123456789 8 23.1 Option (a): • Production rate set via setpoint of wA flow controller • Level of R1 controlled by manipulating wC • Ratio of wB to wA controlled by manipulating wB • Level of R2 controlled by manipulating wE • Ratio of wD to wC controlled by adjusting wD Options (b)-(e) are developed similarly. See table below. Option a
b
c
d
e
•
Production Rate Set With wA wA wA wA wA wB wB wB wB wB wC wC wC wC wC wD wD wD wD wD wE wE wE wE wE
Control Loop # 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
Type of Controller Flow Ratio Level Ratio Level Flow Ratio Level Ratio Level Flow Ratio Level Ratio Level Flow Ratio Level Ratio Level Flow Ratio Level Ratio Level
Controlled Variable wA,m wB,m HR1 wD,m HR2 wB,m wA,m HR1 wD,m HR2 wC,m wB,m HR1 wD,m HR2 wD,m wC,m HR1 wB,m HR2 wE,m wD,m HR2 wB,m HR1
Manipulated Variable wA (V1) wB (V2) wC (V3) wD (V4) wE (V5) wB (V2) wA (V1) wC (V3) wD (V4) wE (V5) wC (V3) wB (V2) wA (V1) wD (V4) wE (V5) wD (V4) wC (V3) wA (V1) wB (V2) wE (V5) wE (V5) wD (V4) wC (V3) wB (V2) wA (V1)
Subscript m denotes “measurement”. Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
23-1
In options c, d, and e, valves 1 and 2 can be used interchangeably. Thus, a total of 8 options can be developed.
Advantages and Disadvantages: Each option is equivalent in the sense that 5 control loops are required: 1 flow, 2 level, and 2 ratio. Since there is no cost or complexity advantage with any option, the production rate should be set via the actual product rate, wE , i.e. option e.
WB
RC 2
A
FT 1
B
WD
V2
RC 4
Ratio
A
FT 2
B Ratio
FC 1
V4
FT 4
FT 3
WA WC
LC 3
V1
V3
LT 2
LT 1
LC 5
R1
FT 5
R2
WE
P1
P2
Figure S23.1. Solution for option a)
23-2
V5
23.2
a)
The level in the distillate (HD) will be controlled by manipulating the recycle flow rate (D), and the level in the reboiler (HB) via the bottoms flow rate (B). Thus, HD and HB are pure integrator elements. Closed-loop TF development assuming PI controller: τI s + 1
GCL ( s ) = (
τI )s2 + τI s + 1 Kc K p
Relations:
τI = τ2 Kc K p
τ I = 2ζτ
K p = −1 for both HD and HB loops.
12345 = 1 for a critically damped response Initial settings: K c = −0.4 τ I = 10 Final tuning: changed to proportional control only to obtain a faster response K c = −1
b)
The distillate composition (xD) will be controlled by manipulating the reflux flow rate (R), and the bottoms composition (xB) via the vapor boilup (V). Use a step response to determine an approximate first-order model (calculations are shown on last page). xD 0.0012 = R 2.33s + 1 xB −0.000372 = V 2.08s + 1
23-3
Using the Direct Synthesis method: K τs + 1 1 τ Gc ( s ) = (1 + ) τc K τs G (s) =
Choosing τc =
c)
1 τ , the settings are: 4
xD - R loop
{
xB - V loop
{
K c = 3333.3 τ I = 2.33 K c = −10649.6 τ I = 2.08
The reactor level (HR) will be controlled by manipulating the flow from the reactor (F). HR is a pure integrator element. Using the same relations as in part a, initial controller settings are: K c = −0.4 τ I = 10 After tuning: K c = −1 τI = 5
