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Process Control /Lec. 1 1Fourth Class

Introduction to process control

Temp.indicator

Hot water

WaterHeater

Steam

Cold water

Figure (1) Open loop system

Figure (2) Manual Control system

Figure (3) Automatic Control system (Closed Loop)


Control System Objectives

o Economic Incentiveo Safetyo Equipment Protectiono Reduce variabilityo Increase efficiencyo Ensure the stability of a processo Elimination of routine

Definitions:

System: it is a combination of components that act together and perform a certainobjective.

Plant: it is the machine of which a particular quantity or condition is to be controlled.

Process: is defined as the changing or refining of raw materials that pass through orremain in a liquid, gaseous, or slurry state to create end products.

Control: in process industries refers to the regulation of all aspects of the process.Precise control of level, pH, oxygen, foam, nutrient, temperature, pressure and flow isimportant in many process applications.

Sensor: a measuring instrument, the most common measurements are of flow (F),temperature (T), pressure (P), level (L), pH and composition (A, for analyzer).Thesensor will detect the value of the measured variable as a function of time.

Set point: The value at which the controlled parameter is to be maintained.

Controller: A device which receives a measurement of the process variable,compares with a set point representing the desired control point, and adjusts its outputto minimize the error between the measurement and the set point.

Error Signal: The signal resulting from the difference between the set pointreference signal and the process variable feedback signal in a controller.

Feedback Control: A type of control whereby the controller receives a feedbacksignal representing the condition of the controlled process variable, compares it to theset point, and adjusts the controller output accordingly.

Steady-State: The condition when all process properties are constant with time,transient responses having died out.

Transmitter: A device that converts a process measurement (pressure, flow, level,temperature, etc.) into an electrical or pneumatic signal suitable for use by anindicating or control system.


Heater(Heating)

Transmitter(Thermocouple)

Temperature

Controlled variable: process output which is to be maintained at a desired value byadjustment of a process input.

Manipulated variable: process input which is adjusted to maintain the controlledoutput at set point.

Disturbance: a process input (other than the manipulated parameter) which affectsthe controlled parameter.

Process Time Constant( ): Amount of time counted from the moment the variablestarts to respond that it takes the process variable to reach 63.2% of its total change.

Block diagram: it is relationship between the input and the output of the system.It iseasier to visualize the control system in terms of a block diagram.

X(s) Y(s)Input Output

Block diagram

Transfer Function: it is the ratio of the Laplace transform of output ( responsefunction) to the Laplace transform of the input ( driving force) under assumption thatall initial conditions are zero unless that given another value.

e.g. the transfer function of the above block diagram is G (s) = Y(s)/X(s)

Closed-loop control system: it is a feedback control system which the output signalshas a direct effect upon the control action.

Final control element

Advantage: more accurate than the open-loop control system.Disadvantages: (1) Complex and expensive

(2) The stability is the major problem in closed-loop control system

Transfer functionG(s)


Open-loop control system: it is a control system in which the output has no effectupon the control action. (The output is neither measured nor fed back for comparisonwith the input).

(Timer) Motor

Advantages:(1) Simple construction and ease of maintenance.(2) Less expensive than closed-loop control system.(3) There is no stability problem.

Disadvantages:(1) Disturbance and change in calibration cause errors; and output may be

different from what is desired.(2) To maintain the required quality in the output, recalibration is necessary

from time to time

Note: any control system which operates on a time basis is open-loop control system,e.g. washing machine, traffic light …etc.

Process Control /Lec. 2 5FourthClass

Part 1 Laplace Transforms

Revision of Laplace Transforms

1. The Laplace transforms

1.1 The definition of the laplace Transforms

The laplace transform of a dunction f(t) is defined to the f(s) according to the

equation:

0

)()]([)( dtetftfLsF st

Example:Find the Laplace Transform 1)( tf

Solution:

r

st

r

dteLsF0

.1]1[)( limss

est

rs

sr

1

0

lim

Example:Find the Laplace Transform atetf )(

Solution:

0

)(][)( dteeLsF tasat

asas

e tass

s

1)(

0

Example: Find the Laplace Transform of )cos()(&)sin( wttgwt

Solution: Again, from Example 3 and Theorem above

i

eeLwtL

iwtiwt

2)][sin(

iwsiwsi

11

2

122 ws

w

Similarly,22

)][cos(ws

swtL

Example: Find the Laplace Transform of

)sinh()(&)cosh()( wttgwttf

Solution: From Example 2 and Theorem above,

2)][cosh(

wtwt eeLwtL

wsws

1

2

11

2

122 bs

s

Similarly,22

)][sinh(ws

wwtL


1.2 Shifting Theroem

)(])([ asFetfL at .

