Transcript No Slide Title

Process Control: Designing Process and Control
Systems for Dynamic Performance
Chapter 4. Modelling and Analysis for Process Control
Copyright © Thomas Marlin 2013
The copyright holder provides a royalty-free license for use of this material at non-profit
educational institutions
CHAPTER 4 : MODELLING &
ANALYSIS FOR PROCESS CONTROL
When I complete this chapter, I want to be
able to do the following.
• Analytically solve linear dynamic models
of first and second order
• Express dynamic models as transfer
functions
• Predict important features of dynamic
behavior from model without solving
CHAPTER 4 : MODELLING &
ANALYSIS FOR PROCESS CONTROL
Outline of the lesson.
• Laplace transform
• Solve linear dynamic models
• Transfer function model structure
• Qualitative features directly from model
• Frequency response
• Workshop
WHY WE NEED MORE DYNAMIC MODELLING
I can model this;
what more do
I need?
I would like to
• model elements
individually
• combine as needed
• determine key
dynamic features
w/o solving
T
A
T
A
WHY WE NEED MORE DYNAMIC MODELLING
I would like to
• model elements
individually
• This will be a
“transfer function”
Now, I can combine
elements to model
many process
structures
T
A
WHY WE NEED MORE DYNAMIC MODELLING
Now, I can combine
elements to model
many process
structures
Even more amazing,
I can combine to
derive a simplified
model!
THE FIRST STEP: LAPLACE TRANSFORM

 st
L( f (t ))  f ( s )   f (t )e dt
0


Constant: L(C )  Ce
0
 st
C st
dt   e
s
t 
t 0
C

s
Step Change at t=0: Same as constant for t=0 to t=
f(t)=0
t=0
THE FIRST STEP: LAPLACE TRANSFORM

We have seen
this term
often! It’s the
step response to
a first order
dynamic system.
 st
L( f (t ))  f ( s )   f (t )e dt
0


1
  e (1/ s )t dt 
e (1/ s )t
s  1/
0



0
0
0





0
L((1  e t / ))  (1  e t / )e st dt  e st dt   e t / e st dt
 1/ s
1
1
1

1
 
 

s s  1 /  s  s  1 s( s  1)
1
s  1/
THE FIRST STEP: LAPLACE TRANSFORM
Let’s learn a new
dynamic response
& its Laplace
Transform
Let’s consider plug flow through a pipe. Plug flow has no
backmixing; we can think of this a a hockey puck
traveling in a pipe.
What is the dynamic response of the outlet fluid property
(e.g., concentration) to a step change in the inlet fluid
property?
THE FIRST STEP: LAPLACE TRANSFORM
Let’s learn a new
dynamic response
& its Laplace
Transform
What is the value of
dead time for
plug flow?
Xout
 = dead time

Xin
time
THE FIRST STEP: LAPLACE TRANSFORM
Y, outlet from dead time
Let’s learn a new
dynamic response
& its Laplace
Transform
• Is this a
dead time?
1
0.5
• What is the
value?
0
-0.5
0
1
2
3
4
5
time
6
7
8
9
10
1
2
3
4
5
time
6
7
8
9
10
X, inlet to dead time
1
0.5
0
-0.5
0
THE FIRST STEP: LAPLACE TRANSFORM
Let’s learn a new
dynamic response
& its Laplace
Transform
Our plants have
pipes. We will
use this a lot!
The dynamic model for dead time is
Xout (t )  X in (t   )
The Laplace transform for a variable after dead time is
L( X out (t ))  L( X in (t   ))  e
 s
X in ( s)
THE FIRST STEP: LAPLACE TRANSFORM
We need the Laplace transform of
derivatives for solving dynamic models.
I am in desperate
need of examples!
constant
First
derivative:
General:
 df (t ) 
L
 sf (s )  f (t ) t 0

 dt 
constant
n 1


 d n f (t ) 
df
(
t
)
d
f
(
t
)
n
n

1
n

1
 s f (t )

L

s
f
(
s
)


s

....


n
n 1
t 0


dt t 0
dt
 dt 
t 0 

SOLVING MODELS USING THE LAPLACE
TRANSFORM
Textbook Example 3.1: The CSTR (or mixing tank)
experiences a step in feed composition with all other
variables are constant. Determine the dynamic response.
dC'A
V
 F(C'A0  C'A )  VkC' A
dt
dC A'
V
F
'
'

