1
2013/14 Answers Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
UNIT OPERATIONS A
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
1 Section A
S2
Robin Westacott
Page 1
Of 1
Diagram Required:
Other Data/ Special Stationery Required:
Di Preistler chart and No
equation sheet
(attached)
A valve tray distillation column has a total condenser and a total reboiler. The column
operates at 200 kPa pressure and there is an estimated pressure drop of 10 kPa.
Using the information below:
a) Determine the temperature in the condenser
[6]
b) Estimate the temperature of the utility fluid for the rebolier,
[6]
c) Using an appropriate method, determine the minimum number of stages for
this column.
[5]
d) The estimated minimum reflux ratio is 0.8. Estimate the actual number of
stages in this column
[4]
The column is fitted with valve trays. Comment on what would be the impact of
changing these to sieve type trays?
[4]
Component
n-Pentane
n-Hexane
n-Heptane
n-Octane
Feed Flowrate
kmol/hr
(Sat liquid)
40
50
35
35
Recovery in Distillate
100%
98%
1%
0.1%
Chart for K value data, Gilliland correlation graph and a sheet of preliminary design
equations are supplied separately. Note that not all the equations are needed for the
present calculations.
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2013/14 Answers Distillation
K Value Chart
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2013/14 Answers Distillation
Shortcut Distillation Column Design Equations
Fenske Equation:
Nmin
 x   x  
ln A   B  
 x B D  x A  W 
= 
ln α Ave, A,B
Or
Nmin
1st Underwood Equation
i =n
α i,r x Fi
∑ (α
i=1
i,r
− q)
 d  w
ln A  B
 w A  dB
= 
ln α Ave,A,B



2nd Underwood Equation
i=n
= (1 − q)
R min + 1 =
α i,r x Di
∑ (α
i=1
i,r
− θ)
Kirkbride’s Correlation
W  x
m

ln  = 0.206 ln  HK
D x
p

  LK
2
  ( x LK ) W  

 
 
