Chemical Process Controls:
PID control, part I
By Peter Woolf (pwoolf@umich.edu)
University of Michigan
Michigan Chemical Process
Dynamics and Controls
Open Textbook
version 1.0
Creative commons
Heater Example
F, Tin
Goal: Heat a stream to a
desired set point temperature,
Tset
Assumptions:
• All liquid in lines and tank,
thus Fin=Fout=F
• Flow is constant
• Fluid does not boil
• No reactions
• Tank is well stirred
• Heater has no lag
F, T
F, Tin
Simplified Energy Balance:
TÝ Qfeed  Qexit  Qheat
F, T
How should we control
T to Tset using Qheat?

Option 1: On/off control (bangbang control)
IF TC1>TC1set THEN:
set Qheat to zero
ELSE
set Qheat to max
Assume we can control
Qheat via the heating coil
Advantages of on/off control:
• Relatively simple to design
and execute
• Binary sensors and
actuators are generally more
reliable and less expensive
F, Tin
F, T
How should we control
T to Tset using Qheat?
Option 1: On/off control (bangbang control)
IF TC1>TC1set THEN:
set Qheat to zero
ELSE
set Qheat to max
Problems with on/off control:
• Inefficient (like driving with full
gas and full brakes)
• Can generate noise when
seeking stability
• Physical: Wear on valves and
switches
• Physical: Hydraulic hammer
F, Tin
F, T
How should we control
T to Tset using Qheat?
Option 2: Continuous control
IF TC1>TC1set THEN:
adjust down Qheat
ELSE
adjust up Qheat
Partial Answer: PID Control
Questions:
• What does “adjust up” and
“adjust down” mean exactly?
• What information do you
want to use to decide your
adjustment?
• All past measures?
• Just current measures?
• Estimate of future
states?
• All of the above?
• What if you can’t provide
more heat?
P, I, D, PI, PID
• P: Proportional control. Responds in
proportion to the current deviation.
Philosophy of P
Deviations will be affected in proportion to
their magnitude.
A temperature deviation of 10 degrees
will receive 10x the response that a 1
degree deviation will cause.
P, I, D, PI, PID
• P: Proportional control. Responds in
proportion to the current deviation.
Example:
TÝ Qfeed  Qexit  Qheat
Qheat  K c (Tset  T)  c
What is the
steady state?
Proportional gain
P, I, D, PI, PID
At steady state
Set dT/dt=0 and
solve
Q feed  Qexit  c  K cTset
Tss 
Kc
Observations:
1. At steady state we will always get an offset with
P-only control.
2. Larger Kc means smaller offset
3. Larger Kc values bring you near to steady state
faster. Increasing the speed of response is the
primary benefit of P- control.
4. Larger Kc requires greater controller effort
P, I, D, PI, PID
• I: Integral control. Responds proportionally
to the integral of the deviation.
Philosophy of I
Deviations will be affected in proportion to
the cumulative sum of their magnitude.
A temperature deviation of 1 degree for 10
time steps will receive 10x the response that
a 1 degree deviation for 1 time step will
cause.
P, I, D, PI, PID
• I: Integral control. Responds proportionally
to the integral of the deviation.
Example:
TÝ Q feed  Qexit  Qheat
Qheat 
What is the
steady state?
1
I
t
 (T
set
 T)dt
0
Integral control gain
P, I, D, PI, PID
.
At steady state
Set dT/dt=0 and
solve
Observations:

1. Key Advantage:
Integral control
eliminates offset

2. Can destabilize
controller
3. Integrator
windup

T  Qfeed  Qexit 
t
 (T
1
I
 I Qfeed  Qexit 
set
 T)dt  0
0
t
 (T
set
 T)dt
0

d
d  t
 I Qfeed  Qexit    (Tset  T)dt
dt
dt  0


Tss  Tset

P, I, D, PI, PID
• D: Derivative control. Responds
proportionally to the change of the
deviation.
Philosophy of D
Deviations will be affected in proportion to
their Speed.
A temperature change of 10
degrees/second will receive 10x the
response that a 1 degree/second
deviation will cause.
P, I, D, PI, PID
• D: Derivative control. Responds
proportionally to the change of the
deviation.
Example:
TÝ Q feed  Qexit  Qheat
Qheat
What is the
steady state?
d(Tset  T)
 D
dt
Derivative control gain
P, I, D, PI, PID
At steady state
Set dT/dt=0 and
solve
d(T

T)
set
TÝ Q feed  Qexit   D
dt
TÝ Q feed  Qexit   D TÝ
Q

Q
feed
exit
Ý
T
1  D

Therefore derivative control alone
does not yield a steady state
temperature--it only resists
change
P, I, D, PI, PID
• D: Derivative control continued:
– Observations:

Corrects for fast responses and oscillations.
Reducing oscillations in feedback systems is the
key advantage of derivative control.
• Does not eliminate offset
• Slows the response
P, I, D, PI, PID
• D: Derivative control continued:
• Derivative kick: if we have a setpoint change, we will get
a spike from our D controller causing derivative kick. This
is removed by replacing the derivative term with just -T’,
instead of (Tset-T).
• Calculating derivatives: We never have infinite
measurements from a real system, so how to calculate
derivatives? Simplest way is finite difference: d(x(i)-x(i1))/d(t), or lagged over a window.
P, I, D, PI, PID
• PID: All control elements combined to yield
the strengths of each method.
Philosophy of PID
Deviations will be affected in proportion to
their magnitude, cumulative sum of
Magnitude, and Speed .
P, I, D, PI, PID
• PID: All control elements combined to yield
the strengths of each method.
e.g.
TÝ Qfeed  Qexit  Qheat
Qheat
d(Tset  T)
 K c (Tset  T) 
(Tset  T)dt   D
c

