1
CN3132
CN 3132
NATIONAL UNIVERSITY OF SINGAPORE
EXAMINATION FOR THE DEGREE OF B. ENG. (CHEMICAL)
CN 3132 – SEPARATION PROCESSES
(Semester 1: 2021/2022)
29 Nov 2021 – Time Allowed 2.5 Hours
Online Examination and Academic Integrity
This is a F2F online examination. You are to abide by NUS Honour Code at all times during the
entire duration of the examination.


Please read the NUS Code of Student Conduct (http://nus.edu.sg/osa/resources/code‐of‐
student‐conduct). You will be required to declare your adherence to the Code of Student
Conduct before each exam.
Any form of communication with anyone other than with the module
lecturers/invigilators is STRICTLY PROHIBITED.
Instructions
1. Section wise distribution of questions is provided below. ANSWER ALL QUESTIONS.
Section Topic
1 to 3
4
5
6 to 7
8
9 to 10
Binary flash distillation
Number of Mark allocation
Questions
8
10 for ten fill in the blanks
/MCQ
6
11 for nine fill in the blanks
Absorption rate‐based
design
Cooling tower
Binary multistage distillation
8
15
Binary absorption
Partially miscible extraction
7
8
10 for eight fill in the blanks
16 for sixteen fill in the
blanks
8 for eight fill in the blanks
13 for eleven fill in the
blanks
2. Multiple attempts are not allowed.
3. You will have to complete the sections in sequence. Once a section is completed, you will
not be allowed to go back to that section.
4. Remember to click “submit” option at the end of the examination.
1
CN3132
CN 3132
NATIONAL UNIVERSITY OF SINGAPORE
EXAMINATION FOR THE DEGREE OF B. ENG. (CHEMICAL)
CN 3132 – SEPARATION PROCESSES
(Semester 1: 2021/2022)
29 Nov 2021 – Time Allowed 2.5 Hours
Online Examination and Academic Integrity
This is a F2F online examination. You are to abide by NUS Honour Code at all times during the
entire duration of the examination.


Please read the NUS Code of Student Conduct (http://nus.edu.sg/osa/resources/code‐of‐
student‐conduct). You will be required to declare your adherence to the Code of Student
Conduct before each exam.
Any form of communication with anyone other than with the module
lecturers/invigilators is STRICTLY PROHIBITED.
Instructions
1. Section wise distribution of questions is provided below. ANSWER ALL QUESTIONS.
Section Topic
1 to 3
4
5
6 to 7
8
9 to 10
Binary flash distillation
Number of Mark allocation
Questions
8
10 for ten fill in the blanks
/MCQ
6
11 for nine fill in the blanks
Absorption rate‐based
design
Cooling tower
Binary multistage distillation
8
15
Binary absorption
Partially miscible extraction
7
8
10 for eight fill in the blanks
16 for sixteen fill in the
blanks
8 for eight fill in the blanks
13 for eleven fill in the
blanks
2. Multiple attempts are not allowed.
3. You will have to complete the sections in sequence. Once a section is completed, you will
not be allowed to go back to that section.
4. Remember to click “submit” option at the end of the examination.
2
CN3132
Section 1
(3 marks)
In an interview for an internship position at the Shell Refinery at Bukom Island, you are
presented with the x – y plot for benzene – 1 butanol binary system at 101.3 kPa and the
corresponding T – x plots for the same system. The plots presented to you are shown in
Figure S1.1. You are asked to answer the following questions on the operation of a two‐stage
flash distillation unit shown in Figure S1.2. Both units operate at 101.3 kPa. In Figure S1.2, A
is benzene and B is 1 butanol.
v
2
1. The purpose of the two‐stage flash distillation process is
.
Choose from the following options:
1 To increase purity of the heavy component in the bottom product;
2 To increase purity of the light component in the top product;
3 To increase flow rate of the bottom product;
4 To increase flow rate of the top product.
Input the number of the option you think is appropriate. For example, input 3 if you
think the answer is To increase flow rate of the bottom product.
2. You are told that the operating temperature of unit 1 is 90oC.
xA1 = 0.35 and yA1 = 0.775 .
