Chapter 2
Solution 2.1
Basis for calculation: 100 kmol dry gas
CO + 0.5O2  CO2
H2 + 0.5O2  H2O
CH4 + 2O2  CO2 + 2H2O
C2H6 + 3.5O2  2CO2 + 3H2O
C6H6 + 7.5O2  6CO2 + 3H2O
Reactions:
REACTANTS
O2
CO2
CO
H2
CH4
C2H6
C6H6
N2
Nat. Gas
4
16
50
15
3
2
10
Totals
100
88.5
8
25
30
10.5
15
PRODUCTS
CO2
4
16
15
6
12
H2O
50
30
9
6
10
53
95
If Air is N2:O2 = 79:21
N2 with combustion air
= 88.5 x 79/21 = 332.9 kmol
Excess O2
= 88.5 x 0.2 = 17.7 kmol
Excess N2
=17.7 x 79/21 = 66.6 kmol
Total
= 417.2 kmol
(i)
Air for combustion = 417.2 + 88.5 = 505.7 kmol
(ii)
Flue Gas produced = 53 + 95 + 10 + 417.2 = 575.2 kmol
(iii)
Flue Gas analysis (dry basis):
N2
CO2
O2
N2
409.5 kmol
53.0 kmol
17.7 kmol
480.2 kmol
85.3 mol %
11.0 mol %
3.7 mol %
100.0 mol %
Solution 2.2
Use air as the tie substance – not absorbed.
H2 O
0.05 % NH3
200 m3 s-1
760 mm Hg
20oC
H2O
5 % NH3 © 2009 Elsevier Ltd. All rights reservedNH3
10
Partial volume of air = 200(1 - 0.05) = 190 m3 s-1
Let the volume of NH3 leaving the column be x, then:
0.05
x

100 190  x
0.05(190 + x) = 100x
x
9.5
 0.0950 m3 s-1
(100  0.05)
(i) Flow rate of gas leaving column = 190 + 0.0950 = 190.1 m3 s-1
(ii) The volume of NH3 adsorbed
= (200)(0.05) – 0.0950
= 9.905 m3 s-1
If 1 kmol of gas occupies 22.4 m3 at 760 mm Hg and 0oC,
273
 9.905 
Molar Flow = 
 0.412 kmol s-1

 22.4  (273  20)
Mass Flow = (0.412)(17) = 7.00 kg s-1
(iii) Let the water flow rate be W, then:
1
7.00

100 W  7.00
W = 700 – 7 = 693 kg s-1
Solution 2.3
OFF-GAS
REFORMER
3
-1
2000 m h
2 bara
35oC
H2 + CO2 + unreacted HC’s
At low pressures vol% = mol%
(i) Basis: 1 kmol of off-gas
Component
CH4
C2H6
C3H8
C4H10
mol%
77.5
9.5
8.5
4.5
MW
16
30
44
58
mass (kg)
12.40
2.85
3.74
2.61
 21.60
So the average molecular mass = 21.6 kg kmol-1
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(ii) At STP, 1 kmol occupies 22.4 m3
5
 2000  2 x 10  273


Flow rate of gas feed = 
 156.248 kmol h-1

5 
 22.4  1.013 x 10  (273  35)
Mass flow rate
= (156.248)(21.60) = 3375 kg h-1
(iii) Basis: 100 kmol of feed
Reaction (1): CnH2n+2 + n(H2O)  n(CO) + (2n + 1)H2
Component
CH4
C2H6
C3H8
C4H10
n
1
2
3
4
Amount
77.5
9.5
8.5
4.5
If the conversion is 96%, then:
CO
77.5
19.0
25.5
18.0
 140.0
H2
232.5
47.5
59.5
40.5
380.0
H2 produced = (380.0)(0.96) = 364.8 kmol
CO produced = (140.0)(0.96) = 134.4 kmol
Reaction (2): CO + H2O  CO2 + H2
If the conversion is 92%, then: H2 from CO = (134.4)(0.92) = 123.65 kmol
Total H2 produced = 364.8 + 123.65 = 488.45 kmol/100 kmol feed
If the gas feed flow rate = 156.25 kmol h-1, then
 488.45 
H2 produced = 156.25
  763.20 kmol h-1  (763.2)(2) = 1526 kg h-1
100


