Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary...
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Transcript of Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary...
![Page 1: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/1.jpg)
Chapter 6
Dynamic Behavior of Ideal Systems
![Page 2: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/2.jpg)
Overall Course Objectives
• Develop the skills necessary to function as an industrial process control engineer.– Skills
• Tuning loops
• Control loop design
• Control loop troubleshooting
• Command of the terminology
– Fundamental understanding• Process dynamics
• Feedback control
![Page 3: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/3.jpg)
Ideal Dynamic Behavior
• Idealized dynamic behavior can be effectively used to qualitatively describe the behavior of industrial processes.
• Certain aspects of second order dynamics (e.g., decay ratio, settling time) are used as criteria for tuning feedback control loops.
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Inputs
A A
t
a
A
P
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First Order Process
• Differential equation
• Transfer function
• Note that gain and time constant define the behavior of a first order process.
)()()(
tuKtydt
tdypp
1)(
s
KsG
p
pp
![Page 6: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/6.jpg)
First Order Process
u
y (t )
0.95 AK p
0.63 AK p
0 p 3
p
t
y
u
![Page 7: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/7.jpg)
Determine the Process Gain and Process Time Constant from Gp(s)
85.0
directlydeterminedbecanandThen
15.0
8)(G
formstandardtoRearrange
2
16)(
p
p
p
pp
p
K
K
ss
ssG
![Page 8: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/8.jpg)
Estimate of First-Order Model from Process Response
4timesettling
p
p u
yK
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In-Class Exercise
• By observing a process, an operator indicates that an increase of 1,000 lb/h of feed (input) to a tank produces a 8% increase in a self-regulating tank level (output). In addition, when a change in the feed rate is made, it takes approximately 20 minutes for the full effect on the tank to be observed. Using this process information, develop a first-order model for this process.
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Second Order Process
)()()(
2)(
2
22 tuKty
dt
tdy
dt
tydppp
• Differential equation
• Transfer function
• Note that the gain, time constant, and the damping factor define the dynamic behavior of 2nd order process.
12)(
22
ss
KsG
pp
pp
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Underdamped vs Overdamped
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Effect of on Overdamped Response
0
0.2
0.4
0.6
0.8
1
0 4 8 12t/ p
y(t)
/AK
p
=1
=3
=2
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Effect of on Underdamped Response
0
0.5
1
1.5
2
0 4 8 12t/ p
y(t)
/AK
p
=1.0
=0.1
0.4
0.7
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Effect of on Underdamped Response
-2
-1
0
1
2
3
4
0 4 8 12t /n
y( t
)/A
Kp
z=-0.1
z=0
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Characteristics of an Underdamped Response
y(t)
trise
D
B C
T
±5%
trt
Time
• Rise time• Overshoot (B)• Decay ratio
(C/B)• Settling or
response time• Period (T)
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Example of a 2nd Order Process
PT
PC
Vent
P sp
C.W.
• The closed loop performance of a process with a PI controller can behave as a second order process.
• When the aggressiveness of the controller is very low, the response will be overdamped.
• As the aggressiveness of the controller is increased, the response will become underdamped.
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Determining the Parameters of a 2nd Order System from its Gp(s)
75.02
3
224
Then
134
2)(
formstandardtheintogRearrangin
5.05.12
1)(
2
2
p
pp
p
p
K
sssG
sssG
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Second-Order Model Parameters from Process Response
p
Data: PI controller with 20% overshoot and
with a period of oscillation equal to 5 min.
Solution: PI controller yields K 1. With Equation 5.15
20% overshoot yields ζ 0.456. Then, Equation 5.17
with the period of oscillation yields 0.p
2
708 min
1( )
0.0502 1.29 1pG ss s
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High Order Processes
Time
y
n=15
n=3
n=5
• The larger n, the more sluggish the process response (i.e., the larger the effective deadtime)
• Transfer function:
np
pp
s
KsG
1)(
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Example of Overdamped Process
AT
LC
LC
AT
DL
B
V
• Distillation columns are made-up of a large number of trays stacked on top of each other.
• The order of the process is approximately equal to the number of trays in the column
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Integrating Processes
0 20 40 60 80 100Time (seconds)
Fout
Ls
• In flow and out flow are set independent of level
• Non-self-regulating process
• Example: Level in a tank.
• Transfer function:
sAsG
cp
1)(
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Deadtime
F
C A 0
F
FT
FCF spec
AT
L
• Transport delay from reactor to analyzer:
• Transfer function:
FALtCtC cs /where)()(
sp esG )(
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FOPDT Model
• High order processes are well represented by FOPDT models. As a result, FOPDT models do a better job of approximating industrial processes than other idealized dynamic models.
Time
FOPDT Model
5th OrderProcess
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Determining FOPDT Parameters
Timet1/3 t2/3
1/3 y
2/3 y
y
0
• Determine time to one-third of total change and time to two-thirds of total change after an input change.
• FOPDT parameters:
u
yKt
ttpppp
4.0
7.0 3/13/13/2
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Determination of t1/3 and t2/3
616
615
314414
112313
4212
2011
6000
3/2
3/1
3/2
3/1
t
t
y
y
y
yut
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In-Class Exercise
• Determine a FOPDT model for the data given in Problem 5.51 page 208 of the text.
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Inverse Acting Processes
Time
y(t)
u(t)
• Results from competing factors.
• Example: Thermometer• Example of two first
order factors:
pppp
p
p
p
pp
andKK
s
K
s
KsG
11
)(
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Lead-Lag Element
1
1)(
lg
s
ssG ld
Time
y(t
)
ld> lg
ld< lg
1.0
0.0
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Recycle Processes
Feed
Product T f
T o T r
• Recycle processes recycle mass and/or energy.
• Recycle results in larger time constants and larger process gains.
• Recycles (process integration) are used more today in order to improve the economics of process designs.Energy Recycle
![Page 30: Chapter 6 Dynamic Behavior of Ideal Systems. Overall Course Objectives Develop the skills necessary to function as an industrial process control engineer.](https://reader036.fdocuments.in/reader036/viewer/2022062619/5516d9d15503469d338b4873/html5/thumbnails/30.jpg)
Mass Recycle Example
Steam
TT
C Product
LC
LC
LC
TT
Fresh BFeed
Fresh AFeed
PT
Steam
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Overview
• It is important to understand terms such as:– Overdamped and underdamped response– Decay ratio and settling time– Rectangular pulse and ramp input– FOPDT model– Inverse acting process– Lead-Lag element– Process integration and recycle processes