3
CN3132
Section 2
(1 mark)
7
In the second phase of the interview at Shell Refinery, you are shown Figure S2.1. You are
asked to carefully look at the operating lines given in Figure S2.1 and answer the following
question.
Two of the three operating lines shown in Figure S2.1 (and numbered 1, 2, 3) represent the
two‐stage flash distillation process shown in Figure S2.2. Choose the correct answer from the
following options:
o Lines 1 and 2 are the operating lines for units 1 and 2, respectively.
o
- Lines 1 and 3 are the operating lines for units 1 and 2, respectively.
o Lines 2 and 3 are the operating lines for units 1 and 2, respectively.
o Lines 2 and 1 are the operating lines for units 1 and 2, respectively.
o Lines 3 and 2 are the operating lines for units 1 and 2, respectively.
4
CN3132
Section 3
(6 marks)
The next stage of the interview is to test your ability to use the operating lines to carry out
some design calculations for the two‐stage flash distillation process. Please be reminded that
in engineering, correct calculations are also important. The unit number is provided in each
operating line to avoid confusion. There is no partial mark for procedure.
Calculate the following using the operating lines in Figure S3.1 and the process schematic in
Figure S3.2.
-
1. Flow rate of L1 = 1571.4
kmol/h.
2. Flow rate of V2 = 306.
3. zA2 =
0.62
kmol/h.
0.38
and zB2 =
z 0.4
=
x=
0.2
L= 30
y 0.62
F = V+ 1
30c0
=
.
3000(0.4)
4. (L1xB1)/(FzB1) = 0.6984 .
5. (V2yA2)/(FzA1) =
8. 21427
V
-
=
V(0.62)
=
0.62V +
=
300
0.32V
=
1200
=
L(0.2)
0.22
=
1200 0.62V +
.
+
0.2(3000
600
+
-
-
V)
0.2V
0.42V
w1428.6
=
2
=
300
-
14286
=
1571.4
of
2. Slowrate
YA1
=
U2
3A2 = 0.84
EA2 = 0.66
7)
Fz v
=
=
A2 0.56
1428.6
=
1428.620.62) V(0.84)
=
885.732
=
=
83.732
+
0.84V
0.28 VG
=
1
=
=
1
-
-
=
0.8
(1571.4)(0.83
=
(3000)(1
-
306.1
xA)
0.2
0.4)
(306.1)(0.84)
4
3000(0.4)
0.6984
0.21427
=
=
1478.6EVathz
ha(0.56)
0.56(1428.6
0.84V2+800
v=
3((B)
+
F Vz
+ 2
-
-
0.56V2
V2)
5
CN3132
Section 4
(11 marks)
Consider the absorption of benzene from a gas stream using wash oil solvent in a packed
column at 1 atm and 25 °C. The equilibrium constant under these conditions is equal to 0.095.
The benzene mole fraction in the entering gas stream is 0.06 and the target removal is 92%.
The entering wash oil solvent is pure. Operation is at Ls/Gs = 0.2. Dilute assumption
is not valid
E
here.
-
- . .
- -
Please be reminded that in engineering, correct calculations are also important. There is no
partial mark for procedure.
U.00507
1. Benzene mole fraction at the exit gas stream of the absorber is __________.
0.2269
2. Benzene mole fraction at the exit liquid stream of the absorber is __________.
3. In the calculation of NTU for an absorber using overall mass transfer coefficient, we
determine ____________
(liquid/gas) mole fraction in equilibrium with bulk
gas
ligul
____________ (liquid/gas) mole fraction.
- *
H.04
4. The nOG for this absorber is ________.
-0
5. If Ls/Gs increases, you will expect nOG to _________ (increase/decrease) as the
___________ (driving
force/efficiency) __________ (increases/decreases).
C
6. Your friend comments that it is possible to plot the operating line and equilibrium curve
both in terms of mole ratios and calculate the nOG by graphical solution, that is, by drawing
vertical lines connecting operating line to equilibrium curve. Do you agree with your
O
friend? ____________ (yes/no).
