ch23 - 123456789 8 23.1 Option (a): • Production rate set...

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Unformatted text preview: 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 = 1 τ , the settings are: 4 xD - R loop xB - V loop c) { { 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) 2 1 D 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 0 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 20 xB (mole fraction A) 1800 10 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 30 180 500 10 0 10 340 320 HB (lb-mol) B (lb-mol/hr) 520 0 480 300 460 440 280 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 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 0.016 V (lbmol/hr) 1800 0.014 1600 0.012 0 10 20 30 40 50 0.01 700 400 HD (lb-mol) D (lb-mol/hr) 650 300 600 200 550 500 0 xB (mole fraction A) 2000 1400 0.94 50 0 10 20 30 40 100 50 340 500 320 HB (lb-mol) B (lb-mol/hr) 520 480 300 460 440 280 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 9 8 50 200 HD (lb-mol) D (lb-mol/hr) 550 500 150 450 400 100 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) 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 0 1590 10 20 30 0 10 20 30 0 10 20 30 0 10 20 30 0.01 0.0095 0 10 20 30 510 190 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 30 0.0105 1595 450 20 0 xB (mole fraction A) V (lbmol/hr) 1600 470 10 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 (mole fraction A) 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 xD (mole fraction A) (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 = UA(T − T ) Q R C 1 = CP F0 (T0 − TRef ) + CP D (TD − TRef ) − C P F (TR − TRef ) − H R λ kz − Q 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 0.4 60 30 40 50 60 30 40 50 60 1200 1150 TC (R) 594 593 1100 592 1050 591 1000 590 30 40 50 60 0.341 950 700 650 D (lb-mol/hr) 0.3405 KR (lb-mol/hr) 60 1250 595 50 1300 596 40 F (lb-mol/hr) 597 30 600 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 ...
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This homework help was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.

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