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

Princess Sumaya Univ.
Electronic Engineering Dept.
3442
Industrial Instruments 2
Chapter 9
Controller Principles
Dr. Bassam Kahhaleh
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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)
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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
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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.
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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.
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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.
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9: Controller Principles
Control System Parameters
 Error
e = r – b.
 Percentage
cp 
c  c min
c max  c min
x 100 %
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9: Controller Principles
Control System Parameters
p: percentage
u: actual
ep 
 Error
 Control
Parameter
rb
b max  b min
uu
p
u
max
min
u
min
x 100 %
x 100 %
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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:
Speed (RPM)
a) Current corresponding
600
to 310 RPM
310
b) The value of (a) in
140
percent.
I (mA)
4
20
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9: Controller Principles
Control System Parameters
Example
Sp = m I + So
140 = 4 m + So
Speed (RPM)
600
310
600 = 20 m + So
140
 m = 28.75 rpm/mA
I (mA)
So = 25 rpm
4
20
310 = 28.75 I + 25
9 . 91  4
 I = 9.91 mA
p
x 100 %  36 . 9 %
20  4
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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.
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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.
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9: Controller Principles
Discontinuous Controller Modes

Two-Position Mode
 0%
p 
100 %
Neutral Zone
ep  0
ep  0
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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
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9: Controller Principles
Discontinuous Controller Modes

Two-Position Mode
I (mA)
Example
20
H = K I + HO
2 = K (4) + HO
3 = K (20) + HO
4
 K = 0.0625 m / mA
HO = 1.75 m
2
3
HH = 0.0625 * 12 + 1.75 = 2.5 m
Displacement (m)
HL = 0.0625 * 10 + 1.75 = 2.375
The neutral zone = HH – HL = 2.5 – 2.375 = 0.125 m
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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.
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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
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9: Controller Principles
Discontinuous Controller Modes

Multiposition Mode
0%


p   50 %
 100 %

e p   e1
 e1  e p  e 2
e 2  ep
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9: Controller Principles
Discontinuous Controller Modes

Multiposition Mode
Example
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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.
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9: Controller Principles
Discontinuous Controller Modes

Floating-Control Mode

Single Speed
 K F
 
0
dt

dp
ep   eP
ep   eP
p   K F t  p (0)
ep   eP
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9: Controller Principles
Discontinuous Controller Modes

Floating-Control Mode

Single Speed
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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.
p   K F t  p (0)
Solution
100 % = (2 %/s * t) + 25 %
t = 37.5 s
ep   eP
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9: Controller Principles
Discontinuous Controller Modes

Floating-Control Mode

Multispeed
  K Fi
 
0
dt

dp
e p  e Pi
ep  eP1
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9: Controller Principles
Continuous Controller Modes

Proportional Control Mode
p  K P ep  pO
ep 
rb
b max  b min
x 100 %
Direct (- K) & Reverse (+ K) Action
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9: Controller Principles
Continuous Controller Modes

Proportional Control Mode
p  K P ep  pO
% per %
Proportional Band:
PB 
100
KP
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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 600 m3/h
Calculate:
The new controller output
The new offset error
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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 600 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 %
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9: Controller Principles
Continuous Controller Modes

Integral Control Mode
t
p ( t )  K I  e p dt  p ( 0 )
0
% per sec %
Integral Time: TI = 1 / KI
sec
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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 = - 0.15 % per second per % error.
If the speed jumps to 13.5 rpm,
calculate the controller output after 2 s for a
constant ep.
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9: Controller Principles
Continuous Controller Modes

Integral Control Mode
ep 
rb
x 100 %
b max  b min
Example
setpoint = 12 rpm (range of 10 – 15 rpm)
P(O) = 22%
Solution
KI = – 0.15
eP = (12–13.5)/(15–10)*100 %.
speed = 13.5 rpm
eP = – 30 %
constant ep
P(t) = KI eP t + P(0)
2 seconds time
t
p ( t )  K I  e p dt  p ( 0 )
0
P(t) = (– 0.15) * (– 30) * 2 + 22
P = 31 %
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9: Controller Principles
Continuous Controller Modes

Derivative Control Mode
p (t )  K D
de p
dt
sec
Not used alone because it provides no output
when the error is constant.
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9: Controller Principles
Continuous Controller Modes

Derivative Control Mode
Example
p (t )  K D
de p
dt
 po
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9: Controller Principles
Composite Control Modes

Proportional – Integral Control (PI)
t
p ( t )  K P e p  K P K I  e p dt  p ( 0 )
0
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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
t
p ( t )  K P e p  K P K I  e p dt  p ( 0 )
0
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9: Controller Principles
Composite Control Modes

PI
Example
KP = 5
KI = 1 sec-1
P(0) = 20%
t
p ( t )  K P e p  K P K I  e p dt  p ( 0 )
0
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9: Controller Principles
Composite Control Modes

Proportional – Derivative Control (PD)
p (t )  K P ep  K P K D
de p
dt
 po
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9: Controller Principles
Composite Control Modes

PD
Example
KP = 5
KD = 0.5 sec
Po = 20%
p (t )  K P ep  K P K D
de p
dt
 po
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9: Controller Principles
Composite Control Modes

Three – Mode Controller (PID)
t
de p
0
dt
p ( t )  K P e p  K P K I  e p dt  K P K D
 p (0)
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9: Controller Principles
End of Chapter 9