 C A  KCA0 with  
and K 
dt
F  kV
F  kV
I hope we get the same
answer as with the
integrating factor!
F
CA0
CA
V
A B
 rA  kC A
(We’ll solve this in class.)
SOLVING MODELS USING THE LAPLACE
TRANSFORM
Textbook Example 3.2Two isothermal CSTRs are initially
at steady state and experience a step change to the feed
composition to the first tank. Formulate the model for
CA2.
dC' A1
 F( C'A0  C' A1 )  V1k1C' A1
dt
dC' A2
V2
 F( C'A1  C' A2 )  V2 k 2C' A2
dt
V1
Much easier than
integrating factor!
F
CA0
dC A' 1
1
 C A' 1  K1C A' 0
dt
dC A' 2
2
 C A' 2  K 2C A' 1
dt
CA1
V1
CA2
V2
A B
 rA  kC A
(We’ll solve this in class.)
SOLVING MODELS USING THE LAPLACE
TRANSFORM
Textbook Example 3.5: The feed composition experiences a
step. All other variables are constant. Determine the
dynamic response of CA.
Non-linear!
F
CA0
A B
 rA 
2
kC A
(We’ll solve this in class.)
CA
V
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
Let’s rearrange the Laplace transform of a dynamic model
Y(s) = G(s) X(s)
X(s)
G(s)
Y(s)
A TRANSFER FUNCTION is the output variable, Y(s),
divided by the input variable, X(s), with all initial
conditions zero.
G(s) = Y(s)/X(s)
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
G(s) = Y(s)/ X(s)
X(s)
• How do we achieve zero initial
conditions for every model?
• We don’t have “primes” on the
variables; why?
• Is this restricted to a step input?
• What about non-linear models?
• How many inputs and outputs?
G(s)
Y(s)
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
G(s) = Y(s)/ X(s)
X(s)
Some examples:
C A (s)
Mix ingtank :
 G( s )  ?
C A0 ( s )
C A2 ( s )
Two CSTRs :
 G( s )  ?
C A0 ( s )
G(s)
Y(s)
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
G(s) = Y(s)/ X(s)
X(s)
G(s)
Y(s)
Why are we doing this?
• To torture students.
• We have individual models that we can
combine easily - algebraically.
• We can determine lots of information
about the system without solving the
dynamic model.
I chose the
first answer!
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
Let’s see how to
combine models
3
F0 ( s)
m
/s
Gvalve ( s) 
 .10
% open
v( s )
T ( s)
Gtank1 ( s)  1

F0 ( s)
 1.2 K
m 3 /s
250s  1
Tmeasured ( s )
Gsensor ( s ) 
T2 ( s )
T ( s ) 1.0 K / K
Gtank2 ( s )  2

T1 ( s )
300s  1
T

1.0 K / K
10s  1
(Time in seconds)
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
The BLOCK DIAGRAM
v(s)
F0(s)
Gvalve(s)
T1(s)
Gtank1(s)
T2(s)
Gtank2(s)
It’s a picture of the model equations!
• Individual models can be replaced easily
• Helpful visualization
• Cause-effect by arrows
Tmeas(s)
Gsensor(s)
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
Combine using BLOCK DIAGRAM ALGEBRA
v(s)
F0(s)
Gvalve(s)
T1(s)
Gtank1(s)
v(s)
T2(s)
Gtank2(s)
G(s)
Tmeas(s)
Gsensor(s)
Tmeas(s)
 Tmeas ( s )   T2 ( s )   T1 ( s )   F0 ( s ) 
Tmeas ( s )
 G( s )  




v( s )
 T2 ( s )   T1 ( s )   F0 ( s )   v( s ) 
 Gs ( s )GT 2 ( s )GT 1 ( s )Gv ( s )
TRANSFER FUNCTIONS: MODELS VALID FOR
ANY INPUT FUNCTION
Key rules for BLOCK DIAGRAM ALGEBRA
QUALITATIVE FEATURES W/O SOLVING
FINAL VALUE THEOREM: Evaluate the final valve of the
output of a dynamic model without solving for the entire
transient response.
Y (t ) t   lim sY(s)
s 
Example for first order system
C A (t ) |t   lim s
s 0
C A0 K p
s( s  1)
 C A0 K p
QUALITATIVE FEATURES W/O SOLVING
What about dynamics
can we determine
without solving?
We can use partial fraction
expansion to prove the following
key result.
Y(s) = G(s)X(s) = [N(s)/D(s)]X(s) = C1/(s-1) + C2/(s-2) + ...
With i the solution to the denominator of the transfer
function being zero, D(s) = 0.
1t
Y (t )  A0  A1e
 2t
 A2 e
 qt
...  [C1 cos(t )  C2 sin(t )]e
Real, distinct i
 pt
 ...  ( B0  B1t  B2t  ..)e
2
Complex i
q is Re(i)

