4 17 14 102.86 52.00 18.00 27.';13 27.26 18.96 (118.56) (135.24)
5 17 14 85.71 52.00 18.00 2726 27.26 17.31 (101.24) (107.99)
6 1714 6857 52.00 18.00 27.26 27.26 15.81 (85.43) (80.73)
7 1714 51.43 52.00 18.00 27 ';13 21.26 14.44 (70.99) (53.41)
8 17 14 34.29 52.00 18.00 27 ';13 27.26 13.19 (57.80) (26.21)
9 17 14 1714 52.00 18.00 27';13 21.26 12.04 (4576) 1.04
10 1714 0.00 52.00 18.00 2726 27.26 11.00 (34.76) 28.30
11 0.00 52.00 18.00 20.40 20.40 7.52 (27.24) 48.70
12 0.00 52.00 18.00 2040 2040 6.87 (20.38) 69.10
13 0.00 52.00 18.00 20.40 20.40 6.27 (14.11) 89.SO
13 4250 13.06 (105) 13200

10.10 The spreadsheet and cash flow diagrams for each case are shown on the following pages.
(a) MACRS method for 5 years
INPV= -$0.12 million I
(b) Straight Line depreciation over 7 years
INPV= -$0.33 million I
10-6

Cash Flow Diagram (Five year MACRS)
0.0
~ -1.0
.!!
-2.0(5
"...0 -3.0
f1I
'" -4.0
~
i -5.0
., -6.0
"0;
-7.0>
1:)
.~ -8.0
a. -9.0
-10.0·j
-1 0 2 3 4 5 6 7 8 9 10 11 12 13
Project Life (Years)
>-'
oI
§EfIorateCFD
-.....l
Discounted Profitibility Criterion Non-Discounted Profitibility Criteria
i
Net Present Value (millions) (0.12) Cumulalive Cash Position (milrtOns) 8.58
Discounted Cash Row Rate of Roturn 10.72"'!' Rate of Return on Invasbnent 11.14%
Discounted Payback Period (years) 6.5 Payback Period (years) 3.9
Cash Flow
(Non- Cash Flow Cumulative Cash Cumulative Cash Flow
Year Investment dk FCll-Sdk R COMd R-COM...dk)'(l-t)+d discounted) (discounted) Flow (discounted) (Non-discounted)
0 0.00 7.70 0.00 0.00 0.00 0.00
0 0.80 7.70 (0.80) (0.80) (0.80) (0.80)
1 5.01 7.70 (5.01) (4.51) (5.31) (5.81)
2 2.70 7.70 (2.70) (2.19) (7.50) (8.50)
2 2.00 7.70 (2.00) (1.62) (9.12) (10.50)
3 1.54 6.16 4.10 1.90 1.94 1.94 1.42 (7.70) (8.56)
4 2.46 3.70 4.10 1.90 2.31 2.31 1.52 (6.19) (6.26)
5 1.48 2.22 4.10 1.90 1.91 1.91 1.13 (5.05) (4.35)
6 0.89 1.33 4.10 1.90 1.67 1.67 0.90 (4.16) (2.67)
7 0.89 0.45 4.10 1.90 1.67 1.67 0.81 (3.35) (1.00)
8 0.45 4.10 1.90 1.50 1.50 0.65 (2.70) 0.50
9 4.10 1.90 1.32 1.32 0.52 (2.18) 1.82
10 4.10 1.90 1.32 1.32 0.46 (1.72) 3.14
11 4.10 1.90 1.32 1.32 0.42 (1.30) 4.46
12 4.10 1.90 1.32 1.32 0.38 (0.92) 5.78
12 2.80 0.80 (0.12) 8.58

>-'
oI
00
Cash Flow Diagram (Straight Line Depreciation over 9.5 years)
0.0
" -1.0
~
-2.0"0
"0
'0 -3.0
..c -4.0
~
1 -5.0
., -6.0::l
~ -7.0
t!
.[ -8.0
c. -9.0 .
H
-10.0
-1 0 2 3 4 5 6 7 8 9 10 11 12 13
Project Life (Years)
Q.enomte CFD
Discounted Profitibility Criterion Non-Discounted Profitibility Criteriai
Not Present Value (millions) (0.46) Cumulative Cash Position (mafions) 8.74
Discounted Cash Row Rate or Raturn 10.03% Rato of Roturn on Invesbnont 11.35%
Discounted Payback Period (years) 7.7 Payback Period (years) 4.7
Cash Flow
(Non- Cash Flow Cumulative Cash Cumulative Cash Flow
Year Investment dk FCIL-Sdk R COM, R-COM,,-dk)*(1-t)+d discounted) (discounted) Flow (discounted) (Non-discounted)
0 0.00 7.70 0.00 0.00 0.00 0.00
0 0.80 7.70 (0.80) (0.80) (0.80) (0.80)
1 5.01 7.70 (5.01) (4.51) (5.31) (5.81)
2 2.70 7.70 (2.70) (2.19) (7.50) (8.50)
2 2.00 7.70 (2.00) (1.62) (9.12) (10.50)
3 0.81 6.89 4.10 1.90 1.64 1.64 1.20 (7.92) (8.86)
4 0.81 6.08 4.10 1.90 1.64 1.64 1.08 (6.83) (7.21)
5 0.81 5.27 4.10 1.90 1.64 1.64 0.98 (5.86) (5.57)
6 0.81 4.46 4.10 1.90 1.64 1.64 0.88 (4.98) (3.92)
7 0.81 3.65 4.10 1.90 1.64 1.64 0.79 (4.19) (2.28)
8 0.81 2.84 4.10 1.90 1.64 1.64 0.71 (3.47) (0.63)
9 0.81 2.03 4.10 1.90 1.64 1.64 0.64 (2.83) 1.01
10 0.81 1.22 4.10 1.90 1.64 1.64 0.58 (2.25) 2.65
11 0.81 0.41 4.10 1.90 1.64 1.64 0.52 (1.73) 4.30
12 0.81 4.10 1.90 1.64 1.64 0.47 (1.26) 5.94
12 2.80 0.80 (0.46) 8.74

10.11 (a) DCFROR = i, where NPV = 0
Year
1
2
3
4
5
6
7
8
9
10
11
12
13
Nondiscounted
Cash Flow ($106
/yr)
-10.000
-15.000
-15.000
7.015
6.206
6.295
6.852
6.859
7.218
5.954
5.459
5.789
15.898
Discounted Cash
Flow ($106
/yr)
-9.097
-12.412
-11.291
4.803
3.866
3.567
3.532
3.216
3.079
2.310
1.927
1.859
4.643
IDCFROR = i = 0.099 = 9.9% I
Cumulative Discounted
Cash Flow ($106
/yr)
-9.097
-21.509
-32.800
-27.997
-24.131
-20.564
-17.032
-13.817
-10.738
-8.428
-6.501
-4.643
0.000
(b) The DCFROR is very close to 10%; therefore, it is a borderline decision.
10.12 The spreadsheet and cash flow diagrams are shown on the following pages.
(a) Payback period (PBP) ~ 2.54 years
(b) Cumulative cash position (CCP) = $360.60 million
(c) Rate of return on investment (RORI) = 78%
(d) Discounted payback period (DPBP) = 2.47 years
(e) Net present value (NPV) = $130.81 million
(f) Discounted cash flow rate of return (DCFROR) = 32.8%
10-9

Fel 80
Z0.1
a
Total Annual Net Profit After-Tax Cumulative Discounted Disc. Cum. ~
Year Investment Depreciation Revenue
Costs
Net Profit Income Tax
AfterTax Cash Flow Cash Flow Cash Flow Cash Flow
~0 ($10) ($10.00) ($10.00) ($10.00) ($10.00)
1 ($25) ($25.00) ($35.00) ($22.73) ($32.73) 0
2 ($25) ($25.00) ($60.00) ($20.66) ($53.39)
a3 ($35) ($35.00) ($95.00) ($26.30) ($79.68)
4 $8.57 $60.00 $50.00 $10.00 $3.80 $6.20 $14.77 ($80.23) $10.09 ($69.60)
00
00
5 $8.57 $120.00 $92.00 $28.00 $10.64 $17.36 $25.93 ($54.30) $16.10 ($53.50) g6 $8.57 $120.00 $47.00 $73.00 $27.74 $45.26 $53.83 ($0.47) $30.39 ($23.11)
@7 $8.57 $120.00 $50.00 $70.00 $26.60 $43.40 $51.97 $51.50 $26.67 $3.56
8 $8.57 $120.00 $60.00 $60.00 $22.80 $37.20 $45.77 $97.27 $21.35 $24.91 ......
9 $8.57 $120.00 $51.00 $69.00 $26.22 $42.78 $51.35 $148.62 $21.78 $46.69 l:;j
10 $8.57 $120.00 $40.00 $80.00 $30.40 $49.60 $58.17 $206.79 $22.43 $69.12
.-+
§.
11 $120.00 $40.00 $80.00 $30.40 $49.60 $49.60 $256.39 $17.38 $86.50 .......
12 $120.00 $40.00 $80.00 $30.40 $49.60 $49.60 $305.99 $15.80 $102.30 ~
13 $120.00 $40.00 $80.00 $30.40 $49.60 $49.60 $355.59 $14.37 $116.67 ~
13 $15 $15.00 $370.59 $4.34 $131.02 ~
......l:;j
...... S......
0 ......
I
................... 0
0 l:;j
00
0
H)
0-
0.............
~
;n

Non-discounted Cash Flow Diagram
400
<0 300
<
0
~ 200~
0
u:
.s:::
100III
..U
~
~ 0
:; 1~E
::J
-54.3u -100 -60
-95 -80.23
-200
Year
Discounted Cash Flow Diagram
150
121.02
<0 100<
0
~
~
0
50u:
.s:::
III
co
U
~
0
-1
11
:;
E
-50::J
u
-100 I
Year
10-11

10.13 (a) From year 4
Taxation rate, t = income tax / net profit = 3.8/10 = 38%
I T=38% I
(b) Total Fixed Capital Cost (FCh) = 80 - 20 = $60 million
IFCh = $60 million I
(c) Depreciation method corresponds to Straight Line depreciation over 7 years.
dSL = 6017 = $8.53 million
IStraight Line depreciation I
(d) COM =Total Annual Costs - Depreciation
Year
Depreciation Total Annual Costs COM
($1061
1yr) ($1061
Iyr) ($1061
1yr)
0
1
2
3
4 8.57 50.00 41.43
5 8.57 92.00 83.43
6 8.57 47.00 38.43
7 8.57 50.00 41.43
8 8.57 60.00 51.43
9 8.57 51.00 42.43
10 8.57 40.00 31.43
11 40.00 40.00
12 40.00 40.00
13 40.00 40.00
10-12

10.14 (a) NPV= -FC!+'ICFlP / F,i,n)
Process 1
i=6%
NPV =-$15+ $3 + $8 + $7 + $5 + $2 =$6.28 million
(1.06)1 (1.06)2 (1.06)3 (1.06)4 (1.06)5
Process 1
i= 18%
NPV $15 $3 $8 $7 $5 $2 $1 00 '11'=- + + + + + =. ml IOn
(1.18i (1.18)2 (1.18)3 (1.18)4 (1.18)5
Process 2
i=6%
NPV $15 $5 $5 $5 $5 $5 $6 06 '11'= - + + + + + =. m! IOn
(1.06)1 (1.06)2 (1.06)3 (1.06)4 (1.06)5
Process 2
i=18%
NPV $15 $5 $5 $5 $5 $5 $064 '11'= - + + + + + =. ffil Ion
(1.18)1 (1.18)2 (1.18)3 (1.18)4 (1.18)5
IRecoffilnend Process 1 I
(b) set NPV=0 in the above equations and calculate i
Process 1: DCFROR = 21%
Process 2: DCFROR = 19.9%
Process 1 is better.
(c) Payback Period
Cumulative cash flows.
Year Process 1
1 3.00
2 11.00
3 18.00
4 23.00
5 25.00
10-13
Process 2
5.00
10.00
15.00
20.00
25.00

Process 1
PBP = 2+(15-11)/(18-11) = 2.57 years
Process 2
PBP = 3 years
PBP1 = 2.57 years
PBP2 = 3 years
Process 1 is better.
(d) Parts a, b, and c all concluded with Process 1 being the better choice.
10-14

10.15 Compare using EAOC.
EAOC=FCl(AlP,i,n) + fOC
EAOCA =15((OtXl+~.09)' )+4 =$9.93thousand/yr
1+0.09' -1
EAOCB =25((OtXl+~.09)' )+3 =$9.43 thousand/yr
1+0.09 -1
EAOCe =40((OtXl +~.09)' )+2 =$9.95 thousand/yr
1+0.09 -1
IChoose Equipment B I
10.16 (a) EAOC = FCl(AlP,i,n) + fOC
EAOCA = 5((OtX1 +~.15)' )+2 = $3.49thousand/yr
1+0.15 -1
EAOCB = 10((OtXl+~/5)' )+1 = $3.40thousand/yr
1+0.15 -1
IChoose Equipment B I
(b) WhenEAOCA -EAOCE= 0; i = 0.1735
When i = 18%; EAOCA = -$3.60 thousandlyr
EAOCE= -$3.62 thousand/yr
Ii= 18% I
10-15

10.17 Compare using EAOC
EAOC=FCl(AIP,i,nj + YOC
The table below only includes equipment that differs in cost between the two options.
10.18
Option
FCI YOC
($thousand) ($thousand/yr)
1 70 15
2 100 9
Choose Option 2
CC =P (l+it
(l+it -1
Purchased cost ofpump = $35,000
Installation cost = (0.75)($35,000) = $26,250
P = $35,000 + $26,250 = $61,250
i = 10%, n = 3 years
CC = 61,250 (1+0.1)3 = $246,295
(1+0.IY -1
Capitalized Cost = $246,2951
Years AlP
10 0.15
15 0.13
10.19 Compare using EAOC
EAOC =FCl(AlP,i,nj + YOC
(
(O.l1Xl + 0.11)6 JEAOCRVF = 15 ( )6 + 3 = $6.55 thousand/yr
1+0.11 -1
[
(0.11)(1+0.11)8J
EAOCFP =10 8 +5 =$6.94 thousand/yr
(1+0.11) -1
(
(0.11)(1 + O.l1io J
EAOCHe = 25 10 + 2 = $6.25 thousand/yr
(1+0.11) -1
IChoose Hydrocyclone & CentrifugeI
10-16
EAOC
($thousand/yr)
25.38
22.09

10.20 Incremental analysis: INPV= -FCI+ (PIA,i,n)YS
INPV=-250+70 (1+0.07Y -1 =$127.25 thousand
0.07(1 +0.07Y
INPV= $127.25 thousand
Purchase and install the baghouse filter.
10.21 INPV=-FCI + (PIA,i,n)YS
Project
FCI Cash Flow
PIA
INPV
($ million) ($ million) ($ million)
A 80 11 7.54 2.91
B 100 14 7.54 5.53
C 120 16 7.54 0.60
A&B 180 25 7.54 8.44
A&C 200 27 7.54 3.52
B&C 220 30 7.54 6.13
ALL 300 41 7.54 9.04
Projects A and B provide the greatest NPV and meets the capital ceiling for investment of
$250 million.
IChoose Projects A and B I
10.22 INPV= -FCI+ (PIA,i,n)YS
Project
FCI Yearly Savings
Years PIA
INPV
($ thousands) ($ thousands) ($ thousands)
Solar 25 2 15 9.40 -6.19
Insulation 5 0.9 15 9.40 3.46
Both 30 2.5 15 9.40 -6.49
IChoose insulation only. I
Tax credit needed for solar collector:
When INPVSolar = 0, FCIsolar = $18.8 thousand.
Tax credit = 1-(18.8/25) = 0.248
Tax credit = 25%
10-17

10.23 (a) ROROI1= Incremental Yearly Savings / Incremental Investment
Case
FeI Cash Flow
ROROn($ million) ($ million)
Base 75
1 15
2 25
3 30
IBase case is the best option.I
(b) 1NPV=-FC1+ (PIA,i,n)YS
FC!
($ million)
Case
Cash Flow
($ million)
base 75 19
1 15
2 25
3 30
ICase 3 is the best option. I
10.24 (a) NPV= -FC1+ICFlP I F,i,n)
INPV= -$47.7 thousand; Do not invest.I
(b) When NPV= 0, i = 5.01%
Ibreak-even i = 5.01% I
3
5
7
19 0.253
3 0.244
5 0.240
7 0.248
PIA
INPV
($ million)
5.02 20.36
5.02 20.41
5.02 20.45
5.02 25.49
(c) When NPV = 0, for profitable investment FC1 = $452.3 thousand
IFC1 = $452.3 thousand I
10-18

10.25 Look at incremental cases.
ROROII= Incremental Yearly Savings / Incremental Investment
ROROIIBase +l-Base = (0.77-0.75)/(5.1-5) = 0.20
See table for complete ROROII calculations.
Case
FCI
($ million)
Base 5
Base + option 1 5.1
Base + option 1 + option 2 5.3
Base + option 3 5.2
Base + option 1 + option 3 5.4
IChoose Base case + Option 11
10.26 EAOC = -FCI(AlP,i,n) +YS
Cash Flow
($ milliol!l
0.75
0.77
0.79
0.782
0.81
{
0.12(1+0.12Y2 JEAOC =(- 49.6 ( y2 + 8 =-$0.007 thousand/yr
1+0.12 -1
IEAOC = -$0.01 thousand/yr; Borderline investment
10-19
ROROII
0.200
0.100
0.120
0.133

10.27 (a) Base Case
NPV= -FC]+CFlP / A,i,n)
NPV =-21+(12-5.1-1.3l( (I+
X
.18Y - ") = $1.63 million
0.18 1+0.18
!NPV= $1.63 million; Recommend construction I
(b) ]NPV= ~FC] +YS(P / A,i,n)
Alternative 1
INPV = -2.15 + (O.08{ ( (I + X-
18Y
- .) =-$1.82 million
0.18 1+0.18
INPV = -1.35 +(o.oi((I +X-
18Y
- .) =-$1.19 million
0.18 1+ 0.18
IDo not recommend alternatives I
(c) INPV= 0 = -3.26+4.08YS
!YS= 0.80 I
10-20

10.28 (a) All values are in $ millions.
9 9 9 9 9 13
~6r~----,----,-ff~---,--l,O~ 1 2 3 4 5 6 7 8 9 10 11 12
-3
-15 -13
(b) All values are in $ millions.
o 1 2 3 4 5 6 7 8 9 10 11 12
t
-3
-15 1
~ ~ -1~.4 -8~5-13.8 -12.0
-18.6 -16.0
-23 9' -21.6
29 8
-26.6 .
-33.6 - .
(c) NPV=-$8.54million
Do not recommend construction of plant.
10-21

10.29 INPV= -FeI+YS(P / A,i,n)
LNPV ~-1.7S+(YS{ (1+0.18)' -I J~o
(0. 18X1 + 0.18Y
IYS = $0.429 millionlyr I
LNPV = -Fe! +(2.2S{ (I +O.1S)' -I J~ 0
(0.lSX1+0.15)6
IFeI= $8.515 million I
{CI+i)' -IJINPV =-5.1 + (0.9 (iX1 + iY =0
Ii<10.4% I
10-22

10.32
Cash Flow
Year
($ million)
0 -5.00
1 -4.00
2 2.00
3 2.00
4 2.00
5 2.00
6 2.00
7 2.00
8 2.00
9 2.00
10 2.00
11 2.00
12 2.00
13 3.00
NPV= $0.24 million
Recommend investment
Discounted
Case Flow
($ million)
-5.00
-3.42
1.46
1.25
1.07
0.91
0.78
0.67
0.57
0.49
0.42
0.36
0.30
0.39
10-23
Cumulative
Cash Flow
($ million)
-5.00
-8.42
-6.96
-5.71
-4.64
-3.73
-2.95
-2.28
-1.71
-1.23
-0.81
-0.46
-0.15
0.24

10.33 (a) ROROII= Incremental Yearly Savings I Incremental Investment
Alternative
FeI Yearly Savings
PIA ROROII
INPV
($ million) ($ million) ($ million)
1 2.25 0.65 3.50 0.29 0.02
2 3.45 0.75 3.50 0.22 -0.83
1&2 5.70 1.40 3.50 0.25 -0.80
All ROROIIvalues are greater than 18%; therefore, all alternatives are good.
IAll alternatives will work. I
(b) INPV= -FeI+YS(P / A,i,n)
See above table for INPV values.
IRecommend only Alternative 1·1
(c) Recommend only Alternative 1 because the INPV value is positive.
IRecommend only Alternative 1.1
10-24

10.34 (a) All values are in $ millions.
4 4 4 4 4 4 4 4 4 5
t t t t t t t t t j
oJ II 2
...,
4 5 6 7 8 9 10 11.:)
-10
-11
(b) All values are in $ millions.
0.65
0 1 2
...,
4 5 6 7 8 9 10 t 11.:)
t ~ ~ ~ +
-2.08 -0.79
-10 -6.95 -5.l4 -3.52
-11.2 -8.97
-19.8 -16.6 -13.8
(c)
Cash Flow
Discounted Cumulative Disc.
Year Cash Flow Cash Flow
($ million)
($ million} ($ millionl
0 -10 -10.00 -10.00
1 -11 -9.82 -19.82
2 4 3.l9 -16.63
3 4 2.85 -13.79
4 4 2.54 -11.24
5 4 2.27 -8.97
6 4 2.03 -6.95
7 4 1.81 -5.14
8 4 1.62 -3.52
9 4 1.44 -2.08
10 4 1.29 -0.79
11 5 1.44 0.65
INPV= $0.65 million I
(d) For the NPV=$2 million, profit =$4.27 million
(e) !NPV= -Fe!+YS(P / A,i,n)
lNPV =-3+YS( (1+0.16)' -1 )=0
0.16(1 +0.16Y
IYS= 0.916 millionlyr I
10-25

10.35 EAOC = FCI(AJP,i,n) + YOC
(
0.14(1+0.14Y)EAOCA =5 ( )3 +5.5 = $7.65thousand/yr
1+0.14' -1
(
0.14(1 + 0.14)5)
EAOCs =15 ( )5 +3.5=$7.87thousandlyr
1+0.14 -1
(
0.14(1 + 0.14Y)EAOCc =5 ( )9 +2.5 = $7.15thousandlyr
1+0.14 -1
IChoose Alternative C I
10.36 EAOC = FCI(AJP,i,n) - YS
(
0.15(1 + 0.15)8)
EAOCA =88 ( Y -(15+25+33)=-$53.39thousand/yr
1+0.15 -1
EAOCB =125(°(1+ °tl' )-(15+45+25)= -$57.14thousand/yr
1+0.15 -1
EAOCc = 250(Ot(1+~15l' )-(15+ 75+18)= -$52.29thousandlyr
1+0.15 -1
IChoose Alternative B I
10-26

10.37 (a) ROROII= Incremental Yearly Savings / Incremental Investment
ROROII =($32.5 -$15) = $17.5 =0.17
A $103.7 $103.7
ROROII =($65-$30) = $35 =0.22
B $162.4 $162.4
ROROII = ($74.75-$34.5) = $40.25 =0.19
c $216.1 $216.1
From above, choose Case B. However, can also do incremental comparison
between cases.
Use Case A as base case because it has the lowest FCI and evaluate ROROII going
from Case A to Case B.
ROROII = ($35-$17.5) =0.298
A-B ($162.4-$103.7)
ROROII is greater than 15%, now use Case B as base case and compare to Case C.
ROROII = ($40.25-$35) = 0.098
B-C ($21~.1-$162.4)
Reject Case C because ROROIIis less than 15%.
IChoose Case B I
(b) EAOC=FCI(AIP,i,n)-YS
(
(0.05X1+ 0.05Y J
EAOCA =$103.7 ( )7 -$17.5 = $0.42 thousandlyr
1+0.05 -1
EAOCB =$162.4((OtX1 +);05)' J-$35 = -$6.93 thousandlyr
1+0.05 -1
EAOCc =$216.1(((·05Xl+~.05)' J-$40.25 =-$2.90 thousand/yr
1+0.05 -1
IChoose Case B I
10-27

(c) EAOC = FCI(AIP,i,n) - YS
(
(0.05X1+ 0.05Y5 )EAOCA =$103.7 ( )15 -$17.5=-$7.51 thousand/yr
1+0.05 -1
EAOCB = $162.4(Ot)(J+~;05Y' )-$35 = -$19.35 thousand/yr
1+0.05 -1
(
(0.05X1+0.05f J
EAOCc =$216.1 ( )15 -$40.25 =-$19.43 thousand/yr
1+0.05 -1
IChoose Case C I
(d) EAOC = FCI(AIP,i,n) - YS
(
(0.05X1+ 0.05Y J
EAOCA = $103.7 ( )5 - $17.5 = $6.45 thousand/yr
1+0.05 -1
EAOCB =$162.4(OtXl+~.05Y J-$35 =$2.51 thousand/yr
1+0.05 -1
EAOCc =$216.1(OtX1 + ~.05Y)-$40.25 =$9.66 thousand/yr
1+0.05 -1
None ofthe cases provide an acceptable EAOC.
10-28

10.38 Individual solutions may differ slightly because ofthe Monte Carlo method being used.
Probable Variation of Key Parameters over Plant Lift
lowerLimtt Ugger Limit Base Value
FCll (JOk (JOk $ 120,000,000
Price of Product -13% 3% $ 52,000,000
Working Capital (JOk 0% $ 35,000,000
Income Tax Rate' (JOk 0% 4(JOk
Interest Rate' (JOk 0% 9%
Raw Material Price -11% 19% $ 18,000,000
Salvage Value (JOk 0% 1
, Please note that variations for percentages are a percent of a percent.
For example, a 1(JOk variance on a 12% interst rate would imply a 1.2% uncertainty
Run Econorric Analysis
Net Present Value Data
lowNPV -24.8 (f) 1000
High NPV 10.5 C
'0
Il.
Bins UllllerValue # Ilointslbin Cumulative ..;§ 750
0 -24.8 0 0
1 -21.2 8 8 b
2 -17.7 23 31 t.c 500
3 -14.2 49 80 E
"4 -10.6 104 184 z
5 -7.1 174 358 ~ 250
6 -3.6 193 551 'iii
"9
7 -0.1 218 769 E
8 3.5 150 919 "u
9 7.0 67 986 ·30 -25 -20 ·15 -10 -5 0 10
10 10.5 14 1000
Net Present Value (millions of dollars)
Discounted Cash Flow Rate of Return Data
'" 1000
Low DCFROR 0.06 C
High DCFROR 0.11
'0
Il.
.I!I 750
'"Bins Ullper #/bin Cumulative Cl
0 0.06 0 0 'E~
1 0.07 8 8 Q)
.c 500
2 0.07 17 25 E
::>
3 0.07 44 69 z
4 0.08 99 168
Q)
,i; 250
5 0.08 149 317 '":;
6 0.09 205 522 E
::>
7 0.09 203 725 U
8 0.10 187 912
0
000 0.02 0.Q4 0.06 0.06 010
9 0.10 69 981
DCFROR
10 0.11 19 1000
IWith at least a 75% chance that the NPV<O, this might not be a good decision.I
10.39
Lowest NPV = -$24.8 million ~ approximately 1%
HighestNPV = $10.5 million ~ approximately 10%
10-29
15
012

10.40
Monte Carlo Analysis for (using MACRS 5 yrsl
Probable Variation of Key Parameters over Plant Liff
Lower Limn Upper Limit Base Value
FCIL ·14% 21% S 7,700,000
Price of Product ·15% 10% S 4,100,000
Worldng Capital ook 0% $ 2,000,000
Income Tax Rate" ook 0% 40%
Interest Rate' ·14% 9% 11%
Raw Material Price ·11% 32% $ 1,900,000
Salvage Value 0% 0% 1
'Please note that variations for percentages are a percent of a percent.
For example, a 100k variance on a 12% interst rate would Imply a 1.2% uncerialnty
Run Economc AnalySIS
LowNPV
High NPV
Bins
o
1
2
3
4
5
6
7
8
9
10
·3.6
1.9
Upper Value
·3.6
·3.0
·2.5
·1.9
.1.4
·0.8
·0.3
0.3
0.8
1.4
1.9
# pointslbin
o
4
19
62
141
202
246
197
99
26
4
Cumulative
o
4
23
85
226
428
674
871
970
996
1000
., 1000
C
'15
n.
~ 750
'0
t;;
.., 500
E
~
:;
~
-4
From the above figure, the chance ofgetting an
·3 ·2 ·1 0 1
NPV < 0 is 75-80%. Probably not a wise investment opportunity.
10-30

10.41 (a)
A i(l+it
P - (1+it-1
A. = $250,000 (0.085X1+0.085ys =$30,105.12
HIgh (1+0.085YS-1
A = $250,000 (0.06X1 +0..06Ys = $25,740.69
Low (1 + 0.06)b -1
AHigh = $30,105.12
ALolI' = $25,740.69
(b)l(c)
Variable Rate Simulation
Year i
Payment Interest Principal Paid
($/y) (Sly) (Sly)
1 0.0765 24,795 19,114 5,681
2 0.0664 23,010 16,230 6,779
3 0.0727 24,078 17,271 6,807
4 0.0757 24,577 17,471 7,106
5 0.0694 23,583 15,526 8,057
6 0.0720 23,968 15,521 8,447
7 0.0734 24,164 15,196 8,968
8 0.0611 22,528 12,108 10,420
9 0.0600 22,397 11,271 11,126
10 0.0836 25,170 14,761 10,409
11 0.0802 24,788 13,325 11,463
12 0.0717 23,925 11,100 12,825
13 0.0670 23,488 9,505 13,983
14 0.0783 24,432 10,022 14,410
15 0.0623 23,252 7,071 16,180
16 . 0.0807 24,421 7,850 16,570
17 0.0786 24,312 6,352 17,960
18 0.0704 23,951 4,423 19,528
19 0.0727 24,028 3,147 20,880
20 0.0660 23,878 1,479 22,399
Total Payment: $478,742
10-31
Remaining Balance
($/y)
244,319
237,539
230,732
223,626
215,569
207,121
198,154
187,733
176,607
166,198
154,735
141,910
127,927
113,517
97,337
80,767
62,807
43,279
22,399
0

Fixed Rate Simulation
Year i
Payment Interest Principal Paid Remaining Balance
J$/y) ($/y) ($/y) ($/y)
1 0.075 24,523 18,750 5,773 244,227
2 0.075 24,523 18,317 6,206 238,021
3 0.075 24,523 17,852 6,671 231,349
4 0.075 24,523 17,351 7,172 224,178
5 0.075 24,523 16,813 7,710 216,468
6 0.075 24,523 16,235 8,288 208,180
7 0.075 24,523 15,613 8,910 199,270
8 0.075 24,523 14,945 9,578 189,693
9 0.075 24,523 14,227 10,296 179,396
10 0.075 24,523 13,455 11,068 168,328
11 0.075 24,523 12,625 11,898 156,430
12 0.075 24,523 11,732 12,791 143,639
13 0.075 24,523 10,773 13,750 129,889
14 0.075 24,523 9,742 14,781 115,107
15 0.075 24,523 8,633 15,890 99,217
16 0.075 24,523 7,441 17,082 82,136
17 0.075 24,523 6,160 18,363 63,773
18 0.075 24,523 4,783 19,740 44,033
19 0.075 24,523 3,302 21,221 22,812
20 0.075 24,523 1,711 22,812 0
Total Payment: $490,461
10-32

10.42
Monte Carlo Analysis for Example 10.1 (using MACRS 5 yrs)
Probable Variation of Key Parameters over Plant Life
Lower Limit UggerLimit Ba~Value
FCIL -20% 35% $ 150,000,000
Price of Product -25% 10% $ 75,000,000
Working Capital 0% 0% $ 30,000,000
Income Tax Rate' 0% 0% 45%
Interest Rate' 0% 0% l00h
Raw Material Price -10% 25% $ 30,000,000
Salvage Value 0% 0% 10000000
• Please note that variations for percentages are a percent ofa percent.
For example, a 10% variance on a 12% interst rate would imply a 1.2% uncertainty
Bun economic Analysis I
LowNPV
High NPV
Bins
o
1
2
3
4
5
6
7
8
9
10
-63.9
43.3
UggerValue
-63.9
-53.2
-42.5
-31.8
-21.0
-10.3
0.4
11.1
21.8
32.5
43.3
# pointslbin
o
7
16
51
108
137
172
211
190
84
24
Cumulative
o
7
23
74
182
319
491
702
892
976
1000
~ 1000
G;
~ 500
:J
Z
.,
.i; 250
til
'5
E
8
.so
10-33
·60 ·40 ·20 0 20
40 60

10.43 (a) NPV=-FC!+(~'i,n}R-Cwr-CIIM-CUT)
NPV = -5+( (I+X-
15YO
-~OJ(4.84-0.506- 2.2-1.32)= -$0.914 million
0.15 1+ 0.15
1This process is not profitable. 1
(b) When NPV = 0, number of batches = 26.9 batches/yr
127 batches/yr 1
(c) When NPV = 0 and number ofbatches = 26/yr, i = 14.1%
li=14.1%1
10.44 (a) INPV= -FC!+YS(P / A,i,nj or EAOC= FC!(AlP,i,nj - YS
INPV = -750,000+(1,321,125 -950,000)( (I +x-17
)' -)5J(1.25) =$734,196
0.17 1+0.17
EAOC = (750,000{(0.17X1 +0.17Y J(1.25)= -$229,483
(1+0·17Y-1
IRecommend the process I
(b) INPV =-750,000+ (1,321,125 -950,000)( (1+X-17)5-)5J(x) = $734,196
0.17 1+0.17
x = 1.12
ICapacity increase of 12% I
(c) When FC! = $750,000, capacity increase is 25%, and NPV= 0
i = 55%
IDCFROR is reached I
10-34

10.45 (a) INPV= -FCI +YS(P I A,i,n) or EAOC =FCI(AIP,i,n) - YS
INPV = -5.1 + (2.75 -1.5)(t;_2)'-;,)(1.20)= -$0.11 million
0.2 1+0.2
EAOC ~(5./ (~.2Xl+r)6 )(1.2) = $0.034 million
1+0.2 -1
IDo not recommend this process. I
(b) When INPV = 0 or EAOC = 0, FCI = $4.99 million
IFCI= $1.66 million/tank I
(c) When INPV = 0 or EAOC = 0 and FCI = $2 million/tank, i = 0.13
IDCFROR = 13%
10-35

Chapter 11
11.1 For the ethylbenzene process shown in Appendix B, check the design specifications for the
following three pieces of equipment against the appropriate heuristics, P-301, V-302, T-
302. Comment on any significant differences that you find.
P-301 - From Table B.2.3
P-301 AlB, Carbon steel- positive displacement, Actual power = 15 kW, Efficiency 75%
From Table 11.9, Heuristic 1
Power for pumping liquids: kW = (1.67)[Flow(m3/min)][i1P(bar)]/8
E = Fractional Efficiency = Esh (see Table 11.5)
density of benzene at 58.5°C = 875 kg/m3
mass flow ofbenzene through pump = 17952.2 kg/h (Stream 3 - Table B.3.1)
Flowrate = 17952.2/875 = 20.5 m3
/h = 0.342 m3/min
M> = 2000 -110 = 1890 kPa = 18.9 bar
Ptheoretical = (1.67)(0.342)(18.9) =10.8 kW
From Table 11.9 - Heuristic 7
Efficiency 70% at 7.46 leW (10 hp), 85% at 37.3 kW (50 hp)
8sft ~ 75%
Pactllal =10.8/0.75 = 14.4 kW
Good agreement with heuristics
V-302 - From Table B.2.3
V-302, Carbon steel with SS demister, vertical, LID = 3, V= 10 m3
, Maximum operating
pressure = 250 kPa
Optimum length/diameter = 3, but the range 2.5 to 5 is common.
I UseL/D=3
Holdup time is 5 min for half-full reflux drums and gas/liquid separators, 5-10 min for a
product feeding another tower.
Since LlV separator feeds a tower use a hold up of liquid equal to 10 min for half-full
drum
11-1

Properties of Stream 15
Vapor
Flowrate = 1038 kg/h
Density = 2.169 kg/m3
Vol flow = 478.6 m3
/h
Liquid
Flowrate = 24345.9lQih
Density = 821,3 kg/mJ
Vol flow = 29.64 m3
/h
. (29.64)(10)
Volume of drum V = = 10m3
, (60)(0.5)
T-302 - From TableB.2.3
T-302, carbon steel, 76 SS sieve trays plus reboiler and total condenser, 45% efficient trays,
feed on tray 56, additional feeds ports on 50 and 62, reflux ratio =0.6608, 15 in tray spacing,
column height 28.96 m, diameter = 1.5 m, maximum pressure rating of300 kPa
From Tables 11.13 and 11.14- Heuristics
Table 11.13
Rule 5: Optimum reflux in the range of 1.2 - 1.5 Rl1lill
Rule 6: Optimum number of stages approximately 2NI1lill
Rule 7: Nl1lill = In{ [xl(l- x)]ovhi[x/(l- x)hot}/ln a
Rule 8: Rl1lill = {FID}/(a - 1)
Rule 9: Use a safety factor of 10% on number oftrays
Table 11.14 ---+ Rule 2: Fs = up"O.5 = 1.2 -+ 1.5 m/s(kg/m3)o.5 and Rule 4: Stra)' = 60 - 90 %
For our case we have
Stream 18 19 20
Temp °C 145.4 139.0 191.1
Pres kPa 120.0 110.0 140.0
Vapor mole
fraction 0.0 0.0 0.0
Total kmollh 101.1 89.9 11.3
Total kg/h 11024.5 9538.6 1485.9
Flowrates in
kmollh
Ethylene 0.00 0.00 0.00
Ethane 0.00 0.00 0.00
Propylene 0.00 0.00 0.00
Benzene 0.17 0.17 0.00
Toluene 0.00 0.00 0.00
Ethylbenzene 90.63 89.72 0.91
1,4-DiEthBenzene 10.35 0.0001 10.35
11-2

Key components are ethylbenzene and 1,4-DiEthylbenzene. The formulae for Nllli1h etc.
should be based on these key components, i.e., use a benzene free basis.
Xovhd = 89.72/89.7201 = 0.999999, Xbo! = 0.91111.3 = 0.08053, Uovhd = 3.83, UbOl= 3.19
ugeol11 {Jve = (Uovh{[UboaO.
5
= 3.50
Mllil1 = In{ [0.999999/(1 - 0.999999)]/[0.08053/(1 - 0.08053)]} /In (3.50) = 12.9
Rl11il1 = (101.1I89.9}/(3.5 -1) = 0.453
Range of R = (1.2 ~ 1.5)Rlllill = 0.544 ~ 0.680
1 RT-302 =0.6608 1 Within range
Nfheorefical;:::; (2)(12.9) = 25.8
8tray = 0.60 - 0.90
ET-302 = 0.45 ILower than typical range
Nacfllal;:::; (25.8/0.45)(1.1) = 63 trays
NT-302 =76
Pv = 3.546 - 3.879 ~ use 3.71 kg/m3
u = (1.2 ~ 1.5)/3.71°.5= 0.62 ~ 0.78 m/s
Vapor flow rate (stream 19) = 9539 kg/h
Vol. flow rate, v = 0.714 m3
/s
High by 13 trays
Dtower = [4v/nu]0.5 = [(4)(.714)/(3.142)/(0.62 ~ 0.78)]°·5 = 1.08 ~ 1.21 m
IDT-302 =1.5 m I Higher than above range for D
Overall, for T-302, the agreement with the heuristics is fair. This is probably due, in
part, to the use of a fairly low reflux ratio that is at the bottom of the typical range
given in the heuristics. This tends to increase the number of trays. In addition, the tray
spacing used in T-302 is only 12 or IS" which is significantly lower than the standard
spacing of24". This has the effect of reducing the tray efficiency and requires a larger
diameter column to stay away from flooding. These differences probably account for
the discrepancies in the above results.
11-3

11.2 For the styrene process shown in Appendix B, check the design specifications for the
following tlu'ee pieces of equipment against the appropriate heuristics, E-401, C-401, T-
402. Comment on any significant differences that you find.
8:
~I-
E-401 - From Table B.3.3
carbon steel, A = 541 m2
, boiling in shell, condensing in tubes, 1 shell - 2 tube passes, Q
= 29,695 MJ/h
From Table B.3.1 and Figure B.3.1, it can be seen that E-401 heats Stream 2 from 116°C
to 225°C using high-pressure steam at 254°C. The T-Q diagram from Chemcad is given
below:
240
Zl5
230
225
220
215
210
205
200
195
190
185
180
175
1711
165
160
155
150
145
140
135
130
125
120
115
110
-t-
.~ I I I I I
·11~+rl I~
I I I I I I I.. I I
+II I -LLH I
- -
- .
I I! I I I I
-
It- I I ! I I
, II I I I -,I
II I I
II I
II I ,
I ! I ! I I I I I I ! :
I. I I I I I I r I
I I I I 1.1 I I
I I II II I , I I I I
-I
I I III II II I I I I
I I
II II ! I I
I t
l I I I I J I I, ~
I I
I I.
I I I ! I I I II I
11
I
I I I I I , I
I V I I I , ! I I
I I I I I I
I
I I I T
I I,
-,
I I I
II I I I I I I I
-1 i I I
ffi~
I I
-+
I I
I I I I I I I I I
I I I I I I I I I I I I I , I I
o 1000 2500 4000 5500 7000 8500 10000 11500 13000 14500 16!XXJ 17500 lSOOJ 20500 22000 23500 25000 25500 28000 29500 31000
This is a 3 zone HX.
QJ = 5,500 MJ/h
Q2 = 23,000 - 5,500 = 17,500 MJ/h
Q3 = 29,700 - 23,000 = 6,700 MJ/h
Delta H MJJh
Heat transfer coefficients for estimating purposes, W/m2
°C: water to liquid, 850;
condensers, 850; liquid to liquid, 280; liquid to gas, 60; gas to gas 30; reboiler 1140.
Maximum flux in reboiler 31.5 kW/m2.
So choose Uj = 500 (liq - cond steam). U2 = 850 (boiling liq - condensing steam, U3 =
60 (condensing steam - gas)
11-4

[(254-115) -(254-163)] °
/5,.1/111-1 = (254 -115) = 113.3 C
In-'------'-
(254-163)
/5,.TiI11-2 = (254-163) = 91°C
_ [(254-163)-(254-225)] _ 0c
/5,.Tim-3 - (254 -163) - 54.2
In-'----~
(254-225)
Using a value ofF = 1 - since all zones have a phase change, we get
Al = Ql = (5,500x10
6
)
U1/5,.Tim-1 (3600)(500)(113.3)
6
A2 = Q2 = (17,500x10 ) =62.8m2
U2/5,.Tim-2 (3600)(850)(91)
A3 = Q3 = (6,700x10
6
) = 572.3m2
U3/5,.Tim-3 (3600)(60)(54.2)
3
Atotal =LAi =662.1m
2
i=l
IAE-401 =541 m
2
Agreement is within 20% of heuristic - OK
C-401- From Table B.3.3
carbon steel, W= 364.2 kW, 80% adiabatic efficiency
Feed stream = Stream 16
T= 65°C, P = 0.75 bar, 111 = 2682 kg/h, mw = 12.4 kg/kmol
POll! = 2.4 bar
Theoretical reversible adiabatic power = mzIRT1[({P2IP1}(/ -1)]la
where TI is inlet temperature, R = Gas Constant, Zj = compressibility, 111 = molar flow
rate, a = (k-l)lk and k = C/Cv, R = 8.314 .T/mol K
assume k = 1.4, a = 0.2857
P . = 2682x10 (1)(8.314)(273.2+65) 2.4 -1 =233085W=233kW3 [( )0.2857 ]
rev-adtab (12.4)(3600) 0.2857 0.75
11-5

Efficiencies oflarge centrifugal compressors is about 76 - 78%
PaCflla/= 233/0.77 = 303 kW
&C-401 =0.80
PC-401 =364 kW
Compressor appears to be somewhat oversized (~20%), although the estimate is in
the ballpark. This may be to allow for future expansion or could be an error in the
design calculations.
T-402- From Table B.3.3
carbon steel, total condenser (E-409), feed at location equivalent to tray 36, reflux ratio =
25.8, structured packing, Cf = 1, diameter = 4.1 m, HETP = 0.3 m, height = 34.5 m
From Tables 11.13 and 11.14- Heuristics
Table 11.13
Rule 5: Optimum reflux in the range of 1.2 - 1.5 Rmil1
Rule 6: Optimum number of stages approximately 2N,llill
Rule 7: N,lIill = In{ [x/(1 - X)]ovhc/[X/(l - x)]bof}/ln a
Rule 8: Rmill = {F/D}/(a - 1)
Rule 9: Use a safety factor of 10% on number of trays
Table 11.14
Rule 2: Fs = upvO.
5
= 1.2 ---+ 1.5 m/s(kg/m3)o.5
Rule 4: 8frc~1' = 60 - 90 %
Height equivalent to theoretical stage (HETS) for vapor-liquid contacting is 0.4-0.56 m
for 2.5 cm (1 in) pall rings and 0.76-0.9 m. for 5 cm (2 in) pall rings.
Structured packing is generally more efficient than loose packing so an HETP (HETS) =
O.3m is reasonable.
Assuming a headspace of 1.5 m above the packing and a liquid level of 3m at the bottom
Number ofequivalent theoretical plates = (34.5 - 4.5)/0.3 = 100 plates
Using data from Table B.3.3 for the feed and product streams for T-402, we get the
following
11-6

Stream No. 22
Temperature (OC) 119.5
Pressure (kPa) 60
Vapor Mole 0
Fraction
Total Flow (kg/h) 47905
Total Flow (kmol/h)
Component Flows
Water
Ethylbenzene
Styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
Using a toluene free basis, we have
Xovhd = 332.66/(332.66+1.20) = 0.9964
Xhot= (0.34)/(119.3 + 0.34) = 0.002842
453.9
333.0
120.53
0.33
23 24
105 124.5
210 60
0 0
35473 12432
334.2 119.7
332.66 0.34
1.20 119.3
0.33
From the Chemcad output we get the following values for relative volatility
atop =1.19
abot = 1.22
aave = [(1.19)(1.22)]Y2 = 1.205
Nlllil1 = In{ [x/(l - X)]ovhd/[X/(l - x)hot}/ln a
In xovhd (1- Xbot) In (0.9964)(1- 0.002842)
N
min
= (1~oVhd)Xbot - (1-0.:64)60.002842) =61.7
aave 1.2 5)
Rlllil1 = {F/D}/(a - 1)
R . = (453.9) =663
111111 (334.2)(1.205 -1) .
Using heuristics we get
R = (1.2 - 1.5)Rmin = 7.95 - 9.94
N = 2Nmin = 123
RT-402 =25.8
NT-402 = 100
11-7

The reflux ratio for this column is much higher than the value from the heuristic
while the number of trays is lower by ...,20%. This column appears to be running
inefficiently and should be investigated since the reboiler and condenser duties are
probably very high and wasteful in utilities. One possible explanation is that the
higher value ofR is chosen to keep the height of the tower down but this seems to be
overkill. Tower ....eeds investigating!
11-8

11.3 For the drying oil shown in Appendix B, check the design specifications for the following
three pieces of equipment against the appropriate heuristic V-501, P-501, H-501. Comment
on any significant differences that you find.
P-501 AlB - From Table B.4.3
Centrifugal, Carbon steel, Power = 0.9 kW (actual), 80% efficient, NPSHR at design flow
= 14 ft ofliquid
From Table 11.9
1. Power for pumping liquids: kW = (1.67)[Flow(m3
/min)][.6..P(bar)]/s, s = Fractional
Efficiency = Esh (see Table 11.5)
2. Net positive suction head (NPSH) of a pump must be in excess of a certain number,
depending upon the kind ofpumps and the conditions, if damage is to be avoided.
NPSH = (pressure at the eye of the impeller -vapor pressure)/(pg). Common range is
1.2-6.1 m of liquid
4. Centrifugal pumps: Single stage for 0.057-18.9 m3
/min, 152 mmaximum head;
multistage for 0.076-41.6 m3
/min, 1675 m maximum head. Efficiency 45% at 0.378
m3
/min, 70% at 1.89 m3
/min, 80% at 37.8 m3
/min.
From Table 11.8
4. Control valvesrequire at least 0.69 bar (10 psi) drop for good control.
Inlet stream is Stream 2 from Figure B.4.1 and Table B.4.1
Liquid
Flowrate = 10,703 kg/h
Vol flow = 13.49 m3
/h = 0.2248 m3
/min
p] = 105 kPa = 1.05 bar
P3 = 230 kPa = 2.30 bar (Stream 3)
Note: .6..Pplllllp = P3 - p] + Mer
P= 1.67V.6..P = (1.67)(13.49/60)(2.30-1.05+0.69) = 0.728
kW
&
From Table 11.9 - heuristic 4, the efficiency is ~ 45%. This is much lower than the
actual efficiency of 80%. Using 8 = 80% we get:
P =0.728 =0.91kW
0.8
11-9

Excellent agreement with actual power, but the efficiency looks to be too high for
such a small pump. This may be a mistake or a high efficiency pump has been
specified for this service.
H-50l - From Table B.4.3
total heat duty required = 13219 MJ/h =3672 kW, design capacity =4000 kW, Carbon
steel tubes, 85% thermal efficiency
Fired heaters: radiant rate, 37.6 kW/m2; convection rate, 12.5 kW/m2; cold oil tube
velocity = 1.8 m/s approximately equal transfer in the two sections; thermal efficiency
70-90% based on lower heating value; flue gas temperature 140-195°C above feed inlet;
stack gas temperature 345-51 O°C
Only thing to check here is the thermal efficiency which lies in the acceptable range,
namely 70%<85%<90% - good agreement with heuristics.
V-SOl - From Table B.4.3
V-50l, Horizontal, Carbon steel, LID = 3, V= 2.3 m3
Optimum length/diameter = 3, but the range 2.5 to 5 is conUllon.
! useL/D=3!
Holdup time is 5 min for half-full reflux drums
Liquid
Flowrate = 10,703 kg/h
Vol flow = 13.49 m3
/h
(13.49)(5)
Volumeofdrum V= =2.25m3
, (60)(0.5)
11-10

Chapter 12
12.1 When assigning this problem, the students should be told whether a generic
answer or a· more specific answer is requested. If the more specific answer is required,
the problem involves literature searches and involves digging into the process
simulator's databank. For some process simulators, the latter is easy, whereas for others
it is nearly impossible.
a) The physical property data needed are:
• pure-component heat capacities
• pure-component densities
• equation ofstate (and its parameters) and/or liquid-state activity coefficient model
(and its parameters)
1& thermal conductivities
• viscosities
• pure-component vapor pressures
b) Sources ofthese data are:
• process simulator databank or handbooks
• process simulator databank. or handbooks for critical properties and for some
activity coefficient model parameters. Literature data for vapor-liquid equilibria (or
liquid-liquid equilibria) from which model parameters can be regressed.
• process simulator databank. or handbooks
12-1

12.2 The U.S. patent literature can be searched at any patent and trademark depository
library on-line (1971 to present) or in the hardcopy indices (1790 to present). By
Internet, one can use http://patents.uspto.gov/ to search the front page of U.S. patents
from 1976 to the present or http://patent.womplex.ibm.comlibm.htm1to search U.S.
patents 1971 to present (full text and images). Some similar sources are available for
patents granted by other countries. Note that many ofthe useful patents are more than a
few years old.
Most such patents include only sketchy experimental data, and then only the bare
minimum. For example, the size of the reactor (and catalyst charge), total inlet fiowrate
and composition, and incomplete productivity data may be presented. For multiple
runs, the temperature and/or the inlet composition may have been varied-but maybe
not. Space velocities are common.
The process simulator, on the other hand, will require the order with respect to each
reactant, the frequency factor (pre-exponential), and the activation energy for each
reaction. These data are almost never given in patents. Thus, the student must regress
these parameters from the data presented.
12-2

12.3 There are many heuristics for reactor design, including:
• Design for heat transfer, not reaction kinetics, as the former is usually controlling.
• Use a CSTR when control ofthe reactor is essential.
• For exothermic reactions, use a minimum temperature difference between the
reactor and the heat-transfer fluid, to increase controllability.
• For endothermic reactions, operate at the lowest possible temperature to minimize
heating costs.
• For exothermic reactions, operate at the highest possible temperature to maximize
the utility value ofthe hot stream produced.
• For endothermic reactions, operate at the highest possible temperature to maximize
conversion.
• For exothermic reactions, operate at the lowest possible temperature to maximize
converSIOn.
• If the byproduct reaction has a higher activation energy than does the desired
reaction, operate at low temperatures.
• If the byproduct reaction has a lower activation energy than does the desired
reaction, operate at high temperatures.
• Expect a rapid increase in temperature near the entrance to a tubular reactor for
exothermic reactions. Cut the catalyst with inactive support to create a uniform heat
release per unit length ofreactor.
• Increase selectivity at the expense of yield, as the former can drastically reduce
separation costs.
• Design for increased reactor temperature over the life ofthe catalyst, to compensate
for loss ofcatalyst activity.
Be sure to point out that heuristics are fallible and often mutually inconsistent.
12-3

12.4
a) From a capital-costs viewpoint, one wants to minimize the heat-transfer area. This is
accomplished ifthe approach temperature is increased. However, very low temperature
utilities can involve higher equipment costs.
b) From the operating-costs viewpoint, the greater the approach temperature, the
smaller the flowrate of cooling medium can be (i.e., the inlet and outlet temperature of
the cooling medium can be very different). However, the lower the cooling-medium
temperature, the lower is the value of the "cooling" medium as a heat source for heat
integration. For example, we might be able to raise high-pressure steam (at high
temperature, low approach temperature) or low-pressure steam (at low temperature,
high approach temperature). The high-pressure steam is worth more towards lowering
operating costs. Thus, there are competing effects on operating costs.
c) From the operability viewpoint, a smaller approach makes the temperature control
more stable. Ifthe approach is one degree, for example, a temperature excursion ofone
degree would double the heat transfer rate, bringing the reactor back to its design
temperature. If the temperature in the reactor suddenly decreased by two degrees, the
"cooling" medium would actually heat the reactor back up. On the other hand, if one
wanted to change the setpoint for the reactor temperature, a small approach temperature
would be a considerable problem. The temperature ofthe cooling medium would need
to be changed.
12.5 The operability of a process involving an exothermic reaction is a serious safety
concern. If the cooling system is unable to remove the heat of reaction, the reactor
temperature will rise, resulting in an increase in reaction rate. This increased reaction
rate will lead to an increased heat release rate. The exponential temperature rise that
results can lead to a catastrophic event called a ""runaway."
Thus, one tries to design an inherently stable process control system. In this case, the
small approach temperature design is preferred. In fact, if there is a sudden failure of
the control system, the reactor temperature will remain stable, at least for a little while.
12-4

12.6 The following are referred to figure numbers.
a) B1: Stream 13
B2: Stream 22
B3: Stream 14
C8: Stream 11
b) B1: Stream 16
B2: Stream 18
B3: Stream 10
B4: Streams 9, 12
C3: Streams 18,21
12.7The following are referred to figure numbers.
a) B2: Streams 5,8
B4: five cooling loops on R-501
C1: Stream 4
reflux in each distillation column
b) Bl: pump-aroundofP-201AIB
There are many other examples, but most involve similar pump-arounds for positive-
displacement pumps
c) B1: Streams 16,17
B2: Streams 15, 18
B3: Streams 10, 13
B4: Streams 9,12
C3: Streams 15, 18,21
d) return ofcondensate to boiler in every flowsheet
B1: preheater E-202
C5: bfw/steam between E-703 and E-702
Use ofsteam raised in process for the hot utility (several processes)
12-5

12.8 Recycles can also be used for:
• Reduction ofwaste streams
• Mixing ofcomponents in a vessel
• Maintaining a consistent temperature in a vessel
• Reduction ofhumidity in the air (as in the recirculation mode ofan air conditioner)
• Seeding ofa crystallizer circuit
• Seeding of microorganisms in biological unit operations.
• Control ofparticle size distribution
12.9 The required information for utilities includes:
• Cost per unit ofenergy transferred (e.g., $/GJ)
• Temperature ofutility
• Range ofacceptable utility return temperatures
• Pressures ofutility
• Maximum flowrate ofutility available
• Thermal conductivity, viscosity, and heat capacity of utility or equivalent
information to allow for heat-transfer calculations
• Heat ofvaporization and vapor pressure for phase-change utility streams
• Cost ofpumping ofutility
• Availability ofutility at site
• Credit for utility returned but unused
• Safety characteristics ofthe utility
• Operability ranges ofthe utilities
• Dependability of the utilities (i.e., freedom from interruption, temporal consistency
oftemperatures and pressures)
• Purity requirements for returned utility stream
• Corrosion properties ofutility stream
12-6

12.11 £6(. 10 , I
[z.{ v-,)] !
IJl (AI-I) (
61 so _
AI I - - ~ s:
~, .3; - f:,
12-7

!.'. 10, I12.12 (J'
~" (2 (,.I-I) !
~ ~ AiV~ ~ t-f l-f..,r",.:fw-.
t.., ?w,~:;:t I I
12-8
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12-9
. f,~ ?t.
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~) ld-e' '" { r' -t~ rH~d4 0';y.AJV.
vV'll fv'h bk-b uY. If ~t4a~~ '1 f)r~ (.,c h.w-
b) AlflfL pY'(:-l«L~ rJ~-~ ~
tA-.ttU. ~
C) IJvlf1U cL-, ;w::t r u.. 1~t.~
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d) ~tt.- ~i6~~~' Sf/}) 1t-p~
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12-10

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"* XV Data
Job: chapter
10 binary
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Date:
07/12/2001
Time: 13:51 :59

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Job: chapter
10 binary
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Date:
07/12/2001
Time: 11 :31 :55

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Job: chapter
10 binary
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Date:
07/12/2001
Time: 11 :33:01

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10 binary
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Date:
09/12/2002
Time: 13:57:57

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Job: chapter
10 binary
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Date:
09/1212002
Time: 13:58:23

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Date:
09/12/2002
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Date:
07/1212001
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Date:
07/12/2001
Time: 11 :37:02

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Date:
07/12/2001
Time: 11 :37:51

12.15 7<-'j tAIJ~~'C/~c-..,I CJt-- ~~
U ~'1 UtJI 01 VA <-
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-_ L ~ - JII.'1> ~ ~~ - /..I¥fJvl./wcU yu."-f
_. tvtrL - S')M I ~ f~ rP (/All tV r/Ac-
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12-20

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Diethyl Ether I Water at 50.00 kPa By UNIQ
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10 binary
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Date:
07/12/2001
Time: 12:45:12

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Date:
07/12/2001
Time: 13:49:02

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Date:
07/12/2001
Time: 12:44:36