Solution 2.4
ROH (Selectivity = 90 %)
RCl
ROR
(Conversion = 97 %)
Basis: 1000 kg RCl feed
Relative molecular masses:
CH2=CH-CH2Cl
76.5
CH2=CH-CH2OH
58.0
(CH2=CH-CH2)2O
98.0
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RCl feed
=
1000
76.5
= 13.072 kmol
RCl converted = (13.072)(0.97)
ROH produced = (12.68)(0.9)
ROR produced = 12.68 – 11.41
Mass of allyl-alcohol produced
Mass of di-ally ether produced
= 12.68 kmol
= 11.41 kmol
= 1.27 kmol
= (11.41)(58.0) = 661.8 kg
= (1.27)(98.0) = 124.5 kg
Solution 2.5
Basis: 100 kmol nitrobenzene feed.
The conversion of nitrobenzene is 96% and so 100(1 - 0.96) = 4 kmol are unreacted.
The selectivity for aniline is 95% and so aniline produced = (96)(0.95) = 91.2 kmol
Therefore, the balance is to cyclohexylamine = 96 – 91.2 = 4.8 kmol
From the reaction equations:
C6H5NO2 + 3H2  C6H5NH2 + 2H2O
1 mol of aniline requires 3 mol of H2
C6H5NO2 + 6H2  C6H11NH2 + 2H2O
1 mol of cyclohexylamine requires 6 mol of H2
Therefore, H2 required for the reactions = (91.2)(3) + (4.8)(6) = 302.4 kmol
A purge must be taken from the recycle stream to maintain the inerts below 5%. At
steady-state conditions:
Flow of inerts in fresh H2 feed = Loss of inerts from purge stream
Let the purge flow be x kmol and the purge composition be 5% inerts.
Fresh H2 feed = H2 reacted + H2 lost in purge
= 302.4 + (1 – 0.05)x
Inerts in the feed at 0.005 mol fraction (0.5%)
= (302.4  0.95 x)
0.005
1  0.005
= 1.520 + 4.774 x 10-3x
Inerts lost in purge = 0.05x
So, equating these quantities: 0.05x = 1.520 + 4.774 x 10-3x
Therefore: x = 33.6 kmol
The purge rate is 33.6 kmol per 100 kmol nitrobenzene feed.
H2 lost in the purge = 33.6(1 – 0.05) = 31.92 kmol
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Total H2 feed = 302.4 + 31.92 = 334.3 kmol
Therefore: Total fresh hydrogen feed including inerts =
334.3
= 336.0 kmol
1  0.005
(iii) Composition at the reactor outlet:
Stoichiometric H2 required for aniline = 300 kmol
H2 feed to the reactor = (300)(3) = 900 kmol
Fresh feed H2 = 334.3 and so recycle H2 = 900 – 334.3 = 565.7 kmol
Inerts in Fresh Feed = (334.3)(0.005) = 1.6715 kmol
 0.05 
Inerts in Recycle (at 5%) = 565.7 
 = 29.77 kmol
 1  0.05 
Therefore, total inerts = 1.6715 + 29.77 = 31.445 kmol
Aniline produced = 91.2 kmol
Cyclohexylamine produced = 4.8 kmol
If 302.4 kmol of H2 are reacted, then H2 leaving the reactor = 900 – 302.4 =
597.6 kmol
H2O produced = (91.2)(2) + (4.8)(2) = 192 kmol
Composition:
kmol
mol %
Aniline
91.2
9.90
Cyclo-hexalymine
4.8
0.52
H2O
192
20.85
H2
597.6
64.89
Inerts
31.45
3.41
4
0.43
Nitrobenzene
921.05
100.00
Solution 2.6
AN
Cyclo
H2O
H2
Inerts
NB
950
10
1920
5640
300
40
H2
5640
Inerts 300
Pressure 20 psig = 1.38 barg
Assumptions: H2 and inerts are not condensed within the condenser.
Temp. = 270oC
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Need to assume a condenser temperature. If cooling water is used as coolant then:
Temp. of the gas at the condenser outlet = 50oC and return the cooling water at 30oC
(20oC temp. difference).
Antoine coefficients (from appendix C):
Aniline
16.6748, 3857.52, -73.15
Nitrobenzene
16.1484, 4032.66, -71.81
H2O
18.3036, 3816.44, -46.13
Vapor pressures at 50oC:
ln( P o )  18.3036 
H2O:
3816.44
323  46.13
Po = 91.78 mm Hg = 0.122 bar
ln( P o )  16.6748 
Aniline:
(From Steam Tables = 0.123 bar)
3857.52
323  73.15
Po = 3.44 mm Hg = 0.00459 bar
Nitrobenzene: ln( P o )  16.1484 
4032.66
323  71.81
Po = 1.10 mm Hg = 0.00147 bar
NB. The cyclo-hexalymine is ignored because it is present in such a small quantity.
Mol fraction =
partial pressure
total pressure
If the total pressure is 2.38 bara (20 psig):
H2O
=
0.122
= 0.0513
2.38
AN
=
0.00459
= 0.0019 = 0.19 %
2.38
NB
=
0.00147
= 0.00062 = 0.06 %
2.38
Total
= 5.13 %
5.38 %
Take H2 and the inerts as tie materials.
Flow (H2 and inerts) = 5640 + 300 = 5940 kmol
Mol fraction (H2 and inerts) = 100 – 5.38 = 94.62 %

