Assignment#5

Assignment#5 (CHE446, Winter 2011)
Problem 7.7
(a)
Goverall (s) =
Ke−θs
(τ1 s + 1)(τ2 s + 1)
From Figure E7.7:
time delay: θ = 4 min
10 − 0
Gain: K =
= 10
1−0
Smith’s method (for obtaining the time constants):
10 ∗ 0.2 = 2 ⇒ t20% + θ ≈ 5.6 ⇒ t20% = 1.6 min
10 ∗ 0.6 = 6 ⇒ t60% + θ ≈ 9.1 ⇒ t60% = 5.1 min
t60%
5.1
t20%
1.6
Figure 7.7
= 0.314 −−−−−→ ξ = 1.63 ⇒
= 3.1 ⇒ τ =
= 1.645
=
t60%
5.1
τ
3.1
p
τ1 = τ ξ + τ ξ 2 − 1 = 4.81 min
p
τ1 = τ ξ − τ ξ 2 − 1 = 0.56 min
(b)
Goverall (s) = GHE ∗ (s)Gpipe (s)GT C † (s) =
10e−4s
(4.81s + 1)(0.56s + 1)
Assuming plug-flow in the pipe with constant-velocity:
Gpipe (s) = e−θp s
3
1
Distance
⇒ θp =
∗
= 0.1 min
V elocity =
time
0.5 60
⇒ Gpipe (s) = e−0.1s
Knowing from the problem statement that the thermocouple has a first order behavior and its time
constant is considerably smaller than the heat exchanger and also assuming it has unit gain and no time
delay:
GT C (s) =
∗
†
KT C
1
⇒ GT C (s) =
τT C s + 1
0.56s + 1
HE: Heat Exchanger
TC: Thermocouple
1
From the problem statement it is known that the heat exchanger has a FOPTD behavior; GHE (s) can
be obtained from the remaining portion of Goverall (s):
GHE (s) =
KHE e−θHE s
;
τHE s + 1
⇒ GHE (s) =
θ = θpipe + θHE ⇒ θHE = 4 − 0.1 = 3.9 min
10e−3.9s
4.81s + 1
Problem 7.10
Gain:
K=
◦
C
∆Y
890 − 850
=
= −0.8
∆U
950 − 1000
cf m
Assume that the input change in air flow rate is made at t = 2+ min so that the observed input first
changes at t = 3 min ; the output first changes at t = 5 min. This means that the time delay is two
sampling periods, i.e., θ = 2 min.
890
T (oC)
880
Observed Output
first Changes
870
860
850
840
0
1
2
1
2
3
4
5
6
7
Time(min)
8
9
10
11
12
13
3
4
5
6
7
Time(min)
8
9
10
11
12
13
1000
q (Input change)
990
980
970
960
950
0
Observed Input
first Changes
T63.2% = 850 + (890 − 850) ∗ 0.632 = 875.3◦ C
Interpolating between t = 7 min and t = 8 min gives‡ :
t63.2% =
‡
7−8
(875.3 − 873) + 7 = 7.46 min
873 − 878
Time constant can also be determined using graphical methods.
2
⇒ t63.2% = τ + θ + t(0)§ → τ = 7.46 − 2 − 3 = 2.46
G(s) =
−0.8e−2s
2.46s + 1
Problem 8.4
In system I the control valve is A-O meaning that an increasing pressure signal from the controller will
open the valve more, thus increasing the flow rate:
P¯ ↑⇒ U¯ ↑
(1)
In system II the control valve is A-C meaning that an increasing pressure signal from the controller will
close the valve more, thus decreasing the flow rate:
P¯ ↑⇒ U¯ ↓
(2)
(a)
System I (Air-to-Open valve):
P¯ ↑⇒ U¯ ↑⇒ Kv > 0
System II (Air-to-Close valve):
P¯ ↑⇒ U¯ ↓⇒ Kv < 0
(b)
System I (Air-to-Open valve):
If E = Fsp − F ¶ < 0 ⇒ F > Fsp or F increases ⇒ Need to close the valve (U¯ ↓) and since the valve is
A-O ( i.e: (1) ) need to decrease the controller output (P¯ ↓) ⇒ Reverse-acting controller.
System II (Air-to-Close valve):
If E = Fsp − F < 0 ⇒ F > Fsp or F increases ⇒ Need to close the valve (U¯ ↓) and since the valve is
A-C ( i.e: (2) ) need to increase the controller output (P¯ ↑) ⇒ Direct-acting controller.