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Date:
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Time: 12:44:16

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Date:
07/12/2001
Time: 12:43:53

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Job: ethers
Date:
09/12/2002
Time: 14:10:01

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10 binary
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Date:
07/12/2001
Time: 13:47:59

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Date:
07/12/2001
Time: 12:49:46

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Date:
07/12/2001
Time: 13:48:27

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Date:
07/12/2001
Time: 12:48:58

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Job: chapter
10 binary
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Date:
07/12/2001
Time: 12:48:13

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Job: chapter
10 binary
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Date:
07/1212001
Time: 13:47:25

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0.9
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0.5
0.4
0.3
0.2
0.1
0
1
I~!f 1/
~
I
././V'
Idv
I ~ .~
V
I
~ 1'--- .....
~
I /
J 1/
I I /
, ,
'I :/1
E2:--------i--------J. I
o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
X1 Mole Frac
*" XY Data
Job: chapter
10 binary
azeotropes
Date:
07/12/2001
Time: 12:47:36

12.16
a) fl.t (..-,- C(,., ....c •..1 "'" ~ 1i<JW'(j f <r-
b) ~ q)A- V~~ 1bov~~ ~~t.r
(}.~ .
~~.~~~ 7£10.1
wi- E1"-- ~ rwt: /JMAu. ~ ~ ~
g,! - 1)fT ~~.~~ ']
'PfI'~ DJ - VfI1 G" r~ ~ ~
>-t~
12-34

'i:'
Q)
.t::
ill
>-
:5Q)
E
is.........
c
o:p
()
ro
.t::
Q)
"0
~
>-'
N
I
W
VI
Residue Curve Map, Ethanol/Dimethyl Ether / Diethyl Ether by UNIQ
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
:;~~}~;l<
~~:;:~,;:;:;:>~""e',
'~l:''-;''~>'~£::~J;:::'''''::::::::""
J ,,/:<'1<> '.
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'~"
....
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r·.
........... "" T'':,.
0.2
···.1 "'"
",
0.1 ".
" .
",
0
...._' .....
I
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Mole fraction (Ethanol)
Bubble point at 1500.00 kPa Binary Azeotrope
Ethanol = 167.697 C (Ethanol,Diethyl Ether) = (0.145,0.855) at 139.331 C
Dimethyl Ether = 61.416 C
Diethyl Ether = 140.555 C
Job: chapter
10 binary
azeotropes
Date:
07/1212001
Time: 13:05:51

() >/~ ~~:f., ht EltJ.l h.:t~.~.
~~d1f-v'
'et
::: p~
o ,. .Llt-~~l
D1. -
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(~~~~/dlTPM-~~
~~ ~ ti--J 1{'11) lRp.:;. )
~ u-. tb ~~ ~~ lv)a fc~'1 ~~
12-36

e-ll)
..c.....
w
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Q)
E
is'-"
t:
0
::;:::>
(,)
til
'-
"-
Il)
"0
~
.......
N
I
w
-....)
Residue Curve Map, Ethanol I Dimethyl Ether I Diethyl Ether by UNIQ
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
I
0
~)~:>.,!'~~~:::~~' ..........
~.::::~":~". .
.~. ',::.."': :~, ''''.
':'i:.':::::'~'<>:'! ",
.:'.:~:~~~:::<'::~~: ;:»~:" '
':;~;::.',!;;~~'::;,::>L:,'·'·,·,
<r:>·'<>.",:'::::':,::::::::;,.,:
' ··<':1···.,.""::J:"::>::.:::::."
1',:,'.~~::~.,~•••••:.~~::,:":::~:~.,~~>:>,,
, ,..c".. '.' I'<>'.c.l: ,............'<
I ....···.1.....,... '" '.'
......... ...."". .....,........>..:---...,"........::::::::.:2.:., ,.,.... ". :::..
"
0.1 0.2 0,3 0.4 0.5 0.6
Mole fraction (Ethanol)
Bubble point at 500.00 kPa Binary Azeotrope
Ethanol = 125.342 C
Dimethyl Ether = 19.231 C
Diethyl Ether = 88.807 C
'.
'.
0.7 0.8 0.9
Job: chapter
10 binary
azeotropes
Date:
07/12/2001
Time: 13:09:09

r
T
~t'~
&lt6tv'6t<.
t"t tJ.,....... I
~ "fAM~" ~~~
12-38

c-
o
c:
cu
.J::.
W'-'
c:
o:;::l
(.)
cu
~
<P
o
:2:
>-'
N
I
V..l
0
Residue Curve Map, Water I Ethanol I Benzene by UNIQ
0.6
0.5
0.4
0.2
0.1
o
o
~
0.1 0.2
Bubble point at 101.32 kPa
Water = 100.000 C
Ethanol = 78.290 C
Benzene = 80.090 C
0.3 0.4 0.5 0.6 0.7 0.8
Mole fraction (Water)
Binary Azeotrope
(Ethanol,Benzene) = (0.436,0.564) at 67.441 C
(Water,Benzene) =(0.158,0.842) at 79.861 C
(Water,Ethanol) = (0.100,0.900) at 78.170 C
0.9
Job: chapter
10 binary
azeotropes
Date:
07/12/2001
Time: 13:28:01

10.18 Equation 12.1 applies for any "theoretical simple sequence" separation, i.e.,
where each separation unit makes a perfect split between components. These
sequences can be for any separation type, e.g., distillation, filtration, liquid-liquid
extraction, leaching, etc. In fact, these separation types can be mixed in the
sequence with, for example, one unit being extraction and one being distillation.
10.19 In the derivation ofEquation 12.1, it is assumed that each unit is a perfect
separator, with any given component only appearing in one ofthe two outlet
streams. The sequence is assumed to contain only such separators, and it is
assumed that there is no mixing or splitting ofstreams between separation units.
10.20 Non-simple separation units include:
a. Distillation columns with multiple side-draws. Capital costs could go
down because the number ofcolumns decreases. Operability and safety
may improve because ofthe reduced number ofreboilers, but operability
might degrade because ofthe interconnectedness ofa complex column
compared with multiple columns with built-in surge capacity (reboiler
pots and condenser tanks). Operating costs may decrease through lower
heat losses.
b. Batch columns, in which the overhead is sent to more than one
downstream unit on a schedule during the cycle time. These units can
decrease capital costs while they increase the complexity ofthe operation.
Operating costs may increase or decrease.
c. Columns with non-perfect splits, such as the first column in Figure
12.1(c). The advantages of such a column are described on page 400.
d. A multi-stage filtration column, which has outlets on each stage. Capital
costs may decrease because fewer units are required. Operability and
safety may be compromised somewhat because the failure ofany filter
requires the shutdown ofthe entire column. Operating costs are likely to
decrease as solids handling is simplified.
10.21 Equation 12.1 applies to batch and semi-batch processes ifand only if each unit is
a perfect separator. Ifa single unit is used to produce three or more outlets, or if
the separation is sloppy for any component, the assumptions ofthe derivation of
Equation 12.1 are violated. Note that the multiple "outlets" in the batch process
may refer to a single physical outlet port that produces multiple outlet streams
scheduled during the cycle time.
10.22 Although this process would result in pure C (in the limit), this is an advantageous
strategy only ifbreaking the A+B azeotrope is not desired. Also, not all of
component C is captured in such a process, as C will leave with the azeotropic
mixture. The more pure the pot becomes, the more moles of C that are lost.
(Note that a batch stripping column with total reflux is not the model for the
residue curve map.)
12-40

10.23 The composition ofa ternary, minimum-boiling azeotrope can be an unstable
node or a saddle point. Because compositions close to this ternary azeotrope have
higher boiling points, it cannot be a stable node. Whether it is an unstable node or
a saddle point depends on the boiling points ofthe pure components and binary
azeotropes.
10.24 Typical heuristics include:
a. Iflimited kinetics modeling and/or kinetics data are available, use a batch
reactor at the same conditions as the experiment.
b. If extrapolation/interpolation risk is too great (often the case in pharmaceutical
and other biological production), use a batch reactor.
c. Ifparticles are being grown from seed, try a semi-batch reactor.
d. If large throughputs are required, try a continuous reactor.
e. If average properties ofthe reactor product are important, but variations in
properties between small samples ofproduct are not, try a continuous reactor.
10.25 Biological systems tend to have a finite (often short) lifetime before deactivation
through age or mutation. Some biological products are only produced during
specific time intervals ofthe organism's life cycle. Materials can slowly poison
biological systems or become substrates for cultures that result in undesirable
growth characteristics. Biological products often degrade over time, through
growth or oxidation that make them undesirable.
a. Reaction feed preparation must match reaction input schedule, but the time-
dependent property changes noted above make the customary surge strategies
(e.g., large tanks) unusable.
b. Similarly, separator feed preparation must often minimize holding times in
surge tanks to maintain quality. Biologicals are often extremely sensitive to
small temperature changes.
c. Biological processes often use semi-batch separators such as adsorption or
chromatography because separations often involve removal oftrace
contaminants from streams or capture ofproducts that are trace in reactor
outlets.
d. When either reactors or separators are unsteady state, the recycle may be
unsteady. Another factor to consider is the speed ofanalysis ofthe stream, the
results ofwhich may be needed before recycle commitment.
e. Environmental control is often unsteady, even when continuous processes are
upstream. But environmental control is essentially unsteady when the
upstream process is unsteady.
12-41

10.26 Azeotropic mixtures can be used as solvents to simplify recycle ofthe solvent
from overhead streams (U.S. patent 6,551,978), as refrigerants (U.S. patent
6,221,273), as solvents in other processes such as printing (U.S. patent
6,682,877), and in art restoration (Augerson, C., "The Use of Azeotropes in
Treating the Elaborate Polychromy ofthe Sleigh "The Turtle" (1832), in the
Coach Museum, Chateau ofVersailles," Wooden Artifacts Group session, 2004
Annual Meeting ofthe American Institute ofConservation, Portland, Oregon). In
each case, the similarity ofthe boiling/condensing characteristics ofazeotropes
and pure substances is key to the application. Azeotropes can be produced
similarly to the way "pure" components are produced, i. e., they will be the
limiting composition in a simple distillation column. A minimum-boiling
azeotrope would exit at the top ofa column; a maximum-boiling azeotrope would
exit at the bottom ofa column.
10.27 In biological reactor systems, small temperature differences between the reactor
and the heating/cooling surfaces are especially important. Large temperature
differences can lead to uneven temperature distributions in the reactor, as well as
slow response to temperature excursions. Although a large temperature
difference can lead to lower costs for the reactor, the extreme temperature
sensitivity ofmost biological systems usually dominates.
10.28 Continuous sterilization is often used for process equipment when sufficient cell
lysis throughout the process equipment surfaces can be assured. However, batch
sterilization is usually preferred, especially for small equipment and for reactor
feedstock, where even a very small number of surviving organisms can lead to
large cultures and gross contamination.
12-42

Chapter 13
'3 .. ~or all of the problems given in this chapter, the results ofcomputer simulations
using the CHEMCADTM III (Chern Stations, Houston, TX) simulation package are
given. Small differences between the data given in Tables 1.5, 1.7, and Appendix B
and the results ofthese simulations are not uncommon. Any significant differences
will be noted in the solutions and reasons for the differences will be given. When
using other simulators results will differ from those given here. However, these
differences should be small. Ifyou cannot get similar answers to those given here be
sure to check (1) the thermodynamics package used, (2) the topology ofthe
flowsheet, (3) the equipment specifications. Hint: remove non-condensables from
feed when simulating short-cut columns - otherwise you get very low temperatures at
the top ofthese columns. In reality the non-condensable gases would be vented from
the overhead reflux drum - but this cannot be simulated using the short-cut methods.
Toluene Hydrodealkylation Process
The CHEMCADTM simulation results are given on the following pages.
Comparison ofthe stream compositions shows excellent agreement with the
conditions given in Table 1.5 ofthe text. Differences of 1 or 2°C in temperature
and 0.2 kmollh in flowrates are not unusual. Note that the duties ofthe heat
exchangers from the simulation are all within a few percent, the largest
discrepancy being E-105 which is nearly 5% high (in the simulation). The reactor
volume is not calculated unless specific kinetics are used (see Problem 13.5). The
results for tower, T-101 give 24.5 theoretical stages or 24.5/0.6 = 40.8 actual
trays - Table 1.7 gives 42 trays. The power ofthe feed pump, P-101, and the
recycle compressor, C-101, are both within a few % ofthe values in Table 1.7.
Table 1.7 Simulation (following pages) %diff
E-I01 15,190 MJ/h 15,047 MJ/h 0.9
E-I02 46,660 MJ/h 46,650 MJ/h 0.0
E-103 1,055 MJ/h 1,052 MJ/h 0.3
E-I04 8,335 MJ/h 8,387 MJ/h -0.1
E-I05 1,085 MJ/h 1,138 MJ/h -4.9
E-I06 9,054 MJ/h 9,172 MJ/h -1.3
P-IOI 14.2 kW 14.2 kW 0.0
C-IOI 49.1 kW 50.2 kW -2.2
T-lOl 42 Trays 24.5/0.6 = 40.8 Trays 2.3
H-I01 27,040 MJ/h 27,044 MJ/h 0.0
Overall the results ofthe simulation are all well within the expected accuracy.

ChemCAD 3.20.16-386 License: WEST VIRGINIA UNIVERSITY
Job Code: CHP18-1 Case Code: CHP18-1 Date: 06-26-97
Date: 06-26-97 Time: 22:02
FLOWSHEET SUMMARY
Equipment Label Stream Numbers
1 PUMP 91 -2
2 HTXR 93 -4
3 FIRE 4 -6
4 REAC 105 -9
5 DIVI 8 -101 -19
6 FLAS 94 -8 -95
7 FLAS 96 -17 -18
8 HTXR 18 -97
9 SHOR 10 -"100 -11
10 MIXE 11 1 -91
11 MIXE 5 3 -92
12 COMP 101 -102
13 DIVI 102 -7 -5
14 MIXE 7 6 -105
15 VALV 95 - 96
16 HTXR 9 -94
17 MIXE 17 21 -99
18 VALV 19 -21
19 MIXE 92 2 -93
20 CSEP 97 -98 -10
21 HTXR 100 -15
22 MIXE 98 99 -16
Stream Connections
Page: 1
Time: 22:02
Stream Equipment Stream Equipment Stream Equipme
From
1
2 1
3
4 2
5 13
6 3
7 13
8 6
9 4
10 20
Calculation mode
Flash algorithm
To
10
19
11
3
11
14
14
5
16
9
Sequential
Normal
11
15
16
17
18
19
21
91
92
93
)3- '2-
From To From
9 10 94 16
21 95 6
22 96 15
7 17 97 8
7 8 98 20
5 18 99 17
18 17 100 9
10 1 101 5
11 19 102 12
19 2 105 14

Job Code: CHP18-1 Case Code:
Equipment Calculation Sequence
10 1 11 19 2 3
18 21 17 22
Equipment Recycle Sequence
10 1 11 19
Recycle Cut Streams
11 5 7
2 3
14
14
CHPI8-1 Date:
4 16 6 15 7
4 16 6 15 7
Recycle Convergence Method: Direct Substitution
Max. loop iterations 40
Recycle Convergence Tolerance
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-004
1.000E-004
1.000E-004
1.000E-004
1.000E-004
Recycle calculation has converged.
COMPONENTS
ID # Name
1 1 Hydrogen
2 2 Methane
3 40 Benzene
4 41 Toluene
THERMODYNAMICS
K-value model SRK
Enthalpy model SRK
Liquid density API
06-26-97
5 8
5 8
Page: 2
Time: 2:- 2
20 12 13
20 12 13

~r·:·
.$"
G.•.•
Job Code: CHP18-1
Overall Mass Balance
Hydrogen
Methane
Benzene
Toluene
Total
Case Code: CHP18-1
kmol/h
Input
286.000
15.000
0.000
108.700
409.700
Output
177.964
123.036
108.036
0.664
409.700
13~:'-(
Date: 06-26-97
kg/h
Input
576.519
240.645
0.000
10015.726
10832.890
Page: 3
Time: 22:02
Output
358.739
1973.871
8439.145
61.156
10832.912

ChemCAD 3.20.16-386 License: WEST VIRGINIA UNIVERSITY Page: 4
Job Code: CHP18-1 Case Code: CHP18-1
EQUIPMENT SUMMARIES
Date: 06-26-97 Time: 22:02
Equip. No.
Name
Output pressure bar
Efficiency
Calculated power kw
Calculated Pout bar
Head m
Vol. flow rate m3/h
Equip. No.
Name
Pressure drop 1 bar
TOut Str 1 C
Calc Ht Duty MJ/h
Str1 Pout bar
Equip. No.
Name
Pressure drop 1 bar
Tout Str 1 C
Calc Ht Duty MJ/h
Str1 Pout bar
Equip. No.
Name
Temperature Out C
Pressure Drop bar
Thermal Efficiency
Heat Absorbed MJ/h
Fuel Usage (SCF)
Pump Sunnnary
1
25.8295
0.7500
14.1959
25.8295
294.0007
16.0060
Heat Exchanger Sunnnary
2
0.1400
225.0000
15047.3799
25.3595
21
0.1400
38.0000
-1138.0046
2.5199
8
0.1400
90.0000
1052.0774
2.6599
Fired Heater Sunnnary
3
599.9999
0.3400
0.9000
27044.5293
31645.8867
16
0.1400
38.0000
-46652.5000
23.8600

Job Code: CHPI8-1 Case Code: CHPI8-1
EQUIPMENT SUMMARIES
Equip. No.
Name
specify thermal mode:
C
Key Component
Frac. Conversion
Heat of Reaction
(kJ/kmol)
Reactor Pressure bar
Calc H of Reac.
(kJ/kmol)
Stoichiometries:
Hydrogen
Methane
Benzene
Toluene
Equip. No.
Name
Flow rate units
Output stream #1
Output stream #2
Equip. No.
Name
Flash Mode
Param 1
Param 2
Heat duty MJ/h
K values:
Hydrogen
Methane
Benzene
Toluene
Reactor Summary
4
1
653.7601
4
0.7500
-41930.0000
24.0000
-41930.0000
-1.000
1. 000
1.000
-1.000
Divider Summary
Flash
5
1
0.7280
0.2720
Summary
6
2
38.0000
23.9000
-0.5951
130.395
15.155
0.012
4.020E-003
Date: 06-26-97 Time: 22:02
13
o
0.0531
0.9469
7
0
1113.123
123.718
0.086
0.027

EQUIPMENT SUMMARIES
Date: 06-26-97 Time: 22:02
Shortcut Distillaton Summary
Equip. NO.
Name
Select mode:
Colm pressure drop
(bar)
Light key component
Light key split
Heavy key component
Heavy key split
Number of stages
R/Rmin
Lower bound R/Rmin
Upper bound R/Rmin
Number of stages, minimum
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
Reflux ratio, minimum
Reflux ratio, calculated
9
2
0.1700
3.0000
0.9900
4.0000
0.0100
24.5070
1.5000
1. 0100
2.0100
13.4679
12.7535
-8386.7939
9171.5449
1.1620
1.7430
Mixer Surmnary
Equip. No.
Name
Output Pressure bar
Equip. No.
Name
Equip. No.
Name
Type of Compressor:
Pressure out bar
Efficiency
Actual power kW
Cp/Cv
Ideal Cp/Cv
Calc Pout bar
Theoretical power kW
10
17
Compressor Surmnary
12
1
25.4995
0.7500
50.2107
1.3772
1.3405
25.4995
37.6580
11
19
14
25.0000
22

EQUIPMENT SUMMARIES
Equip. No.
Name
Pressure out bar
Valve Summary
15
2.7999
Date: 06-26-97
18
2.5000
Component Separator Summary
Equip. No.
Name
Component No. 1
Component No. 2
20
1.0000
1.0000
Page: 7
Time: 22:02

Job Code: CHP18-1
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total krnol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Case Code: CHP18-1 Date: 06-26-97 Time: 22:02
1 2 3
25.0000 60.0250 25.0000
1.9000 25.8295 25.4995
1296.7 2576.9 -1121.8
0.00000 0.00000 1.0000
108.7000 144.3381 301.0000
0.0000 0.0000 286.0000
0.0000 0.0000 15.0000
0.0000 1. 0617 0.0000
108.7000 143.2763 0.0000
4 5 6
225:0000 45.0450 599.9999
25.3595 25.4995 25.0195
-5098.4 -21601. 21946.
1.0000 1. 0000 1. 0000
1204.2986 758.9607 1204.2986
735.3435 449.3436 735.3435
317.1072 302.1072 317.1072
7.8405 6.7788 7.8405
144.0074 0.7310 144.0074
7 8 9
45.0450 38.0000 653.7601
25.4995 23.9000 24.0000
-1211.3 -31584. 20735.
1. 0000 1.0000 1.0000
42.5608 1100.9911 1246.8594
25.1982 651. 8431 652.5054
16.9415 438.2538 442.0850
0.3801 9.8337 116.2570
0.0410 1.0605 36.0121
10 11 15
90.0000 149.6958 38.0000
2.6599 2.8299 2.5199
6969.7 1229.1 5387.4
0.00000 0.00000 0.00000
141.0997 35.6381 105.4617
0.0000 0.0000 0.0000
0.0000 0.0000 0.0000
106.1741 1. 0617 105.1124
/;-1

f· Job Code: CHP18-1 Case Code: CHP18-1 Date: 06-26-97 Time: 22:02.. FLOW SUMMARIES
Toluene 34.9256 34.5763 0.3493
Stream No. 16 17 18
Stream Name
Temp C 35.5823 38.3049 38.3049
Pres bar 2.5000 2.7999 2.7999
Enth MJ/h - 8851. 6 -197.45 5863.2
Vapor mole fraction 1. 0000 1. 0000 0.00000
Total kmol/h 304.2383 3.8714 141. 9969
Flowrates in kmol/h
Hydrogen 177.9636 0.6412 0.0211
Methane 123.0363 2.9551 0.8761
Benzene 2.9239 0.2491 106.1741
Toluene 0.3145 0.0260 34.9256
Stream No. 19 21 91
Stream Name
Temp C 38.0000 35.3601 59.0413
Pres bar ·23.9000 2.5000 1.9000
Enth MJ/h -8590.8 -8590.7 2525.8
Vapor mole fraction 1.0000 1.0000 0.00000
Total kmol/h 299.4696 299.4696 144.3381
Flowrates in kmol/h
Hydrogen 177.3013 177.3013 0.0000
Methane 119.2050 119.2050 0.0000
Benzene 2.6748 2.6748 1. 0617
Toluene 0.2885 0.2885 143.2763
Stream No. 92 93 94
Stream Name
Temp C 39.6650 48.2868 38.0000
Pres bar 25.4995 25.4995 23.8600
Enth MJ/h -22723. -20146. -25917.
Vapor mole fraction 1. 0000 0.875653 0.883029
Total kmol/h 1059.9605 1204.2986 1246.8594
Flowrates in kmol/h
Hydrogen 735.3436 735.3435 652.5054
Methane 317.1072 317.1072 442.0850
Benzene 6.7788 7.8405 116.2570
Toluene 0.7310 144.0074 36.0121

Job Code: CHP18-1
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Case Code: CHP18-1
95
38.0000
23.9000
5665.7
0.00000
145.8684
0.6623
3.8312
106.4233
34.9516
98
90:0000
2.6599
-63.423
1. 0000
0.8972
0.0211
0.8761
0.0000
0.0000
101
38.0000
23.9000
-22993.
1. 0000
801. 5214
474.5417
319.0487
7.1589
0.7720
Date: 06-26-97 Time: 22:02
96 97
38.3049
2.7999
5665.7
0.0265406
145.8684
0.6623
3.8312
106.4233
34.9516
99
35.3990
2.5000
-8788.2
1.0000
303.3410
177.9425
122.1602
2.9239
0.3145
102
45.0450
25.4995
-22812.
1.0000
801.5214
474.5417
319.0487
7.1589
0.7720
90.0000
2.6599
6915.3
0.00501726
141.9969
0.0211
0.8761
106.1741
34.9256
100
115.1665
2.6599
6525.4
0.00000
105.4617
0.0000
0.0000
105.1124
0.3493
105
588.5500
25.0000
20735.
1.0000
1246.8594
760.5417
334.0487
8.2207
144.0484
/1~/(

·~"--. .
t. 11.2. Simulation ofthe DME process, Appendix B, Figure B. Land Tables B.Uand B~2
Again the simulation results are shown in the output following this page. It is
worth noting that due to non-ideal behavior ofthe waterlDME/methanol system a
UniquaclUnifac thermodynamic model was chosen for this simulation. Again the
reactor was modeled with a simple stoichiometric reactor module. Therefore no
reactor volume is calculated in the output. The required minimum input
information for this simulation is given below:
1. Topology (see first page offollowing simulation)
2. Specification offeed streams - stream l(meoh)
3. Equipment specifications
P-20l output pressure (efficiency not required for H&M balance)
E-201 stream outlet temperature
E-202 outlet temperature offeed to reactor.
R-20l conversion ofkey component (no kinetics req'd for H&M balance)
E-203 stream outlet temperature.
T-201 identification ofkey components, recovery ofkey components, RIRmin
4. Conversion tolerance for recycle loop etc.
note: that for the simulation used here, exchanger E-202 was included. However, to
ease the recycle calculation, E-202 could be eliminated ifdesired (and added back
when recycle convergence was obtained). Also note that reflux pumps are not
. included anywhere in the simulation since the towers are modeled with the reboiler
and condenser included and no estimates ofthe reflux pumps are given
Equipment
P-20l
E-201
E-202
E-203
E-204
E-205
E-206
E-207
E-208
T-201
T-201
Table B.1
7.2kW
14,400 MJIh
2,030 MJIh
12,420 MJIh
2,490 MJIh
3,140 MJIh
5,790 MJIh
5,960 MJIh
1,200 MJIh
22 trays
26 trays
Simulation Results
7.1 kW
14,361 MJIh
2021 MJIh
12,766 MJIh
2,833 MJIh
3,105 MJIh
5,725 MJIh
5,899MJIh
1,190 MJIh
15.2/0.7 = 22 trays
18.110.7= 26 trays
1.4
0.3
0.4
-2.8
-13.8
1.1
1.1
1.0
0.8
0.0
0.0
Simulation result for E-204 is somewhat higher than in Table 8.1 - but condenser duty (E-205) is OK.

Job Code: CHP18-2 Case Code: CHP18-2 Date: 06-27-97 Time: 10.~9
Date: 06-27-97 Time: 10:19
FLOWSHEET SUMMARY
1 PUMP P-201 1 -2
2 MIXE 13 2
3 HTXR E-201 3 -4
4 HTXR E-202 4 6
5 REAC R-201 5 -6
6 HTXR E-203 7 -8
7 VALV 8 -9
8 SHOR T-201 9 -10
9 VALV 11 -'12
10 SHOR T-202 12 -13
11 HTXR E-208 14 -16
12 VALV 16 -15
Stream Connections
Stream Equipment Stream
From To
1 1 7
2 l 2 8
3 2 3 9
4 3 4 10
5 4 5 11
6 5 4 12
Calculation mode : Sequential
Flash algorithm : Normal .
1 2 3 4 5 6 7 8
-3
-5 -7
-11
-14
Equipment
From. To
4 6
6 7
7 8
8
8 9
9 10
9 10 11 12
2 3 4 5 6 7 8 9 10
Recycle Cut Streams
13 6
Stream Equipme
From
13 10
14 10
15 12
16 11

:~. ::",::".,
Job Code: CHPI8-2 Case Code: CHPI8-2
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-002
1.000E-002
1.000E-002
1.000E-002
1.000E-002
COMPONENTS
1
2
3
ID #
117
133
62
THERMODYNAMICS
Name
Methanol
Dimethyl Ether
Water
Date: 06-27-97 Time: 10:19
K-value model UNQC ( Uniquac with Unifac Rand Q )
LLV Three phase K Values
Enthalpy model
Liquid density
Mixed Model
Library
UNIQUAC Parameters:
I
1
1
2
J
2
3
3
Uij-Ujj
-218.911
-365.948
-371.327
Uji-Uii
582.494
589.393
972.438
/Jr-ILf
Aij
0.000
0.000
0.000
Aji
0.000
0.000
0.000

Job Code: CHP18-2
Methanol
Dimethyl Ether
Water
Total
Case Code: CHP18-2
kmol/h
Input
259.697
0.000
2.497
262.195
Output
1.286
129.206
131.703
262.195
. ,.
11,--lS
Date: 06-27-97 Time: 10.19
kg/h
Input
8321.220
0.000
44.990
8366.210
Output
41.193
5952.383
2372.633
8366.209

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 10:19
Equip. No.
Name
Output pressure bar
Efficiency
Calculated power kW
Calculated Pout bar
Head m
Vol. flow rate m3/h
Equip. No.
Name
Output Pressure bar
Equip. No.
Name
Pressure drop 1 bar
TOut Str 1 C
VF Out Str 1
Calc Ht Duty MJ/h
LMTD (End points) C
LMTD Corr Factor
Str1 Pout bar
Str2 Pout bar
Equip. No.
Name
TOut Str 1 C
Calc Ht Duty MJ/h
Str1 Pout bar
Equip. No.
Name
Specify thermal mode:
C
Key Component
Frac. Conversion
Pump Summary
1
P-201
15.5000
0.6000
7.1171
15.5000
187.2383
10.6017
Mixer Summary
2
15.2000
Heat Exchanger Summary
3 4 6
E-201 E-202 E-203
0.1500 1.1500
250.0000 100.0000
1. 0000
14361.1562 2021.1194 -12765.6055
121.8691
1.0000
15.0500 15.0500 13.9000
15.0500
11
E-208
50.0000
-1190.7069
7.5000
Reactor Summary
5
R-201
1
366.9053
1
0.8000
/3-lb

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 10:1~
Heat of Reaction
(Btu/lbmol)
Calc H of Reac.
(Btu/lbmol)
Stoichiometrics:
Methanol
Dimethyl Ether
Water
Equip. No.
Name
Pressure out bar
Pressure drop bar
-5273.0000
-5273.0000
-2.000
1.000
1. 000
valve Summary
7
10.4000
9
3.4000
Equip. No. 8 10
Name
Select mode:
T-201 T-202
Colm pressure drop
(bar)
Light key
Light key
Heavy key
Heavy key
Number of
R/Rmin
component
split
component
split
stages
Lower bound R/Rmin
Upper bound R/Rmin
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
2
0.2000
2.0000
0.9900
1. 0000
0.0100
15.1800
1.2000
1.0100
2.0100
5.8186
8.0900
-3105.6807
2833.1018
0.2946
0.3535
/3-/1
2
0.3000
1.0000
0.9900
3.0000
0.0100
18.1301
1.3000
8.8926
9.5651
-5899.2832
5725.7480
1.3073
1.6994
12
1.2000

ChemCAD 3.20.16-386 License:
~~- ... WEST VIRGINIA UNIVERSITY Page: 6
~, :;
Job Code: CHP18-2 Case Code: CHP18-2 Date: 06-27-97 Time: 10:19
FLOW SUMMARIES
Stream No. 1 2 3
Stream Name meoh feed
Temp C 25.0000 25.3177 45.3236
Pres bar 1. 0100 15.5000 15.2000
Enth MJ/h -63735. -63709. -78921.
Total kmol/h 262.19 262.19 328.15
Flowrates in kmol/h
Methanol 259.70 259.70 323.01
Dimethyl Ether 0.00 0.00 1.31
Water 2.50 2.50 3.83
Stream No. 4 5 6
Stream Name
Temp C 153.5873 250.0000 366.9053
Pres bar 15.0500 15.0500 15.0500
Enth MJ/h -64560. -62539. -62521.
Total kmol/h 328.15 328.15 328.15
Flowrates in kmol/h
Methanol 323.01 323.01 64.60
Dimethyl Ether 1. 31 1.31 130.51
Water 3.83 3.83 133.03
Stream No. 7 8 9
Stream Name
Temp C 280.5589 100.0000 88.0901
Pres bar 15.0500 13.9000 10.4000
Enth MJ/h -64542. -77308. -77310.
Total kmol/h 328.15 328.15 328.15
Flowrates in kmol/h
Methanol 64.60 64.60 64.60
Dimethyl Ether 130.51 130.51 130.51
Water 133.03 133.03 133.03
Stream No. 10 11 12
Stream Name
Temp C 46.1822 153.2090 138.6080
Pres bar 10.4000 10.6000 7.2000
Enth MJ/h -26161. -51422. -51421.
Total kmol/h 129.85 198.29 198.29
Flowrates in kmol/h
Methanol 0.65 63.96 63.96
Dimethyl Ether 129.21 1.31 1. 31
Water 0.00 133.03 133.03
Ii-If

Job Code: CHP18-2
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Methanol
Dimethyl Ether
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Methanol
Dimethyl Ether
Water
Case Code: CHP18-2
13
120.2759
7.2000
-15211.
0.00000
65.95
63.32
1.31
1.33
16
50.0000
7.5000
-37574.
0.00000
132.34
0.64
0.00
131.70
(3-rq
Date: 06-27-97 Time: 10:1~
14 15
166.8828 50.0000
7.5000 1.2000
-36384. -37574.
0.00000 0.00000
132.34 132.34
0.64 0.64
0.00 0.00
131.70 131.70

~r;:-~"
,,"0:- "'3.3 Simulation ofthe IPA to acetone process, Appendix B, Figure S.C). Prf-'b Tf6.es.
(6.0.' B.O.2
Again the simulation results are shown in the output following this page. It is
worth noting that due to non-ideal behavior ofthe Acetone/WaterlIPA system a
Unifac thermodynamic model was chosen for this simulation. Again the reactor
was modeled with a simple stoichiometric reactor module. Therefore no reactor
volume is calculated in the output. The required minimum input information for
this simulation is given below:
1. Topology (see first page offollowing simulation)
2. Specification offeed streams - Stream l(IPA), Stream 8 (water)
P-II01 output pressure (efficiency not required for H&M balance)
R-II01 conversion ofkey component (no kinetics req'd for H&M balance)
E-I102 stream outlet temperature
V-I102 temperature and pressure offlash
T-llOl number oftrays, location offeeds - this module must be simulated with a
rigorous tray-to-tray method because it is an absorber and no short-cut
methods exist for this type of unit.
T-I102 identification ofkey components, recovery ofkey components, RIRmin
4. Conversion tolerance for recycle loop etc.
Equipment
P-I101
E-II01
E-1102
E-1103
E-II04
E-I105
E-1106
E-1107
E-1108
T-1101
T-1102
T-ll03
Table 8.10;).
0.43 kW
3,550 MJ/h
3,260 MJ/h
563 MJ/h
3,095 MJ/h
3,500 MJ/h
7,340 MJ/h
7,390 MJ/h
174 MJ/h
2.5 m packing
66 trays
19 trays
Simulation Results % diff.
0.42 kW 2.3
3,550 MJ/h 0.0
3,337 MJ/h 2.4
506 MJ/h -2.8
2,897 MJ/h 3.5
3,306 MJ/h 5.5
644 MJ/h -91.2
689 MJ/h -90.7
174 MJ/h 0.0
5 theor. stages HETP=0.5 m
46.7/0.7 = 67 trays -1.5
12.410.7 = 18 trays 5.3

All the above results are in good agreement except for those associated with column,
T-4lC>"3.This column was simulated using a short-cut routine in CHEMCADTM III.
This algorithm uses average relative volatilities to make estimates ofminimum reflux
ratio and minimum number oftrays and then applies a correlation to estimate the
number of trays required for a given RIRmin value. However, the top product from
this column is a water-IPA azeotrope (or close to the azeotrope).;rlii~eans that at
the top of the column the relative volatilities of water and IPA aPproach unity. Thus
the separation will be much more difficult than predicted by this short-cut method.
This accounts for the much higher reflux ratio (condenser and reboiler duties) in
Table B. c. J.
It may be a good exercise to get the students to resimulate this column using a
rigorous column package. The results should be much closer than the short-cut
results.
Note that the results for T- b 1- the absorber seem reasonable giving a HETP of0.5
m or about 20 inches.

Date: 06-27-97 Time: 11:48
FLOWSHEET SUMMARY
1 MIXE 1 14 -98
2 PUMP P-401 98 -100
3 HTXR E-401 4 -97
4 REAC R-401 97 -3
5 HTXR E-402 3 -96
6 FLAS V-402 90 -5 -99
7 TOWR T-401 8 5 -7 -6
8 MIXE 6 99 -9
9 SHOR T-402 92 -'11 -12
10 SHOR T-403 12 -14 -15
11 HTXR E-408 15 -94
12 CSEP 9 -91 -92
13 HTXR E-403 96 -90
14 VALV 100 -4
Stream Connections
Stream Equipment Stream Equipment Stream Equipme
From To
1 1 9
3 4 5 11
4 14 3 12
5 6 7 14
6 7 8 15
7 7 90
8 7 91
Flash algorithm : Normal
1 2 14 3 4 5
1 2 14 3
Recycle Cut Streams
14
4 5
13
13
From To From
8 12 92 12
9 94 11
9 10 96 5
10 1 97 3
10 11 98 1
13 6 99 6
12 100 2
6 7 8 12 9 10 11
6 7 8 12 9 10
f3-:2'-

Job Code: CHPI8-3 Case Code: CHPI8-3 Date: 06-27-97 Time: 11. 8
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-003
1.000E-003
1.000E-003
1.000E-003
1.000E-003
COMPONENTS
1
2
3
4
ID #
1
140
62
145
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
* UNIFAC main groups
Name
Hydrogen
Acetone
Water
Isopropanol
UNIFAC
Latent Heat
Library
62 and
62 and
62 and
9,
7,
5,
No interaction parameters.

f· '...~.........
,,:
Job Code: CHP18-3
Hydrogen
Acetone
Water
Isopropanol
Total
Case Code: CHP18-3
kmol/h
Input
0.000
0.000
37.147
34.816
71.963
Output
34.773
34.773
37.147
0.043
106.736
-/8-2/f
Date: 06-27-97 Time: 11:48
kg/h
Input
0.000
0.000
669.207
2092.278
2761.485
Output
70.095
2019.614
669.207
2.562
2761.478

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 11~~8
Equip. No.
Name
Equip. No.
Name
Output pressure bar
Efficiency
Calculated power kW
Calculated Pout bar
Head m
Vol. flow rate m3/h
Equip. No.
Name
Pressure drop 1 bar
TOut Str 1 C
Calc Ht Duty MJ/h
Strl Pout bar
.Equip. No.
Name
Pressure drop 1 bar
TOut Str 1 C
Calc Ht Duty MJ/h
Strl Pout bar
Equip. No.
Name
C
MJ/h
Key Component
Frac. Conversion
Calc H of Reac.
(J/kmol)
Mixer Surrunary
1
Pump Surrunary
P-401
2
2.8000
0.4000
0.4170
2.8000
22.8841
3.3526
8
3
E-401
0.1400
234.0000
3549.4170
2.1600
13
E-403
0.1400
20.0000
-505.5067
1. 8800
5
E-402
0.1400
45.0000
-3337.0029
2.0200
Reactor Surrunary
4
R-401
2
350.0000
2732.1438
4
0.9000
5.5270e+007
11
E-408
0.1400
45.0000
-174.1102
1.2600

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 11:48
Stoichiometries:
Hydrogen
Acetone
Isopropanol
Equip. No.
Name
Flash Mode
Param 1
Param 2
Heat duty MJ/h
K values:
Hydrogen
Acetone
water
Isopropanol"
Equip. No.
Name
TOP pressure bar
Colm pressure drop
(bar)
No. of stages
1st feed stage
2nd feed stage
Est. dist. rate
(kmol/h)
Est. stage 1 T C
Est. bottom T C
Equip. No.
Name
Select mode:
Colm pressure bar
Colm pressure drop
(bar)
Light key component
Light key split
1.000
1.000
-1.000
Flash Summary
V-402
6
2
20.0000
1.6300
23.8612
20410.104
0.195
0.030
0.041
Towr Rigorous Distillation Summary
T-401
7
1. 5000
0.1300
5
1
5
36.9400
25.0000
26.0000
9 10
T-402 T-403
2 2
1.2000 1.2000
0.2000 0.2000
2.0000 4.0000
0.9950 0.9990
13-26·

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 11:
Heavy key component
Heavy key split
Number of s"tages
RjRmin
Feed stage
Condenser duty MJjh
Reboiler duty MJjh
4.0000
0.0060
46.6768
1.2000
21. 4049
23.4403
-2896.6062
3306.5957
1.7167
2.0601
3.0000
0.0500
12.3589
1.5000
6.7886
4.4531
-644.3007
688.5560
1.1573
1.7359
Equip. No.
Name
Component No. 1
Equip. No.
Name
Pressure drop bar
12
1.0000
Valve Summary
14
0.5000
3-,27

t;' Job Code: CHP18-3 Case Code: CHP18-3 Date: 06-27-97 Time: 11:48
FLOW SUMMARIES
Stream No. 1 3 4
Stream Name
Temp C 25.0000 350.0000 31.6765
Pres bar 1.0100 2.1600 2.3000
Enth MJ/h -15955. -11399. -17697.
Total kmol/h 51.9628 92.6039 57.8310
Flowrates in kmol/h
Hydrogen 0.0000 34.7730 0.0000
Acetone 0.0000 34.9339 0.1609
Water 17.1472 19.0334 19.0334
Isopropanol 34.8156 3.8637 38.6366
Stream No. 5 6 7
Stream Name
Temp C 20.0000 26.7269 33.2741
Pres bar 1.6300 1. 6300 1. 5000
Enth MJ/h -1110.0 -5918.5 -905.75
Total kmol/h 39.7823 20.9410 38.8413
Flowrates in kmol/h
Hydrogen 34.7707 0.0006 34.7702
Acetone 4.4803 1.7341 2.7462
Water 0.4146 19.1053 1.3093
Isopropanol 0.1168 0.1010 0.0157
Stream No. 8 9 11
Stream Name
Temp C 25.0000 21.5080 61.1912
Pres bar 2.0000 1. 6300 1. 2000
Enth MJ/h -5713.9 -20026. -7816.5
Total kmol/h 20.0000 73.7626 32.0499
Flowrates in kmol/h
Hydrogen 0.0000 0.0028 0.0000
Acetone 0.0000 32.1877 32.0268
water 20.0000 37.7241 0.0000
Isopropanol 0.0000 3.8479 0.0231
Stream No. 12 14 15
Stream Name
Temp C 90.3397 82.7982 109.2365
Pres bar 1.4000 1.2000 1.4000
Enth MJ/h -11799. -1743.1 -10012.
Total kmol/h 41.7099 5.8682 35.8418
~-::- Flowrates in kmol/h
~..
Hydrogen 0.0000 0.0000 0.0000::...;.
Acetone 0.1609 0.1609 0.0000
water 37.7241 1.8862 35.8379
t 3-.28'

ChemCAD 3.20.16-386 License: WEST VIRGINIA UNIVERSITY Page,: 8
Job Code: CHP18-3
FLOW SUMMARIES
Case Code: CHP18-3 Date: 06-27-97 Time: 11:48
Isopropanol
Stream No.
Stream Name
Temp C
Pres' bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Acetone
water
Isopropanol
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Acetone
Water
Isopropanol
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Acetone
Water
Isopropanol
3.8249
90
20.0000
1.8800
-15241.
0.421707
92.6039
34.7730
34.9339
19.0334
3.8637
94
45.0000
1. 2600
-10186.
0.00000
35.8418
0.0000
0.0000
35.8379
0.0038
98
31. 4815
1.0100
-17699.
0.00000
57.8310
0.0000
0.1609
19.0334
38.6366
13-29
3.8210
91
21.5080
1.6300
-0.000144186
1.0000
0.0028
0.0028
0.0000
0.0000
0.0000
96
45.0000
2.0200
-14736.
0.528497
92.6039
34.7730
34.9339
19.0334
3.8637
99
20.0000
1. 6300
-14107.
0.00000
52.8216
0.0023
30.4536
18.6188
3.7469
0.0038
92
21.5080
1.6300
-20026.
0.00000
73.7598
0.0000
32.1877
37.7241
3.8479
97
234.0000
2.1600
-14148.
1.0000
57.8310
0.0000
0.1609
19.0334
38.6366
100
31.6765
2.8000
-17697.
0.00000
57.8310
0.0000
0.1609
19.0334
38.6366

Comparison ofshort-cut and rigorous simulation algorithms for the Benzene
I?;·lr Recovery Colurnn, T-1 01.
The results ofthe simulations for the two algorithms are given in the attached
print outs. Comparison of Streams 1, 2, and 3 (short-cut results) with Streams
11, 12, and 13 (rigorous results) show that the input and output streams for the
two simulations are essentially identical. A comparison ofother pertinent output
is given below - note that the same number of trays are used in both simulations.
result short-cut ngorous % diff.
reflux (kmollh) 184.6 166.457 9.8
reflux ratio 1.7498 1.5782 9.8
condenser duty (MJ/h) 8,397 7,874 6.2
reboiler duty (MJ/h) 9,196 8,673 5.7
Although the differences between the two simulations are not large it is worth
noting that this system (benzene - toluene) is often used as the standard example
for ideal (constant molal overflow) systems. Even for this system we have some·
error in this assumption - clearly the more non-ideal the system becomes the
greater these differences are likely to become. This is well illustrated in Problem
13.3 above for T-·1I0~.
When using rigorous simulations several other useful options are available such as
tray-to-:-tray profiles offlows and temperatures etc. The liquid and vapor profiles
for 1~101 are included in the output and it can be seen that the constant molal
overflow assumption is a very good one (but not exactly correct even for this
"ideal" system).
Change in L (above feed)
Change in V (above feed)
Change inLN
Change in L' (below feed)
Change in V' (below feed)
Change in L'N'
166.21 to 160.91 kmollh
271.93 to 267.21 kmollh
0.611 to 0.602 (1.5 %)
326.68 to 311.79 kmollh
291.05 to 275.72 kmollh
1.122 to 1.131 (0.7 %)
/3-30

Job Code: CHPI8-4 Case Code: CHPI8-4 Date: 06-27-97 Time: 16;30
Date: 06-27-97 Time: 16:30
FLOWSHEET SUMMARY
1 SHOR T-I0l shtcut
2 TOWR T-I0l rigor
1 -2 -3
11 -12 -13
Stream Connections
Stream
1
2
Equipment
From To
1
1
Stream
3
11
1 2
No recycle loops in the flowsheet.
COMPONENTS
1
2
ID #
40
41
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Name
Benzene
Toluene
SRK
SRK
Library
/3-3/
Equipment
From To
1
2
Stream
12
13
Equipme
F~.om
2
2

'c~~... <
Job Code: CHP18-4
Benzene
Toluene
Total
Case Code: CHP18-4
kmol/h
Input
212.340
69.860
282.200
Output
212.340
69.860
282.200
13-32
Date: 06-27-97 Time: 16:30
kg/h
Input
16586.726
6436.970
23023.695
Output
16586.732
6436.968
23023.700

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 16.
Equip. No.
Name
Select mode:
Colm pressure drop
(bar)
Light key component
Light key split
Heavy key component
Heavy key split
Number of stages
R/Rmin
1
T-l0l shtcut
2
0.1700
1.0000
0.9900
2.0000
0.0100
24.5637
Feed stage
1. 5000
13.5053
12.7818
-8397.1709
9196.0742
1.1665
Condenser duty MJ/h
Reboiler duty MJ/h
Reflux ratio, calculated 1. 7498
Equip. No.
Name
TOP pressure bar
Colm pressure drop
(bar)
No. of stages
1st feed stage
Select condenser mode:
Condenser spec.
Condo comp i
Select reboiler mode:
Reboiler spec.
Rebl. comp i
Est. dist. rate
(kmol/h)
Est. reflux rate
(kmol/h)
Est. stage 1 T
Est. bottom T
Calc cond duty
Calc rebr duty
Reflux ratio
C
C
MJ/h
MJ/h
Reflux mole kmol/h
Reflux mass kg/h
2
T-l0l rigor
2.7000
0.1700
25
13
7
0.9900
1
6
0.9702
2
100.0000
150.0000
100.0000
130.0000
-7874.4390
8673.1680
1.5782
166.4570
13010.6465
13-33

--.,
'........:
FLOW SUMMARIES
Stream No. 1 2 3
Stream Name
Temp C 90.0000 115.7703 150.3135
Pres bar 2.7000 2.7000 2.8700
Enth MJ/h 6969.6 6534.9 1233.6
Total kmol/h 141.1000 105.4576 35.6424
Flowrates in kmol/h
Benzene 106.1700 105.1083 1. 0617
Toluene 34.9300 0.3493 34.5807
Stream No. 11 12 13
Stream Name
Temp C 90.0000 115.7743 150.3140
Pres bar 2.7000 2.7000 2.8700
Enth MJ/h 6969.6 6535.3 1233.1
Total kmol/h 141.1000 105.4709 35.6292
Flowrates in kmol/h
Benzene 106.1700 105.1083 1. 0617
Toluene 34.9300 0.3626 34.5674
/3.-34-

Job Code: CHP18-4 Case Code: CHP18-4 Date: 06-27-97 Time: 16 :30
DISTILLATION PROFILE
unit type : TOWR Unit name: T-101 rigor Eqp # 2
* Net Flows *
Temp Pres Liquid Vapor Feeds Product 'Du
Stg C bar kmol/h krnol/h kmol/h kmol/h MJ
1 115.8 2.70 166.46 105.47
2 115.9 2.70 166.21 271. 93
3 116.1 2.71 166.03 271. 69
4 116.4 2.71 165.79 271.50
5 116.7 2.72 165.48 271.26
6 117.1 2.73 165.09 270.96
7 117.6 2.74 164.59 270.56
8 118.2 2.74 164.00 270.06
9 118.8 2.75 163.31 269.47
10 119.6 2.76 162.55 268.78
11 120.4 2.77 161. 7'4 2,68.02
12 121.3 2.77 160.91 267.21
13 122.2 2.78 326.68 266.38 141.10
14 122.6 2.79 326.24 291.05
15 123.3 2.80 325.24 290.61
16 124.4 2.80 323.64 289.61
17 126.1 2.81 321.12 288.01
18 128.8 2.82 318.05 285.49
19 132.4 2.83 314.85 282.42
20 136.6 2.83 312.42 279.22
21 140.9 2.84 311.16 276.79
22 144.5 2.85 311. 02 275.53
23 147.2 2.86 311.35 275.39
24 149.1 2.86 311. 79 275.72
25 150.3 2.87 276.16 35.63
Mole Reflux ratio 1.578
'3-35

.cO
·~O
3.5 The simulation ofthe reactor system is shown in the attached print out. The
reactor was simulated as two adiabatic packed beds with a intermediate "cold
shot" (Stream 7) at the inlet ofthe second bed.
R-IOIA
1+---( 7 >---
R-IOIB
The simulation shows that the overall conversion in the reactor is the desired
75%. The temperature profiles for each section ofthe reactor (R-I0IA and R-
101B) show that the maximum temperatures reached in each section are
653.35°Cand 653.74°C, which are close to but do not exceed the 655°C limit.
The total combined volumes ofeach section ofthe reactor are
Vlotal = VRIOIA + VRI01B = 30.11 + 11.45 =41.56 m3
from Table 1.7 we have Vtotal =1tD2L14 =1t(2.3)2(10)/4 = 41.55 m3
Thus the simulation and Table 1.7 are in excellent agreement.

Job Code: CHP18-5
Date: 06-27-97
FLOWSHEET SUMMARY
Equipment Label
1 KREA R-101A
2 MIXE
3 KREA R-101B
Stream Connections
Case Code: CHP18-5
Time: 17:13
Stream Numbers
6 - 98
7 98 -99
99 -9
Stream Equipment
From To
6 1
7 2
9 3
98 1 2
99 2 3
1 2 3
No recycle loops in the flowsheet.
COMPONENTS
ID #
1 1
2 2
3 40
4 41
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Name
Hydrogen
Methane
Benzene
Toluene
SRK
SRK
Library
Date: 06-27-97 Time: 17:11

;~..
~:
..:"".";
t.' :
~,
Job Code: CHP18-5
Hydrogen
Methane
Benzene
Toluene
Total
Case Code: CHP18-5
kmol/h
Input
760.600
334.250
7.970
144.040
1246.860
Output
6.52.551
442.299
116.019
35.991
1246.860
13 "-3'8
Date: 06-27-97
kg/h
Input
1533.218
5362.372
622.569
13271.990
20790.148
Page: 2
Time: 17:13
Output
1315.412
7095.798
9062.695
3316.265
20790.171

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 17:~j
Kinetic Reactor Summary
Equip. No. 1 3
Name R-101A R-101B
1 1No. of Reactions
Pressure Drop bar
Specify reactor type:
1.0000
Specify reaction phase
2
1
2
2
1
2
Tout C 653.3050 653.7204
Specify calc. mode: 1 1
Vol 30.1058 11.4480
4 4Key
Conversion
Time Unit:
0.6000 0.3760
Volume Unit:
Activation E/H of Rxn
Molar Flow Unit:
Unit:
2
1
1
2
1
5 5
1
Overall Ht of Rxn
(MJ/h)
-3621.1519 -908.4697
Reaction stoichiometrics and Parameters for unit no. 1
Reaction 1
Rate const = 2.8330e+007 Act.
Comp stoich. Exp.factor
1 -1.00e+000 5.0000e-00l
4 -1.00e+000 1.0000e+000
2 1.00e+000 O.OOOOe+OOO
E = 1.4811e+002 Hrxn =
Adsorb Fac. Adsorb E
O.OOOOe+OOO O.OOOOe+OOO
Reaction Stoichiometrics and Parameters for unit no. 3
Reaction 1
Rate const = 2.8330e+007 Act.
Comp stoich. Exp.factor
1 -1.00e+000 5.0000e-00l
4 -1.00e+000 1.0000e+000
E = 1.4811e+002 Hrxn =
Adsorb Fac. Adsorb E
O.OOOOe+OOO O_OOOOe+OOO
Plug Flow Profile for unit no. 1
Vol
m3
0.00
1. 51
3.01
Temp
C
600.02
602.46
604.91
Press
bar
25.00
24.95
24.90
Total
kmol/h
1204.29
1204.29
1204.29
Mole frac
Hydrogen
6.107E-001
6.074E-00l
6.040E-00l
O.OOOOe
Adsorb E
O.OOOOe+
O.OOOOe+
O.OOOOe+
O.OOOOe+
O.OOOOe
Adsorb E
O.OOOOe+
O.OOOOe+
O.OOOOe+
O.OOOOe+
Mole fra
Metha
2.635E 0
2.667E-00
2.701E-OO

EQUIPMENT SUMMARIES
Date: 06-27-97 Time: 17:13
4.52
6.02
7.53
9.03
10.54
12.04
13.55
15.05
16.56
18.06
19.57
21.07
22.58
24.08
25.59
27.09
28.60
30.11
Vol
m3
0.00
1.51
3.01
4.52
6.02
7.53
9.03
10.54
12.04
13.55
15.05
16.56
18.06
19.57
21.07
22.58
24.08
25.59
27.09
28.60
30.11
Vol
m3
0.00
0.57
1.14
607.46
610.02
612.63
615.24
617.91
620.63
623.35
626.13
628.91
631. 68
634.46
637.24
640.02
642.74
645.46
648.13
650.74
653.29
Temp
C
600.02
602.46
604.91
607.46
610.02
612.63
615.24
617.91
620.63
623.35
626.13
628.91
631.68
634.46
637.24
640.02
642.74
645.46
648.13
650.74
653.29
24.85
24.80
24.75
24.70
24.65
24.60
24.55
24.50
24.45
24.40
24.35
24.30
24.25
24.20
24.15
24.10
24.05
24.00
Press
bar
25.00
24.95
24.90
24.85
24.80
24.75
24.70
24.65
24.60
24.55
24.50
24.45
24.40
24.35
24.30
24.25
24.20
24.15
24.10
24.05
24.00
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
Total
kmol/h
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
1204.29
Plug Flow Profile for unit no. 3
Temp
C
640.68
641.41
642.07
Press
bar
24.00
24.00
24.00
Total
kmol/h
1246.88
1246.88
1246.88
/3-:/fO
6.007E-001
5.972E-001
5.937E-001
5.901E-001
5.865E-001
5.829E-001
5.792E-001
5.755E-001
5.718E-001
5.680E-001
5.643E-001
5.605E-001
5.568E-001
5.531E-001
5.495E-001
5.459E-001
5.424E-001
5.389E-001
Mole frac
Benzene
6.311E-003
9.581E-003
1.291E-002
1.630E-002
1.975E-002
2.325E-002
2.681E-002
3.041E-002
3.405E-002
3.773E-002
4.144E-002
4.517E-002
4.891E-002
5.266E-002
5.641E-002
6.013E-002
6.382E-002
6.747E-002
7.107E-002
7.459E-002
7.804E-002
Mole frac
Hydrogen
5.407E-001
5.398E-001
5.388E-001
2.735E-00
2.769E-00
2.804E-00
2.840E,...00
2.876E~00
2.912E-00
2.949E-00
2.986E-00
3.023E-00
3.061E-00
3.098E-00
3.136E-00
3.173E-00
3.210E-00
3.246E-00
3.282E-00
3.318E-00
3.352E-00
Mole fra
Toluene
1.196E-00
1.163E-00
1.130E-00
1.096E-00
1.061E-00
1.026E-00
9.908E-00
9.548E-00
9.183E-OO
8.815E-00
8.444E-OO
8.071E-OO
7.697E-OC
7.322E-OC
6.948E-OC
6.575E-OC
6.206E-OC
5.841E-OC
5.481E-OC
5 .129E- OC
4.785E-OC
Mole fre
Methane
3.374E-0<
3.383E-0(
3.393E-0<

EQUIPMENT SUMMARIES
l. 72
2.29
2.86
3.43
4.01
4.58
5.15
5.72
6.30
6.87
7.44
8.01
8.59
9.16
9.73
10.30
10.88
11.45
Vol
m3
0.00
0.57
l.14
l. 72
2.29
2.86
3.43
4.01
4.58
5.15
5.72
6.30
6.87
7.44
8.01
8.59
9.16
9.73
10.30
10.88
11.45
642.79
643.52
644.18
644.91
645.57
646.24
646.91
647.57
648.24
648.85
649.52
650.13
650.74
651.35
651.96
652.57
653.13
653.74
Temp
C
640.68
64l. 41
642.07
642.79
643.52
644.18
644.91
645.57
646.24
646.91
647.57
648.24
648.85
649.52
650.13
650.74
65l. 35
651.96
652.57
653.13
653.74
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
Press
bar
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
24.00
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
Total
kmol/h
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
1246.88
/3-1;-1
Date: 06-27-97 Time: 17:
5.379E-00I
5.370E-00l
5.360E-00l
5.351E-00l
5.342E-00I
5.333E-00l
5.324E-00l
5.315E-00l
5.307E-00l
5.298E-00l
5.290E-00l
5.281E-00l
5.273E-00l
5.265E-00l
5.257E-00l
5.249E-00l
5.241E-00l
5.234E-00I
Mole frac
Benzene
7.567E-002
7.662E-002
7.757E-002
7.852E-002
7.945E-002
8.037E-002
8.129E-002
8.220E-002
8.309E-002
8.398E-002
8.486E-002
8.573E-002
8.659E-002
8.743E-002
8.827E-002
8.910E-002
8.991E-002
9.071E-002
9.150E-002
9.228E-002
9.305E-002
3.402E-00
3.411E-00
3.420E-00
3.430E-00
3.439E-00
3.448E-00
3.457E-00
3.465E-00
3.474E-00
3.483E-00
3.491E-00
3.499E-00
3.508E-00
3.516E-00
3.524E-00
3.5.32E-00
3.540E-00
3.547E-00
Mole fra
Toluene
4.625E-00
4.529E-
4.434!£- 0
4.340E-00
4.247E-00
4.154E-00
4.062E-00
3.972E-00
3.882E-00
3.793E-00
3.705E-00
3.619E-OQ
3.533E-00
3.448E-00
3.364E-00
3.282E-00
3.200E-00
3.120E-00
3.041E-00
2.963E-00
2.887E-00

EQUIPMENT SUMMARIES
Mixer Summary
Equip. No. 2
Name
13-/.f2-
Date: 06-27-97
Page: 6
Time: 17:13

Job Code: CHP18-5
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Hydrogen
Methane
Benzene
Toluene
Case Code: CHP18-5
6
600.0000
25.0000
21897.
1. 0000
1204.2999
735.4000
317.3000
7.6000
144.0000
98
653:3050
24.0000
21896.
1.0000
1204.2999
649.0215
403.6784
93.9783
57.6216
13-43
Date: 06-27-97 Time: 17:~3
7 9
40.8000
25.0000
-1218.7
1.0000
42.5600
25.2000
16.9500
0.3700
0.0400
99
640.6738
24.0000
20678.
1. 0000
1246.8599
674.2215
420.6284
94.3483
57.6616
653.7204
24.0000
20678.
1.0000
1246.8598
652.5510
442.2987
116.0188
35.9912

The results will differ for different simulators and their databases of BIP's. The
following is for ChemCAD.
ChemCAD BIPs BIPS set to zero
No. of stages 23.3 26.2
Reflux ratio 1.745 2.147
3.7
In Appendix B, each ofthe four processes has a section on the thermodynamic models
used. They are summarized below:
Process Model used Reason Another model Page
DME UNIQUAC/ Highly NRTL, ifVLE
UNIFAC nonideal. for all binaries is 946
VLEData regressed
mIssmg
Acrylic UNIFAC LLEData User-input 957
Acid missing distribution
coefficients
based on LLE
Acetone UNIQUAC BIPs NRTL 965
available.
azeotrope
Heptenes SRK Well PR 976
behaved.