mol fraction other
Flow of other components in vapor = 
 mol fraction (H 2  inerts)
H2O
=

 flow (H 2  inerts)

5.13
x 5940 = 322.0 kmol
94.53
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AN
=
0.19
x 5940 = 11.9 kmol
94.53
NB
=
0.06
x 5940 = 3.8 kmol
94.53
Composition of the gas stream (recycle):
kmol
vol %
H2
5640
89.84
Inerts
300
4.78
H2O
322.0
5.13
AN
11.9
0.19
NB
3.8
0.06
Cycl.
Trace
--
Total
6277.7
100.00
Composition of the liquid phase:
Liquid Flow = Flow In – Flow in Gas Phase
kmol
kg
vol %
w/v %
H2
0
--
--
--
Inerts
0
--
--
--
H2O
1920 - 322
1598
28764
61.9
23.7
AN
950 – 11.9
938.1
87243
36.3
71.8
NB
40 – 3.8
36.2
4453
1.4
3.7
10
990
0.4
0.8
121,450
100.0
100.0
Cycl.
Total
2582.3
This calculation ignores the solubility of nitrobenzene in the condensed aniline.
Note: H2O in the recycle gas would go through the reactor unreacted and would add
to the tie H2O in the reactor outlet. But, as the recycle gas depends on the vapor
pressure (i.e. the outlet temp.) it remains as calculated.
The required flows of nitrobenzene and aniline are therefore:
Inlet Stream:
kmol
vol %
AN
950
10.34
Cycl.
10
0.11
2242
24.42
H2O
1920 + 322
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NB
40
0.44
H2
5640
61.42
Inerts
300
3.27
9182
100.00
Total
An iterative calculation could be performed but it is not worthwhile.
Solution 2.7
Basis: 100 kg feed
AQUEOUS
H20
AN
NB
Cycl
23.8
72.2
3.2
0.8
30oC
100.0
ORGANIC
Minor components such as nitrobenzene and aniline will be neglected in the
preliminary balance.
Let the flow rate of aqueous stream be F kg per 100 kg of feed.
Mass balances:
IN
Aqueous
Organic
xA (100 – F)
Aniline
72.2
=
0.032F
Nitrobenzene
3.2
=
3.2 
Cyclohexylamine
0.8
=
0.0012F
0.01(100 – F)
Water
23.8
=
xW F
0.0515(100 – F)
1
301
3.2 
300
301
Clearly the problem is over-specified, as we need to solve for 8 flow rates and we
have 4 mass balances and 5 compositions specified. We can solve for F from the
cyclohexylamine balance and then solve for xA and xW, but the resulting aqueous and
organic stream flow rates will not necessarily sum to F and (100 – F):
From cyclohexylamine balance:
(0.01 – 0.0012) F = 1 – 0.8
F = 22.73
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Hence substituting in the other balances:
IN
Aqueous
Organic
Aniline
72.2
=
0.727
71.47
Nitrobenzene
3.2
=
0.0106
3.189
Cyclohexylamine
0.8
=
0.0273
0.773
Water
23.8
=
19.82
3.980
Total
100
20.59
79.41
So the aqueous phase does not sum to the previously determined value F = 22.73.
If we iterate and substitute F = 20.6:
IN
Aqueous
Organic
Aniline
72.2
=
0.659
71.54
Nitrobenzene
3.2
=
0.0106
3.189
Cyclohexylamine
0.8
=
0.0247
0.794 X
Water
23.8
=
19.71
4.089
Total
100
20.4
79.61
So the phase mass balances are now more or less correct, but the mass balance on
cyclohexylamine does not close.
It is more important that the mass balance closes properly than that the compositions
exactly match those specified in the problem statement. In the case of
