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Process Control: Designing Process and Control
Systems for Dynamic Performance
Chapter 5. Typical Process Systems
Copyright © Thomas Marlin 2013
The copyright holder provides a royalty-free license for use of this material at non-profit
educational institutions
CHAPTER 5 : TYPICAL PROCESS
SYSTEMS
When I complete this chapter, I want to be
able to do the following.
• Predict output for typical inputs for
common dynamic systems
• Derive the dynamics for important
structures of simple dynamic systems
• Recognize the strong effects on process
dynamics caused by process structures
CHAPTER 5 : TYPICAL PROCESS
SYSTEMS
Outline of the lesson.
• Common simple dynamic systems
- First order
- Dead time
-Second order
- (Non) Self-regulatory
• Important structures of simple systems
- Series
- Recycle
• Workshop
- Parallel
- Staged
SIMPLE PROCESS SYSTEMS: 1st ORDER
The basic equation is:

dY ( t )
K = s-s gain
 Y (t )  K X (t )
 = time constant
dt
Output is smooth, monotonic curve
tank concentration
1.8
1.6
Maximum
slope at
“t=0”
1.4
1.2
1
» 63% of steady-state DCA
At steady state
DY = K DX

0.8
0
20
40
60
80
100
120
80
100
120
time
Output changes immediately
inlet concentration
2
1.5
DX = Step in inlet variable
1
0.5
0
20
40
60
time
Would this be
easy/difficult to
control?
SIMPLE PROCESS SYSTEMS: 1st ORDER
These are simple
first order systems
from several
engineering
disciplines.
SIMPLE PROCESS SYSTEMS: 2nd ORDER
Would this be
easy/difficult to
control? 
The basic equation is:
2
2
d Y (t )
dt
2
 2
dY ( t )
dt
 Y (t )  K P X (t )
KP = s-s gain ,  = time constant ,  = damping factor
overdamped
underdamped
1.5
0.8
Controlled Variable
Controlled Variable
1
0.6
0.4
0.2
0
0
10
20
30
40
Time
50
60
70
0
20
40
60
80
100
Time
120
140
160
180
200
0
20
40
60
80
100
Time
120
140
160
180
200
1
0.8
Manipulated Variable
Manipulated Variable
0.5
0
80
1
0.6
0.4
0.2
0
1
0
10
20
30
40
Time
50
60
70
80
0.8
0.6
0.4
0.2
0
SIMPLE PROCESS SYSTEMS: 2nd ORDER
These are simple
second
order systems
from several
engineering
disciplines.
SIMPLE PROCESS SYSTEMS: DEAD TIME
 = dead time
X out ( t )  X in ( t   )
X out ( s )  e
 s
X in ( s )
Would this be
easy/difficult to
control?
Xout

Xin
time
SIMPLE PROCESS SYSTEMS: INTEGRATOR
Level sensor

dV
dt
  A
dL
dt
  F in   F out
Fin ( t )  f ( L )
Liquid-filled
tank
Fout ( t )  f ( L )
pump
valve
Plants have many inventories whose flows in and out do
not depend on the inventory (when we apply no control
or manual correction).
These systems are often termed “pure integrators”
because they integrate the difference between in and out
flows.
SIMPLE PROCESS SYSTEMS: INTEGRATOR
Level sensor

dV
 A
dt
dL
dt
  Fin   F out
Liquid-filled
tank
pump
valve
Plot the level for this scenario
Fin
Fout
time
SIMPLE PROCESS SYSTEMS: INTEGRATOR
Level sensor

dV
 A
dt
dL
dt
  Fin   F out
Liquid-filled
tank
pump
valve
Level
Fin
Fout
time
SIMPLE PROCESS SYSTEMS: INTEGRATOR
Level sensor
Liquid-filled
tank
pump
valve
• Non-self-regulatory variables
tend to “drift” far from
desired values.
• We must control these
variables.
Let’s look ahead
to when we
apply control.
STRUCTURES OF PROCESS SYSTEMS
NON-INTERACTING SERIES
• The output from an element
does not influence the input to
the same element
• Common example is tanks in
series with pumped flow
between
T
• Block diagram as shown
v(s)
F0(s)
Gvalve(s)
T1(s)
Gtank1(s)
T2(s)
Gtank2(s)
Tmeas(s)
Gsensor(s)
STRUCTURES OF PROCESS SYSTEMS
NON-INTERACTING SERIES
v(s)
F0(s)
Gvalve(s)
In general:
T1(s)
Gtank1(s)
Y (s)
X (s)
With each
element a first
order system:
Y (s)
X (s)
T2(s)
Gtank2(s)
Tmeas(s)
Gsensor(s)
n
  Gi (s)
i 1
• overall gain is
product of gains
n
Ki
i 1
( i s  1 )

