3442 industrial instruments 2 chapter 9 controller principles dr. bassam kahhaleh princess sumaya...

34423442Industrial Instruments 2Industrial Instruments 2

Chapter 9Chapter 9Controller PrinciplesController Principles

Dr. Bassam KahhalehDr. Bassam Kahhaleh

Princess Sumaya Univ.Princess Sumaya Univ.Electronic Engineering Dept.Electronic Engineering Dept.

Princess Sumaya UniversityPrincess Sumaya University 3442 - Industrial Instruments 23442 - Industrial Instruments 2 22 / 40 / 40

9: Controller Principles9: Controller PrinciplesProcess CharacteristicsProcess Characteristics Process EquationProcess Equation

A process-control loop regulates some A process-control loop regulates some dynamicdynamic variablevariable in a process. in a process.

Example:The control of liquid temperature in a tank.

The controlled variable is The controlled variable is the liquid temperature Tthe liquid temperature TLL

TTLL is a function: is a function:

TTLL = F(Q = F(QAA, Q, QBB, Q, QSS, T, TAA, T, TSS, T, TOO))


9: Controller Principles9: Controller PrinciplesProcess CharacteristicsProcess Characteristics Process LoadProcess Load

Identify a set of values Identify a set of values for the process for the process parameters that results parameters that results in the in the controlledcontrolled variablevariable having the setpoint having the setpoint value.value.

This set = This set = nominalnominal setset..Process loadProcess load = all parameter set – the = all parameter set – the controlledcontrolled variablevariable


9: Controller Principles9: Controller PrinciplesProcess CharacteristicsProcess Characteristics Process ChangeProcess Change

Process Load Process Load ChangeChangeA parameter changes value from its nominal value A parameter changes value from its nominal value causes the controlled value to change from its causes the controlled value to change from its setpoint.setpoint.

TransientTransient Change ChangeA temporary change of a parameter value.A temporary change of a parameter value.


9: Controller Principles9: Controller PrinciplesProcess CharacteristicsProcess Characteristics Process LagProcess Lag

The time it takes for the process to respond The time it takes for the process to respond after a process load or transient change after a process load or transient change occurs, to ensure that the occurs, to ensure that the controlled variablecontrolled variable returns to the setpoint.returns to the setpoint.


9: Controller Principles9: Controller PrinciplesProcess CharacteristicsProcess Characteristics Self RegulationSelf Regulation

The tendency of some processes to adopt a The tendency of some processes to adopt a specific value of the controlled variable for specific value of the controlled variable for nominal load with no control operations.nominal load with no control operations.


9: Controller Principles9: Controller PrinciplesControl System ParametersControl System Parameters ErrorError

e = r – b.e = r – b.

PercentagePercentage

%100minmax

minx

cc

cccp


9: Controller Principles9: Controller PrinciplesControl System ParametersControl System Parameters

ErrorError

Control ParameterControl Parameter

%100minmax

xbb

brep

%100minmax

minx

uu

uup

p: percentage

u: actual


9: Controller Principles9: Controller PrinciplesControl System ParametersControl System Parameters

4 20

I (mA)

Speed (RPM)

140

600

ExampleExample

A controller outputs a 4 – 20 mA signal to control A controller outputs a 4 – 20 mA signal to control motor speed from 140 – 600 RPM with a linear motor speed from 140 – 600 RPM with a linear dependence. Calculate:dependence. Calculate:

a)a) Current corresponding Current corresponding to 310 RPMto 310 RPM

b)b) The value of (a) in The value of (a) in percent.percent.

310


9: Controller Principles9: Controller PrinciplesControl System ParametersControl System ParametersExampleExample

SSpp = m I + S = m I + Soo

140 = 4 m + S140 = 4 m + Soo

600 = 20 m + S600 = 20 m + Soo

m = 28.75 rpm/mAm = 28.75 rpm/mA

SSoo = 25 rpm = 25 rpm

310 = 28.75 I + 25310 = 28.75 I + 25

I = 9.91 mAI = 9.91 mA

4 20

I (mA)

Speed (RPM)

140

600

310

%9.36%100420

491.9

xp


Control System ParametersControl System Parameters Control LagControl Lag

The time it takes for the final control element to adopt a The time it takes for the final control element to adopt a new value (as required by the process-control loop new value (as required by the process-control loop output) in response to a sudden change in the output) in response to a sudden change in the controlled variablecontrolled variable..

Dead TimeDead TimeThe elapsed time between the instant a deviation (error) The elapsed time between the instant a deviation (error) occurs and when the corrective action first occurs.occurs and when the corrective action first occurs.

