Colorado School of Mines CHEN403
Feedback Control Numerical Example
Feedback Control System Numerical Example
Numerically Solving Feedback Control Problems
Most of the emphasis in this course is in the analytic solutions for fairly simple processes.
We can obviously do direct numerical solutions, too, though. There may be some difficulty
in defining the integral & derivative terms for the controller however.
Let’s consider a constant volume, well-mixed stirred tank with heating provided by a steam
coil. The tank operates with 2,000 L (2.0 m³) of liquid. The inlet flow rate is 250 L/min
(0.25 m³/min) at a temperature of 20°C. At steady state 350°C steam is sufficient to heat
the effluent to 80°C. Let’s further assume the liquid density is a constant 1 kg/L and the
heat capacity is a constant 1 kcal/kg·°C.
The heat balance around the tank will be:
Cˆ pV
dT1
 F0Cˆ p T0  T1   UA Ts  T1  .
dt
The traditional method to analyze the control scheme is to linearize the ODE:
Cˆ pV
dT1
 F0*Cˆ p T0  T1   F0Cˆ p T0*  T1*  UA Ts  T1 
dt


and put into a form that leads to transfer functions:




Cˆ pV
F0*Cˆ p
UA  T   Cˆ p T0*  T1* F 
dT1




T

T

1
0
s
0
F0*Cˆ p  UA dt
F0*Cˆ p  UA
F0*Cˆ p  UA
F0*Cˆ p  UA
 Cˆ pV

Cˆ p T0*  T1*
F0*Cˆ p
UA 

 *
s  1 T1  *
T0  *
Ts 
F0
 F Cˆ  UA

F0 Cˆ p  UA
F0 Cˆ p  UA
F0*Cˆ p  UA
 0 p

T1 
K0
Ks
K
T0 
Ts  F F0
s  1
s  1
s  1
Cˆ pV
where:   *
F0 Cˆ p  UA
F0*Cˆ p
K0  *
F0 Cˆ p  UA
John Jechura ([email protected])
April 23, 2017
-1-
© Copyright 2017
Colorado School of Mines CHEN403
Ks 
Feedback Control Numerical Example
UA 
F Cˆ p  UA
*
0

Cˆ p T0*  T1*
KF 
F *Cˆ  UA
0

p
Typically we would manipulate the steam temperature to deal with any disturbances from
the inlet temperature and/or flowrate. The following is an information diagram for a PI
controller scheme.
Tsp
+

-
F0
KF
s  1
T0
K0
s  1

1  C
Kc 1 
Ga  1

I s 

T1m
Ts
Ks
s  1
+
+
T1
Gm  1
What if we want to look at the response to a change in the inlet flow rate. If we use analytic
techniques we get an answer to the linearized ODE. We could use numerical techniques to
determine how closely this predicted response is to what we would expect from the nonlinearized ODE.
First, let’s use the initial steady state condition to determine the UA value (which we will
hold constant). The heat balance at the steady state is:



0  F Cˆ p T0*  T1*  UA  Ts*  T1*
*
0

 UA 



F0*Cˆ p T1*  T0*

T T
*
s
*
1
10.251 80  20
.
350  80
10.251 80  20
350  80
 0.05556 kcal/°C·min
John Jechura ([email protected])
April 23, 2017
-2-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
This screen shot shows the input to POLYMATH to numerically solve this problem of when
the controller is turned off ( K c  0 and 1/ I  0 ). Note that there is a term ErrSum that
integrates the difference between the measured outlet temperature and the set point and is
needed for the integral control.
Let’s calculate the dynamic response when the inlet flow rate is doubled to 500 L/min
(0.50 m³/min). From the original steady state equation we would expect the new steady
state outlet temperature to be:
John Jechura ([email protected])
April 23, 2017
-3-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
UA Ts*   F0*Cˆ p T0*
UA    F0*Cˆ p 
0.05556 350  10.50120 

0.05556   10.501
0  F Cˆ p T0*  T1*   UA  Ts*  T1*   T1* 
*
0
 53 °C
85
Outlet Temperature (°C)
80
75
70
65
60
55
50
0
5
10
15
20
25
Time (min.)
This chart shows the response – the new outlet temperature is about 53°C and it is reached
in about 15 minutes. This new temperature matches what we would expect from the steady
state equation.
John Jechura ([email protected])
April 23, 2017
-4-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
130
500
Kc = 10
120
Outlet Temperature (°C)
100
400
Kc = 1
90
80
350
70
Kc = 10
60
Kc = 1
50
300
No Control
Steam Temperature (°C)
450
110
250
40
30
200
0
5
10
15
20
25
Time (min.)
Let’s look at the effects of various amounts of proportional control. This chart shows both
the outlet temperature and the steam temperature for various values of K c . As expected
there is an offset that gets smaller the larger the controller gain. However, notice that the
larger the gain, the larger the steam temperature.
John Jechura ([email protected])
April 23, 2017
-5-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
900
Kc = 5,  I = 0.1
800
Outlet Temperature (°C)
700
Kc = 5,  I = 1.
130
600
110
500
90
Kc = 5,  I = 0.1
70
400
300
Kc = 5,  I = 1.
Steam Temperature (°C)
150
200
50
No Control
100
30
0
0
5
10
15
20
25
Time (min.)
Let’s look at the effects of various amounts of proportional control integral control. This
chart shows both the outlet temperature and the steam temperature for various values of
I when K c  5 . As expected there is zero offset. We have also picked controller settings
that turn this 1st order system into an underdamped 2nd order system.
John Jechura ([email protected])
April 23, 2017
-6-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
One other thing that can easily be done in a numerical solution is to put in non-traditional
constraints. For example, what if our steam system is limited to 650°C? This screen shot
shows how we can easily add this constraint. Now we have a variable TsMan which is the
steam temperature that the controller would like us to use & the actual value from the
system is Ts.
John Jechura ([email protected])
April 23, 2017
-7-
© Copyright 2017
Colorado School of Mines CHEN403
Feedback Control Numerical Example
900
Kc = 5,  I = 0.1
700
Kc = 5,  I = 0.1, Tsmax = 650
130
Outlet Temperature (°C)
800
600
110
500
90
70
Kc = 5,  I = 0.1
400
Kc = 5,  I = 0.1, Tsmax =
300
Steam Temperature (°C)
150
200
50
No Control
100
30
0
0
5
10
15
20
25
Time (min.)
This figure shows the true response when steam temperature cannot get above 650°C. This
constraint on the steam temperature actually helps the response, not in the time that it
takes to get to the new steady state but it prevents the response from getting too far away
from the set point.
John Jechura ([email protected])
April 23, 2017
-8-
© Copyright 2017