The following results are from ChemCAD.
NRTL
Stages required: 17.0
DME Methanol Water
DME 139.995,
a=0.3
Methan 153.78, -24.4933,
01 a=0.3 a=0.3001
Water 307.166,
a=0.3001
Wilson
DME Methanol Water
DME 54.7165
Methanol 564.363 -52.605
Water 620.63
UNIQUAC
DME Methanol Water
DME 361.266
Methanol 32.7318 95.259
Water -10.377
13-1p

There are many sets ofmethan011water data, 39 ofwhich are given in the DECHEMA
Vapor-Liquid Equilibrium Data Collection, volume 1, part I, FrankfurtlMain, 1977.
Below is a set for 760 Torr from page 48 ofthis reference.
T [0C] x[methanol] y[methanol]
100.00 0 0
92.40 0.05 0.2780
87.70 0.10 0.4250
81.70 0.20 0.6020
78.00 0.30 0.6920
75.40 0.40 0.7520
73.20 0.50 0.7980
71.20 0.60 0.8380
69.40 0.70 0.8780
67.70 0.80 0.9150
66.00 0.90 0.9600
64.60 1.00 1.00
The BIPs obtained are:
Methanollwater Water/methanol
UN1QUAC 95.259 -10.377
ChemCAD
database
UN1QUAC 30.035 40.863
regression
UNIFACIUNIQ -364.12 582.59
UAC
The results for the data (circles) and the UNIQUAC regression are shown in the
figure on the following page.
13-4-fJ

IIEthanoIJWateVLEat 760Torr
110~----------------------------------------------------,
100
70
OO~--~-----+----~----r---~-----r----~----r---~----~
0.0 0.2 0.4 0.6 0.8 1.0
rmlefractioo rmthand
13-4-7

The following results are from ChemCAD at 760 Torr.
13.'
Henry's Law
PR with
ChemCAD
BIPs
816.8
79.0
676.0
102.1
Different simulators handle supercritical components differently. ChemCAD
switches to its database Henry's constants for supercritical components ifthese
Henry's constants are available. Otherwise, it uses the vapor pressure equation, even
though it has no meaning physical meaning above the critical point. The vapor
pressure equation used is the DIPPR equation, unless only the Antoine's constants are
available. If only the Antoine's constants are available and the reduced temperature is
greater than 2, the vapor pressure used is the vapor pressure calculated by Antoine's
equation at a reduced temperature of2.
13-4--<3

13.12 Simulation ofthe Ethylbenzene process, Appendix B, Figure B.2.1 and Table B.2.1. The
simulation results are shown in the output following this page. The UNIFAC model was
used for the phase equilibrium calculations, and the enthalpy calculations used the
ChemCAD "Latent Heat" method, which is equivalent to setting heats ofmixing to zero.
The required minimum input information for this simulation is given below:
1. Flowsheet topology
2. Specification offeed streams (Stream 1 and Stream 2)
H-301 inlet pressures, output temperatures
R-301, R-302, R-303, R-304 conversion ofkey component
E-301, E-302, E-303, E-304, E-305 outlet temperatures .
.V-302 pressure
T-301, T-302 identification ofand recovery ofkey components
4. Calculational tolerances
13.13 Simulation ofthe Styrene process, Appendix B, Figure B.3.1 and Table B.3.1. The
simulation results are shown in the output following this page. The SRK model was used
for the phase equilibrium and the enthalpy calculations. The required minimum input
information for this simulation is given below:
H-401 output temperature
R-401, R-402 conversion ofkey component
E-401, E-402, E-403, E-404, E-405 outlet temperatures
V-401 pressure
T-401, T-402 identification ofand recovery ofkey components
C-401 outlet pressure, efficiency
13.14 Simulation ofthe Maleic Anhydride process, Appendix B, Figure B.5.1 and Table B.S.1.
The simulation results are shown in the output following this page. The "ideal vapor
pressure" model (Raoult's Law) was used for the phase equilibrium and "latent heat"
model for enthalpies (zero heat ofmixing). The required minimum input information for
this simulation is given below:
H-601 output temperature
R-601 conversion ofkey component
E-601, E-602, E-603 outlet temperatures
T-601, T-602 identification of and recovery ofkey components
C-601 outlet pressure, efficiency

CHEMCAD 5.6.0 Page 1
Job Name: Ethyl.benzene Date: 12/14/200B Time: 13:38:22
FLOWSHEET SUMMARY
Equipment Label. stream Numbers
1 FIRE 127 -107
2 !<REA 8 -9
3 !<REA 11 -12
4 HTXR 118 -132
5 MIXE 7 5 -109
6 NIXE 9 10 -111
7 KREA 140 -13
8 FLAS 119 -126 -145
9 SHOR 19 -22 -123
10 DIVI 2 -103 -4
11 DIVI 103 -5 -10
12 NIXE 4 107 -6
13 VAIN 135 -119
14 PUMP 123 -21
15 FIRE 134 -140
17 NIXE 12 13 -14
18 HTXR 111 -11
19 CSEP 145 -137 -15
20 HTXR 14 -118
21 HTXR 132 -135
22 DIVI 18 -117 -3
23 NIXE 3 21 -134
24 NIXE 117 1 -131
25 !<REA 6 -7
26 PUMP 131 -127
27 PUMP 17 -18
28 NIXE 137 126 -16
29 TOWR 15 -17 -19
30 TOWR 41 -42 -43
31 HTXR 109 -8
stream Connections
stream Equipment stream Equipment Stream Equipment
From To From To From To
1 24 15 19 29 117 22 24
2 10 16 28 118 20 4
3 22 23 17 29 27 119 13 8
4 10 12 18 27 22 123 9 14
5 11 5 19 29 9 126 8 28
6 12 25 21 14 23 127 26 1
7 25 5 22 9 131 24 26
8 31 2 41 30 132 4 21
9 2 6 42 30 134 23 15
10 11 6 43 30 135 21 13
11 18 3 103 10 11 137 19 28
12 3 17 107 1 12 140 15 7
13 7 17 109 5 31 145 8 19
14 17 20 111 6 18
~ ... 50

CHEMCAD 5.6.0
Job Name: Ethylbenzene Date: 12/14/2008 Time: 13:38:22
Calculation mode
Flash algorithm
Sequential
Normal
10 11 30 1 12 25 5 31 2 6 18 3 17 20
19 29 27 22 24 26 9 14 23 15 7 28
1 12 25 5 31 2 6 18
22 24 26 9 14 23 15 7
Recycle Cut streams
127 13
3 17 20 4 21 13
Recycle Convergence Method: Wegstein
Wegstein lower bound -5.00
Acceleration frequency 4
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-004
1.000E-003
1.000E-003
1.000E-003
1.000E-003
Wegstein upper bound
~ -5
Page 2
4 21 13 8
8 19 29 27
0.00

CHEMCAD 5.6.0
Overall Mass Balance kmol/h kg/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Total
COMPONENTS
ID #
1 40
2 45
3 22
4 383
5 3
6 41
7 23
8 62
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Input
97.169
90.688
93.000
10.424
7.000
2.000
0.000
0.000
300.281
Name
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
UNIFAC
Output
8.683
181.160
0.502
10.437
7.000
0.000
2.000
0.000
209.782
Formula
C6H6
C8H10
C2H4
C10H14
C2H6
C7H8
C3H6
H20
Input
7590.270
9628.111
2609.022
1399.075
210.490
184.283
0.000
0.000
21621.249
No correction for vapor fugacity
Latent Heat
Library
Std vapor rate reference temperature is 0 C.
Atmospheric pressure is 101.3250 kPa.
* Component ID 22 does not have UNIFAC subgroups.
3 - 52
Page 3
Output
678.268
19233.250
14.072
1400.857
210.490
0.003
84.161
0.000
21621.100

CHEMCAD 5.6.0
Job Name: Ethylbenzene
EQUIPMENT SUMMARIES
Equip. No.
Name
Pressure Drop kPa
Temperature Out C
Heat Absorbed MJ/h
Fuel Usage(SCF)
Date: 12/14/2008 Time: 13:38:22
Fired Heater Summary
1
15.0000
400.0000
17295.6367
24285.9785
15
10.0000
500.0000
5057.1372
7101.0703
Equip. No. 2 3
Name
Reactor type 2 2
Reaction phase 1 1
Thermal mode 2 2
Pressure Drop kPa
7
2
1
2
Tout C 453.8678 449.6506 497.8943
Reactor volume m3 15.00'00 18.0000 1.0000
Conversion 0.9990
Key 0 0 4
No. of Reactions 3 3 1
Molar Flow Unit 1 1 1
Activ. E/H of Rxn Unit 7 7 7
Volume Unit 1 1 1
Time Unit 2 2 2
Overall IG Ht of Rxn -3438.4460 -3438.2615 -10.7212
(MJ/h)
Mass unit 1 1 1
Reaction 1
RateConst = 1.6700e+006 Act.E = 2.2500e+004 Hrxn
Comp stoich. Exp.factor AdsorbFac. AdsorbE
1 -1.00e+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3 -1.00e+000 1.0000e+000 O.OOOOe+OOO O.OOOOe+OOO
2 1.00e+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
Comp Stoich. Exp. factor AdsorbFac. AdsorbE
2 -l.OOe+OOO 1.0000e+OOO O.OOOOe+OOO O.OOOOe+OOO
3 -1.00e+OOO 1.0000e+000 O.OOOOe+OOO O.OOOOe+OOO
Reaction 3
=
=
RateConst = 3.0000e+008 Act.E = 2.0000e+004 Hrxn =
Comp stoich. Exp. factor AdsorbFac. AdsorbE
3-53
Page 4
25
2
1
2
50.0000
443.4468
12.0000
0
3
1
7
1
2
-2636.6072
1
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 13:38:22
2
7
1.00e+OOO
1.00e+000
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
Reaction stoichiometrics and Parameters for unit no. 3
Reaction 1
RateConst = 1. 6700e+006 Act.E = 2.2500e+004 Hrxn =
1 -1.00e+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3 -1.00e+OOO 1.0000e+OOO O.OOOOe+OOO O.OOOOe+OOO
2 1.00e+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
3
4
-1.00e+OOO
1.00e+OOO
Reaction 3
1.0000e+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
3
2
7
-2.00e+OOO
1.00e+OOO
1.00e+OOO
2.0000e+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
Reaction 1
RateConst = 1.3000e+OO7 Act.E = 2.5000e+OO4 Hrxn
Comp Stoich. Exp.factor AdsorbFac. AdsorbE
4 -1.00e+OOO 1.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
1 -1. OOe+OOO 1.0000e+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 1
RateConst = 1. 6700e+006 Act.E = 2.2500e+004 Hrxn
=
=
1 ~1.00e+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3
2
-l.OOe+OOO
1.00e+OOO
Reaction 2
1.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
RateConst = 1.OOOOe+006 Act.E = 2.2500e+004 Hrxn
2 -l.OOe+OOO 1.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3
4
-l.OOe+OOO
1.OOe+OOO
Reaction 3
1.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
Page 5
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
RateConst 3.0000e+008 Act.E = 2.0000e+004 Hrxn = O.OOOOe+OOO
Comp Stoich. Exp.factor AdsorbFac. AdsorbE AdsorbExp.
3 -54-

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 13:38:22
6
3
7
2
-1. OOe+OOO
-2.00e+000
1.00e+000
1.00e+000
Equip. No.
Name
1st Stream dp kPa
1st Stream T Out C
Calc Ht Duty MJ/h
LMTD Corr Factor
1st Stream Pout kPa
Equip. No.
Name
1st Stream TOut C
Calc Ht Duty MJ/h
LMTD Corr Factor
1st stream Pout kPa
Equip. No.
Name
Equip. No.
Name
1.0000e+000
2.0000e+000
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
4
170.0000
-12355.7969
1.0000
1915.0000
31
380.0000
-1938.1415
1. 0000
1935.0000
Mixer Summary
5
23
18
3BO.0000
-2614.3660
1.0000
1935.0000
6
24
~-SS
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
20
20.0000
2BO.OOOO
-10097.7314
1.0000
1915.0000
12
2B
Page 6
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
21
BO.OOOO
-4939.4751
1.0000
1915.0000
17

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES .
Equip. No.
Name
Flash Mode
Heat duty MJ/h
K values:
Benzene
Ethylbenzene
Ethylene
1,4-DiEthBenzen
Ethane
Toluene
Propylene
vlater
Equip. No.
Name
Mode
Light key component
Light key split
Heavy key component
Heavy key split
R/IUnin
Number of stages
Min. No. of stages
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
Colm pressure kPa
Calc. Reflux ratio
Colm pressure drop
(kPa)
Equip. No.
Name
Split based on
Output stream #1
Output stream #2
Date: 12/14/2008 Time: 13:38:22
Flash Summary
8
2
-0.0003
0.746
0.118
19521. 078
0.019
77708.516
0.276
34.217
82.861
9
2
2.0000
0.9900
4.0000
1.0000e-005
1.2000
35.2053
14.1985
25.4263
-4973.1953
4975.6084
110.0000
0.4662
0.5594
10.0000
Divider Summary
10
0
0.7000
0.3000
11
0
0.5000
0.5000
22
1
40.0000
Page 7

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 13:38:22
Valve Summary
Equip. No.
Name
Pressure out kPa
Equip. No.
Name
Output pressure kPa
Efficiency
Calculated power kW
Calculated Pout
Head m
Vol. flow rate
Mass flow rate
Equip. No.
Name
Component No. 3
Component No. 5
Component No. 7
Equip. No.
Name
No. of stages
1st feed stage
kPa
m3/h
kg/h
Top pressure kPa
Colm pressure drop
(kPa)
Condenser mode
Condenser spec.
Condo comp i
Reboiler mode
Reboiler spec.
Reboiler comp i
Iterations
Initial flag
Calc cond duty MJ/h
Calc rebr duty MJ/h
Est. Dist. rate
(kmol/h)
Est. Reflux rate
13
110.0000
Pump Summary
14 26
2000.0000 2000.0000
1.0000 1.0000
1. 0876 11.2529
2000.0000 2000.0000
267.8507 230.1106
2.0811 21.4187
1489.5172 17938.9434
Towr
19
1.0000
1.0000
1.0000
Rigorous
29
21
9
110.0000
10.0000
7
0.9990
1
7
0.9900
2
100
1
-7256.7793
9129.1543
500.0000
1000.0000
Distillation
30
36
26
110.0000
10.0000
7
0.9500
2
7
1. 0000
4
100
1
-5086.6245
5065.9414
100.0000
200.0000
~-57
27
2000.0000
1.0000
8.6112
2000.0000
237.3644
16.3906
13308.2090
Summary
Page 8

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
(kmol/h)
Est. T top C
Est. T bottom C
Tray type
Column diameter m
Tray space m
Thickness (top) m
Thickness (bot) m
No of sections
Calc Reflux ratio
Calc Reflux mole
(kmol/h)
Calc Reflux mass kg/h
Page 9
Date: 12/14/2008 Time: 13:38:22
82.0000 152.0000
145.0000 185.0000
0 3
1.3716
0.6096
0.0016
0.0103
0 1
0.3876 0.6617
65.9067 57.1168
5158.1973 6060.7837

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Date: 12/14/2008 Time: 13:38:22
1
25.0000
110.0000
4789.5
0.00000
99.0000
97.0000
0.0000
0.0000
0.0000
0.0000
2.0000
0.0000
0.0000
5
25.0000
2000.0000
1441.3
1.0000
35.0000
0.0000
0.0000
32.5500
0.0000
2.4500
0.0000
0.0000
0.0000
9
453.8678
1935.0000
30317.
1.0000
236.2702
174.8614
49.8812
0.6834
4.2941
4.5500
0.0025
1.9975
0.0000
2
25.0000
2000.0000
4117.9
1.0000
100.0000
0.0000
0.0000
93.0000
0.0000
7.0000
0.0000
0.0000
0.0000
6
383.2898
1985.0000
30814.
1.0000
259.0393
226.3389
0.7004
27.9000
0.0000
2.1000
2.0000
0.0000
0.0000
10
25.0000
2000.0000
1441.3
1.0000
35.0000
0.0000
0.0000
32.5500
0.0000
2.4500
0.0000
0.0000
0.0000
3
84.1704
2000.0000
2292.4
0.00000
40.0000
39.7846
0.2154
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
7
443.4468
1935.0000
30814.
1.0000
233.9761
203.9239
24.1042
1. 0183
0.8297
2.1000
0.1815
1. 8185
0.0000
11
380.0000
1935.0000
29144.
1.0000
271.2702
174.8614
49.8812
33.2334
4.2941
7.0000
0.0025
1.9975
0.0000
Page 10
4
25.0000
2000.0000
1235.4
1.0000
30.0000
0.0000
0.0000
27.9000
0.0000
2.1000
0.0000
0.0000
0.0000
8
380.0000
1935.0000
30317.
1.0000
268.9761
203.9239
24.1042
33.5683
0.8297
4.5500
0.1815
1.8185
0.0000
12
449.6506
1935.0000
29144.
1.0000
238.5409
148.1625
70.5547
0.5016
10.3221
7.0000
0.0000
2.0000
0.0000

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Date: 12/14/2008 Time: 13:38:22
13
497.8943
1990.0000
7082.7
1.0000
51.2870
29.4758
21.7397
0.0000
0.0715
0.0000
0.0000
0.0000
0.0000
17
82.9686
110.0000
9714.2
0.00000
170.0401
169.1243
0.9158
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
'22
139.0379
110.0000
1015.2
0.00000
89.9306
0.1692
89.7613
0.0000
0.0001
0.0000
0.0000
0.0000
0.0000
14
458.4666
1935.0000
36226.
1.0000
289.8279
177.6382
92.2945
0.5016
10.3936
7.0000
0.0000
2.0000
0.0000
18
84.1704
2000.0000
9745.2
0.00000
170.0401
169.1243
0.9158
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
41
145.4355
120.0000
740.82
0.00000
101.2812
0.1691
90.6884
0.0000
10.4237
0.0000
0.0000
0.0000
0.0000
3-bO
15
73.6193
110.0000
8583.7
0.00000
271.2577
169.2935
91.5837
0.0000
10.3804
0.0000
0.0000
0.0000
0.0000
19
145.4251
120.0000
742.05
0.00000
101.2176
0.1692
90.6679
0.0000
10.3804
0.0000
0.0000
0.0000
0.0000
42
139.0300
110.0000
974.96
0.00000
86.3230
0.1691
86.1538
0.0000
0.0001
0.0000
0.0000
0.0000
0.0000
Page 11
16
73.6193
110.0000
258.23
1.0000
18.5702
8.3447
0.7108
0.5016
0.0132
7.0000
0.0000
2.0000
0.0000
21
185.5938
2000.0000
-266.83
0.00000
11.2870
0.0000
0.9067
0.0000
10.3803
0.0000
0.0000
0.0000
0.0000
43
170.3402
120.0000
-254.89
0.00000
14.9581
0.0000
4.5345
0.0000
10.4236
0.0000
0.0000
0.0000
0.0000

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
103
25.0000
2000.0000
2882.5
1.0000
70.0000
0.0000
0.0000
65.1000
0.0000
4.9000
0.0000
0.0000
0.0000
117
84.1704
2000.0000
7452.7
0.00000
130.0401
129.3397
0.7004
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
126
73.6193
110.0000
244.18
1.0000
17.9549
8.3447
0.7108
0.5012
0.0132
6.9986
0.0000
1.3864
0.0000
107
400.0000
1985.0000
29578.
1.0000
229.0393
226.3389
0.7004
0.0000
0.0000
0.0000
2.0000
0.0000
0.0000
118
280.0000
1915.0000
26129.
1.0000
289.8279
177.6382
92.2945
0.5016
10.3936
7.0000
0.0000
2.0000
0.0000
127
60.5602
2000.0000
12283.
0.00000
229.0401
226.3397
0.7004
0.0000
0.0000
0.0000
2.0000
0.0000
0.0000
'3-b
109
423.2435
1935.0000
32255.
1.0000
268.9761
203.9239
24.1042
33.5683
0.8297
4.5500
0.1815
1.81.85
0.0000
119
73.61.93
110.0000
8833.4
0.061950
289.8279
177.6382
92.2945
0.5016
10.3936
7.0000
0.0000
2.0000
0.0000
131
59.3422
110.0000
12242.
0.00000
229.0401
226.3397
0.7004
0.0000
0.0000
0.0000
2.0000
0.0000
0.0000
Page 12
111
434.4776
1935.0000
31758.
1.0000
271.2702
174.8614
49.8812
33.2334
4.2941
7.0000
0.0025
1.9975
0.0000
123
184.5116
120.0000
-270.75
0.00000
11.2870
0.0000
0.9067
0.0000
10.3803
0.0000
0.0000
0.0000
0.0000
132
170.0000
1915.0000
13773.
0.040547
289.8279
177.6382
92.2945
0.5016
10.3936
7.0000
0.0000
2.0000
0.0000

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
l,4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Benzene
Ethylbenzene
Ethylene
1/4-DiEthBenzene
Ethane
Toluene
Propylene
Water
Page 13
Date: 12/14/2008 Time: 13:38:22
134 135 137 140
119.8999 80.0000 73.6193 500.0000
2000.0000 1915.0000 110.0000 1990.0000
2025.6 8833.4 14.045 7082.7
0.00000 0.027016 1.0000 1. 0000
51.2870 289.8279 0.6153 51.2870
39.7846 177.6382 0.0000 39.7846
1.1221 92.2945 0.0000 1.1221
0.0000 0.5016 0.0004 0.0000
10.3803 10.3936 0.0000 10.3803
0.0000 7.0000 0.0014 0.0000
0.0000 0.0000 0.0000 0.0000
0.0000 2.0000 0.6135 0.0000
0.0000 0.0000 0.0000 0.0000
145
73.6193
110.0000
8589.2
0.00000
271.8730
169.2935
91.5837
0.0004
10.3804
0.0014
0.0000
0.6135
0.0000
13- h2.

CHEMCAD 5.6.0
Job Name: styrene_JAS 2008 Date: 12/14/2008 Time: 15:25:58
FLOWSHEET SUMMARY
Equipment
1 LLVF
2 MIXE
4 HTXR
5 HTXR
6 VALV
7 MIXE
8 KREA
9 HTXR
10 KREA
11 HTXR
12 HTXR
13 HTXR
14 CSEP
15 CSEP
16 COMP
17 VALV
18 TOWR
19 TOWR
20 PUMP
35 PUMP
36 PUMP
37 PUMP
Label
Stream Connections
stream
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Equipment
From To
2
2 5
5 7
4
4 6
6
7
8
9
10
11
12
9
1
7
8
9
10
11
12
13
Stream Numbers
7 -29 -40 -30
25 1 -2
4 -5
2 38 -3 -39
5 -8
3 8 -9
9 -10
10 6 -11 -31
11 -12
12 32 -13 -33
13 34 -14 -35
14 36 -15 -37
15 -50 -18
50 -16 -19
16 -17
19 -20
20 -21 -22
22 -23 -24
23 -25
21 -26
24 -27
18 -28
stream Equipment
15
16
17
18
19
20
21
22
23
24
25
26
27
28
From To
13 14
15 16
16
14 37
15 17
17 18
18
18
19
19
20
35
36
37
35
19
20
36
2
Calculation mode
Flash algorithm
Sequential
Normal
stream
29
30
31
32
33
34
35
36
37
38
39
40
50
Page 1
Equipment
From To
1
1
9
11
12
13
5
1
14
11
12
13
5
15
1 4 6 2 5 7 8 9 10 11 12 13 14 15 17 18 19 20
16 35 36 37
3-63

CHEMCAD 5.6.0
Date: 12/14/2008 Time: 15:25:58
2 5 7 8 9 10 11 12 13 14 15 17 18 19 20
Recycle Cut streams
25
Recycle Convergence Method: Wegstein
Wegstein lower bound -5.00
Acceleration frequency 4
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-003
1.000E-003
1.000E-003
1.000E-003
1.000E-003
Wegstein upper bound
'3-6tt
Page 2
0.00

CHEMCAD 5. 6. 0
Input
Water 551241.000
Ethylbenzene 528.236
styrene 121.251
Hydrogen 89.004
Benzene 28.686
Toluene 34.967
Ethylene 25.086
Methane 31.035
Total 552099.000
output
551241.000
351.970
240.576
177.052
54.351
66.246
50.751
62.314
Input
9930598.077
56081.208
12628.489
179.414
2240.765
3221. 900
703.758
497.900
Output
9930598.077
37367.607
25056.488
356.902
4245.613
6103.988
1423.781
999.709
552244.000 10006153.335 10006151.521
Page 3

CHEMCAD 5.6.0
COMPONENTS
ID #
1 62
2 45
3 178
4 1
5 40
6 41
7 22
8 2
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Name Formula
Water H2O
Ethylbenzene C8H10
styrene C8H8
Hydrogen H2
Benzene C6H6
Toluene C7H8
Ethylene C2H4
Methane CH4
SRI<
Water/Hydrocarbon immiscible
SRI<
Library
Page 4

CHEMCAD 5.6.0
Job Name: styrene_JAS 2008
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:25:58
Equip. No.
Name
Flash Mode
Param1
Equip. No.
Name
Output Pressure kPa
Equip. No.
Name
1st Stream dp kPa
1st Stream T Out C
2nd Stream TOut C
2nd stream VF Out
Calc Ht Duty MJjh
LMTD (End points) C
LMTD Corr Factor
Utility Option:
1st stream Pout kPa
2nd Stream Pout kPa
Equip. No.
Name
1st Stream dp kPa
1st stream TOut C
2nd Stream TOut C
2nd Stream VF Out
Calc Ht Duty MJ/h
LMTD (End points) C
LMTD Corr Factor
Utility Option:
1st Stream Pout kPa
2nd Stream Pout kPa
Three Phase Flash Summary
1
3
75.0000
Mixer Summary
2
200.0001
Heat Exchanger
4
35.0000
802.0000
201698.6250
1.0000
0
565.0002
12
15.0000
180.0000
1.0000
35230.5078
100.1799
1. 0000
1
105.0001
600.0002
7
170.0001
Summary
5
20.0000
225.0000
253.0000
29913.7852
69.2964
1.0000
1
180.0001
4237.0020
13
15.0000
65.0000
40.0000
396136.0625
75.7415
1.0000
1
90.0001
400.0002
[3-67
9
15.0000
650.0000
700.0000
19014.2578
117.4589
1.0000
1
145.0001
571.0001
Page 5
11
15.0000
270.0000
1. 0000
163175.7813
270.2108
1.0000
1
120.0001
4237.0020

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:25:58
Equip. No.
Name
Pressure out kPa
Equip. No.
Name
Reactor type
Reaction phase
Thermal mode
Pressure Drop kPa
Tout C
Reactor volume m3
Concentration Flag
Conversion
Key
No. of Reactions
Molar Flow Unit
Valve Summary
6 17
180.0001 60.0000
S 10
2 2
1 1
2 2
10.0000 10.0000
609.9968 640.2557
1000.0378 700.0264
1 1
0.5000 0.5000
2 2
4 4
2 2
Activ. E/H of Rxn Unit 6 6
Volume Unit 1 1
Time Unit 2 2
Overall IG Ht of Rxn 10382.5908 4624.7280
(MJ/h)
Reaction 1
RateConst = 4.2400e+OO6 Act.E = 2. 1708e+004 Hrxn =
2 -1. OOe+OOO 1.0000e+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
RateConst = 7.5500e-001 Act.E = -7.8040e+003 Hrxn =
Reaction 3
RateConst = 7.2150e+008 Act.E = 4. 9675e+004 Hrxn
5
7
1.OOe+OOO
1.00e+000
Reaction 4
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
1'3-&8
O.OOOOe+OOO
O.OOOOe+OOO
=
Page 6
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O;OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:25:58
RateConst = 1.7230e+003 Act.E = 2. 6857e+004 Hrxn =
Reaction 1
2 -1.00e+000 1.0000e+00O O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
RateConst = 7.5500e-001 Act.E = -7.8040e+003 Hrxn =
Reaction 3
2 -1.00e+000 1.0000e+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 4
Comp Stoich. Exp.factor AdsorbFac. Ads0 rbE
4 -1.00e+000 1.0000e+00O O.OOOOe+OOO O.OOOOe+OOO
6 1.OOe+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
Equip. No. 14 15
Name
Top Temp Spec 65.0000 65.0000
Bottom Temp Spec 65.0000 65.0000
Press Drop kPa 15.0000
Component No. 1 0.0089 1.0000
Component No. 2 1.0000
Component No. 4 1.0000 1.0000
Component No. 7 1.0000 1.0000
Component No. 8 1.0000 1.0000
Page 7
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO

CHEMCAD 5.6.0
Equip. No.
Name
Pressure out kPa
Type of Compressor
Efficiency
Actual power kW
Cp/ev
Ideal Cp/ev
Calc Pout kPa
Calc. mass flowrate
(kg/h)
Compressor Summary
16
240.0001
1
0.8000
282.9636
1.3282
226.3709
1.3209
240.0001
2686
3-70
Page 8

CHEMCAD 5. 6. 0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:25:58
Equip. No.
Name
No. of stages
1st feed stage
Top pressure kPa
Colm pressure drop
(kPa)
Condenser mode
Condenser spec.
Condo comp i
Reboiler mode
Reboiler spec.
Reboiler cOmp i
Calc cond duty MJ/h
Calc rebr duty MJ/h
Est. Dist. rate
(kmol/h)
Est. Reflux rate
(kmol/h)
Est. T top C
Est. T bottom C
Calc Reflux ratio
Calc Reflux mole
(kmol/h)
Equip. No.
Name
Output pressure kPa
Efficiency
Calculated power kW
Calculated Pout ltPa
Head m
Vol. flow rate m3/h
Mass flow rate kg/h
18
25
14
40.0000
20.0000
7
0.9900
6
7
0.9900
2
19
80
25
40.0000
20.0000
7
0.9990
2
7
0.9900
3
-28362.5488 -345656.0625
33415.6992 344720.0000
300.0000
780.0000
106.0000
146.0000
11.5209
737.8981
400.0000
3200.0000
128.0000
150.0000
25.8749
8741. 8809
64115.9258 927916.0000
Pump Summary
20 35
210.0001 200.0001
0.8000 0.8000
2.6776 0.3739
210.0001 200.0001
21. 9120 19.7192
45.3298 6.7262
35861.6133 5565.1875
13-71
36
200.0001
0.8000
0.7455
200.0001
17.5688
15.3248
12452.6650
Page 9
37
200.0001
0.8000
6.3287
200.0001
13.0031
145.7092
142833.7188

CHEMCAD 5.6.0
FLOW SUMMlIRIES
Date: 12/14/2008 Time: 15:25:58
stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Water
Ethylbenzene
styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Water
Ethylbenzene
Styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Totalkmol/h
Flowrates in kmol/h
Water
Ethylbenzene
Styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
1
136.0000
210.0001
2007.7
0.00000
183.6000
0.0000
180.0000
0.0000
0.0000
1.8000
1.8000
0.0000
0.0000
5
802.0000
565.0002
-1.7045E+006
1.0000
8000.0000
8000.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
9
632.7648
170.0001
-1. 6713E+006
1.0000
8521.4582
8000.0000
516.3207
1.2031
0.0000
1.8000
2.1342
0.0000
0.0000
2
116.2373
200.0001
3312.3
0.00000
521.4581
0.0000
516.3207
1.2031
0.0000
1.8000
2.1342
0.0000
0.0000
6
BOO.OOOO
571.0001
-9. 7846E+005
1.0000
4590.4279
4590.4279
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
10
609.9968
160.0001
-1. 6713E+006
1.0000
8618.8442
8000.0000
402.1696
87.1901
69.2212
13.1984
18.9001
11.3984
16.7659
3-72
3
225.0000
180.0001
33226.
1.0000
521. 4581
0.0000
516.3207
1.2031
0.0000
1.8000
2.1342
0.0000
0.0000
7
65.0000
90.0001
-2.2469E+006
0.027598
8674.6644
8000.0000
348.2358
121.2506
89.0037
26.8858
33.1671
25.0858
31. 0353
11
650.0000
145.0001
-1. 6523E+006
1.0000
8618.8442
8000.0000
402.1696
87.1901
69.2212
13.1984
1B.9001
11.3984
16.7659
Page 10
4
lps feed
158.9001
600.0002
-1.9062E+006
1.0000
8000.0000
8000.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
8
801.2046
1BO.000l
-1.7045E+006
1.0000
8000.0000
8000.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
12
640.2558
135.0001
-1. 6523E+006
1.0000
8666.4510
8000.0000
340.0488
120.5307
88.0486
27.4656
33.4132
25.6656
31.2790

CHEMCAD 5.6.0
FLOW SUMMARIES
Date: 12/14/2008 Time: 15:25:58
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Water
Ethylbenzene
Styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Water
Ethylbenzene
styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
Stream No.
stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Water
Ethylbenzene
styrene
Hydrogen
Benzene
Toluene
Ethylene
Methane
13
270.0000
120.0001
-1.8J.55E+006
1.0000
8666.4510
8000.0000
340.0488
120.5307
88.0486
27.4656
33.4132
25.6656
31.2790
J.4
180.0000
105.0001
-1. 8507E+006
1.0000
8666.4510
8000.0000
340.0488
120.5307
88.0486
27.4656
33.4132
25.6656
31.2790
17 18
198.1571 65.0000
240.0001 75.0001
-16980. -2.2438E+006
1.0000 0.00000
216.3932 7928.5997
71.4000 7928.5997
0.0000 0.0000
0.0000 0.0000
88.0486 0.0000
0.0000 0.0000
0.0000 0.0000
25.6656 0.0000
31.2790 0.0000
21 22
66.1419 119.5297
40.0000
2115.5
0.00000
64.0487
0.0000
3.4006
0.1033
0.0000
27.4657
33.0791
0.0000
0.0000
60.0000
17048.
0.00000
457.4101
0.0000
336.6483
120.4275
0.0000
0.0000
0.3342
0.0000
0.0000
3-73
15
65.0000
90.0001
-2.2468E+006
0.027633
8666.4510
8000.0000
340.0488
120.5307
88.0486
27.4656
33.4J.32
25.6656
31.2790
19
65.0000
75.0001
14111.
0.00000
521.4584
0.0000
340.0488
120.5307
0.0000
27.4656
33.4132
0.0000
0.0000
23
105.1237
40.0000
1295.0
0.00000
337.8520
0.0000
336.3146
1.2031
0.0000
0.0000
0.3342
0.0000
0.0000
Page 11
16
65.0000
75.0001
-17998.
1.0000
216.3932
71.4000
0.0000
0.0000
88.0486
0.0000
0.0000
25.6656
31. 2790
20
65.0077
60.0000
14111.
0.00000
521. 4584
0.0000
340.0488
120.5307
0.0000
27.4656
33.4132
0.0000
0.0000
24
124.5388
60.0000
14822.
0.00000
119.5560
0.0000
0.3337
119.2223
0.0000
0.0000
0.0000
0.0000
0.0000

Job Name: styrene_JAS_2008 Date: 12/14/2008 Time: 15:25:58
FLOW SUMMARIES
Stream No. 25 26 27 28
Stream Name
Temp C 105.1895 66.2044 124.5951 65.0379
Pres kPa 210.0001 200.0001 200.0001 200.0001
Enth MJ/h 1304.6 2116.8 14825. -2.2438E+006
Vapor mo~e fraction 0.00000 0.00000 0.00000 0.00000
Total kmol/h 337.8520 64.0487 119.5560 7928.5997
F~owrates in kmol/h
Water 0.0000 0.0000 0.0000 7928.5997
Ethy~benzene 336.3146 3.4006 0.3337 0.0000
Styrene 1. 2031 0.1033 119.2223 0.0000
Hydrogen 0.0000 0.0000 0.0000 0.0000
Benzene 0.0000 27.4657 0.0000 0.0000
Toluene 0.3342 33.0791 0.0000 0.0000
Ethy~ene 0.0000 0.0000 0.0000 0.0000
Methane 0.0000 0.0000 0.0000 0.0000
Stream No. 29 30 31 32
Stream Name bfw-hps
Temp C 63.4602 63.4602 700.0000 90.0000
Pres kPa 75.0000 75.0000 571.0001 4237.0017
Enth MJ/h -18276. -2.2405E+006 -9. 9747E+005 -1.0521E+006
Vapor mole fraction 1.0000 0.00000 1.0000 0.00000
Tota~ kmol/h 266.2101 7914.5552 4590.4279 3742.6837
F~owrates in kmol/h
Water 83.0414 7913.8349 4590.4279 3742.6815
Ethy~benzene 20.2240 0.2604 0.0000 0.0000
styrene 6.0709 0.1590 0.0000 0.0000
Hydrogen 88.9275 0.0198 0.0000 0.0000
Benzene 7.9881 0.1837 0.0000 0.0000
Toluene 4.3474 0.0609 0.0000 0.0000
Ethy~ene 24.7121 0.0268 0.0000 0.0000
Methane 30.8988 0.0107 0.0000 0.0000
stream No. 33 34 35 36
Stream Name bfw-~ps ow
Temp C 253.7945 90.0000 158.9001 30.0000
Pres kPa 4237.0017 600.0002 600.0002 400.0002
Enth MJ/h -8. 8893E+005 -2. 3122E+005 -1. 9599E+005 -1. 4999E+008
Vapor mo~e fraction 1.0000 0.00000 1.0000 0.00000
Total kmo~/h 3742.6837 822.5403 822.5403 525103.0000
F~owrates in kmo~/h
Water 3742.6815 822.5393 822.5393 525103.0000
Ethy~benzene 0.0000 0.0000 0.0000 0.0000
Styrene 0.0000 0.0000 0.0000 0.0000
Hydrogen 0.0000 0.0000 0.0000 0.0000
Benzene 0.0000 0.0000 0.0000 0.0000
To~uene 0.0000 0.0000 0.0000 0.0000
Ethy~ene 0.0000 0.0000 0.0000 0.0000
Methane 0.0000 0.0000 0.0000 0.0000
13-7Lt

CHEMCAD 5.6.0
Job Name: styrene_JAS_2008
FLOW SUMMARIES
Date: 12/14/2008 Time: 15:25:58
Stream No. 37 38 39
stream Name hps
Temp C 40.0000 253.7945 253.0000
Pres kPa 400.0002 4237.0017 4237.0017
Enth MJ/h -1. 4960E+008 -2. 3330E+005 -2. 6321E+005
Total kmol/h 525103.0000 982.2584 982.2584
Flowrates in kmol/h
Water 525103.0000 982.2587 982.2587
Ethylbenzene 0.0000 0.0000 0.0000
Styrene 0.0000 0.0000 0.0000
Hydrogen 0.0000 0.0000 0.0000
Benzene 0.0000 0.0000 0.0000
Toluene 0.0000 0.0000 0.0000
Ethylene 0.0000 0.0000 0.0000
Methane 0.0000 0.0000 0.0000
Stream No. 50
Stream Name
Temp C 65.0000
Pres kPa 75.0001
Enth MJ/h -2516.7
Vapor mole fraction 0.34226
Total kmol/h 737.8516
Flowrates in kmol/h
Water 7.1.4000
Ethylbenzene 340.0488
Styrene 120.5307
Hydrogen 88.0486
Benzene 27.4656
Toluene 33.4132
Ethylene 25.6656
Methane 31.2790
Page 13
40
63.4602
75.0000
11962.
0.00000
493.8982
3.1242
327.7514
115.0207
0.0564
18.7141
28.7588
0.3469
0.1258

Job Name: Malei.c Anhydri.de_JAS 2008 Date: 12/14/2008 Ti.me: 15:53:41
FLOWSHEET SUMMARY
Equi.pment Label Stream Numbers
1 TOWR 9 8 -103 -102
2 KREA 6 15 -7 -16
3 HTXR 1 -110
4 MIXE 5 110 -100
5 FIRE 100 -6
6 HTXR 7 -8
7 COMP 4 -5
8 CSEP 102 -104
9 SHOR 11 -13
10 MIXE 104 103
11 REAC 105 -11
12 MIXE 14 10
13 KREA 91 -93
14 MIXE 92 93
15 DIVI 2 -91
Stream Connecti.ons
Stream Equi.pment Stream
From
1
2
3 14
4
5 7
6 5
7 2
8 6
9 12
Calculati.on mode
Flash algori.thm
To
3
15
7
4
2
6
1
1
Sequenti.al
Normal
10
11
12
13
14
15
16
91
92
Equi.pment Calculati.on Sequence
3 7 15 4 5 13 14 2
Equi.pment Recycle Sequence
1 8 11 9 12
Recycle Cut Streams
9
-105
-14
-12
-9
-3
-92
Equi.pment
From To
12
11 9
10
9
9 12
2
2
15 13
15 14
6 1 8 11
Recycle Convergence Method: Di.rect Substi.tuti.on
Max. loop i.terati.ons 150
3-76
Stream Equi.pment
From To
93 13 14
100 4 5
102 1 8
103 1 10
104 8 10
105 8 11
110 3 4
9 12 10

Job Name: Maleic Anhydride_JAS_2008 Date: 12/14/2008 Time: 15:53:41
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-003
1.000E-003
1.000E-003
1.000E-003
1.000E-003
Overall Mass Balance kmol/h
Maleic Anhydride
DibutylPhthalate
Nitrogen
Water
Oxygen
Benzene
Quinone
Carbon Dioxide
Maleic Acid
Sodium Nitrite
Sodium Nitrate
Total
COMPONENTS
ID #
1 272
2 910
3 46
4 62
5 47
6 40
7 1327
8 49
9 445
10 975
11 976
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Input Output
0.000 47.447
0.110 0.096
4410.000 4410.000
0.000 181.576
1170.000 739.222
84.600 6.634
0.000 1.438
0.000 265.381
0.000 1.000
2065.630 2065.630
2934.369 2934.369
10664.710 10652.793
Name Formula
Maleic Anhydride C4H203
DibutylPhthalate C16H2204
Nitrogen N2
Water H2O
Oxygen 02
Benzene C6H6
Quinone C6H402
Carbon Dioxide CO2
Maleic Acid C4H404
Sodium Nitrite NNa02
Sodium Nitrate NNa03
Ideal Vapor Pressure
Latent Heat
Library
Equip # 1 K-value : VAP
kg/h
Input
0.000
30.618
123541. 739
0.000
37438.830
6608.444
0.000
0.000
0.000
142518.161
249406.739
559545.000
'3-77
Output
4652.532
26.633
123541.727
3271.083
23654.369
518.187
155.468
11679.422
116.070
142518.161
249406.739
559540.378

CHEMCAD 5.6.0
Job Name: Maleic Anhydride_JAS_2008
EQUIPMENT SUMMARIES
Page 3
Date: 12/14/2008 Time: 15:53:41
Equip. No.
Name
No. of stages
1st feed stage
2nd feed stage
Top pressure kPa
Cond pressure drop
(kPa)
Colm pressure drop
(kPa)
Condenser type
Condenser mode
Condenser spec.
Condo comp i
Reboiler mode
Reboiler spec.
Reboiler comp i
Iterations
Calc cond duty MJ/h
Calc rebr duty MJ/h
Est. Dist. rate
(kmol/h)
Est. Reflux rate
(kmol/h)
Tolerance
Est. T top C
Est. T bottom C
Column diameter m
Tray space m
No of sections
Calc Reflux ratio
Calc Reflux mole
(kmol/h)
1
7
1
7
75.0000
2.0000
5.0000
1
5
0.5000
1
5
1.0000
4
200
-86766.0000
18620.3184
1500.0000
1200.0000
0.0010
100.0000
260.0000
4.5720
0.6096
1
0.1901
531.9780
Calc Reflux mass kg/h 140737.7344
Equip. No.
Name
Reactor type
Reaction phase
Thermal mode
Pressure In kPa
Pressure Drop kPa
Tout C
Q MJ/h
Reactor volume m3
Length of Tubes m
Diameter of Tubes m
Number of Tubes
Kinetic Reactor
2
2
1
5
250.0000
30.0000
609.0915
-72332.6094
34.5576
3.2000
0.0250
22000.0000
Summary
13
2
1
1
250.0000
10.0000
600.0000
-74636.0547
22.0008
'3-1 8

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:53:41
Number of steps
Tolerance
Conversion
Key
No. of Reactions
Molar Flow Unit
Activ. E/H of Rxn Unit
Volume Unit
Time Unit
U W/m2-K
Util dir
Util T at L C
T ref for HtRcn C
Overall IG Ht of Rxn
(MJ/h)
Edit Reaction No.
Mass unit
1000.0000
1.0000e-004
0.9500
7
4
1
7
1
2
100.0000
1
562.0779
25.0000
o
4
1
7
1
2
o
-87301.2031 -87216.9063
-1 -1
1 1
Reaction 1
5 -4.50e+000 1.0000e-010 O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
Reaction 3
RateConst = 2.3300e+004 Act.E = 2.142ge+004 Hrxn =
4
8
1.00e+000
4.00e+000
Reaction 4
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
'b-79
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO

Job Name: Maleic Anhydride_JAS 2008
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:53:41
Reaction 1
RateConst = 7.7000e+006 Act.E = 2.5143e+004 Hrxn = O.OOOOe+OOO
6 -1.00e+000 1.0000e+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
5 -4.50e+000 1.0000e-010 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
1
8
4
1.00e+000
2.00e+000
2.00e+000
Reaction 2
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
8
4
6.00e+000
3.00e+000
Reaction 3
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
RateConst = 2. 3300e+004 Act.E = 2. 142ge+004 Hrxn =
Reaction 4
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
AdsorbExp.
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO
O.OOOOe+OOO

Job Name: Maleic Anhydride_JAS 2008
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:53:42
Equip. No.
Name
1st Stream dp kPa
1st Stream T Out C
1st Stream VF Out
Calc Ht Duty MJ/h
1st Stream Pout kPa
Equip. No.
Name
OUtput Pressure kPa
Equip. No.
Name
Pressure Drop kPa
Temperature Out C
Heat Absorbed MJ/h
Fuel Usage (SCF)
Equip. No.
Name
Pressure out kPa
Type of Compressor
Efficiency
Actual power kW
Cp/Cv
Ideal Cp/Cv
Calc Pout kPa
Calc. mass flowrate
(kg/h)
3 6
10.0000 5.0000
260.0000
1.0000
1737.9625 -33195.7969
250.0000 215.0000
Mixer Summary
4 10
250.0000
Fired Heater Summary
5
15.0000
460.0000
28938.4980
40634.5117
Compressor Summary
7
250.0000
1
0.8000
2524.9128
1.3987
2019.9303
1.3974
250.0000
80490
12 14

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 15:53:42
Equip. No.
Name
Component No. 3
Component No. 5
Component No. 6
Component No. 8
Equip. No.
Name
Mode
Light key component
Light key split
Heavy key component
Heavy key split
R/Rmin
Number of stages
Min. No. of stages
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
Colm pressure kPa
Reflux ratio, ml.nl.mum
Calc. Reflux ratio
Colm pressure drop
(kPa)
Equip. No.
Name
Thermal mode
Temperature C
Key Component
Frac. Conversion
Calc H of Reac.
(J/kmol)
Stoichiometrics:
Maleic Anhydrid
Water
Maleic Acid
8
1.0000
1.0000
1.0000
1.0000
9
2
9.0000
0.9990
2.0000
8.0000e-005
50.0000
16.2351
15.9805
9.7945
-84058.4922
136038.0469
70.0000
1.2052
60.2580
10.0000
Reactor Summary
11
1
194.1640
4
1.0000
-3.5680e+007
-1.000
-1. 000
1.000
'3-82

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Page 8
Date: 12/14/2008 Time: 15:53:42
Divider Summary
Equip. No.
Name
Output stream #1
Output stream #2
15
0.9000
0.1000
'3-83

CHEMCAD 5.6.0
Job Name: Mal.eic Anhydride_JAS 2008
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mol.e fraction
Total. kmol./h
Fl.owrates in kmol./h
Mal.eic Anhydride
Dibutyl.Phthal.ate
Nitrogen
Water
Oxygen
Benzene
Quinone
Carbon Dioxide
Mal.eic Acid
Sodium Nitrite
Sodium Nitrate
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mol.e fraction
Total. kmol./h
Fl.owrates in kmol./h
Mal.eic Anhydride
Dibutyl.Phthal.ate
Nitrogen
Water
Oxygen
Benzene
Quinone
Carbon Dioxide
Mal.eic Acid
Sodium Nitrite
Sodium Nitrate
1
30.0000
260.0000
2107.0
0.00000
42.3000
0.0000
0.0000
0.0000
0.0000
0.0000
42.3000
0.0000
0.0000
0.0000
0.0000
0.0000
5
144.7681
250.0000
9497.8
1.0000
2790.0000
0.0000
0.0000
2205.0000
0.0000
585.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
Page 9
Date: 12/14/2008 Time: 15:53:42
2
460.0000
235.0000
42281.
1.0000
2832.2999
0.0000
0.0000
2205.0000
0.0000
585.0000
42.3000
0.0000
0.0000
0.0000
0.0000
0.0000
6
460.0000
235.0000
42281.
1.0000
2832.2999
0.0000
0.0000
2205.0000
0.0000
585.0000
42.3000
0.0000
0.0000
0.0000
0.0000
0.0000
3
586.1242
235.0000
-32355.
1.0000
2828.4993
22.0952
0.0000
2205.0000
90.5157
370.4695
4.2890
0.7110
135.4191
0.0000
0.0000
0.0000
7
609.0915
220.0000
-30051.
1.0000
2825.1976
26.3516
0.0000
2205.0000
92.0598
368.7527
2.3447
0.7272
129.9619
0.0000
0.0000
0.0000
4
30.0000
101.0000
408.08
1.0000
2790.0000
0.0000
0.0000
2205.0000
0.0000
585.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
8
260.0000
215.0000
-63247.
1.0000
2825.1976
26.3516
0.0000
2205.0000
92.0598
368.7527
2.3447
0.7272
129.9619
0.0000
0.0000
0.0000

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Maleic Anhydride
DibutylPhthalate
Nitrogen
Water
Oxygen
Benzene
Quinone
Carbon Dioxide
Maleic Acid
Sodium Nitrite
Sodium Nitrate
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Maleic Anhydride
DibutylPhthalate
Nitrogen
Water
Oxygen
Benzene
Quinone
Carbon Dioxide
Maleic Acid
Sodium Nitrite
Sodium Nitrate
9
329.1962
80.0000
-3.2707E+005
0.00000
501.6278
0.0000
501.6268
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0010
0.0000
0.0000
13
189.2050
70.0000
-11363.
0.00000
26.2568
24.8517
0.0401
0.0000
0.0000
0.0000
0.0000
0.3650
0.0000
1.0000
0.0000
0.0000
Page 10
Date: 12/14/2008 Time: 15:53:42
10
320.0000
100.0000
-72.616
0.00000
0.1100
0.0000
0.1100
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
14
329.1982
80.0000
-3.2700E+005
0.00000
501.5177
0.0000
501. 5167
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0010
0.0000
0.0000
'3-85
11
194.1640
82.0000
-3.9034E+005
0.00000
527.7746
24.8517
501.5569
0.0000
0.0000
0.0000
0.0000
0.3650
0.0000
1.0010
0.0000
0.0000
15
418.5000
200.0000
-2.0055E+006
0.00000
5000.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
2065.6302
2934.3694
12
83.6666
75.0000
-68118.
1.0000
2798.0366
0.4999
0.0556
2205.0000
91. 0598
368.7527
2.3447
0.3622
129.9620
0.0000
0.0000
0.0000
16
562. 0779
200.0000
-1. 9331E+006
0.00000
5000.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
2065.6302
2934.3694

CHEMCAD 5.6.0
Job Name: Male:i..c Anhydr:i..de_JAS 2008
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fract:i..on
Total kmol/h
Flowrates :i..n kmol/h
Male:i..c Anhydr:i..de
D:i..butylPhthalate
N:i..trogen
Water
Oxygen
Benzene
Qu:i..none
Carbon D:i..ox:i..de
Male:i..c Ac:i..d
Sod:i..um N:i..tr:i..te
Sod:i..um N:i..trate
Stream No.
Stream Name
Temp C
Pres kPa
Enth MJ/h
Vapor mole fract:i..on
Total kmol/h
Flowrates :i..n kmol/h
Male:i..c Anhydr:i..de
D:i..butylPhthalate
N:i..trogen
Water
Oxygen
Benzene
Qu:i..none
Carbon D:i..ox:i..de
Male:i..c Ac:i..d
Sod:i..um N:i..tr:i..te
Sod:i..um N:i..trate
91
460.0000
235.0000
38053.
1.0000
2549.0700
0.0000
0.0000
1984.4998
0.0000
526.5000
38.0700
0.0000
0.0000
0.0000
0.0000
0.0000
102
194.0467
82.0000
-3.9034E+005
0.00000
528.8044
25.8517
501.5569
0.0025
1.0000
0.0005
0.0266
0.3650
0.0002
0.0010
0.0000
0.0000
Page 11
Date: 12/14/2008 T:i..me: 15:53:42
92
460.0000
235.0000
4228.1
1.0000
283.2300
0.0000
0.0000
220.5000
0.0000
58.5000
4.2300
0.0000
0.0000
0.0000
0.0000
0.0000
103
83.6769
75.0000
-68121.
1.0000
2798.0069
0.4999
0.0556
2204.9976
91.0598
368.7522
2.3181
0.3622
129.9618
0.0000
0.0000
0.0000
93
600.0000
240.0000
-36583.
1.0000
2545.2694
22.0952
0.0000
1984.4998
90.5157
311.9694
0.0590
0.7110
135.4191
0.0000
0.0000
0.0000
104
194.0467
82.0000
2.6245
1.0000
0.0298
0.0000
0.0000
0.0025
0.0000
0.0005
0.0266
0.0000
0.0002
0.0000
0.0000
0.0000
100
143.0405
250.0000
13343.
1.0000
2832.2999
0.0000
0.0000
2205.0000
0.0000
585.0000
42.3000
0.0000
0.0000
0.0000
0.0000
0.0000
105
194.0467
82.0000
-3.9034E+005
0.00000
528.7746
25.8517
501.5569
0.0000
1.0000
0.0000
0.0000
0.3650
0.0000
0.0010
0.0000
0.0000

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No. 110
Stream Name
Temp C 112.6301
Pres kPa 250.0000
Enth MJ/h 3844.9
Vapor mole fraction 1.0000
Total kmol/h 42.3000
Flowrates in kmol/h
Maleic Anhydride 0.0000
DibutylPhthalate 0.0000
Nitrogen 0.0000
Water 0.0000
Oxygen 0.0000
Benzene 42.3000
Quinone 0.0000
Carbon Dioxide 0.0000
Maleic Acid 0.0000
Sodium Nitrite 0.0000
Sodium Nitrate 0.0000
Page 12
Date: 12/14/2008 Time: 15:53:42
~-'37

13.15 Simulation ofthe Ethylene Oxide process, Appendix B, Figure B.6.I and Table B.6.1.
The simulation results are shown in the output following this page. The PSRK model
was used for the phase equilibrium (except for towers T-701 and T-702 in which the
UNIFAC model was used), and the SRK model was used for enthalpies. The required
minimum input information for this simulation is given below:
2. Specification offeed streams (Streams 1,2, and process water)
R-70I, 702 conversion ofkey component
E-701, 702, 703, 704, 705, 706, 707 outlet temperatures
T-703 identification ofand recovery ofkey components
T-70I,702 number oftrays
C-70I, 702, 703, 704 outlet pressures, efficiencies
13.16 Simulation ofthe Formalin process, Appendix B, Figure B.7.l and Table B.7.1. The
simulation results are shown in the output following this page. The Elliot-Suresh-
Donahue Equation of State (ESDK) model was used for the phase equilibrium, and the
"latent heat" model (zero heat ofmixing) was used for enthalpies. The required
minimum input information for this simulation is given below:
2. Specification offeed streams (Streams 1,2, and 11)
R-801conversion ofkey component, outlet temperature
E-80I, 802, 803, 806 outlet temperatures
T-802 identification of and recovery ofkey components
T-80I number oftrays
C-80I outlet pressure, efficiency
13.17 The batch aspects vary by simulator. For Chemcad, these are Batch Reactor, Batch
Column, Tank UnitOp, Dynamic Vessel UnitOp, and Scheduling.
1~-8'8