CyclingCyclingThe cycling of the variable above and below the The cycling of the variable above and below the setpoint value.setpoint value.

9: Controller Principles9: Controller Principles


Control System ParametersControl System Parameters Controller ModesController Modes

• Continuous / DiscontinuousContinuous / DiscontinuousSmooth variation of the control parameter versus Smooth variation of the control parameter versus

ON / OFF.ON / OFF.

• Reverse / Direct ActionReverse / Direct ActionAn increasing value of the controlled variable An increasing value of the controlled variable

causes an decreasing / increasing value of the causes an decreasing / increasing value of the controller output.controller output.



Discontinuous Controller ModesDiscontinuous Controller Modes Two-Position ModeTwo-Position Mode

Neutral ZoneNeutral Zone


0

0

%100

%0

p

p

e

ep




ExampleExampleA liquid-level control system linearly converts a A liquid-level control system linearly converts a displacement of 2 – 3 m into a 4 – 20 mA control displacement of 2 – 3 m into a 4 – 20 mA control signal. A relay serves as the two-position controller signal. A relay serves as the two-position controller to open or close an inlet valve. The relay closes at to open or close an inlet valve. The relay closes at 12 mA and opens at 10 mA. Find:12 mA and opens at 10 mA. Find:

The relation between displacement level and The relation between displacement level and currentcurrent

The neutral zoneThe neutral zone




2 3

I (mA)

Displacement (m)

4

20

ExampleExampleH = K I + HH = K I + HOO

2 = K (4) + H2 = K (4) + HOO

3 = K (20) + H3 = K (20) + HOO

K = 0.0625 m / mAK = 0.0625 m / mA HHOO = 1.75 m = 1.75 m

HHHH = 0.0625 * 12 + 1.75 = 2.5 m = 0.0625 * 12 + 1.75 = 2.5 m

HHLL = 0.0625 * 10 + 1.75 = 2.375 = 0.0625 * 10 + 1.75 = 2.375

The neutral zoneThe neutral zone = H = HHH – H – HLL = 2.5 – 2.375 = 0.125 m = 2.5 – 2.375 = 0.125 m




ExampleExampleAs a water tank loses heat, the temperature drops As a water tank loses heat, the temperature drops by 2 K per minute. When a heater is on, the system by 2 K per minute. When a heater is on, the system gains temperature at 4 K per minute. A two-position gains temperature at 4 K per minute. A two-position controller has a 0.5 min control lag and a neutral controller has a 0.5 min control lag and a neutral zone of ± 4 % of the setpoint about a setpoint of zone of ± 4 % of the setpoint about a setpoint of 323 K.323 K.

Plot the tank temperature versus time.Plot the tank temperature versus time.




ExampleExample

Neutral zone = 310 – 336 KNeutral zone = 310 – 336 K

Temp. gain = 4 K per minuteTemp. gain = 4 K per minute

Setpoint = 323 KSetpoint = 323 K

Temp. loss = 2 K per minuteTemp. loss = 2 K per minute

Control lag = ½ minuteControl lag = ½ minute


Discontinuous Controller ModesDiscontinuous Controller Modes Multiposition ModeMultiposition Mode


p

p

p

eeeee

eep

2

21

1

%100%50

%0


Discontinuous Controller ModesDiscontinuous Controller Modes Multiposition ModeMultiposition Mode


ExampleExample


Discontinuous Controller ModesDiscontinuous Controller Modes Floating-Control ModeFloating-Control Mode

If the error is zero, the output does not change If the error is zero, the output does not change but remains (floats) at whatever setting it was but remains (floats) at whatever setting it was when the error went to zero.when the error went to zero.




Single SpeedSingle Speed


Pp

PpF

ee

eeK

dt

dp0

PpF eeptKp )0(



Single SpeedSingle Speed





ExampleExampleSuppose a process error lies within the neutral Suppose a process error lies within the neutral zone with p = 25%. At t = 0, the error falls below zone with p = 25%. At t = 0, the error falls below the neutral zone. If K = +2% per second, find the neutral zone. If K = +2% per second, find the time when the output saturates.the time when the output saturates.