23-4
Figure S23.2a. Simulink-MATLAB block diagram for case a)
1
D
2
0.3408
xD
KR
DxD/HR
KRz/HR
1 s 3 F0
5 F
Integrator z
1 s flow sum
z
Level integrator Sum-z
4 z0
2
F0z/HR F0z0/HR
Dz/HR 1 HR
Figure S23.2b. Simulink-MATLAB block diagram for the CSTR block
23-5
2
R
R
V
23.5
xD
Hs
Hs
2
xD
1
D
D
y (i+1)
x(i-1)
1
HD
HD
Top 13-20
x(i-1) L yi
3
F
F
4
z
z
V LF H
0
xi
xi
y (i+1)
FeedTray 12
x(i-1)
yi
Hs R F
5 V
xB
V
6
B
B
3 xB
HB
Bottom 1-11
4 HB
Figure S23.2c. Simulink-MATLAB block diagram for the Tower block
23-6
600
300 280 260
560
HD (lb-mol)
D (lb-mol/hr)
580
540
240 220
520
500
200
0
5
10
15
20
25
180
30
460
5
10
15
20
25
30
0
5
10
15 time (h)
20
25
30
280 260
440
240 420 HB (lb-mol)
B (lb-mol/hr)
0
400
380
360
220 200 180
0
5
10
15 time (h)
20
25
30
160
Figure S23.2d. Step change in V (1600-1700) at t=5
23-7
0.955
xD (mole fraction A)
1250 1200 R (lbmol/hr)
0.95
1150
0.945
1100 1050
0.94 0
10
20
30
20
30
0
10
20
30
20
30
20
30
0.02
1750
0.018 0.016
V (lbmol/hr)
1700 1650
0.014
1600 1550
10
xB (mole fraction A)
1800
0
0.012
0
10
20
30
0.01
time (h) 560
260
D (lb-mol/hr)
240 HD (lb-mol)
540
220
520
500
200
0
10
20
30
180
340
500
320
480
0
10
300
460 440
10
HB (lb-mol)
B (lb-mol/hr)
520
0
280
0
10
20
30
time (h)
260
time (h)
Figure S23.2e. Step change in F (960-1060) at t=5
23-8
xD (mole fraction A)
1160
0.952
1140 R (lbmol/hr)
1120
0.948
1100 1080
0.95
0
10
20
30
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0.014
V (lbmol/hr)
0.013
1650
0.012
1600 1550
0.011
0
10
20
30
0.01
540
240 HD (lb-mol)
D (lb-mol/hr)
530
220
520
200
510 500
0
10
20
180
30
290
470
285
HB (lb-mol)
B (lb-mol/hr)
475
465
280
460 455
275
0
10
20
270
30
1020
2440 2430
HR (lb-mol)
F (lb-mol/hr)
0
xB (mole fraction A)
1700
0.946
1000
2420
980 960
2410
0
10
20
30
time (h)
2400
time (h)
Figure S23.2f. Step change in F0 (460-506) at t=5
23-9
* Calculation of First-Order Model Parameters for xD and xB Loops xD -R: A step change in the reflux rate (R) of +10 lbmol/hr is made and the resulting response is used to fit a first-order model: K τs + 1 ∆x 0.9624 − 0.950 K= D = = 0.0012 ∆R 10
G (s) =
672489 4 432427 7243 44 0.632(∆xD ) = (0.632)(0.012) = 0.007584 τ = time( xD = 0.957584) = 12.33 − 10 = 2.33 xB -V: Similarly, a step change in the vapor boilup (V) of +10 lbmol/hr is made: K=
∆xB 0.00678 − 0.0105 = = −0.00372 ∆V 10
Use 63.2% of the response to find 0.632(∆xB ) = (0.632)(−0.00372) = −0.00235 τ = time( xB = 0.00815) = 12.08 − 10 = 2.08 0.966
0.964
0.962
xD (mol fraction A)
0.96
0.958
0.956
0.954
0.952
0.95
0.948
0
5
10
15
20
Time (hr)
Figure S23.2g. Responses for step change in the reflux rate R
23-10
25
-3
11
x 10
10.5
10
xB (mol fraction A)
9.5
9
8.5
8
7.5
7
6.5
0
5
10
15
20
25
Time (hr)
Figure S23.2h. Responses for step change in the vapor boilup V.