(a)
(b)
a1
y,
Es
Y
1
=
-
z
0.00507
i
0.0051
=
f ↓
↓IGS
=
0.2
-
xN
->
Lo:
-N: 0.2269
GN+1
Yn+ 1
=
LS:
0.06
XN:0.2935
GS
YN+ 1
=
0.06
1
-
=
Yout
:
0.06
0.0638
Yin21-FR)
=
I
0.0638(1
-
Ucoto-ESCo.oot-oos
Lo"
=
-
G5(Y1 Y(n + 1) hs(X0 XN)
-
0.92)
0.2935
=
6)=
xx
is
(c)
we
y*
need
Min
Yarg
0.2935
0.0638
0.03445
Your
0.0051
ESs
Xarg:
-
x
X
Y
(a)
8.
14675
O
(Yin-You)
a(...)=
o
y 0.095x
=
0.2669
0.02535
0.1279
0.01215
Δ
+ Xin
y*
u
3
0.06
I
of
0.033347.28
0.00507
197.238
sompson:
nog:z.0.0638
H.O4
-
0.005)(28.8 4(47.28)
-
+
+
197.238]
6
CN3132
Section 5
(10 marks)
You have made it to the final round of interviews for the role of a process engineer in a utilities
plant. In this round, you are given the schematic of a water cooling tower (Figure S5.1), and
the accompanying McCabe Thiele diagram (Figure S5.2). Your task is to interpret the given
information and answer the following questions. The specific heat capacity of water can be
taken to be 4.187 kJ/kg K.
Figure S5.1. Schematic of the cooling tower.
Figure S5.2. McCabe Thiele diagram for the cooling tower of Figure S5.1.
7
CN3132
Please be reminded that in engineering, correct calculations are also important. There is no
partial mark for procedure.
0-014
1. The absolute humidity of the entering air stream is _____________
kg water vapor/kg dry
air.
33.5
2. The dry‐bulb temperature of the exiting air stream is __________
°C.
0.033 kg water vapor/kg dry air.
3. The absolute humidity of the exiting air stream is ___________
25
19
-
4. The wet‐bulb temperature approach for this cooling tower is _________ °C.
b
1.88
, the value of the constant α in this case is __________.
5. Given that
6. The number of overall gas‐enthalpy transfer units (nOG) required to achieve the desired
1.98
cooling is _________.
7. If water flow rate can be assumed to be constant at L = 200 kg/s, the required Gs is
216.38
_____________
kg dry air/s.
Holl
8. The amount of water that has evaporated in this cooling tower is ____________
kg/s.
3.
opline
L.CAL. Hoat-Hin
->
120-54.6
=
-
42-2S
=)
Irin-twon
G
=3,87
200
-(n
openo
↑
-
54.2
=
=
7.2-4
20
Gsmm
T
1.88
IG5
GS
=
-
1.85
5
min
65: 1.65 Gsmn
Ha
6.
Hi
3
12 0
=
>128
12 87.1
=
n3
186
>77
54.2
=
s rmpsun
nog:
1.98
7. 3.87=200
(4.167)
As
55
=
216.38
8 water
evaporsed
=
(Yont
-
Yon's
216.3820.033 0.044)
-
=
4.11 (t)
&....
VasudevanS/CN3132/Cooling Tower
27
8
CN3132
Section 6
(5 marks)
The importance of external mass balance equations for a binary multi‐stage distillation
column were discussed and illustrated in this module with many examples. The important
process variables are shown in Figure S6.1 where values are given for some variables. Now it
is time to show your understanding by calculating the ones not given. Note that the mole
fraction values shown are for the lighter component in that stream.
93.74 mol% of the lighter component in the feed is recovered in the distillate stream.
Answer the following questions:
478.26
1. Flow rate of D =
kmol/h.
2. Flow rate of B= 521.74 kmol/h.
0.6
3. L/V =
7 43
/ =
4.
.
.
5. The constant relative volatility of the lighter component over the heavy component = 3.
yN+1 = 0.1607 .
3.