SolutionSolution100 % = (2 %/s * t) + 25 %100 % = (2 %/s * t) + 25 %t = 37.5 st = 37.5 s

PpF eeptKp )0(



MultispeedMultispeed


10 Pp

PipFi

ee

eeK

dt

dp


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Proportional Control ModeProportional Control Mode

Direct (- K) & Reverse (+ K) ActionDirect (- K) & Reverse (+ K) Action

OpP peKp

%100minmax

xbb

brep



OpP peKp

Proportional BandProportional Band::

PKPB

100

% per %% per %



ExampleExampleValve A: 10 mValve A: 10 m33/h per percent./h per percent.PPOO = 50 % = 50 %

KKPP = 10 % per % = 10 % per %

Valve B changesValve B changesfrom 500 mfrom 500 m33/h/hto 600 mto 600 m33/h/h

Calculate:Calculate:The new controller outputThe new controller outputThe new offset errorThe new offset error



ExampleExampleValve A: 10 mValve A: 10 m33/h per percent./h per percent.PPOO = 50 % = 50 %

KKPP = 10 % per % = 10 % per %

Valve B changesValve B changesfrom 500 mfrom 500 m33/h/hto 600 mto 600 m33/h/h

Calculate:Calculate:The new controller outputThe new controller outputThe new offset errorThe new offset error

SolutionSolutionQQAA must go up to must go up to

600 m600 m33/h/h

QQAA = 10 m = 10 m33/h * P/h * P

P = 60 %P = 60 %

60 = 10 e60 = 10 ePP + 50 + 50

eePP = 1 % = 1 %


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Integral Control ModeIntegral Control Mode

Integral Time: Integral Time: TTII = 1 / K = 1 / KII

)0()(0

pdteKtpt

pI

% per sec %% per sec %

secsec


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Integral Control ModeIntegral Control ModeExampleExampleAn integral controller is used for speed control with An integral controller is used for speed control with a setpoint of 12 rpm within a range of 10 – 15 rpm. a setpoint of 12 rpm within a range of 10 – 15 rpm. The controller output is 22% initially.The controller output is 22% initially.The constant The constant KKII = - 0.15 % per second per % error. = - 0.15 % per second per % error.

If the speed jumps to 13.5 rpm,If the speed jumps to 13.5 rpm,calculate the controller output after 2 s for a calculate the controller output after 2 s for a constant constant eepp..


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Integral Control ModeIntegral Control Mode

ExampleExamplesetpoint = 12 rpm (range of 10 – 15 rpm)setpoint = 12 rpm (range of 10 – 15 rpm)P(O) = 22%P(O) = 22%KKII = – 0.15 = – 0.15

speed = 13.5 rpmspeed = 13.5 rpmconstant constant eepp

2 seconds time2 seconds time

SolutionSolutioneePP = (12–13.5)/(15–10)*100 %. = (12–13.5)/(15–10)*100 %.

eePP = – 30 % = – 30 %

P(t) = P(t) = KKII eePP t + P(0) t + P(0)

P(t) = (– 0.15) * (– 30) * 2 + 22P(t) = (– 0.15) * (– 30) * 2 + 22

P = 31 %P = 31 %

%100minmax

xbb

brep

)0()(0

pdteKtpt

pI


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Derivative Control ModeDerivative Control Mode

Not used alone because it provides no output Not used alone because it provides no output when the error is constant.when the error is constant.

dt

deKtp

pD)(

secsec


9: Controller Principles9: Controller PrinciplesContinuous Controller ModesContinuous Controller Modes Derivative Control ModeDerivative Control Mode

ExampleExample

op

D pdt

deKtp )(


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes Proportional – Integral Control (Proportional – Integral Control (PIPI))

)0()(0

pdteKKeKtpt

pIPpP


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes PIPI

)0()(0

pdteKKeKtpt

pIPpP ExampleExampleKKPP = 5 = 5

KKII = 1 sec = 1 sec-1-1

P(0)P(0) = 20% = 20%Plot a graph of the Plot a graph of the controller output as controller output as a function of timea function of time


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes PIPI

)0()(0

pdteKKeKtpt

pIPpP

ExampleExampleKKPP = 5 = 5

KKII = 1 sec = 1 sec-1-1

P(0)P(0) = 20% = 20%


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes Proportional – Derivative Control (Proportional – Derivative Control (PDPD))

op

DPpP pdt

deKKeKtp )(


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes PDPDExampleExampleKKPP = 5 = 5

KKDD = 0.5 sec = 0.5 sec

PoPo = 20% = 20%

op

DPpP pdt

deKKeKtp )(


9: Controller Principles9: Controller PrinciplesComposite Control ModesComposite Control Modes Three – Mode Controller (Three – Mode Controller (PIDPID))

)0()(0

pdt

deKKdteKKeKtp

pDPpIPpP

t



End of Chapter 9End of Chapter 9

3442 industrial instruments 2 chapter 9 controller principles dr. bassam kahhaleh princess sumaya...

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