Example: Find the Laplace Transform of )cos()( btetf at

Solution: Since ,][cos22 bs

sbtL

then

22)(]cos[

bas

asbteL at

Example: Find the Laplace Transform of ttetf 2)(

Solution:2

2

)2(

1][

steL t

Example: Find the Laplace Transform of tettf 5).3sin()(

Solution:9)5(

3]).3[sin(

2

5

setL t

Example: Find the Laplace Transform of tettf ).2cos()(

Solution:4)1(

)1(]).2[cos(

2

s

setL t

1.3 Laplace Transform of the Derivative

)0()(])(

[ fssfdt

tdfL

)0()0()(])(

[ 22

2

ffsfsdt

tfdL

)0(''f)0(fs)0(fs)s(fs]dt

)t(fd[L 23

3

3

)0(...)0()0()()]()(

[ )1(21 nnnnn

n

ffsfssfstdt

tfdL

.


Example: Find the Laplace Transform of the function )(tx which satisfy the

following differential equation and initial condition:-

22542

2

3

3

xdt

dx

dt

xd

dt

xd0

)0()0()0(

2

2

dt

xd

dt

dxx

Solution: By taken the laplace transform for both sides of the equation

ssxxssxxsxsxsxxsxssxs

2)(2)]0()([5)]0()0()([4)]0(")0()0()([ 223

ssxssxsxssxs

2)(2)(5)(4)( 23

sssssx

2]254)[( 23

)254(

2)(

23

sssssx

1.4 Laplace Transform of the integrs

s

sFdttfL

t )()(

0

Example: Find )(sx for the following equation

3)0(])([0

xtdttxLdtdx

Lt

.

Solution:

2

1)()0()(

ss

sxxssx

2

1)(3)(

ss

sxssx

13)()( 23 sssxsxs

13)()1( 23 ssxss

)1(

13)(

2

2

ss

ssx )s)(s(s 1112


transform ofaplaceLFind theExample:

ht

hth

t

)t(f

0

01

00

As inabovefigure,it is clear that f(t) may be represented by the difference of twofunctions.

)ht(u)t(uh

)t(f 1

Whereu(t - h) is the unit-step function translated h units to the right. We may nowusethe linearity of the transform and the previous theorem to write immediately

s

e

h)s(f

hs

11

This result is of considerable value in establishing the transform of the unit-impulsefunction.

Solved problems1.Use Definition to determine the Laplace transform of the given function

(a) 23

2{ }L t

s

(b) 3 3 (3 ) (3 )2 2

0 0 0

1 1{ } lim ( )

3 (3 ) ( 3)

N

t st t s t s t

N

tL te e te dt te dt e

s s s

(c)2 2

1 1{ ( )}

seL f t

s s s

(d)3( 2) 3

32

0 3

1{ ( )}

2

s sst t st e eL f t e e dt e dt

s s

2. Use Table of Laplace Transform to determine

(a) 2 23

5 1 12{5 6 }

2tL e t

s s s

(b) 23 2 2

2 3 6{ 3 2 sin 3 }

( 1) 9tL t t e t

s s s

(c) 2 2 22 3

2 2{ cos 3 }

( 2) 3 ( 2)t t sL e t t e

s s

0 h Time


A Short List of Important Laplace Tranforms

0),( ttf F(s) Domain F(s)

1 1

s

1 0s

2 t2s

1 0s

3 tn

0n integer 1ns

!n

0s

4 eat

as

1

as

5 btcosh22 bs

s

bs

6 btsinh22 bs

b

bs

7 btcos22 bs

s

0s

8 btsin22 bs

b

0s

9 btcoseat22 b)as(

as

as

10 bteat sin22)( bas

b

as


1.5 Inverse of Laplace Transform

Inversion by partial fraction

The inverse of the laplace could be obtained from tables for simple functions but for

more complicated function partial fraction is needed

)()(1 tfsfL

1- Partial fraction

Example:

?