I
recovey
93.74%
->
0.9374
=
+
1200: 2
2.F B
=
10w
v=
721.64
475.26
+ 1)
=
↓
0.6
L=
B:
521.74
=0.6
=
478.26
0.6x1200
E. r
720
=
I 720+ (13 (1000)
=
1720
=
vV
=
-
x1
-
q).F
1200
=
-
45
-
1)(0.06)
0.1607
1200
H
1 +
↓(3
=
1000 (0.5)
=
3
3 10.06)
YA=
478.26
↓ = 721.24
=
r
x
=
3v L + 1)
D.20.95)
0.9374
(1.H
=
1.43
9
CN3132
Section 7
(11 marks)
A multi‐stage distillation process is used to separate a binary mixture of ethyl acetate – acetic
acid. The current process uses a total condenser and a partial reboiler. It produces two top
a1
and bottom products enriched in ethyl acetate and acetic acid, respectively. A saturated liquid
side stream is also withdrawn from the 2nd stage from top. Feed flow rate (F) is 2000 kmol/h.
Demand for the product withdrawn from the side stream has lately dropped.
=
Task assigned to the process engineers
The plant manager has tasked the process engineers to re‐design the process without the side
stream such that top and bottom product purities, and feed location remain unchanged. The
two process schematics are shown in Figure S7.1. The process engineers have produced two
McCabe Thiele plots for the two processes, which are shown in Figure S7.2.
10
CN3132
Your task is to compare the changes in the operating parameters of the two processes. Please
carefully check all the information provided in the two figures.
Please be reminded that in engineering, correct calculations are also important. There is no
partial mark for procedure.
When coordinates of more than two points on a straight line are shown, always use the two
furthest points to calculate the slope. Slopes based on the other intermediate points will
lead to inconsistent/unacceptable results.
Fill in the following blanks:
83
O
&
1. Process 2 in Figure S5.2 is the process with side stream.
(true/false)
O
2. In the process with side stream, V’‐L’=D+S.
(true/false)
O
3. In process 1, feed is introduced at the optimum location.
(true/false)
4. In process 2, feed is introduced at the optimum location.
(true/false)
O
5. In process 1, distillate flow rate (D) = 418.5 kmol/h.
Flow rate of the side stream (S) = 334.8
kmol/h.
6. D changes from 418.5 kmol/h in process 1 to
0.8
7. L/V changes from
in process 1 to
kmol/h in process 2.
0.67
in process 2.
8. Vapor flow rate returned to the column in process 1,
=
kmol/h.
9. Vapor flow rate returned to the column in process 2,
=
kmol/h.
10. Latent heat of evaporation of the bottom product is 51600 kJ/kmol.
Compared to process 1, reboiler heat duty in process 2 reduced by
5.
0.83
0.31=
+ 5.x5
D.xx
=
x+ y
=. gradient
x
D 0.93
in
1
(Ton]
=
xs
0.95-0.19
0.95
0.68
=
0.83 0.93(417 5)
=
-
9(0.68)
+
H15.3 +)
-
gradies
in 1
0.95
0.3)
-
=
(ToL
=
0.831418.5
+
3)
=
397.575
+
0.685
347.353 0.836: 397.575 + 0.685
+
0.156
3
=
=
50.22
334.8
0.8
0
0.95
0.67
x 106 kJ/h.
8.
I
11
CN3132
Section 8
(8 marks)
An air stream containing 10 mol% ammonia is treated in an absorber to bring the ammonia
concentration down to 1 mol% using fresh‐water solvent. The equilibrium relationship for the
ammonia‐water system is given by yNH3 =0.33xNH3. The process schematic and mole ratio plot
of the equilibrium data are shown in Figure S8.1. Two linear regression lines are also shown
in the equilibrium data plot. These are the required m values (i.e., linearized slope of a
nonlinear equilibrium data) for analytical solution of an absorption process. Air flow rate at
the absorber inlet (GN+1) is 15 kmol/h. McCabe Thiele plots for the absorption process for two
different columns are shown in Figure S8.2. Same equilibrium data as in Figure S8.1 are used.
.
=
0.0101
15
6
-= 0.11
.
12
CN3132
Fill in the following blanks.
1. Solvent flow rate used in process 1 is
2. Solvent flow rate used in process 2 is
3.
x
3.99
4 99
kmol/h.
kmol/h.
Absorption factor calculated from the Kremser’s equation for N=5 is
.
