Transcript Chapter 5

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
Dt
A
A
P
A
a
First Order Process
dy (t )
p
 y (t )  K p u (t )
dt
Kp
G p ( s) 
 ps  1
• Differential equation
• Transfer function
• Note that gain and
time constant define
the behavior of a first
order process.
First Order Process
0.95 AK p
y (t )
0.63 AK p
D
y
Du
u
0

t
p
3
p
Determine the Process Gain and
Process Time Constant from Gp(s)
16
G p (s) 
s2
Rearrange t o st andard form
8
G p ( s) 
0.5 s  1
T hen p and K p can be det ermineddirectly
 p  0.5
Kp  8
Estimate of First-Order Model
from Process Response
Dy
Kp 
Du
settling time
p 
4
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
d
y (t )
dy(t )
2
p
 2  p
 y (t )  K p u (t )
2
dt
dt
• Differential equation
G p ( s) 
Kp
 s  2  p s  1
2 2
p
• Transfer function
• Note that the gain,
time constant, and the
damping factor define
the dynamic behavior
of 2nd order process.
Underdamped vs Overdamped
Effect of  on Overdamped
Response
1
 =1
y(t)/AKp
0.8
 =2
 =3
0.6
0.4
0.2
0
0
4
t/ p
8
12
Effect of  on Underdamped
Response
2
 =0.1
y(t)/AKp
1.5
0.4
0.7
1
 =1.0
0.5
0
0
4
t/ p
8
12
Effect of  on Underdamped
Response
4
y (t )/AKp
3
2
1
=0
0
-1
=-0.1
-2
0
4
t /n
8
12
Characteristics of an
Underdamped Response
C
y(t)
B
±5%
T
D
trise
trt
• Rise time
• Overshoot (B)
• Decay ratio
(C/B)
• Settling or
response time
• Period (T)
Time
Example of a 2nd Order Process
Psp
PC
Ve n t
PT
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)
1
G p ( s)  2
2s  1.5s  0.5
Rearranging into the standard form
2
G p ( s)  2
4s  3s  1
T hen
p  4  2
 
3
2 p
 0.75
Kp  2
Second-Order Model Parameters
from Process Response
Data: PI controller with 20%
overshoot
and
with a period of oscillation equal to 5 min.
Solution: PI controller yields K p  1.
20% overshoot yields ζ  0.456.
With Equation 5.15
Then, Equation 5.17
with the period of oscillation yields  p  0.708 min
G p ( s) 
1
0.0502 s 2  1.29 s  1
High Order Processes
• The larger n, the more
sluggish the process
response (i.e., the larger
the effective deadtime)
• Transfer function:
y
n=3
n=5
n=15
G p ( s) 
Time

Kp
p
s  1
n
Example of Overdamped Process
LC
L
D
AT
V
LC
AT
B
• 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
Ls
Fout
0
20
40
60
80
Time (seconds)
100
• In flow and out
flow are set
independent of
level
• Non-self-regulating
process
• Example: Level in
a tank.
• Transfer function:
1
G p ( s) 
 Ac s
Deadtime
FC
Fspec
FT
F
CA0
AT
L
• Transport delay from reactor to analyzer:
Cs (t )  C(t  ) where    L Ac / F
• Transfer function:
Gp (s)  e s
F
FOPDT Model
5th Order
Process
FOPDT Model
Time
• 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.
Determining FOPDT Parameters
2/3 D y
Dy
1/3 D y
0
t1/3
t2/3
Time
• Determine time to one-third of total change and
time to two-thirds of total change after an input
change.
• FOPDT parameters:
t 2 / 3  t1/ 3
Dy
p 
 p  t1/ 3  0.4 p
Kp 
0.7
Du
Determination of t1/3 and t2/3
t u
y
0 0 0
1 1 0
Dy  6
y1/ 3  2
2 1 2
y2 / 3  4
3 1 3
t1/ 3  2  1  1
4 1 4
t2 / 3  4  1  3
5 1 6
6 1 6
In-Class Exercise
• Determine a FOPDT model for the data
given in Problem 5.51 page 208 of the text.
Inverse Acting Processes
• Results from competing
factors.
• Example: Thermometer
• Example of two first
order factors:
y(t)
u(t)
G p ( s) 
Time
Kp
 ps  1
K p  K p

and
K p
 p s  1
 p   p
Lead-Lag Element
y (t )
 ld s  1
G( s) 
 lg s  1
 ld> lg
1.0
 ld< lg
0.0
Time
Recycle Processes
Product
To
Tf
Tr
Fe e d
Energy Recycle
• 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.
Mass Recycle Example
Fre sh A
Fe e d
Fre sh B
Fe e d
LC
PT
LC
S te am
TT
TT
LC
S te am
C Produ ct
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