?

)()(

)()())((

1

1

1

o

o

o

asbsA

bsasbsas

A

Example:

221

2 )()()())(( bsbsasbsas

A o

Example:

)()())(( 2221

22 ws

s

asasws

A o

Example: Solve the IVP .1)0(,1)0(, yytyy

Solution:

Step 1:Take Laplace Transform:2

2 1)()1)((s

sYssYs (subsidiary equation)

Step 2:Solve for )(sY11)1)(1(

1

1

11

)(22

23

2

2

s

D

s

C

s

B

s

A

sss

ss

s

ss

sY

Then: 1)1s)(1s(

1ssA

0s

23

,2

3

)1s(s

1ssC

1s

2

23

,

2

1

)1(

1

12

23

sss

ssD and lastly, to find B, substitute ,2s we get

6

1

2

3

24

1

12

13

Band obtain .0B


Step 3: Solve for )(ty :11

1)( 2

123

2

ssssY

tt eetsYLty 21

231 )]([)(

Example:

Find ]e4x2dt

dx

dt

xd2

dt

xd[L t2

2

2

3

31

where

1)0(''x,0)0('x,1)0(x

Solution:

)2s(

1

s

4

)s(x2)]0(x)s(sx[)]0('x)0(sx)s(xs[2)]0(''x)0('sx)0(xs)s(xs[ 223

)2(

14121)](2)()(2)([ 223

sssssxssxsxssxs

22)2(

14)(]22[ 223

ss

sssxsss

)2(

)2)()(22()2(4

)22(

1)(

2

23

ss

ssssss

ssssx

)s)(s)(s)(s(s

sss)s(x

1212

896 24

)s()s()s()s(s)s(x o

12124321

)s)(s)(s(s)s)(s)(s(s)s)(s)(s(s

)s)(s)(s(s)s)(s)(s)(s(sss o

212112122

1211212896

432

124

assume s=0

))()()((o 12128

24

8

o

3/11-1=sassume 2

12/17-2=sassume 3

3/21=sassume 4

12/12=sassume 1


1.6 Final Value Theorm

)s(sflim)t(flimst 0

1.7 Initial Value Theorm

)s(sflim)t(flimst

0

Example:

Find the final value of the function x(t) for which the laplace inverse is:-

)sss(s)s(x

133

123

1133

1

133

1

230

2300

)sss(lim

)sss(s

slim)s(sxlim)t(xlim

s

sst

1.8 Special Functions

1- Step function

s

A)s(f

tA

t)t(f

0

00

.

If A=1 the change is called unit step change

s)s(f

t

t)t(f

1

01

00

1

3/2

2

12/17

1

3/11

2

12/12)(

ssssssx

tttt eeeetx3

2

12

17

3

11

12

12)( 22

A

0 Time


Step function with Time Delay

ases

A)s(f

atA

at)t(f

0

2. Pulse function

)1(

)(

0

0

00

)(

as

as

es

A

es

A

s

Asf

at

atA

t

tf

if A=1unit Pulse

3. Impulse function

tAarea)s(f

tt

ttA

t

)t(f

0

0

00

4. Ramp function

2

0

00

s

A)s(f

tAt

t)t(f

0 a Time

A

0 a Time

A

A Area=1

Unit Impulse Time

A

F(t) δt

Time

Time

f(t)Slope=A

f(t)


Ramp function with time delay

ases

A)s(f

atAt

at)t(f

2

0

5. Sine function

fT

f

s

A)s(f

atwtsinA

at)t(f

1

2

0

22

1.9 Disturbance function (complex function)

80

872

762

650

542

412

102

t

tt

tt

t

tt

t

tt

)t(f

ssssss es

es

es

es

es

ess

)s(f

)t()t()t()t()t()t(t)t(f

8

2

7

2

6

2

5

2

4

222

2422222

82722625242122

a Time

f(t)Slope=A

f(t)

