Post on 28-Mar-2015
transcript
Chapter 6
Dynamic Behavior of Ideal Systems
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
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.
Inputs
A A
t
a
A
P
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
First Order Process
u
y (t )
0.95 AK p
0.63 AK p
0 p 3
p
t
y
u
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
Estimate of First-Order Model from Process Response
4timesettling
p
p u
yK
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.
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
Underdamped vs Overdamped
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
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
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
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)
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.
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
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
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)(
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
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)(
Deadtime
F
C A 0
F
FT
FCF spec
AT
L
• Transport delay from reactor to analyzer:
• Transfer function:
FALtCtC cs /where)()(
sp esG )(
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
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
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
In-Class Exercise
• Determine a FOPDT model for the data given in Problem 5.51 page 208 of the text.
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
)(
Lead-Lag Element
1
1)(
lg
s
ssG ld
Time
y(t
)
ld> lg
ld< lg
1.0
0.0
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
Mass Recycle Example
Steam
TT
C Product
LC
LC
LC
TT
Fresh BFeed
Fresh AFeed
PT
Steam
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