 ...
Real, repeated i
QUALITATIVE FEATURES W/O SOLVING
With i the solutions to D(s) = 0, which is a polynomial.
1t
Y (t )  A0  A1e
 2t
 A2 e
 pt
 ...  ( B0  B1t  B2t  ..)e
2
 qt
...  [C1 cos(t )  C2 sin(t )]e

 ...
1. If all i are ???, Y(t) is stable
If any one i is ???, Y(t) is unstable
2. If all i are ???, Y(t) is overdamped
(does not oscillate)
If one pair of i are ???, Y(t) is
underdamped
Complete statements
based on equation.
QUALITATIVE FEATURES W/O SOLVING
With i the solutions to D(s) = 0, which is a polynomial.
1t
Y (t )  A0  A1e
 2t
 A2 e
 pt
 ...  ( B0  B1t  B2t  ..)e
2
 qt
...  [C1 cos(t )  C2 sin(t )]e
 ...
1. If all real [i] are < 0, Y(t) is stable
If any one real [i] is  0, Y(t) is unstable
2. If all i are real, Y(t) is overdamped (does not
oscillate)
If one pair of i are complex, Y(t) is underdamped

QUALITATIVE FEATURES W/O SOLVING
A B
1
dC A' 1
 C A' 1

K1C A' 0
dt
dC A' 2
2
 C A' 2  K 2C A' 1
dt
 rA  kC A
F
CA0
CA1
V1
CA2
V2
1. Is this system stable?
2. Is this system over- or underdamped?
Without
solving!
3. What is the order of the system?
(Order = the number of derivatives
between the input and output variables)
4. What is the steady-state gain?
(We’ll solve this in class.)
QUALITATIVE FEATURES W/O SOLVING
FREQUENCY RESPONSE:The response to a sine input of
the output variable is of great practical importance. Why?
Sine inputs almost never occur. However, many
periodic disturbances occur and other inputs can be
represented by a combination of sines.
For a process without control, we want a sine input to
have a small effect on the output.
QUALITATIVE FEATURES W/O SOLVING
Amplitude ratio = |Y’(t)| max / |X’(t)| max
Phase angle = phase difference between
input and output
P
output
Y, outlet from system
0.4
0.2
B
0
-0.2
-0.4
0
1
X, inlet to system
3
time
4
5
6
2
3
time
4
5
6
P’
1
input
2
A
0.5
0
-0.5
-1
0
1
QUALITATIVE FEATURES W/O SOLVING
Amplitude ratio = |Y’(t)| max / |X’(t)| max
Phase angle = phase difference between
input and output
For linear systems, we can evaluate directly using transfer function!
Set s = j, with = frequency and j = complex variable.
Amp. Ratio  AR  G( j )  Re( G( j )) 2  Im(G( j )) 2
Im(G( j )) 