(c)
System I: Reverse-acting controller ⇒ Kc > 0.
System II: Direct-acting controller ⇒ Kc < 0.
Example 11.4
In order to reproduce Figure 11.23, either the closed-loop transfer function (11-76) or the feed-back
control system in Figure 11.8, can be used.
§
¶
t(0) corresponds to the time when the input first changes.
F: Flow
3
Figure 1: SIMULINK Model
EFFECT OF CONTROLLER GAINS ON CLOSED−LOOP RESPONSE
TO A UNIT STEP CHANGE IN SET POINT
3
2.5
2
1.5
Y
1
0.5
0
Ysp
−0.5
Kc=2
Kc=6
−1
K =15
c
−1.5
0
2
4
6
8
10
Time(min)
12
14
Figure 2: Y vs. Time
4
16
18
20
Example 12.8
The identified FOPDT transfer function is used to reproduce Figure 12.16:
0
1.54e−1.07s
Xm
(s)
= G(s) =
P 0 (s)
5.93s + 1
Figure 3: SIMULINK Model
P(%)
43
30
0
5
10
15
Time (min)
(a) Controller output
20
25
30
5
10
15
20
25
Time (min)
(b) Response of the estimated process tranfer function
to a step change in the controller output
30
65
xm(%)
55
45
35
25
15
0
Figure 4: simulation result
5
Since the xm vs. time plot in Figure 4 is produced using the estimated transfer function G(s), it is
slightly different from Figure 12.8 which is obtained from the real data.
Problem 12.5
G(s) = GIP (s)Gv (s)Gp (s)Gm (s) =
Ke−θs
τs + 1
GIP (s) = KIP
Gv = Kv
Gp (s)Gm (s) can be estimated using the given data in the table:
16.9 − 12
mA
= 2.45
20 − 18
psig
Time delay: θ = 2 min
Time constant:
12 + (16.9 − 12) ∗ 0.632 = 15.05
Gain: Kp Km =
Interpolating between t = 5 and t = 6 mink :
t63.2% =
5−6
(15.05 − 14.8) + 5 = 5.4 min
14.8 − 15.4
t63.2% = θ + τ + t(0) ⇒ τ = 5.4 − 2 − 0 = 3.4 min
Gp (s)Gm (s) =
k
2.45e−2s
3.4s + 1
Model parameters for Gp Gm can also be identified using graphical methods.
6
16.9
17
16.5
16
15.5
15.05
T2m (mA)
15
14.5
14
13.5
13
12.5
12
0
1
2
3
4
5 5.4
Observed output
first changes
6
t (min)
7
8
9
10
11
12
Figure 5: T2m vs. time
⇒
G(s) = GIP (s)Gv (s)Gp (s)Gm (s)
2.45e−2s
= KIP Kv
3.4s + 1
psi
mA
e−2s
psi
∗ 0.9
∗ 2.45
∗
= 0.75
mA
psi
psi 3.4s + 1
|
{z
}
K=1.65
−2s
G(s) =
(a)
Since
12.1:
θ
τ
=
2
3.4
1.65e
3.4s + 1
≈ 0.6 > 0.25, a conservative choice of τc = 21 τ = 1.7 min is used. From case H in Table
τ + 2θ 1
3.4 + 22 1
=
= 0.99
1.7 + 22 1.65
τc + 2θ K
θ
2
τI = τ + = 3.4 + = 4.4 min
2
2
τθ
3.4 ∗ 2
τD =
=
= 0.77 min
2τ + θ
2 ∗ 3.4 + 2
Kc =
7
(b)
From Table 12.3, the ITAE PID settings for a set-point change:
2 −0.85
)
⇒ Kc = 0.92
3.4
τ
2
= 0.796 − 0.1465( ) ⇒ τI = 4.79 min
τI
3.4
τD
2
= 0.308( )0.929 ⇒ τD = 0.64 min
τ
3.4
KKc = 0.965(
(c)
From Table 12.3, the ITAE PID settings for a step disturbance:
KKc = 1.357(
2 −0.947
)
⇒ Kc = 1.36
3.4
2
τ
= 0.842( )−0.738 ⇒ τI = 2.73 min
τI
3.4
τD
2
= 0.381( )0.995 ⇒ τD = 0.76 min
τ
3.4
8

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