flow rate control system second order plus dead time model april 20, 2006 u.t.c engineering 329
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![Page 1: FLOW RATE CONTROL SYSTEM SECOND ORDER PLUS DEAD TIME MODEL April 20, 2006 U.T.C Engineering 329](https://reader035.vdocuments.us/reader035/viewer/2022062714/56649d485503460f94a23f04/html5/thumbnails/1.jpg)
FLOW RATE CONTROL SYSTEM
SECOND ORDER PLUS DEAD TIME MODEL
April 20, 2006
U.T.C
Engineering 329
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Yellow Team
• Jimy George
• Jeff Lawrence
• Taylor Murphy
• Jennifer Potter
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Outline
• Flow System Background
• SOPDT System Theory
• Model Results
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Outline
• Proportional Controller
• Comparison of FOPDT & SOPDT results
• Conclusion
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Flow System Setup
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Block Diagram
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Laplace Domain
R(s) C(s)
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Time Domain
1)()(
1
12
2)(
1
12
10
02
01
tttt
eettUAKtc
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Modeling ObservationsExperimental Versus Model For Step Up Response
66
70
74
78
82
86
90
24 25 26 27 28Time (s)
Inp
ut (
%)
15
16
17
18
19
20
21
Output (lb/min)
Experimental Input
Model InputExperimental Output
Model Output
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Parameters involved
A = 15Input Baseline = 70
Output Baseline = 16K = 0.24t0 = 0.4
tau1 = 0.22tau2 = 0.18
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Modeling Observations RecapExperimental Versus Model For Step Up Response
66
70
74
78
82
86
90
24 25 26 27 28Time (s)
Inp
ut (
%)
15
16
17
18
19
20
21
Output (lb/min)
Experimental Input
Model InputExperimental Output
Model Output
tau1 = 0.22tau2 = 0.18
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Negative Feedback Loop
Kc
R(s) M(s) C(s)
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Closed Loop Transfer Function
111
11
21
21
0
0
ss
eKK
ss
eKK
CLTFst
c
st
c
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Characteristic Equation
01222220
2102
210
20
130
21
KKs
tKK
ts
tts
tcc
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Characteristic Equation
For
024.01048.06.012.00079.0 23 cc KsKss
K = 0.24t0 = 0.4
tau1 = 0.22tau2 = 0.18
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Solving CE for Kc
Direct Substitution
Set s = iωU
Set like terms equal to zero
Imaginary part:
[0.0079 ωU3-(0.6 +0.048Kcu) ωU]i=0i
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Direct Substitution (cont’d)
Real part:
0.24Kcu - 0.12 ωU2 + 1 = 0
ωU = 5 => fu = 0.8
Kcu = 8.3 %/(lb/min)
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Corresponding Frequency Experiment
Response at 0.8Hz frequency
65
70
75
80
85
90
95
10 11 12 13 14 15
Time(sec)
Inpu
t(%
)
15
16
17
18
19
20
21
Out
put
Input Value(%)
Output(lb/min)
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Observations
• Phase Angle = -1800
• Amplitude Ratio = 0.12
• Kcu calculated = 8.3
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Comparison of fu
SOPDT
Bode Plots Luyben MethodRouth/Direct
Substitution MethodsDirect Substitution
Method
0.87 0.67 1.2 0.8
fu HzFOPDT
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Comparison of Kc
SOPDT
Bode Plots Luyben MethodRouth/Direct
Substitution MethodsDirect Substitution
Method
10 9.6 10 8.3
FOPDTKc (%/(lb/min))
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Conclusion
• Kc = 8.3 %/(lb/min)
• SOPDT more accurate than FOPDT
• Always scope for improved results