Phase angle    G( j )  tan 
 Re( G( j )) 
1 
These calculations are tedious by hand but easily performed in
standard programming languages.
QUALITATIVE FEATURES W/O SOLVING
Example 4.15 Frequency response of mixing tank.
Time-domain
behavior.
Bode Plot - Shows
frequency response for
a range of frequencies
• Log (AR) vs log()
• Phase angle vs log()
QUALITATIVE FEATURES W/O SOLVING
CA2
F
CA0
CA1
Sine disturbance with
V1
CA2
V2
Must have
fluctuations
amplitude = 1 mol/m3
frequency = 0.20 rad/min
Data from 2 CSTRs
< 0.050 mol/m3
Using equations for the frequency response amplitude ratio
Kp
| C A2 |
| G( j ) |
| C A0 |
(1   2 2 )
| C A2 || C A0 |
Kp
(1   2 2 )
| C A2 | (1.0)( 0.12)  0.12  0.050
Not acceptable. We need
to reduce the variability.
How about feedback
control?
OVERVIEW OF ANALYSIS METHODS
Transfer function and block diagram
We can determine
individual models
and combine
1. System order
2. Final Value
3. Stability
4. Damping
5. Frequency response
We can determine
these features without
solving for the
entire transient!!
Flowchart of Modeling Method
Goal:
Assumptions:
Data:
Combining Chapters 3
and 4
Variable(s): related to goals
System : volume within which variables are indepen dent of position
Fundam ental Ba lance: e.g. material, energy
DOF = 0
Is model linear?
Yes
Check
D.O.F.
No
DOF ¹ 0
Anothe r equa tion:
-Fundamental balance
-Constitutive equations
Expand in Taylor Series
Express in deviation var iables
Group param eter s to evalua te [ga ins (K), time-constants (), dead-times()]
Take Laplace transform
Substitute specific input, e.g.,
step, and solve for output
Ana lytical solution
(ste p)
Num erical solution
Ana lyze the m odel for:
- causality
- order
- stability
- damping
Combine severa l m odels into
integrated system
We can use a
standard modelling
procedure to
focus our
creativity!
Too small to read here - check it out in the textbook!
CHAPTER 4: MODELLING & ANALYSIS WORKSHOP 1
Example 3.6 The tank with a drain has a continuous flow in
and out. It has achieved initial steady state when a step
decrease occurs to the flow in. Determine the level as a
function of time.
Solve the linearized model using Laplace transforms
CHAPTER 4: MODELLING & ANALYSIS WORKSHOP 2
T
A
1. System order
2. Final Value
3. Stability
4. Damping
5. Frequency response
The dynamic model for a nonisothermal CSTR is derived in
Appendix C. A specific example
has the following transfer
function.
T (s)
( 6.07 s  45.83)
 2
Fc ( s ) ( s  1.79s  35.80)
Determine the features
in the table for this
system.
CHAPTER 4: MODELLING & ANALYSIS WORKSHOP 3
Answer the following using the MATLAB program
S_LOOP.
Using the transfer function derived in Example 4.9,
determine the frequency response for CA0 CA2. Check
one point on the plot by hand calculation.
F
CA0
CA1
V1
CA2
V2
CHAPTER 4: MODELLING & ANALYSIS WORKSHOP 4
We often measure pressure for process monitoring and control.
Explain three principles for pressure sensors, select one for P1
and explain your choice.
T6
Feed
T1
T2
T5
F1
T4
T3
F2
P1
Vapor
product
L1
F3
Process Steam
fluid
A1
L. Key
Liquid
product
CHAPTER 4 : MODELLING &
ANALYSIS FOR PROCESS CONTROL
When I complete this chapter, I want to be
able to do the following.
•
Analytically solve linear dynamic models of first
and second order
•
Express dynamic models as transfer functions
•
Predict important features of dynamic behavior
from model without solving
Lot’s of improvement, but we need some more study!
• Read the textbook
• Review the notes, especially learning goals and workshop
• Try out the self-study suggestions
• Naturally, we’ll have an assignment!
LEARNING RESOURCES
•
SITE PC-EDUCATION WEB
- Instrumentation Notes
- Interactive Learning Module (Chapter 4)
www.pc-education.mcmaster.ca/
- Tutorials (Chapter 4)
•
Software Laboratory
- S_LOOP program
SUGGESTIONS FOR SELF-STUDY
1. Why are variables expressed as deviation variables when
we develop transfer functions?
2. Discuss the difference between a second order reaction
and a second order dynamic model.
3. For a sine input to a process, is the output a sine for a
a. Linear plant?
b. Non-linear plant?
4. Is the amplitude ratio of a plant always equal to or
greater than the steady-state gain?
SUGGESTIONS FOR SELF-STUDY
5. Calculate the frequency response for the model in
Workshop 2 using S_LOOP. Discuss the results.
6. Decide whether a linearized model should be used for
the fired heater for
PIC
1
a. A 3% increase in the
fuel flow rate.
b. A 2% change in the
feed flow rate.
c. Start up from ambient
temperature.
d. Emergency stoppage
of fuel flow to 0.0.
AT
1
feed
FT
1
PI
4
TI
1
PI
5
TI
5
TI
2
TI
6
PT
1
TI
3
TI
7
TI
9
TI
4
TI
8
FT
2
PI
2
air
TI
10
FI
3
PI
3
TI
11
PI
6
fuel