F  ( x HK )D  

Gilliland Chart
Distribution of Non-key components
d 
ln i  = A ln(α Ave,i,r ) + C
 wi 
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2013/14 Answers Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations A
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
1 Solution
S2
GBT
Page 1
Of 3
Diagram Required:
Other Data/ Special Stationery Required:
Gililand chart,
No
equation sheet
a) Temperature in condenser
Molar compositions for distillate and bottoms product
N-PENTANE
N-HEXANE
N-HEPTANE
N-OCTANE
Feed Recovery Distillate Bottoms
(kmol/hr)
(kmol/hr) kmol/hr
40
100
40
0
50
98
49
1
35
1
0.35
34.65
35
0.1
0.035 34.965
160
89.385
Mole Fraction Composition
xf
xd
xb
0.2500 0.4475 0.0000
0.3125 0.5482 0.0142
0.2188 0.0039 0.4907
0.2188 0.0004 0.4951
70.615
1
1
1
Total condenser
Distillate
Composition
N-PENTANE
N-HEXANE
N-HEPTANE
N-OCTANE
mol frac
0.448
0.548
0.004
0.000
1.00
Ki=y/x
y=Kx
Ki=y/x
y=Kx
P (atm)
2
Ki
0.7893
0.2756
0.1027
0.0411
T deg C
50
yi/Ki
0.3532
0.1511
0.0004
0.0000
0.5047
P (atm)
2
T deg C
74
yi/Ki
0.6941
0.3218
0.0009
0.0000
1.0169
1.5510
0.5871
0.2377
0.1024
Pressure drop across the column is 10 kPa so the bottom pressure is
200+10=210 kPa
Ki=y/x
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y=Kx
Ki=y/x
y=Kx
March 2016 v1
Comment [RW1]: Did you get this
from the de Priester chart? I agree
with the number, but you can’t get
it to 4dp from the chart!
ANS 4dps are for a correlation –
close to the numbers here.
5
2013/14 Answers Distillation
Bottoms Conditions
Composition
N-PENTANE
N-HEXANE
N-HEPTANE
N-OCTANE
Average Volatilities
N-PENTANE
N-HEXANE
N-HEPTANE
N-OCTANE
0.000
0.014
0.491
0.495
P (atm)
2.10
Ki
2.7855
1.1372
0.4978
0.2294
T deg C
100
y=Kx
0.0000
0.0161
0.2443
0.1136
0.3740
P (atm)
2.10
Ki
6.0886
2.7290
1.3151
0.6592
T deg C
138
0.0000
0.0386
0.6453
0.3264
1.0104
Avg Alpha
5.50
2.26
1.00
0.46
Key components
d
LK
HK
N-HEXANE
N-HEPTANE
b
0.548
0.004
Average volatility of LK/HK
Nmin
0.014
0.491
2.26
10.39
Minimum reflux is said to be 0.8. Use the standard multiplier (1.2x Rmin but
anything reasonable will be acceptable) R=0.96.
Actual number of stages –
(R-Rmin)/(R+1) = 0.081
From the Gilliland chart, (N-Nmin)/(N+1) = 0.55
Actual number of stages is then 24
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2013/14 Answers Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
2
S2
Page 1 of
2
Diagrams Required:
Other Data/Special Stationery Required:
A 100 kmol/hr mixture, composition given below, is distilled in a column
operating at 1 bara. A short cut calculation predicts this column will have 41
minimum stages and a minimum reflux ratio of 8.5. The feed is saturated
liquid.
a) Using this information determine the:
i) Composition and temperature of the liquid streams leaving the first and
second trays as numbered from the top of the column.
[12]
ii) Estimate the vapour flow to the second tray
[3]
iii) The actual number of stages, assuming a suitable reflux multiplier [3]
iv) Vapour and liquid flows internally in the column.
[2]
b) Outline what you consider is the reason that this column appears to have a
large number of stages, and a proportionally large reflux ratio. List in the
order of priority, recommendations you would to adjust these numbers.
Explain the background to your selection.
[5]
Feed
(kmol/hr)
N-PENTANE
20
BENZENE
20
CYCLOHEXANE
30
TOLUENE
30
100
Distillate Bottoms
(kmol/hr) kmol/hr
19
1
18
2
3
27
1
29
41
59
Equilibrium constants are given in the accompanying table. Molkanov’s
expression for actual number of stages is:
where
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2013/14 Answers Distillation
Pressure
Temp
deg C
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
K Value Table
1 atm
N-PENTANE
BENZENE
CYCLOHEXANE
TOLUENE
0.32483
0.39694
0.48170
0.58071
0.69569
0.82848
0.98104
1.15543
1.35385
1.57857
1.83199
2.11662
2.43503
2.78993
3.18408
3.62034
4.10163
4.63095
5.21136
5.84599
6.53799
7.29057
8.10699
8.99052
9.94446
10.97213
0.06242
0.07876
0.09858
0.12244
0.15098
0.18488
0.22492
0.27191
0.32677
0.39046
0.46405
0.54865
0.64547
0.75578
0.88095
1.02240
1.18163
1.36023
1.55984
1.78219
2.02908
2.30235
2.60393
2.93580
3.30001
3.69866
0.06335
0.07966
0.09938
0.12305
0.15126
0.18467
0.22402
0.27007
0.32368
0.38577
0.45731
0.53936
0.63303
0.73950
0.86004
0.99595
1.14862
1.31951
1.51011
1.72200
1.95682
2.21625
2.50203
2.81596
3.15988
3.53568
0.01998
0.02575
0.03290
0.04168
0.05240
0.06537
0.08096
0.09960
0.12173
0.14785
0.17852
0.21433
0.25594
0.30406
0.35943
0.42288
0.49527
0.57754
0.67067
0.77570
0.89373
1.02592
1.17348
1.33768
1.51985
1.72137
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2013/14 Answers Distillation
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March 2016 v1
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2013/14 Answers Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number: Exam Diet:
Checked by:
Co-ordinator:
2 Solution
Robin Westacott
Page 1 of
1 Diagrams
Other Data/Special Stationery
Required:
Required:
Solution
Part a)
N-PENTANE
BENZENE
CYCLOHEXANE
TOLUENE
Feed Recovery Distillate Bottoms
(kmol/hr)
(kmol/hr) kmol/hr
20
95
19
1
20
90
18
2
30
10
3
27
30
3
1
29
100
41
59
Composition
xf
xd
0.200
0.463
0.200
0.439
0.300
0.073
0.300
0.024
1
1
Need to find the condenser temperature
Distillate
Composition
N-PENTANE
BENZENE
CYCLOHEXANE
TOLUENE
mol frac
0.463
0.439
0.073
0.024
1.00
Ki=y/x
P (atm)
1
Ki
1.5786
0.3905
0.3858
0.1479
y=Kx
T deg C
50
yi/Ki
0.7315
0.1714
0.0282
0.0036
0.9348
Ki=y/x
P (atm)
1
1.7270
0.4334
0.4275
0.1657
y=Kx
T deg C
53
yi/Ki
0.8003
0.1903
0.0313
0.0040
1.0259
Now work out the top tray temperature
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xb
0.017
0.034
0.458
0.492
1
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2013/14 Answers Distillation
TOP TRAY
Vapour
mf
N-PENTANE
0.4634
BENZENE
0.4390
CYCLOHEXANE 0.0732
TOLUENE
0.0244
Temp
Liquid
Ki
2.5725
0.6879
0.6740
0.2744
67
x=y/Ki
0.180144403
0.638209199
0.108559911
0.088895993
1.015809506
Min reflux ratio
Actual Reflux ratio (use 1.2xRmin)
Distillate rate
Liquid rate back to column
8.5
10.2
41
418.2 kmols/hr
Vapour rate from top plate
459.2 kmols/hr
Using the liquid rate back to the column and the composition of that liquid we
get a mass balance round the 1st tray is then
Distillate Liquid from tray
N-PENTANE
19
0.18x418=75
BENZENE
18
0.64x418=267
CYCLOHEXANE
3
45
TOLUENE
1
37
41
425
Vapour to top tray
19+75=94
18+267=285
48
38
466
Constant molar overflow would suggest internal liquid rate is 418, vapour rate
460.
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2013/14 Answers Distillation
SECOND TRAY
Rate
N-PENTANE
BENZENE
CYCLOHEXANE
TOLUENE
Temp
Ki
3.2679
0.9079
0.8859
0.3714
mf
94
285
48
38
466
0.2025
0.6116
0.1039
0.0820
76
x=y/Ki
0.0620
0.6737
0.1173
0.2206
1.0736
Answer: 67 deg for the 1st tray, 76 deg C 2nd tray.
Material balance
Distillate Liquid from 2nd tray
N-PENTANE
19
26
BENZENE
18
282
CYCLOHEXANE
3
49
TOLUENE
1
92
Vapour to
next tray
45
300
52
93
490
Answer to Part i)
N-PENTANE
BENZENE
CYCLOHEXANE
TOLUENE
Compositions on each tray
Tray 1
Tray 2
Liquid
Vapour
Liquid
Vapour
0.1801
0.4634
0.0620
0.2025
0.6382
0.4390
0.6737
0.6116
0.1086
0.0732
0.1173
0.1039
0.0889
0.0244
0.2206
0.0820
Temperature
67
76
ii) Vapour flow to 2nd tray is 490 kmol/hr
iii) Number of stages
Reflux multiplier
Rmin
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2013/14 Answers Distillation
R actual
10.2
X
0.151785714
(N-Nmin)/(N+1)
0.473589348
Nmin
Nactual
41
79
iv) Internal flowrates
Based on an assumption that below tray 2 that the liquid and vapour rates are
constant, the internal flows will be:
c) The large number of trays is due to the close volatility of benzene and
cyclohexane – there is not much of a difference across all temperature
ranges at this pressure. The difference in relative volatility should
become higher if the pressure were reduced. So in order to reduce the
number of stages, the 1st recommendation is to run the column under
vacuum conditions.
The second recommendation is to separate the benzene and
cyclohexane in their own column. This may allow an extractive type
distillation to be used that would preferentially be attractive to benzene
for example. Although the overall size of the column may be reduced,
there could be more than one column to separate this mixture out.
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2013/14 Answers Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