Job Name: ethy~ene oxide with recyc~e Date: 12/14/2008 Time: 14:37:10
FLOWSHEET SUMMARY
1 COMP 1 -3
2 HTXR 3 -4
3 COMP 4 -5
4 HTXR 5 -6
5 COMP 6 -7
6 MIXE 7 9 -10
7 VAIN 2 -8
8 HTXR 10 -11
9 KREA 11 -12
10 HTXR 12 -13
11 COMP 13 -14
12 sens 15 14 -16 -17
13 HTXR 16 -18
14 VAIN 18 -19
15 KREA 19 -20
16 HTXR 20 -21
17 COMP 21 -22
18 sens 23 22 -24 -25
19 DIVI 24 -27 -26
20 VAIN 27 -28
21 MlXE 8 28 -9
22 MIXE 17 25 -29
23 HTXR 29 -30
24 VAIN 102 -31
25 sens 31 -32 -33
100 VAIN 30 -101
101 CSEP 101 -103 -102
200 SHOR 90 -91 -92
201 SHOR 93 -94 -95
Stream Connections
Stream Equipment Stream Equipment Stream Equipment
1 1 15 12 29 22 23
2 7 16 12 13 30 23 100
3 1 2 17 12 22 31 24 25
4 2 3 18 13 14 32 25
5 3 4 19 14 15 33 25
6 4 5 20 15 16 90 200
7 5 6 21 16 17 91 200
8 7 21 22 17 18 92 200
9 21 6 23 18 93 201
10 6 8 24 18 19 94 201
11 8 9 25 18 22 95 201
12 9 10 26 19 101 100 101
13 10 11 27 19 20 102 101 24
14 11 12 28 20 21 103 101
3,.;.CC?9

Job Name: ethylene oxide with recycle Date: 12/14/2008 Time: 14:37:10
Calculation mode
Flash algorithm
Sequential
Normal
1 2 3 4 5 7 200 201 6
17 18 19 20 21 22 23 100 101
8 9
24 25
10 11 12 13 14 15
6 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Recycle Cut Streams
9
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-003
1.000E-003
1.000E-003
1.000E-003
1.000E-003
Overall Mass Balance kmol/h
Input OUtput
Ethylene 712.911 336.269
Ethylene Oxide 708.001 1068.702
Carbon Dioxide 0.000 31.731
Oxygen 3281.353 3052.653
Nitrogen 14100.092 14096.652
Water 80058.749 80090.472
kg/h
Input
20000.000
31189.564
0.000
105000.002
395000.000
1442258.306·
Output
9433.694
47079.518
1396.474
97681.858
394904.000
1442829.798
Total 98861.109 98676.483 1993447.974 1993325.051
COMPONENTS
ID #
1 22
2 129
3
4
5
6
49
47
46
62
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Name
Ethylene
:Et:llyleru.~ 9Jti,ci!;!
Carbon Dioxide
Oxygen
Nitrogen
Water
l?SRK
IPl? flag on
SRK
Library
Formula
C2H4
C2H40
C02
02
N2
H20
16

Equip :If 12 K-value : UNIF
Equip :If 18 K-value : UNIF
Std vapor rate reference temperature is o C.
Atmospheric pressure is 1.0132 bar.
* UNIFAC main groups 2 and 1058, No interaction parameters.
'$ -9'

Job Name: ethylene oxide with recycle
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:37:10
Equip. No.
Name
Pressure out bar
Type of Compressor
Efficiency
Actual power MJ/h
Cp/Cv
Theoretical power
(MJ/h)
Ideal Cp/Cv
Calc Pout bar
Equip. No.
Name
Pressure out bar
Type of Compressor
Efficiency
Actual power MJ/h
Cp/Cv
Theoretical power
(MJ/h)
Ideal Cp/Cv
Calc Pout bar
Equip. No.
Name
1st Stream dp bar
1st Stream T Out C
Calc Ht Duty MJ/h
1st Stream Pout bar
Equip. No.
Name
1st Stream dp bar
... 1st Stream T Out C
Calc Ht Duty MJ/h
1st Stream Pout bar
Compressor Summary
1
3.0000
1
0.8000
68510.1172
1.3990
54808.0977
1.3966
3.0000
17
30.0000
1
0.8000
19666.7070
1.4272
15733.3662
Heat
1.3679
30.0000
Exchanger
2
0.3000
45.0000
-58486.8594
2.7000
13
0.3000
240,0000
229646.0781
29.7000
3
9.0000
1
0.8000
82629.3125
1.4008
66103.4531
1.3949
9.0000
Summary
4
0.3000
45.0000
-83202.0703
8.7000
16
0.3000
45.0000
-212483.3281
25.4500
3-92
5 11
27.0000 30.0000
1 1
0.8000 0.8000
77204.5703 19803.6094
1.4093 1.4259
61763.6602 15842.8877
1.3898 1.3663
27.0000 30.0000
8 10
0.3000 0.3000
240.0000 45.0000
147433.5781 -215064.4844
26.5000 25.4500
23
0.3000
45.0000
-21591.4238
89.7000

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:37:10
Mixer Summary
Equip. No. 6 21
Name
Output Pressure bar 26.8000
Valve Summary
Equip. No. 7 14
Name
Pressure out bar 27.0000 26.5000
Equip. No. 100
Name
Pressure out bar 20.0000
Equip. No.
Name
Reactor type
Reaction phase
Thermal mode
Pressure Drop bar
Tout C
Q MJ/h
Reactor volume m3
Concentration Flag
Specify calc. mode
Conversion
Key
No. of Reactions
Molar Flow Unit
Volume Unit
Time Unit
Overall IG Ht ofRxn
(MJ/h)
Mass unit
9
2
1
1
0.7500
240.0000
-33103.3867
201.8075
1
1
0.2000
1
3
2
6
1
2
-32822.8203
1
15
2
1
1
0.7500
240.0000
-26179.0000
201.5797
1
1
0.2000
1
3
2
6
1
2
-26013.8945
1
22
90.0000
20
27.0000
Reaction 1
RateConst = 1. 9600e+000 Act.E = 2.4000e+003 Hrxn =
1 -1.00e+000 1.0000e+000 9.8000e-004 -1. 1200e+004
4 -5.00e-001 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3 -'13
24
10.0000
O.OOOOe+OOO
AdsorbExp.
1.0000e+000
O.OOOOe+OOO
O.OOOOe+OOO

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:37:10
Reaction 2
RateConst = 9.3600e-002 Act.E = 6.4000e+003 Hrxn = O.OOOOe+OOO
1 -1.00e+000 1.0000e+000 9.8000e-004 -1. 1200e+004 1.0000e+000
4 -3.00e+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
3 2.00e+000 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 3
RateConst = 4.2768e-001 Act.E = 6.2000e+003 Hrxn = O.OOOOe+OOO
2 -1.00e+000 2.0000e+000 3.3000e-005 -2.1200e+004 2.0000e+000
Reaction 1
RateConst = 1. 9600e+000 Act.E = 2.4000e+003 Hrxn = O.OOOOe+OOO
1 -1.00e+000 1.0000e+000 9.8000e-004 -1. 1200e+004 1.0000e+000
4 -5.00e-001 O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO O.OOOOe+OOO
Reaction 2
RateConst = 9. 3600e-002 Act.E = 6.4000e+003 Hrxn = O.OOOOe+OOO
1 -1.00e+000 1.0000e+000 9.8000e-004 -1. 1200e+004 1.0000e+000
Reaction 3
RateConst = 4.2768e-001 Act.E = 6.2000e+003 Hrxn = 0.0000e+000
2 -1.00e+000 2.0000e+000 3.3000e-005 -2.1200e+004 2.0000e+000

Scds Rigorous Distillation Summary
Equip. No.
Name
No. of stages
1st feed stage
2nd feed stage
Condenser mode
Condenser spec
Cond comp i pos.
Reboiler mode
Reboiler spec.
Reboiler comp i
Co1m press drop bar
Est. mst. rate
(kmol/h)
Est. reflux rate
(kmol/h)
Est. T top C
Est. T bottom C
Top pressure bar
Calc Cond duty MJ/h
Calc Reblr duty MJ/h
Calc Reflux mole
(kmol/h)
Calc Reflux ratio
Tray type
12
5
1
5
o
o
o
o
18000.0000
25.0000
25.0000
20059.8223
362247.1563
3
18
5
1
5
o
o
o
o
18000.0000
25.0000
25.0000
25
25
12
o
6
0.9990
2
3
182.3000
2
0.5000
284.0000
240.0000
92.0000
182.0000
10.0000
-14398.4043
438665.1250
20054.5645 313.9723
362040.4063
3
3 -95
0.8909
13823.2471
3

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:37:10
Column diameter
Tray space m
Thickness (top)
Thickness (bot)
No of sections
No of passes (Sl)
Weir side width
Weir height m
System factor
Equip. No.
Name
Output stream #1
Output stream #2
Equip. No.
Name
Component No. 1
Component No. 3
Component No. 4
Component No. 5
Equip. No.
Name
Mode
m
m
m
m
Light key component
Light key split
Heavy key component
Heavy key split
··········R/Rmin
Number of stages
Min. No. of stages
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
Colm pressure bar
Reflux ratio, m:Ln:Lmum
Calc. Reflux ratio
Number of points
Lower bound R/Rmin
Upper bound R/Rmin
Colm pressure drop
(bar)
4.8768
0.6096
0.0953
0.0953
1
1
0.8827
0.0508
1.0000
Divider Summary
19
0.5000
0.5000
4.7244
0.6096
0.0889
0.0889
1
1
0.8446
0.0508
1.0000
101
1.0000
1.0000
1.0000
1.0000
200 201
2 2
2.0000 2.0000
0.9990 0.9990
6.0000 6.0000
1.0000e-005 0.0010
1:5000 ···1.5000
15.7341 17.5054
8.8737 6.8893
10.2085 9.2527
-24506.5488 -9023.3018
448850.5313 9181.1094
10.0000 10.0000
1:4672 0.1220
2.2008 0.1829
0 10
1.1000
2.0000
0.5000 0.5000
3 -9b
5.4864
0.6096
0.0333
0.0333
1
1
0.9779
0.0508
1.0000

FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
1
25.0000
1.0132
-115.73
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
5
206.1111
9.0000
92537.
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
9
26.3315
27.0000
31440.
1.0000
18258.1383
1047.9103
6.4718
31.7102
3049.7688
14091.2997
30.9775
2
25.0000
50.0000
35030.
1.0000
712.9108
712.9108
0.0000
0.0000
0.0000
0.0000
0.0000
6
45.0000
8.7000
9334.8
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
10
106.7495
26.8000
1.1798E+005
1.0000
35635.3116
1047.8481
6.4621
31. 7042
6330.3852
28187.9521
30.9611
'3 -Cfl
3
159.1927
3.0000
68394.
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
7
195.2089
27.0000
86539.
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
11
240.0000
26.5000
2. 6541E+005
1.0000
35635.3116
1047.8481
6.4621
31. 7042
6330.3852
28187.9521
30.9611
4
45.0000
2.7000
9907.5
1.0000
17381.4452
0.0000
0.0000
0.0000
3281.3525
14100.0925
0.0000
8
-6.2991
27.0000
35030.
1.0000
712.9108
712.9108
0.0000
0.0000
0.0000
0.0000
0.0000
12
240.0000
25.7500
2.3231E+005
1.0000
35535.1456
838.5940
206.7925
49.5512
6203.4489
28187.9521
48.8082

FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
13
45.0000
25.4500
17246.
1.0000
35535.1456
838.5940
206.7925
49.5512
6203.4489
28187.9521
48.8082
17
51.9139
30.0000
-5. 6901E+006
0.00000
20181.7757
0.7036
191.3511
0.0091
1.4518
2.6777
19985.5828
21
45.0000
25.4500
-2271.2
1.0000
35273.1783
670.5651
175.8233
63.4319
6100.9706
28185.2748
77.1108
14
63.7223
30.0000
37050.
1.0000
35535.1456
838.5940
206.7925
49.5512
6203,4489
28187.9521
48.8082
18
240.0000
29.7000
2.3639E+005
1.0000
35353.3685
837.8904
15.4427
49.5421
6202.0000
28185.2748
63.2209
22
63.7760
30.0000
17396.
1.0000
35273.1783
670.5651
175.8233
63.4319
6100.9706
28185.2748
77 .1108
15
25.0000
30.0000
-5. 7204E+006
0.00000
20000.0003
0.0000
0.0000
0.0000
0.0000
0.0000
20000.0003
19
239.9476
26.5000
2.3639E+005
1.0000
35353.3649
837.8904
15.4427
49.5421
6202.0000
28185.2748
63.2209
23
25.0000
30.0000
-5.7204E+006
0.00000
20000.0003
0.0000
0.0000
0.0000
0.0000
0.0000
20000.0003
16
30.2942
30.0000
6746.0
1.0000
35353.3720
837.8904
15.4413
49.5421
6202.0000
28185.2766
63.2260
20
240.0000
25.7500
2.1021E+005
1.0000
35273.1783
670.5651
175.8233
63.4319
6100.9706
28185.2748
77.1108
24
30.0814
30.0000
-7180.0
1.0000
35090.4550
669.9992
12.9436
63.4204
6099.5377
28182.5994
61.9550

FLOW SUMMARIES
Date: 12/14/2008 Time: 14:37:10
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
25
52.2611
30.0000
-5. 6959E+006
0.00000
20182.7218
0.5659
162.8796
0.0115
1.4329
2.6751
20015.1567
29
52.0733
90.0000
-1.1386E+007
0.00000
40364.5011
1.2695
354.2293
0.0206
2.8846
5.3527
40000.7448
33
182.3000
10.5000
-1.0958E+007
0.00000
40002.5698
0.0000
2.1765
0.0000
0.0000
0.0000
40000.3975
26
30.0814
30.0000
-3590.0
1.0000
17545.2275
334.9996
6.4718
31.7102
3049.7688
14091.2997
30.9775
30
45.0000
89.7000
-1.1408E+007
0.00000
40364.5011
1.2695
354.2293
0.0206
2.8846
5.3527
40000.7448
90
45.0228
10.0000
-1.1408E+007
0.00000
40355.0000
0.0000
354.2293
0.0000
0.0000
0.0000
40000.7483
1~-99
27
30.0814
30.0000
-3590.0
1.0000
17545.2275
334.9996
6.4718
31. 7102
3049.7688
14091.2997
30.9775
31
45.0228
10.0000
-1. 1408E+007
0.00000
40355.0000
0.0000
354.2293
0.0000
0.0000
0.0000
40000.7483
91
86.4069
10.0000
-25596.
0.00000
354.2750
0.0000
353.8749
0.0000
0.0000
0.0000
0.4000
28
29.4755
27.0000
-3590.0
1.0000
17545.2275
334.9996
6.4718
31.7102
3049.7688
14091.2997
30.9775
32
86.4025
10.0000
-25451.
0.00000
352.4049
0.0000
352.0528
0.0000
0.0000
0.0000
0.3524
92
182.3307
10.5000
-1.0958E+007
0.00000
40000.7058
0.0000
0.3542
0.0000
0.0000
0.0000
40000.3514

CHEMCAD 5.6.0
FLOW SUMMARIES
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
Stream No.
Stream Name
Temp C
Pres bar
Enth MJ/h
Vapor mole fraction
Total kmol/h
Flowrates in kmol/h
Ethylene
Ethylene Oxide
Carbon Dioxide
Oxygen
Nitrogen
Water
93
92.3400
10.0000
-41538.
0.00000
411.7658
0.0000
353.7715
0.0000
0.0000
0.0000
57.9943
102
45.0200
20.0000
-1. 1408E+007
0.00000
40355.0000
0.0000
354.2293
0.0000
0.0000
0.0000
40000.7448
Page 12
Date: 12/14/2008 Time: 14:37:10
94
86.3659
10.0000
-25468.
0.00000
353.4758
0.0000
353.4178
0.0000
0.0000
0.0000
0.0580
103
45.0200
20.0000
62.558
1.0000
9.5275
1.2695
0.0000
0.0206
2.8846
5.3527
0.0000
3 -I DO
95 101
178.2449 45.0200
10.5000 20.0000
-15912. -1. 1408E+007
0.00000 0.00000
58.2900 40364.5011
0.0000 1.2695
0.3538 354.2293
0.0000 0.0206
0.0000 2.8846
0.0000 5.3527
57.9363 40000.7448

Job Name: formalin JAS 2008 Date: 12/14/2008 Time: 14:45:38
FLOWSHEET SUMMARY
1 PUMP 3 -4
2 COMP 1 -5
3 HTXR 4 -6
4 HTXR 5 -7
5 MIXE 6 7 -8
6 REAC 8 -68
7 REAC 68 -69
8 HTXR 69 -9
9 HTXR 9 -10
10 SCDS 11 10 -12 -13
11 SHOR 70 -14 -15
12 PUMP 15 -16
13 HTXR 16 -17
14 VAIN 14 -18
15 FLAB 21 -22 -23
16 FLAB 24 -25 -26
65 MIXE 12 64 -65
66 MIXE 18 2 -3
67 CSEP 13 -64 -70
71 TOWR 72 -73 -74
75 SCDS 76 -77 -78
99 KREA 98 -99
101 FLAB 100 -101 -102
102 SCDS 103 101 -104 -105
103 MIXE 105 102 -106
Stream Connections
Stream Equipment Stream Equipment Stream Equipment
1 2 16 12 13 73 71
2 66 17 13 74 71
3 66 1 18 14 66 76 75
4 1 3 21 15 77 75
5 2 4 22 15 78 75
6 3 5 23 15 98 99
7 4 5 24 16 99 99
8 5 6 25 16 100 101
9 8 9 26 16 101 101 102
10 9 10 64 67 65 102 101 103
11 10 65 65 103 102
12 10 65 68 6 7 104 102
13 10 67 69 7 8 105 102 103
14 11 14 70 67 11 106 103
15 11 12 72 71
Calculation mode Sequential
Flash algorithm Normal
'6- D)

CHEMCAD 5.6.0
2 4 15 16 71 75 99 101 102 103
67 11 14 66 12 13 65
1 3
1 3 5 6 7 8 9 10 67 11 14 66
Recycle Cut Streams
3
Flow rate
Temperature
Pressure
Enthalpy
Vapor frac.
1.000E-003
1.000E-003
1.000E-003
1.000E-003
1.000E-003
5 6 7
Page 2
8 9 10

Input OUtput Input OUtput
Methanol 256.662 109.329 8223.975 3503.127
Oxygen 1630.813 1558.081 52184.399 49857.026
Formaldehyde 250.558 397.891 7523.257 11947.076
Water 628.299 773.764 11318.800 13939.361
Hydrogen 1.655 3.523 3.336 7.101
Nitrogen 6249.559 6249.560 175075.155 175075.168
Total 9017.547 9092.147 254329.000 254329.000
3-l03

CHEMCAD 5.6.0
COMPONENTS
!D #
1 117
2 47
3 114
4 62
5 1
6 46
THERMODYNAMICS
K-value model
Enthalpy model
Liquid density
Name
Methanol
Oxygen
Formaldehyde
Water
Hydrogen
Nitrogen
ESDK
Latent Heat
Library
Formula
CH40
02
CH20
H2O
H2
N2
ESDK Pa~ameters:
I J Kij Kji KijT
1 3 -0.15000 -0.15000 0.00000
1 4 0.01550 0.01550 0.00000
1 5 -0.08500 -0.08500 0.00000
1 6 0.17270 0.17270 0.00000
2 6 -0.00300 -0.00300 0.00000
3 4 0.34000 0.34000 -111.00000
4 5 -0.00700 -0.00700 0.00000
4 6 0.24500 0.24500 0.00000
5 6 -0.02220 -0.02220 0.00000
'3- 04-
KjiT
0.00000
0.00000
0.00000
0.00000
0.00000
-111.00000
0.00000
0.00000
0.00000
Page 4

CHEMCAD 5.6.0
Job Name: formalin JAS 2008
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:45:38
Equip. No.
Name
Output pressure kPa
Efficiency
Calculated power MJ/h
Calculated Pout kPa
Head m
Vol. flow rate m3/h
Mass flow rate kg/h
Equip. No.
Name
Pressure out kPa
Type of Compressor
Efficiency
Actual power MJ/h
Cp/Cv
Theoretical power
(MJ/h)
Ideal Cp/Cv
Calc Pout kPa
Calc. mass flowrate
(kg/h)
Equip. No.
Name
1st Stream dp kPa
1st Stream T Out C
Calc Ht Duty MJ/h
LMTD Corr Factor
1st Stream Pout kPa
Equip. No.
Name
1st Stream dp kPa
1st Stream T Out C
Calc Ht Duty MJ/h
LMTD Corr Factor
1st Stream Pout kPa
Pump Summary
1 12
300.0000 350.0000
0.8000 0.7500
0.9941 1.5183
300.0000 350.0000
25.9730 29.7754
4.0002 5.6897
3120.1714 3897.0942
Compressor Summary
2
300.0000
1
0.7000
657.5833
1.4003
460.3083
1. 3989
300.0000
4211
3
35.0000
150.0000
4110.0771
1.0000
265.0000
13
35.0000
35.0000
-1169.7457
1.0000
315.0000
4
35.0000
200.0000
76.7449
1.0000
265.0000
'3 - 05
8
35.0000
200.0000
-8928.3965
1. 0000
185.0000
Page 5
9
35.0000
100.0000
-983.1221
1.0000
150.0000

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:45:38
Mixer Summary
Equip. No. 5 65 66
Name
OUtput Pressure kPa 255.0000 101.3250
Reactor Summary
Equip. No. 6 7
Name
Thermal mode 1 1
Temperature C 934.7972 923.0992
Key Component 2 1
Frac. Conversion 0.9950 0.0500
Reactor Pressure kPa 220.0000
Calc H of Reac. -313560.0000 85040.0000
(kJ/kmol.)
Stoichiometrics:
Methanol. -1. 000 -1.000
Oxygen -0.500 O.OOOE+OOO
Formaldehyde 1.000 1.000
Water 1.000 O.OOOE+OOO
Hydrogen O.OOOE+OOO 1.000
Scds Rigorous Distill.ation Summary
Equip. No. 10 75 102
Name
No. of stages 20 24 20
1st feed stage 1 14 1
2nd feed stage 20 0 20
Condenser mode 0 7 0
Condenser spec 0.9500
Cond comp i pos. 0 1 0
Reboiler mode 0 7 0
Reboiler spec. 0.9500
Reboiler comp i 0 4 0
Colm press drop kPa 10.0000 20.0000 10.0000
Est. dist. rate 320.0000 20.0000 120.0000
(kmol/h)
Est. refl.ux rate 260.0000
(kmol/h)
Est. T top C 150.0000 71.0000 50.0000
Est. T bottom C 50.0000 106.0000 50.0000
Top pressure kPa 140.0000 130.0000 105.0000
Calc Cond duty MJ/h -32465.8008
Calc Reblr duty MJ/h 32748.3359
Iterations 0 100 0
Calc Reflux mole 157.4460 864.0048 71.6312
3-06
Page 6
103

EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:45:38
(Janol/h)
Calc Reflux ratio
Column type
Column diameter m
Thickness (top) m
Thickness (bot) m
No of sections
Calc. tolerance
Equip. No.
Name
Mode
Light key component
Light key split
Heavy key component
Heavy key split
R/Rmin
Number of stages
Min. No. of stages
Feed stage
Condenser duty MJ/h
Reboiler duty MJ/h
Colm pressure kPa
Calc. Reflux ratio
Number of points
Lower bound R/Rmin
Upper bound R/Rmin
Colm pressure drop
(kPa)
2876.5950
1
0.8718
0.0016
0.0143
1
0.0012
37.4236
24629.4746
o
o
0.0012
11
3
1.0000
0.9500
4.0000
0.0500
1.3000
18.0345
9.7624
9.6609
-6779.3984
7062.0737
130.0000
5.4205
7.0467
10
1.1000
2.0000
20.0000
1302.0605
o
o
0.0003
Shortcut Distillation # 11 Case Studies:
R/Rmin Reflux ratio No. of stgs Feed stg Qcond Qreb
MJ/h MJ/h
1.10 5.96 23.1 12.2 -S.866E+003 6.149E+003
1.20 6.51 19.9 10.6 -6. 323E+003 6.606E+003
1.30 7.05 18.0 9.7 -6. 780E+003 7.063E+003
1.40 7.59 16.8 9.0 -7.236E+003 7. 518E+003
1.50 8.13 15.8 8.5 -7. 692E+003 7. 975E+003
1. 60 8.67 15.1 8.2 -8. 149E+003 8. 432E+003
1. 70 9.22 14.6 7.9 -8.606E+003 8. 889E+003
1. 80 9.76 14.1 7.7 -9.063E+003 9. 346E+003
1. 90 10.30 13.8 7.5 -9. 520E+003 9.803E+003
2.00 10.84 13.5 7.3 -9. 977E+003 1.026E+004
3-107

CHEMCAD 5.6.0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:45:38
Equip. No.
Name
Pressure out kPa
Equip_ No.
Name
Flash Mode
Param 1
Param 2
Heat duty MJ/h
K values:
Methanol
Oxygen
Formaldehyde
Water
Hydrogen
Nitrogen
Equip. No.
Name
Component No. 2
Component No. 5
Component No. 6
Equip. No.
Name
No. of stages
1st feed stage
Top pressure kPa
Colm pressure drop
(kPa)
Condenser mode
Condenser spec.
Condo comp i
Reboiler mode
Reboiler spec.
Reboiler comp i
Iterations
Calc cond duty MJ/h
Valve Summary
14
120.0000
Flash Summary
15 lEi
0 0
1.184 1.325
1163.982 4151.597
0.040 1.023
0.468 0.986
5425.908 22992.754
4822.877 19785.629
67
1.0000
1.0000
1.0000
101
2
50.0000
115.0000
-5083.4517
0.226
5964.569
0.020
0.164
40958.988
30731.762
71
24
14
130.0000
20.0000
7
0.9500
1
7
0.9500
4
100
-32370.5156
3 - Dca
Page 8

CHEMCAD 5. 6. 0
EQUIPMENT SUMMARIES
Date: 12/14/2008 Time: 14:45:38
Calc rebr duty MJ/h 32654.2363
Est. Dist. rate 20.0000
(kmol/h)
Est. Reflux rate 260.0000
(kmol/h)
Est. T top C 70.0000
Est. T bottom C 103.0000
Tray type 3
Column diameter m 2.2860
Tray space m 0.6096
Thickness (top) m 0.0016
Thickness (bot) m 0.0032
No of sections 1
Calc Reflux ratio 37.3041
Calc Reflux mole 861.2657
(kmol/h)
Calc Reflux mass kg/h 24551.4297
Equip. No.
Name
Reactor type
Reaction phase
Thermal mode
Pressure In kPa
Tout C
Q MJ/h
Reactor volume m3
Concentration Flag
Specify calc. mode
Conversion
Key
No. of Reactions
Molar Flow Unit
Volume Unit
Overall IG Ht of Rxn
(MJ/h)
Mass unit
Partial P unit
99
2
1
1
250.0000
354.2749
-11896.6875
32.4871
1
1
0.9000
1
2
2
6
1
-13222.2568
2
2
3-109
Page 9

FLOW SUMMARIES
Stream No. 1 2 3 4
Stream Name
Temp C 25.0000 30.0000 40.6637 40.7822
Pres kPa 101.3250 120.0000 101.3250 300.0000
Enth MJ/h 3. 6966E-005 -18349. -24019. -24018.
Total kmol/h 145.94 76.92 99.92 99.92
Flowrates i.n kmol/h
Methanol 0.00 76.92 94.11 94.11
Oxygen 30.66 0.00 0.00 0.00
Formaldehyde 0.00 0.00 0.00 0.00
Water 0.00 0.00 5.81 5.81
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 115.28 0.00 0.00 0.00
Stream No. 5 6 7 8
Stream Name
Temp C 183.0128 150.0000 200.0000 171.9390
Pres kPa 300.0000 265.0000 265.0000 255.0000
Enth MJ/h 657.58 -19908. 734.33 -19173.
Total kmol/h 145.94 99.92 145.94 245.86
Flowrates in kmol/h
Methanol 0.00 94.11 0.00 94.11
Oxygen 30.66 0.00 30.66 30.66
Formaldehyde 0.00 0.00 0.00 0.00
Water 0.00 5.81 0.00 5.81
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 115.28 0.00 115.28 115.28
Stream No. 9 10 11 12
Stream Name
Temp C 200.0000 100.0000 30.0000 84.5653
Pres kPa 185.0000 150.0000 150.0000 140.0000
Enth MJ/h -28102. -29085. -40801. -24935.
Total kmol/h 278.02 278.02 143.00 224.15
Flowrates in kmol/h
Methanol 31.44 31.44 0.00 13.35
Oxygen 0.15 0.15 0.00 0.15
Formaldehyde 62.67 62.67 0.00 0.04
Water 66.82 66.82 143.00 93.68
Hydrogen 1.65 1. 65 0.00 1. 65
Ni.trogen 115.28 115.28 0.00 115.28
3-10

FLOW SUMMARIES
Stream No. 13 14 15 16
Stream Name
Temp C 89.8453 75.4568 106.6348 106.7102
Pres kPa 150.0000 130.0000 150.0000 350.0000
Enth MJ/h -44950. -5669.7 -38998. -38996.
Total kmol/h 196.87 23.00 173.87 173.87
Flowrates in kmol/h
Methanol 18.10 17.19 0.90 0.90
Oxygen 0.00 0.00 0.00 0.00
Formaldehyde 62.63 0.00 62.63 62.63
Water 116.14 5.81 110.33 110.33
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 0.00 0.00 0.00 0.00
Stream No. 17 18 21 22
Stream Name
Temp C 35.0000 73.3569 75.5718 75.5718
Pres kPa 315.0000 120.0000 130.0000 130.0000
Enth MJ/h -40166. -5669.7 -5682.9 -0.18970
Vapor mole fraction 0.00000 0.0052328 4.0129E-005 1.0000
Total kmol/h 173.87 23.00 23.10 0.00
Flowrates in kmol/h
Methanol 0.90 17.19 17.20 0.00
Oxygen· 0.00 0.00 0.00 0.00
Formaldehyde 62.63 0.00 0.09 0.00
Water 110.33 5.81 5.81 0.00
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 0.00 0.00 0.00 0.00
Stream No. 23 24 25 26
Stream Name
Temp C 75.5718 106.6357 106.6357 106.6357
Pres kPa 130.0000 150.0000 150.0000 150.0000
Enth MJ/h -5682.7 -33615. -33615. 0.00000
Total kmol/h 23.09 173.77 173.77 0.00
Flowrates in kmol/h
Methanol 17.20 0.91 0.91 0.00
Oxygen 0.00 0.00 0.00 0.00
Formaldehyde 0.09 62.54 62.54 0.00
Water 5.81 110.32 110.32 0.00
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 0.00 0.00 0.00 0.00
1'3-111

FLOW SUMMARIES
Stream No. 64 65 68 69
Stream Name
Temp C 89.8453 84.5669 934.7972 923.0992
Pres kPa 150.0000 140.0000 255.0000 220.0000
Enth MJ/h 0.0083595 -24935. -19173. -19173.
Total kmol/h 0.00 224.16 276.37 278.02
Flowrates in kmol/h
Methanol 0.00 13.35 33.10 31.44
Oxygen 0.00 0.15 0.15 0.15
Formaldehyde 0.00 0.04 61.01 62.67
Water 0.00 93.68 66.82 66.82
Hydrogen 0.00 1. 65 0.00 1.65
Nitrogen 0.00 115.28 115.28 115.28
Stream No. 70 72 73 74
Stream Name
Temp C 89.8453 89.8453 75.5764 106.6356
Pres kPa 150.0000 150.0000 130.0000 150.0000
Enth MJ/h -44950. -44950. -5681.2 -38985.
Total kmol/h 196.86 196.86 23.09 173.78
Flowrates in kmol/h
Methanol 18.10 18.10 17.19 0.90
Oxygen 0.00 0.00 0.00 0.00
Formaldehyde 62.63 62.63 0.09 62.54
Water 116.14 116.14 5.81 110.33
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 0.00 0.00 0.00 0.00
Stream No. 76 77 78 98
Stream Name
Temp C 89.8453 75.5772 106.6358 350.0000
Pres kPa 150.0000 130.0000 150.0000 250.0000
Enth MJ/h -44950. -5680.9 -38986. 55571.
Total kmol/h 196.86 23.09 173.78 7713.00
Flowrates in kmol/h
Methanol 18.10 17.19 0.90 94.00
Oxygen 0.00 0.00 0.00 1600.00
Formaldehyde 62.63 0,09 62,54 0.00
Water 116.14 5.81 110.33 0.00
Hydrogen 0.00 0.00 0.00 0.00
Nitrogen 0.00 0.00 0.00 6019.00
3-,2

FLOW SUMMARIES
Stream No. 99 100 101 102
Stream Name
Temp C 354.2749 100.0000 50.0000 50.0000
Pres kPa 250.0000 150.0000 115.0000 115.0000
Enth MJ/h 43675. -29085. -3209.2 -30959.
Vapor mole fraction 1. 0000 1.0000 1.0000 0.00000
Total kmol/h 7755.44 278.02 132.22 145.81
Flowrates in kmol/h
Methanol 9.34 31.44 5.36 26.09
Oxygen 1557.77 0.15 0.15 0.00
Formaldehyde 84.66 62. (57 1.11 61.56
Water 84.45 66.82 8.66 58.16
Hydrogen 0.21 1.65 1.65 0.00
Nitrogen 6019.00 115.28 115.28 0.00
Stream No. 103 104 105 106
Stream Name
Temp C 30.0000 42.5463 42.3507 47.7261
Pres kPa 150.0000 105.0000 115.0000 115.0000
Enth MJ/h -19992. -3029.3 -20172. -51131.
Total kmol/h 70.07 130.12 72.17 217.98
Flowrates in kmol/h
Methanol 0.00 1.39 3.96 30.05
Oxygen 0.00 0.15 0.00 0.00
Formaldehyde 0.00 0.00 1.11 62.67
Water 70.07 11.64 67.09 125.25
Hydrogen 0.00 1.65 0.00 0.00
Nitrogen 0.00 115.27 0.00 0.01
13-113

13.18 The Thermodynamic models used are:
Process Enthalpy Phase Equ. Alternate Notes
DME Latent heat UNIFAC UNIQUACIUNIFAC 2 liquid
phases
EB Latent heat UNIFAC UNIQUAC/UNIFAC Lack of
BIPs
Styrene SRK SRK PR HC
Maleic Latent heat Raoult's SRK Fairly ideal
Anhydride Law mixtures
Ethylene SRK PSRK+ SRK+ UNIFAC Simple
Oxide UNIFAC gases
except in
columns
Formalin Latent heat ESDK Polynomial data Hydrogen
bonding
Acetone Latent heat UNIFAC UNIQUAC with Azeotrope
BIPs regressed from
data
Acrylic Latent heat UNIFAC UNIQUAC with LLEand
Acid BIPs regressed from VLE
data
SRK SRK PR HC
Heptenes
13.19 The following results are from Chemcad for vapor-liquid equilibrium ofwater and air:
T Henry's Law SRK
IDoC 11.4 ppm 02 (mass) 0.413 ppm O2 (mass)
27°C 7.60 ppm O2 (mass) 0.659 ppm 02 (mass)
Note that the Henry's Law results are close to reality, but the SRK results are not only offby
more than an order ofmagnitude and their temperature-dependence is inverse. This is to be
expected, since cubic equations ofstate do poorly for liquid phases and for water.
3-l1 4

13.20 The following is from the Chemcad simulation ofT-lOl:
SRK PR PR Bips = zero
Number of 23.3 23.5 23.5
stages
Reboiler duty 8.78 MMBtu/hr 8.72 MMBtu/hr 8.72 MMBtu/hr
Condenser duty 5.15 MMBtu/hr 5.15 MMBtu/hr 5.15 MMBtu/hr
Reflux Ratio 1.75 1.77 1.77
SRKBIPs Hydrogen Methane Benzene Toluene
Hydrogen 0 -0.00900 0.32358 0.32358
Methane -0.00900 0 0 0
Benzene 0.32358 0 0 0
Toluene 0.32358 0 0 0
PRBIPs Hydrogen Methane Benzene Toluene
Hydrogen 0 0.0156 -0.50000 -0.50000
Methane 0.0156 0 0.03630 0.04000
Benzene -0.50000 0.03630 0 0
Toluene -0.50000 0.04000 0 0
Note that the BIPs are very different for the SKR and PR models, even differences in
sign! Note also that there are more binaries represented in the PR database on
ChemCAD for this system than for SRK.
However, the major outcome is that it makes no noticeable difference for the column.
This is because the important BIPs (between the heavy and light keys, benzene and
toluene) are zero in all three cases.
As noted, T-101 was modeled with the shortcut column module (heavy and light key
recoveries in distillate of0.01 and 0.99, RlRmin of 1.5, partial condenser) followed
by a splitter for the distillate. The feed stream to the column was specified, as well as
the top and bottom pressures.
The BIPs for PR in ChemCAD can be changed only by creating "cloned"
components. The BIPs for the original components in the database cannot be
changed.
l3-S

Chapter 14
14.1 Describe a Pareto analysis. When is it used?
Strictly speaking, a Pareto analysis is statistical technique but we use it here in the more
common form of the 20-80 principle. Namely, for many things, 80% of the information is
associated with only 20% of the issues. Thus when performing an optimization of a process
or product, if we itemize all the contributing costs then most often approx. 80% of those
costs are associated with only about 20% of the variables. This is very useful to do early in
an optimization analysis since it focuses our attention on the important decision variables.
14.2 What is the difference between parametric optimization and topological optimization? List
one example of each.
Parametric optimization focuses on adjusting operating (decision) variables in order to
improve the objective function. Examples include, adjusting the T and P at which a reactor
operates, or adjusting the surface area of a heat exchanger or number of trays for a
distillation column.
Topological optimization focuses on adjusting the layout or topology of the flowsheet in
order to improve the objective function. Examples include, changing the order in which a
separation sequence is implemented, looking at the effect of adding a heat recovery
exchanger, or changing a utility (cw to refrigerated water).
14.3 What is an objective function? Give two examples of one.
An objective function (OF) is a mathematical relationship that one wishes to minimize
(cost) or maximize (profit). For chemical processes, the OF will most often be a function
of variables that relate to the economics of the processes. For example, maximizing the net
present value (NPV) or minimizing the equivalent annual operating (EAOC) cost are
examples of common objective functions. We can also talk about maximizing conversion
or yield but such OFs may not lead to the economic optimum that is the commercial goal of
all processes.
14-1

14.4
Optimal cooling water exit temperature = 39°C
14-2

14.5 – Background simulations from CHEMCAD
14-3

14.5 – Background Information from Chemcad
14-4

14.5: Using previous background information
we find that the value of the products is much
greater than refrigerated water. This fact drives
the solution to maximizing the recovery of
acetone/IPA. In reality, we would choose some
minimum ΔT ~ 2-3°C.
14-5

14.7 – Optimum Pipe Diameter
Optimum pipe diameter = 12”
14-7

14.8 – Biological Reactor
Optimum Temperature = 40°C
14-8

14.9 – Distillation Column
R/Rmin,opt = 1.05
14-9

14.10 – Brine Fouling
Optimum cleaning time is ~ 6 months
14-10

14.11 – Optimal Cycle Time
x= $350
y = $200
V = 7m3
CA0 = 3kmol/m3
Cclean = $800
tclean = 1.0 h
k = 0.153 h-1
A1 = $ 6,550
B1 = $11,550
C1 = 1.5 h
Substituting values into Equation (14.18), we get:
0)( 11111 =−++
−
AeBktBkCB optkt
opt (14.18)
Solving we get, topt=10.53 h
14-11

14.12 – Second order reactor – optimum cycle time
2
k
A B→
A material balance at time, t gives: 0(
2
)A A
B
V C C
N
−
= and for a 2nd
order reaction we have
0
01
A
A
A
C
C
C kt
=
+
2
0 0
0
0 02 1 2 1
A A
B A
A A
C CV V
N C
C kt C kt
⎡ ⎤ ⎡ kt ⎤
∴ = − =⎢ ⎥ ⎢
+ +⎣ ⎦ ⎣
⎥
⎦
The amount of remaining A is 0
01
A
A A
A
VC
N VC
C kt
= =
+
The objective function (OF) can now be written as:
2
0 0
0 02 1 1
A A
clean
A AB A clean
clean
C kt VCxV
y C
C kt C ktxN yN C
OF
t tθ
⎡ ⎤ ⎡ ⎤
− −⎢ ⎥ ⎢ ⎥
+ +− − ⎣ ⎦ ⎣ ⎦= =
+
2
0
0 0
0
2
0
1 1 0 1 1 1
(1 )
2
(1 )( )
, , , ,and
2
A
A clean A
A clean
A
A clean clean Ao A
xVC kt
yVC C C kt
OF
C kt t t
let
xVC k
A B yVC C C D C C k E
− − +
=
+ +
= = = = = 0C k
1 1 1 1 1 1 1 1 1 1 1 1
2
1 1 1 1
( ) ( ) ( ) ( )
( )(1 ) ( )(1 ) ( 1)clean clean clean clean
At B C D t A D t B C A D t B C
OF
t t E t t t E t E t E t t t
− − − − − + − − +
= = =
+ + + + + + +
Differentiating the OF with respect to t and setting =0 gives
[ ]
[ ] [
2
1 1 1 1 1 1 1 1 1 1
2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1
( )( ( 1) ) (2 1) ( ) ( ) 0
collecting terms we get
t ( ) 2 ( ) ( )( 1) ( )( 1) 2 ( )
( ) ( 1)(
clean clean clean
clean clean
clean clean
A D E t E t t t E t E t A D t B C
A D E E A D t A D E t A D E t E B C
A D t E t B C
− + + + − + + − − + =
− − − + − + − − + + +
− + + +
[ ]2
1 1 1 1 1 1 1 1 1 1 1
) 0
( )t 2 ( ) ( ) ( 1)( ) 0clean cleanE A D E B C t A D t E t B C
=
]+
∴ − − + − − + + + =
a b c
2
4
and
2
opt opt clean
b b ac
t t
a
θ
− ± −
= = t+
14-12

14.13 – Two product batch sequencing
Optimum Solution is
to just make Product B
14-13

14.14 –Three product batch sequencing
14-14

14.15Optimization ofT-202
The approach used here follows that developed in the chapter for T-201. The base case
conditions are given below:
Feed Temperature = 139°C
Feed Pressure 7.4 bar .
Feed Composition
DME = 1.4 kmol/h
MeOH = 64.3 kmol/h
Water = 132.9 kmoVh
R!Rmin = 1.3
Utility Costs Used
Cooling Water = 14.8 $11000m3
MP steam = 13.71 $/1000 kg
Electricity = 0.06 $/kWh
FCl = total module cost from Capcost
Recovery ofmethanol in top product = 0.989
Recovery ofwater in top product = 0.01
Tray efficiency
Tray spacing = 18 inches
Top temperature = 116°C (in Figure B.1 top temperature is given as 121°C. The top
temperature apparently is quite sensitive to the thermodynamic package used for the k
values. The current results use the Uniquac/unifac option in CHEMCAD and this gives
slightly different results from the previous version - hence the different top temperature)
Bottom temperature = 167°C
Spread sheets on the next page give summaries ofthe results for different pressures and RlRmin
values. The equipment considered are T-202, E-206, E-207, E-208, V-202, and P-203A1B. All
equipment specifications are based on the following assumptions:
1. Height ofcolumn = (number ofactual trays times)(tray spacing-I8"). An additional 9 to 15
feet should be added to take into account ofthe skirt, vapor space at top oftower, and liquid
storage at bottom ofthe tower. This was not considered here.
2. Overall heat transfer coefficient for condenser = 850 W/m2a
C
3. Overall heat transfer coefficient for reboiler = 1140 WIm20C
4. Overall heat transfer coefficient for E-208 = 280 W/m20
C
5. Efficiency ofP-203 = 40% and destination pressure = 15.5 bar (to return to front end)
6. Medium pressure steam and cooling water used as utilities.
A graph showing the results is given on the page after the spreadsheets. The optimum reflux
ratio is between 1.01 and 1.12 for the three cases considered and the optimum pressure is
approximately 7.0 bar. It should be noted for the 5.0 bar case, that the high temperature utility
could be changed to low-pressure steam (a topological change). This has the effect of decreasing
the steam cost but increasing the size and cost of the reboiIer, E-206. The net result is that the
5.0 bar case becomes cheaper that the 7.0 bar case. The full optimization has not been carried
out here but should proceed to investigate lower pressures using Ips as the heating medium for
the column.
14-15

,.......
-+::-I
,.......
0
Bivariate Optimization for Column T-202 in DME Process
RlRmin
1.01
1.12
1.29
1.51
E-206 P
5.3 bar
5.3 bar
5.3 bar
5.3 bar
RlRmin
1.01
1.12
1.29
1.51
E-206 P
7.6 bar
7.6 bar
7.6 bar
7.6 bar
R/Rmin
1.01
1.12
1.29
1.51
E-206 P
9.2 bar
9.2 bar
9.2 bar
9.2 bar
Rmin
1.600
1.600
1.600
1.600
R
1.616
1.792
2.064
2.416
E-207 DT E-207 Area
65.9 26
65.9 28
No. Trays
37.8
20.8
17.5
15.2
E-207 P
5.0 bar
5.0 bar
65.9 30 5.0 bar
65.9 33 5.0 bar
P =7.3 bar
Rmln
1.358
1.358
1.358
1.358
R
1.371
1.521
1.751
2.050
E-207 DT E-207 Area
80.4 22
80.4 23
80.4 25
80.4 28
P =9.0 bar
Rmln
1.301
1.301
1.301
1.301
R
1.314
1.457
1.678
1.965
No. Trays
38.3
21.4
17.9
15.6
E-207 P
7.3 bar
7.3 bar
7.3 bar
7.3 bar
No. Trays
39.3
21.8
18.3
15.9
E-207 OT E-207 Area E-207 P
88.3 19 8.9 bar
88.3 20 B.9 bar
88.3 22 B.9 bar
88.3 24 8.9 bar
Qcond
5245
5564
6044
6684
V-202 D
0.83
0.84
0.87
0.90
Ocond
5310
5648
6155
6830
V-202 D
0.81
0.82
0.85
0.88
Qcond
5048
5364
5837
6468
V-202D
0.76
0.77
0.80
0.82
Oreboll
5043
5363
5842
6481
V-202 H
2.5
2.5
2.6
2.7
Oreboil
4843
5180
5688
6363
V-202 H
2.4
2.5
2.5
2.6
Qreboil
5281
5596
6069
6700
V-202 H
2.3
2.3
2.4
2.5
Nactual
54.0
29.7
25.0
21.7
V-202 P
5.0 bar
5.0 bar
5.0 bar
5.0 bar
N aclual
54.7
30.6
25.6
22.3
V-202 P
7.3 bar
7.3 bar
7.3 bar
7.3 bar
N actual
56.1
31.1
26.1
22.7
V-202 P
8.9 bar
B.9 bar
8.9 bar
B.9 bar
P = 5.0 bar
TIop
101
101
101
101
T-202 D
0.92
0.95
0.99
1.05
TIop
115.5
115.5
115.5
115.5
T-202 D
0.81
0.83
0.87
0.91
Ttop
123.4
123.4
123.4
123.4
T-2020
0.76
0.78
0.81
0.86
T bol
153
153
153
153
T-202 H
30
16
14
12
Tbot
167
167
167
167
T-202 H
30
17
14
12
T bot
175
175
175
175
T-202 H
31
17
14
12
P power cooling w sleam
5.8 15553 289957
6.2 16499 308356
6.8 17922 335897
7.6 19820 372637
elect
2913
3109
3411
3803
$0.354/GJ $6.87/GJ $0.06/kWh
T·202 P
5.3 bar
5.3 bar
5.3 bar
5.3 bar
P-203 E-208 P E-208
5.8 21.4 5.0 bar
6.2 21.4 5.0 bar
6.8
7.6
21.4
21.4
P power COOling w steam
4.2 15746 278457
4.5 16748 297834
4.9
5.4
T·202 P
7.6 bar
7.6 bar
7.6 bar
7.6 bar
P power
3.4
3.6
3.9
4.3
T-202 P
9.2 bar
9.2 bar
9.2 bar
9.2 bar
18251
20253
P-203
4.2
4.5
4.9
5.4
cooling w
14969
15906
17309
19180
P-203
3.4
3.6
3.9
4.3
327042
365853
E-208
22.8
22.8
22.8
22.8
steam
303641
321752
348949
385229
E-208
23.6
23.6
23.6
23.6
5.0 bar
5.0 bar
elect
2121
2255
2461
2728
P E-208
7.6 bar
7.6 bar
7.6 bar
7.6 bar
elect
1698
1803
1965
2175
P E-208
9.2 bar
9.2 bar
9.2 bar
9.2 bar
total utils E-206 Area
308422 40
327963 42
357231 46
396261 51
FCITM
471000
381000
375000
375000
NPV
-3.064
·3.017
-3.206
-3.475
296324 69
316836 74
347755
388834
FCITM
462000
399000
377000
378000
82
91
NPV
-2.962
-2.977
-3.145
-3.430
320308 143
339461
368222
406584
Fel TM
509000
429000
409000
412000
152
164
181
NPV
-3.22138
-3.19258
-3.3501
-3.6197
RlRmin
1.01
1.12
1.29
1.51
R/Rmin
1.01
1.12
1.29
1.51
R/Rmin
1.01
1.12
1.29
1.51

>-'
~
I
>-'
-...)
Problem 12.3 - Solution
-2.8
U) _ 9 High temperature utility =mps ~
5 2. Low temperature utility =cw ~... ~ ~
component recoveries are constant~
~ -3.1
:>
Q.. -3.2
~
cb -3.3
:::s
~ -3.4
.....
as -3.5
loio..,. ...
I .. ~ -, ... II I I 1- -I· J--I--.oJ I I , r I i j
:""'iiiii
..
[ii".
:""Iii ~
p =7.~ h:u
I I I I
P =5.0
i~:::11111111111111111111111111- rrrn1 1.1 1.2 1.3 1.4 1.5 1.6
Ratio of reflux to minimum reflux for T-202, RlRmin

14.16 Optimization ofDME separation section with water removal from the first tower.
A PFD for the rearranged separation section of the DME process is shown on the next page.
Water is removed as the bottoms product from the first column and DME is removed as the top
product from the second column. A base case was taken as a starting point and the operating
conditions for the base case are shown in the figure. The costs for the base case for different
RJRmin values for both towers are shown in the accompanying spreadsheet.
The results of this preliminary optimization are shown in the figure following the spreadsheet,
where NPV is plotted as a function of RlRmin. It should be noted that the reflux ratio of each
tower can be varied independently but this is not done for the base case shown here. The lowest
NPV for the base case is $-6.411xl06 using a reflux ratio of 1.09 times the minimum.
The costs associated with T-202 for Problem 12.3 gave an optimum NPV of $-2.96x106 again
for a reflux of 1.1 times the minimum. At this stage, the results given in Table 12.4 should be
reworked for current (2002 prices). These can then be used in conjunction with Problem 12.3 to
compare the results ofProblem 12.4.
14-18

......
..j:::..
I
......
0
ct-0 ~().(
41°C
T-201
17 J 0(.
<6-.3 ;~
P-201 NB
(Y)ps
l 0-,3 ~O-<"
/+(, °c
T-202
P-202 NB
DME product
Qps
E-206
5 -s I:x.M-
10.6 bo..r
methanol recycle
'3 0 °c... P-203 NB
50°C-
waste water
Base Case for Problem 12.4 - DME Process with Water removal from first Column, T-201

Base Case forT-201+ T-202 in DME Process - Problem 12.4
Tower T-201 - water bottom, DME + MeOH top
E-204
R/Rmin Rmin R No. Trays Qcond Qreboil N actual T top T bot cooling w steam elect total utils Area
1.01 0.337 0.341 60.6 7110 5446 87 46.9 171 21083 313128 1283 335494 102
1.09 0.337 0.368 24 7244 5580 35 46.9 171 21481 320833 1309 343622 105
1.16 0.337 0.391 21.8 7378 5714 32 46.9 171 21878 328537 1331 351746 107
1.24 0.337 0.418 20.5 7512 5848 30 46.9 171 22275 336242 1357 359874 110
1.31 0.337 0.442 19.6 7646 5982 29 46.9 171 22673 343946 1380 367999 112
E-205
E-204 P E-205 DT Area E-205 P V-201 0 V-201 H V-201 P T-201 0 T-201 H T-201 P P-202 R/Rmin
8.3 bar 11.2 208 8.0 bar 1.06 3.2 8.0 bar 1.09 48 8.3 bar 2.6 1.01
8.3 bar 11.2 212 8.0 bar 1.06 3.2 8.0 bar 1.11 19 8.3 bar 2.6 1.09
8.3 bar 11.2 216 8.0 bar 1.07 3.2 8.0 bar 1.12 17 8.3 bar 2.6 1.16
8.3 bar 11.2 220 8.0 bar 1.08 3.2 8.0 bar 1.13 16 8.3 bar 2.7 1.24
...... 8.3 bar 11.2 224 8.0 bar 1.08 3.2 8.0 bar 1.14 16 8.3 bar 2.7 1.31
../:>.I
tv
Tower T-202 - water bottom, DME + MeOH top0
E-206
R/Rmin Rmin R No. Trays Qcond Qreboil N actual T top T bot cooling w steam elect total utils Area P-204
1.01 0.122 0.123 84.6 2576 3085 121 45.7 130 7639 327612 983 . 336233 25 1.13
1.09 0.122 0.133 17.1 2598 3106 25 45.7 130 7704 329842 986 338532 25 1.13
1.16 0.122 0.142 14.4 2619 3127 21 45.7 130 7766 332072 990 340828 25 1.13
1.24 0.122 0.151 13.4 2640 3148 20 45.7 130 7828 334302 993 343124 26 1.13
1.31 0.122 0.160 12.9 2661 3169 19 45.7 130 7891 336532 996 345419 26 1.13
E-207
E-206 P E-207 DT Area E-207 P V-202D V-202 H V-202 P T-202D T-202 H T-202 P P-203 E-208 P E-208 A FCITM NPV R/Rmin
10.6 bar 9.9 85 10.30 bar 0.91 2.7 10.30 bar 0.60 67 10.6 bar 0.8 10.6 bar 20 1400000 -7.422 1.01
10.6 bar 9.9 86 10.30 bar 0.92 2.8 10,30 bar 0.60 14 10.6 bar 0.8 10,6 bar 20 860000 -6.411 1.09
millions

Base Case Calculation for Problem 12.4·· T-201 and T-202 reversed
-6.200 .......
1 - ------,.--- --- -,-----
1
-6.400 .It..
/1
7
-I--.
1
-6.600 / ! """/ 1
- /
II) / 1
c /
0
:: -6.800 I
.-S
/
/ 1
~
- ;:- -7.000
..p..
~I
N
/ :
1 1
1
/
/ 1
- -7.200 f
Recoveries in both columns were set at 0.989 and 0.01/ 1
-7.400
/
Water is removed as bottom product from T-2011
/ !
DME is removed as top product from T-202./ 1
.. 1
-7.600
1
-' - - ,-- -
1 1.1 1.2 1.3 1.4
RlRmin for T-201 and T-202

14.17 Use ofLow Pressure Steam vs Medium Pressure Steam in E-204.
The data given in Table 12.6 can be used to find the break-even pressure for the switch to Ips
from mps. The idea is to evaluate the FeI associated with the heat exchanger, E-204, with the
cost of the steam. It is assumed here that the overall heat transfer coefficient for the reboiler is
fixed at 1140 W/m2°C for all comparisons. The costs of mps and Ips are taken from Table 6.3
and the capital costs of the exchanger are taken from CAPCOST using a floating head
exchanger.
The NPV is evaluated as NPV = -(2.005xFCI + 6.871xCUT). Since the only things that change
are the size and cost of E-204 and the cost of the steam, these are the only costs that should be
included in the NPY. A spreadsheet showing the comparisons is given on the next page and a
figure showing the two cost curves as a function ofT-20l operating pressure is also given. From
the Figure, the operating pressure at which the switch to Ips should take place is about 11.5 bar.
14-22

Problem 19.11 - Comparison of Low Pressure to Medium Pressure Steam Useage in E-204
E-204-lps E-204-mps E-204-lps E-204-mps E-204-lps E-204-mps
Pres T-201 T reboil E-205 Duty cost Ips cost mps Area Area Cost Cost NPV NPV
bar C MJ/h $1,000 $1,000 ml2 ml2 $1,000 $1,000 $1,000 $1,000
11.5 157 2689 285.6 30B.8 21B.4 24.3 151 72.2 -2265 -2266
10.3 153 2549 270.7 292.7 88.7 20.0 94.3 72.1 -2049 -2156
9 147 2352 249.8 270.1 44.1 15.5 77 73.2 -1871 -2002
7.5 140 2127 225.9 244.2 25.9 11.8 71.6 75.8 -1696 -1830
'-'
-/:>.
Nw

-1500
S -1600
0
0
T'" -1700
~
-6: -1800
Z
cD -1900
=-~ -2000
....
- c+>- CI) -2100I
tv
'"+>-
eIl. -2200
....CI)
Z -2300
-2400
Comparison of LPg and MPS useage in E-204
NPV = 2.005(FCI) + 6.871 (cost of steam)
low-pressure steam -
-
-- use of low pressure steam
-- advantageous below this operating~
--....... pressure
--==medium-pressure steam .- .....
---.
--
• ~
--
7 8 9 10 11
Operating Presuure of T·201, bar
12

Chapter 15
15.1 Costs affected by changing the min temp approach for a HEN are
Cost ofutilities - hot and cold
Cost ofheat exchangers in the network, including process-process and process-utility
exchangers
15.2 By decreasing the /!,.Tmin in a HEN, the process-process heat exchangers at the pinch will
require larger areas and will therefore be more expensive. However, the amount and cost
ofhot and cold utilities for the HEN will be reduced.
15.3 (a) For streams requiring an MOC cheaper than CS, the film heat transfer coefficient, h,
for the stream should be increased. This means that smaller heat exchange areas will be
computed for these streams and this will lead to lower capital investment. This is
equivalent to using a material factor, FM< 1 (for CS).
(b) The reverse is true for streams requiring an MOC more expensive than CS. These h
values should be reduced, leading to larger more expensive heat exchangers. This is
equivalent to using a material factor, FM > 1 (for CS).
15.4 As /!,.Tmin increases the utility requirements increase (yearly operating costs) but the
process-process exchangers become smaller and less expensive. However, the costs ofthe
utility exchangers will increase. The costs for utilities and fixed capital investment
(exchangers) are illustrated below. An optimum /!,.Tmin will exist.
, ;
, Equivalent Annual Op Cost, , ,
EAOC ofHEN , ,
/ ""... '...."
...
/!,.Tmin, opt
,
Minimum temperature approach, /!,.Tmin
Utility costs, $/y
I
Cost ofexchanger, $
15.5 At the pinch, for streams above the pinch always match streams such that lilCp,hOf ~ lilCp,cold
Using this criterion avoids violating the /!,.Tmin criterion set at the start ofthe problem.
15-1