• no longer first order
system
• slower than any
single element
STRUCTURES OF PROCESS SYSTEMS
NON-INTERACTING SERIES
v(s)
F0(s)
0.10/(5s+1)
T1(s)
T2(s)
-1.2/(5s+1)
Tmeas(s)
1/(5s+1)
3.5/(5s+1)
Controlled Variable
0
Step Response
-1
-2
• Looks as though some
dead time occurs
-3
-4
-5
0
10
20
30
40
50
Time
Manipulated Variable
10
• Slower than any
element
8
6
4
• K =  (Ki)
2
0
60
70 monotonic,
• Smooth,
not first order
0
10
20
30
40
Time
50
60
70
STRUCTURES OF PROCESS SYSTEMS
NON-INTERACTING SERIES
v(s)
F0(s)
Gvalve(s)
T1(s)
Gtank1(s)
With each
element a first
order system
with dead time:
Y (s)
X (s)
n

i 1
K ie
 i s
( i s  1 )
T2(s)
Gtank2(s)
Tmeas(s)
Gsensor(s)
Guidelines on step response
• Sigmoidal (“S”) shaped
• t63% » (i + i) [not rigorous!]
• K =  (Ki)
[rigorous!]
• Usually, some “apparent dead
time” occurs
STRUCTURES OF PROCESS SYSTEMS
Class Exercise: Sketch the step response for the system
below.
?
=2
=2
STRUCTURES OF PROCESS SYSTEMS
Class Exercise: Sketch the step response for the system
below.
DYNAMIC SIMULATION
Controlled Variable
5
4
3
2
1
0
0
5
10
15
20
25
15
20
25
Time
Manipulated Variable
5
4
3
2
1
0
0
5
10
Time
STRUCTURES OF PROCESS SYSTEMS
Class Exercise: Sketch the step response for each of the
systems below and compare the results.
Case 1
=2
=2
=2
=2
Case 2
=2
=2&=2
=1
=1
Two plants can have different intermediate variables and have
the same input-output behavior!
case 1 responses
4
Case1
3
2
1
0
Step
0
2
4
6
8
10
time
12
14
16
case 2 responses
4
18
20
Case2
3
2
1
0
0
2
4
6
8
10
time
12
14
16
18
20
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURES result from more than one
causal path between the input and output. This can be a
flow split, but it can be from other process relationships.
Example process systems
Block diagram
ABC
G1(s)
X(s)
G2(s)
Y(s)
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURES
G1(s)
X(s)
Y(s)
G2(s)
If both elements are first order, the overall model is
Class exercise:
Derive this
transfer function
Y (s)
X (s)

K p ( 3 s  1 )
( 1 s  1)(  2 s  1)
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURES can experience complex
dynamics. Parameter is the “zero” in the transfer function.
Sample step response at t=0
1.5
G1(s)
Y(s)
X(s)
output variable, Y’(t)
4
3
1
2
1
Which would
be difficult/easy
to control?
0.5
0
-1
G2(s)
0
-0.5
0
-2
1
2
3
4
5
time
6
7
8
9
10
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURE
Class exercise: Explain the dynamics of the outlet
temperature after a change to the flow ratio, with the total
flow rate constant.
T
STRUCTURES OF PROCESS SYSTEMS
93
93
92
92
Why an overshoot?
senor output
mixing temperature
PARALLEL STRUCTURES: Explain the dynamics of the
outlet temperature after a step change to the flow ratio.
91
90
89
0
5
10
15
20
25
time
91
90
89
0
5
10
15
20
25
time
fraction by-pass
0.7
0.6
0.5
T
0.4
0
5
10
15
time
20
25
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURE
Class exercise: Explain the dynamics of the outlet
concentration after a step change to the solvent flow rate.
reactant
FA
CA0
solvent
FS
CAS=0
CA1
V1
CA2
V2
STRUCTURES OF PROCESS SYSTEMS
PARALLEL STRUCTURE
Class exercise: Explain the dynamics of the outlet
concentration after a step change to the solvent flow rate.
tank 2 concentration
0.43
Why an inverse response?
0.42
0.41
0.4
0.39
0
10
20
30
40
50
60
0
10
20
30
time
40
50
60
0.1
solvent flow
0.09
0.08
0.07
0.06
0.05
STRUCTURES OF PROCESS SYSTEMS
RECYCLE STRUCTURES result from recovery of
material and energy. They are essential for profitable
operation, but they strongly affect dynamics.
Process example
Block diagram
T0
T3
T4
STRUCTURES OF PROCESS SYSTEMS
RECYCLE STRUCTURES
T1(s)
T0(s)
T3(s)
GH1(s)
GR(s)
T2(s)
GH2(s)
T 4( s )
T 0( s )