3
Robin Westacott
Page 1 of
2
Diagrams Required:
Other Data/Special Stationery Required:
HYSYS output
The product stream from the drier and quench tower following an ethylene
cracker is to be separated into essentially pure components. Composition of
this mixture is given below.
a) Two options for separating the hydrogen are proposed; i) to distil the
mixture, ii) to use a series of 4 flash separation stages. Discuss the
relative merits of both process stages, mentioning the impact on
refrigeration duty needed. [4]
b) The company selects option 2 which removes hydrogen first. The
decision then is to find out the sequence of columns to separate the
remaining mixture.
i) Using the technique of heuristics, outline the sequence of
columns you would recommend
[4]
ii) Using the principles of marginal vapour flow, validate your chosen
sequence against the direct sequence for all columns. Assume all
columns will operate at 5 atm pressure, at the mid range
temperature of -100 deg C. [10]
c) A de-methaniser used to separate methane from the other components
has been run as a rigorous column in HYSYS, the key output of which
is shown separately. By analysing the output, answer the following:
i) What condition is the feed to this column?
[2]
ii) Comment on the number of stages in this column and what
adjustments you would recommend to optimise the design.
[5]
K values for this question are found in the accompanying graphs.
Name
Hydrogen
Methane
Ethylene
Ethane
Propene
Propane
1-Butene
n-Butane
Mole %
0.0083
0.0826
0.3306
0.1653
0.2066
0.0826
0.0496
0.0744
Table of Feed Composition
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2013/14 Answers Distillation
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2013/14 Answers Distillation
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2013/14 Answers Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
3 Solution
2S2 2011-2012
Page 1 of
1
Diagrams Required:
Other Data/Special Stationery Required:
a) The principle issue of distillation with the hydrogen feed is to reduce the temperature
of the column sufficiently low as to condense the methane and potentially the
hydrogen. But given that the hydrogen is the most volatile of all components, it will
not be possible to pull the temperature sufficiently low enough for any condenser. In
any case, to cool the mixture sufficiently low, a special refrigeration loop would be
needed.
The best option will be to use a series of flash separation stages to drive of the
majority of the hydrogen before it reaches the column. Even with this column, you
would need a refrigerant that can be pulled down low enough to condense methane.
(in practice you can expand high pressure hydrogen or high pressure methane to
drop its temperature which may provide the right level of temperature).
rd
Note: Students have encountered this type of distillation before in 3 year HYSYS
modelling class.
b) I) Heuristics would say that you separate the mixture based on
a. Flow rate – sepatate the components with the largest flow first – in this case
you would separate the ethylene methane from the rest of the mixture.
b. Components which are difficult to separate should be left in their own column
– in this case, separating the C2’s and C3’s and C4’s in their own columns
would be preferable
c. Separate those that are fouling – we have a relatively clean system here
operating at low temperatures. This is unlikely to cause issues due to fouling.
iii) Marginal vapour flowrates
Compare the direct sequence with the one recommended by heuristics
Assume the sequence will be done at the mid range temperature of -100 deg
Step 1: Collect the flowrates and K values. Work out the relative volatility wrt the least
volatile component
HYDROGEN METHANE ETHYLENE ETHANE PROPYLENE PROPANE 1-BUTENE N-BUTANE
Feed
K values
5
75.37333
10
200
50
5.14541 0.24281 0.10262
Rel Vol 119668.0473 8169.223 385.504 162.920
60
10
30
0.00938 0.00708 0.00086
0.00063
14.894
30
11.249
1.365
Now work out the marginal flows for all possible key components
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2013/14 Answers Distillation
HYDROGEN METHANE ETHYLENE ETHANE PROPYLENE PROPANE 1-BUTENE N-BUTANE
Feed
5
K values
75.37333
10
200
50
60
5.14541 0.24281 0.10262
Rel Vol 119668.0473 8169.223 385.504 162.920
A
B
10
30
0.00938 0.00708 0.00086
30
0.00063
14.894
11.249
1.365
1.000
C
D
E
F
G
H
theta term
1.2136
0.1278
0.0140
0.0053
0.0002
0.0005
63918.6350
1.9799
0.2097
0.0791
0.0032
0.0070
4277.3631
3.4462
1.2833
0.0500
0.1098
274.2119
4.3454
0.1559
0.3413
88.9073
1.1660
2.4852
13.0715
5.6531
6.3068
lk/hk
A/B
B/C
5.1853
C/D
5.0115
10.3473
D/E
5.0037
10.1100 259.9517
E/F
5.0005
10.0160 207.0195 54.3616
F/G
5.0003
10.0077 203.3264 52.0135 104.0651
G/H
5.0000
10.0014 200.6154 50.3656 65.1743
33.5242
1.1825
Now work out the marginal flow for the direct sequence
A
B
A/BCDEFGH
C
D
E
F
G
H
Marginal Flow
1.2136 0.1278 0.0140 0.0053 0.0002 0.0005
1.3613
1.9799 0.2097 0.0791 0.0032 0.0070
2.2788
3.4462 1.2833 0.0500 0.1098
4.8893
4.3454 0.1559 0.3413
4.8426
B/CDEFGH
C/DEFGH
D/EFGH
E/FGH
1.1660 2.4852
3.6512
F/GH
5.6531
5.6531
G/H
0.0000
Total
0.0000
22.6763
Now for the heuristic sequence
A
B
ABC/DEFGH
5.0115 10.3473
AB/C
5.1853
C
D
E
F
G
H
Marginal Flow
3.4462 1.2833 0.0500 0.1098
20.2481
5.1853
A/B
0.0000
E/FGH
F/GH
1.1660 2.4852
3.6512
5.6531
5.6531
0
0.0000
G/H
Total
34.7378
This shows that the direct sequence is the preferred option –
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2013/14 Answers Distillation
c)
i)
ii)
There is no information on the HYSYS printout which says the state of the feed.
This is deliberate. Students have to read through the column profile pages to
work out the details. The “Column Profile Flows” – which is in several places to
work out that show that the feed is 2055 kmols/hr, but the liquid flow to the next
tray is 2447, liquid from the top section is 518.7. This means that 2447518.7=1928 must come from the feed and must be liquid. Vapour from the bottom
section is 591.9 and through the top section is 718.7 which means 718.7-591.9 =
127 comes from the feed. So the feed is a two phase mixture. Feed enters on
plate 7 as indicated in the column profile summary
The printout contains tables but no profile pots so the task here is to see if the
students realise that they can visualise the column profile from the given data set.
Analysing the tables however will give them sufficient information:
a. The overall liquid and gas rates don’t change very much from plate 9-15. This
is not surprising due to constant molar overflow. The temperatures however
remain very similar over the same range -57 on plate 9 and -57 on plate 14.
b. The composition profile (mapping the liquid/gas composition for each
component) shows that over this range there is little activity in the column
and hence no separation taking place, To further optimise this column, the
recommendation is to reduce the number of trays in this section.
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2013/2014 Exam Answers Control II
Model answers control II
Model answer Question 1
a) Process interface unit: computer unit in distributed control system that does
most of the basic /regulatory control in a typical plant
b) Throughput manipulator: control system that sets that flow that control the
production rate of a plant
c) Derivative kick: sudden change in controller output when setpoint changes,
causing de/dt to suddenly increase- and hence the derivative part of a PID
controller.
d) Integral windup: saturation of controller output by integral part of PID
integrating to above 100%.
e) Selective control: control system in which the control system takes in and
compares data from various places but bases the controller output on a
selection.
Level 4
Business planning/logistics
Advanced control
Supervisory control
Basic/ Regulatory control
Level 3
Level 2
Level 1
Critical safety systems
Model answer Question 2
a)
Time (seconds)
-2
-1
0
1
2
Value of controlled
variable (oC)
100.00
100.00
100.00
100.20
100.29
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e i (%TO)
∆m (%CO)
m i (%TO)
10
10
10
9.8
9.71
0.5
0.29
0.4
50
50
50.5
50.79
51.19
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2013/2014 Exam Answers Control II
3
4
5
100.35
100.39
100.42
9.65
9.61
9.58
0.42
0.44
0.45
51.61
52.05
52.50
PI controller so use

