Process Dynamics and Control Solutions

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

FT 1

WD

V2

RC 4

V4

FT 2 FC 1

FT 4

FT 3

WC

LC 3

V1

V3

LT 2

LT 1

LC 5

R1

FT 5

R2

WE

P1

P2

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

HB

Bottom 1-11

23-6

3 xB 4 HB

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

* 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

xD (mole fraction A)

0.96

0.95

0.94

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

xB (mole fraction A)

0.016 0.014 0.012 0.01

HD (lb-mol)

400 300 200 100 340 HB (lb-mol)

320 300 280 260

2480 2460 2440 2420 2400

23-13

xD (mole fraction A)

0.955

R (lbmol/hr)

1150 1100

0.95

1050

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 xB (mole fraction A)

20

x 10

10 9 8

200 HD (lb-mol)

10

150 100 50 290 285

HB (lb-mol)

0

280 275 270

2410 2400

HR (lb-mol)

1000

2390 2380 2370

time (h)

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

xB (mole fraction A)

-3

11

x 10

10 9 8

0

20

40

60

0

20

40

60

0.96

0.95

0.94

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