Control Control Station Station
Software For Process Control Analysis, Tuning & Training
Simplifying PID Control. Software Plant Performance. Optimizing For Process Control Analysis, Tuning & Training
Hands-on Workshop Series
Providing Real-World
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7.1
In the absence of more accurate data, use a first-order transfer function as T '( s ) Ke -s = Qi '( s ) s + 1
o T () - T (0) (124.7 - 120) F = = 0.118 qi 540 - 500 gal/min = 3:09 am 3:05 am = 4 min
K=
Assuming that the operator l
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20.1
a)
The unit step response is 1 2e - s 5 20 -s 1 Y ( s ) = G p ( s )U ( s ) = (10 s + 1)(5s + 1) s = 2e s + 5s + 1 - 10s + 1 Therefore, y (t ) = 2 S (t - 1) 1 + e - (t -1) / 5 - 2e - (t -1) / 10 For t = 1.0, S i = y
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18.1
McAvoy has reported the PI controller settings shown in Table S18.1 and the set-point responses of Fig. S18.1a and S18.1b. When both controllers are in automatic with Z-N settings, undesirable damped oscillations result due to the c
Revised: 1-3-04
Chapter 15
15.1 For Ra=d/u
Kp = Ra d =- 2 u u
which can vary more than Kp in Eq. 15-2, because the new Kp depends on both d and u.
15.2 By definition, the ratio station sets (um um0) = KR (dm dm0) Thus
u - u m0 K 2u 2 K 2 u =
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13.1
AR = G ( j) =
3 G1 ( j) G2 ( j) G3 ( j)
= 3 2 + 1 4 2 + 1
=
3 (-) 2 + 1 (2) 2 + 1
From the statement, we know the period P of the input sinusoid is 0.5 min and, thus,
= 2 2 = = 4 rad/min P 0 .5
Substituting the numerical val
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11.1
11.2
1 G c ( s ) = K c 1 + s I The closed-loop transfer function for set-point changes is given by Eq. 11 1 36 with Kc replaced by K c 1 + s , I
H ( s ) H sp ( s )
H ( s ) H sp ( s )
1 1 K c K v K p K m 1 + s
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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 casca
Revised: 1-3-04
Chapter 14
14.1 Let GOL(jc) = R + jI where c is the critical frequency. Then, according to the Bode stability criterion | GOL(jc)| = 1 = R 2 + I 2 GOL(jc) = - = tan 1 (I/R) Solving for R and I: R = -1 and I = 0 Substituting s = jc in
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24.1
a)
i.
First model (full compositions model):
Number of variables: NV = 22 w1 w2 w3 w4 w5 w6 w7 w8 Number of Equations: NE = 17 Eqs. 2-8, 9, 10, 12, 13, 15, 16, 18, 20(3X), 21, 22, 27, 28, 29, 31 Number of Parameters: NP = 4 VR,
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19.1 From definition of xc, 0 xc 1 f(x) = 5.3 x e (-3.6x +2.7) Let three initial points in [0,1] be 0.25, 0.5 and 0.75. Calculate x4 using Eq. 19-8,.
x1 0.25 f1 8.02 x2 0.5 f2 6.52 x3 0.75 f3 3.98 x4 0.0167
For next iteration, select
Rev: 3-30-04
Chapter 21
21.1 No. It is desirable that the minimum value of the output signal be greater than zero, in order to readily detect instrument failures. Thus, for a conventional electronic instrument, an output signal of 0 mA indicates tha
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3.1 a) 1e
[
- bt
sin t = e
0
]
- bt
sin t e dt = sin t e - ( s + b )t dt
- st 0
[- (s + b) sin t - cos t ] = e - ( s + b ) t ( s + b ) 2 + 2 0 = ( s + b) 2 + 2
b)
1 e
[
- bt
cos t = e
0
]
- bt
cos t e dt = c
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4.1
a) b) c) d)
iii iii v v
4.2
a) b) c)
5 10 10 s (10 s + 1) From the Final Value Theorem, y(t) = 10 when t Y (s) = y(t) = 10(1-e-t/10) , then y(10) = 6.32 = 63.2% of the final value.
d) e)
5 (1 - e - s ) (10s + 1) s From the Fina
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5.1
a)
xDP(t) = hS(t) 2hS(t-tw) + hS(t-2tw) h xDP (s) = (1 - 2e-tws + e-2tws) s Response of a first-order process, K h -t s -2t s Y (s) = (1 - 2e w + e w ) s + 1 s or Y(s) = (1 - 2e-tw s + e-2tw s) 1 + 2 s s + 1 Kh Kh 1 = = K
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6.1
a)
By using MATLAB, the poles and zeros are: Zeros: (-1 +1i) , (-1 -1i) Poles: -4.3446 (-1.0834 +0.5853i) (-1.0834 0.5853i) (+0.7557 +0.5830i) (+0.7557 -0.5830i) These results are shown in Fig E6.1
Figure S6.1. Poles and zeros of G
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8.1
a)
For step response,
M s D s +1 s +1 Ya ( s ) = K c D U ( s ) = K c M s ( D s + 1) D s + 1
input is u (t ) = M
,
U ( s ) =
Ya (s ) =
K c M D KcM + D s + 1 s ( D s + 1)
Taking inverse Laplace transform
y (t ) = a
Revised: 1-3-04
Chapter 12
12.1 For K = 1.0, 1=10, 2=5, the PID controller settings are obtained using Eq.(12-14):
1 1 + 2 15 = c K c D = 1 2 = 3.33 1 + 2 Kc =
,
I = 1+2=15
,
The characteristic equation for the closed-loop system is
1
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10.1
Rev: 12-6-03
According to Guideline 6, the manipulated variable should have a large effect on the controlled variable. Clearly, it is easier to control a liquid level by manipulating a large exit stream, rather than a small stream.
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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
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22.1
Microwave Operating States Condition Open the door Place the food inside Close the door Set the time Heat up food Cooking complete Fan OFF OFF OFF ON OFF Light Timer ON OFF OFF ON OFF OFF OFF OFF ON OFF Rotating Microwave Base Gene
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1.1
a) b) c) d) e)
True True True False True
1.2
QL
Q
T
TC
ON/OFF SWITCH
Controlled variable- T (house interior temperature) Manipulated variable- Q (heat from the furnace)
Solution Manual for Process Dynamics and Control, 2nd e