∆t 
∆m = K c  en − en −1 + en 
τI 

t = 0:
e n-1 = 10
e n = 10

∆t 
so ∆m = K c  en − en −1 + en  = 1(10 − 10 + 1 / 20 * 10) = 0.5
τI 

m n = m n-1 +∆m= 50+1=50.5
t=1
e n-1 = 10oC=10%TO
e n = 9.8

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.8 − 10 + 1 / 20 * 9.8) = −0.2 + 0.49 = 0.29
τI 

m n = m n-1 +∆m= 50.5+0.29=50.79
t=2

∆t 
so ∆m = K c  en − en −1 + en  = 1(10 − 10 + 1 / 20 * 10) = 0.5
τI 

m n = m n-1 +∆m= 50+1=50.5
t=1
e n-1 = 10oC=10%TO
e n = 9.8

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.8 − 10 + 1 / 20 * 9.8) = −0.2 + 0.49 = 0.29
τI 

m n = m n-1 +∆m= 50.5+0.29=50.79
t=2
e n-1 = 9.8oC=9.8%TO
e n = 9.71

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.71 − 9.8 + 1 / 20 * 9.71) = −0.09 + 0.49 = 0.40
τI 

m n = m n-1 +∆m=50.79 + 0.40=51.19
t=3
e n-1 = 9.71
e n = 9.65
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2013/2014 Exam Answers Control II

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.65 − 9.71 + 1 / 20 * 9.71) = −0.06 + 0.48 = 0.42
τI 

m n = m n-1 +∆m=51.19 + 0.42=51.61
t=4
e n-1 = 9.65
e n = 9.61

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.61 − 9.65 + 1 / 20 * 9.61) = −0.04 + 0.48 = 0.44
τI 

m n = m n-1 +∆m= 51.61+0.44=52.05
t=5
e n-1 = 9.61
e n = 9.58

∆t 
so ∆m = K c  en − en −1 + en  = 1(9.58 − 9.61 + 1 / 20 * 9.58) = −0.03 + 0.48 = 0.45
τI 

m n = m n-1 +∆m= 52.05+0.45=52.50
Model answer question 4
60
50
%CO/%TO
40
Controller output
30
Humidity
20
10
0
0
250
500
750
1000
1250
1500
1750
2000
Time (seconds)
a) K=(50-25)/(20-15)=25/5=5%TO/%CO
b) ok is 250 sec-50sec=200 sec. 150-200 from Fit 3 is also ok (see part c))
c) Fit3: 0.283*25+25 = 32.1; t1= 350sec
0.632*25+25 = 40.8; t2=550 sec
Tau=3/2(550-350)= 300 sec
to=550-300=150.
d) Kc = 0.9/5*300/150=0.36; tauI=3.33*to=500.
Model answer question 5:
A and B: local consistent
C: consistent but not local consistent
D: not consistent (nor control over flow rate in recycle)
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2013/2014 Exam Answers Control II
Model answers question 6
RC
FA
FC
FC
FB
V2
LC
PC
V1
FC
LC
V3
LC
AC
b
V2
V1
FC
LC
c
FC
V2
V1
LC
LC
AC
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2014/15 Exam Answers, Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
UNIT OPERATIONS A
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
1 Section A
S2
Page 1
Of 1
Diagram Required:
Other Data/ Special Stationery Required:
equation sheet
No
(attached)
A saturated liquid mixture containing 40 mol% n-hexane, 40 mol% cyclohexane, 20
mol% n-heptane is distilled, into a distillate containing 95% n-hexane and 5 mol%
cyclohexane. The column operates at 2 bara pressure and has a total condenser
fitted operating at 94ºC with the rebolier at 134ºC . Short cut calculation suggests
there are 18.1 minimum stages and a minimum reflux ratio of 6.
a) Determine the actual reflux ratio and actual number of stages in this column.
[5]
b) What would you expect to happen to the condenser and reboiler temperature
if the column pressure were to increase? As a consequence, what would you
expect to happen to the distribution of the heptane and the size of the
column?
[5]
c) Determine the composition of the light and heavy key components in the
bottoms product.
[15]
K values for the three components are shown graphically. Gilliland correlation graph
and a sheet of preliminary design equations are supplied separately. Note that not
all the equations are needed for the present calculations.
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2014/15 Exam Answers, Distillation
K Value Chart
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2014/15 Exam Answers, Distillation
Shortcut Distillation Column Design Equations
Fenske Equation:
Nmin
 x   x  
ln A   B  
 x B D  x A  W 
= 
ln α Ave, A,B
Or
Nmin
1st Underwood Equation
i =n
α i,r x Fi
∑ (α
i=1
i,r
− q)
 d  w
ln A  B
 w A  dB
= 
ln α Ave,A,B