15.6 Composite temperature-enthalpy (T-Q) diagram
hot streams
cold streams
Cumulative Enthalpy
15.7 A cascade diagram illustrates the amounts ofavailable energy, in excess ofthat required
by process streams, at each temperature level in the temperature interval diagram. The
cascade diagram is useful for visualizing how excess energy is cascaded downwards to
heat lower temperature streams. It is also useful for identifying where the network pinch
occurs.
15.8 Minimum number ofexchangers = n + m + # ofutilities (1- for hot utility above pinch)-
1 = n + m. The actual number ofexchanger may be lower than this ifexact matches of
energy are possible.
15.9 For streams that change phase, two approaches are used
(i) ifthe stream is pure and the phase change occurs at a single temperature then we
may assume some arbitrary, small I:l.T of say 1°C and use an equivalent mcp for
h 111[kg/s]A.[kJ/kg] d' 1 h ft at stream, where (I11Cp )eqlliv = an A, IS the atent eat 0
WC]
vaporization.
(ii) lfthe stream has several components and the heating curve is non-linear, then the
curve can be approximated by two straight lines, the slopes ofwhich represent
equivalent mcp values.
15.10 MUMNE =minimum utility, minimum number of exchangers heat exchanger network. It
is the HEN that gives the minimum number ofexchangers and minimum hot and cold
utility duties for a system ofprocess streams that are exchanging energy for a given
(chosen) minimum temperature approach for the network.
15-2

15.11 (a)
mCp
(BTIJ'WF)
Stream
Number
Temperature
COF)
600
500
470
360
320
280
(b)
4
1
400
120
440
160
1
HU
20
4 3
2 3
A 300
B 90
440 C 330
160 D
~160 E
1
20
..
A
100
~ 100
B
-120
OJ
rTol
~
I
E 1
280
• [CUl
. 160. ~
15-3
5
4
150
550
200
Temperature Q
COF) (BTU/hr)
580
100
480
-120
450
0
340
120
300
160
260
260
Pinch at 470-450 of
HU = 20 BTU/hr
CU = 280 BTU/hr

(c)
1
Cf,]Above 520
130
20
3 exchangers 390
1 20
I 3~O I QJ
1 2
Below 600
270
760
4 exchangers 330
1 1480
280
QJ 17~O I [;]280
15-4

Stream Number
mCp (BTqmoF)
1
4
2
4
3
3
-, 580
4
5
<2) HU Q3 = 20 BTU/hr
600 - -r- 573.3
Q2 = 370 BTU/hr
507.5 ~ -'- 450 -,480
Ql = 150 BTU/hr
C!) t-----------------tC!)
470 -'-
Or split stream 1
Stream Number
mCp (BTqmoF)
1
4
/
3 1
600 -r
2
4
3
3
-,580
17 Q2 = 390 BTU/hr 17
~~--~~----------~~
470 - _ 450
600 -r
-450
4
5
--480
HUQ3=
~-20BTU/hr
-- 476
Ql = 130 BTU/hr
C!)~--------~----------------tC!)
470- -450
15-5

(e) Below pinch mCpH 2:: mCpe
Stream Number 1 2 3 4
mCp (BTIJ'ht'P) 4 4 3 5
/ /
1 3 1 4
470 - _450
@
Q4 = 330 BTU/hr
@
450
470 - -f- 360 -'- 340 --
(3)
Qs = 150 BTUIhr
(3)
320 -- -"-
300
-r--
350 -r
(J) r-+ CU Q7 = 280 BTUIhr
280 -'-
47°@I Q6 = 600 BTU/hr I@45O
320 300
OR
/
1 4
47°@I Q4 = 600 BTU/hr I@45O
320 __ 480 300
470 -,
(3)
Qs = 330 BTU/hr (3)
381.5 r
-340 _ 450
@ Q6 = 150 BTUIhr @
350 - -300
(J) ---+ CU Q7 = 280 BTU/hr
280 -
15-6

15.12 (a)
Stream
Number
Ihep
{BTI}'ht'F}
Temperature
(OF)
400
320
300
180
100
(b)
1
2
t160
2
4
A
B
480 e
320 D
HU 60
60
3 4
3 2
Temperature Q
(OF) (BTU!hr)
390
160
~ 310
60 40 -100
290
360 240 -120
170
240 80
90
20
A
160
~ 16o
B
-100
~ 60
e
-120
Pinch T 180-170o
e
15-7

(c)
Above
4 Exchangers
Below
2 Exchangers
1 2 HU
160 480 60
160 26i7~20 ~O
~4-~-0--1 I 2~0 I
2
320
2L ~
3
240
15-8
CU
80

(d) Above pinch mCpH ::::; mCpe
Stream Number 1
mCp (BTlf'hf'F) 2
400 -,
2
4
/
3 1
3
3
4
2
--310
~~________Q~3_=~1~60~B_T_U~/=ill____________~~
Stream Number
mCp (BTqhf'F)
320-
1
2
--
HU Q4 = 60 BTUIill ---+ @)
300 -r- -r 230
(2) t--_-->J)_____2=-=12~OFB;..,;;oT-"-U'-=/ill"'---____i(2)
180 -'- -'- 120
300 -, -r 290
CD QI = 360 BTU/hr CD
180 -'-
2
4
-170
3
3
180 -, --170
(3) Qs = 240 BTU/ill (3)
120 -- -'-90
@ ~ CU Q6 = 80 BTU/ill
100 '-
15-9
4
2

15.13 (a)
Stream
Number
mCp
(BTIJ'ht>F)
Temperature
(OF)
180
160
120
100
90
80
(b)
1
3
60
120
r 60
2
5
A
B
100 C
50 D
E
HU 60
60
3
3
120
60
30
A
20
~ 20
B
.. -80
C
60
~ 60
D
20
~ 80
E
50
15-10
4
2
Temperature
40
80
40
130 _I cu
. 130
COF)
160
140
100
80
70
60
Q
(BTU/hr)
20
-80
60
20
50
70

(c)
HU 1
Above 60 180
3 exchangers 60
1
60
1120
3 4
120 120
1 2
Below 60 200
4 exchangers
6/~ ~30
3 4 CU
90 40 130
Cd) Above pLllCh !hCpH ~ !hCpc
rilCp (BTIJ'ht'F) 3 5 3 2
--170
-r HU Q3 = 60 BTUIhr 0
-,...130
Q2 = 60 BTU/hr
@ @
-,-100
160 -I- --140
CD
QI = 120 BTU/hr
CD
120 -'- -,-100
(e)
15-11

Below pinch mCpH :2: mCpe
Only one utility stream
Stream Number
mCp (BTlf'hf>F)
120-
1
3
2
5
3
3
-100
Q4 = 60 BTU/hr
~t-----"'-'--------I®
100 120 r 80
4
2
- 100
Qs = 40 BTU/hr
~r---~-+-----~~
112 r
® Q6 = 30 BTU/hr ®
106 -
CU Q7 = 130 BTU/hr - (J)
For two utility streams
Stream Number
mCp (BTlf'hf>F)
1
3
120 -
80 --
2
5
3
3
70
4
2
80
- 100
Q4 = 40 BTU/hr
~t---------------------------~~
106.67
CUQ6=20 -®
BTU/hr
100
120 ,... - 100 -- 80
~ Qs = 90 BTU/hr ~
102 r .... 70
80 -
15-12

15.14 (a)
Stream
Number
mep
(BTIJ'bf>F)
Temperature
(OF)
500
450
430
380
360
340
(b)
1
4
200
80
200
80
2
4
A
B
200 C
80 D
E
I
HU
I
10
10
3
3
150
60
150
A
50
50
1
B
.. -60
C
50
50
D
80
130
E
80
15-13
4
4
Temperature
(OF)
490
80
440
420
200
210
370
80
350
330
IcUl
~
Q
(BTU/hr)
50
-60
50
80
80
200

(c)
1 HU
280 10
Above pinch
103 exchangers 210 70
3 4
210 80
1 2
Below pinch 280 360
4 exchangers
lsi l3~)SD
Exact Match Solution
Below pinch
3 exchangers
3
150
1
280
4
280
~.---~-,
3 4
150 280
15-14
2
360
~ID
CD
210
~o
CD
210

(d) Above pinch rhCPH S; rhCpc
Stream Number
rhCp (BTlJ'hr"F)
1
4
2
4
3
3
--490
4
4
0- HU Q3 = 10 BTU/hr
500 -- -- 486.67
Q2 = 200 BTU/hr
01------------10
450 -- -- 420 -;- 440
Ql=80BTU/hr
01-----------------~0
430 -'- ~420
(e) Below pinch rhCpH ;::: rhCpe
Stream Number
rhCp (BTlJ'hr"F)
1
4
430 --
2
4
430 -,-
3
3
4
4
-;- 420
Q4 = 280 BTU/hr
(i)1-----------;(i)
360 -I- -r- 440 -~ 350
Qs = 150 BTU/hr
~t--------t--------I~
392.5 -I-
CUQ7=
130 BTU/hr--f(J)
360 -'-
-- 420
® - CU Q6 = 80 BTU/hr
340 --
15-15

Exact match solution
Stream Number
mCp (BTqWF)
1
4
430 -r-
2
4
3
3
430 -, -- 420
<!> Q4 = 150 BTU/hr <!>
39205 -r- -- 370
~ ~ CU Qs = 210 BTU/hr
340 -'-
4
4
-r- 420
® 1--_ _ _ _Q=6_=_2_80_B_T_U_/l_llo_____--I ®
360 -'- -'- 350
15-16

15.15 (a)
Stream
Number
mCp
(BTqtbJ:OF)
Temperature
(OF)
250
210
200
100
40
(b)
1
3
120
30
2
5
A
B
500 C
300 D
I
HU
I
120
~
120
3
4
40
400
240
A
120
120t
B
-40
80 1
C
-200
D
60
15-17
4
3
Temperature Q
(OF) (BTU/hr)
240
120
200
30 -40
190
300 -200
90
60
30
-60
6o ..1cUl
~

(c)
Above pinch
4 exchangers
Below pinch
2 exchangers
HU 1 2
120 150 500
1~~0°3
440
2
300
4
330
61 ,0
CU 3
60 240
15-18

(d) Above pinch mCpH ~ mCpe
mCp (BTqtht'F) 3 5 4 3
!
3.2 1.8
250 -, -- 200
(!)
Q4 = 150 BTU/hr
(!)
-,- 200
200 -'- HU Q3 = 120 BTU/hr (2) - ~ 150
200-- -I- 170
8 Q2 = 320 BTU/hr
0
100-,- -'- 90
200-,
CD
QI = 180 BTU/hr
CD
100 -'- 190
Or
mCp (BTqtht'F) 3 5 4 3
!
2.9 2.1
250 -- -r- 200 -,- 200
(!) Q4 = 150 BTU/hr (!)
(2) HUQ3=
200 - f- 162.5 --120 BTU/hr-
200--
0 Q2 = 290 BTU/hr
0 -~ 160
100-- --90
200--
CD
QI = 210 BTU/hr
CD
100 -'- -'- 90
15-19

(e) Below pinch mCpH ;::: mCpe
Stream Number
mCp (BTlJ'hf>F)
CU
4
3
-r- 90
Q5 = 240 BTU/hr (2)
-'- 30

15.16 (a)
From interval D From interval A
(mcp)2(20) - (3+5)(20)=-100 (4)(50) - (mcp)3(50) = 50 ~ (mcp)3 = 3
(mcp)2 = 3
/Stream
~2 4
mCp 4 ? ? 5 kWfC Q
300 --- -- ---------- ---- --- ---------- 280
A 50
250 --- -- ---------- ---- -- ---------- 230
IB ?
230 --- -- ------- --- ---- -- ---------- 210
II
CI ?
180
____1____J___---- -- ------ --- 160
D -100
160 ------------- --- ---- -- ------ --- 140
E ?I
140 --- -- - - - --- --"--- ---- ------------- 120
- -
?kW
- -
QB = (4 + 3 - 3)(20) = 80 kW
Qc = (4 + 3 - 3 - 5)(50) = - 50 kW
QE = (3)(20) = 60 kW
15-21

(b)
HU
Q=20
(c) Tpinch, hot = 160°C
Tpinch,cold ::;: 140°C
(d) Above Pinch
1
480kW
20
A
Q=50
50
B
Q=80
130
C
Q=-50
80
D
Q = -100
2
270kW
HU
20kW
42~~7~' 20/
3 ~r----"L--4--"-----'
Below Pinch
420kW 350kW
2
60kW
~ 60
CD
60kW
15-22
Min hot utility = 20 kW
Min cold utility = 60 kW
Number of exchangers
above pinch = 4
Number of exchangers
below pinch = 1

I Design above pinch
I
Number of Exchangers Remaining fO
(i' Process Exchanger(at or away from the Pinch)
,~.
Utility Exchanger
Exchange,l Duty IDT Viclaticrl Area ....
1 270 No 675
2 350 No 362.Q -
3 20 Hot Util Ex 26.2
4 130 No 259.7 ....
300.0 280.0
~I f ~J
0 (0 Enter ExohangerDuty
(positive numbelS only)
236.7
. ,
(0
- Split streams
I267.5 210.0
0 0
,------.----. ... '--"11 .§tartAgain
Il_..~'.:'.1!E.~.~~~PLi
Hot Streams- Total Cold Streams- Total
180.0 140.0 Enthalpy Remaining Enthalpy Remaining
250.0 230.0
0-0 Stream Enthalpy ....
~
Enthalpy ....
Total 0 ~
-0---
160.0 140.0 Utility 0 - 3 0 -
1 0 4 0
2 0
stream No. 1 2 3 4
MCp 4.0 3.0 3.0 5.0
1 -
.... 1 - ....
Number of Exchangers Remaining IT
I I
,- Process Exchanger (at or away from the Plnoh)
Design below pinch r.- Utility Exchanger
Exchange,l Duty IDT Violatiorl Area ....
1 60 Cold Utll Ex 79.6
160. 140. -
....
11 I ~I-
160.0 Enter ExchangerDuty
(2)
(positive numbelS only)
- Spl~Streams
I140.0
140. 120.
IrJ~!~~I6!~pji-]1 .§tartAgain
I
Hot Streams - Total Cold Strums - Total
Enthalpy Remaining Enthalpy Remaining
I~ Enthalpy ....
~
....
Total
-0---
Total
-0---
1-:;-- 0 - ~o -
stream No. 1 2 3 4 I~ 0 ~O
~ 0
MCp 4.0 3.0 3.0 5.0 -
-
i -
i -
1 -
.... 1 - ....
15-23

15.17 (a)
Stream
1 2 3 4
Number
mCp
2 4 5 4
(BTIJ'bf>F)
Temperature Temperature Q
eF) (OF) (BTU/hr)
400 380
200 A 250 -50
350 330
40 80 B 100 20
330 310
220 440 C 440 220
220 200
400 D 400 0l I
120 100
190
(b)
1
HU 1 50
·1 -~o 150
B
20
20 1
C
220
240
1
OJ 240.[;]
240
15-24

(c)
Above pinch
2 exchangers
Below pinch
4 exchangers
lOy
3
100
HU
50
50
1
260
2
200
/ao3
250
2
920
16~/680
4
840
~40
CU
240
(d) Above pinch rilCpH ~ rilCpc
Stream Number
rilCp (BTlJht'F)
1
2
2
4
HU Q2 = 50 BTU/hr
400 -,.-
3
5
--380
--~~'0
-~ 370
(!) Ql =200 BTU/hr (!)
350 _I- _I- 330
15-25
4
4

(e) Below pinch mCpH ~ mCpe
Stream Number
mCp {BTlJ'hr'F}
CUQs=
240 BTUIhr
1
2
350 --
340 -I-
220 -~
2
4
Q3=20BTUIhr
350 --
3
5
!
4 1
-- 330
330 -- _~ 310
~ Q4= 80 BTU/b ~
330 -- 310 -~
4
4
310 --
Q6 = 840 BTUIhr
r-----------------~®
120-- 100 _L-.
15-26

15.18
File Worksheet System Help
Table of Resutts
Minimum Temperature Approach = 20'F
Hot stream Data
Mass Flow Cp Temp In Temp Out stream Enthalpy Film Heat Transf. Coef
M-Ibm EttuAbf'F 'F 'F MEttum Ettum1fl2f'F
1.000
6.000
3.000
1.000
1.000
1.000
620.0
420.0
420.0
320.0
120.0
220.0
300.0
1800.
600.0
Cumulative Hot stream Energy Available = 2700.0 MEttum
Cold stream Data
75.00
25.00
10.00
Mass Flow Cp Temp In Temp Out stream Enthalpy Film Heat Transf. Coef
M-Ibm EttuAbf'F 'F 'F MEttum EttuJ!1lfl2f'F
5.000
2.000
4.000
1.000
1.000
1.000
400.0
200.0
100.0
600.0
300.0
400.0
-1000.
-200.0
-1200.
Cumulative Cold stream Energy Available = -2400.0 MEttum
Data for Generating Temperature Interval Diagram
Number of Temperature Intervals = 4
Interval Temperature Range Excess Heat
'F 'F MEttUlh
A 620.0 420.0 -800.0
B 420.0 320.0 600.0
C 320.0 220.0 300.0
D 220.0 120.0 200.0
Pinch Temperature - Hot =420'F Part (b)
Pinch Temperature - Cold = 400°F
Hot Utility Requirement = 800 MEttum
Part (a)Cold Utility Requirement = 1100 MEttum
45.00
30.00
30.00
Cummulative Q
MEttUlh
-800.0
-200.0
100.0
300.0
Minimum Number of Exchanger Required to Accomplish Minimum Utility Loads
In Special Circumstances the Minimum Required may be Lower than Indicated Below
Number Above the Pinch =2
Number Below the Pinch =5
Part (C)
Data for Composite Enthalpy - Temperature Diagram
Temperature Hot stream Enthalpy Temperature Cold stream Enthalpy
OF MEttum OF MEttum
120.0 .0000 100.0 1100.
220.0 600.0 200.0 1500.
320.0 1500. 300.0 2100.
420.0 2500. 400.0 2500.
620.0 2700. 600.0 3500.
Heat Transfer Area for Process Exchangers in Network = 2220 ft2
15-27

Help Print Return to Main Menu Exchanger Design Report
Number of Exchangers Remaining ro-
(~ Process Exohanger (at or away from the Pinoh)
(ii Utility Exohang(u
Exch,nge'l Duty IDT VlolaliorlAle,
1 200 No 122
2 800 Hoi Ulil Ex 774.2
600.0
• I ~I
0
620.0 440.0
Enter ExchangerDuty
(positive numbers only)
C) C)
- Spl! Streams
420.0 400.0
I[1r!!er-~~~~i]I .!l.larlAgain
Hot Streams· Total Cold Streams - Total
Stream No. 2 3 4 5 6
Enthalpy Remaining Enih.alpy Remaining
Stream Enthalpy ..... Stream Enlhalpy ...
MCp 1.0 6.0 3.0 5.0 2.0 4.0
Tolal 0 To1al 0
Utility 0 4 0
0 (5 0
2 0 6 0
3 0
Help Print Return to Main Menu Exchanger Design Report
Number of Exchangers Remaining ro-
(- Process Exohanger (at or away from the Pinch)
r. Utility Exchanger
420.0 400.0
420.
G) G)
400.
420.0 300.0 Exohangerl Duty IDT Vlolallor! Area
220.0
G) Q 100.0 1 1200 No 1Q70.Q
420.0 2 200 No 2<10.1
G) 353.3 200.0 3 100 Cold Util Ex 70.7
4 400 Cold Uti! Ex 1486.6
320. 320.0
Q
300.
;1 1 ~I
220.0 Enler ExchangerDuty
(posilive numbers only)
@
- Split Streams
220. 120.0 200.
Jl.larlAgain
Hot Sireams· Total Cold Streams· Total
Enthalpy Remaining Enthalpy Remaining
Stream Enthalpy ... Stream
120. 100. ----Total 0 Total 0
1 0 Utility 0
2 0 4 0
3 0 (5 0
6 0
Stream No. 2 3 4 5 6
MCp 1.0 6.0 3.0 5.0 2.0 4.0
15-28

15.19 Using the heat transfer coefficients given, use the Sensitivity Plot feature of HENSAD to
generate the following data and plot
From Problem 15.18, the heat transfer area for the network is 2220 ft2
rintBeturn to Main Menu ~nerate Data Plot B!'ea and Utilities Plot f.AOC e.
Minimum Value of Approach Temperature j1 OF
Maximum Value of Approach Temperature J101 OF
Number of points (maximum =100) r--so
Hot Lltiity FUm Coefficient 1177 BtUhlrJtt2rF
Cold LltHity FHm Coefficient J142 BtUhlrJtt2rF
Temperature Driving Force in Hot ~ OF
Lltiity Exchangers I ~
Temperature Driving Force in Cold ~ OF
Lltiity Exchangers I ~
Price for Hot Litility
Price for Cold Litiity
Time Period for Cost Analysis
Interest Rete
Exchanger Price
log(Totsl Module Cost, $) = Kl+K2log(A)
109=log10 A = Area, ft2
j18.76i $lmiUion Btu
j3735 $lmiHion Btu
IS years
r--w %p.e.
K1 I 2.8346
K2 I 0.5731
~ B.otum to Previous Screen Return to Main Menu
:
MlnTemp Exch
Approsch Ares
OF ft2
1.000 7671.6
3.000 6598.5
5.000 6126.3
7.000 5831.4
9.000 5622.7
11.00 5464.9
13.00 5340.6
15.00 5240.2
17.00 5157.5
19.00 5088.5
21.00 5035.6
23.00 4994.1
25.00 4957.4
27.00 4925.1
29.00 4896.7
31.00 4871.8
33.00 4850.0
35.00 4831.0
37.00 4814.4
39.00 4800.1
41.00 4787.9
4:-lnn 4777S
EAOC - Minimum Temperature Approach Plot
EAOC ($1000tyr)
300
270
240
210
180
150
120
90
160 /::,.Tmil1. opt = 8°F
30
0
Resufts
HoI LItH Cold util
Load Load
MBtuhl MBtuhl
705.0 1005.
715.0 1015.
725.0 1025.
735.0 1035.
745.0 1045.
755.0 1055.
785.0 1065.
775.0 1075.
785.0 1085.
795.0 1095.
804.0 1104.
812.0 1112.
820.0 1120.
826.0 1126.
836.0 1136.
844.0 1144.
852.0 1152.
860.0 1160.
868.0 1168.
876.0 1176.
884.0 1184.
R~'n 11~
0 10 20 30 40 50 60 70 80 90
Minimum Approach Temperature (OF)
15-29
EAOC Numberot
Exchangers
(1000)$/y
162.5 8.000
177.9 8.000
176.7 8.000
176.4 8.000
176.6 8.000
177.1 8.000
177.9 8.000
178.8 8.000
179.6 8.000
160.8 8.000
161.9 6.000
182.8 8.000
183.8 8.000
184.8 8.000
165.8 8.000
166.9 8.000
188.0 6.000
189.1 8.000
190.2 8.000
191.3 8.000
192.5 8.000
1~:-lR Rnnn
100 110
";
!
G§
0,
~,
(v,

15.20
ho FM had/ho hadj
BTUlhr/ft2°F Fig 15.16 BTU/hr/ft2°F
75 1.00 1.00 75.00
25 2.00 0.45 11.25
10 3.00 0.28 2.80
45 7.00 0.09 3.83
30 2.50 0.35 10.50
30 1.00 1.00 30.00
Using the adjusted heat transfer coefficients from the above table we get the following
results.
The adjusted heat transfer area for the network is 5765 ft2
The EAOC and minimum approach temp is shown below:
Erint /ieturn to Previous Screen Return to Main Menu
EAOC - Minimum Temperature Approach Plot
EAOC ($1000/yr)
300
270
240
210
180
150
120
90
60
!1Tmill.opf = 13OF
30
0
0 10 20 30 40 50 60 70
Minimum Approach Temperature (OF)
15-30
80 90 100 110

15.21 S~.... 2- "3
fr. ~f ;Z
"'5
~ .t..r
Z'I';"(4'J-MJC.]
0.1 0 0./ K
o. D f>
0.01 0./')...
0.1 ~
0.' L{ O. () ~ -D.t) 1..-
/).fJ J
/) f.> If·
-0.1.1'-/
0.6 t.
().o ,
/).OL
D.Of>
_ o. J) r
-.-0 , '0
~p,o&
~]I'').1./
~
~"k -e~ IA' ~ r~""''-<'"
o.ll.·
C:p~"1
15-31

2- ?
j. '1 4
0.10
~
11, bG
D,II;
P"QZ~C'. g"1
(),IO
f";{Lu
&.1
c!:-- "3
0.01..
15-32

15.22
() .'1
A-
Il"
Iv(
2.{
?4, __ ~.l.(
0, ()'lr
~.& _______fl
C.
~."
-"
-
.<," Al
~I
_ _-~--- tp) tJtf.(
15-33

'P, QL- L,
1 ..').l
0.011
.
~
~~
&~
M
0.( 1- fO.rf'l
-GG
O.ObS0.0'
~1·' ~/~ ..fW"",
- 9}f.:;t
O.()lfS-
Lv
15-34

15.23
0, ,
() (u G
0,0:;
().olt
4
J t?.4
- ,t'$'
~,.",
t
-
-
I
A
~
I(j.() l C
~'()~ 1>
I
J ~
"

~ -=---
'-- - -
15-35
) If
If 1M(.li'J~'
- 0, I r
O,OIf
I' -~,
'I'-
.,. It. ..",~~
I~
o,tn>
-t),
I [
c;>,ot' o,~
O.I~
()." 'V
0
".0'" ~
-
PftJCI-t
PINc.H H~

_ O,I~
,.. j 0.04-'2,. ...::-- . (L..,.)
0, ( +- O,I{"
~t)
o,ul 'f 0, I'l-
b~
l "l. 1
l
~
~
~ 4
, Lf
,
~ ~
~ 3 'l.-
,
0.01
O~t
0. 0 4 f~
or!)
;l 0.10
;~1
0. 11 4 £"~
It 0" 0
LJ ~.v, f'" O,I'-.
0, pc.
100&-()rd"f
'().()<
15-36

15.24 2- S Y
,
Lf b 4 (
~ ..;, (/)'1-~'"X )/VI
I
f), I0
0,/1.{
{1,O,{ /~ 0,. '1
I'"
b,llrO.o~ II'
o.()~ i~ O,Il- -".~l.i
o·~ ') o.{)'J~
0.(.1'1 /).0 <. ( o,o~ tJ.~Lf
0,0 l -*"- - I'
. o,og
- ,!i':0(. I)
0.06
0,0110 _~, 61 "
0,06 (
D.O<
E... O,o6~ _OIDI."
.1
0
O. OOb')
-
-o,OY
Q".c)I-1
-15
}_O,I1Ll
<:: YI't r ,,,1l.1 " r~~' 70)" i J/I"I"t-
i .,oa
L
i~.t,..;t ;. IJVI~ ~ V -AJI~j fl.u
O..·IA k j>11I<d!l)/..t 4. 1W ~~
...... .J0,0'1 M L-u~
~, -0.01< >~
..... JO.IJV)
....
bJ {;~~'J
15-37

'l- S: Lf
•
tvs 4 & 4 I
1"" 0,11..-
__O,~
~o,oq<r.J. D,I1"1
0,0)
T
l.~
f!) I '1..--
:t'Ob~
0.0 t
O,oLf
--0.0&
0 ~
- 0.00
15-38

15.25
0.10
~f 0'1
0, if)

fW
O,U$
p.o
Ol~V
~.CI
,1/
t:J,f) '{

A
ti!
c.ob (
O.o'V-'l
~
Ii
€
0"'-;;,41
~
(,..
15-39

Q. ''-1
r' -v
O,!)~
6,
f 1'1('
_0.0'-
0•.>1-
~
(J ,6
0.-' O.Ob ().o'-
(),
Q.ov- A O".-_....
(),
------+o.o~
pi...~

Q,tf)
~--
0,0{,. --
~
:l
r;.b"lI:n
15-40
O,O~S'
'
o.oi

Mfo~~~
t... ~
,
b;....
h3 ~
- O,l) ~.o~ '
+a.l>
-
t.~__ /).rv-
04)
1~,:T O,Ot"
qr
S Lv
0"11 £I ,IT'-""
0.0'
Lv b-- ,
10·0(.
15-41

Chapter 16
16.1 (a) Fluidized Bed
Input ~ Output
Frictional loss
(b) Turbine
Input ~ ~ Output
Work
(c) Pump
Work
Input ~ Output
(d) Stripper
Liquid Vapor
Input
~ I Output
Possibly heat
....
added
Vapor
T I Liquid
Input Output
16-1

(e) Adiabatic Batch Reactor
Process Input___..~I_______~---+'~process Output
(at T = x) _ (at fixed T)
(f) Semibatch Reactor with Heat Removal
Process
Input
Utility Input
(i.e. cooling water)
1Utility
Output
+---__..Process Output
(at fixed T)
(g) Heat Exchanger between Two Process Streams (no utilities)
Process Input A'-'--__~~I
(at T =x)
Process Input B
1Process Output B
(Heat)
16-2
-I--_ _...Process Output A
(at fixed T)

16.2 (a) Fluidized Bed
Equipment Independent: mass and energy (ifneeded) balance
Equipment Dependent: pressure-drop equation for fluidized bed
(b) Turbine
Equipment Independent: energy balance
Equipment Dependent: none
(c) Pump
Equipment Independent: energy balance
Equipment Dependent: pump curve
(d) Stripper
Equipment Independent: mass and energy balance
Equipment Dependent: flooding relationship
(e) Adiabatic Batch Reactor
Equipment Dependent: design equations
(f) Semibatch Reactor with Heat Removal
Equipment Dependent: <;lesign equations
(g) Heat Exchanger between Two Process Streams (no utilities)
Equipment Independent: Q= mepllT without phase change
Q= mA, with phase change
Equipment Dependent: Q= UAIlT;mF
16-3

16.3 Distillation Column
F
v
L
~
V'
L'
A = volatile component
(Assume binary)
D
B
lFZAF =DXAD +BXAB JEquipment Independent: Column
FHF +QR =DHD +BHB +Qc
[
V =L +D ] Condenser
Q=(rnep!J.T)cw =(rnA)process
[
LI=B+VI ]
Q= (rnA) = (rnA) Reboiler
steam process
Equipment Dependent:
bY relationship
Column
flooding relationship
[Q =UA!J.TLM] Condenser and Reboiler
16-4

16.4 (a)
P-301 AlB
work
H-301 E-301l2
Benzene
~
Ethylene
~
r
EB
LPS
~
r
heat
1
(b)
E-401
HPS
11
H-401
heat
HPS
11
E-403
HPS
11
E-303
HPS
E-404
LPS
E-304
LPS
11
E-305
cw
E-307
cw
~
E-308
HPS
~
E-306 P-302A1B
LPS work
11 ~ 1
Ethylbenzene (EB)
Benzene Recycle
Recycle and React
P-303 AlB P-305 AlB
work work
E-309 t-304AlBcw work
11 1
Fuel ~a
EB
~
f--
C-401
work
E-406 P-402 AlB E-409 P-404 AlB P-406 AlB
LPS work cw work work
~
s
E-405
r0
1AIB
E-407l E-408 P-403AlBr05AlBlcw work
n1T
work work
11 1 1 1
hydrogen
~
benzene
toluene
Styrene ~
styrene
waste
water
r--
ethylbenzene recycle
16-5

(c)
cetylateda
castor oil
~
~
(d)
DBP
~
air
benzene
r--+
H-501
heat
P-501 AlB jwork
t
E-502 P-502 AlB
heat work
E-501 jE-503
j
E-504
1T
cw HPS
tr tr
Drying Oil
E-505
cw
E-506
LPS
P-503 AlB P-504 AlB
work work
t t
acetylated castor oil recycle
P-601 AlB H-601 E-602
HPS
E-603 P-604 AlB E-606 P-606 AlB
workwork heat LPS work cw
C-601 E-601 P-602AIB P-603 AlB E-604 E-605 P-605 AlB
work LPS work work cw Ips work
1 tr 1 1 lr lr t Ir
Maleic Anhydride (MA)
dibutyl phthalate recycle
16-6
acetic
acid
drying
oil
gum
r---
off
gas
r::--+
IMA~
I--

(e)
C-701
work
process
terwa
--.
leneethy
--.
ira
--.
r--+
1
(f)
deionized
water
--.
methanol
--.
air
--.
r
E-701
cw
E-702
cw
E-703
hps
E-704 E-705 E-706 E-707 E-709
C-702 C-703
work work
l ~ l
P-801 AlB
work
C-801
work
E-801
mps
lr
cw hps cw hps cw
R-701 C-704 R-702 C-705 E-708
bfw -+ mps work bfw -+ mps work hps
~
E-802
hps
lr , l lr , 1 lr
Ethylene Oxide (EO)
ethylene recycle
E-803
cw
R-801
bfw-+ mps
E-804
mps
lr lr
Fonnalin
methanol recycle
16-7
E-805
cw
P-803 AlB
work
P-802AIB
work
E-806
cw
lr
P-701 AlB
work
lrfuel
gas
-=---..
light
~
EO
f-+
waste
water
~
I---
off-
gas
formalin
II

Chapter 17
17.1 Second order reaction: -r=kC~; k=Ae-E
/
RT
Ifideal Cj =hTifideal
220°C = 493K
250°C = 523K
E( 1 1)2 = e R 523K 493K
E =5957.35 K
R
(a) r2=k2(C~2)2 =(Aj)e-5957.35K(51~K 49~K)[(PA/RT)iJ
1i 1G CAl ~. (PA / RT)1
r2 = 1.581~)2(1;)2 = 1.587(498K)2 = 1.47
1i IlIA! 1; 518K
(b)
17-1

17.2 Prank = 3 atm = p..; P2 =5 atm
3
VI =12.5~
h
Assume turbulent flow
17-2

17.3
Tube
Organic
..
1.25" 16 BWG
Do= 31.75 mm
Di =28.45 mm
Ro = 15.875 mm
Ri = 14.225 mm
In(Ro/) In(15.875 [mmJ/ )
---,--,1----,RR'I-,,-' = /14.225 [mmJ = 0.002438
k 45 [o/mK]
1 1 l~R;.z;J Ro 1
-=-+ +--
Uo ho k Ri hi
1 _,_1_+0.002438+ 15.875 1
Uil U02 2000 14.225 700
-1- = UOI = 1 + 0.002438 + 15.875 1
Ui2 2174.95 14.225 700
U02
=1.009 ~0.9% change
UOI
17-3
hi = 700 W/m2
K
ho = 2000 W/m2
K
Shell
Cooling Water

17.4
Assume thin walls: _1_ = ~+RWall & fouling +~
Uo ho hi
hi = 1000 W/m2
K
ho = 500 W/m
2
K
1) 1 1 1 2 =0.00035 m2
K/W
.L'"wall & fouling = 300 W/m2K 1000 W/m2K 500 W/m K
(assumed unchanged)
1 _1_ + 0.00035 + ~
UOi U02 hOI hil
-l-=U= 1 1
01 -+0.00035+-
U02 h02 hi2
(a) h02 =500(0.8)°·6 =437.34 W/m2
K
1 = 1 + 0.00035 m2K/W + 1
U02 437.34 W/m2
K 1000 W/m2
K
(b) hrJ. =1000(0.8t
8
= 836.5 W/m
2
K
1 = 1 + 0.00035 m2K/W + 1
U02 500 W/m2
K 836.5 W/m2
K
(c)
17-4

17.5
Gas
Tubes
..
ho >> hi assumed since ho is condensing and hi is process gas
17-5
Steam
Shell

17.6 rxr
g-
M =2pjLu
2
=32jLrh
2
D pn2
D5
2.5" sch 40 Dil = 62.71 nun
3" sch 40 Di2 = 77.92 nun
M2 = D1: =(77.92)5 =2.96
~ D2 62.71
M2 -~ = 2.96-1 =1.96
~ 1
W
· .M
=m-
p
W2 ~ i M'2Pj ~1.96
Wi IMlh
!= (elec. cost)(W) =>I:::::; double elec. Cost I
17-6

17.7
8~
(a) t 3x
(b) V2 =(D2J4 =(52.50)4 =5.03 t
vI DI 35.05
(c)
(d)
. 1 v2 L I l t
voc-=>-=-=-=10
L vI L2 0.1
(e) VOC MR4 => ~2 = M2 (D2 J4 = 3(5.03) =15.1
vI LVi DI
(f)
(g)
17-7

17.8 (a)
160
150
4014
1~----------------------~30
(b)
~----------------------~254
240
(c)
L-__~ 187
64.25
30
17-8

17.9
_ _ 8"sch401.5 kg/s ( ()
Di = 7.981 in = 202.7 mm
A = 322.7 X 10-4 m2
100m
At STP . = kmol 29 kg = 1.295 kg
POir 22.4 m3 kmol m3
2jLpu?
M =-"-'----
D
m= 1.5 kg/s = pAu = (1.295 :~ )(0.03227 m
3
)u
m
u=35.9 -
s
Dup (.2027 [mJ)(35.9 [mls])(1.295 [kg/m
3
])
Re=-= =547900
~ 1.72x10-5 [kg/ms] ,
s= 4.6 X lO-5 m for commercial steel
!-.= 4.6 x 10-
5
m = 0.00023
D .2027 m
f =0.00391
2(0.00391)(100 [mJ)(1.295 [kglIn3
])u2
2
A1'= =4.99u
0.2027 [m]
17-9

17.11
Need more information; limitations ofmethod.
17-11

Chapter 18
18.1 A pump curve shows the relationship between pressure head and volumetric flowrate
through a pump
For a centrifugal pump
Head
across
pump
/ Pump curve
r----L_ /_;- System curve
--------------
,,,
I
I
I
I
I
I
~ Typical region ofoperation
Volumetric flowrate through pump
18.2 All depends on the system curve, so the statement is false since parallel or series
arrangements could give higher scale up. An additional consideration is that when
placing two pumps in series, the maximum generated head is twice that for a single pump
or 2 pumps in parallel and this may cause overpressure in downstream units.
18.3 For the pump and system curve (Figure a) the intersection represents the maximum flow
through the pump. Operation to the right ofthis point is not possible. For the NPSH curve
(Figure b), operation to the right ofthe intersection point may be possible but will cause
the pump to cavitate.
18.4 The film heat transfer coefficient, for turbulent flow inside a tube, increases with flowrate
to the 0.8 power. For shell side flow, for laminar flow, and for phase changes the
relationships are different.
18.5 The only way to fix the cavitation problem for this case is to raise the NPSHA curve.
Therefore, by running the pumps in parallel, the flow is split between the pumps and the
suction line friction loss will be reduced, thus raising the NPSHAcurve.
NPSH / ~---;. /' NPSHA - 2 pumps in parallel
NPSHA /
, '--;---. NPSH__ R
-----------
Volumetric flowrate through pump ___-..
18-1

18.6 For laminar flow, the flow ofliquid is governed by the Hagen-Poiseuille equation given
as:
For turbulent flow the frictional pressure drop is given by
2ffPV
2
Lp
-Mf =
D
where the friction factor is a weak function ofReynolds number and hence ofviscosity.
Therefore, the laminar flow case will be much more sensitive to changes in temperature
than the turbulent flow case.
18.7 The relative change in pressure and hence vapor specific volume in the two columns will
be 10/9.6 = 1.04 and 1/004 = 2.5. The vapor velocity is proportional to the specific
volume (l/density) and hence the 1 bar column is much more likely to flood than the 10
bar column.
18.8 (a) An increase in column pressure increases the vapor density and decreases the
superficial vapor velocity in the tower. Hence the column will have a reduced tendency to
flood.
(b) The increase in column pressure will bring the equilibrium line closer to the x-y line.
This makes the separation harder. Hence a higher reflux ratio will be required to obtain
the desired separation.
(c) The number of stages will increase since, again, the separation is harder and the efflux
and top and bottom purities are fixed.
(d) The overhead condenser temperature will increase as predicted by Antoine's equation.
18.9 From Equation (18.21), the NPSHA is given by
_ 2pjLu
2
*NPSHA - Prank + pgh - p
D
For this case, the pressure in the tank is the vapor pressure and the 1st
and last terms on
the right hand side cancel. Therefore, changes in ambient temperature will have
negligible effect on the tendency of the pump to cavitate. This is in contrast to not
saturated liquids whose vapor pressure will change with ambient temperature.
18-2

18.10 Assuming turbulent flow, we can write:
M'2 ~ Q~ => Qi ~ M'2 Ql => Q2 ~ QI~M'2 ~(35)J(5-1)1(3-1) ~49.5 m3Jh
LVI QI LVI LVI
18.11 (a) Only the 2-pumps in parallel set up will give you the desired flow. The desired
operating point lies to the right ofthe single and 2-pumps in series set up and hence is not
a viable operating point for those arrangements.
(b) There is not enough information to determine ifthe pumps will cavitate. You would
need to look at the NPSH curves.
18-3

ProbLefY' 18.12
EttUdli::uns I1lvolved.: 9::UA~~ -:: VltoUC-fA1"otl. :; ~sA
a.) <iI:: ~it -= ,"So1{I~beo ..U ;:: Lt.'t(?ltlO
b
~T/n
('325" -c.S3) - ("2.Q3 -253)
.6.T :: -= S'f# '1 (Ie
~ [t"?>2S"-2.5"-:!Ycz<l"3 -l'.5"3)
lJ ".! g ::: 1.f.'t2'to,tf~JI 1. I 1 :::'3S1S~J
AA1J.". . l'l 22. rm'l 5 ~I lj. ~c. h m;z: ~c.
~) eLe 0. n. T-u.be.$:
_ 1 ' 0 rtlGQc I" I
~"T - UCT
A -= '2> 5'15". ~T 2..2.b rr/-
=,23 Y-fD"'S- hc./~J
18-4

Problem 18.13
. "'''''' ~ ..CIe d~ t'tDe.4.: Rr =R,,: -+ Ro ~ Roof "'" Hi t'}- .i
Fi.q. I .1 5-':i4te-m.~ Ror =R..c: -fo Rc. -+ R "'" R.f r,..l'.l
E,..ii':E'j-": : (RT-R:)' (R.. -R:)+(Ro- ~+ (R-I"-I'!~)
+(Rf - Ps) ~ ii.i.
FroYrL Pr'G.b if.l: R~ -- .'Z~ 'l' l6 S" . Yl 6 C/ Jt J
Fru"'M cl.~t a. in Fitj €.
, 1 [ n 0 c."l .r " /,
Rl ~ U 9 :: 2.2..6 vn7.:f 25-90 11 =.71~lb h t/~1
(R,.-tC'~') = (1."-l:t:')~lV': O,'tS.,.,o·S"16c./..Qr
A,,£ume. ~ Lit1lc: ""teet1'). -,iJe~o'U.~ (tt~"f.ed. LU"dterl R,::J?4,.L
R .boi.li'11.~ uTdle. " l t'I.4e-wa..4LtiA1"e -1:0 .flo", (R: ~'R6
For lt~hi oil ~ R I"V t/~ ~ 1/m
o
.&
Fo't' CJe.4-n t uhe..l~ t See. PYob.'S. ~1.)
m- '= % :s. ...,I.I'ZY.'6"~1r jQ °c -
. c~ 'n I 2.~t.i J (~lr-~l·t
~or 1=., ":. l ': s~,soo~/t1
.m,. 3.'I?,"l-IO'"~ r Ij!'C I I
"" "2.lJi ~l 1(~~S-30G~
; ss.~oo ~,/~
..~ R.... -= ~~C>1 t<.;.
1:.n li-rvt i.ti. n. '1 c:Q..4e UJ"he roe. 0 it ye c;(tA Y.c. e. con-ero l s~~'I ...
Pi", -= t?,.
11f ': G.4Js ~J6"< "" 0 ..ou7 ~ ,.'Z3~lOS -::. C.'R it !.O1
~Imfrc.".eJ <'!J>iiw..4te. obia,neJ b~ "., heat
tr<2.n...ofe.t" rOot 'oob to 'f'epLd.e~ ibi... J( M i i (S~ ~. ,eJi.
18-5
/

Pro.hlem 18.14
r<>,
~- ':: F0 to efe6l. nt",~e..& :
RT
:: R0 • R.,. -'- ~f .t. R.., E9; ).'
R5- :lI 0 ( ~ ltot!t' t 'U..be..&. )
A.of ~ 0 t .. ..&J.'l.L1'n.e.a)
R / RL : /Z
E'b j .bec G"mt""
R,. -= Ro + 2. R0 :r '3 Ro :. 0.IG3,c,b"S"
h ~C/~l
h.. ':& ,I
- - R Ilo
" I ~Jl 'o.q.1 ')(Ie; 5 t-:'c. 2Z.' rr?
':' 10, iOO ..Ii J /h 0c yn'Z.
~,i !:r_~- = ~J' l
R,L A. C..i-z. ~ ret!' h·C 'Z'Z ....rn:z
-::. ), '100 AJ /., "c. m.7.
18-6

Pro.ble.m 18.15
l-iedi Ff4-:t litn,t dtLOl'1 (TclbLe '1.11. "tte",cg')'j I.r~w / tyn?"
1'1"')1, Q ::.'31.5'~~/1 ~r ~bOO.L IZ'2.b'WI
t
=.
tn' LPtW...4 l I h
2.r ,'It10' Jtl/n
Tei'nf. Limita.tion (Ta..ble I~18JTteln 14) Ttna.x: 315'°C
FIa.v Ratio: M
6.0
1.3
5.5 m(kg/h)- 55,3OOM
1.2
s~~t.<:h I~·.q
Steam Tempetlll.rnt- 253·C (sat.)
1.1
'C'
5.0
Area.: A- 22.6m2 1.0
Ihdi.(a.te u7lti
~ 4.5
0
.for
~
4.0 -
Q~ 1=Lu.'t.
.N~-Operati1uJ
3.5
. ~e II i G)'j.
3.0
Z.3 1:' f l'
2.5
2.0 3JS
310 320 330 340 350 360
I
Temperature. T1 ("C)
a.) See S'k.et.ch: Md:X. T L"h ~era:tLn.l. ~lS °c
M.0 'X.. Q'= 2..S",'11tJ' IIh
Jr) 0 i. ( te"""e e oee,""ti.o11 el.ceecL.. ~ .,r0C ("ed..c.he....3(5~
.!. On hot .A1Ly.fa..ee. PGt~",eriZQti.o77/d.eCO')rtpositia7'l
e~~ed8d .
.ii. E~..:eecL.4 hoY"mo.L fLu~ LeAreL .far l£~ ht Gil.
.c) T-= -arz,o~) ~:().9(® :. o.~ )t. $"~~oo) ': If'f.a9Q~!h
18-7

a)
m..) Cf,oiL, TIN t Te are..the AC,'"e f.,,, .b&.bB ~aAe
; (BC-) t hew Cc2~e (v£w)
E'D' ..i reJ.~c~~-to
Q..,r+II
Qee
-- ( - VK ) tl5At
( l - 1/14 ) a.c.
.1,.) _/~_kCW__ = '25.1 -= '1.11
'Z2.~
c)
-:Cn..c...eGL.6e i 1' Q fe..u ihd~ i'h~re6l.A.e 111 A
to 1'" ~
Q=)f~A
NeW' ...ca.-6.e hds cloA.er QPft"OcLCh te'mpe"'Qiu,.~
lec2J~ln, to .AtmdJ Lie.f' inc:rt'!d..le I.." Q rtlt~o
18-8

Proh l~ln 18.17
.
IV'
E4.&. h ?,ump deli lITe....b
.Ja."na.e flCY'td I r....I..4ure.
tl.rcf- FLows Arc ddJ'tLP"e.
Tof" A,ca.lc (.fl o'W') J... 2..'bottorYJ
Ss:.ale .
V4e tot' AC,d.{e #' Jt:'ft .J".co.le
hc.r~ . po..rcdLe L pumP'"
C) .IV .. 3.0
a.
,l,e
i
.,
J.)
e4.,c It "U1'nF de L//rcS"
..to."Me i!P.
P"':"U~(3" J.,.,op.s are ",Jditive.
La.f.t "'(dle. ~ '2)C ri~ht ..ueJ.e
)..u ..i.~ ht ..!CAle r b~tto')n
S~ClLe t(HO t-uro ..AerLe.. fU?I f-'
I
.fJ :: l.r' }.r., 3.0
LShY' 0
GUfYti 0
",1m 0
18-9

~: Problem 18.18
~>}"'.
ExchCL!.Cj!r
p= Iba....
StorQ.~e Z-: 20-m
F"ure lb.> . pro·tnJe.6. b.Ae Co.~e t 11.'1 the pumr cu....N'e
It ( ' . 1/.
",ew .,.lour t 11:. . 'JI t.S- • L,5" IT. I tnl.lI
7..
FV-t..GtLol&<Rl 10A. ~ h~ " 20." m ~ ~,"S') -::: 4'.'1 m.
I
Ele1T4t L6lt Ch4l'c:jc: A h! ': 3Z - 2G ': 12.u m
Sia.ti.c Pr42.4.J u.toe Chd.~e: (1:2- t.1)'J(fOS'/(7r'O¥,,gI) :' '2..7"h1
a.. Hea.J. re~uLrec:l..(t'Oln f'U:m[3 ='1(,,4 -10 J:t.c, ... 2..1:: 'J~I111
Jr. @ h -= ",J1n ir =l.bS- rJImilJ)
PUlnP S'j.Ate",..A ~h(r7l1'" D'h l=i,-u.re 17.'1 Ccil1l1ot
ft'ov..r:de .fLoW' I fre...A4fJ.t"e.
b1
h
.N-
.c. No P'"1'np. ..4~..6te")'ro wi LL oferdt.t!! if PLa.cecl.
a.i POL")'I t 2 • P'"e.4.J.tire ~t Lon Let re«t 4 ~ yeJ iA
~e,.t J,....,...e ~ t-Jot Pa..ll.4 ibIe If
PLci.ce pump ot. eo~~t i
18-10

Pro.h Le'YYl 18.19
Bett.
S"hcr.H
h6ft4-
!
i
M<:lior Moi: Or' . PU1nrD
PlIlley pu LL Co":1
'a50urp1R
,II , "J 5-0" 't If
lid. .7 L4.
d..:a.. olia.
(3tl.,4e Cue: ,trdc:~ ')')0 t6t" { '''eli ca.. pU7np fulLie..a.
Pu.'h1f rf'hl.: NZ -::. N, (0./D2 )
... id.en/;i ~ie.&. b'a.lue.4 obt12..tned fro?n F'g 12.'1
(7.1.lt. into pelLe.t')
Hea.cl.: h2'~' ( 0 1 / D2. )2-
FwW": 11"1.:''';; ( D, /D-z.2.
,,". ~lle.lf hz.= h~ (~/,)2: I,Ooh~
tUz ~ Ir~ ( ,;,) -; nrf
*. '2. h*'S" P-uLle4: h =' hi (&'/S') :: J.4,# I
J 'Z .¥i 3 »
1r4" ~ ((./sY :. 1.7Zg-u;
4
11
PuLLeJ:f h"l, -::. ~ ((,,/q': 2.2S h7
17~ :. ;r;(f./q)3 =- '3#37S-Nt
SAYt1.pLe ('Cllc.ul.a..tio~~ (5 " Pulle~)
~ *'~ro~ J-lo, 1o,.Cf e 1:,= S-O 1'1 .' 'lr.,'= o~S'm._
;J "tn L1
18-11

·: ~"
o....
0.
a
N
oQ
-
18-12
o
• ...u
~
.~
....,.....
"-
...s
~
~
~
"-i:.....
0
"!d
Q..
Fa
&
i..
0
'+-
~
~..;s
'-'
t.I
U
Po
~
~
'-()
"+0
So.
~
Q.
cr
-"
if)
-
~
QJ......,
...Q
e0-

18.20 At 208°C, p* = 78.96 kPa
NPSHA =~lIle( - p. =79 kPa-78.96 kPa =0.04 kPa
u - 18,500 kg/h - 4 m/s
- ~78.6kg/m3XO.001313m2X3600s/h)-
For Figure 18.9 NPSHR ~ 0.2 kPa
NPSHA < NPSHR =>Not suitable
Lower T - 1°C would be enough
*P = 77 kPa and NPSHA = 2 kPa
18-13

18.21 (a) 1.25 in sch 80 ~ Di = 0.03246 m
2P.fL u2
NPSHA
= ~ank + pgh _ eq - p'
D
NPSH
A
=1.10325 bar + 870(9.81X2ViO-S)bar- 2(870XO.005X
6
)Q0-sX
U2
)
~ 0.03246 bar
-0.172 bar
to locate A: u =~. u oc_
1
_
pA' D2
u = 5.73 mlID~J = 5.73 mli0.02664)2 =
D; 0.03246 3.86 mls
l'u-=-3-.8-6m1----,~
(b) T= 40°C
InP* = 10.97 4203.6
T(K)
InP' =10.97- 4203.6
40+273
p. = 0.0854 bar
NPSH
A
= 1.10325 bar + 870(9.81X2XlO-s)bar _2(870XO.005X6XlO-
5
Xu
2
)
0.02664 bar
-0.0854 bar
NPSHA =1.189-0.019595u2
1.4 ~--------- --,
1.2 t:~~~--------------~
"t::-
its
e. 0.8 r-------~~;:-------------J
~
-o-Question 18.21 (a)
_Question 18.21 (b)
~ 0.6 r----------~~---------J
0.4 r-----------~~------J
0.2 r--------------=~}-----J
o 2 4 6 8 10
velocity, u (m/s)
18-14

18.22 (a) at 125 Mg/h and 3500 rpm
Pin =2 bar
Pout =?
From Figure 18.11 P,m! ~ 3.6
P,,,
P,m! = P,m! =3.6
P,,, 2 bar
Ipoll! = 7.2 barI
(b) at 125 Mg/h and 2200 rpm
Pin =4 bar
Pout = 9 bar
From Figure 18.22 P,m! = 1.8
P,,,
IPOIII = 7.2 bar => Not possible I
(c) at 125 Mg/h
Pin = 2 bar
Pout = 24 bar
POll! =12
p;"
P
At 3500 rpm: ~ ~ 3.6 => use 2 stages
P;/1
p. 1 =2 barIn,
Pout,! =(2 barX3.6)= 7.2 bar =P,,,,2
POllt 2 = (7.2 barX3.6)= 25.92 bar => Slightly past 24 bar
P
At 2200 rpm: ~ = 1.8 =>use 5 stages
P,,,
Pin,l = 2 bar
POllt ,! = (2 bar)(1.8)= 3.6 bar = P,,,,2
Pout,2 =(3.6 barX1.8) =6.48 bar =P;",3
POllt ,3 =(6.48 barXL8)=11.664 bar =P;1l,4
Pollt,4 =(11.664 barX1.8) = 21.0 bar =P;1l,5
POlit 5 = (21.0 barX1.8) = 37.79 bar => Much greater than 24 bar, big waste.
. IUse 3500 rpm with 2 stage~
18-15

18.23 NtoG = 20
G= 80 kmollh
L = 20 kmollh
i.=0.25
G
YOIII = 0.006 = 0.6
Yin 0.01
L 20
A=O.4=-:::>m= ( X )=0.625
mG 80 0.4
If L is at 85% A = ~ = (0.85X°.4)= 0.34
mG
YOIII =0.56
Yout = 0.56(0.01)
IYout =0.00561
18-16

18.24
For a Packed tower, lise Eqn (18.23)
.I/•.1m "- .l!A.:'>Jt
.t,.~.;" .:f 1 I ;t,l!
(LA)
( 1. A)
1.10.:!i;
.1l•.J" ...:.f .'!AL'U:
N= 101 Iu
y'A,out = 0
YA,in 0.02
YA,out = 0.001
G= 80
L= 40
Solving 18.23 for A we get A= 1.1326
III = LlAG 0.44
yA,in A G
100 90 80 70 60
0.0110 1.00 44.1 39.7 35.317 30.9 26.49
0.0138 1.05 46.4 41.7 37.0828 32.45 27.81
0.0173 1.10 48.6 43.7 38.8487 33.99 29.14
0.0216 1.15 50.8 45.7 40.6145 35.54 30.46
0.0268 1.20 53 47.7 42.3804 37.08 3'1.79
0.0402 1.30 57.4 51.7 45.9121 40.'17 34.43
0.0584 1.40 61.8 55.6 49.4438 43.26 37.08
0.0821 1.50 66.2 59.6 52.9754 46.35 39.73
0.1117 1.60 70.6 63.6 56.5071 49.44 42.38
0.1477 1.70 75 67.5 60.0388 52.53 45.03
0.1903 1.80 79.5 71.5 63.5705 55.62 47.68
0.2397 1.90 83.9 75.5 67.1022 58.71 50.33
0.2958 2.00 88.3 79.5 70.6339 61.8 52.98
100
-.:::
90
80
+----+----+----+------1-----1---- G = 'j 00
+----+----+----+---=~~---I---==G=90
--'0
E
.:.::
--.l
.u
-!U
...~
0
u:::
~
::I
0-
::i
70
60
50
40
30
20
10
0
I
+----+---~~--+-~-=~----I-~~~G=80
I
+-_~~_=~~-=_~~-+_--~~~~G=70
-I--,L..".c::.J-...~=--_k_~+=-----_+___~-~=:::::. G =60
0.00 0.05 0.10 0.15 0.20 0.25 0.30
Inl:et Gas-phase mole fraction. Y A,itt
18-17
0.35