I 0
dt
1
P
t
I
D
Test your PID intuition!!!
P
signal K c 
  F  Fset

I
1
I
D
t
 dt  
0
D
d
c
dt
Test your PID intuition!!!
P
signal K c 
I
1
I
D
t
 dt  
0
D
d
c
dt
  F  Fset
Step:
P  +
I
= PI
=
PI
+
D
= PID
=
Response to a
step then a drop
Results from
PID.example.2.xls
Response to
a sinusoid
Results from
PID.example.2.xls
When to use which controller?
Estimate
When
to use
Examples
P
present
Systems with
slow responses,
t olerant to
offset
I
back
Not often used
alone, as is t oo
slow
Example use:
float valves,
thermostats,
humidistat.
Example use:
used for very
noisy systems
D
forward
Not used alone
because is t oo
sensitive t o
noise and does
not have
setpoint
Example use:
none
PI
P resent & back
Often used
P ID
All time
Often used,
most robust, but
can be noise
sensitive
Example use:
thermostats,
flow control,
pressure control
Examples:
Cases where
the system has
inertia that
could get out of
hand:
temperature and
concentration
measurements
on a react or for
example.
Avoid runaway.
PID vs Feed forward, Feed
back control
• P, I, and D look at different time windows
– P responds to current error
– I responds to accumulated past error
– D anticipates future error
• Feedback control: uses changes in the control
variable to control the control variable.
• Feed forward control: uses changes in an upstream
variable to control the control variable.
• BUT FF and FB can use P, I, PI, or PID control
Basic PID modeling
Tank example: Imagine we have a tank where the exit flow is governed by a P ID
controller to maintain the level in the tank.
LC
1
Questions:
What will this do?
LC1
Fo
How do we model it?
TK 1-001
V1
dh
 Fo  c * v1 * h
dt
1
d
v1  Fos K C    dt   i
i
dt
A
  h  hset
Fo
TK 1-001
Basic PID modeling
V1
dh
 Fo  c * v1 * h
dt
1
d
v1  Fos K C    dt 
d i
i
dt
A
  h  hset
Substitute in the error term
d d(h  hset ) dh


dt
dt
dt
Create a new variable
xI

dxI

dt
Fo
TK 1-001
Basic PID modeling
V1
dh
 Fo  c * v1 * h
dt
1
d
v1  Fos K C    dt 
d i
i
dt
A
  h  hset
Substitute
new terms in
dh
A
 Fo  c * v1 * h
dt
1
dh
v1  Fos KC  x I   d
i
dt
dxI
   h  hset
dt
Basic PID modeling

dh 1 
1
dh 
  Fo  c * Fos KC  x I   d  h 
dt A 
i
dt  

dxI
   h  hset
dt
Substitute and rearrange
dh
A
 Fo  c * v1 * h
dt
1
dh
v1  Fos KC  x I   d
i
dt
dxI
   h  hset
dt
Basic PID modeling

dh 1 
1
dh 
  Fo  c * Fos KC  x I   d  h 
dt A 
i
dt  

dxI
   h  hset
dt
Simplify and move dh/dt to the left
Solve numerically
Or analytically!

1 
1  
 Fo  c * Fos KC  x I  h 
 i  

dh A 

c
dt
1 *  d h
A
dxI
 h  hset
dt
PID modeling complexities
• Heaters, valves, and switches don’t
have infinite range.
– Solve with IF.. THEN.. Type logic
• Responses become unstable with some
parameter values
Heater Example
F, Tin
Goal: Heat the output stream to
a desired set point temperature,
Tset
Assumptions:
• All liquid in lines and tank,
thus Fin=Fout=F
• Flow is constant
• Fluid does not boil
• No reactions
• Tank is well stirred
• Heater has no lag
• Heater has finite range
F, T
idealized behav ior
Case: Heated CSTR
Starting at 120, and
Tset=130. Tfeed=100.
135
(see PID.example.xls)
130
temperature
125
120
idealized res pons e
T feed
limited ac tion
115
110
105
100
0
0 .5
1
1 .5
2
t ime
2 .5
3
3 .5
idealized behav ior
Case: Heated CSTR
Starting at 120, and
Tset=130. Tfeed=100.
140
135
(see PID.example.xls)
130
temperature
125
idealized res pons e
T feed
limited ac tion
120
115
Here we use a
smaller
KI to show
integrator
windup
110
105
100
0
1
2
3
4
t ime
5
6
7
idealized behav ior
140
Case: Heated CSTR
Starting at 120, and
Tset=130. Tfeed=100.
135
(see PID.example.xls)
145
temperature
130
125
idealized res pons e
T feed
limited ac tion
120
115
110
Here Ki and Kc are
smaller.
105
What happened??
100
0
1
2
3
4
t ime
5
6
7
idealized behav ior
140
Case: Heated CSTR
Starting at 120, and
Tset=130. Tfeed=100.
135
(see PID.example.xls)
145
temperature
130
125
idealized res pons e
T feed
limited ac tion
120
115
110
Here Ki and Kc are
even smaller.
105
What happened??
100
0
1
2
3
4
t ime
5
6
7
Outstanding questions
• How should we choose Kc, Ki, Kd?
• Can we know what parameter sets will
make our system stable?