2nd Underwood Equation
i=n
= (1 − q)
R min + 1 =
α i,r x Di
∑ (α
i=1
i,r
− θ)
Kirkbride’s Correlation
W  x
m

ln  = 0.206 ln  HK
p
D x
 

  LK
2
  ( x LK ) W  

 
 
x
)
(
F  HK D  

Gilliland Chart
Distribution of Non-key components
d 
ln i  = A ln(α Ave,i,r ) + C
 wi 
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2014/15 Exam Answers, Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations A
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
1 Solution
S2
GBT
Page 1
Of 3
Diagram Required:
Other Data/ Special Stationery Required:
equation sheet
No
a) Actual reflux ratio from iterative procedure
Average volatilities are
Volatilities wrt HK.
Component
Distillate
Hexane
1.01
Cyclohexane
0.73
heptane
0.44
Bottoms
1.56
1.13
0.7
Alpha Dist
1.38
1
0.6
Alpha Bottom
1.38
1
0.62
Alpha Avg
1.38
1
0.61
st
1 Underwood equation says
i =n
α i,r x Fi
∑ (α
i=1
i,r
− q)
= (1 − q)
For saturated liquid, q=1
Trial and error- guess theta which should lie between average alpha for LK and HK
Component
Avg AlphaFeed CompGuess thetaGuess thetaGuess thetaGuess theta
N-HEXANE
1.15
1.3
1.164
1.170
1.38
0.400
2.361
6.551
2.515
2.576
CYCLOHEXANE 1.00
0.400
-2.667
-1.333
-2.435
-2.359
N-HEPTANE
0.200
-0.226
-0.177
-0.220
-0.218
-0.531
5.041
-0.140
0.000
0.61
The minimum reflux is therefore
Minimum reflux ratio is therefore
Component
Avg AlphaDist. CompTheta
1.170
N-HEXANE
1.38
0.950
6.119
CYCLOHEXANE
1.00
0.050
-0.295
N-HEPTANE
0.61
0.000
0.000
Rmin+1=
5.824
Rmin
4.824
b) If the column pressure were increased, then the condenser and reboiler temperatures
would also have to increase to keep the same distribution of light and heavy keys. From the K
value graph, the relative volatility of the heptane would reduce and therefore there would be
no change in the distribution. The column may have to be larger though to accommodate the
cyclohexane that would be in the distillate.
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2014/15 Exam Answers, Distillation
c) Basis of 100 kmols of feed
Component
Hexane
Cyclohexand
heptane
Feed
kmols
40
40
20
100
The fenske equation is Nmin
Dist Comp
Dist Rates
0.95
0.05
0
1
D
Bottom
Comp
xlkB
xhkB
x
Bottom Rate
20
B
 x   x  
ln A   B  
 x B D  x A  W 
= 
ln α Ave, A,B
Since we know the minimum number of stages and the top and bottom temperatures. We
have
Volatilities wrt HK.
Component
Hexane
Cyclohexand
heptane
Distillate
1.01
0.73
0.44
Bottoms
1.56
1.13
0.7
Alpha Dist
1.38
1
0.6
Alpha Bottom
1.38
1
0.62
Alpha Avg
1.38
1
0.61
(Note - the answers will depend on the value for this exponential term)
eq1
Balance for each component
Replace the x hkB term using eq 1
, and
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2014/15 Exam Answers, Distillation
from
, multiply 19 to give
and we have
Therefore
Hence
Hence
Therefore the key composition of the bottoms products are
Therefore the heptane in the bottoms is
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2014/15 Exam Answers, Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
2
S2
Page 1 of
2
Diagrams Required:
Other Data/Special Stationery Required:
The HYSYS output, attached separately, shows details of a first attempt at a rigorous
column separating a mixture of propane, butane, pentane and hexane. Use this
information to answer the following:
a)
i)
Determine the average liquid and vapour rates inside the column and hence
provide a sketch diagram of the column showing all the key flowrates.
[4]
ii) Sketch a diagram of stages 9,10 and 11 and include the internal liquid and
vapour flowrates.
[4]
iii) Verify the thermal state of each feed stream by calculating the q value.
[4]
b) Assuming a combined feed that is saturated liquid, calculate what the minimum
reflux ratio should be for this column.
[5]
c) Calculate the minimum number of stages needed for this column
[4]
d) Construct a temperature and key component ratio plot for this column and hence
outline what further changes could be done with the column
[4]
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2014/15 Exam Answers, Distillation
Selected HYSYS Output
Stage
Temperature
(C)
Pressure
(kPa)
Net Liq
(kgmole/h)
Net Vap
(kgmole/h)
Net Feed
(kgmole/h)
Net Draws
(kgmole/h)
Condenser
6.176
150
6.983
---
---
40.2
1
7.842
150
6.87
47.18
---
---
2
8.385
150
6.807
47.07
---
---
3
8.648
150
6.776
47.01
---
---
4
8.777
150
40.47
46.98
35
---
5
13.85
150
40.54
45.67
---
---
6
16.51
150
39.83
45.74
---
---
7
20.11
150
38.66
45.03
---
---
8
25.43
150
37.41
43.87
---
---
9
31.87
150
36.05
42.61
---
---
10
39.65
150
111.2
41.25
65
---
11
42.23
150
111.5
51.38
---
---
12
45.06
150
112
51.7
---
---
13
47.72
150
112.7
52.23
---
---
14
49.97
150
113.1
52.87
---
---
15
52.07
150
113
53.35
---
---
Reboiler
55.47
150
---
53.2
---
59.8
Table 1: Column Flow Profiles
Stage
Propane
n-Butane
n-Pentane
n-Hexane
Condenser
0.0653
0.8417
0.093
0
1
0.0554
0.8108
0.1337
0
2
0.0543
0.7904
0.1553
0
3
0.0539
0.7799
0.1662
0
4
0.0537
0.7747
0.1716
0
5
0.0101
0.7861
0.2038
0
6
0.0018
0.7272
0.2709
0.0002
7
0.0003
0.6105
0.3876
0.0017
8
0
0.456
0.5329
0.011
9
0
0.3098
0.6344
0.0558
10
0
0.1991
0.5958
0.2052
11
0
0.1557
0.6363
0.208
12
0
0.1116
0.6776
0.2108
13
0
0.0731
0.7131
0.2138
14
0
0.0437
0.7359
0.2204
15
0
0.0233
0.7324
0.2443
Reboiler
0
0.01
0.6555
0.3344
Table 2 Column Composition Profiles (Liquid)
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2014/15 Exam Answers, Distillation
Tray Number
Propane
n-Butane
n-Pentane
n-Hexane
Condenser
3.81
0.8689