A _______T_______

-A

01 4 5 6 7 8

2

Slope=2

-2


Example:

ssss es

es

es

ess

)s(f

)t()t()t()t(t)t(f

5

2

4

2

3

222

12211

542321

Example:

asas es

be

s

b

s

b)s(f

)at(b)at(bbt)t(f

2

222

2

22

Example:

s.sss.s. es

.e

se

s

.e

s

.e

ss

.)s(f

).t(.)t()t(.).t(.).t(.)t(f

53325150 5015050150

535031250515050150

3

2

1

0 1 3 4 5T

0.5

0 0.5 1.5 2 3 3.5

-.5

0 a 2a

F(t)

Slope=b Slope=-b


Solved problems

1. Determine the inverse Laplace transform

(a) 1 1 2 5 /23 3

3 3 1 3

(2 5) 8 ( 5 / 2) 16tL L t e

s s

(b) 1 /4 /42

1 1 47 5 47cos( ) sin( )

2 3 2 4 42 47

t ts t tL e es s

2. Determine the partial fraction expansion

(a)2

2

8 5 9 1 2 11

( 1)( 3 2) 1 1 2

s s

s s s s s s

(b)2 2 2

5 36 1 32 2 3

( 2)( 9) 2 9 9

s s

s s s s s

(c)2 2

4 3 3 3 2

3 5 3 3 5 3 3 2 1 1

( 1) ( 1)

s s s s

s s s s s s s s

(d)2 2

1 1 1 1 1 1

( 1)( 1) 2 ( 1) 4 ( 1) 4 ( 1)

s

s s s s s

2. Determine1{ ( )}L F s

(a)

1 3

11 1 1( ) 3 2

( 1)( 3) 1 3

[ ( )] 3 2t t

sF s

s s s s

L F s e e

(b)3 2

3 3

1 2 2

7 2 3 6 3 1 6( )

( 2) 2

3[ ( )] 1 6

2t

s s sF s

s s s s s

L F s t e


Part IILinear Open-Loop Systems

Response of first order systems

2. Dynamic behavior of first order system

Before studying the control system it is necessary to become familior with the

response os some of simple basic systems (i.e study the dynamic behaviour of the

first and second order systems).

2.1 The transfer function:

The dynamic behaviour of the system is described by transfer function(T.F)

ܨܶ. =ܽܮ ݈ ܽܿ ܽݎݐ݁ ݂ݏ݊ ݉ݎ ݂ ݁ݎ)ݐݑݐݑℎ݁ݐ ݊ݏ ܿ݁ )

ܽܮ ݈ ܽܿ ܽݎݐ݁ ݂ݏ݊ ݉ݎ ݂ ℎ݁݅݊ݐ )ݐݑ ܿݎ݂ ݅݊ ݃ ݂݊ݑ ݊݅ܿݐ ݀ ܾܽݑݐ݅ݏ ݊ܿ݁ )

X(s) Y(s)Input Output

Forcing function ResponceBlock diagram

ܨܶ. = (ݏ)ܩ =(ݏ)ݕ

(ݏ)ݔThis definition is applied to linear systems

2.2Development of T.F for first order system:

Mercury Thermometer:

It is a measuring device use to measure the temperature of a stream.

Consider a mercury in glass thermometer to belocated in a flowing stream of fluid for which thetemperature x varies with time.

The opject is to calculate the time variation of thethermometer reading y for a particular change of x

The following assumptions will be used in this analysis:-

1. All the resistance to heat transfer resides in the film surrounding the bulb(i.e.,the resistance offered by the glass and mercury is neglected).

2. All the thermal capacity is in the mercury. Furthermore, at any instantthemercury assumes a uniform temperature throughout.

