1. 3442 Industrial Instruments 2 Chapter 9 Controller Principles
Princess Sumaya University
Industrial Instruments 2
Princess Sumaya Univ. Electronic Engineering Dept.
3442 Industrial Instruments 2 Chapter 9 Controller Principles
Dr. Bassam Kahhaleh
Dr. Bassam Kahhaleh

2. 9: Controller Principles
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9: Controller Principles
Process Characteristics
Process Equation
A process-control loop regulates some dynamic variable in a process.
Example:
The control of liquid temperature in a tank.
The controlled variable is the liquid temperature TL
TL is a function:
TL = F(QA, QB, QS, TA, TS, TO)

3. 9: Controller Principles
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9: Controller Principles
Process Characteristics
Process Load
Identify a set of values for the process parameters that results in the controlled variable having the setpoint value.
This set = nominal set.
Process load
= all parameter set – the controlled variable

4. 9: Controller Principles
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9: Controller Principles
Process Characteristics
Process Change
Process Load Change
A parameter changes value from its nominal value causes the controlled value to change from its setpoint.
Transient Change
A temporary change of a parameter value.

5. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Process Characteristics
Process Lag
The time it takes for the process to respond after a process load or transient change occurs, to ensure that the controlled variable returns to the setpoint.

6. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Process Characteristics
Self Regulation
The tendency of some processes to adopt a specific value of the controlled variable for nominal load with no control operations.

7. 9: Controller Principles
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9: Controller Principles
Control System Parameters
Error
e = r – b.
Percentage

8. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Control System Parameters
Error
Control Parameter
p: percentage
u: actual

9. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Control System Parameters
Example
A controller outputs a 4 – 20 mA signal to control motor speed from 140 – 600 RPM with a linear dependence. Calculate:
I (mA)
Speed (RPM)
140
600
Current corresponding to 310 RPM
The value of (a) in percent.
310

10. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Control System Parameters
Example
Sp = m I + So
140 = 4 m + So
600 = 20 m + So
 m = rpm/mA
So = 25 rpm
310 = I + 25
 I = 9.91 mA
I (mA)
Speed (RPM)
140
600
310

11. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Control System Parameters
Control Lag
The time it takes for the final control element to adopt a new value (as required by the process-control loop output) in response to a sudden change in the controlled variable.
Dead Time
The elapsed time between the instant a deviation (error) occurs and when the corrective action first occurs.
Cycling
The cycling of the variable above and below the setpoint value.

12. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Control System Parameters
Controller Modes
Continuous / Discontinuous
Smooth variation of the control parameter versus ON / OFF.
Reverse / Direct Action
An increasing value of the controlled variable causes an decreasing / increasing value of the controller output.

13. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Two-Position Mode
Neutral Zone

14. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Two-Position Mode
Example
A liquid-level control system linearly converts a displacement of 2 – 3 m into a 4 – 20 mA control signal. A relay serves as the two-position controller to open or close an inlet valve. The relay closes at 12 mA and opens at 10 mA. Find:
The relation between displacement level and current
The neutral zone

15. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Two-Position Mode
I (mA)
Displacement (m) 4 20
Example
H = K I + HO
2 = K (4) + HO
3 = K (20) + HO
 K = m / mA
HO = 1.75 m
HH = * = 2.5 m
HL = * = 2.375
The neutral zone = HH – HL = 2.5 – = m

16. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Two-Position Mode
Example
As a water tank loses heat, the temperature drops by 2 K per minute. When a heater is on, the system gains temperature at 4 K per minute. A two-position controller has a 0.5 min control lag and a neutral zone of ± 4 % of the setpoint about a setpoint of 323 K.
Plot the tank temperature versus time.

17. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Two-Position Mode
Example
Neutral zone = 310 – 336 K
Temp. gain = 4 K per minute
Setpoint = 323 K
Temp. loss = 2 K per minute
Control lag = ½ minute

18. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Multiposition Mode

19. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Multiposition Mode
Example

20. 9: Controller Principles
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9: Controller Principles
Discontinuous Controller Modes
Floating-Control Mode
If the error is zero, the output does not change but remains (floats) at whatever setting it was when the error went to zero.

21. 9: Controller Principles
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9: Controller Principles
Discontinuous Controller Modes
Floating-Control Mode
Single Speed

22. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Floating-Control Mode
Single Speed

23. 9: Controller Principles
Princess Sumaya University
Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Floating-Control Mode
Example
Suppose a process error lies within the neutral zone with p = 25%. At t = 0, the error falls below the neutral zone. If K = +2% per second, find the time when the output saturates.
Solution
100 % = (2 %/s * t) + 25 %
t = 37.5 s

24. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Discontinuous Controller Modes
Floating-Control Mode
Multispeed

25. 9: Controller Principles
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9: Controller Principles
Continuous Controller Modes
Proportional Control Mode
Direct (- K) & Reverse (+ K) Action

26. 9: Controller Principles
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9: Controller Principles
Continuous Controller Modes
Proportional Control Mode
% per %
Proportional Band:

27. 9: Controller Principles
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9: Controller Principles
Continuous Controller Modes
Proportional Control Mode
Example
Valve A: 10 m3/h per percent.
PO = 50 %
KP = 10 % per %
Valve B changes
from 500 m3/h
to m3/h
Calculate:
The new controller output
The new offset error

28. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Continuous Controller Modes
Proportional Control Mode
Example
Valve A: 10 m3/h per percent.
PO = 50 %
KP = 10 % per %
Valve B changes
from 500 m3/h
to m3/h
Calculate:
The new controller output
The new offset error
Solution
QA must go up to
600 m3/h
QA = 10 m3/h * P
P = 60 %
60 = 10 eP + 50
eP = 1 %

29. 9: Controller Principles
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9: Controller Principles
Continuous Controller Modes
Integral Control Mode
Integral Time: TI = 1 / KI
% per sec %
sec

30. 9: Controller Principles
Princess Sumaya University
Industrial Instruments 2
9: Controller Principles
Continuous Controller Modes
Integral Control Mode
Example
An integral controller is used for speed control with a setpoint of 12 rpm within a range of 10 – 15 rpm. The controller output is 22% initially.
The constant KI = % per second per % error. If the speed jumps to 13.5 rpm,
calculate the controller output after 2 s for a constant ep.

31. 9: Controller Principles
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9: Controller Principles
Continuous Controller Modes
Integral Control Mode
Example
setpoint = 12 rpm (range of 10 – 15 rpm)
P(O) = 22%
KI = – 0.15
speed = 13.5 rpm
constant ep
2 seconds time
Solution
eP = (12–13.5)/(15–10)*100 %.
eP = – 30 %
P(t) = KI eP t + P(0)
P(t) = (– 0.15) * (– 30) *
P = 31 %

32. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Continuous Controller Modes
Derivative Control Mode
Not used alone because it provides no output when the error is constant.
sec

33. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Continuous Controller Modes
Derivative Control Mode
Example

34. 9: Controller Principles
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9: Controller Principles
Composite Control Modes
Proportional – Integral Control (PI)

35. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Composite Control Modes PI
Example
KP = 5
KI = 1 sec-1
P(0) = 20%
Plot a graph of the controller output as a function of time

36. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Composite Control Modes PI
Example
KP = 5
KI = 1 sec-1
P(0) = 20%

37. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Composite Control Modes
Proportional – Derivative Control (PD)

38. 9: Controller Principles
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Industrial Instruments 2
9: Controller Principles
Composite Control Modes PD
Example
KP = 5
KD = 0.5 sec
Po = 20%

39. 9: Controller Principles
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9: Controller Principles
Composite Control Modes
Three – Mode Controller (PID)

40. 9: Controller Principles
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9: Controller Principles
End of Chapter 9