23-11
23.3
The same controller parameters are used from Exercise 23.2
Figure S23.3a. Simulink-MATLAB block diagram
23-12
0.96
R (lbmol/hr)
xD (mole fraction A)
1300 1200
0.95
1100 1000
0
10
20
30
40
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
V (lbmol/hr)
0.014
1600
0.012
0
10
20
30
40
50
0.01
700
400 HD (lb-mol)
D (lb-mol/hr)
650
300
600
200
550 0
10
20
30
40
100
50
340
500
320
HB (lb-mol)
B (lb-mol/hr)
520
480
300
460 440
10
0.016
1800
500
0
xB (mole fraction A)
2000
1400
0.94
50
280
0
10
20
30
40
260
50
2480 HR (lb-mol)
1200 F (lb-mol/hr)
2460
1100
2440
1000 900
2420 0
10
20
30
40
50
2400
time (h)
time (h)
Figure S23.3b. Step change in F0 (+10%) at t=5
23-13
xD (mole fraction A)
0.955
R (lbmol/hr)
1150 1100
0.95
1050 1000
0
10
20
30
40
50
0.945
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
-3
11
V (lbmol/hr)
xB (mole fraction A)
1700 1600
10
1500 1400
0
10
20
30
40
HD (lb-mol)
D (lb-mol/hr)
200
500
150
450
100
0
10
20
30
40
50
50
290
470
285
HB (lb-mol)
B (lb-mol/hr)
475
465
280
460 455
275
0
10
20
30
40
270
50
1000
2410 2400
HR (lb-mol)
F (lb-mol/hr)
9 8
50
550
400
x 10
950
2390
900 850
2380
0
10
20
30
40
50
2370
time (h)
time (h)
Figure S23.3c. Step change in z0 (-10%) at t=5
23-14
23.4 The flow controller on F, the column feed stream, should be simulated in MATLAB as a constant flow. The controller parameters used are taken from those derived in Exercise 23.2.
0.952
R (lbmol/hr)
xD (mole fraction A)
1140 1120 1100 1080
0
10
20
30
40
50
0.95
0.948
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0
10
20
30
40
50
0.011
V (lbmol/hr)
xB (mole fraction A)
1610 1600
0.0105
1590 1580
0
10
20
30
40
50
0.01 200
500
180 HD (lb-mol)
D (lb-mol/hr)
520
480
160
460 440
140
0
10
20
30
40
120
50
340
500
320
HB (lb-mol)
B (lb-mol/hr)
520
480
300
460 440
280
0
10
20
30
40
520
3000 HR (lb-mol)
500
2800
480 460
260
50
F0 (lb-mol/hr)
a)
2600
0
10
20
30
40
50
time (h)
2400
time (h)
Figure S23.4a. Step change in F0 (+10%) at t=5
23-15
Using the approximate relation (23-17), a +10% step change in F0 will result in a reactor holdup of: z0 0.90 = = 2800 lbmol 1 1 1 1 k R ( − ) 0.34( − ) 506 960 F0 F
HR ≈
Using the exact relation (23-6, rearranged): F 0 z 0 − Bx B (506)(0.9 − 0.0105) = = 2910 lbmol (0.34)(0.455) kR z
HR =
The value taken from the graph (2910) matches up with the expected value from the equation without the approximation. 3000
2900
2800
HR (lb-mol)
b)
2700
2600
2500
2400
0
5
10
15
20
25
30
35
40
45
time (h)
Figure S23.4b. Step change in F0 (+10%) at t=5
23-16
50
23.5 a),b) Feedforward control is implemented using the HR-setpoint equation: H R (t ) ≈
z0 (t ) 1 1 − ) kR ( F0 (t ) F
Empirical adjustment of the feedforward equation is required because it is not exact: H R (t ) ≈
z0 (t ) + 70 1 1 kR ( − ) F0 (t ) F
This adjustment matches the initial values of HR (i.e., with and without feedforward control). Parts a and b are represented graphically.
