Solutions_F11_ECH157_HW6

Solutions_F11_ECH157_HW6 - 14.8 Kcu and c are obtained...

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Unformatted text preview: 14.8 Kcu and c are obtained using Eqs. 14-7 and 14-8. Including the filter GF into these equations gives -180 = 0 + [-0.2c tan-1(c)]+[-tan-1(Fc)] Solving, c = 8.443 c = 5.985 for for F = 0 F = 0.1 Then from Eq. 14-8, 2 1 1 K cu 2 2 2 c 1 F c 1 Solving for Kcu gives, Kcu = 4.251 Kcu = 3.536 for for F = 0 F = 0.1 for for F = 0 F = 0.1 Therefore, cKcu = 35.9 c Kcu= 21.2 Because cKcu is lower for F = 0.1, filtering the measurement results in worse control performance. 1 14.9 (a) Using Eqs. 14-7 and 14-8, 5 1 1 (1.0) AR OL K c 1 2 2 2 25 100 1 1 = tan-1(-1/5) + 0 + (-2 tan-1(10)) + (- tan-1()) 14-11 Bode Diagram 2 10 1 AR/Kc 10 0 10 -1 10 -2 10 -100 Phase (deg) -150 -200 -250 -300 -350 -2 10 -1 0 10 10 1 10 2 10 Frequency (rad/sec) Figure S14.9a. Bode plot (b) Set = 180 and solve for to obtain c = 0.4695. Then AR OL = 1 = Kcu(1.025) c Therefore, Kcu = 1/1.025 = 0.976 and the closed-loop system is stable for Kc 0.976. (c) For Kc = 0.2, set AROL = 1 and solve for to obtain g = 0.1404. Then g = = -133.6 g From Eq. 14-12, PM = 180 + g = 46.4 (d) From Eq. 14-11 14-12 GM = 1.7 = From part (b), Therefore, 1 1 = Ac AR OL AR OL c c = 1.025 Kc 1.025 Kc = 1/1.7 or Kc = 0.574 Bode Diagram 2 10 1 AR/Kc 10 0 10 -1 10 -2 10 -150 Phase (deg) -180 -200 -250 -300 -350 -2 10 -1 0 10 10 1 10 2 10 Frequency (rad/sec) Figure S14.9b. Solution for part b) using Bode plot. Bode Diagram 2 10 1 AR/Kc 10 0 10 -1 10 -2 10 -150 Phase (deg) -180 -200 -250 -300 -350 -2 10 -1 0 10 10 1 10 2 10 Frequency (rad/sec) Figure S14.9c. Solution for part c) using Bode plot. 14-13 2 14.12 (a) Schematic diagram: TC Hot fluid TT Cold fluid Mixing Point Sensor Block diagram: (b) GvGpGm = Km = 6 mA/mA GTL = e-8s GOL = GvGpGmGTL = 6e-8s If GOL = 6e-8s, | GOL(j) | = 6 GOL (j) = -8 rad Find c: Crossover frequency generates 180 phase angle = radians -8c = - or c = /8 rad/s 14-19 2 16s /8 c 1 1 Find Kcu: Kcu = 0.167 | G p ( j c ) | 6 Find Pu: Pu = 2 Ziegler-Nichols ¼ decay ratio settings: PI controller: Kc = 0.45 Kcu = (0.45)(0.167) = 0.075 I = Pu/1.2 = 16/1.2 = 13.33 sec PID controller: Kc = 0.6 Kcu = (0.6)(0.167) = 0.100 I = Pu/2 = 16/2 = 8 s D = Pu/8 = 16/8 = 2 s (c) 1.4 1.2 1 y 0.8 PID control PI control 0.6 0.4 0.2 0 0 30 60 90 120 t Fig. S14.12. Set-point responses for PI and PID control. 14-20 150 (d) Derivative control action reduces the settling time but results in a more oscillatory response. (a) From Exercise 14.10, 14.13 Gv ( s ) 5.264 0.083s 1 2 (0.432s 1)(0.017 s 1) 0.12 Gm ( s) (0.024s 1) 1 The PI controller is Gc ( s ) 51 0.3s Hence the open-loop transfer function is G p ( s) GOL Gc Gv G p Gm Rearranging, GOL 6.317 s 21.06 1.46 10 s 0.00168s 4 0.05738s 3 0.556s 2 s 5 5 14-21 123456789 8 3 16.1 The difference between systems A and B lies in the dynamic lag in the measurement elements Gm1 (primary loop) and Gm2(secondary loop). With a faster measurement device in A, better control action is achieved. In addition, for a cascade control system to function properly, the response of the secondary control loop should be faster than the primary loop. Hence System A should be faster and yield better closed-loop performance than B. Because Gm2 in system B has an appreciable lag, cascade control has the potential to improve the overall closed-loop performance more than for system A. Little improvement in system A can be achieved by cascade control versus conventional feedback. Comparisons are shown in Figs. S16.1a/b. PI controllers are used in the outer loop. The PI controllers for both System A and System B are designed based on Table 12.1 (τc = 3). P controllers are used in the inner loops. Because of different dynamics the proportional controller gain of System B is about one-fourth as large as the controller gain of System A System A: Kc2 = 1 System B: Kc2 = 0.25 τI=15 τI=15 Kc1=0.5 Kc1=2.5 0.7 Cascade Standard feedback 0.6 0.5 Output 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 100 time Figure S16.1a. System A. Comparison of D2 responses (D2=1/s) for cascade control and conventional PI control. Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp 16-1 In comparing the two figures, it appears that the standard feedback results are essentially the same, but the cascade response for system A is much faster and has much less absolute error than for the cascade control of B 0.7 Cascade Standard feedback 0.6 0.5 Output 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 time 60 70 80 90 100 Figure S16.1b. System B .Comparison of D2 responses (D2=1/s) for cascade control and conventional PI control. Figure S16.1c. Block diagram for System A 16-2 Figure S16.1d. Block diagram for System B 16.2 a) The transfer function between Y1 and D1 is Y1 = D1 Gd 1 ⎛ ⎞ Gc 2Gv 1 + Gc1 ⎜ ⎟ G p Gm1 ⎝ 1 + Gc 2GvGm 2 ⎠ and that between Y1 and D2 is G p Gd 2 Y1 = D2 1 + Gc 2Gv Gm 2 + Gc 2Gv Gm1Gc1G p using Gv = Gp = 5 s +1 , Gd 2 = 1 , 4 , Gm1 = 0.05 , (2s + 1)(4s + 1) 16-3 Gd 1 = 1 , 3s + 1 Gm 2 = 0.2 The complex closed-loop poles arise when the characteristic polynomial is factored. This polynomial is (s2 + s + 0.313) = (s + 0.5 + 0.25 i) (s + 0.5 −0.25i) 1+ 6 ⎛ τ s +1 ⎞ s +1 s +1 + Kc ⎜ I =0 ⎟6 s −3 ⎝ τI s ⎠ s − 3 1 ( τ I + 6τ I + K c 6 τ I ) s 2 + (−3τ I + 6τ I + 6τ I K c + 6 K c ) s + Kc 6 = 0 The poles are also the roots of the characteristic equation: Hence, the PI controller parameters can be found easily: K c = 0.052 τ I = 0.137 4 16.7 Using MATLAB-Simulink, the block diagram for the closed-loop system is shown below. Figure S16.7a. Block diagram for Smith predictor 16-14 represents the time-delay term e-θs. where the block The closed-loop response for unit set-point and disturbance changes are shown below. Consider a PI controller designed by using Table 12.1(Case A) with τc = 3 and set Gd = Gp. Note that no offset occurs, Servo response Regulatory response 1.2 1 Output 0.8 0.6 0.4 0.2 0 0 5 10 15 time 20 25 30 Figure S16.7b. Closed-loop response for setpoint and disturbance changes. 16.8 The block diagram for the closed-loop system is Figure S6.8. Block diagram for the closed-loop system ⎛ 1 + τI s ⎞ where Gc = K c ⎜ −θs ⎟ ⎝ 1 + τI s − e ⎠ 16-15 and Gp = K p e−θs 1 + τs 15.3 (a) The block diagram is the same as in Fig. 15.11 where Y H2, Ym H2m, Ysp H2sp, D Q1, Dm Q1m, and U Q3. b) (A steady-state mass balance on both tanks gives 0 = q1 – q3 or Q1 = Q3 (in deviation variables) (1) From the block diagram, at steady state: Q3 = Kv Kf Kt Q1 From (1) and (2), Kf = 1 K v Kt (2) c) 5 (No, because Eq. 1 above does not involve q2. (b) From the block diagram, exact feedforward compensation for Q1 would result when 15.4 Q1 + Q2 = 0 Substituting Gf = Q2 = KV Gf Kt Q1, 1 K v Kt 15-2 (c) Same as part (b), because the feedforward loop does not have any dynamic elements. (d) For exact feedforward compensation Q4 + Q3 = 0 From the block diagram, (1) Q2 = KV Gf Kt Q4 (2) Using steady-state analysis, a mass balance on tank 1 for no variation in q1 gives Q2 Q3 = 0 (3) Substituting for Q3 from (3) and (2) into (1) gives Q4 + KV Gf Kt Q4 = 0 Gf = or 1 K v Kt For dynamic analysis, find Gp1 from a mass balance on tank 1, A1 dh1 q1 q2 C1 h1 dt 15-3 Linearizing (4), noting that q1 = 0, and taking Laplace transforms: A1 dh C q2 1 h1 dt 2h 1 or (2 h1 / C1) H1( s) Q2 ( s) (2 A h / C ) s 1 111 q3 C1 h1 Since q3 C1 2 h1 or h1 (5) Q3 ( s ) C 1 H1 ( s) 2 h 1 (6) From (5) and (6), Q3 ( s) 1 GP1 Q2 ( s) (2 A h / C ) s 1 111 Substituting for Q3 from (7) and (2) into (1) gives Q4 ( or 1 ) K v G f Kt Q4 0 (2 A1 h1 / C1 ) s 1 Gf 1 [(2 A1 h1 / C1 ) s 1] K v Kt 15.5 (a) For a steady-state analysis: Gp=1, Gd=2, From Eq.15-21, Gf = (b) Gd 2 2 Gv Gt G p (1)(1)(1) Using Eq. 15-21, 15-4 Gv = Gm = Gt =1 (7) ...
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This note was uploaded on 01/28/2012 for the course ECH 157 taught by Professor Palagozu during the Fall '08 term at UC Davis.

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