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Unformatted text preview: 14.8 Kcu and c are obtained using Eqs. 147 and 148. Including the filter GF
into these equations gives
180 = 0 + [0.2c tan1(c)]+[tan1(Fc)]
Solving,
c = 8.443
c = 5.985 for
for F = 0
F = 0.1 Then from Eq. 148, 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. 147 and 148, 5
1
1 (1.0)
AR OL K c 1 2 2
2
25 100 1 1 = tan1(1/5) + 0 + (2 tan1(10)) + ( tan1()) 1411 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 closedloop 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. 1412, PM = 180 + g = 46.4
(d) From Eq. 1411 1412 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. 1413 2
14.12 (a) Schematic diagram:
TC
Hot fluid
TT Cold fluid
Mixing Point Sensor Block diagram: (b) GvGpGm = Km = 6 mA/mA
GTL = e8s
GOL = GvGpGmGTL = 6e8s
If GOL = 6e8s,
 GOL(j)  = 6 GOL (j) = 8 rad Find c: Crossover frequency generates 180 phase angle = radians
8c =  or c = /8 rad/s 1419 2 16s
/8
c
1
1
Find Kcu: Kcu = 0.167
 G p ( j c )  6 Find Pu: Pu = 2 ZieglerNichols 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. Setpoint responses for PI and PID control. 1420 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 openloop 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 1421 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 closedloop performance than
B.
Because Gm2 in system B has an appreciable lag, cascade control has the
potential to improve the overall closedloop 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 onefourth 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 161 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 162 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) 163 Gd 1 = 1
,
3s + 1 Gm 2 = 0.2 The complex closedloop 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 MATLABSimulink, the block diagram for the closedloop system
is shown below. Figure S16.7a. Block diagram for Smith predictor 1614 represents the timedelay term eθs. where the block The closedloop response for unit setpoint 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. Closedloop response for setpoint and disturbance changes. 16.8 The block diagram for the closedloop system is Figure S6.8. Block diagram for the closedloop system ⎛ 1 + τI s ⎞
where Gc = K c ⎜
−θs ⎟
⎝ 1 + τI s − e ⎠ 1615 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 steadystate 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 152 (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 steadystate 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 153 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 steadystate analysis:
Gp=1, Gd=2, From Eq.1521,
Gf =
(b) Gd
2 2
Gv Gt G p (1)(1)(1) Using Eq. 1521, 154 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.
 Fall '08
 PALAGOZU

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