0.2139
0.05292
1
3.998
0.9221
0.2297
0.05761
2
4.061
0.94
0.235
0.05921
3
4.091
0.9488
0.2377
0.06
4
4.107
0.9531
0.239
0.06039
5
4.734
1.135
0.2947
0.07749
6
5.09
1.241
0.3279
0.08796
7
5.603
1.396
0.3776
0.104
8
6.428
1.65
0.4614
0.1319
9
7.543
2.004
0.5822
0.1737
10
9.071
2.505
0.7596
0.2378
11
9.626
2.691
0.8271
0.2629
12
10.26
2.905
0.9061
0.2928
13
10.88
3.119
0.9858
0.3234
14
11.44
3.308
1.057
0.3512
15
11.97
3.493
1.128
0.3789
Reboiler
12.87
3.808
1.249
0.4273
Table 3 Column K-Values Profile
Feed Compositions
Propane
n-Butane
n-Pentane
n-Hexane
Feed 1
0.2857
0.5714
0.1429
0.0000
Feed 2
0.0000
0.1538
0.5385
0.3077
Table 4 Feed Specifications (mol fraction)
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2014/15 Exam Answers, Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number: Exam Diet:
Checked by:
Co-ordinator:
2 Solution
Page 1 of
1 Diagrams
Other Data/Special Stationery
Required:
Required:
The HYSYS output, attached separately, shows details of a first attempt at a rigorous
column separating a mixture of propane, butane, pentane and hexane. Use this
information to answer the following:
a)
i)
Determine the average liquid and vapour rates inside the column and hence
provide a sketch diagram of the column showing all the key flowrates.
[4]
ii) Sketch a diagram of stages 9,10 and 11 and include the internal liquid and
vapour flowrates.
[4]
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2014/15 Exam Answers, Distillation
iii) Verify the thermal state of each feed stream by calculating the q value.
[4]
Feed stream at tray 10.
Feed stream flowrate : 65 kmols/hr
Liquid from tray 9 : 36.05 kmols/hr
65+35.05=100.05 which means that vapour has to have been condensed from tray
11. Therefore the feed at tray 10 is subcooled.
Molar flowrate of vapour condensed is 51.38 – 41.25 = 10.13
Feed stream at tray 4
Feed stream flowrate: 35 kmols/hr
Liquid from tray 3: 6.776 kmols/hr
35+6.776 = 41.776 kmols/hr
Actual liquid from tray 4 is 40.47 kmols/hr
This means that some of the liquid in the feed is vapourised
b) The rigorous column shows a reflux ratio of 0.174. Using appropriate methods,
calculate what the minimum reflux ratio should be for this column.
[5]
Short cut calculations deal with 1 feed to the column so work out the feed
composition from the HYSYS data provided
Composition
Molar Rates
Combined Feed
Feed 1
Feed 2
Feed 1
Feed 2
Rate
Comp
Propane
0.2857
0.0000
10.0
0.0
10.0
0.1
Butane
0.5714
0.1538
20.0
10.0
30.0
0.3
Pentane
0.1429
0.5385
5.0
35.0
40.0
0.4
Hexane
0.0000
0.3077
0.0
20.0
20.0
0.2
35
65
35
65
100
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2014/15 Exam Answers, Distillation
Now use the 1st Underwood equation. From the K-value table, extract the K values at
the condenser and re-boiler. Table 2 shows that propane and hexane are cleanly
split so that the LK and HK must be butane and pentane.
Condenser
Reboiler
Average
Rel Vol
Propane
3.81
12.87
7.00
13.55
Butane
0.87
3.81
1.82
3.52
LK
Pentane
0.21
1.25
0.52
1.00
HK
Hexane
0.05
0.43
0.15
0.29
1st Underwood Equation
Guess Theta
Feed Comp
Rel Vol
Sum Terms
Propane
0.10
13.55
0.114
Butane
0.30
3.52
0.565
Pentane
0.40
1.00
-0.615
Hexane
0.20
0.29
-0.043
Sum
0.020
2nd Underwood equation
1.65
Rel Vol
Dist Comp
Sum term
Propane
0.0653
13.55
0.074
Butane
0.8417
3.52
1.585
Pentane
0.093
1.00
-0.143
Hexane
0
0.29
0.000
Rmin+1=
1.516
Rmin
0.516
c) Calculate the minimum number of stages in this column
Minimum number of stages
Propane
Butane
Pentane
Hexane
Distillate
0.0653
0.8417
0.093
0
Avg Vol
Nmin
3.52
5.08
Bottoms
0
0.01
LK
0.6555 HK
0.3344
d) Construct a temperature and key component ratio plot for this column and
hence outline what further changes could be done with the column
[4]
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2014/15 Exam Answers, Distillation
KLK/HK Component Plot
10
9
LK/HK Ratio
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17
Stage
Temperature Plot
Temperature
60
50
40
30
20
10
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17
Stage
The key component ratio plot for the top of the column looks as if that section is ok – the key
plot flattens out around the feed stage and steppes either side. The fact that key component
ratios are changing on either side of the feed tray indicates the column is working ok. Below
nd
the 2 feed stage however, the column is not really doing much.
The temperature profile shows 3 distinct sections which you would expect. The issue is that
the end points at the reboiler and condenser are not probably as steep as normal.
Since the split is around the inner two components – then its likely this column needs more
stages. Looking at the compositions of each feed stream, from that view point it seems that
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2014/15 Exam Answers, Distillation
the feed stages could be adjusted – The next iteration should try the feeds closer together –
the molar composition of the butane and pentane match with stage 7 & 8 of the column rather
than being too far apart.
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2014/15 Exam Answers, Distillation
Examination Question
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
3
Page 1 of
2
Diagrams Required:
Other Data/Special Stationery Required:
HYSYS output
As part of a liquidfied natural gas process, methane is to be recovered by a
distillation column operating at 30 bar pressure. The feed stream and expected