You may use the following Kremser’s equation (Use Excel only for this problem if you
want):
4. You are now required to calculate the solvent flow rate for N=5 from the absorption factor
1
calculated in question 3 above using both the m values given in Figure S8.1. Solvent flow
rate for m=0.2627 is
kmol/h and for m=0.2497 is
kmol/h.
5. Solvent (fresh water) costs $0.0.0015/kg. Compared to process 2, saving from solvent cost
in process 1 is 0.027
$/h. Use the solvent flowrates obtained in questions 1 and
2.
6. For the same separation performance, using more stages decreases solvent demand.
(true/false)
O
7. For analytical solution, all available data points should be used to obtain the regressed
linear slope.
(true/false)
8
1.9S 15(1
-
=
0.1)
13,5
=
↓
As (0.0101
20
-
0.111]
:Ls,
0.341)
-
3,9943 Knolla
=
2. 1.5
(0.0101
-
0.111)
:
20
=
-
0.273)
Is r
4.98
4.99-3.99=1
5
1
x
18
km
in
x
0.0271n
&)
0.0015
saved
s
xy
13
CN3132
Section 9
(8 marks)
The right triangular diagram for a countercurrent extraction cascade is shown in Figure S9.1.
Analyze the figure carefully and answer the following questions.
Figure S9.1. Right triangular diagram for solute A and solvent S.
1. Three points a, b and c are shown. Identify the solvent, extract and raffinate points (just
give the alphabet in the blank).
C
Solvent, S: Point ______
b
Extract, EN: Point ______
a
Raffinate, R1: Point ______
2. Part of the Hunter and Nash graphical procedure to find the number of stages is shown.
3
The number of equilibrium stages that have been illustrated so far is ______.
b
3. When stepping off the stages, what would be the target end point? Point ______
(a or b
or F).
4. Your lecturer is explaining the significance of inverse lever‐arm rule in class and states that
if the feed flow rate turns out be higher than the extract flow rate, the operating point
will still lie on the right. ______ (true/false).
F E 0
&
=
-
5. We wish to determine the minimum solvent flow rate. The tie line coinciding with
operating line at minimum solvent flow rate is expected to be _______ O
(above/below) the
tie line labelled as uv.
6. The operating point at minimum solvent flow rate is expected to be on the ______
(left/right) of the current operating point O.
O
14
CN3132
Section 10
(5 marks)
The same right triangular diagram for a countercurrent extraction cascade is shown in Figure
S10.1. Analyze the figure carefully and answer the following questions.
Figure S10.1. Right triangular diagram for solute A and solvent S.
200
1. If the sum of the feed and solvent flow rates is 500 kg/h, the feed flow rate is _______
300
kg/h and solvent flow rate is ______ kg/h.
52.63 kg/h.
2. The flow rate of the extract exiting stage 1 (i.e., E1) is _______
.... IEE
SF 500
=
xAF.
F
(0.4)(500
200
YAS.S
+
-
-
8)
0
+
=
5
-
80
300
300. 3as
200
YAe.En
+
-0.09)(500 En)
-
4)5
-
<El
s
=
30(0)
=
0.09En
+
+
0.16(500)
=
10.753)En
=
0.733En=
0 663ENI
&
R 2+E0
x)
R, tEN
xAk.R1
7
R2
Rx(((2) 1,.cin
=
+
=
-
-
43
↑
En=52.63
= 500
R
=>
-
52.63
447:37
80
s0
+
E.CYAE1)
(300 E0)20.09) 447.37(
300
F= 500
5
00 =
L
<
EO
0.16(500)
0.43
=
R,
2.
XAm. M
=
=
0.46
120
=
END OF THE PAPER
=M
RITEn
=
500
-
447.37
=
52.63
=
R1t E1
=
300
14
CN3132
Section 10
(5 marks)
The same right triangular diagram for a countercurrent extraction cascade is shown in Figure
S10.1. Analyze the figure carefully and answer the following questions.
Figure S10.1. Right triangular diagram for solute A and solvent S.
1. If the sum of the feed and solvent flow rates is 500 kg/h, the feed flow rate is _______
kg/h and solvent flow rate is ______ kg/h.
2. The flow rate of the extract exiting stage 1 (i.e., E1) is _______ kg/h.
END OF THE PAPER