3. The glass wall containing the mercury does not expand or contract duringthetransient response.

TransferfunctionG (s)


It is assumed that the thermometer is initially at steady state. This meansthat,before time zero, there is no change in temperature with time. At time zerothethermometer will be subjected to some change in the surrounding temperaturex(t).(i.eat t


( +߬ݏ (ݏ)ܻ(1 = (ݏ)ܺ

(ݏ)ܻ

(ݏ)ܺ= (ݏ)ܩ =

1

+߬ݏ 1… … … … … … (2.2)

ܨܶ. =(ݏ)ܻ

(ݏ)ܺ= (ݏ)ܩ =

ܶ.ܮ ݂ ℎ݁݀ݐ ݅݁ݒ ݊݅ܽݐ ݅݊ ℎݐ ݉݁ݎ ݉ ݁݁ݐ ݁ݎݎ ܽ݀ ݅݊ ݃

.ܮ .ܶ ݂ ℎ݁݀ݐ ݅݁ݒ ݊݅ܽݐ ݅݊ ݊݅݀݊ݑݎݑݏ ݃ܶ݁݉ ݁ ܽݎ ݁ݎݑݐ

Any system has a T.F of the form of equation (2.2) it is called first order systemwhich is a first order differntial equation (Linear).

1.3 Properties of transfer functions

A T.F relates two variables in a physical process. One of these is (Forcing or Inputvariable) and the other is the effect (Reponce or Output).

ܨܶ. =(ݏ)ܻ

(ݏ)ܺ= (ݏ)ܩ

If we select a particular input variation x(t) for which the L.T is x(s) then the reponce.

(ݏ)ܻ = (ݏ)ܺ.(ݏ)ܩ … … … … … … . (2.3)

(ݏ)ଵܻିܮ = (ݐ)ܻ = (ݏ)ܺ.(ݏ)ܩଵିܮ

If G(s) is 1st order of a thermometer

(ݏ)ܻ = (ݏ)ܺ.(ݏ)ܩ =1

+߬ݏ 1(ݏ)ܺ. … … … … … … … . . (2.4)

x(s) y(s)

1.4 Transient response for different changes

(ݏ)ܻ =1

+߬ݏ 1(ݏ)ܺ.

Y(t)=? For different types of x(t)

1-Step Change

s

AsX )(

1.

1

1)( 1

sss

A

ssY o

s)s(A o 11

S=0 Ao

/s 1

1

1 1)/(A o then Ao

G(s)

A ----------x(t)

t


/1/1

/1

11)(

A

s

A

s

A

s

A

s

A

s

AsY

)1()( tt eAAeAtY …………….. (2.5)

Several features of this response, worth remembering, are

The value of y(t) reaches 63.2 % of its ultimate value when the time elapsed isequal to one time constant .߬

When the time elapsed is 2 ,߬ 3 ,߬ and 4 ,߬ the percent response is 86.5%, 95%,and 98%, respectively.

Where the ultimate value is find steady state value

)s(sylim)t(ylimV.Ust 0

…..(2.6)

Example (5.1): A thermometer having a time constant of 0.1 min is at a steady state

temperature of 90 Fo. At time t = 0, the thermometer is placed in a temperature bath

maintained at 100°F. Determine the time needed for the thermometer to read 98 Fo.

Solution:At s.s.xs=ys=90 F

o

Step Change s

A)s(X

A=100-90=10

s)s(X

10

10101010

10

110

1010

110

1

1

1

s

B

s.

A

)s(s.)s.(sss.s

A

s)s(Y

101010 )s.(B)s(A

110

100 As

1010 Bs

10

1010

10

10

10

1

ssss.)s(Y

By taken lablace inverse for the equation

)e(e)t(Y tt 1010 1101010

Substitute Y(t)=y(t)-ys =98-90Y(t)=8

)e( t101108 te. 10180

).ln()eln( t 2010

).ln(t 2010

1020 .).ln(t

t=0.161 min

A ----------Ultimat valuey(t)

t


2-Impulse Input

A)s(X

11

1

s

AA

s)s(Y

1s

A)s(Y

ts e

Ay)t(y)t(Y

st ye

AtY

)(

2-Sinsoidal input

0 twtsinAx)t(x s

wtsinAx)t(x s

wtsinAx)t(x)t(X s ……………….(2.9)

22 ws

Aw)s(X

])s)(ws(

[Aw)s(ws

Aw)s(Y

1

1

1

12222

This equation can be solved for y(t) by means of a partial fraction expansion as

described in previous lectures.