23-17
R (lbmol/hr)
xD (mole fraction A)
1120 1110 1100 1090
0
10
20
0.95
30 xB (mole fraction A)
V (lbmol/hr)
1600
1590
30
0
10
20
30
0
10
20
30
0
10
20
30
0
10
20
30
0.0095 0
10
20
30 190 HD (lb-mol)
D (lb-mol/hr)
500
180
490
170
480 0
10
20
160
30
490
300 HB (lb-mol)
B (lb-mol/hr)
480
290
470
280
460 0
10
20
270
30
500
2500 HR (lb-mol)
F0 (lb-mol/hr)
2400
400
2300
300 200
20
0.01
510
450
10
0.0105
1595
470
0 0.011
1605
1585
no control FF control
2200
0
10
20
30
2100
time (h)
time (h)
Figure S23.5a. Step change in z0 (-10%) at t=5
23-18
c)
The controlled plant response is much faster with the feedforward controller (~10 hours settling time versus ~20 hours without it).
d)
Advantage: Faster response. Disadvantage: Have to measure or estimate two flow rates and one concentration, therefore significantly more expensive.
a)
Use a flow controller to keep F constant (make F a constant in the simulation).
b)
Use ratio control to set F. The ratio should be based on the initial steady state values (960/460). Therefore, as F0 changes, F will be controlled to the corresponding value set by the ratio.
23.6
Parts a) and b) show very different results for the two alternatives. With alternative # 3, feedforward control is necessary to keep the level in the distillate receiver from integrating. However, in alternative # 4, the control structure without the feedforward loop is superior to that with feedforward control. Responses are displayed with controlled variables adjacent to their corresponding manipulated variable.
23-19
1500
HD (lb-mol)
R (lb-mol/hr)
2500 2000 1500 1000
0
20
40
0
40
60
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0.016
V (lbmol/hr)
0.014 0.012
0
20
40
60
0.01 0.52
D (lbmol/hr)
z (mole fraction A)
600
500
0.5
0
20
40
60
0.48
2800 HR (lb-mol)
1100 1050
F (lbmol/hr)
250
x B (mole fraction A)
20
2000
2600
1000
2400
0
20
40
60
2200 1 x D (lbmol/hr)
F0 (lbmol/hr)
520 500
0.95
480 460
40
HB (lb-mol)
B (lb-mol/hr)
0
2500
950
20
300
3000
400
0
350
500
1500
500
60
550
450
No control (a) FF control (b)
1000
0.9 0
20
40
60
time (h) time (h)
Figure S23.6a. Alternative #3 (with and without FF controller). Step change in F0 (+10%) at t=10
23-20
400
1200
300
HD (lb-mol)
R (lb-mol/hr)
1300
1100 1000
0
20
40
100
60
0
20
40
250
60
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
x B (mole fraction A)
V (lbmol/hr)
0.02
1800
0.015
1600
0.01
0
20
40
60
0.005 0.98
D (lbmol/hr)
550
x D (mole fraction A)
600
0.96
500
0
20
40
60
0.94
3000
HR (lb-mol)
1100 1050
2500
1000 950
0
20
40
60
0.6
F0 (lb-mol/hr)
520 500 480 460
2000
z (mol fraction A)
F (lb-mol/hr)
40
300
2000
450
20
HB (lb-mol)
500
1400
0
350
B (lb-mol/hr)
550
450
No control (a) FF control (b)
200
0.5
0
20
40
60
time (h)
time (h)
Figure S23.6b. Alternative #4 (with and without FF controller). Step change in F0 (+10%) at t=10
23-21
23.7
Parts a) and b) can be satisfied by combining two or more of the previous simulations into one to compare the results together. To compare how the alternatives match up, in terms of the snowball effect, a set of arrays has been constructed. All arrays are of the form: D F 0 D z 0
HR F0 HR z0
where the response of D or HR is analyzed as a result of a step change in F0 or z0. In the notation below: S represents the occurrence of the snowball effect (>20% change in steadystate output for a 10% change in input). A represents an acceptable response (~10% change in steady-state output). B represents the best possible response (no change in steady-state output). Alternative #1 S B S B
Alternative #2 B S B A
Alternative #3 A A B A
Alternative #4 A A B A
These results indicate that Alternative #2 still exhibits a snowballing characteristic, but in HR instead of D. Alternatives #3 and #4, on the other hand, eliminate the effect altogether.