18.25
For a Packed tower, lise Eqn (18.23)
[1;,,<,, ,,';...11 I .;,; (I A)
'I.'''"' ~. I r.1 ,~)
N= 10
y' A,oul = 0
YA,ln 0.02
YA,oul =0.001
G = 80 Fixed
L = 40
Solving 18.23 for Awe get A = 1.1326
m =VAG 0.44
m= exp(10.92 - 3598/T[K]) T = 306.5 K
Y~~ A T
43.3 38.3 33.3 28.3 23.3
0.01'/0 1.00 5'1.2 42.6 35.3 29.1 23.8
0.0138 1.05 53.7 44.8 37.-1 30.5 25.0
0.0173 1.10 56.3 46.9 38.8 32.0 26.1
0.0216 1.15 58.8 49.0 40.6 33.4 27.3
0.0268 1.20 6'1.4 51.2 42.4 34.9 28.5
0.0402 1.30 66.5 55.4 45.9 37.8 30.9
0.0584 1.40 71.6 59.7 49.4 40.7 33.3
0.0821 '1.50 76.8 64.0 53.0 43.6 35.7
0.1117 '1.60 81.9 68.2 56.5 46.5 38.0
0.1477 1.70 87.0 72.5 60.0 49.4 40.4
0.1903 1.80 92.1 76.7 63.6 52.3 42.8
0.2397 1.90 97.2 81.0 67.1 55.2 45.2
0.2958 2.00 102.3 85.3 70.6 58.1 47.5
120.0 - .•...-~---.~.~.-.
r_ 100.0 T =43.3';'C
.s:::::
"'-
'0
E.::c:
-......
I1i....m...~
0
u::
:E
::I
C"
::J
80.0
60.0
40.0
20.0
0.0
T=33.3':·C •
I
+--'7"'---t::....-=--+----=>"-i""'=-----i----t---T = 28.3·:·C '
T= 23.3·"C
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35
Inlet Gas-phase mole fraction, YA.III
18-18

18.26 See Section 22.2 and Example 22.1
18.27 Increase capacity by 25%
Q=mCpI1T
Q=mA
Q=UAI1TLM
Process
..
If A= constant then Q2 =m2,p =1.25
Ql ml,p
Cooling Water
Q2 =m2 I1T2 =M(T-30)=>1.25=M(T-30)
Ql ml I1Tl 15 15
Q2 =U2 I1TLM,2
Q1 ~ I1TLM,l
/
(75 - 30)-(75 - T)
J75-30J
1.25 =MO.8 __~_7_5_-_T__
37
_[MO.8 T-30
37 / 5
=>1.25- - - ~
" 75-T
Solve equations (1) and (2) simultaneously
T = 44.4°C no penalty
M= 1.31 an increase of31%
18-19

18.28
tA/Q, M T ("()
-- -0, 0 01'1]1 'f7, I ()
0,'" r ~'II
I{r.~
j. t,)() J I.f~ IV
r ?".( J,1 J '-I'/, tp
/, <;. 0> I, ~ 4- 43. K
Exit Cooling Water Temperature, °C
43 44 45 46 47 48
2+-------~--------~------~--------~------~
I.-
Q)
en
c:Q)
"'0
c:
0
<..,)
l.-
.e 1I.-
0
1:5
m
u..
C)
.£;
J1r.n
o+---------------------~--------------------~
o 1 2
CW Mass Flowrate Ratio
18-20

Pro.bJ~m 18.29
Vt'fer lTelo~i~.I:f Li-mit (..fLoodi",,!): AJ4- =I.(1)Y1.i6
(See f~e >&~ )
L0 tv P"e Lo Co " t)j li)'n.it ( Effi eit! 1le'1 E • o.,~~
Fro"Yn Fi~'U.1'"e rz.fY Lirtlit (L(,'W'~r) r'nL'Ull ::'n~J
h
D '= 3D k-mo fin
<? ()-ffe,.. JV"eLoci.tj Li:W£ t : V,.,,~:. ")0 ""07 :: f~1 ."-mol.lh
A.4~u.')'rli 'h, velacit"1 (jJ. <i vapor .fLOo"," (Y
Note: V~ IA dW'~..l" littLe l4.,.~e"" th4n V-.J+I
(se(" ta.bLcP,».1,) :. ~L.ooJ.i.'hj "m6..e L;jLe~
at top crt <:oL~'m.1t
18-21

~.~ F!. ..
......... 0
~ .-.)
tcf
1 ~ -..,. !:..
J
ll'I
a:
0
,.
c..
~
C
IJ.. ~
~ N
0 - ~
~ 0
0
oJ
.-' ~
~
'I:S
-.I
....
-.
--~
......,
u .:SoJ
0.~
0 ,..U
0
-d ~
" "QJ
LL
~
;:t
u
.,u
?
'd
~
~
oJ
.00
.::to ('tl (l
. .
- - - -: 0 a:- c:i?
- ..; 0 0
~
"~
18-22

Pt"ohte"h' 18.30
tA.4U"YY e Fi 'J"U-re i'.Cf i4 CD rr"e c.t
@ Hi~he1r. Li-mi t
V~L -:: 07 +'30 ::" .,7 ~"Wul/h
@ Lc,w€v- LL"'ml t
VLl ":: 41.+30 :. 7 Z ~~ol/h
Ra.tio Ill. /V'-o1'w :: lZ II~ 7 ::: D.S'Z.
Ta..ble P I'b.l i revedl.~ V. (O"er-hea.cl»VIJ+~(h6tL·1J.,p)
Eloa.yYL ple @ r:" O.q ) "i '= 0 ..2..5"
O"c~leQ.j , ,"q t:" tn. q v J '= "., .;J j VV·H :. 138
V. L - '+'1.5J.
I/Va.)"I- /1'3~ '::tl.,,~
ALL CG'hJ( tt01,h Aho"UJ" vati.o to h~:: 1.0S"
Boi.l- up l'-m it.&·,
HL~her Li.m i.i :- "?>7 / 1.0S -:. (~O k."h'ol!h..
Lowe- LL'nlL t :' 7('/lfJ 'S := £q ~"""oL/h-t"
".
18-23

-..
:J
0....
.~
•--.I
• ...t
o
cO
//
3.'35
reed. CO~fc~i tL01'.
Pt"ohle-m I~·?>O Pet'tor~a.")ce Oi~"'c1-nt .fat
Di.A ti LLa.tL011. Tow et
18-24

Pt"obLe-m. 18.31
•
A H~::: Q", /VtJ~ I
SeLect 'h1iJ- ra:n~e C'G'?'dL tia1t4
.. l!'=- 0.'3) r=o.<1 -where Ta,hle PI.IS
9 Lve.-4
VN... ': H2.7 .,Q·,.,u:)L /h
<it' ': 3.'I c:j / h
. .
.6.1-4 =- 3.(,1 Cl/h {I'Z.7 jl?'nCJlk =0.63? c.Y -,/1. ~1Iol
J:'Gr v-«Lt(.e..4 of (VJI•• )~A" a (!~+, )HlJ :from
fJ robLe"')lyl. '''.1q
t4i<3he- li.-m.i.t·. VJ~l:: 1"30 ..Q."noL/h~,
LowC!t- Ltbit·, VJJ-+- 1 u(P<t...QrncJ/ht-
18-25

...- S.1.
~
"'" 't.g...,
c::.!)
...,;
~ '-t.'!
0
- '+.0 .
:r-
~. "3-'
0
l..
3.2.
..,
-.J
<".t....c
~ t.,-t
2.~
cr
z,.o
o;z5 0:30 (;.35 O.'fO
feed. CG""'"f ~ , ~t i.G-n.
PrcbL<:-m. g-'31 ?e~fo'(1'no..l..c.e Pi.4't"CL1"n .for
Di.4t ;{La.tlC"t r: 01J.Tey
18-26

Cl..)
I
r---, - D ~ 3D .9nnJ / h
~p =- O.g~S--
F:::
({JF :z 0:25
B-:..1 - - - -
~8 :::.
r F tfJF =~O D ': (0. ~8.5 )(3D) : 2b.5'S" ~'h1.Ol/h
F:; 0 2'.S"S/[O:z.s)Co.q)-:: IIg ~mollh
B = F" - 0 = (US -3-0) ";:; gg ~--rnol/h.
~ ~B -= F t'P
F r :I (((6)(O;2S)(t-O.Cf) :2.qs-Jt'rnli1!h
!Va :: 2.Qslgg -:0.0'3'35"
.Q,.) Fro~ Ta..bLe p _~.~q
Lo ':: IIS. '1 it--moL/V 'J VtJ~ I :: rag kL I h
C') Not. V'e c01n'i:ne-ncl.ec:l:~ FLOW..4 L11. ~LoodL?Lea
re~i.G'h
18-27

Probte1"1 18.33
a.) Develope c'U.rttr~ of rM~~ ( r ~"1' iLooJl1'c;l)
Obta.i.')1ecl @t11 tet;.etLO'"t of r LV i t h .'
1'na."1. L1nLl'W refL1L'X. 4 :: t07, ...beoe· Fi:J fg·2~
r 1'f,F' .~ J'#.-F
(),CfS 0.?,()q O.Zgg
O.'}o O.26lt 012bL
See plot
(j·B5 ".
'"
O.ZlfS 0.208
O.se 0. ~"30 0.18 4
,.. e"Y..trcipo l a. ted ClJ r,u-eS'
@APp :.' 0 , '3? 1 rM~~ :: 0.'1'6
A.) F r tYF ':. D tf-o -= GO) (0.885 =- 25·S6 ]7no LIn.
.!) 1.
2..
3.
F~ 26.56/( r)(N-p
Vl =La .... D -= J()l-+:aO :" 13 7 ]7rtO LIn
.~r: r F V, Lo
O· 'Z.c~8 ~g:5' . l'j7 .~O1
O·2'C(
,
11
1"1.&
C.2'tS 1'22.1
1O.c..~o I '3 &.<1
-I L~h. '"i..b~ of fLocJL19
HL~ '" utLL, t~ ~Mt~ (Ho:)(. reftu.~ lL~ec1.
ItL<Jh er duiieA m.d'j 'hOt bef0..o-*ihle i.....
e'tl..a.tL"S epuLp'W.e11.t•.
18-28

l.O
O,'lb
O.CJ5
o.qo
r
O.~5
O.So
0.75
6.10
1"37
.,--
< l"O-.I
0
~-......)
- ICY7..;: 00.,..
J
-u-
cao
'<!
0.25
0:25
,,;
o,~
aF
0·30
'aF
18-29
~
.....
0.32.1
0.35
Lo
F

PyO b. 18 I' ~4$ B.3$ ~ u. fF{ e 'h'le nta. L In.forn.a..t LO")'
~
Mole Fractions
1
~~ X1 Y1
0.00000 0.00000
~- D~:gO 0.05000 0.09915
5
O. 10000 O. 18960
/'1.0 'ao.90 0.15000 0.27223
0.20000 0.34783
0.25000 0.41709
F: 100 ...
0.30000 0.48065
0.35000 0.53907
0.40000 0.59284
0.45000 0.64242
r
B 7 ' 0.50000 0.68821
'--___~..._':: 0 O. 55000 O. 73055
~r.l,=o.()m 0.60000 0.76977
,~ 0.65000 0.80616- ,
Op'eya.til1~ Li'he ReL~ti.o~~ hip-4
Lewe,,: ~"" =(L/V')N"" ,(f31'y} ~B
0.70000 0.83997
0.7500Q 0.87142
'0.80000 0.90074
0.85000 0.92809
0.90000 0.9536g
0.95000 0.97758
1 . 00000 1 . 00000
Upper; ~tt''' (L/v),1-m .. ( DIv)I)! 0 '
~~'3' rt It, ~ hG~.4 .feed. LA. not LtlttoodUrc.ed. O~ pl.o.te tJ-,.t
~c.cT'e.A. ~o.'X.L""'U"'" .4£pa.nitio"". ihL..A' Fr;lI..e: WdoA
-aleta-e:no e~ c.lAoillCJ a.. eroee A"" A l')n.uL4-[ 0," (STU: ~he,....f
pa.~~ ).
SoL'Ut ZO".A. ~..o'U':clcc:l ,'U.Aed. 0.. trCl'CcJ,e ..Thi.ele cliCL~ ro.1It
Q.,.,,,i ~,.'uU141'1'l dot4. ..4~"l' a.!1cnrC'.
Fo ~ t "e bC'L..o.c. CQ.M (~': 2.~8
L :: ~CO :: (2.'3')('0) -= 71.4
'1-= "",1):: 1LlI ,+''3u ':: l(n.Cf
-V-=" ::' Ic,l.Cf
-l :. 1..+ F :: 71.'1.160 ' lil.'1
<:( -= f:/D =l6G/30 ='3.a3~.
Note: ?A4U1fr.ft C.O"A ~..oeJ. i" '!"Of'uea.{ a,"'4t'J.4aA.
f"()-rn ye(lot"...", ,too.fhoA La.)ft ~t the ')1c''Merc.c..L ,..e.4 u!to ..-
o.tft4.1"4C:'1. Thet.t t.re'Vctu.Q.rLe in ~reolC:ct,""1.4j tr,.e."J.,. Q,bd
.a.i.J. L')' U.hJu..tta..~d."1«3. lh H~'A ~rol,lG'W the ~reph"c"l
«'nell", ALA ti.-ncL. Ii{·'1 etou1li b..iu.1In ..ota.,CA &~i......uf.c.tc.", JS.?"
18-30

Pro hLe'W' 18.34 (-4e~ .AVPF JttWldal to Prob'. s.?>lt- -a "e.35)
10 t~, ..oV"e Ot3er'aiL"~ ,ft!~cJ 1a.t.CI&.i-io'h. i...-.. ehe.,~U ttl ~e
o~ti"Ml.c.~ pL4f~. No Ct4.1~'A t') ~ Ln1:ld (I" out-Pu.T. ~
a.ueft4J,le. AeA'L'U.~ LA tHe cn~ N"'tM-icz.b{e th~ ~4.1 Gte
cJ.f:~c;4. -
f1e-lhit,,,, ofe..a:tC:~ Li.1l~4 i.n te.t-1WA of R ~iV~.A.
(}Pfc...: --':1., = ~/(R. ,,' /'Ian .... o.88~/(~I)
lc,w-et: 1m ~ (:!r. )~h.1 4- (~ ~i )~~
Fein. ba...-... ea...A4 ( n ;:I' 2.1@
~ff'er: ~iO. 76'1 1Vo" .... 0.'1(" It; Lc.wc."':~m ':. ,.•ct "'~-O.S'ZI7
The. Gfe....6.t(')'1 1~')1e.A .(O'" .....e.venJ. U4lu.~ 0" R Flottec:l
c:a.Lc.~ wc:tfl ep'U.'L~~ ..,u.". cu.rp'~. A.b. op_~rc.t'~ U"eA
Q.ftrtlCL." -,u.i.Ul>-iu..')r, ~v~ CLtwl ftu.~po- 0'" ..ot.,•.A C:"ereA.A-C.
At ~.,.1..O,," GO "'14"",~e" '" .A.t""c:A. 4l.t"'e re~ ~c: red.. -
SolldC4"" )tra.t~'1! . '
.i... Ha.ke .e....t e.AtL 'Wa.tc.~; R
..i...i.. ~'t ett{a., ra..1M
W. -; tcp off ....t~e.4 .
.Us. ecm..,a..P" ~o 14 .'Z .AiQ.o~eA (~ !~tetn 'lJOlu.e)
1':1. 1." .,;t:. o.oCl..ptA.bL ~ ~o io ...tep J." r, Q,&4G~"Le
-tc,1"ml'7a.te
-SCllo~f le .1' evo.lu...of lo-k .
t' 'Z' '2.11.1
~" =l2.q/~..~¥)~~ ..{~4i.f,S')/t,.".} 2 6.1IN' Nn ... c:u,$'tJ
...........
~ '" ~tLt.~'f"".1'/l.'''1,",,"· - O.18~3,,'''' :: tc.q, ~ -o.S'3l1
. ....c.....c......
STItt¢CS ':::. ,.... ~ S'Ef."o PL6r'
Sta.~,e.. If.z, balcu.l4ted 14.2. a.uo,L4.h fe
~ e cepi.« Itt eO.
.It. DtLt~ ~ cae V
..... V = (Ja4.. ' 0 ~.1'i ..... (3D} : IOG.1.
~ l'......c ': lCVI
~
% 0 RCf"CAot c ':$ ( '
0 l.Ci -1(,f.,"Z-)10() '::I t,-z.%
, ICll.N
18-31

- tAtl'irt<1te CurY'e-nl Ste<L'n't tGs'l (lew' Pt-eA.flu.,.e )
P",o-m Ta.kLe '.If Cut ~ ~ 3.J1/Cj
tro-vn 14.~le. P~.~cr Dut«-, ::3.0Gi/Y-T"
eUrt"-'It <'6.4i: 3.c, Cj !,If"~ l'3~6 ~ 1(3.17,;175". GO<>Iv,.
. hiY c1CL1i "'1" GJ
j4trin~4 :' (0.&'2) (i~,66G) -=fCJ,060/'t'"
18-32

~______________~~____________________~2
C» 0".
- .:>
f1".~____________~______~~______________IO
~
6
,
;:J? ,... ~ 0 ~ ~ ~ -.¢ ci I,) 0 0 a ~ 0
-1'
18-33

Prob/.e-m 18.35 (Aee .A.uPr-l(eme.ni to Ptols.4 16.Si+a~.'3s.)
10 impro'lG ~CFQ~io71 4eer} m'(l~~cJ 'f, Cf1i:mu,7h [oeGlle1't. 7h'''''
a.lier..,4 vca.LU:e crf "0 t XJIi'
tidero.L B4La:nte:
. 0"'0 -t 8"""e := FNt'
F.,,,,, 0= 3Q,S =-70 , F=lbC f""p::o..321 •
"''9 ::O!-f5l" - G. II Z96 4 c. .
-
Ftew..it£b9 ~UQ.lL~" fo," ).ffer 0rero.t~1'~ l'n<e i.l
t'e~ of e ~L&e., .
~m. '=(RI(~.'))-X~"*~Q liC.3,}
(1=0" R:: 2:31) .ljm.:J 0.701.1 19-", '" h-o/3J.a
1h.. 0f'l~a.Ci~ li....~ ,for .AJeJU-......!"L v-,u c " df ~ ~ fLoi~eJ
a.La~, ",,&. ~ h t he e.~iJ,..~ u.~ ~UtATe. AA Al>1) i.Mrea..4 U
th.e. 0Fert2iU~ ll11e CLff'roo..ok.eA. the. equi1c:~.iorn litlt.
SoLu.ti.o'n Sfra.te.,,~:
..;. )Ita.k'. ~e...J ....tc:'WWlot~ ctf ~n
, ,
~. PLc,t ope,.c:Lc.ra, l ~~
.Jf-o Step o,,f 04. t~~..a.
.AII'. C'o-ntpa.T"< tc. ~Lf.l.
IV. 1:",.i4 Qceept~'" e tet-ml1l.4t.e. 1..f .boT
V"~tUJ't" 1.0 .d.~.~;
~1nFle rArGl~(an .
..). .
J..i.i.
.J..~
Ilf-Q = C.q -:,~~ o:::.G.LfS"86 -aQ'fll4." -= O.Cf12Q
oe- era.t,·n. ~ LLn e : -'1",:: 0 ..70'1 "'"~ + 0.'13.Zi ..
N~. 6f ~ta..se~:
oe e~A.~o,,: {vot
0.70'1/11.. 0 + .O~7.,q
li.2 YS 14.1..
c:r,eee Ffc..bllJ "'0 iA Low cr
18-34

(;#7
o:l..
G.'
Pro g. 35 CG'l1
Opero.t i n~ L(.'hel
R= (.'as
, I '
0.1 0.' O.~ 6M 0.5 0." 6." G.Cl ~.O
ty,
1- C~t to.p4lo.fe to p..fi,J c. crt ~Q
@> AD 1r o.se;S' .'" .N l.$'t~c.al :r n.t
Q) "foo ':z ,0.'1 N(st...., eli::- J&,2..
, ~ AN'!' llf.1. L4Lt~ l'''c!a..p e~t"o.o£4.ti,", ~rO.iiblf
C'lLI'te1It "'~vnk.t ~ .;oJ:;eJ 0 o.iiii' ~·tI"""J&1ICI 71'"t B""sel2~
.. f:na..l 0 1~...." ~,~ e
, ~:'Z. ~ B~1'l" enC IZq.tit" {~~0e., J to·s.O -=f,:z'ttto"
( ~ Sca,~ne. 1"-, l:'j.... J.L ~a.,.~.e /"1,
"'.'11# r e1f~U.e :: (~4"".c. c"ltu.La.'t,.,. u.a.cs. -1'0a 0 .. ge61.J )
I" , i 6'
, =7:l.~rJG~v
5Q."i~A :' cr.13-7..2Z)/'ttl .,'0' ~ fr,064/"'f..
, 18-35

~
~
I::S
0
rn
~
~
~~
ci
Q
N
.0
'n
-d
la
t",cP .~
- ~Q)
V"
£~
0
cl:
q
-
18-36

18.36
Process Gas Process Gas
~
100°C 50°C
Cooling Water Cooling Water
.. ..
°30 C
Design Conditions
QI = Q2 since MP2 = Mpl and process temperature remains the same
Cooling Water Energy Balance
McwPPI (40°C - 30°C)=MCW2CP2 (T - 30°C)
T= 300
C+(MCW1
)(400C- 30°C)=30°C +~=36.667°C
MCW2 1.5
Performance Equation
Q2 =1= U2 A2 b.TLAf2
QI UI Al b.TLAl1
u (*+~r= hphf _ hf/hp _ H . A, =1
~= (~J (hp+hf Xhpf1+hr/hp -1+H' A,
_ (100°C - 36.667°C)- (50°C - 30°C) _ °
b.T1M2 - IJ1000C-36.6670CJ -37.59 C
1 500C- 300C
b. =(100°C - 40°C)- (50°C - 30°C) = °
TLMI IJ100°C _400CJ 36.41 C
1 500C _300C
1= H 37.59°C => 0.9685 =~
1+H 36.41°C l+H
H = 0.9685 = 30.74
1-0.9685
Ihf = 30.74hp I
ifhp ~ 30 W/m2
K,1 ht'~ 920 W/m
2
K I
18-37

18.37
Process Process
70°C 35°C
Cooling Water Cooling Water
<II
°40 C °30 C
Design Conditions
mn = 0.7mpl
Process Stream
Cooling Water
Q2 =0.7 =mCW2Cpw2 (t2 - 30°C)
Q1 mCWPpwl (40°C - 30°C)
0.7= M (2 -30°C
cw 100C
18-38

(a)
Reduce Mew to 43.5% ofthe design flowrate
(b) I(2 = 46.08°C I
18-39

18.38
Process Process
50°C; 2 kg/s hi = 200 W/m2
K 100°C
Steam
ho = 5000 W/m2
K
Steam
III
°160 C °160 C
Design Conditions
Q2 = mSleom2 .ILz ~ Q= M (Equation 1)
Q
~ Sleam
I mSleaml /'1
Process Stream
(Equation 2)
1 1 1 / 2
= / 2 + / 2 ~ UI = 192.31 W m K
UI 5000 W mK 200 W mK
_1 = 1 + 1 ~ U =214.08 W/m2
K
U2 5000 W/m2
K 200(1.15)°·8 2
(h = constant h- = M, O.S)a , I P
U2 = 214.08 W/m
2
K = 1.113
UI 192.31 WIm2
K
18-40

Q- (11131 50°C J- . (82.49°C)ln(T - 50°C/T-100°C)
0.6746
Q = ---r-----,
1 T-50°C
(Equation 3)
T-lOO°C
Combine Equation 2 and 3
1 15 = 0.6746
. In T-50°C
T-100°C
18-41

Chapter 19
19.1 From Figure B .10.1, it can be seen that the molten salt is heated by a fired heater in the
circulating loop. This implies that heat is added to the reactor, i.e., the reaction is
endothermic. At start-up, ifthe catalyst activity is lower than designed then the amount
ofreaction taking place will be less and consequently the amount ofheat needed to be
added to the reactor should increase in order to increase the temperature and the
conversion. Therefore, the flow ofcirculating molten salt should be increased in order
to compensate for the higher heat load.
19.2 Pressure at the bottom of a column decreases.
The pressure drop across the column is due to the pressure drop caused by the gas
flowing upwards through the layer ofliquid on the trays. This pressure drop is due mainly
to the weir height which will not change. Thus, the pressure at the top ofthe column will
also decrease.
As pressure decreases both the top and bottom temperatures also decrease in keeping
with Antoine's equation.
As the pressure drop decreases, the vapor density will decrease and the superficial
velocity will increase - this will tend to increase the tendency to flood. As the
temperature at the bottom ofthe column decreases the driving force in the reboiler will
increase causing more vapor to move up the column and again increasing the tendency to
flood (although more steam would have to flow into the reboiler).
19-1

r"~'-:
'C-
....- -
q:3 gI~ t~,1 /) if t!--~~17 ~ 7, 7 w-J~ j>1"I'>?"tft!
-hiJ (j j ;; 1ft"??, S1' if) ~/lte-L ~ A~/~
~~ ¥
r:;:c. ~ ~?. 6 ? 7 '7. €7. 4 ,. /o?.
p~~ 'i~S'2, &Yo '16/J?,6/) '1)St(,6fo fI5"J,7)r ill(!6.1Yo WO,7&"
P"'II~ "It/lf, rib 2A1,~~1 5"12.8/,1 (r:;I.!~ VVf,ii~ 'IJ&1,7)1.
ahfW vd- ~ /q/L. Ft~) fo I'~
)Vok: 4L, ~D~ ~~ ~ k ~~ wlif" p~r/.A?f'c..
A-J lVAbf4- .f- -- t/...~ 'f1.. j)JAf"- ,t..l./ )W.JN~/J, ,--t
f1.t ~c-t-H) j)J~ tk ~,,?,t:-t. .f'Pue- llif4c to 7k
~ ,c.~~, -;:"..tJw.. J~J~)~ ~ ~
_~~,
~~~-~~~
'Ef-?~ c
r;.At ~4
, L- -1"S/2.-9o
cJ/.., 'J.~ 1~ ~ '~~
~ ..~.. /, I -j?f~-f;::'f<f!N~~
-~f.. T:-'35?/lvc- (o//~/~.f,Q-
19-2

5
6
7
8
9
10
110
108
-.c
- 106(5
E
.::t:.
- 104Q)
-~
c: 102
0
''8
:::J 100
ea. 98Q)
c:Q)
E 96
:::J
u
94
92
99.42 102.39 104.72 106.49
98.28 101.24 103.56 105.32
97.15 100.09 102.40 104.16
96.02 98.95 101.24 103.00
94.90 97.80 100.09 101.84
93.77 96.66 98.94 100.68
----_._---------
5 6 7 8
107.79 108.73
106.62 107.56
105.46 106.39
104.29 105.23
103.13 104.06
101.97 102.90
P =3000 kPa
dotted line at 357.4°C
------
9 10
400°C
390°C
380°C
370°C
mass fraction propylene
19-3 .
11

5 99.42 100.62 101.74 102.79 103.75 104.63 105.44
6 98.28 99.47 100.58 101.61 102.56 103.44 1.04.24
7 97.15 98.33 99.42 100.44 101.38 102.25 103.04
8 96.02 97.18 98.27 99.27 100.21 101.06 101.84
9 94.90 96.04 97.11 98.11 99.03 99.88 100.65
10 93.77 94.91 95.97 96.95 97.86 98.70 99.47
J
1 Or!» 5D~
P t·i,. t p~

1--1'4--
..It"'.,
108
106
-..c::::::
0 104
E~
-Q)
102-~
c:
0
100
~
::J
'C 3300 kPa
e 98a.
<D
3250 kPa
3200 kPa
c:
<D 96
E
::J
3150 kPa
3100 kPa
0
94 3050 kPa
3000 kPa
92
5 6 7 8 9 ,10 11
19-4

Ilj ihlft¥06#ttA iJ .f()/~",I VII) fk CkWt~ ?~j
(;;U<//~~. f'flIc ~~JVt*l'h J,?V!~ 7k
d~... ~ Nklv~ ~ ~ d 7,r ~.
1fdY' hXJ1 ~ bo./-tp ~1;.,.;I / f9. 51 ,th#(//~
wrfi1k 4~~1t.c; ~ fold )h'M6-1~ t~
A L(iL Irl;f'HUlt;'" ~ ~LJ ~ tkf?d.tJ-~___
tAIuM",. 1L;;...,.~f"Jj;k,·l~. 11.< IJ*e, 1I-.A
WL-7l/j~ w'- 7):: 17.1' ~_ol/~ )...t¥~ ).0-9.
t"c.J.. -Jcw..1rt. kfill4-~,
'f-trll)'t;· D~ I'.JC. 1l__1/t...
~=- S"$.l(4i kIM.IIL.
r:;: 7 I). 6Jk~~)/f....
(t~~I?,,"f
-!It.. ,V',j,-l ~.IJ )
Lt>=- /6t.61 t~I/1... 1L/l. ~ '1,£1-; (I~~/,.)
~-II =/ff, f'J IM.//'"
LAI -:: 'l- &{ i.4fI" r,.,.,JIt.
Zip;: d."lttt;
"XSIl f). ()f) b1&
19-5
le..2t1l># fi."oJt t: /.. J.!
6fHlfWl IW.J~ k; J.. 6fJl/o"....

·.
-
(t.t1.10(:.
boHo--~ ~.......
r"Jf ...N< ~~I'( )
/04. (..
I- - -- .• - - - ""1
T 1 _ ~".,
1v~- - I
• I.
/"2.._
- -~
I
1+- G
MO. •
_ bl2..
-- -4-1, '?
hl>.o;:.. ~.:
h .. , k ....l L.,
M:::
~) -r::: I'10. ~ 0 c. , ....IJ I~~ ~........ T
~!r~ r-k.,.-T '-t .. Jht...f ~
p~ :- /, t.{ ./:,/).. -) P4d '10.~·c.
.bP=t~:: 22.I..(JtP~ T
Pry::- /S-/, ~ "" ~
L:::--
T.J.r"; q1 t) c
, 19-6 -

19·9
-
------.-.---'-----------~ -.... .......
350
300
system curve
base case
-... 250
S
;.... 200
0
...CD 2 pumps parallel
~ 150
-"0 pump curve
ns
CD 100.c
50 j
J I
- systeqn curve
0~~=:~L-____~L2~ex~c~h~an~g~~jrrS~i~n~p:ar~a:lle:IJ
o 40 80 got 'tV "120 11'60
flowrate (gaUmin)
rYl6-;... Fl"..., n. ill (1"'/w.~'... )
(,..> ~1~4-~p)
H'It ~~
b~ ~~
,,~ (I~. ') 1.) /,1tJ (It:)() ~)
19-7 .

ft~~JV~
tJ)t (·n
U~ t v,> fl'J. ~~
E I~.~...
t: . ceJ
f 11· a.J.. I<. f 'lJ.-f
1£.f.t~
?"......~,tL-j, :z Z~~
I
t-/){ b~ U'lt ~~ 't!'I~ U~"-1"_ IW
~ (tiC) Jo/'6, ( 5?--7, g' >'-1') ,7- 112,0
r40t (ftc ) ]qrz. t.t ] 6S, ( ]g? V 5' .1. D
tlaw 1.. r(~ (~-"f IS'. 3 I~
}". ~ ~~1
Ii. t: t .. 1.,-,-,(>
1. 07 /. 33 I,/~ II J..-t)
..(:-,,/.;
~
19-8

/9,9 !;cr·
- fl.'£r J{~
~,,~ H>'-
Cvo~
~(I) ()i.-
- -cr, -
~(~)
~4- _
~ -
tA_
-- -C<,
<;;1~44 11114 f
(-#(/SH""""'_
J'" t... N-e. U "jAl"
I
-
Qz.. _
C(, ,.
tP~ i"';~ (Tv-7J) (I )
{{:: LlA {j'f, ..... (?.)
/(= L{A b7i.. c})
II=- jM'?, (tfS"-5 0) (f )
cv
,
M (r4- TJ)
JfA(,)1-
'PQ",1f..-- ~,r...:Ltf..1;:- ~
Mrn
5U
;; ('4-Tl )
( ..l)~ ~O,Tl
1~I).,,*,if ~ -t MO'! f,o. ':1,",
:2 C(r;-~o)- (Tv-'1S)]
W1.'L [1-.• + ;!;,:.•Jt..
;: =JI ~
M(.w <;/fltU IV'A11II-ht ... hT?<,... =- 1<'0<::-
/. f, 1'''1'~ r1'~",.1>1 Ie. -t). Ii~ flA. i T: i./r-(.
R=Mw~ t 01.-
t1;: :xif ~,o Dc:
1Y ~ 747,1. 4),
I PM; 1.5'&'<i'
fX;; Mtw ~ 1,20
TJ ~ ~ '(0 ,OJ"
T"(:- 71)1 Y 4Je.
Ct'~~· W- - ~ ~ ~ A,..j l;i7~IJVT V-~~
19-9

Iq.. 0
~J~ s+-:i-Lfi !,·~...d...- h. J1. b --J J7. j
P...J N
·l! "2- p....r>'t/0 f?~~1 l.s~
~Ib~ W')I. ~/~ ,",)(b~ ~'I- p~ II.J..
-r~ (0c.) >'-13. S 11-1.1- 33 r.~. ""330.1e>
TI.1 (0, ) 1 ~f.4 )~5',r 31.). ~ . 1b3',[
~1. I )"/} roo 5',<, g-~/9
~-"'1'
& $C~ '-fJ L'.lS 't'2.S- I, C~ I,'?.
~
~
'<'1. II
---
19-10

/:::-01 J
rr= ;V; Cp( 7;,,, - ", )
-.::>")
-
S-f...-~
("('tit "" "l-(..,}-(r,,~1!f )
0. -;; "l Ar. k- T~.- H'.::' :!)
Tc, --~)""
t,{~ til::. At.d. f f,.'-A. fit/ f Is-. t ~
h ~1".1j,; rn,f"bfJl/:A-
19-11

'r~r/' )15
~1"D~ ~(_<t. fZ. y.<)J~r;~
?~~
f1. T(,,{O() ,,~I o( ) M
--
D,fo
1~. tot 'Z bl c> 0.'1 r-
] >,'?) ~b?, <( 0,1
O.~~
I, Do
-; $'"1). t:> t ~. r..:> }. 00
f t ?'l-
19-12

bP fN
"1' , ::- ~y.o- 0 -"11) (IS,~) -1-0, OS;-l-:;-(I~'):'().OIO(15.r)
?,/P ~ 30I .l.r.::.
It- PI::: )oo.{y"
p2- ;: if()I .it. p,-
>J ~ p~:- 2 -zr )&f -.
'7)vttA P-s, ~r,:; 27.)" )ePe..
o '2.~6 lei?..r, ::
p~;:: >'i ~ Jr.. Pc
9",:- I o~ )R'Pc.
1'1 b.y> jAr ___ p"':; t..'J C- 1.~
'.. 1
t~C;;--: '"S"2.. - ,ti'1..'-'f ~ 4' O. ~ '>0'- ~ -- 0.0 IJ ~ ~
" 19-13

I'=f.I iJ.
---~) H,::: /QO '?
?(t..,~.. ~ £..-I~/) ~ 1/.1' ; / ..
pt. J'F~~ '7 .l~ 4~r) ::. ??'7, 16..,lf/..
Pi f)<--t 0 I) =12.'7,?" Jd?.. _~ p~ '~I ~ p~
hj {' 2
f (t....f>-..';' £"-1.,) =-A1.-H -I 10 ..;-~{)
[)P"'~
, ... -;
(j>{~ut 1 f·I",) =- q-;, l' -+(7,,-1'10 ) (~~)
~.. 6t~ ~ fuw.. ~ 4$<.A ~ f Pll.vf ,0'
()
f( ~ ?P"/~) ):; /OI.AP,
r~.,.. 3+"'':''S""J''~<+'''''''':' ~ ~,U~P....
~I -::- I 0 () t;: A I"'"A.,.I
19-14

bf/r,e"d---:: I, U rs; r ~~""
e-Pl"," "32:1'- [1- (firf] r"'
~ vJ-.-< ~~ r-; tP", =.. ,d.... ~th"D
t
tJ~:; ID',for--•.,,-
19-15

Note:
This addition to Problem 19.15 will appear in subsequent printings of this
edition.
C. Assume that all the resistance to heat transfer is on the process fluid
side and that the temperature of the condensing steam remains
constant. Determine the temperature ofthe fluid entering Tank 2.
kol.J
i:: I.rl-Ie.t p~tt"r;f f;!-i""'p
now ""j:tltO""",..,
19-16

>,ulw (JJ.<3) >/M'J/j;,...e'Q:Jft.. ? Ii-&(
ty..- +hIr •N... {AI(- t. J uk~ A t.A#I t"1 h;..:.. I''r ;.....; ~{Jilt IJJ
i! /'" ·b-v·{.. #fvlit.f,·~
( )(),; ~
0,( ~ -:: ~ ~-"""JI-o-
I~ 82.5 D ~
~ IbP~"
19-17

'ct.,6
&,) r..uJ+~
~¥::-u ~u ~0
r-.)...;6. ro'V",'.L··.)
Q) PIr:: API- '" .j h rz. -1 + h p)-IA .f belf-~ +p, Pr--' = Ii PI-l.
= 10 -I- tto ~ 10 ..j. Y. 0 -t 1 ~ -:;; Iso v~
I,~t ,'" qv..vrtu.,.~ I..t l?cJ,P. ~lr~/,II-)S
OtN.~I" "..,. aO#'-r~-r,,-( b-et'~ -+~'--..... ~ >-0
~~ J MAK-fMV ..... ~"...J:;-
19-18

~
~ {L{!! hP(J.;?.J
-----/.0 7.'7..-1 c0 V-=O.~-L/S
(J,t /t;{vt fY'1l1L /tftJ >]io1..t:A JMI V~
c/, (, ~/r
IJ,IA JL
IJ ,v q
0 0
part b
300
280
co 260a...::.::
- 240
..-
0 220..-I
a. 200
en
en 1800l.-
t)
160co
())
140en
co
()) 120l.-
t)
c 100
())
80I.-
:J
en 60en())
I.-
40a.
20
a
'"~/
7
/
V
/
I
--~
/ '"'- /pc: rta
/ ,/ '
"-
V ~
/ "
/ -'
./ f
---V
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Flow of Liquid through P-1 01, Us
Solution to Problem 1Y.1 b
19-19

4PoI.I"i'- .,.)::- :2. 0 - 2-1;; -I ~ 'I
f 5 ~ h;: $'2- ~ 7: 2.-L( p: ~" - 17. rJ i
2ft '
~~~ ::-S-°5r'"
b~~Mr ~ ;'1 f>;
bI?c.v ~ IOr'-
>0 b Ffv"'-P::- l> P. r ~ t. k "i' t:. ~L t 1:/1{A.I
I 1.;1 ,/{<J T
s y = I J +-' ~ Pfr -t 'D
~IJ,:; I? r$ ;
.
M£.lI. V:: DU)f:
-
l) ~t S..5".... ~,H,.: II ft;; So ~ p,1
fJ'fJHft;;'- PrAUIL -+ f~ 6~ - ~?/r - /'1
- .;1 V· l'k - J - r":: ?-L/- r*"
(fl
, Y
,-IJllt1 (IIV'I~j ?.- "-/- P =- IIPJHn: $"1 Sr;/
f
"t '
~ li,~ f'
19-20

60
en 50
0.. part c
0..
E
::J 40a...
en
en
0
'-
()
30«Q)
en
~
Q) 20'-
::J
en
enQ)
'-
/
L
-~
part a /
/N
/
V ~~
~
"""---------a... 10
o
o 10 20 30 40 50 60 70 80
Flowrate, gpm
18
/15
"'0 12
'5
0"
t-
'/part b __
/~
o
.:= 9
:::c-
C/)
Q
<: 6
3
o
o 10 20 30 40 50 60 70 80
Flowrate, gpm
Solution to Problem 1q.1.7
19-21

c) It CtJ cl<.Js.~ - fl1) ~
t-f~7 ~ A ~v" =- 4'1-fJ :
Ptd t=-fv/ ,-lIe l =: lnJNk -Ib rr-~ + !:> f (11.17
== 2 I -+- Lf -J 4- V -;;:; 6. 1"fS ;
1 .
.Ek
2~
F'
19-22

19.18 (a) Acrylic Acid reboiled at 89°C = 362 K
InP' . =17.31 5264.4 =17.31- 5264.4 =15.92 kPa
acrylic T(K) 362K
(b) PlOP = Pbottom - M
PlOP =15.92 kPa-9.5 kPa = 6.42 kPa
• 4981.5
InPacelic = 17.54 T(K)
In(6.42 kPa)= 17.54 _ 4981.5
T
IT = 317.68 K =44.68°C I
(c) T-Q Diagrams
Condenser
49.68°C+---------I
Q
Q= UA!J.T;m
Q =mcwCp,cw!J.Tcw
Q=mpAp
If Qand melY are constant
!J.TClv = constant => Tin = 35°C
Tuu{= 45°C
UA is constant => !J.Tlm is
constant
IT,.."II<' = 49.68°C => PIOI' = 8.18 kPa I
19-23
Reboiler
120°C1 - - - - - - - - - 1
89°C 1 - - - - - - - - - 1
Q
Pbottolll =8.18 kPa+9.5 kPa=17.68 kPa
IT;WI/OIII = 91.6°C I
Since Q=UA!J.T and Qand UA are
constant

19.19 (a)
T-Q Diagram
Q=mpCpllT
Q=UAllT;m
ForM= 1.15
Q
~( Mo.
8
y ~ I
M 300C = 32.740CA)n((1400C-900C)/(1400C-T)))
L15 (1.15
0
.
8
y 1 J
30°C = 32.74°CAln((140°C-90°C)/(140°C-T))
In 50°C =0.891
140°C-T
IT= 119.5 °C I
19-24

(b) M =a + bv2
to get system curve
a =250+ pgh =250 kPa+ QOOOkg/m3X9.81m/sX-3m)
= 250 kPa - 29.43 kPa
=220.57 kPa
at v=35m3/h and M=220.57 kPa+20 kPa=240.57 kPa
240.57 = 220.57 + b(35y
b =0.01633
M = 220.57 + 0.01633';
35m3/h(1.15)= 40.25m3
/h
at 40 m3
Jh M =247.03 kPa => No problems
-----------.0. __ ......__
-'----..._---.-
..~.-..•..•.-...-..
..._,...-.__..,
...-..:.....,...~ .
-1
_ - -t--
--:-. _ - - - .- - - - - .- - - - Old New
3C
volumetric flow-rate (m';h)
Figure P19,19 Pllmp Ctlr'.'c· for Problem '10
(c) Desuperheater - Q= UA!1T
If Q increases !1T increases => must raise steam pressure or lower column
pressure
(d) Liquid phase reaction => no effect on pressure => bad idea
19-25

19.20
With valve closed U = UBC, -MJ,BC = PI - P2, VOL Flow = QBC
With valve open U = U2, -I1PJ,2, Q2
(a) For fully developed turbulent flow, -I1Pj ocg
IFlow increases by 30% I
(b) 1.69 -1.3 = 0.39 or 39% ofnew flow through bypass
(c) U=(~+~J-l ~(~J-l =ho
hi ho ho
U2 =~=(!2)O.6 =1.17
UBC hBC 1
1117% OfUBC
19-26

19.21 (a) M=pgh+M[
M =(775kg/m3~.81m/s2X42 m-6 m)+130,000 Pa
M =403420 Pa = 403 kPa
Look at pump curve at 7 Lis => above pump curve - single pump will not work;
could try pump in series (parallel will not work).
Ift1P is high, maybe pipe diameter is too small.
(b) Q= UAt1T
.l= 1 + 1 => U=1875W/m2K
U 5000W/m2
K 3000W/m2
K
6000 x103W =Q875W/m2K~00m2pT
t1T=10.67°C
~Ieam =10.67°C+ 87°C = 97.7°C
ITSfeam = 97.7°C I
(c) Flooding - Will column perform this separation without flooding (may need large
reflux ratio ifN is much lower than needed?
N - Are there an efficient number oftrays to achieve desired separation without
very high reflux ratio to cause flooding?
Feed Location - Does feed location cause trays to be "wasted?" This means that
there would be trays that are not doing much separation.
Condenser and Reboiler - Are these sufficient to handle heat transfer?
If oversized - will they perform as needed?
19-27

19.22 (a)
TWall
(OC)
110
120
130
140
Tsteam = 160°C
Tprocess = 70°C
/::,.T =160°C - 70°C =90°C
In order to find Twall need hprocess => Trial and En-or
Q=UA/::,.T
~ = U/::,.T = hp{Twall -70°C)
Twall = 70°C+~90°C
hp
/::,.T hprocess ksteam
(OC) (W/m20
C) (W/m
2
oC)
40 2500 3000
50 2300 3000
60 2160 3000
70 2100 3000
Twall is approximately 120°C and U= 1301 W/m2
°C
Qnell' = Unew/::"T;'I!II' =1301W/m2
oC 90°C =1.95
Qbase Ubase/::"1;;ase 1500W/m2
oC 40°C
IQnell' = 95% Qbase I
U TWall
(W/m2
oC) (OC)
1364 119.09
1301 120.91
1256 122.33
1235 122.93
(b) The reboil rate will increase; therefore, pressure in column will increase.
19-28

19.23 (a)
(b)
~2 = P2 -~ = 101 kPa-200 kPa=-99 kPa
M pllmp =(Mf + Mev)- ~2
M pulIIP
from pump curve
Mev =225 kPa-(250 kPa-99 kPa)
IMcv=74 kPa
4P(kPa}
100(1.5)k=225
50(1.5t=112.5
25(1.5t=56.25
50(1.5)l=112.5
Pin at E-2904 = 607.3 kPa
Pin at E-2905 = 438.5 kPa
P(kPa) Location/Equipment
101 TK-2901
Pipe
326
. E-2905
438.5 Inlet to E-2905
Pipe
494.8
E-2904
607.3 Inlet to E-2904
Note to instructor: There is no part c for this problem. What is labeled part c in the book
is the question for part b.
19-29

19.24 (a) AD·2LlIfocm
:.(~1.2)2 = (Mt,2)= (52 kPa)
mil Mil 40kPa, ,
m =m ~52kPa =2(kg)~52kPa1,2 1,1 40 kPa s 40 kPa
Im1.2 = 2.280 kg/s I
AD·2LlIfocm
:.(~2'2)2 = (M2,2)=(52kPa)
m21 M21 40kPa, ,
m =m ~52kPa =1.5(kg)~52kPa2,2 2,1 25 kPa s 25 kPa
Im2.2 =2.163 kg/s I
(b) Exchanger 1
IU2 = 217.3 W/m
2
K I
(c) Exchanger 2
19-30

(d) Exchanger 1
Q2 = U2 211Tfm,2 = ril2 p, T2 = 2.28 = 1.14
QI U AII1Tfm,1 ril [' ,11111 2
(
. ) (T-SO)-(T-100)/1n( T-SO)
1.14 = 217.3 7 T - 100
200 72.13
Ij T-SO )_(217.31 so 1 1 )
IT-100 - 200 72.13 1.14
IT= IS3.4°C I
(e) Exchanger 2
Q2 U2 211Tfm.2 ril2 p, I1T2 2.163
-= . = =--=1.442
Ql U AII1Tfm,1 .Ie ,11111 1.5
( )
(T-SO)-(T-100)/ln( T-SO)
1.442 = 191.3 7 T -100
ISO 72.13
Ij T-SO )_(191.31 so 1 1 )
'T-100 - 150 72.13 1.442
IT= 1S9.1°C I
(f) M2 =M,.
40(3~1J=2S(~:~J
m2 =(¥)O.S)~~~
1m2 = 2.941 kg/s I
19-31

Chapter 20
20.1 The main reason is the requirement for large reactors in order to obtain very high
conversions. Ifthe reaction rates are very high for the conditions used (for example the
combustion ofa fuel in air or oxygen) then conversions >99% can be achieved. But for
many chemical processes the reaction rates are moderate and as conversion increases so
does the volume ofreactor required. For example, for a 1st order irreversible plug flow
reactor the ratio ofreactor volumes for conversions of 50%, 90%, 95%, 99%, and 99.9%
are as follows:
1:3.3:4.3:6.6:10
In addition, for high conversions the flow pattern and mixing in the reactor become
increasingly important. Even small deviations from ideal plug flow cause large changes
in reactor volume at very high conversions. The final decision on what single-pass
conversion to use is based on the economics, but we get decreasing returns as the
converSIOn mcreases.
20.2 This statement is not necessarily true. Ifthe reaction is at equilibrium, then according to
Le Chatelier's principle - an increase in reaction temperature will lead to a decrease in
conversion. However, ifthe reaction is not at equilibrium then an increase in temperature
increases the rates ofreaction and this will move the reaction to the right, increasing
converSIOn.
20.3 Since the reaction rate doubles for a temperature change from 250°C to 270°C, we may
write that
-r, e-E1R(273+270) -EIR(~-~)
270 - 2 - => e 543 523
-- - - e-E1R(273+250)
-r250
:.EIR= I
ln
(2)1 =9,842
(543 - 523)
(a) For a change from 250°C to 260°C, we get
-r. x e-E1R(533)
~ = - = ---,.,..,.,.--,---
10 e-EIR(523)
-r250
-9842[~-~J
:. x = 10e' 533 523 = 14.23 => a 42.3% increase
(b) Assuming that the feed composition is constant then the gas phase concentrations
increase by 15%
-r1.l5P = ~ = (1.1 ~)3 => x = 1.52 a 52% increase
-rp 10 (1)"
20-1

20.4 (a) The reaction is exothermic, since heat is removed after the first packed bed section.
(b) The reaction is run at high temperature because
(i) a high temperature is needed to obtain high enough reaction rates to give a
"reasonably" sized reactor.
(ii) the reaction must take place in the vapor phase and the only way to obtain a vapor
is to increase the temperature (as opposed to creating a vacuum that would reduce
the concentration and hence the reaction rate).
20-2

20.5
20.6
A+B=C
As Tt,Xt
Aspt,Xt
x
} 2
aA + bB-+pP-+uU
3
jJB-+vV
Low B minimizes V
Low x (conversion) minimizes U
T
E} > E2 > E3 =>high T maximizes reaction 1
20-3
Increasing P

20.7
ACO~DO+AA
2DO k2 >Gum
E1 < E2 => low T, CDO low, low conversion
20.8 C2H4 + 0.502 ---) C2H40
C2H4 + 302 ---) 2C02 +2H20
C2H40+2.502 -+2C02 +2H20
E +0 ---) EO -+ CO2 + H20
"'- .J
Low conversion
Ethylene excess => low O2 - flammable limit issue
Low temperature
20-4
AA
E
·~----....O

20.9 (a) Countercurrent flow ofHTM
T
Q
(b) Cocurrent flow ofHTM
T
Q
Large ll.T desired at entrance where reaction rate highest to keep reaction from
quenching.
20-5

20.10 (a) P+B~C
!? B C
I-x I-x x s= 0.5
I-x I-x x
I-0.5x 1-0.5x 1-0.5x
-r= kC2 (I-xi
Po (1-0.5xY
C -C l-x-
( JP - Po 1-0.5x
c -C I-x
( JB - Bo 1-0.5x
w x
w x
Fpo = 2 ( 1- x )2-kCp
o 1-0.5x
1 x2 (1- xlY (1- 0.5x2Y
1.25 = Xl (1- x2Y (1- 0.5xlY
1 x2 (1- 0.68Y (1- 0.5x2Y
1.25 :::: 0.68 (1- x2Y Q.- (0.5XO.68)Y
20-6

(b)
(c)
(d)
125% increase 1
fof FPol = Ii-rpl
I FPo2 7Xl -rP2
IT2 = 579 K = 306°C I
!YL FPoi =(X2)( - rpi ) J2!li (f;.O,/i! {l-xl)Y(I-0.5x2f
JV; FPo2 xl - rA2P 7~I k12 (?Po21 )2 (1- xt12 (1- O.5XI)2
1 _ YA I (3 MPa)
( )
2
12s- ~ P,
IP2 = 3'.35 Mpa I

1
t=-ln(l-x)
k
(a)
(b)
t2 _ kJ
In(l- x2) _ In(l- 0.75)
t;- k2 In(l-xl ) -In(I-0.50)
(c) t=C
A
1 dx =_1___1_ =_1___1__1 =_1_~x [ ]
o 0 kC~o(1-xy kCAo (l-x)o kCAo (I-x) kCAo I-x
1 x
t=----
kCAo I-x
!i=!2[I-~IJ=(0.75X 1-0.5)= 3
tl Xl l-x2 0.50 1-0.75
(d) Since this is not a linear function, it takes much longer to approach 100%
conversion since approach is asymptotic.
20-8

B€.11 ~ e'l'le Prod1.L C. ti.o1" (Pl"ob.:2o.1 - ..20.Ib)
Rea.~tor S"1nU.iCLti.O"n Set..u.p Va.L1.Le..6 ~
a..) Che-mica.l Rea..c.1.L.Gll: C, WSC.H~ i HZ --"'C,H6 -t CHq
1.D. ~ c>. (1) ('2) ()). (4 )
Jr) Ra.te E-x.p...-eMi.011. ~
-~:.se{C1CZ.
. - T. D. S-toic.h..
CO~~'he,.,.t~ 'NU111he.r Coeff{~ie,.,t
d.) kinet i'<:A: n:: n -lE./R1]
..R ~o e
jo lJla'h1°t;tm3 ] ,:: 3.0'!.IO!D ~
E l~ ca.l / ~'moL 1;:: 5,6Ci)C fO 4 ¥:
kL~etic
E1.P01)ebt·
- ~
~) Trd.1t.6port a:nd The1"''ma.L Propert(e.4:
. ~ l 'C
E~t ha.lp,,: SRt< ':vo. 'Ue.A: SR<
5) Rea.ctLc'n pha..o.e: ('10..4
20-9

FLol.U {~1'ro(/h )
H'1dr~e1t 7bO.b
1'1ethQ.1.e. 3311.'3
Bell,)ene. i, 0
Tob.te1tt! lLfq,O
*
~ea,.eto'" Specs, :
Rea.cto~ T~pt.:
~ Co.tc:u la.ted
Va.tu. e~
Thermo1. NoclG: Ad ~~j,oi.~c.
Si YI14,Lati.o")'! Speelf~ C(n~I"erAio11-t>- Ca.lc. VoLu1')tf!
Sa.~e 5imulG1.tio<n(ba..b.e ca.Ae):
Set contrerS;011. ":: C.l{
SirnvLo.t lO') ReAult..6. '0
T-=~[3()C
FL"WA (It-mol/n1
H~d rOCJe", bS"Z.b
M.dh Q.:n e. Lf42."3
ae)1~e 'h. ~ Jb.O·
TOL1Lc-ne 3b,(j
V ':: 'i:u,. 1M3
20-10

Pt" ob. ;2.0. ?.
l.O
o
/" L{.O
Prob.20. ~
CC"'n.•dll.1t 3eed
@ Ba...cl!. C<&'.o.e
20-11

Proh. 20·3 Cota.rt a~ pa.qe rrececl rn<:}
ReQ,(!tor
T= 750 ·c
p: Z5.3:
? I. V", 4.76 . r T~ )0
F'low [j)7)Cltl h1
- ....;- Hljdroge? va:f".
. Met~a:n.e 33Q,3
Be-n ~e'he ~.O
10Lt( e71e {44~
P..ob ,20·12)
Re.o.c to r Spec..o.
-It'
Ca.le.ulQ.teJ.
Rea.c.h:~1" 1'jpe: PL1.LLJ FLc.'U.r
The.~1'n a.l Mode: AJia.ba.tic
Va..lu.e~
S{1'YlULa.tL01": S1iec.Lf~ }Gl1.L-me. -t:>- Ca. Lc:u la.ie. LGn
Fo'" tr~t"ious va.L'lle.4 o-f H~Jroqel .flow.
CaLtu lctte. 'tCl ")(5 Hi. in-e'l't oh!a.i1'l outpU'T STreQ.~'
Sa-m pLe Co.L Co u ld..l: lO'h.
l=ot". 1 'idro~e'h =:) 25"0 ~ moUn .
t-x.ceJ>4 H'to '" l(Z.~c -l'f'l)/Wn "100 =§
Set J ':: q.7~fTT'-~ I5i:rnu lo.t LCrl ReA.uL-tAo:
To:: ~~ I ~( ~
Flcu.- (Jl.mGlin' 'J€P ~,,6
H'1Jro<)e:'l1. I Zl. r
kethCL.1Le 4&2.<6
Be-n.,}e1te r~(..s
Tob.J.~'l'e. t~.S'
20-12

o.....
~
~ o.ss
F
o
v
~
p:
QJ
~
--'
~ 075
Pro!:'. ~o·3
IVol1.1."n.e:: '1.1(" rn!
tOO "2.00 300
% E-x.ce.A..4 Hz.
"
20-13

Prot. 2~'4- (..4ta.tl O"n pa...ge .before Proh.20.12)
R,ea.ct6,.. 1 Rea.c.f..ot C
FlDW (Jk1'lWLAt1
H~d.rogell 766.6
Meth.a.11e 334.3
Be')1.,}e71 e 8.0
Tol'U.en.e 441.4
I hp"'"t
Rea.eior-_,Spee..4. Rea.~tDr 1 Rea.ctor 2.
Rea.e.to" -r'-jf~
Th e.r'"ma.l MocJt:
Stirred To.;lk PlU9 FLoW'
Adio.ba.tie Ad to.ba. t ,:c
Two Step Si'm'U.la..t~O')1
St~r 1.) Rea.et c'r 1: Set 'lol-U:1'ne -i>- Ca.lc:ul.aie..,
. CQ~11e~bl~
Step 2.} Re.o..c.tc.t" 2: Set COlrtrer.4l01-1>-CoLe. 'Jotuble
Sa.Tnple Co.leu.la.tLo,1'.: Re.a.eto,," 'VoLu:m~:: 2.7S'rm'3
Si"rtluta:l~()ll RUUltA.: Step1
T~ 7~O °c P-:Z.S.ba.r
FLoWA (.irm.oLlh)
H~ J t-0<l e.1
Metha.'l'e
Be11.')e~ e
Tt).:U. e"he
20-14
1'£.&

Pr-oh 10.ILj·('o-n{
Tclu.e-ne CG'n'U'etAl()7 L"t1 Re:t.etot' 1== (1t;'t-7lf.~)jf'fq =0;48
Tol1J..e''f! Co"'t1"erh..i.o~ 'nceJeJ i.n. {(eQ.ctor Z *0 protrc.'Je
otret-dLl ~C:"l1'r-e.rAi.o"h :: (7ltf:,-3t.)/1'1.~ -:. 0.5Z
_ 5.S
rn
~ S.b
"-'
~ 5.4
..:; 5.2
a
> 5,0
i..
o
~ li.6
'0
Q.J
0:
.......
d
-+oJ
~ 4.2
Pl1"nuLafLcr)'! Re'AultA: St~ 2(CO)I}. =-0.$'1)
Teme:' gr~0 c p:: 2S ho.~
Flc;U1..ll. :,,(~~ Aa)
H" cl ro~ e~
J;f ~ t ha..'"n f!
13e..,..'1 0'1 e.
To;u. e,"'tl e
~
V =- 1.72 rN1.
Co"h...bTo.'h. t J eed
~ Ba.4e. Ca..4 e
I
I (~::!L<2.!~~ ~ __
r.D
20-15
~
0.8.3
0.7 ~QJ
O.S ~
o
0.5"0
er:.
0.4 l-
V')
03 U
0.'2
0.1
6.0