G R ( s )G H 1 ( s )
1  G R ( s )G H 2 ( s )
T4(s)
STRUCTURES OF PROCESS SYSTEMS
RECYCLE STRUCTURES
Class exercise: Determine the effect of recycle on the
dynamics of a chemical reactor (faster or slower?).
• Exothermic
reaction
T0
• feed/effluent
preheater
T3
T4
G H 1 ( s )  0 . 40
K /K
G H 2 ( s )  0 . 30
K /K
G R ( s )  3 /( 10 s  1)
STRUCTURES OF PROCESS SYSTEMS
Class exercise: Determine the effect of recycle on the
dynamics of a chemical reactor (faster or slower?).
T4 is a deviation variable
T4 without recycle
2.5
2
Without recycle,
faster and smaller
effect
1.5
1
0.5
0
0
5
10
15
20
25
time
30
35
40
45
50
T4 with recycle
25
20
Different
scales!
With recycle, slower
and larger effect
15
10
5
0
0
50
100
150
200
250
time
300
350
400
450
500
STRUCTURES OF PROCESS SYSTEMS
STAGED STRUCTURES
Liquid
Vapor
xD
FR
Tray n
FV
xB
Liquid
Vapor
STRUCTURES OF PROCESS SYSTEMS
STAGED STRUCTURES
XD (mol frac)
0.985
0.98
0.975
0.025
XB (mol frac)
“Steps” because analyzer
provides new measurement
only every 2 mintes.
0.99
0.97
0.965
0.02
0.015
0.01
0.005
0
10
20
30
40
Time (min)
50
0
8532.5
1.37
8532
x 10
10
20
30
40
Time (min)
50
10
20
30
40
Time (min)
50
4
1.365
V (mol/min)
R (mol/min)
Complex structure,
smooth dynamics
8531.5
8531
1.36
1.355
8530.5
8530
1.35
0
10
20
30
40
Time (min)
50
0
OVERVIEW OF PROCESS SYSTEMS
Even simple elements can yield complex dynamics when
combined in typical process structures.
We can
• Estimate the dynamic
response based on
elements and
structure
• Recognize range of
effects possible
• Apply analysis
methods to yield
dynamic model
CHAPTER 5: PROCESS SYSTEMS WORKSHOP 1
Four systems experienced an impulse input at t=2. Explain what you can
learn about each system (dynamic model) from the figures below.
(a)
(b)
3
3
2
output
output
2
1
1
0
0
-1
0
5
10
15
20
25
30
0
5
10
(c)
20
25
30
20
25
30
(d)
3
2.5
2
output
2
output
15
1
0
1.5
1
0.5
-1
0
0
5
10
15
time
20
25
30
0
5
10
15
time
CHAPTER 5: PROCESS SYSTEMS WORKSHOP 2
Using the guidelines in this chapter, sketch the response of
the measured temperature below to a +5% step to the valve.
G valve ( s ) 
F0 ( s )
3
 . 10 m / min
v (s )
% open
G tank1 ( s ) 
T1 ( s )
3

 1 . 2 K /( m / min)
250 s  1
F0 ( s )
G sensor ( s ) 
G tank2 ( s ) 
T2 ( s )
T1 ( s )

1 .0 K / K
300 s  1
T
T measured ( s )

T2 ( s )
1 .0 K / K
10 s  1
(Time in seconds)
CHAPTER 5: PROCESS SYSTEMS WORKSHOP 3
Sensors provide an estimate of the true process variable
because the measurement is corrupted by errors.
• Discuss sources of noise in a measurement.
• Define the following terms for a sensor
- Accuracy
- Reproducibility
• Explain some process measurements needing (a) good
accuracy and (b) good reproducibility
• Suggest an approach for operating a process when a key
material property (composition, etc.) cannot be measured
using an onstream analyzer.
CHAPTER 5: PROCESS SYSTEMS WORKSHOP 4
We are designing the following reactor with recycle. We
have two choices for the conversion (X) in the reactor. Will
the plant dynamics be affected by the selection?
Pure
product
Fresh
feed
flow is
constant
X = 50%
or
X = 95%
Pure,
unreacted
feed
CHAPTER 5 : TYPICAL PROCESS
SYSTEMS
When I complete this chapter, I want to be
able to do the following.
•
Predict output for typical inputs for common
dynamic systems
•
Derive the dynamics for important structures of
simple dynamic systems
•
Recognize the strong effects on process dynamics
caused by process structures
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!
CHAPTER 5: LEARNING RESOURCES
•
SITE PC-EDUCATION WEB
- Instrumentation Notes
- Interactive Learning Module (Chapter 5)
www.pc-education.mcmaster.ca/
- Tutorials (Chapter 5)
•
Software Laboratory
- S_LOOP program
•
Textbook
- Chapter 18 on level modelling and control
- Appendix I on parallel structures
CHAPTER 5: SUGGESTIONS FOR SELF-STUDY
1. Extend textbook Figure 5.1 for new input functions
(additional rows): impulse and ramp.
2. Determine which of the systems in textbook Figure 5.3
can be underdamped.
3. Explain the shape of the amplitude ratio as frequency
increases for each system in textbook Figure 5.1.
4. Discuss the similarity/dissimilarity between self
regulation and feedback.
5. Explain textbook Figure 5.5.
6. Discuss the similarity between recycle and feedback.
CHAPTER 5: SUGGESTIONS FOR SELF-STUDY
7. Discuss how the dynamics of the typical process
elements and structures would affect our ability to
control a process. Think about driving an automobile
with each of the dynamics between the steering wheel
and the direction that the auto travels.
8. Formulate one question in each of three categories (T/F,
multiple choice, and modelling) with solution and
exchange them with friends in your study group.