23-22
400
R (lb-mol/hr)
1400
HD (lb-mol)
300
1200
200
1000
0
20
40
100
60
60
Alternative #1 Alternative #2 Alternative #3 Alternative #4
300
0
20
40
250
60
0
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
40
60
1800
xD (mole fractionxB A) (mole fraction A)
0.02
V (lbmol/hr)
2000
0.015
1600 1400
40
HB (lb-mol)
500
450
20
350
B (lb-mol/hr)
550
0
0.01
0
20
40
60
0.96
D (lbmol/hr)
700
0.005
600
0.94
z (mole fraction A)
500 400
0
20
40
60
3000 HR (lb-mol)
0.6
0.5
0.4
0.92
2500
0
20
40
60
2000 1200
500
1100
F (lb-mol/hr)
F0 (lb-mol/hr)
520
480 460
time (h)
1000
0
20
40
60
900
0
20
Figure S23.7a. Step change in F0 (+10%) at t=10
23-23
300
HD (lb-mol)
R (lb-mol/hr)
1150 1100
200
1050 1000
100
0
20
40
0
60
40
60 Alternative Alternative Alternative Alternative
290 280
0
20
40
60
V (lbmol/hr)
1700 1600 1500
0
20
40
60
11
#1 #2 #3 #4
0 -3 x 10
20
40
60
0
20
40
60
0
20
40
60
0
20
40
60
40
60
10 9 8
0.96
D (lbmol/hr)
550
270
xB (mole fraction A) xD (mole fraction A)
460
1400
20
HB (lb-mol)
480
440
0
300
B (lb-mol/hr)
500
500
0.95
450 400
0
20
40
60
2600 2400
HR (lb-mol)
z (mole fraction A)
0.6
2200
0
20
40
60
0.9
2000
time (h)
1000 F (lb-mol/hr)
z0 (mol fraction A)
0.5
0.85
0.8
0.94
0
20
40
60
950 900 850
0
20
Figure S23.7b. Step change in z0 (-10%) at t=10
23-24
23.8
Begin with a dynamic energy balance on the reactor: CP
d ( H R (TR − TRef ))
dt 1 Q = UA(TR − TC )
= CP F0 (T0 − TRef ) + CP D (TD − TRef ) − C P F (TR − TRef ) − H R λ kz − Q1
This model can be simplified using the mass balance: dH R = F0 + D − F dt
And, rearranging to get an equation for modeling the reactor temperature: 1 dTR = [ F0CP (T0 − TR ) + DCP (TD − TR ) − UA(TR − TC ) − H R λ kz ] dt CP H R
It is clear from the following figures that the temperature loop is much faster than the interconnected level-flow loops. This characteristic allows the reaction rate multiplier to settle before it can affect the other variables.
23-25
616.45 Thermal model Constant temperature
616.44
z (mol fraction A)
0.55
TR (R)
616.43 616.42
0.5
0.45
616.41
616.4 30
40
50
1300
596
1250
595
1200
40
50
60
30
40
50
60
30
40
50
60
TC (R)
1150
593
1100
592
1050
591
1000
590
30
F (lb-mol/hr)
597
594
30
40
50
60
0.341
950
700
650 D (lb-mol/hr)
0.3405 KR (lb-mol/hr)
0.4
60
600
0.34
550
0.3395 500
0.339
30
40
50
60
450
Time (hr)
Figure S23.8a. Step change in F0 (+10%) at t=30 (Constant temperature simulation does not include a thermal model)
23-26
616.45 Thermal model Constant temperature
616.44
z (mol fraction A)
0.55
TR (R)
616.43 616.42
0.5
0.45
616.41
616.4 30
40
50
0.4
60
600
30
40
50
60
30
40
50
60
30
40
50
60
1000
F (lb-mol/hr)
598
TC (R)
596
594
950
900
592
590
30
40
50
60
0.341
850
550
500
D (lb-mol/hr)
KR (lb-mol/hr)
0.3405
0.34
450
0.3395
0.339
30
40
50
60
400
Time (hr)
Figure S23.8b. Step change in z0 (-10%) at t=30 (Constant temperature simulation does not include a thermal model)
23-27