/
.~ .....:
P1"'01. 20. 5 C4tQ.tt 0." ~<Je freceJiilj Pt"o..b. 20'-'2)
f?ea..ctor
"<:. } [ V::'i./e tm~ ! ~.
IT ~750·(
P-= 25 bCl.t"
MoL .frad.i.01'lA
H~Jroqell D.'! "I' -
Metha."e D.ZGg ~,
8e11')eYe o.ooblt -¥
l()lue~e 0, 1{b ~
- ,.~ Tot0. L VlG"W" VA2la. ~'le ~
Rea.~ to~ Spe~A: )j.Cal (11 La.tecl. Vo.ltL6A
Rea.c:tol' ~ pe: PL'U.j FLour
Ther1'tl.a.L MoJe = AJ.~~.btL,
S(mv:l~tlO'h: Spe.~i+tj VoilL'Yr}e -t>- C,,-Lc.uLste
COll1.r.
5a.1'nfl e Ca.t(.u.{0. t ~ O'Y:
To ta. L Flow :: 7~D (.iwroLlh )
V::: Lt.7,-m,3
T-= '~tOc
FlOU/A (.imol/hr
H~ cl t'G<3e;,. 37l1·0
Metna.1'te 284.S
Bel1'}en e i~·3
To Ll,.C.e~e 3.'S
Tolu.e.ne ConV'er.A.~oll :: (0.' '&)(75'0) -3.15' =O.'l6'3
(0.1 It;,) (75'0)
Be"Yl~e')1t.Ge'nera.i:e.J. ': (0.11" (7S"C) (0·'1'3) ': G'1S~~
. ( ~n",i' " ..,.I
20-16

Prob. 2()·S <!!u-nl.
1.0 Prob. 20.15
~ O.q(.)
'-
O.6 ,---,--,,---:-,:-:---,--~I:-:---I.-.--:7-:-:--....l-~-:--....l----2'---l
400 bOO sao 1000 12DO }400
Feed Ra.te C2n""ol/h)
20-17
-0
<U
~
100~
o
s-
o...
90 oJ
F
CI
~
~
so ~

_)III..... T -=. va."". , T • 1
po:: 25 ba. yo p:: 2SPll'"
F Low [.A1"t'oVhJ
r~Cjd.~()q en 7'D.6
Met ha.11.e jj'f.3
8 e')1~ e1le 8.0
To LUe on e l~ li.O~----~~~~~~~-------- -
R ea.etG'" $p~CA" * ca(cula..ted
Re.a.~to'" TCjpe: P(u.~ FLour "'tIo.hc.e.o.
The.rhCLl Mode; ~cl.LQ.l,Q.tc:C.
ShnulCl.t l(i~ Sfe(it~ VoLu.-me-t>- CC1lcu.lClote.
S C 1 l
· C'~nvet.b(Ci'h
01. M f lea. c.u. a.tLull.;
T -= 770·C
V':: Ltl' W3
Sirn.u la.ttvV't Re..b.ltA:
T -:; ~qS.5" 0 c
FLoWA (ltrnoLlh'
H~ J. ~o it. Y. bC.5".2.
MethGl.ne 4b~.1
Be",~ehe Ltj.4
Tol u.e.he 8.G,~
ConverACon'= i -·S.b~/l'iL{ -;:: D.9Lf
(CG11t.)
:.lO-~

[00
~ 90.....
~
~
?
F 80o
v
v
~
Q) 70
~
j2.
60
Prob. r2c. b
[v=4.7f.~ I
I I I
'730 7140 7~6 760 110 180
~~rQ'(a.tl..l.re ( ec. )
20-19

IV
o
too
t~:':,,' '
a.
c)
d'
e
.f)
r-""~
" "l
C'U 1'1 e1"e Pro d.u ctlGl'l
Rea.ciar Sf'tt1uLa.fLG" Sct..up Values
( P ","ob 20.7-~O.20)
,
Che'h'li.ca.1 RQQ.etlo~~: Rea ctLG1't 1 Rea.ctiol1 2
C3 H6 t C,H6 ~ Ct H,;
(l} (2..) (3)
(31 t C( H,z~ CI~HI8"
(I) 3) (Lf)
Si. 1l1U: ldtiG'h Va.lue","
-yo. ~ ~ I Ct
Cz. ¥1 - t2.= ~lC.,C3 *
I.P. StoLen .- kinetIc 1.D IStOLCh. !Kineti.(:.
NU1'n.ber Coeffide'b t IE1U)O')Jnt N(.1nber Coeffi.c{e1J t tlp01le'ht
Ptop'jt~lle (C3 HG) a) 11 -1 ¥' 1 'II' (I) ~ -1~ i¥
13e111} e-ne (C." "6) l2 II- -1 'WI' 1 ....
Lu.1l1e"h.e (C~ -in) (3) ~ +1"" ~ (1Y.r.· - i ~ itt'-
DIP aft('
,
(4 )~ ... i~
'It
'(li) -
Kllle'tttA ~ - I -t ..il, ~~.le-t7RT , ~ =~ e-E/12T
,JfI~~-m0Yd] ~1
. 'l.. Cl.J..
2..•8~ 10 7 ' 2.32'/./0q I
"
f. lJt. Cdll...a ')'not1 2't,SCf1 ~ 3S;670~
TrCLl1~port fTllef'"m Q l Entho.lp~: g~~ j (- v-a.LueA : Sr2~
Pt'oper t i. eA
RpQ..t t i.Q~ Pha...b.e 'aa.....b.
l In",o r'"'ma.t to""" !>1.lpe1i.ed (Reo. e t i.u"Yl.. ~fOe('c. fie) I ~"p- Dll~oprop'11heh1e-lle

..
(I' Prop~te:ne.
( p)
('l..) Lv:me1le
(~)
~ Be11'?J e-ne ~ C'u:m ei'L e
(8) ~ (C)
of P",oP"1 Le'1'l e ~ DIP8
(p) (D)
r, ~ ~C.P C6 i 'r, ':f ~t Cf C<!
r~ -:: '{-, - r, :: j I Cp l s - Jt'L CP Cc:
rp ':: ~ -:: ~'- C'p Cc:
n(Ae.Le~l; ;'1'" t ~) :: n-= .9a. CB CP - ~1. CP Cc
"'0 Jt-z.Cp Cc
n:: Jt C. 8_ 1
Jt-z. ~~
F,,~ i.Jea.( C..c: -::. "Yl..l./V -= tt..i/RT =1T~i/{n
q.a,.,A
h. ""..IiI (rf..!f'e/U :- i :: ~I.t:~!t _ 1
Je1.(;r(-tdP1 1~~~~ -
" -E./R-r
~, -::. ~I.O e ':: -i l CJ (E7.-c"Yf?:T
..i ~ -E1.112,. ---.:l e't z,u e. ~~,v
L~ t ~~~o / Ji, ::. 1(. t he11.
"h. =KlelEr.-E,)j121l.Aj~1!1 - i
K = 2.~ '1-lb/2. 3'2 rl(j'1 ::: O.6{2.
~_ E. :. "35,C/l~ -"Z.'1rzr.~7 --'Oi~ ~,o.·L!~mo(
20-21

'-
Prob. ,20.17 (C6'llt.
BCLA.e .~ fbW" . _'I.
r r ,
.-
Dr: H'ere-n t l.£l,L
R.e.ll.e..tor
AJJi..t lG""" yYt.~.J~
-to Q.c!.i'u.A.~ reCl.Cto"
lno 1 .f ro.et i~1..A. .
Tel;Ut> 14).f&LE J:6.. . c UJ1.e~l" cg6l..ee:'T t v I l' "'1
lermA "(,l,
IO-n/12.,.
Sele<:1;v itt... ~oliQ:t.La~ K e• I
!emp t ~
I
4r ,.I
Pr~.ll..u)"e 't ¢ ¢
~Sl" ¢ q:
~p "t ¢ q.
"'i Psz.c,AA"'S 1"- ~ ¢
""tc. t f
.110 ¢. 4'
20-22
..l.4 rah.... n
..I¢ •
Jr..
i.
¢<t !:
I
i } t
~
~
I~
¢'
~
¢

Pr-ob. 20 118' Lrta.rt 0' f'<2'~ e 9receJ~~~ Pt"Ob 21'017)
--------- ---_ .......... _-_ .... _----
: SpLlfte'r Reattor(Pt:' Mixe~ ~
Proce» 1~ ,.[ ~~-~;~hall 4> ~ ~
lnftLt I ( .b~'.. _o...o..o rv ~
~ -~---~~-~-------~~
' Alrr)'l;.lG.ti on t'n od.el fo,. fLu.iJ..hc:d reo.~to"
5pUiter Calc:ula.f:iCtJ1.
S% S~ld. ~l"'lJ.ctio~ to hPt""'Pa...u
Prot! eAA R'la..e.tor 6'1-"fQ..bA
1.VlpU.t Feed
TOe ?J50 ~S'o 35"0
p ~1.Q. 2> 675 3015"" 3075
Bel1')e1'e
Prop'1lene
PropQ,:ne
213,b 202.9
1t().S (0 s.0
$.5'3 S.'lS
lo.bi
5.53
C>.2~
Rea.etB r SfH?(,S'.
-
Rea..etor T'jpe: PLu~ FLo'W"
The",7na.L Made: l~Qiher ma.L
R~Q.e.to '" Simv lOot ~O?l: Sp~ci~'j- If.o tume -:-- CQ.lcvL:4e
CO1'.trer..4 .(.011.
i
P.F. Rea.c:io'" CaLeuLa.tC:c-n. '
5et Vot. u:)'n e -= 7.fq tf'r+w"3
Rea.e.tot" Iteaetor
Feed Pi4cn.a.rQe
350 350 v
307~ 302.S
20-23

·-:. ~~i' •
~~.
~::
Prob, 20.IS tont.
FCoW"s
Be..,,~et1e
Prop~lene
Propa.1te
C.U-me"e
DIPS
FLows
Be.ll~e-ne
Prepq lelle
PrOpCL1'e
. Cu:mene
DfPB
202.'
lOS
'5.15
-
-
RF Reactaw-
Oui~u.t
l'Z.LLt
'(.3,0
5.,5'
~I.O .
C.'1G.
lGLLf
23.0
5.ZS
9" I,0
o.tfq
IO.~
5.sz.
0.28
-
3!)0
?>625
J~'2.,1
2.2.5'
S.S3
S,,0
O."I'l
c)
Cu:m e "'e ~rochL JC.t io-n ch-OpA ~ro7n. q (.l-/ ~ Cj I ({f~)
D1PB f't"oJ u.<:t r:.~,., dropb. -fro-m ,2,(, t6 O.4Q (rst"")
Note', n:,l" eo..eh -moLe of Dr PB produeec! 2tmoL~.4
07 f>r0f'~ le"ll e a.re (.'cn.a'U.~ed 4ca.nncrl be .
reeovereJ 're.c'1-cLeJ -
l-he lLtn-ea.ct ed proP'.S lt1e Ca.1I be reco11'e~ed (
re.ct}elec:l a fl'"O'Ui.d.e hi~ he,," ovet--a.l L pr-0f'jLeh2
':J i.e[clA U-Al'll9 "e"Ur eo..iQ..L~.4 i:.
20-24

Prolo. 20.lg
SE;'e ProJole"M ~O.18 for
a. Plo'W-4heet cJ 'luld. bee! t"~a.cto..
.L: Caleu LQ:t i0 "n 5tr-o.ie~ tj
c. CQ. { c u.l Q, t i() "n .f0 V' V::: 7."1 q m'?,
5trea.,.", lnl'_~ t O,,'pu.t
Toe 350 35"0
p ~Po. 307S" ~O25
FL.01.US
Be-n~enQ C[3,6 1'2.6
P~of~ le"e 10.5 f7. 'f
Propa.he 5.53 5.53
-
'f}.35-
D:tPB - 0.66
C1k1'"r e'h e P.'"odu. ct(G1'. = Ql.35 ~-m.Gt/h
SelectlV'lt'i ': Qr.35"/o.":' l~g
20-25

150
;s-> 140.....
.~
.....
-...GI
Q; 130
VOl
<U
F
QI
~ l20
;:1
v
IDa
8
CCo"l1.A ta.." t R:ed
@ 6c.4..t r:.
I ,
20-26
If. 18
-~-oJ
o
~
~
~
o...)
+-'..,
IOO~
o
ct
gO

Pro.b. 20.20
Se~ Proble-m -:20..IB ~o
a.. FlCl1.VAh.eet Gt .ftui.c! hed 'eo.etCiV"
Jr. CCLlc:.uto..tio'h .st~te~lJ
AA.4U"rn i'rl~ C:d~o.l ~Q..4 *6 1'n4i.llta.L'lt COr$tC.")1t 1.reloC'ti~
'oio l t low (~ew) =ctQ.l FLG'WiBo.-"" Cea..4e) T..e,.c:.
Toto.l .flGW (neUr) =Toto.l-flO1UC..h.4e C4.Ae) :TeJ..aAa S'o..2r4r1l1
T (h.ew)
Sa.'n'l f le: Co.LclLlQ.tL01: 4> LlCJO"c
Fl ( ) - ":l. 2. 9 7 ( ..~5(, "1.7~ ) ~ ; 0 S. 2.
(iW h e1.U' .. ;J • . lfC,,,of" 2, '3
. Set VGlumt::: 1.lf9m~
FLoU/A.
8elljene
PV"oP"tle"e
Propa.-ne
(u.~e~e
DIPe
400
301S-
-
L{OO
302.5"
102.3
5, "
s. ,q
~LLo
.5'0
ClAYn ell Produ.eU 011 -::: q4. ,D Jmol/h
SeLectivl t l1 c 9'- /I.S:, b2.1
tonto
20-27

Prob. 2o.:;' t;) cont.17-:~~.
"'~:..
IVolU "'me '7 "7'fCf d I
I~o
jbO :s
.....
tlto '~
'-.....
120 v
~
.....tJ
100v,
(II
go c
t{'o G
GO ~O
~o
3~D '3bO :'70 3S0 3'jO
TC1'Tlf'era.tlLre (·C )
1100 410
20-28

Prob. 20.2 I ( St a. tot 0 h f r e Co ec;{ " 'rt q fi~e )
ThiA f't"oksle~ cC"nfit""mA vcz.Lue in .AeetLGb.20.5.; lfiiC1t
UA.°e. F{. <1' 20 I Ib a.,4 ba.~ e c'CLJ>e "3
. ChQ.,'no e fro'171 ha...A.e CO-A. e
v
5t r-ea.-rn qI ellte r A.. <2.1 hofto'1t
Strea.1n q2 f:l.itA. o.t tap
Rea.etot' SeecA..:
Rea..etor t'1pe ~ PLuq FLo"W"
No. lub~ 44S Si~ e L./o" ,ofL / ?l"rl
T,er'11a.l Mod.e: Vt i.l L t'1
FLuW": LO'U. "t~· Cu. r-re"Y t
U~ '0 W/m~ 0 (
La leulo.t~(.")1. Moele:~
5 (5.I/"dO .
et folu.'l'r.e -+ Ca.lc.uLo.ie. C(i'ttN'ey~(.a"'h
St ...o.t eq"l U.4.tJ. (Tria.l <err-or :
V<1.r~ PeeJ Ro.t~ --f- Co.{eo CC1'''J'eo
Mo t"
u tt l [ c'o"lrerAio~ 61 O:t i.JI. ohtQ.~ 'h.e.d.
Proee..b-4 Uti Li,t4
StreQ."h' lio. 2 a.. 3 ql qZ
T ·c 2.~4 3Sl~ 407 3lf2
Fl~w (~~oL/h.
H~d roqe-n 0 lfO.Z6
't 't
Ac.e.to"e O.(q .qO.q~ ~ y
WcJet" !'.D6 '2~"Ob ~ "F-
IPA Lf~ 076 4.4{Cf *' *
HTM "/(
~ se 5'8
.
20-29

r:' :,:....:....
Prok" 20.,2. (~o~t.)
'a. ) It<:rt:,,~e L"n feeJ = 67- 57,9:: 9.2,St'Mo(/h
j.) '% Lnerea.A.e -.:: ('t.2/;1.g1 (100), 15"..~ %
20-30

Prob. ;2?;2.2.. c..~tQ...t onpD&.~e ft"'ecedil1.Cj Prob. %.21)
'V.,ae ri~ IQO.lb (l..A h.a...4«: Ca..4~
c..ha:n Q e f r07l1 ba.A e ~a....Q. ~
I n-e.re <1.::e. Sirea:m :'1 f '1>9 So %
FLo1.tr :-(58.' ~f.~S").::. 97~'m6l/~
Re.a.,tot- ~ peeA .:
Rea-totor tlJpe: PLuq FLow .
No. () f tu.be..A .~4S~ ?i"e L/o ::: 20 f¥2.~, .
Therma.L Made· VtlLLt~ .
FLow~ Co .. CVl"'fe-nt
U-= fDO w/m? 0 k
C.al GlJ tC4t lO')' Mocle :
Set vol'U.'n'le (5",(VW~) --- Ca.("u[Q,tiol? cQnUer.olO·f)
PYOee.~A Vtilt:tu..
Stre.Q.11' NC/. 2Q. ~ 91 qz
T 0c. 2.3~ 3bb 4C7 31/
FLow
[-1'1 d rO<Je''' 0 36..18 ", ~
J:c..6 tern e. O~ 16 :'(,.34 .. '#
Wo.tev lCf,Oq t-t04 .,.
"*
TPA 3S.bQ 2.,L( 6 ~ ~
4TM ." ,. <11 81
Total 51.8lf C}1I.02. 81 ~1
a) 1.PA C"lV e~.4 i..07L =f3'~ fi/:'B.b'il -= O..q'36
J,.) %Inerea...4e i.,., Aceto."e Produc.ed
:: (~bI8 -1}1 JDO = 40/0
'3q"7i
20-31

Aceto'ne Prod:u,ctio'h (P..oh ;2.0 •.21-20.:2-6 -)
C': Rea.C!tor Si;nu.l.~tioh Set -up Va.(LLe-l.,
) Che"YnL caL ReaAtLO?~ {C H3)ZCfiOH~(C~)zCo of Hz~
T. D. No. (l) (-z) ('3)
Jr) Ra.teo E'X.pre..b.AlO'h ! _ D. ~
.. r1 -AVJ C,
C) Si1"nu la.tio"n VaLu. eJt.
':r. D. Stoich. Killeftc
CO'l'np6""e'ntA. 'Nu.~.be... C,,~ffi,ie1tt E~7)el1t
- ..,
I.Cfrof~ no~Cf4;)2 CHOM1 (r ¥
Aeeto'h.e r(CH'3)2 co1 ('l) "
... 1 y 1""
. ~
1-1«-iJ rOQ en l -411 (3)
d) v jl := J.e-[E/Ri1
...1 S" '11/
~ [,41 -= 3.5' 16')(10 y:
E[~caVJa~ol1 -:: 7~9 Ml/ltmoL
e) Tra:nAport a.M Thermo.l Propertie.4
Enlho.lp':f: SR~~ <-tra.l(Le..4~ SJ2k
i__ ~ tnfor-mo..tlO"l'. S'lLppl~ed. (Reo.ct.:.o'n S'f-eei.fc:c)
20-32

Prob. ~O.'23 (.At~r1 61 pa.,e p..e~eJ. ~').<3 Pro b. 1..0 .2.1- )
U.,4e Fiq. :zo., 8 a.A. 1a..4e ca...A.e
Cha.,"rq~ ~ro't)i ba.A e CaA e
Ihc. re Q..A e Stt'ea.m <:U bt{ SO~o
~.e. Flow :'~Sg) (1.5) =~7 ~'mo?/h
R.ea.etor Sfec~.
Reec.ctor: PLlLq FLoUT
Nc. 0 f ttLbe.A 4~ 8 Si,e L/o:: 2.0.ft/ 2. .L'h.
The.rM.a.l Mode.: 1JtiLLt~
FLow: Co- cu. yre')'t
U= b()W/m?'°K
CaLcuLa.ti.o'n Mocle~
Set 'lroLu1n.e (S.I m"3)-+ CaLcu.Lo.1e C011tre.r4L01.
Stt"ateCJ'::t U.4ed. (tv-ia.l t errot'") ,
V(J..r~ Ff!~d Ro.te --a, Ca.lt conferA.':o?. to obta.i.')' Q9
Pro (e..b~U t i. Lit4
St '"ea.'1''. No 2.0. 3 ql qz
TOe 2.~4 j-" Li07 ,'~. 5"
FlQ"U1' (Je.'l'n.oLlb)
H ~d vOCJe'l'l 0 ~ ct.Oq 'fo 'f.
Ate.to"h..e o~ fR ~(t'8 'II' ~
W",te.'(' "Z.. t.f() 1.1. 'to *' 'f
tPA It:,.42. 4:~3 ~ ..,.
HT t1 - 'Jr lit R10 ~7.6
TClto.l "s.oe, l"'{.1 ~1.u Yl.O
20-33

pr-ob. 20,.21+ C.ata.ri 6h f~e freeecli.-tl,:) Proh. 2.01~ ~
UAe Flq.20.lb a..A .ba.A.e Ca.Ae
Cha:nqe.A fr011' b4.4e ~a.Ag
1I" t l' e a.A e S-trea:W' 'qt{U~fJ .t~ so·cI
R.ea..c.tor 5f e.e..b.:
Re.a.ttor·, PLu.<3 FLow
No. c).f. tu..heA qe8 si.~e LID -= 20 ~t/2~1"l
The.toM-a. L MeJe: UtiLit Y
t= lc>w'. Co... cvrre-nt.
U-:: bO wirm7. °t<
La.LCoalQ. t l (';,n Mod.e:
Set voLu.m~ (sr. (W'3) --""" Ca.ltulo.te CO')Lve......to1'
St r-ateq'1- (t ~i~t4 ~rr-or'
.i.. Sele.ct feed rate
j.i. C.a.lc.u.Lc.te con'U"£".la.Lo"). C@ 6.'1 ..taoI.u:ti,o.,.. CLC hieve"! )
Pt"oc.e...u Uti l.t~
23l{ 373.7 45'7 383
Ht.:ich··o~e1- 0 S I. 41 ... ~
Ae..e.i",'Yle 0.'2.."1 5"1. bS ,. ")fl:
Water 28.15 2.~.lr ~ ~
IPA 51.1 5.11 '" y("
HTJV1 * '" sa sa-
Tota.l 85;5 ~,.q 5"8 ~8
a.} Pt-oc.e.b~ Feed =? 5.2 (~11loLlh
.Jr ~ flJo ~(.t-ea.~e :: (~~,~ - t) (100) ='+1.7
20-34

Prot. .~ 0.25 (Ata...t Ol F'Q.'3e preced. i 1'l.q Proh :2..0 .,2)
U..6e Fi ~. ,20. b a.A .b4.4 e co.A.Q
ChQ. "r 't e.4. .f,- o...-n. .ba...Aa. eoa.A e
Va. ...':! ot i Li t<j flour
Strea. m 9r
Rea.ctor Spee.6.:
Re Q.. Cot 0 ~ : Plu.~ JZLCIW
No. of tu.he~ .lI~S. 5i~e L/o =20ft./zJKI..
~ Ther N.Q,l Mode : Uti lLt Y
FL()'tU : Co - t u.r r ent
U-= bO WIrm'2 G{<
Ca.lcu la..ti.o'h Mod.e:
5~t volA.-me (S.I rn~)-+ La.lc.uLa.te cG-nArerAtob
St ra.t ec:p:I U4. e c1.:
Select ..4ev-erCll t.(tiLit~ .ft'GW r(J,te~
~or eO. eh. f(c,1.U"
V<l.t"lJ Fee d ~Q.t eo -.a.. Ca.l~ula,t e to 0 btctt-n
0/1 Co,ure:.-..4': 6?'l (b-jh-iol i
E). 0.. -rn el( <;'t" eQ.'" :.2q -2t'"0 l/", erf"Oit.
See. Ftc, u.T "TQ, 1>lC G"Yl 1'C.'" t P0.9 eo
tc,·on.oL-n.o..te.4 on pe."for'Wl ClYlCe CUrAre
'/.. -: 19 A1'1.ol/h
'y -: 4.q.~ ~ m c.l /h
cont.
20-35

Prob. t20.;2S' e<rnt
PrOCt.AA VtLli.t~
St rea:"M No. 2. a. ~ '11 92·
TO c. 2~lt
Fl6"UF
'3~~ 401 ?'3Z.?'
H~ clr 0<3 e."" 0 "26.59 '¥' ~
Ac.e.to-ne O. r, '2~.11 ... C
Wo.te.r It{.S5 Y. SS'
"" '*
IPA 2'l.s"j 2.ili * *
HTM C" ~ 2<1·0 2Cl-O
lota.l 4Lt2 10·S 29.0 '2.q.0
Problem ',2.0.~
70
~
~ &5" -
~
.....I)
~ 60
~
IV. S.I_3 It-l 55'
~
..B 561.1...
~ L10
~
<U
v
E ~O
Q..
I I I I
ZO 30 110 50 60 70 80 'kl 100
UtiHfy Plow l~'mal/h 1
20-36

prob. ~o.z€, .C~ia.,.t <:n ~ge rree~ cJl'Jl~ Pr-ob. :z.o.2J )
VA e Fi~. 7-0·b 0..,4 A ha..Ae ellA e
1h Cr~~.b .; (It ili:!'1{-St reo.m q, blj roc
U =1 ( Feed rca.te)
R~a.cto.. Spe.eA .
Rea.cto~ t'1re·. P-lu..q FloW" .
No. OF tu.beA: 44S Si.e: liD:" 20 f+../z..i.n.
Therma.l Mode~ UtiLi.ty
Flo 'U1 ~ Co - <:u rreni
U =60 Wf me. 0 l<
CaLcu La:ll~'l' Mocl.e:· Set vc>Lu:me. (5.Im.')-:Ca.leu/.a.t.e
Col'l.i~t'~L. O:'h
Stro.ielj ~! __
SeLect fe ed·rate.
E"o.LtLo.ie U
Obta.in COt')erAlO,,
a.. GLven:
AAAume:
1. Pre.A.b ue clrep in V'aa.c:toV' cG1....bla:nt
c. 'Te-mjgero.tut-e fJ.-ofiLe cha.'ttSeA
1lode..aJ
3· I deo L ~aA
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Ne. ....- rZS'f yyrt ~(
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4. Ga.-4 (Proc.ees ..hide) FilTfl CG:"t1tv-oLA
20-37

c:~ " U',';"'h i. = 60 (F/57.Sl.lf>.8 Wo:f.ilm."2.. OK
1-) A.J.4.U?n e' 1= eeeL = 93. 5 .i')7lol/ h
'( o.e
U= bC ~3.S/S"'1. gi,C ,,= ~ g.t ..Q")'noL/h
PLow Tee.ble
PY'Oee..b.4 Uti.l't4
Input 6u.ip1A. t l1'lf'ut 0Lt~u. t
T o( 234 ~70.'1 'iS7 37b.3
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20-38

ChapterQ I
21.1 Cumene Reactor, R-801 - Alternative configuration and control scheme.
a.
Reactor Feed
----~7r_----------~
x
I.
·fI
Propane
R-801
,,
:f!
Reactor Effluent
~------------~8~----·
Consider temperature indicator and controller Y and control valve X. All three
control loops are identical in operation. AB the temperature indicator senses an
increase in the exit temperature from the first catalytic bed, it sends a signal to
open the control valve, thus increasing the flow ofinert material into the first
bed. The increase in flow ofinert has two effects, (i) it dilutes the feed to the bed
thus reducing the reaction rate and hence amount ofheat generated, and (li) it
reduces the residence time in the bed, which also reduces the conversion and
hence the amount ofheat generated. The net effect is that the exit temperature of
the first bed will go down. Ifthe TIC senses a low temperature then the reverse
control action will be initiated.
21-1

b.
Reactor Feed
----~7r_----------~
Propane
R-801
y
Reactor Effluent
~------------'8>-----.
Consider temperature indicator and controller Y and control valve X. All three
control loops are identical in operation. As the temperature indicator senses an
increase in the exit temperature from the first catalytic bed, it sends a signal to
open the control valve, thus increasing the flow of inert material into the second
bed. The increase in flow of inert has two effects, (i) it dilutes the feed to the bed
thus reducing the reaction rate and hence amount ofheat generated, and (ii) it
reduces the residence time in the bed, which also reduces the conversion and
hence the amount ofheat generated. The net effect is that the exit temperature of
the second bed will go down. Ifthe TIC senses a low temperature then the
reverse control action will be initiated.
21-2

c. The control schemes in parts a.and b. above are both examples offeedback
control strategies. Both these control schemes have merit and the better scheme
depends to a large extent on the dynamics ofthe process and the disturbance that
it is desired to control. The cumene producing reaction is quite exothermic so
that the temperature rise over each bed may be quite large. Ifthe temperature of
the catalytic beds is very sensitive to small changes in feed temperature then the
feedback scheme ofpart b. would be preferred, since the bed inlet temperature is
monitored and controlled directly. If, on the other hand, the sensitivity to small
changes in feed temperature is low then the feedback control scheme of part a.
could be used to adjust for changes in catalyst activity occurring during
operation. Note that this feedback scheme does not monitor and control directly
the feed temperature to each bed. Rather the required feed temperature change is
inferred from the exit temperature ofthe bed.
In reality a much more complicated scheme to control the reactor beds would be
employed. This scheme would use both the feedback loops considered above
cascaded together. In addition, many safety features including emergency shut
down procedures would be incorporated into the design ofthe control system.
Most probably, several temperature indicators (thermocouples) would be installed
in each catalyst bed so that the axial (and possibly radial) temperature profile
could be monitored. For this type ofreactor set up, the highest temperature
normally occurs at the exit, however, it is always a good idea to monitor the bed
temperature profiles. One additional feature that would probably be included is
to add a propane feed to the first bed. This is a good idea since the first bed
currently has no control loop associated with it and the reaction rates in this bed
will be the highest due to the high concentrations ofreactants.
21-3

r
r d. A combined cascade control system is shown below.
Reactor Feed
----~7·>---------~
Propane
R-801
Y,
Reactor Effluent
~--------~e~--~
Consider temperature indicator and controllers Y1 and Y2 and control valve X. All
three control loops are identical in operation. As the temperature indicator, Y1,
senses an increase in the exit temperature from the first catalytic bed, it sends a signal
to open the control valve, thus increasing the flow ofinert material into the second
bed. This control loop essentially monitors and controls the inlet temperature ofthe
second bed. The second temperature indicator, Y2' monitors the exit temperature
from the second bed. Let us say that Y2 senses a lower temperature than desired,
indicating less conversion in the second bed perhaps due to a decrease in catalyst
activity. This control loop will now adjust the set-point ofthe first controller, Y1• In
this case it would increase the set-point temperature, causing less propane to be
added with the effect ofincreasing the conversion in the second bed and thus
restoring the desired exit temperature (and conversion). Due to the complexity and
non-linearity ofa system such as this one, the overall control scheme would probably
incorporate computer modeling ofthe reaction system. This would allow the
accurate prediction ofthe changes in the inlet feed stream required to maintain the
desired conversion and the whole process would be monitored by computer.
21-4

The overhead system for T-1Ol is shown in the sketch below.21.2
vent to fuel gas
Ir-'-'---=--~4-~.- . . ··x
t ........ 1
T-101
I
J,
~-(FIC), /
'-'
E-104
V-102
~--<..14.J'-------",_
P-202AIB
a. As the column pressure begins to drop, a signal is sent to control valve X to close
slightly. This in turn causes the flow ofnon-condensables going to fuel gas to
drop. This means that the non-condensables start to build up in the reflux drum,
V-102, and the pressure in the drum starts to increase. As the pressure in V-102
increases this causes the pressure in the column, T-101, to increase. Thus the
desired control action is achieved.
b. Ifthe amount ofnon-condensables fed to the column suddenly increases then
there will be an increase in the amount ofnon-condensables in V-I 02. With no
control action the valve X would not change position and the non-condensables
would build up in V-I 02 causing the pressure to rise in V-102, E-104, and T-101.
Because ofthe control loop, as the pressure in T-101 starts to increase a signal is
sent to open valve X, which in turn allows more flow to the fuel gas and hence
reduces the pressure in V-I02. Thus the desired control action is achieved.
21-5

21.3 Toluene vaporizer - control amount oftoluene and pressure oftoluene leaving the
vaporlZer.
VJ
x
condensate
return header
high pressure
steam header
Y vaporized toluene
to H-IOI
set point adjustment
The above control scheme includes two feedback loops. The first loop consists of
the level indicator controller (LIC) and the control valve X. This loop controls the
amount offlow ofvaporized toluene sent to the fired heater. The loop is a simple
material balance control which regulates the level of liquid toluene in the vaporizer by
opening or closing the valve on the toluene feed line, Stream 2. The second control
loop adjusts the exit control valve Y on the toluene vapor line leaving the exchanger
to adjust the pressure ofthe vapor stream. Ifthe pressure at which the toluene is to
be vaporized is changed, then it will be necessary to adjust the levelofthe liquid
toluene in the exchanger. An automatic set point adjustment scheme is shown as the
square box linking the electrical signal (dotted line) between the PIC and the LIC.
21-6

The reason that the set point adjustment is required is best illustrated by considering
the case when the required pressure (and hence temperature) ofthe toluene vapor is
to be increased (or decreased). In order to increase the pressure in the shell side of
the exchanger, we would increase the set point for the PIC which in turn would cause
control valve Y to close. Without changing the set point on the LIC the flow of
vaporized toluene would decrease. This is because ifthe level of liquid in the
exchanger is kept constant, then from the exchanger performance equation we have
Q= UAIlTlm
By maintaining the level ofliquid constant, the effective A and U remain constant.
Note that the majority ofthe heat transfer occurs in the liquid covered portion ofthe
tubes, i.e., the vapor phase heat transfer above the liquid level is assumed to be
negligible due to the very low gas side film coefficient. Since the pressure ofthe shell
side ofthe exchanger has increased the temperature at which the toluene boils
(vaporizes) increases (recall Antoine's· equation). The temperature driving force,
IlT1m, thus decreases and consequently Qwill also decrease - this results in less
toluene being vaporized. We can remedy this situation by increasing the value ofA
(the portion ofthe tubes exposed to boiling toluene) - thus we adjust the set point of
the LIC so that the level increases. This adjustment in level will be small for small
changes in pressure and could be done manually. Alternatively, there is no direct
flow control on the vapor leaving the exchanger, an alternative scheme would be to
use a cascade control system with a flow element on the toluene exit stream adjusting
the set point on the LIC as shown on the next page.
21-7

.::.~'~' .
condensate
return header
21-8
y
high pressure
steam header
vaporized toluene
to H-lOI
.1---+
c§

21.4 DME reactor feed exchanger, E-202
toll-201 ftomll-20l
B-202
a. Ifthe temperature ofstream 5 increases then the TIC would respond by opening
up the control valve. This would cause less ofStream 6 to flow through E-202
and hence reduce the amount ofheat exchanged in E-202 and thus the exit
temperature, Stream 5, would decrease. For a decrease in the temperature of
stream 5, the reverse control action would occur.
b. Fouling in the heat exchanger - The net effect ofexchanger fouling is to decrease
the overall heat transfer coefficient U. The amount ofheat that can be exchanged
is given by Q=UAATlmF. Since A is constant and F will change very little, the
only way to maintain Qas U ! is to increase ATlm• This can be accomplished by
increasing the flow ofStream 6 through the exchanger, i.e., closing the control
valve. This will result in a higher exit temperature ofthe shell side fluid leaving E-
202 and hence will increase the temperature driving force in the heat exchanger.
21-9

,F"".,.
)-
' Iffouling becomes excessive then the control valve may close shut and further
control action will be lost.
Loss in catalyst activity
The loss in catalyst activity will result in a decrease in conversion and hence the
exit temperature (temperature of Stream 6) will slowly drop. This will result in a
drop in the temperature ofStream 5, and will initiate the TIC to slowly close the
control valve. The net effect ofthis is to force more material through the shell
side ofthe exchanger and this will increase U and increase the exit temperature
of the material leaving the shell side ofE-202. Both these actions will
compensate for the decrease in the '!Tlm which occurs due to the decrease in
temperature of Stream 6.
Unlike the fouling ofthe heat exchanger considered above, the loss in activity has
a major impact on the down stream processing. This is because Stream 7 now
contains less DME and more methanol and columns, T-201 and T-202, will now
be affected.
c. The strategy used is feedback control. It is feedback because the control action is
initiated by changing an input variable (amount offlow of Stream 6 through E-
202) upon measuring a deviation in the output variable (the temperature of
Stream 5 leaving E-202).
21-10

d. There are several ways to do this. The most direct way is to switch the bypass to
the inlet side ofthe exchanger as shown below.
r
6G
N
~
toR-WI fromR-lOI
~-.--
Note that for this set up we can manipulate the amount offeed material passing
through E-202 in order to maintain the exit reactor temperature. For the case
when the catalyst activity changes, this strategy would allow us to control directly
the conversion in the reactor, since the temperature of Stream 6 is a direct
indication ofthe co·nversion.
21-11

21.5 Regulation ofoverhead product purity ofBenzene Tower, T-lOl
E-I04
cw
T-I01
E-I05 YTC:
refractive index
P-I02 AJB
a. The purity ofthe overhead product stream is controlled by a refractive index
monitor which adjusts the set point ofthe flow controller on the reflux line. Thus
the reflux ratio is adjusted (up or down) depending on the purity ofStream 15, as
inferred by the refractive index ofthis stream.
b. This type ofcontrol system is an example ofa cascade control system.
21-12

21.6 Currently the control scheme looks like the following.
H-701
P-701 AlB
C-701
The current control system only regulates the total flow ofmaterial going to the
reactor, in Stream 8. lfthe control valve closes slightly then the pressure in Stream 7
changes and thus the pressure drops in the upper line 1-3-5 and the lower line 2-4-6
must also change. However, the air flow and the naphthalene flow are determined by
the characteristics ofC-701 and P-701 respectively. The pump and compressor
curves will have different head vs flowrate relationships. The result is that as the
total flow through the control valve changes the ratio ofair to naphthalene also
changes.
The obvious remedy to this situation is to place separate flow controllers on Streams
5 and 6 as shown below. In addition, a ratio controller is added which maintains the
desired ratio ofair to naphthalene in Stream 7. Finally, the total flow rate to the
reactor is maintained by adjusting the set point ofthe naphthalene valve. This
strategy allows the total flow to the reactor to be controlled while maintaining the
desired air to naphthalene ratio.
21-13

21.7 Driving an automobile
A responsible person drives an automobile using a combination offeedforward
and feedback control strategies with each ca3'Cading on the other. Consider
driving on a road in a suburban area. As we move down the street, we see a stop
sign 50 yards ahead ofus. We gauge the required braking needed to stop
smoothly at the stop sign and depress the brake pedaL As the car begins to slow
we continually monitor the pressure on the brake pedal and may increase or
decrease the pressure depending on the rate at which the car slows. All ofa
sudden we see out of the comer ofour eye a small child on a bicycle in a
driveway just a few yards ahead. We brake much harder and perhaps come to a
fairly quick stop. Having avoided an accident with the child, we continue to the
stop sign.
As mentioned earlier, we automatically use feed forward (anticipatory) and
feedback (reflective) control strategies. The sighting ofthe stop sign and the
child are feedforward actions while the monitoring ofthe amount of braking is a
feedback strategy.
21-15

21.8 a. Regulate the temperature ofthe reactor
This is easily accomplished by placing a control valve (X) on the discharge of
pump, P-901, and using a temperature indicator controller (TIC) placed either in
the reactor or in Stream 2 to adjust the setting ofthe valve. For example, ifthe
temperature in the reactor was too high, then the TIC would cause the control
valve X to open causing more fluid to circulate through E-901 and hence
increasing the heat removal from the reactor - causing the reactor temperature to
drop. This control loop is shown in the diagram below.
b. Regulate the inventory of the reactor
Again this can be easily accomplished by using a liquid level controller (LIC) in
the reactor to adjust the setting ofa control valve (Y) in the exit line from the
reactor, Stream 2. Ifa change in flowrate of Stream 1 were to take place, say a
decrease in flow. Then the level in the reactor would start to drop, the LIC
would sense this change and close the control valve Y. This in turn would reduce
the flow ofliquid from the reactor thus allowing the level in the reactor to
increase back to its original level. This control loop is shown in the diagram
below.
Note that the rate at which fluid is circulated through the external heat removal
loop (P-901 and E-901) does not affect the level in the reactor.
Feed
i
-¥
I
t----#----
R-901
E-901
cw
y
P-901 AlB
~------------------~2~-------'
Reactor Effluent
21-16

Maintaining constant conversion/residence time
One simple way ofmaintaining a constant residence time in the reactor would be
to use a feed forward signal from a flow element placed in the feed line, Stream
1, to adjust the set point ofthe level controller. Thus ifthe flow ofStream 1
were to drop, the level would be lowered to compensate and ifthe flow were to
increase then the reverse action would be initiated.
R-901
.?--.
f .,
,TIC}-+-]
.,-~." +-
E-901
cw
,'f
~------------------~2~------+
Reactor Effluent
21-17

"21.9 Control ofthe temperature in the reactor
We will assume that the external heat exchanger, E-901, condenses the vapor in
the shell side of the heat exchanger and that the level ofcondensate in the
exchanger can be measured. The control strategy is to adjust the level of
condensate in E-901 to expose more or less heat transfer area to the vapor. Since
it is necessary to control the level ofcondensate, a material balance control loop
is required. The control valve X is adjusted with the signal from the level
controller (UC) on the condensate level in E-901. A temperature control then
adjusts the set point ofthe level indicator to respond to changes in the reactor
temperature. For example, ifthe reactor temperature were to increase, then a
signal would be sent from the temperature sensing element, Y, to the set point of
the level controller. The level set point would be decreased so that more
exchanger surface area is exposed to condensing vapor and hence more
condensation can take place. This would cause the temperature ofthe reactor to
drop. This control scheme is shown in the diagram below.
b. Control ofthe reactor residence time
In order to maintain a constant inventory in the reactor a feedback material
balance control loop is provided for the reactor. A level indicator controller, Z,
controls the position ofthe control valve, W. In order to maintain a constant
residence time the level in the reactor must change in sympathy with the changing
feed flow rate. Therefore, a feedforward control loop is added which senses the
flow of Stream 1 and adjusts the level in the reactor by changing the level set
point, Z. These loops are shown in the diagram below.
21-1~
..---___coo'-'-ling water
Reac!or Eftluent
cooling water
return

c. Explain how tbis system responds to an increase in flow ofStream 1
As Stream 1 increases, the flowrate controller sends a signal to the level indicator
to increase the level set point. Initially, the control valve (W) on the liquid exit
line from the reactor will not change position and the level slowly rises. When
the level in the reactor rises above the new set point, the material balance control
loop comes into play and the control valve (W) starts to open and eventually a
new steady state will be reached. Note that at steady state the liquid feed into
and out ofthe reactor must be the same - all that has happened is that the level in
the reactor, i.e., the liquid inventory, has increased thus maintaining the residence
time ofthe liquid in the reactor.
d. Explain how tbis system responds to fouling ofthe tubes on the cw side ofE-901
As the heat exchange tubes in E-901 begin to foul the amount ofheat transfer
taking place will decrease (for a given condensate level in the exchanger). This
means that the temperature and pressure in the reactor will start to rise since not
enough heat is being removed from the reactor system. As the temperature in the
reactor rises the TIC (Y) sends a signal to adjust the set point of the condensate
level in the exchanger. The level set point is decreased. Tills means that more
tube area is exposed to condensing vapor and more condensation can occur. As
more condensation occurs the temperature and pressure in the reactor decreases.

21.10
A feedforward control loop is implemented for this system in the diagram below.
The idea is to measure the flowrate ofthe recycle stream, Stream 11, and adjust
the control valve (X) on the toluene feed, Stream 1. Thus ifthe recycle stream
increases a signal is sent to the control valve..X to close and vice-versa.
The downside ofthis feedforward system is that the "model" (control algorithm),
describing how the valve X should change with changes in Stream 11, is prone to
be not exactly correct. The consequence ofthis type oferror can be very serious.
For example ifthe valve on Stream 1 did not open quite enough when the flow of
Stream 11 decreased then eventually the feed tank would empty. Iffeedforward
loops are used then care should be taken to ensure that material flows in the plant
will be maintained. It is usually safer to incorporate (cascade) the feedforward
control strategy with some form offeedback loop (as was done in several ofthe
preceding problems).
TK-lOl
P-IOI AlB
11
21-20

Chapter 22
a) M~ ':" /. l/ ~
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22-1

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22-2

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22-3

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22-5

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22-6

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22-7

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22-8

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22-10

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22-14

22.15 (a) Pump and system curves:
vrnax ~ 73 m3/h ~~ 73m
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46% scale-up
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T
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If stored as liquid in equilibrium with vapor Ptank = p*
NPSHA = pgh ~ should not cavitate
22-15

22.16 Drying Oil Facility
Trouble Shooting
(a) Noisy pump
We can do calculations, but these components are not volatile, so the most likely
cause is some fonn ofmechanical problem -worn bearings, etc. A remedy might
be to switch to the back up pump and see if whining stops. Then, the original
pump can be fixed.
(b) Problem with T-501 and the increased Dowthenn A flowrate.
Although a drop in ambient temperature might have an effect, it would be small.
As temperature drops, heat loss increases and inlet cooling water temperature
drops.
Detailed calculations would show that the issue is not with the change in ambient
temperature in October. Look at heat loss:
Q=UA!J.TLM
Q = (mep!J.T)Dawtherm
Since Dowtherm has been increased to compensate to keep purity (specifications)
unchanged, assume that Qhas not changed.
Q(e) =m(t)ep(e)!J.T(-L-)
Q(e) =U(must -L-)A(e)!J.TLM(t)
!J.T decreases
!J.TLM increases
But U will increase because the mass flow of Dowtherm has increased therefore
something in U must have decreased to compensate!
Conclusion - there must be fouling!!
Could be on Dowthelm side, but since Dowthenn is a heat transfer fluid, it is
unlikely that it fouls (unless temperature is too hot -:- max temp is 400°C at higher
temps Dowtherm carbonizes!).
The problem is more likely on process side, since the temperature is 342°C, and
reaction will occur at this temperature. The hottest point in reboiler is on tube
22-16

walls where reaction will occur. Note that there is not much DO here, so gum
formation will be slow, but only a small amount is needed for film heat transfer to
decrease.
Alternatively, the filter may not be working properly, so the gum content in
Stream 7 should be checked.
How to solve the problem?
a) If filter is a problem - recommend regular samples of Stream 7 to identifY
when "gum breakthrough" occurs.
b) Recommend to clean reboiler - but requires shut down and the fouling is on
the shell side.
c) To resolve the problem permanently, we must reduce the bottom temperature
by reducing column pressure.
• As pressure decreases, the column moves towards flooding, because vapor
density drops and vapor velocity increases.
• But as the pressure drops, separation becomes easier since equilibrium line
moves away from xy-line. Thus fewer stages are needed for a given
separation or a lower reflux is needed with a fixed number ofstages. With
lower reflux, we move away from flooding and therefore compensate for
lower pressure - at least qualitatively. The only way to be sure is to do the
calculations.
(c) ACO Feed problem
There are two issues:
First, the control scheme is wrong. The control valve should be using the level
control of V-50l as its signal. As it is now, the flow into V-50l will fluctuate
based on the level in the storage tank, probably being overridden by operators to
maintain flow.
Second, another probable cause is that ambient temperature has dropped causing
the ACO feed's viscosity to increase.
Thus, feed problems may be a combination ofcolder weather and operator error.
If this switch in control scheme is made, we will run into a bottleneck when the
viscosity ofthe oil is so high, that when the CV is wide open, the design flow rate
cannot be maintained.
A solution is to heat feed tank and also heat trace the supply line. The feed tank
may have a steam heater that no one has turned on or maybe it is malfunctioning.
The line is already insulated, but for extended periods oftime this will not help.
22-17

The supply line could be retro-fitted with larger diameter pipe.
(d) Steam venting from E-506
Most probable cause is the addition of the extra pipe and elbows connecting the
exchanger to the low-pressure steam header. This extra pipe causes additional
frictional loss and requires a higher pressure in the shell of E-506 in order to
deliver the steam to the header.
Ifall the steam were to flow to the header then the friction pressure loss in the line
would increase in proportion to the added equivalent length:
A solution is to reset PSV-519 to 60 kPa above header. This can be done because
the Ips header is at 607 kPa, and this would allow the shell to operate at 667 kPA,
which is less than 90% of the rated pressure of 750 kPa for E-506. It is also
possible to re-pipe the connection with 3" pipe, but this is more expensive.
22.17 Scale Down
The equipment that must be considered is as follows:
Exchangers - what are new outlet temperatures
Pressurizing of system - reactor inlet pressure drops
Conversion in reactor - how is it affected by longer space times
Tower flooding/weeping
22-18

22.18
(a) By replacing the 1.5 in pipe with 1 in pipe, the pressure drop probably induces
cavitation in the pump. Calculations would support this. Also, having the valve
before the pump lowers the pressure, further inducing cavitation (left figure). The
figure on the right is a better design.
So, the obvious solution is to retrofit to 1.5 in pipe. Since it worked that way before,
the valve issue is less important. However, in general, it is poor design to put the
valve before the pump.
P-802A1B P-802A1B
E-806
.'1
I .,J;-806
~ r9<J___-~.9-----+
ON
P-803A1B P-803A1B
(b) Something is happening right after plant start up. Two possibilities come to mind. If
the heat exchanger were designed to include a fouling factor, the tubes would
probably not be fouled upon start up. Another possibility is that the catalyst
deactivates with time, but is at full activity at start up. In either case, there would be
better heat transfer, more boiler feed water consumption, and more steam produced.
The downstream steam header is at a constant pressure, so the increase in steam
produced results in an increased pressure upstream, possibly opening the pressure-
relief valve. Calculations show that a fraction of a percent is sufficient to open the
pressure relief valve. This problem should not be ignored, since lost steam is lost
money; although, calculations show that the lost steam value is only hundreds or
thousands per year. Perhaps the pressure-relief valve should be set to a higher
pressure. This requires internal permits for obvious safety reasons.
(c) During the summer, warmer cooling water affects several units, specifically the
absorber, the distillation column, and the product cooler. The increased fuel gas ratc
supports this because if the temperature in the absorber increases, absorption
equilibrium favors the vapor phase.
Similarly, in the distillation column, the log-mean temperature difference must stay
the same, since all flows stay the same, so increasing the cooling water temperature
requires that the condensation temperature in the column condenser increase. This
increases the column pressure, thereby increasing the temperature of the reboiled
vapor. Therefore, the temperature into E-806 increases. So, the outlet temperature
22-19

would increase for two reasons. First of all, the inlet product IS at a higher
temperature. Secondly, the cooling water is at a higher temperature.
Finally, in the calculations, the log-mean temperature correction factor (F) has been
omitted. This turns out to be a lower number than desired, so E-806 is significantly
undersized to begin with. Clearly, this problem is exacerbated in the summer months.
(d) Additionally, the following process improvement might be a good idea. The value of
methanol in Stream 12 is about $2 million/yr. Methanol recovery should be a high
priority. Even under normal operating conditions, the absorber temperature may be
higher than necessary. The absorber pressure could be raised by pressurizing the
entire process (rather than adding a compressor right before the absorber). Additional
cooling andlor separation equipment might be worthwhile.
Two options for better methanol recovery might be:
fuel gas
to T-802
22-20

22.19
(a) Fuel gas has increased in order to maintain Stream 2 at 900°F. This is probably due
to a lower HT coefficient on tube side in furnace. During temporary upset, oil would
sit inside tube and radiant furnace walls may have caused coking on the inside surface
oftubes. This means tube walls are hotter now and so this may be a problem in the
long term.
Solution: Stop oil flow temporarily and inject steam through tubes to "burn off' coke.
(b) Problem same as (a) except coke furnace is greater causing a significant M> increase
across furnace tubes and resulting in reduction in oil flow.
Solution: May need to isolate furnace and burn coke offslowly with steam (same as
part (a) but will take longer.)
(c) Same problem as (a) and (b) except a tube may have ruptured. This would cause a
lower product flow and now the oil is supplying fuel to the furnace causing the 02
level in the stack to drop so air/fuel ratio must be increased.
Solution: Must shut down furnace and replace ruptured tube(s).
22-21

Chapter 23
Problems 1-6 of this chapter are ethics questions for which students should be able to
find several competing rights and responsibilities. Some common issues are given for
the individual problems below. There are several effective ways to use these problems:
• Assign them as regular homework assignments, for which the student prepares a
written response to the questions asked. The typical length of a student response is
one single-spaced printed page, which should include a description of the ethical
issues involved, an analysis ofwhich issues are most important, and a conclusion as
to the best solution.
• Assign students to read and formulate answers to several questions for homework.
Then, in class, students are chosen at random to present their solutions to the class
and to lead a class discussion ofthe problem.
• Assign them as small-group homework assignments. Then, in class, the group
explains the issues and their solution and leads the class discussion.
For additional case studies and decisions regarding ethical dilemmas see the web page
for the Board of Ethical Review (BER) of the National Society of Professional
Engineers (NSPE) at:
http://www.CWfU.edu/affiIlwwwethicsiengcases.html
23.1 Issues that students raise include (but certainly are not limited to) the right ofthe
company to trade secrets, the responsibility of the employee to employer, the
desire to help one's colleagues professional development, and the improbability
that any discussion in class will put the company at a competitive disadvantage.
The students should be able to analyze these competing issues and interests and
to form a rational basis for a response. The problem provides a vehicle to
expose students to confidentiality agreements and to codes of ethics. The
AIChE Code ofEthics and (especially) the NSPE Code of Ethics deal with this
Issue. Students should be able to find the correct sections and use these in their
analysis. Classroom discussion should include what ethical and legal limitations
employers can put on employees divulging information. Finally, students
should realize that· a call to their former employer is likely to clear up the
ambiguity. If students choose to not divulge the information, even if it would be
allowed, this is an ethical issue that should also be discussed.
23.2 This is a common problem for students, and they often will have very strongly
held competing views. This problem is a vehicle to getting students involved in
the issues and to develop a sense of moral autonomy. It is helpful in the
discussion to have different students look at the problem from the different
viewpoints. Would they view the issues differently if they were the interviewer
rather than the interviewee, for example?
23-1

23.3 The issues here atfirst seem many moral rather than ethical, but students often
realize that this kind of an issue can arise throughout one's career-as they
move fromjob to job within an organization or when they learn more about the
organization they work for. After students have discussed the anti-personnel
mines (often with different concerns during wartime and during peace), they
should be encouraged to look at potentially less obvious yet similar issues, such
as: the company's investment policy, other more benign products that one
believes "should not be manufactured." To simulate discussion and to broaden
the problem, one could ask if the student should quit after working for the
company for many years, when the employee realizes that the company
produces such devices.
23.4 This problem involves many competing rights and responsibilities: to one's
employer, employees, government, society, etc. The discussion often fmally
centers on the responsibility of the government to set reasonable standards and
the responsibility of the company to meet those standards. However, the facts
seem to suggest that employee health may not be adequately protected by the
apparent standard. Students then often realize that the problem is more
complex. This is a good time to mention the "general duty clause" ofthe OSHA
Act, as well as the fact that a no-cost or low-cost solution may be available.
23.5 This problem is a good vehicle for "reflection in action," since a the problem is
given in detail and the responses of some chemical engineers are available.
Students should be able to find numerous rights and responsibilities from the
code ofethics that affect this scenario.
23.6 The "falsified data strike back" scenario can be used in a classroom immediately
after the "falsified data" case (Section 20.1.2) is discussed. The point here is to
see what would have happened ifthe data were originally falsified. Once those
consequences are discussed, it is interesting to discuss the outcome if the
original data weren't falsified. The discussion then leads to exactly how should
one have written the first report, and what should be done now, if the data had
been falsified. When students suggest that Jay should contact his former boss,
they should be asked to play through that scene to see whether this strategy is
more likely to clarify the issue or to confuse it.
23.7 It is best if the instructor first contacts the Board to get this information. Most
boards are very helpful in this regard. Students are likely to immediately
question whether all faculty in the department are registered and why not. They
will likely ask for direct advice as to whether they should start the registration
process and why.
23-2

23.8 Codes of conduct can be obtained from companies, but often only if the
company name is removed. The instructor should obtain a couple from
companies that have recently hired graduates from the schooL These codes vary
very widely. Some will include codes of ethics, others will not. Some describe
anti-discrimination and general personnel policies, some do not. Students will
be somewhat uneasy with the omissions.
23.9 There are many such heuristics. The chapter does not focus· heavily on
heuristics related to family responsibilities. Students often are eager to develop
guidelines in this realm. Whistleblowing is another area of interest in which
students often generate several heuristics.
23.10 If students need suggestions for an ethical dilemma, suggest a classical scenario
in which a student observes another student cheating on an exam or on a project.
23.11 The Process Safety Management regulation is a bridge between ethics,
professional responsibility, and safety. At the end of the regulation is a good
synopsis.
23-3

Chapter 24
24.1 Total employee hours worked per year:
40 hlwk)(47 wk/yr)(30,000 employees) = 56.4 x 106
a. Fatalities per year = (56.4 x 106
h)(4 deathsll 08
h) = ~
b. Injuries per year = (56.4 x 106
h)(0.49 incidentsl2 x 105
h) - 2 deaths = 136
a. Fatalities per yea r =2
b. Injuries per year =136
24.2 Students can be taken on a tour ofthe steam plant or expected to fmd it and ask
those in charge appropriate questions.
a. Explosion of the main steam drum from overpressure and explosion of the
furnace after a buildup of unburnt fuel are probably the most obvious, but
students can fmd numerous variants of these or accident scenarios that arise
from problems with raw materials (fuel, water) or product distribution (steam,
power).
b. Students should be able to identify many such safeguards, including: relief
devices, various control-system loops and interlocks, operator training, flame
detection, etc.
24.3 All ofthese MSDS's are readily available from suppliers or the web sites given
in Section 21.2.1.
24.4 Students should be able to find many such regulations, especially through the
Federal Register and Code of Federal Regulations, which can be searched by
keyword (see Table 21.2). These range from EPA regulations for benzene in
motor gasoline to the OSHA PEL. In some states, additional regulations can be
found.
Although all 13 section of the PSM standard are relevant to a unit operations
24.5 fi
laboratory, the most fruitful items for the students' analyses are: process sa ety
information, operating procedures, training, and compliance safety audit.
24-1

24.6
a. The volatile components in paints (both oil- and water-based) tend to evaporate
slowly. In the closed-cup flash point determination, one looks at the equilibrium
vapor concentration. In the open-cup determination, one looks at the steady-
state vapor concentration when the paint is exposed to air. If the vapor pressures
of the components is high enough at room temperature, the equilibrium vapor
concentration could be above the LFL. Ho"wever, if the rate of evaporation is
slow compared to the rate of diffusion of the component from the area
immediately above the liquid to the bulk atmosphere, the steady-state vapor
concentration could be below the LFL.
b. The open-cup flash point is the more useful property, since it is more
. representative of the potential vapor concentrations generated as paints are used.
a.
24·7 This mixture has an octane number of 87 by definition. Using Raoult's Law,
one can calculate the equilibrium concentration ofthe iso-octane and n-heptane.
Psat*(298 K, iso-octane) = 6530 Pa LFL =0.008 UFL = 0.070
Psat*(298 K, n-heptane) = 6050 Pa LFL = 0.010 UFL = 0.084
P = 101,325 Pa
x (iso-octane) = 0.117 x(n-heptane) = 0.883 (from molar volumes)
:. y(iso-octane) = 0.00754 y(n-heptane) = 0.0527
Thus, at the equilibrium, the vapor would be within the flammability limits and
would support combustion.
b. The air-gasoline mixture will be within its flammable limits near the surface of
the liquid. Because the concentration is so much above the LFL, the flammable
vapor cloud will extend a significant distance above the gasoline pool, creating a
serious flammability hazard. Note that, if the ambient temperature is just
slightly above 25°C, the vapor immediately above the gasoline pool will be
above the UFL, and the vapor will not support combustion. However, as the
gasoline vapors mix with the air, the vapor will be within the flammability limits
somewhere in the atmosphere, and a flame could propagate towards the liquid.
These properties make the vapor space in a heated gasoline tank non-flammable,
while making it very unsafe to throw a lit match towards a pool of vaporizing
gasoline.
24-2

24-.8 Many of the options for each of the HAZOP keywords lead to hazardous
conditions for this reactor, when applied to fiowrates and temperatures.
24..q I This problem is similar to Problem" 4 .1
PEL(benzene) = 1 ppm TWA; 5 ppm STEL
Psat* (benzene, 298 K) = 12,700 Pa
P =1.01325 bar
x(benzene) =-0.0025 (assuming all components have approximately the same
molecular weight)
y(benzene) = (Psat*x)/P = 313 ppm
The benzene concentration is well above the PEL at 298 K.. As the temperature
rises, the concentration increases because ofthe increase in the vapor pressure
ofbenzene.
24-3

Cbapter25
General Comments:
In the first printing ofthe third edition, "the ethylbenzene process" appears twice. It will be
removed in subsequent printings.
The suggestions that follow are just a few ideas. The overall themes are:
a. Heat integration saves energy, and is green.
b. Reducing the reactant flowrate is usually green, because most ofthese reactants come
from oil or coal. In a life-cycle analysis, the pollution associated with refining oil or
processing coal is assigned to the raw material. Therefore, anything than increases
overall conversion is green.
c. Ifa solvent is introduced, it should be recycled.
d. Anything that increases selectivity is green. This follows from #2.
e. Most "greening" suggestions are also good ideas from an economic standpoint.
However, the maximum selectivity may not correspond exactly to the economic
optimum.
25.1 ethylbenzene
reduce benzene loss - recover from fuel-gas stream - benzene is made from toluene,
which is isolated from a refinery cut
recover more ethylbenzene from fuel gas
heat integration - preheat feed to fIred heater with hot reactor effluent - this is probably
greener than producing steam
adjust reactor conditions to increase selectivity for ethylbenzene
25-1

25.2 styrene
recycle waste water, purify it (necessary because it might contain some organic residues),
and use it to make steam for the process feed
heat integration where possible
maximize selectivity
recycle the waste benzene and toluene to the companion ethylbenzene plant - virtually all
styrene plants are coupled with ethylbenzene plants
25.3 maleic anhydride
heat integration - hot streams: molten salt, reactor effluent, tower condenser; cold
streams - benzene feed or air/benzene mixed feed stream, reboiler
maximize selectivity
25.4 ethylene oxide
recycle waste water to absorber feeds
maximize selectivity by keeping with low conversion - this is a classic A ~ B ~ C
reaction where B is the desired product (A ~ C is also occurring), so low conversion is
also economical
25.5 formalin
heat integration - use Stream 9 to preheat fired-heater feed to reduce energy use and
emissions from products ofcombustion
recovery ofmethanol from off-gas stream - there is significant loss - millions ofdollars
per year
25.6 DME
selectivity
heat integration
25-2

25.7 acetone
recycle waste water to absorber feed
There is no need to purify the IPAiwater stream for recycle in T-403 to the level shown.
There are several stages doing very little separation. The reflux ratio could be reduced,
since there are too many stages. This saves energy and fossil fuels. Alternatively, fewer
stages column could be used. Or, ifthe reflux ratio is reduced, a smaller diameter
column is needed. This saves on materials, for which manufacture is energy intensive.
heat integration - between reactor effluent and reactor feed - reduces energy requirement
forE-407
25-3

Review ofProject C.l - Allyl Chloride Production
There are two separate problems to be addressed. First, we must increase the production rate of
allyl chloride by as much as possible and as soon as possible. Second, we must look at the
longterm profitability of the plant and determine what can be done and what additional
equipment will be required. Let us look at these two problems separately.
1. Increase the Production ofAllyl Chloride
One key thing to remember is that we do not want to buy new equipment. Since the Alabama
plant is only down for a short while any capital expenditures will be wasted. One could argue
that buying a new pump (off the shelf), for example, could. be justified due to its low cost.
However, ifproduction is to be increased immediately, the installation ofthe new equipment, no
matter how temporary the installation, will take time since safety codes etc. must be followed -
so it is probably better to concentrate on changes in the plant which can be done without new
physical plant. Secondly, the existing equipment has limitations and that it is essential to
determine what are these limitations. Finally. we must be aware that changes in one section of
the process may/will impact other areas.
1.1 Look at Required Pressure Profile through the process for Increased Throughput
As we process more raw materials, the pressure drops over all the equipment will increase. The
feed pressures are determined by the saturation vapor pressures ofthe offsite storage, since both
propylene and chlorine are stored as saturated liquids at 25°C. The pressure information is
confirmed in Table C.l and students should realize that both these materials would be stored as
liquids. The feed pressures ofthe raw materials (Streams 1 and 2) are much larger than required
and the flows are regulated through control valves with fairly large Ms. However, as the flows
increase, and since the downstream pressure ofStream 5 must not exceed 2.1 bar, the pressure of
propylene into the fired heater and chlorine into the reactor must increase to overcome the
increased pressure drop through the equipment. Consider the pressure profile at design
conditions inthe process.
At Current Conditions
Battery limit
E-602
E-603
R-601
J-601
H-601
Pressure at inlet to equipment
(bar)
2.09
2.43
2.77
3.04
3.24
3.58
C.I.- (
aP across equipment
(bar)
0.34
0.34
0.27
0.20
0.34

Now as the flow increases so will the pressure drop, except for AP across the fluidized bed
reactor which will be approximately constant (as stated in the problem). We recall that for a
fluid, the pressure drop across a pipe is given by
2 Leq
-APf =2pv f-
d
For small changes in absolute pressure we can say that APf';:j v2 = (mass flow)2, ie., gas behaves
as an incompressible fluid. Thus we can scale up using /lPoc (mass flow)2.
- @Design @1.2xDesign @1.44xDesign Pmax
Pressure AP Pressure AP Pressure LP
Batery Limit 2.09 bar 2.09* 2.09*
0.34 (0.34)(1.2)2=0.49 0.71
E-603 2.43 2.58 2.80 3.50
0.34 0.49 0.71
E-602 2.43 3.07 3.51 3.50
0.27 0.27 0.27
R-601 2.77 3.34 3.78 4.50
0.20 0.29 0.41
J-601 3.04 3.63 4.19 5.00
0.34 0.49 0.71
H-601 3.24 4.12 4.90 5,00
When we scale-up by 44%, using this analysis, we find that the inlet pressure to E-602 equals the
maximum design pressure (see Table C.2) and we reach a bottleneck. Clearly H-601 is also very
close to its maximum operating pressure. Thus maximum scale-up based on design pressure is
44% ofdesign. The pressure profiles through the process for different scale-up ratios are shown
in the figure on the next page.
IMaximum Scale-up =1.44 or 144% I
·note: we have assumed that the battery limit pressure is constant at 2.09 bar (as given in the
problem statement) despite the increase in flow. This is possible ifStream 5 flows to a pressure
sink, such as a distillation column or flash drum where the system pressure is controlled by the
condensation temperature. Ifthis were not the case and the battery limit pressure increased with
flowrate then the maximum scale-up would be less than 44% based on the analysis ofmaximum
operating pressure. Note also that both raw material streams must be vaporized prior to entering
Unit 600. It is assumed that the vaporizers for the propylene and chlorine can handle the
increased demand - since there is no information given on this point we need only make note of
this assumption in our analysis.
C.l-2.