cyclohexylamine, the composition in the organic phase is much larger than that in the
aqueous phase, so if we exactly match the organic composition then the error in the
aqueous composition will be large. If we exactly match the aqueous composition, the
error in the organic composition will be small. We should therefore use the aqueous
flow rate predicted, and solve for the organic flow rate by difference, giving:
IN
Aqueous
Organic
Aniline
72.2
=
0.659
71.54
Nitrobenzene
3.2
=
0.0106
3.189
Cyclohexylamine
0.8
=
0.0247
0.775
Water
23.8
=
19.71
4.089
Total
100
20.40
79.60
All phases and components now balance, and the error from the compositions defined
in the problem statement is that the cyclohexylamine in the organic phase is 0.97%
instead of the specified 1.0%. This error is acceptable for design purposes.
[Note: At first glance this problem may just appear to be badly formulated, but it is in
fact typical of many design problems, where the engineer may have redundant data
and must resolve inconsistencies in the data.]
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Solution 2.8
Calculation of the feed mol fractions:
w/w
kg/100 kg
MW
kmol/100 kg
mol %
H2O
2.4
3.05
18
0.169
14.1
AN
73.0
92.87
93
0.999
83.2
NB
3.2
4.07
123
0.0331
2.76
Total
78.6
100
1.201
100
Aniline in feed = 0.999 kmol h-1
With 99.9 % recovery, aniline on overheads = (0.999)(0.999) = 0.998 kmol h-1
Overhead composition will be near the azeotrope and so an aniline composition of 95
% is suggested. (NB: Would need an infinitely tall column to reach the azeotrope
composition)
Water composition in overheads = 100 – 95 = 5 mol %
 5 
So water carried over with the aniline = 0.998   = 0.0525 kmol h-1
 95 
Water leaving the column base = 0.169 – 0.0525 = 0.116 kmol h-1
Compositions:
kmol h-1
mol %
TOPS
AN
H2O
NB
0.998
0.0525
Trace
1.505
95.0
5.0
Trace
100
BOTTOMS
AN
H2O
NB
0.001
0.116
0.0331
0.150
0.66
77.73
22.1
100.5
Solution 2.9
Start by looking at the reaction stoichiometry:
C7H8O2
+
C3H6O2
= C10H14O4
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Stoichiometric equation balances, so no need to worry about additional components
Molar weights:
124
74
198
i) 100 kg/day of guaifenesin = 100/198 = 0.505 kmol/day
If plant yield is 87% then feed rate = 0.505/0.87 = 0.5805 kmol/day
glycidine feed rate = 0.5805  74 = 42.96 kg/day
So
guaiacol feed rate = 0.5805  124 = 71.982 kg/day
ii) Yield after reactor = feed rate  0.938 = 0.938 (100/0.87) = 107.82 kg/day
So losses in evaporator = 107.82 – 100 = 7.82 kg/day
iii) It is not necessary to read the patent to understand this part. The clue is that the
catalyst is NaOH. The purpose of the evaporator must therefore be to evaporate the
guaifenesin from the nonvolatile catalyst (which is actually neutralized with acid
before the evaporator). The product that is lost is therefore the product that remains in
the liquid phase in the evaporator.
Possible ways to increase product recovery could include:



Reducing the evaporator pressure
Adding a second stage evaporator operated at lower pressure
Stripping the reactor product using a suitable volatile compound (the patent
says steam is no good, but glycidine could be investigated).
Solution 2.10
It is not necessary to read the patent to answer this question.
i) Start by writing a stoichiometric equation:
C20H21ClN2
NMP-CDHBCP
+ C3H5ClO2
ECF
= C22H23ClN2O2
Loratidine
+X
By difference, X has formula CH3Cl, i.e. is chloroform
ii) Molar weights:
C20H21ClN2
324.5
+ C3H5ClO2
108.5
= C22H23ClN2O2
382.5
+ CH3Cl
50.5
If conversion is 99.9%, then for 16.2g of feed NMP-CDHBCP, 16.2  0.999 =
16.184 g are converted
So required feed of ECF = 16.184 (108.5/324.5) = 5.411 g
Actual feed of ECF = 10.9g , so excess at end of reaction = 10.9 – 5.411 = 5.489 g
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iii) If ECF is not recovered or recycled then 5.489/10.9 = 50.4% is lost to waste
products (spent solvents). Since the selectivity is 100%, almost all of this lost material
could in principle be recovered.
iv) The feed of 16.2 g of NMP-CDHBCP gives a yield of 16.2  0.999 
(382.5/324.5) = 19.076 g of product, so per kg of API we require 200 ml 
1000/19.076 = 10.484 liters of each solvent at each step.
Step
solvent
Reaction
Quench
Wash
Recrystallization
benzene
water
water
isopropyl ether
density
(kg/m3)
879
1000
1000
724
mass
(kg/200ml)
0.176
0.200
0.200
0.145
mass
(kg/kg API)
9.216
10.484
10.484
7.590
In addition to the waste solvent, we also have chloroform produced by the reaction.
Chloroform yield = 50.5 g per 382.5 g of API = 132.0 g/kg of API.
The ECF waste is 5.489 g per 19.076 g product, i.e., 5.489 1000/19.076 = 287.7 g
per kg product.
The trituration step also uses petroleum ether, but the amount of this is much smaller
than the amounts of solvents used. Allow 1 kg per kg product as an initial estimate.
So total mass of waste produced is:
Water
Benzene
Chloroform
Ethyl chloroformate
Petroleum ether
Isopropyl ether
Total
20.968
9.216
0.132
0.288
1.0
7.59
39.2 kg/kg API
Note: this made no allowance for recovery yields on the washing and recrystallization
steps, so the actual waste produced would be higher.
v) 16.2 g of NMP-CDHBCP + 10.9 g of ECF give 19.076 g of product
If 92% of the product is recovered after separation then overall yield = 0.92  19.076
= 17.550 g
So to produce 10kg of API we require:
Feed rate of NMP-CDHBCP = 16.2  (10/17.55) = 9.23 kg/batch
Feed rate of ECF = 10.9  (10/17.55) = 6.21 kg/batch
vi) The reactor should be operated no more than 2/3 full, to allow for swirling of the
contents, vessel internals, etc., so available volume = 3800  2/3 = 2533 liters. Since
the preparation involved decanting the benzene mixture into water, the reaction step
must use only half of the available volume (so that the other equal volume of water
can be added later), so available volume for reaction = 1267 liters
The preparation method describes dissolving 16 g of NMP-CDHBCP in 175g of
benzene (200ml), so the solution is roughly 10 wt%. We have no information on the
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volume of mixing, so as a first approximation we can assume the mixed volume is
200ml. The lab recipe makes a 200ml batch, that ultimately yields 17.55 g of API.
If 200ml gives 17.55 g, then:
1267 liters gives 17.55  (1267/0.2) = 111.2 kg per batch
Reaction time is 18 hours. Estimated times for the other processing steps might be:
Cooling & quench:
2 hours
Decanting
1 hour
Washing & evaporation
2 hours
Trituration
1 hour
Recrystallization
2 hours
Total
8 hours
If all the steps were carried out in the same vessel we would also need to allow time
for cleaning and refilling at the end of the batch, say 2 more hours, giving a total
batch time of 28 hours and production rate of 111.2/28 = 3.97 kg/hr.
If only the reaction step is carried out in the reactor then we still need to allow some
time for filling, emptying and cleaning, say 3 hours, giving a batch time of 21 hours
and a maximum production rate of 111.2/21 = 5.3 kg/hr.
vii) Advantages of carrying out all steps in same vessel:
 Less equipment to clean between batches
 Possibly higher yields during washing and decanting steps
Disadvantages of carrying out all steps in same vessel:
 Could be impractical for trituration
 Could give poor separation during decanting, washing
 Difficult for ice-water quench unless the effect of pouring ice water into
benzene mixture is the same as pouring benzene mixture into ice water (scaleup question).
 Would be difficult to get good recovery in final recrystallization step as
material would coat vessel internals – easiest to send to a crystallizer.
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