recoveries are given in the table below. The column will be fitted with a partial
condenser Assume there is a 0.5 bar pressure drop across the column.
a)
b)
c)
d)
Determine the temperature of the condenser
[5]
Show that the reboiler utility temperature will be close to 50ºC
[5]
Determine the minimum number of stages for this separation
[6]
If the minimum reflux ratio is 0.142, determine the actual number of stages in
this column
[4]
e) Natural gas from a gas reservoir usually contains CO 2 . Explain why CO 2
would need to be removed before this column is used to strip out the
methane.
[2]
f) LNG feed would contain more than 4 components. Outline one method that
could be used to determine the sequence of columns to be used to separate a
typical LNG feed mixture.
[3]
Methane
Ethane
Propane
N-Butane
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Feed
(kmol/hr)
40
50
35
35
B40EA
Recovery in
Distillate
99
0.1
0
0
March 2016 v1
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2014/15 Exam Answers, Distillation
Temp
METHANE
ETHANE
PROPANE
N-BUTANE
-100
0.87252
0.01763
0.00123
0.00011
-95
1.03164
0.02361
0.00178
0.00017
-90
1.20866
0.03113
0.00253
0.00026
-85
1.40418
0.04044
0.00352
0.00039
-80
1.61870
0.05183
0.00483
0.00056
-75
1.85265
0.06560
0.00652
0.00080
-70
2.10635
0.08208
0.00866
0.00113
-65
2.38007
0.10158
0.01136
0.00155
-60
2.67397
0.12447
0.01470
0.00211
-55
2.98816
0.15111
0.01881
0.00283
-50
3.32268
0.18185
0.02379
0.00375
-45
3.67749
0.21709
0.02980
0.00490
-40
4.05250
0.25718
0.03696
0.00634
-35
4.44757
0.30252
0.04542
0.00810
-30
4.86251
0.35348
0.05536
0.01025
-25
5.29707
0.41044
0.06693
0.01285
-20
5.75097
0.47377
0.08031
0.01597
-15
6.22391
0.54384
0.09569
0.01967
-10
6.71553
0.62101
0.11326
0.02404
-5
7.22545
0.70562
0.13322
0.02917
0
7.75328
0.79803
0.15576
0.03514
5
8.29860
0.89854
0.18110
0.04204
10
8.86097
1.00749
0.20944
0.04999
15
9.43992
1.12516
0.24100
0.05908
20
10.03501
1.25186
0.27598
0.06943
25
10.64574
1.38783
0.31461
0.08115
30
11.27165
1.53336
0.35710
0.09437
35
11.91224
1.68866
0.40366
0.10919
40
12.56702
1.85397
0.45451
0.12577
45
13.23551
2.02950
0.50987
0.14421
50
13.91720
2.21542
0.56993
0.16466
55
14.61162
2.41193
0.63490
0.18724
60
15.31827
2.61918
0.70500
0.21211
65
16.03667
2.83730
0.78041
0.23940
Table of K values
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2014/15 Exam Answers, Distillation
Examination Solution
Heriot-Watt University, School of Engineering and Physical Sciences,
Chemical Engineering.
Subject :
Lecturer/Author:
Module Code:
Unit Operations
GW
B40EA
Question Number:
Exam Diet:
Checked by:
Co-ordinator:
3 Solution
2S2 2011-2012
Page 1 of
1
Diagrams Required:
Other Data/Special Stationery Required:
a) Determine the temperature of the condenser
Feed
Recovery
(kmol/hr)
[5]
Distillate
Bottoms
Mole Fraction Composition
(kmol/hr)
kmol/hr
xf
xd
xb
METHANE
40
99
39.6
0.4
0.2500
0.9987
0.0033
ETHANE
50
0.1
0.05
49.95
0.3125
0.0013
0.4150
PROPANE
35
0
0
35
0.2188
0.0000
0.2908
35
0
0
35
0.2188
0.0000
0.2908
39.65
120.35
1
1
1
N-BUTANE
160
Distillate
Composition
Ki=y/x
y=Kx
Ki=y/x
y=Kx
P (atm)
T deg C
P (atm)
T deg C
30
30
-96
mol frac
Ki
yi/Ki
METHANE
0.999
0.9984
0.9971
1.0316
1.0303
ETHANE
0.001
0.0223
0.0000
0.0236
0.0000
PROPANE
0.000
0.0017
0.0000
0.0018
0.0000
N-BUTANE
0.000
0.0002
0.0000
0.0002
0.0000
1.00
-95
yi/Ki
0.9972
1.0304
So the condenser temperature is -95 deg C
b) Show that the reboiler utility temperature will be close to 40ºC
[5]
Ki=y/x
y=Kx
Ki=y/x
y=Kx
P (atm)
T deg C
P (atm)
T deg C
30.00
40
30.00
45
Bottoms Conditions
Composition
Ki
y=Kx
Ki
METHANE
0.003
12.6996
0.0418
12.8328
0.0440
ETHANE
0.415
1.8883
0.7695
1.9229
0.8423
PROPANE
0.291
0.4652
0.1322
0.4761
0.1483
N-BUTANE
0.291
0.1293
0.0366
0.1329
0.0419
0.9800
1.0765
Since the summation is close to 1, then the reboiler temperature is close to 40
deg C.
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2014/15 Exam Answers, Distillation
c) Determine the minimum number of stages for this separation
Composition
[6]
Distillate
xd
Ki
alpha
METHANE
0.999
1.0316
43.69
LK
ETHANE
0.001
0.0236
1.00
HK
PROPANE
0.000
0.0018
0.08
N-BUTANE
0.000
0.0002
0.01
Heavy key K value
0.024
Composition
Bottoms
xb
Ki
alpha
METHANE
0.003
13.2355
6.52
LK
ETHANE
0.415
2.0295
1.00
HK
PROPANE
0.291
0.5099
0.25
N-BUTANE
0.291
0.1442
0.07
Heavy key K value
2.029
Average Volatilities
Avg Alpha
METHANE
16.88
ETHANE
1.00
PROPANE
0.14
N-BUTANE
0.02
Key components
d
b
LK
METHANE
0.999
0.003
HK
ETHANE
0.001
0.415
Average volatility of LK/HK
16.88
Nmin
4.07
d) If the minimum reflux ratio is 0.142, determine the actual number of stages in
this column.
[4]
Set actual reflux = 1.2xRmin
Actual reflux ratio
0.173
(R-Rmin)/(R+1)
0.0246
(N-Nmin)/(N+1)
0.6
Check by correlation
0.673
Actual number of stages
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2014/15 Exam Answers, Distillation
e) Natural gas from a gas reservoir usually contains CO 2 . Explain why CO 2
would need to be removed before this column is used to strip out the
methane.
[2]
CO2 would freeze in the methane column and therefore block internals. Its
normally removed to avoid this.
f)
LNG feed would contain more than 4 components. Outline one method that
could be used to determine the sequence of columns to be used to separate
a typical LNG feed mixture.
[3]
There are 2 potential methods to separate out a mixture of more than 2 components:
• Heuristics
• Minimum vapour flowrate
These would be used to select possible sequences for further investigation by rigorous