Pressure Profiles through Unit 600 - Scale-Up assuming Incompressible Flow
-s-
ns
.c.-
~
::::s
rn
rnQ)
s-
a..
6~--------------------------------------------~
,-----r----,...----r-----.- Maximum Ope ating
Pressure for E uipment
5
4
3
2
1+------.-----.------.-----.------r----~----~
H-601 J-601 R-601 E-602
Equipment
E-603 BL
limiting condition at inlet of E-602
at 144% of design flow.
In the above analysis we have assumed that the density of the fluid (gas) does not vary with
changes in absolute pressure - clearly this assumption is not particularly good for this problem.
We can go back and repeat the analysis assuming that MIX (mass flow)2/p. Ifwe assume ideal
gas behavior, then we can scale the frictional pressure drop using the following equation: MIX
(m/m])2(PlP2)' where m is mass flowrate, P is absolute pressure and 1 and 2 denote design and
scaled-up conditions.
The solution algorithm involves trial and error and an example for a scale-up factor of 55% is
illustrated below:
CI-~

@Design @1.55xDesign @1.55xDesign @1.55xDesign Pmax
Pressure M Pressure M M' Pressure M Pressure
Batery Limit 2.09 bar 2.09* 2.09* 2.09*
0.34 0.74 0.75
E-603 2.43 2.91 2.83 2.84 3.50
0.34 0.64 0.67
E-602 2.43 3.47 3.52 3.50
0.27 0.27 0.27
R-601 2.77 3.74 3.79 4.50
0.48 0.36 0.39
J-601 3.04 4.10 4.17 5.00
0.82 0.57 0.64
H-601 4.67 4.81 5.00
~ ~ ~ ~ -C ~
1st
iteration 2nd iteration 3rd iteration
M'2 = (0.34)(1.55)2 [Y743)1l4(2.09+2.91)]=0.74
Use ratio of[Pave from design / Pave from previous iteration] solution converges after 5
- iterations
~
again the pinch pressme is at the inlet ofE-602, ie., 3.5 bar
IMaximum Scale-Up =1.55 or 155% I
As we will see when we consider other factors, the maximum scale-up is less than 44% so that
the added effort ofusing the trial and error solution to account for compressibility effects is not
justified.
1.2. Look At Individual Equipment
Assume that the reactor conditions are to stay the same, namely use a temperature of 510°C and
assume that small changes in pressure will not affect the selectivity.
1.2.1 Fired Heater, H-601
DesignDuty =4000 MJ/h
Maximum Duty = 5400 MJ/h
C,I-,(

Scale-up = 5400/4000 = 1.35 or 135% ofdesign. The reason for the under utilization ofthe fired
heater at design conditions is not clear from the problem statement. One possible reason is that
the unit was overdesigned in order for start-up of this unit or another unit. Current operation
probably takes place with one burner or a bank of burners not operating. It should be an easy
matter to bring the new burners on line and achieve this level of scale-up. We assume that the
unit can be scaled linearly, i.e. 35% more flow can be processed through the heater with the same
exit temperature of545°C.
IMaximum Scale-Up =1.35 or 135%
1.2.2 Jet Mixer. J-601
This mixer is nothing more than a restriction in the line to provide added pressure drop and
hence ensure good mixing. We need to control the ratio ofpropylene and chlorine feeds to the
reactor in the same ratio as in the design case. The only restriction for this unit will be the
maximum operating pressure, and this has already been accounted for in the analysis given
above.
1.2.2 Reactor. R-601 and the Dowtherm Cooling Loop
R-601
P-601
Two Important Issues
(i) What are input variables & what are out-
put/response variables?
mo (ii) How do we reconcile current operation
with design calculations?
(i) The flow system (exchangers, pumps, valves and pipes) exist. Therefore, all attributes
associated with equipment, e.g., pump charachteristics, pipe sizes, heat transfer areas, etc.
are fixed. In an operating system, such as this one, the only variables that we can control
are the flowrates ofDowtherm (using the control valve) and the flow ofcooling water
through E-601. Thus the only fIxed inputs are the temperature ofthe cooling water into
E~601 (30°C) and the equipment specifIcations. The variables shown on the diagram on
the previous page can be categorized as follows.
Fixed (input) - tA (30°C) + equipment parameters
Variable (input - specffied by us) - mo, I1lcw
Output (to be calculated) - TA, TB, tB, and q (heat exchanged inE-601 and R-601)
C,I~

n. How do we reconcile current operation with the design calculations? The following
points are noted from the problem statement.
(a) Current operations are close to design and have been checked recently (Table C.l,
Table C.2, and the PFD are aCcurate reflections ofcurrent operating conditions)
(b) The heat transfer area for E-601as built (2.6 m2) is significantly different from that
used in design calculations (2.1 m2).
The important point to note here is that ifthe design calculations are correct and the equipment
was built to the specifications given in Table C.2, then the current operations could not be clOSe
to design. In practice, a more rigorous set ofcalculations would have been performed prior to
the construction ofthe plant and the actual plant, as it was built, reflects the results ofthese
calculations. The utility ofthe design calculations provided in the problem statement is,
therefore, not in predicting absolute values ofheat transfer coefficients but rather in identifying
and quantifying the relative magnitude ofthe heat transfer resistances. This will be illustrated
further below. From this discussion it should be clear that the Base Case conditions should be
chosen as the current operation.
Scale-Up Calculations for Dowtherm Loop
We use a ratio analysis to compare the scaled-up case (subscript 2) with the base case (subscript
1). We define the following terms in the analysis:
Q = ratio ofthe heat removed from the reactor in scaled-up case (q2) to the heat removed in the
base case (q/).
MD =ratio ofDowtherm A flowrate in scaled-up case (mm) to Dowtherm A flowrate inbase
case (mD/)'
Mew =ratio ofcooling water flowrate in scaled-up case (mew2) to cooling water flowrate in base
case (mew])'
We further assume that the specific heats ofcooling water and Dowtherm A are the same for the
two cases (this will be a good assumption since temperatures do not change greatly). Finally, q
in the reator must equal q inE-601 for steady state operation.
C./-b
E'~eu6-~
~ At.AraCE
- ()

E-601
EN eaG-"'f BALANC~
_~~;..,.,;;;2~(_t_s_-3_0) ~ Q = Mc.w (teo-30) _ (3)
Mcw C.w I ( 4,.0 -30) 0
(
I
---/ eQVA~ )-1_:1.,_ (lF5S.iANL(?
PEl2R>(2M.ANC~ E"G)ATt~;1'" ~ ¥.
11'1'' A-" """ (-r - 'Q =- u:z. n ~lL.H2 _ 1001'-0°'8' 1ot)Mc:~e le-ts-F+30}
U; I' ATt.h. ~50 n (~) '3;.,.0
1-'- + :- ( ' TA-~t'
Q= 2M~'~ 2.Mc.~fJ Ts-te-1A--30 }
~D .. tt' t(Ts-tsJ/CTA-30J) -~4-)
From the above we see that we have 4 equations in 6 unknowns (Q, MJ> Mew> TB> TA, and tB)' It
is important to realize that we can manipulate or control only two ofthese variables in the plant -
namely M[» and Mew. Intuitively, we should set these as high as possible to obtain the maximum
cooling possible. What are the maximum values possible for these flows (flow ratios)?
Cooling Water Flowrate, Mew
Fromthe problem statement we knowthat the current cooling water velocity is 2.0 mls and that
the maximum cooling water flowrate through any piece ofequipment is 3.5 mls (this limit is to
avoid errosion). Thus we can increase the cooling water flowrate by a factor of3.512.0, that is
the maximum value ofMew is 1.75.
Mew =1.75 (5)
Dowtherm. A Flowrate, MD'
In order to determine the maximum increase in Dowtherm A flowrate it is necessary to use the
system curve with the pump curve provided (Figure C.2). First, locate the current operating
condition on the pump curve. (review Chapter 12, section 12.2 for more details).
At base case conditions Dowtherm A flowrate = 4.62 kg/s = 108 gpm. From design calculations
M'cv:;;; 12.4 psi }
M'friction = 9.9 psi TotalAP =22.3 psi -this point lies on the pump curve at 108 gpm
We note that at a flow of108 gpm the frictional pressure drop = 9.9 psi. Also for a closed loop
there are no hydrostatic head differences so that system curve looks like the following.

Pressure
(psi)
We consider three possible cases
system
curve
108 gpm Flowrate (gpm)
(i) Operate single pump with control valve wide open
(ii) Operate two pumps in series
(m) Operate two pumps in parallel
We can also reconfigure the reactor tubes in parallel which shifts the system curve downwards.
For this case the flowrate through each bank: oftubes would be 113 ofthe flowrate through the
pump, and the equivalent length ofpipe would be 1/3 ofthe original reactor coil.
Now I::.Pfriction, R-6()l, 2 =I::.Pfriction, R-60I, 1 (1/3)2(1/3) =APfriction, R-601, I (1127) =2.9/27=0.107 psi
For this case the base case frictional pressure drop =7+0.107 =7.107 psi., down from 9.9 psi
when reactor tubes piped in series:
The pump curves and system curves for all the cases given above are shown in the figure on the
next page. The maximum flows ofDowtherm A for all combinations ofcases are given by the
intersections ofthe appropriate system and pump curve- which corresponds to the case when
the control valve is wide open. The scale up factors for each case are listed below:
Case Point on Gra)!h Flowrate MD
Reactor Tubes in Series (next page) (gpm)
1 pmnp 1 137 1.27
2 pumps in series 2 150 1.39
2 pumps in parallcl 3 170 1.57
Reactor Tubes in Parallel (next page) (gpm)
1 pump 4 146 1.35
2 pumps in series 5 153 1.42
2 pumps inparallel 6 196 1.81
Clearly ifwe want to maximize flow through the loop we should operate bothpumps in paralleL
The valving will already exist since spared pumps are piped inparallel (see the P&ID in Figure
C. -8

70
65
60
P-601 AlB'w0- 55
ci. Two Pumps in Series
E 50
/::s -
a.. 45
(/)
(/)
400.....
~ 35
Q)
30(/)
0:: 25
~
::s 20(/)
(/)
Q)
15.....
a..
10
5
0
0 20 40 60 80 100 120 140 160 180 200 220 240
Flow of Dowtherm ATM (gpm at 350 °C)
C. -9

1.7). Clearly running both pumps in parallel with the control valves wide open leaves the
operators with nothing to control reactor upsets with. Although this situation would not be
acceptable for long term operations the necessity to increase production as much as possible
might provide the incentive to operate like this for a short time. Even if this operation were
deemed unsafe it does allow a maximum scale-up level to be assessed.
Maximum Scale-up for Reactor Loop
For 2 pumps nroning in parallel and the reactortubes piped in series we put MD = 1.57 and Mew
=1.75 into equations (1) - (4) and solve to give Q = 1.34. Thus a scale-up of approx. 134% is
possible. Ifwe rearrange the reactor tubes in parallel then the form ofequation (2) must be
modified to take account ofthe reduced flow through each tube bank as follows:
... ()-' ..-c·g ')-
(V:! (4.,?O3.n O'!>-;- b .SH+:Sx 0-4 b/3,) (,a-TA) --(;2A)
~Ob ') (33·~) n S'5i~ -~ t
l. s.t- 'E~.}
Solving equations (1), (3), and (4) along with the modified equation (2) above gives us Q=
1.337. Thus a scale-up of134% is possible for this configuration. .From this we see that the
maximum scale-up for the reactor cooling loop is 134%.
IMaximum Scale-up = 1.34 or 134%
1.2.3 Waste heat Boiler. E-602, and the Crude Allyl Chloride Cooler, E-603
We see from the PFD that there is a level controller on the BFW level in E-602. However, in the
problem statement it states that all the heat transfer tubes are covered by BFW so that there is no
immediate fix that we can implement. Note: that if the level in the exchanger were say only
covering 90010 ofthe tubes we could have immediately increased the amount ofheat removal by
11% by simply covering all the tubes.
In addition, another "quick fix" would be to reduce the pressure in the steam header causing a
lower boiling temperature and hence increasing the AT driving force in E-602. However, since
this steam is probably being used elsewhere in the plant, a lower steam pressure would probably
cause problems elsewhere in the plant.
Having eliminate some ofthe "easy" fixes we must concentrate on the heat transfer analysis.
Analysis ofE-602
Q=UAATlm this applies to both zones (desubcooling and boiling)
Note that since all the heat transfer resistance is on the tubeside (process stream is a gas) we can
scale U by (mprocess)O.8. Since we are currently operating at design conditions, that is the exit
temperature from E-602 is 2000C, we conclude that as the process flow increases so will the exit
temperature.
c..-0.

Ofcourse, as the exit temperature from E-602 increases this will directly impact the performance
of £-603 (also currently operating under design conditions, that is exit temperature is 50°C).
However, for E-603 we can increase the cooling water flowrate, although this does not change
the heat transfer coefficient (again we assume all the resistance is on the process gas side), it
does change the llT driving force since the exit temperature ofthe cw decreases.
The equations describing the performance ofE-602 and E-603 are given below.
E- bo2 ~.2. =lONE'S
T .1 ~ .II. 51 .QV
f
bO?
~,.,cet) ~It A
C;wE;N PI2.ESSUQ.~
CW
Ts.1---....
uSE S..,SSt.(LlPT:Z Ruz sc:.frl.t:"" -v(> c..ASE" $ S.>SSt.fZ.lpT 1 =Oa SAse t~E'S(,rI) lAS.£.
fATLO OF- S1l?M- NWS. Msa / rns ,::. Hs . (UI.'fU) cF- {>QOC~S R.owS IVI1.!Mp,
::: Hp . 2A-itu 0 ~ t+f'M ~sPt:"'Vt1>, 't.:tI"r-,': Q .
SNE'Q&-7 s.A-t.AN~ ON ·L?-bl)'l.
Q - 'h ~ ~ ~ MP, (SII-Tc) ct .=;:>
'h MS, ('tp, (sl-.lOo) PI

Q = ct-:r'l :: Ns --(t-{),.8 (~I2) (Tc -T) ~70) -.L
%, NP, AI-I '" (Tc. -£:0) Q3.b
~po,e. (' (t>-bO)
AI,;- 1'2' 4--b (Yl.
---Go)
)N"NOWNS Hp ) H s ) P1Iz1 lI;l, JTc)Tt> a5 6"~~rUl.S
:. 6o",c"", Np (S,t.ALE- up ~.c:rup.)we- ~ SCL.J€".
E-6a3
30
E Ne:Yl&ry SA'-AN(£
Q:= ~::: fV'P2 Cp, 2 IT~-15') ::: Mc.w"Z CPc. "2 (t«:.-30')
9.-, rrP, PPI (2ot>-5a') (4-0 -3 0)
Paz.R:ICZMANC( £Qt'
0. ~ =Mp eTc-Ts) = (f1I'. )""8 eTc-t.. - Ts ~30)1- -(12)
eh SO MP, '" (Tc-t:.'-) b1·33
1'"'5-31:1 ,"
~ we- INCUJ)€" Nf=(). H'lotl t:-oa2 1'"-k:N Fo~ fi oWEN Hp WE'lt.NOv-J
C _'. ~(NlN6- UNtNOWfS AA€ Ts, t (.. *Hc.ON 2s€T 1t+lS~·1S.
tSOl-Ve
C.-IZ'

By solving the equations for E-602 and E-603 we can estimate the value ofthe exit temperature
from E-603 as a function ofscale-up. Assuming that we can increase the cooling water flow by
1.75 (same as in reactor case) we find that we can increase the process flow by 1.06 or 6% and
still maintain the desired exit temperature of50OC. A plot ofexit temperature from E-603 as a
function ofscale-up is shown below. For a 34% increase in process flow the exit temperature
increases to 53.00C. Ifthis temperature change can be accommodated down stream then 34%
scale-up is possible. Otherwise only a 6% scale-up can be achieved.
56
55
()
0
54
U)
E
ttS 53
~.....
en
-520
~
::::J..... 51~
Q)
0-
E 50
~
49
48
Scale-up =1.06 or 106%
[iftemperature ofStrearn 5 must =50°C]
Exit Temperature from E-603 as a Function of Scale-Up
0.9
----
no increase in cooling water flow
1.0 1.06 1.1 1.2
maximum cooling water flow
I
I
I
I
I
I
I
I
I
I
1.3 1.4
Scale-Up Factor (Process Flowratel Design Flowrate)
1.5
We conclude that scale-up is limited to 106% ofcurrent operation ifTs is limited to 500C but can
be increased to 134% (our previous maximum scale-up) ifTs.can be increased slightly.

1
2. Long Term Economic Improvement
Look at variable operating costs.
Raw Materials Unit Cost
Propylene $ 0.47/kg
Chlorine $ 0.18/kg
Propylene Recycle $ 0.47/kg
(100% recovery ofproylene in Stream 5)
Total
Yearly Cost
(3187)(8000)(0.47)= $ 11.98xl06
(1399)(8000)(0.18)= $ 2.01x106
(2439)(8000)(0.47)= ($ 9.17x106)
$ 4.82x106
The raw material costs are usually the biggest cost for a process. The above estimate is
optimistic since it is assumed that all the propylene in Stream 5 is recovered and recycled. Ifthis
were not the case then a recommendation to improve separation of propylene would be
appropriate. The overall conversion ofboth raw materials is very high. However, the selectivity
to allyl chloride is relatively poor with 87.3 % ofthe propylene going to allyl chloride and 79.0%
ofthe chlorine going to allyl chloride. These yields might be improved significantly by using a
catalyst and this would be an excellent suggestion (note additional kinetic infonnation would be
required to quantify the benefits ofchanging to a catalytic process).
Utilities
Natural Gas
CW
Electricity
LPS credit
$2.5/GJ
$ 0.16/GJ
$0.06/kW-h
$ 3.17/GJ*
(4000x106)(8000)(2.5x10-9)/0.9=$88,900
«2188+102S)x106)(8000)
(0.16x10-9)
(2.5)(0.06)(8000)
(2850x106)(8000)(3.17x10-9)
=$ 4,100
=$ 1,200
=($72,300)
• the cost ofBFW is not included since it is assumed that all the steam produced in E-602 will be
returned as condensate.
Although the total utility costs (S94,200/yr) are relatively small compared to the raw
material costs, significant savings may still be obtained.
2.1 Ways to Reduce Utility Costs
2.1.1 Heat Integration
(a) Make more Ips, mps, or bps in Dowthenn A loop instead ofusing cooling water in E-601.
This is a good idea so long as we can sell the extra steam. Ifwe can sell it, then it is unlikely that
we will be able to get anything more than the equivalent fuel credit.
(b) Reduce the natural gas usage in H-601 by preheating propylene.
C.-Ilt

H-601
E-651
R-601
25 349 r--..··--,
~ i 350
I Ipropylene i
400
f !, ,
i I
*
! I
--chl-or-in-e------------- L-
r-' P-601AIB
T
350
25
T-0 Diagram for New Exchanger E-651
400
349
Q
LT1m = (325-51)/ln(325/51) = 148°C
U~60W/m2oC
A= Q/(UMTlm)
=2188xl()6/3600/148/60/0.95
A =72.0 m2 using area we can estimate the fixed capital investment
MOe = SS, Shell and Tube floating head, Operating Pressure = 11.0 bar.
FC! =CnF $ 88,000
Savings in natural gas =(2188xl06X8000)(2.5xlO-9)/0.9 = $ 48,600
EAOC= CTM (0.15)(1.15)5/(1.155-1)-48600 = $ -22,350/yr
A negative EAOC means a positive savings - implement this scheme.
Many other heat savings schemes possible but this may be easiest to implement.
2.1.2 Replace control valves with turbines/expanders
We can not get much work out ofstreams and turbines are expensive. Propylene flow is
too small and the pressure drops are too small.
C. - 5

3. Observations from Plant
3.1 Pump makes a whining noise
3.2 Insulation is falling off
3.3 Relief valve onE-602 is leaking steam
Students should address these points ifonly to bring them to someone's attentioIL Often we find
students do not make any comment about these observations - when asked they say that they
noticed them but couldn't figure out what the reasons could be so they ignored them!! Must
report these potential problems to your supervisor - at the very least!
3.1 Pump makes a whining noise
- Probably a bad bearing. Recommend maintenance crew takes it offline and performs
inspection and needed repairs.
- These repairs should be carried out before scale-up starts since both pumps may be required
to operate in parallel.
- The noise is almost certainly not caused by cavitation
The inlet to the pump is at 350°C at which Dowthenn A has a vapor pressure of5.2 bar. The
minimum pressure ofthe stream entering P-601 is about 10.9 bar which gives a NPSHA=10.9 -
5.2 =5.7 bar. No NPSHR is given for this pump but typical values are given in Chapter 9 and
none are greater than about 0.6 bar. Therefore, very unlikely that pump is cavitating.
3.2 Insulation is falling offline from reactor to E-602.
- We have a small additional heat loss from the system and a very small loss in steam
production from E-602
It is possible that insulation was taken offdeliberately. Ifpipe had a long horizontal run then
the metal could creep and the pipe might begin to sag. By removing the insulation the wall
temperature ofthe pipe would drop significantly and the sagging would cease!
There is no information given as to whether the pipe is sagging but it is worth checking
before we try to reinsulate.
Personnel safety may also be an issue. However, this is only a concern ifan operator could
touch the pipe - this should also be checked.
3.3 Steam leaking from safety reliefvalve (RV) on E-602
Possible causes include:
(i) Some dirt is stuck in the valve seat causing a leak. Remedy - check and fix.
(ii) RV is set to too Iowa pressure. Remedy - check and reset to correct pressure.
(iii)Shell side ofE-602 is at 8 bar (max.. op. press.) - two possible ways that this could occur:
(a) valve on steam line is closed and BFW pump is pressurizing E-602. Remedy - check and fix.
(b) LPS header is at >8 bar and is pressurizing E-602. Remedy- check and fix.
Since no information is given on the steam at the plant, ie., pressure or flowrates have been
fluctuating, (i) and/or (ii) are the most likely causes.
C.-6

Project C.2 - New 20,000 metric tons per year Allyl Chloride
Facility
Look at the base case process, which should be taken as the existing Beaumont facility scaled-up
to 20,000 metric tons per year. The economic breakdown ofthe facility is as follows:
Total utility cost, CUT =$ 3.42x106 per year
Cost ofoperating labor, COL = $ 1.08x106 per year
Raw material costs, CRM = $ 9.43x106 per year
Waste treatment cost, CWT =$ 0.26x106 per year
Grass roots capital cost ofplant, CGR = $ 15.39x106 per year
Annual Revenue, R= $ 37.98x106 per year (34.3 from allyl chloride, and 3.2 from HCI)
Assuming the following: Working capital = $ 3.08x106
Cost ofManufacture, COMd = $ 21.83xl06
i=10%, t=O.15, start-up =2 years, plant life =10 years
Depreciation = MACRS (20, 32, 19.2, 11.52, 11.52, 5.76)
We get a NPV = $ 56.3x1Q6 (note: NPV~$30x 106 when t=Oo4) see attached spreadsheet
Let us analyze the economics ofthe plant - start with the biggest costs and work our way
down. This follows the Pareto analysis discussed in Chapter 19.
Raw Material Costs ($ 9043xlQ6 per year)
Overall conversion ofchlorine ~ 100% } Cl 1 all I will
. ear y sm osses occur
Overall conversIon ofpropylene ~ 100%
Despite the filet that the raw material cost is the highest operating cost there is nothing we
can do to improve the overall utilization of the raw materials, i.e., conversion cannot be
increased. However, the raw material usage could be improved by improving selectivity.
From problem statement we have little control over this element of the design and thus
the raw material usage is already optimized. However, suggestions regarding finding
alternative reaction pathways (e.g. catalytic) are to be encouraged.
Utility Costs ($ 3042xlQ6 per year)
Largest users ofutilities are
Refrigeration Compressors =$ 2.66xlQ6 per year (78% ofCUT)
Propylene Recycle Compressors = $ O.21x106 per year (6% ofCUT)
Fired Heater =$ 0.17x106 per year (5% ofCUT)
=> Look at refrigeration loop
Ct 2--(

SOLUCION Manual de análisis, síntesis y diseño de procesos químicos , tercera edición

Interesting Points about the Refrigeration Loop
/"
./
../
.,.../
Ii
1
t
!
j
!
f
C-602A E-616 C-602B I E-617
...,.-/.........
---~......,i....
....-
Why is thiS-set at 20 bar?
I!
I!
_....-'
Because this is the lowest pressure at which we
can condense propylene with cooling water.
E-613
~
, Why do we not vaporize all the propylene in throttling process?
1
i
II
Because these exchangers have large duties and are used to condense overhead products
in distillation columns. Thus it is better to vaporize cold liquid propylene (at constant
temperature) than to use cold propylene vapor.
I Why do we use a SoC temperature approach for E-604 and E-606? Because there is a
j trade-off between compression costs (to cool the propylene below current low
I temperature) and heat exchanger costs (heat transfer area). Since the electric utility costs
'
I for compression are so high, the use ofa SoC temperature approach (instead ofthe nonnal
i 10°C minimum value) is justified.
I Why is the pressure at the inlet to C-602A set at 0.45 bar?
Saturated liquid @
obar - upstream
ofvalve
Log
Setting this pressure
sets the temperature ~
0.45 bar t--'I-------___..
C z-?

We can optimize the refrigeration loop a little bit but we can't reduce the power requirement by
much. What we need to do is to increase the temperature in the process in order to reduce the
load on the refrigeration loop. Ifwe look at the PFD for the process we see that we need a low
temperature"in the distillation column condensers (namely in T-601 and T-603) and a low inlet
temperature to T-601. The reason for this is so that we can form an L-V mixture. If we refer
back to Chapter 8and look under the tables ofspecial concerns for separators, we see that either
low temperature or high pressure may be required to obtain an L-V mixture suitable for
distillation. Therefore, pressurizing the columns should have a similar effect as using a low
temperature in the condensers. We can pressurize the columns in several ways:
(1) The conservative method is to compress Stream 5.
(2) Alternatively we may pressurize the whole front end of the process by pumping the feed
streams (propylene and chlorine) to a high pressure.
Alternative (1) is conservative because we do not change the pressure ofthe reaction section of
the process and hence the selectivity will not change.
Alternative (2) is potentially cheaper in utility costs since we have eliminated the compressor but
the front-end equipment, e.g., R-601, H-601, E-602 and E-603, must be designed to withstand
higher pressures and will be more costly. Additionally the pressure will change considerably and
the selectivity may change. Although no reaction kinetics are given consider the effect of
increasing pressure on the competing reactions assuming elementary kinetic expressions.
C3H6 +C12 -+ C3I:IsCI + HCI
C3H6 +Cl2 ~ C3HsCl +HCl
C3H6 +2Cl2 ~ CjH4CI2 +2HCI
C3H6 +3Cl2 ~ 3C +6HCl
-rprop = klCpropCCI
-rprop = ~CpropCCI
-rprop =k3CpropC~1
-rprop = k4CpropCti
The effect of increasing the pressure can be seen in the effect that the pressure has on the gas
phase concentration. Assuming ideal gas behavior we have CocPIRT, therefore as pressure
increases the con{'.entration increases proportionally. Looking at the expression for selectivity
we can see that the denominator increases more with increasing pressure than does the
numerator. Thus the selectivity is expected to decrease with increasing pressure. Thus
Alternative (1) may be better.
Why compress Stream 5 and by how much?
It can be argued that little is to be gained by pressurizing Stream 5 since we are replacing one set
ofcompressors (refrigeration compressors C-602A&B) with another compressor(s) in Stream 5.

Also if we do add compressors to what pressure should we compress Stream 5? This is the
subject ofa parametric optimization (see Chapter 19) but let us say that for comparison we will
choose to compress Stream 5 to a pressure which will allow the refrigeration loop to be
eliminated, i.e., so that we can use cooling water in the condensers. This pressure is
approximately 30 bar.
Look at two compressor schemes:
New Compressor in Stream 5 Refrigeration Loop
C-651 (new) C-602A C-602B

2 bar, 95.6 kmoVh, 50°C 30 bar 0.45 bar, 602.5 kmoVh, -46°C 20 bar
If we compare the theoretical power for C-651 to that of C-602 we get approximately the
following relationship (assuming a single stage ofcompression):
W2 = m2T2 [(P2/ 1D2 -1] = (95.6)(323) [(15)°·286 -IJ =0 135
wl ml1i [(P2/ IDf -IJ (602.5)(227)[(44.4)°.286 -1] .
Clearly the option of placing a compressor(s) in Stream 5 to replace the refrigeration
compressors is a good idea
What are the Consequences ofIncreasing the Pressure in Stream 5 to 30 Bar on the Separation?
Obviously we have chosen 30 bar pressure so that we can eliminate the costly refrigeration loop
and use cooling water in the overhead condensers. However, as the top temperature in the
columns increase so do the bottoms temperatures. For the first column, T-601, operating at 30
bar the bottom temperature is approximately 200°C compared to 45°C previously. The question
now arises as to whether allyl chloride is stable at this temperature. The following is a quote
from the Kirk-Othmer encyclopedia, Vol 2, pl06, "Allyl compounds are normally stable at room
temperature even in the absence ofinhibitors. However, prolonged exposure to air, especially in
the presence of light or heat, can cause peroxide formation. Contamination with either
peroxides, or traces ofcatalytically active transitions metals, can present hazards on heating as a
result of rapid and strong exothermic polymerization." The reason that the Beaumont facility
operated at very low temperatures was to avoid the problems associated with this uncontrolled
polymerization!!
The lesson here is that there are large savings to be made in the operating costs ofthe plant by
increasing the pressure ofthe separations section. However, whenever we change the process
conditions by a large amount we must be sure that we are still in a safe operating region and that
we take into account all the potential effects that these changes may have on the process.

Other Issues
Some other issues which should be considered in the optimized design, but which have less
impact on the overall economics are listed below:
(1) Changing the sequence of the columns. Allyl chloride has the largest fiowrate but is
intermediate in volatility (2 chloro -allyl-dichloro). We could change the sequencing as
follows:
2chloro
2 chloro
allyl
(B)
(A)
allyl
dichloro
dichloro
There appears to be little advantage of(B) over (A) so the current sequence is probably OK.
(2) The :first two columns, T-601 and T-603, appear to be doing the same separation. This is
wasteful since we only require a sharp split between the propylenelHCI and the organic
chlorides. We should just make the first column a little taller (more stages) and improve the
separationthus eliminating the need for T-603 (this is a topological optimization).
(3) Since we take the overhead product from T-601 as a liquid then immediately vaporize it in
E-607 it would seem a good idea to use a partial condenser for T-601 and save on utilities.
(4) Why do we recycle propylene as a liquid? This is a nice option to have when shutting down
the process since we can liquefY the recycle and sent it to storage. However, during regular
operations we just vaporize the recycle. We may want to eliminate C-601A and B or not
operate these compressors during normal operation. Note that ifwe operate the separations
section at above 20bar (T-601 and T-602) we can recycle and liquefy the propylene without
the use ofcompressors!
(5) In the front end ofthe process it is worth looking at heat integration
(i) Make high pressure steam in Dowtherm A cooler.
(ll) Drop the temperature into the fluidized bed. Since the reactor is essentially well mixed,
(at least the solids are) we can feed the reactants into the reactor at temperatures

significantly below the reactor operating temperature. This also reduces the duty on the
Dowthenn A cooling loop since the heat of reaction is now used in heating the incoming
feed.
(iii) The heat from the reactor effluent stream can be utilized more efficiently.
There are probably many more cost saving measures that can be implemented, but the major
ones have been addressed above.

Project 3
This project is concerned with determining whether 50% scale-down is possible for
this process, and how it can be accomplished. One possible solution that some students
have considered as an alternative, but have never staked an entire project grade on is to
construct a tank or tanks to store 50% of the product, and to continue to operate the
plant at full capacity for one-half year. An atmospheric pressure, carbon steel tank to
hold 40,000 metric tons of phthalic anhydride (about 40,000 m3) would cost between 2
and 3 million dollars.
Parts of this problem are given as end-of-chapter problems in the text. One challenge
for students is to recognize that determining the conditions for 50% scale-down is
equivalent to defining and solving several of these end-of-chapter problems. A second
challenge is to recognize that, in the context of a chemical process, these end-of-chapter
problems are coupled. The solution to one affects another, and in many cases,
simultaneous solution is required.
Problem 22-13 asks the students to· detennine potential bottlenecks to the scale
down. Detailed calculations not presented elsewhere are presented in that problem.
One bottleneck is the feed section. This is partly due to poor placement of the only
control valve after the mixing point. We call this a caricature problem because it
illustrates the point that it is far better (and standard practice) to place one control valve
after the pump and one either before or after the compressor. The result of having only
one control valve is that only one air flowrate is possible for 50% reduction in
naphthalene. This assumes that there is only one impeller and shaft available on this
compressor, i.e., it only operates at one rpm. This is discussed and analyzed in Section
1~.4. Because the total flow to the reactor is reduced to about 1/3 of the original flow,
maintenance of fluidization in the reactor is an unknown, and another potential
bottleneck. Insufficient information is given in the problem, and what we want is for
students to raise the question ofmaintenance offluidization.
Another bottleneck is the distillation columns. They must be scaled down by 50%.
The coupled heat exchanger-column temperature and pressure analysis, as shown in
Section lQ.2, is presented with Problem .l2.13. Since sieve trays are involved, weeping
may be an issue. A method for avoiding weeping is to increase the reflux ratio in each
tower. Another issue is the low pressure of each tower. Because of the vacuum
conditions, the required pressure increases are significant on an absolute scale. This
results in a significant density increase which could also promote weeping, since the
column diameter would be much larger than it would need to be.
It is also necessary to analyze the molten salt loop to detennine how much reduction
in molten salt is required. It is not true that the molten salt flowrate reduces by 50%. In
fact, the reduction is to 45% ofthe original value. This solution is presented in Problem
CrS-(

Iq .:2.. In this solution, the energy balance and heat exchanger equations for the reactor and
for the molten salt cooler must be solved simultaneously using the methods developed in
Section 1<:.5. This is a good example ofhow analysis of a process involves coupled end-of-
chapter problems. Analysis ofeither the reactor or the molten salt cooler involves solution of
two simultaneous, non-linear equations. In the context ofthe molten salt loop, these two
problems become coupled and the result is a set offive non-linear equations, three ofwhich
are coupled.
When we assign this problem to students, the best projects analyze one or two ofthe above
bottlenecks well but include nothing about the others. Since analysis ofthe molten salt loop
is specifically requested, all students attempt to solve this problem. As might be expected,
there is often difficulty in correctly analyzing the simultaneous equations involved. One
common error is not understanding that the fluidized bed can be assumed to be at a constant
temperature and not accounting for this in the heat transfer analysis for the reactor. This
results in using the reactor feed temperature in the expression for j}.Tlm.

Proiect 4
Let us begin the analysis ofthis problem by examining the reaction network:
maleic anhydride
5
y
o-xylene ~ phthalic anhydride
~
CO2
Intuitive examination of the activation energies suggests that the lowest allowable
temperature will maximize the selectivity for reaction lover reactions 3 and 4. Once
reaction 1 is selected, maximization of the selectivity for phthalic anhydride occurs at
low conversions because reactions 1 and 2 are of the form A ~ B ~ C. Figure E14.3
illustrates this concept. More details are presented in Problem 14.3. The idea that a low
conversion of reactant is a desirable result is always disconcerting to students. In fact,
when we assigned this project, the students who arrived at the conclusion that low
conversion was an optimum solution did so because of exhaustive optimization not
because of the above analysis! The optimum conversion of o-xylene found was in the
30% to 40% range.
Obtaining the above optimum depends upon other topological choices having been
made. One involves whether to operate above or below the flammability limit for 0-
xylene. Due to the enormous cost of the air compressor, operation above the
flammability limit is chosen to minimize the amount of air needed.
The large compressor cost also affects the reactor design. The lowest operating
pressure for the reactor without requiring additional compression before the switch
condensers is optimal. It is also a good idea to arrange multiple reactor units in parallel
(rather than have one large unit) in order to minimize the pressure drop and save on
compression costs.
Because of the low conversion in the reactor, an additional distillation column is
needed to separate and recycle the unreacted o-xylene. In this sequence, the order of
decreasing volatility is o-xylene, maleic anhydride, phthalic anhydride. The question of
distillation sequencing arises, and the optimum, by a very slight margin, is removal of 0-
xylene for recycle first, followed by the maleic/phthalic separation. However, this is such
a close call that slight changes in constraints could alter this result. In terms of
distillation column design, vacuum conditions suggest the possibility of a design
commonly used in practice, but which seems unusual to students. Because of the large
changes in absolute pressure in the column, the gas density is much higher at the bottom
where the pressure is higher. Therefore, the design diameter is smaller at the bottom. A
(.4-1

typical vacuum column for this situation has a larger diameter upper section and a
smaller diameter lower section with the two sections swaged using a frustum.
+- upper section
( frustum
( lower section
Another possibility is to have the same diameter for the entire column but to have less
fractional active tray area in the lower section.
Heat integration is always a good way to reduce costs. However, we always raise
the question of process start-up when a solution with significant heat integration is
presented. The only significant heat integration possible is preheating the boiler feed
water with the reactor effluent. Other heat transfer considerations include placement of
the fired heater. Some solutions find it more economical to place it after the mixing
point in order to save on steam in the other feed heat exchanger. Other solutions
eliminate all other feed heat exchangers and only use a fired heater after the mixing point.
C. .t.j -7

Project 5
Most of the solution to this problem is in Section'l'lA. For the information of the
instructor, the observed "off-spec" conditions in the problem were obtained by running a
simulation with 10 wt% propane impurity at the same total C3 feed mass flowrate, the
conditions that would exist ifthe excess impurity were not observed.
In terms of the troubleshooting aspects, most students can come up with one reason
for the observations. However, most students are happy to stop once they have one
possible solution. OUf goal is to force students to consider as many alternatives as
possible. The solution in Section~~.4 includes numerous alternatives, most of which
cannot be distinguished without rigorous process simulation and/or more process
information. Therefore, students must begin to learn to analyze problems with
incomplete information.
Students are also asked to suggest temporary short-term modifications to return the
process to normal production. A common suggestion is to increase the reaction
temperature. However, as the analysis in Section 't'l..!f. shows, significant temperature
increases may not be possible. Increasing reaction pressure, which is a definite
possibility, is not often suggested because students forget that increasing pressure for a
gas-phase reaction also increases concentration and reaction rate. Given the limitations
on increasing the temperature, some combination of increasing pressure and temperature
may be the best alternative.
The problem with pump cavitation is another "caricature" problem. From the
analysis in Section ~?. ~. Z, it is seen that the propylene feed pump cannot be cavitating.
The observed noise is most likely due to a worn bearing. However, the benzene feed
pump could possibly cavitate ifthe recycle stream temperature were increased slightly by
altering process conditions to compensate for other problems. This analysis is presented
in Problem 2.2.11.

Project 6
We begin by analyzing the reaction network:
C3H6 + C6H6
propylene benzene .. cumene
C3H6 + C9H12 ~ C12H18
propylene cumene p - diisopropyl benzene
As shown in Problem 14.4, low temperature and excess benzene improve the selectivity
for cumene over p-diisopropyl benzene. The actual values for these parameters require
further optimization.
At low temperatures, the reactor increases in size and cost. However, the fired
heater capital and operating costs decrease; and, since a second distillation column is not
needed, there are additional savings in the separation section. However; at the lowest
possible temperature, the reactor size ,begins to dominate. Therefore, the optimum
conditions appear to be a reactor feed temperature in the 330°C range, with a
benzene/propylene mole ratio just below 2. The temperature of 330°C is obtained if the
reactor is treated as isothermal (inlet T = outlet T, but with a temperature spike in
reactor). An even better solution is to feed to the reactor as low as 300°C, to take a
temperature rise in the reactor, and to use the reactor effluent to preheat the reactor
feed, thereby reducing the load on the fired heater. An even more sophisticated solution
is to use staged adiabatic packed beds with intercooling by the feed stream before
entering the fired heater. This also reduces the load on the fired heater. At various
times, all ofthe above alternatives have been developed by different students and student
groups.
While the heat integration discussed above can save significant capital and operating
costs, the issue ofprocess start-up is a concern. We always ask students about this when
solutions with significant heat integration are presented. Since a fired heater may be
required for start-up, ,if one is installed, it would then might more economical to use it
for part ofthe reactor pre-heating rather than let it sit idle and install an "integrated" heat
exchanger for the reactor pre-heating. Analyzing the economics uf this situation is
beyond the scope ofthe economic analysis presented in Section 1.
One time when this problem was assigned, a 20°C maximum temperature rise in the
reactor was a constraint. The result was operation at temperatures above 350°C to
increase the temperature driving force with the 254°C boiling water and increased excess
benzene to maintain a high enough selectivity for cumene to eliminate the need for a
second distillation column.

One feature that students often do not consider is the benzene leaving in the fuel gas
stream from the flash vessel. Optimization of the temperature and pressure in the flash
vessel is an important component of the optimization. However, optimum conditions
vary too much to suggest a range of solutions. In most all cases when cooling water is
used in the heat exchanger prior to the flash vessel, over $1 million of benzene is wasted
as fuel gas. Addition of a second heat exchangerlflash with refrigerated water can
recover on the order of 75% of this benzene. Use of refrigerated water is usually
minimized due to its high cost. In this case, however, very little is needed and the
benzene savings far outweighs the cost of the equipment and the refrigerated water.
CI 6--'"'l.-

;.•.....•"
.,....
Base Case for 20,000 tonnes/yr Allyl Chloride Facility (scaled-up Beaumont design)
Year FCI + WC Revenue COMd it {R-COMd-d}{1-t}+d CF CCF DCF CDCF
0
1 -7.695 -7.70 -7.70 -7.00 -7.00
2 -10.773 -10.77 -18.47 -8.90 -15.90
3 37.98 21.83 3.08 14.19 14.19 -4.28 10.66 -5.24
4 37.98 21.83 4.92 14.47 14.47 10.19 9.88 4.64
5 37.98 21.83 2.95 14.17 14.17 24.36 8.80 13.44
6 37.98 21.83 1.77 13.99 13.99 38.35 7.90 21.34
7 37.98 21.83 1.77 13.99 13.99 52.35 7.18 28.52
8 37.98 21.83 0.89 13.86 13.86 66.21 6.47 34.99
9 37.98 21.83 13.73 13.73 79.93 5.82 40.81
~l
10 37.98 21.83 13.73 13.73 93.66 5.29 46.10
11 37.98 21.83 13.73 13.73 107.39 4.81 50.91
12 3.078 37.98 21.83 13.73 16.81 124.19 5.35 56.27
~
~I (N~Y =_ 56.~71
FCI= 15.39
taxation rate, t= 15%
utilities cost 3.42
raw materials cost = 9.43
waste treament cost = 0.26
revenue = 37.98
discount rate, i = 10%
depreciation = MACRS (20,32,19.2, 11.52, 11.52, 5.76)

Interesting Points about the Refrigeration Loop
wocm (i.}'--I[
y-I ]
,
C-602A E-616 C-602B j E-617
,j
............................/ .........
Why is this··~et at 20 bar?
Because this is the lowest pressure at which we
can condense propylene with cooling water.
E-613
V-607
E-606
Why do we not vaporize all the propylene in throttling process?
Because these exchangers have large duties and are used to condense overhead products
in distillation columns. Thus it is better to vaporize cold liquid propylene (at constant
temperature) than to use cold propylene vapor.
Why do we use a SoC temperature approach for E-604 and E-606? Because there is a
trade-off between compression costs (to cool the propylene below current low
temperature) and heat exchanger costs (heat transfer area). Since the electric utility costs
for compression are so high, the use ofa 5°C temperature approach (instead ofthe normal
10°C minimum value) is justified.
Why is the pressure at the inlet to C-602A set at 0.45 bar?
Saturated liquid @
20 bar - upstream
ofvalve
Log
Setting this pressure
sets the temperature ~
0.45 bar .........r------'__,.
C, t- r

We can optimize the refrigeration loop a little bit but we can't reduce the power requirement by
much. What we need to do is to increase the temperature in the process in order to reduce the
load on the refrigeration loop. If we look at the PFD for the process we see that we need a low
temperature in the distillation column condensers (namely in T-601 and T-603) and a low inlet
temperature to T-601. The reason for this is so that we can fonn an L-V mixture. If we refer
back to Chapter 8 and look under the tables of special concerns for separators, we see that either
low temperature or high pressure may be required to obtain an L-V mixture suitable for
distillation. Therefore, pressurizing the columns should have a similar effect as using a low
temperature in the condensers. We can pressurize the columns in several ways:
(1) The conservative method is to compress Stream 5.
(2) Alternatively we may pressurize the whole front end of the process by pumping the feed
streams (propylene and chlorine) to a high pressure.
Alternative (1) is conservative because we do not change the pressure of the reaction section of
the process and hence the selectivity will not change.
Alternative (2) is potentially cheaper in utility costs since we have eliminated the compressor but
the front-end equipment, e.g., R-601, H-601, E-602 and E-603, must be designed to withstand
higher pressures and will be more costly. Additionally the pressure will change considerably and
the selectivity may change. Although no reaction kinetics are given consider the effect of
increasing pressure on the competing reactions assuming elementary kinetic expressions.
C)H6 + Cl2 ~ C)HSCl + HCl
C)H6 + Cl2 ~ C)HSCI + HCl
C3H6 + 2Cl2 ~ C)H4Cl2 +2HCI
C)H6 +3Cl2 ~ 3C +6HCI
- rprop =klCpropCCI
- rprop = k2C propCCI
- rprop = k3C propCEI
- rprop =k4C propcl;!
The effect of increasing the pressure can be seen in the effect that the pressure has on the gas
phase concentration. Assuming ideal gas behavior we have CocPIRT, therefore as pressure
increases the concentration increases proportionally. Looking at the expression for selectivity
we can see that the denominator increases more with increasing pressure than does the
numerator. Thus the selectivity is expected to decrease with increasing pressure. Thus
Alternative (1) may be better.
Why compress Stream 5 and by how much?
It can be argued that little is to be gained by pressurizing Stream 5 since we are replacing one set
; ofcompressors (refrigeration compressors C-602A&B) with another compressor(s) in Stream 5.
~.

Also if we do add compressors to what pressure should we compress Stream 5? This is the
subject of a parametric optimization (see Chapter 19) but let us say that for comparison we will
choose to compress Stream 5 to a pressure which will allow the refrigeration loop to be
eliminated, i.e., so that we can use cooling water in the condensers. This pressure is
approximately 30 bar.
Look at two compressor schemes:
New Compressor in Stream 5 Refrigeration Loop
C-651 (new) C-602A C-602B
2 bar, 95.6 kmo1/h. 50°C 30 bar 0.45 bar, 602.5 kmoVh, -46°C 20 bar
If we compare the theoretical power for C-651 to that of C-602 we get approximately the
following relationship (assuming a single stage ofcompression):
W2 = m2T2 [(P2 / }D~ -1] = (95.6)(323) [(15)°·286 -1] =0 135
wl mlll [(P2 11Df-l] (602.5)(227)[(44.4)°.286_ 1] .
Clearly the option of placing a compressor(s) in Stream 5 to replace the refrigeration
compressors is a good idea.
What are the Consequences ofIncreasing the Pressure in Stream 5 to 30 Bar on the Separation?
Obviously we have chosen 30 bar pressure so that we can eliminate the costly refrigeration loop
and use cooling water in the overhead condensers. However, as the top temperature in the
columns increase so do the bottoms temperatures. For the first column, T-601, operating at 30
bar the bottom temperature is approximately 200°C compared to 45°C previously. The question
now arises as to whether allyl chloride is stable at this temperature. The following is a quote
from the Kirk-Othmer encyclopedia,.VoI2, p106, "Allyl compounds are normally stable at room
temperature even in the absence of inhibitors. However, prolonged exposure to air, especially in
the presence of light or heat, can cause peroxide formation. Contamination with either
peroxides, or traces of catalytically active transitions metals, can present hazards on heating as a
result of rapid and strong exothermic polymerization." The reason that the Beaumont facility
operated at very low temperatures was to avoid the problems associated with this uncontrolled
polymerization! !
The lesson here is that there are large savings to be made in the operating costs of the plant by
increasing the pressure ofthe separations section. However, whenever we change the process
conditions by a large amount we must be sure that we are still in a safe operating region and that
we take into account all the potential effects that these changes may have on the process.

.~
Other Issues
Some other issues which should be considered in the optimized design, but which have less
impact on the overall economics are listed below:
(1) Changing the sequence of the columns. Allyl chloride has the largest flowrate but is
intermediate in volatility (2 chloro -allyl -<iichloro). We could change the sequencing as
follows:
2 chloro
2 chloro
allyl
(B)
(A)
allyl
dichloro
dichloro
Tqere appears to be little advantage of(B) over (A) so the current sequence is probably OK.
(2) The first two columns, T-601 and T-603, appear to be doing the same separation. This is
wasteful since we only require a sharp split between the propylenelHCI and the organic
chlorides. We should just make the first column a little taller (more stages) and improve the
separation thus eliminating the need for T-603 (this is a topological optimization).
(3) Since we take the overhead product from T-601 as a liquid then immediately vaporize it in
E-607 it would seem a good idea to use a partial condenser for T-601 and save on utilities.
(4) Why do we recycle propylene as a liquid? This is a nice option to have when shutting down
the process since we can liquefy the recycle and sent it to storage. However, during regular
operations we just vaporize the recycle. We may want to eliminate C-601A and B or not
operate these compressors during normal operation. Note that if we operate the separations
section at above 20bar (T-601 and T-602) we can recycle and liquefy the propylene without
the use ofcompressors!
(5) In the front end ofthe process it is worth looking at heat integration
(i) Make high pressure steam in Dowtherm A cooler.
(ii) Drop the temperature into the fluidized bed. Since the reactor is essentially well mixed,
(at least the solids are) we can feed the reactants into the reactor at temperatures
(, b-j

SOLUCION Manual de análisis, síntesis y diseño de procesos químicos , tercera edición

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SOLUCION Manual de análisis, síntesis y diseño de procesos químicos , tercera edición