column simulation tools.
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2014/15 Exam Answers, Distillation
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2014/2015 Exam Answer, Control II
Model answer question 1
a) PID:
Steam
FT1
FC1
TC1
AC1
AT1
TT1
Reactor feed
Reactor
Products
Heat exchanger
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2014/2015 Exam Answer, Control II
Block diagram:
Pfeed
G2
Tfeed
Aset1
G2
F1
Km
GAc1
GTC1
GFC1
Gv
HFT1
HTT1
HAT1
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T1
GHeat exchanger
Greactor
A1
3
2014/2015 Exam Answer, Control II
b)
Design a control system to control:
-
The shell-side pressure in the vertical heat exchanger.
The level in the gas-liquid separator
The composition of the concentrated product stream.
[6 marks]
Condensor
Cooling water
Separator
Condensate
LC
PC
Steam
Evaporator
Feed
Pump
AC
Product
Figure 1b. Forced circulation evaporator system.
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2014/2015 Exam Answer, Control II
c)
All feeds have to be controlled. Design a control system to control:
•
•
•
•
The product flow from the mill into the sump
The torque/speed of the ball mill
The density of the feed to the cyclone separator
The level of the liquid in the sump
[8 marks]
Cyclone
Feedwater
Solid feed
to sump
FC
Mill
LC
Water feed
Motor
AC
SC
Pump
Figure 1c. A wet grinding circuit.
d)
Steam doesn’t affect level so Y2 is level, X1 is flow. Steam and flow affect temperature. Y1 must be
temperature. X2 must be steam.
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2014/2015 Exam Answer, Control II
Model answer Question 2.
a)
Gv ( s ) = 12kg / s / 100%CO = 0.12kg / s %CO
GJ ( s ) =
2 o
C / (kg/s)
2.5s + 1
GR (s) =
0.6 o
C / oC
13s + 1
H R (s) =
100%TO / 50 o C
2
=
%TO / o C
s +1
s +1
H J ( s ) = 100%TO / 100o C = 1%TO / oC
b)
TRsp(s)
R(s)
Gc(s)
Km
C(s)
TRsp(s)
M(s)
E(s)
R(s)
Gv(s)
GR(s)
HR(s)
M(s)
E(s)
Gc(s)
Km
TR(s)
GJ(s)
TR(s)
0.12
C(s)
c) fail open to make sure reactor stays cold.
d) 1+OLTF=0
OLTF= G c G v G J G R H R so 1 + OLTF = 0 = 1 + Gc * 0.12 *
2
0.6
2
*
*
2.5s + 1 13s + 1 s + 1
e) i.
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2014/2015 Exam Answer, Control II
Sketch of cascade control system
TC1
TT1
Reactants
TT2
TC2
∞
Coolant
Product
ii
TRsp(s)
TR(s)
Km
Gc1
0.12
Gc2
1
Block diagram of cascade control system
f. Gc2=2
Slave control system can be replaced by block with function
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2014/2015 Exam Answer, Control II
0.48
2
* 0.12 * 2
0.48
0.48
2. 5 s + 1
=
= 2. 5 s + 1 =
G=
0.48
2
2.5s + 1 + 0.48 2.5s + 1.48
* 0.12 * 2 * 1 1 +
1+
2.5s + 1
2.5s + 1
TRsp(s)
TR(s)
Gc1
Km
Characteristic equation:
1+OLTF=0= 1 + Gc1 0.48 * 0.6 * 2 = 0
2.5s + 1.48 13s + 1 s + 1
1+
Gc1 0.48
Gc1 0.576
0.6
2
=0=
+1
*
*
2.5s + 1.48 13s + 1 s + 1
(2.5s + 1.48)(13s 2 + 14s + 1)
1+
Gc1 0.48
Gc1 0.576
0.6
2
=0=
+1
*
*
2.5s + 1.48 13s + 1 s + 1
(2.5s + 1.48)(13s 2 + 14s + 1)
(2.5s + 1.48)(13s 2 + 14s + 1) + Gc1 0.576 = 0
(2.5s + 1.48)(13s 2 + 14s + 1) + Gc1 0.576 = 0
32.5s 3 + 35s 2 + 2.5s + 19.24 s 2 + 20.72 s + 1.48 + 0.576Gc1 = 0
32.5s 3 + 54.24 s 2 + 23.22 s + 1.48 + 0.576Gc1 = 0
− 32.5iω3 − 54.24ω2 + 23.22iω + 1.48 + 0.576 Kcu = 0
32.5ω2 = 23.22and − 54.24ω2 + 1.48 + 0.576 Kcu = 0
ω = 23.22 / 32.5 = 0.845andKcu = (54.24ω2 − 1.48) / 0.576 = Kcu = 64.7
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2014/2015 Exam Answer, Control II
Model answer question 3
a. h=0.05; a=0.019; T=115/7=16.4=Tu
b. wu=2pi/T=0.38 rad/min
c. kcu 4h/pi a=4*0.05/(3.14128*0.019)=3.35
d.
i. The Ziegler-Nichols closed-loop or on-line tuning method involves the following steps:
1. Switch off the integral and derivative parts of the controller so the controller is only working
in proportional mode.
2. With the loop closed, increase the proportional gain until the output oscillates with constant
amplitude. The value of the gain at which the loop oscillates is the ultimate gain K cu. Obtain
the period of oscillation T u from a time recording of the oscillation. The ultimate frequency
can be obtained from the period of oscillation T u .
ωu =
2π
Tu
ii.
1. Open the loop (this is most easily done by switching the controller to manual). The controller
output is now disconnected, and the output signal should be constant. Make sure the
system is stable and no disturbances can occur whilst the test is being performed.
2. Apply a step change to the controller output.
3. Record the transmitter output signal as a function of time until it reaches steady state.
4. Estimate the process gain by dividing the steady state change in the transmitter output by
the step change in the controller output.
5. Estimate the dead time t 0 and the time constant τ
6. Calculate the settings of the controller parameters using Ziegler-Nichols tuning formulas for
a quarter decay ratio response (i.e. the amplitude of each successive oscillation is ¼ of the
previous one).
i.
T= Tu= (105-2)/6 =17.2; Kcu = 3.39. wu= 2pi/T = 0.366. Pretty close.
e. i. PID Ziegler Nichols Quarter decay based on autotune.
ProportionalK
T
T
K c' = cu
τ 'I = u
τ 'D = u
integral-derivative,
1.7
8
2
PID*
Kc=3.35/1.7=1.97
τ 'I =
Tu
2
16.4/2=8.2 min
τ 'D =
Tu
8
=16.4/8=2.05
Ii Use Tyreus-Luyben
Tyreus-Luyben controller settings for a more conservative response
Kc
τI
τD
PI
0.31K cu
2.2T u
-
Kc=0.31 Kcu=0.31*3.39=1.05; τ I = 2.2Tu=2.2*17.2=37.84 min
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