])s()ws(

s[Aw]

)s)(ws([Aw)s(Y o

11

1 2221

22

11 2221 )ws()s)(s( o

1222

2112 wssss oo

12210 ws (1)

011 os 1o (2)

022 os o2 (3)

By substitution eq.(2) in eq.(3)

212 (4)

By substitution eq.(4) in eq.(1)

12211 w

y(t)

t

x(t)

Area


221 1

1

w

22

2

21 w

220 1 w

])s(

w)ws(

ws

w[Aw)s(Y1

11

1

1 22

2

22

2222

])s()ws(

s[

wAw)s(Y

1

1

1

1 2

2222

])s(w

w

)ws()ws(

s[

w

Aw)s(Y

1

1

1

2

222222

])/s()ws(

w

w)ws(

s[

w

Aw)s(Y

1

1

1 222222

]ewtsinw

wtcos[w

Aw)t(Y t

1

1 22

Using the definition

q

ptanqpr

)sin(rsinqcosp

22

q

ptan 1

w

w

w)

w(r

)w(tan

pw

q

222

2

22

1

111

1

)wtsin(rwtsinw

wtcos 1

)]wtsin(w

we[

w

Aw)t(Y t

22

22

1

1


)w(tan

where

)]wtsin(w

Ae

w

Aw)t(Y t

1

2222 11

As 0 tethent , the first term on the right side of main equation

vanishes and leavesonly the ultimate periodic solution, which is sometimes called thesteady-statesolution

)]wtsin(w

A)t(Y

221 …..(2.10)

By comparing Eq. (2.9) for the input forcing function with Eq. (2.10) forthe ultimateperiodic response, we see that1. The output is a sine wave with a frequency w equal to that of the input signal.

2. The ratio of output amplitude to input amplitude is 11

122

w.

3. The output lags behind the input by an angle . It is clear that lag occurs,for the

sign of is always negative.

loadphase

lagphase

0

0

Example 5.2. A mercury thermometer having a time constant of 0.1 min is placedin a

temperature bath at 100°F and allowed to come to equilibrium with the bath. Attime

t= 0, the temperature of the bath begins to vary sinusoidally about its

averagetemperature of l00oF with an amplitude of 2°F If the frequency of oscillation

is ߨ/10 cycles/min, plot the ultimate response of the thermometer reading as a

function oftime. What is the phase lag?

In terms of the symbols used in this chapter

A X(t) _______T_______

-A


10

21000

1000

10

f

)wtsin()t(xt

yxt

.

ss

Solution

cyclef

T

radfw

min/101

min/1010

22

100202100 tsinx)t(x)t(X stsin)t(X 202

22 20

202

s)s(X

Ultimate responce 0 tethent

)]wtsin(w

A)t(Y

221

o.

)(tan).(tan)w(tan

563

21020 111

Ultimate responseat the above angle

)].tsin().(

)t(Y 5632020101

22

)].tsin()t(Y 563205

2

)].tsin(.)t(Y 563208960 Ultimate response

In general, the lag in units of timeis given by:-

flagphase

1

360

min.

cycle

mincycle.lagphase

05550

10360

563

A frequency ofଵ

గ

௬

means that a complete cycle occurs in (

ଵ

గ)ିଵ ݉ ݅݊ . since cycle

is equivalent to 360o and lag is 63.5o


How to calculate the time constant () for 1st order system1) Mathematical method

Using the definitions

kMixingq

V

klevelLiquidAR

rThermometehA

cpm

tan

tan

2) Exponential method(Step change in the input variable)

A.).(A)e(A)(Ytas

A)e(A)(Ytas

)e(A)t(Y t

6703678011

1

1

1

Time constant ( )߬is the time required for the response to reach 63% of the its utimatevalue.

3. Third method

)e(A)t(Y t 1

tt e

A)(Ae

dt

dy1

Ae

A

dt

dylimt

0

0

Slope of the tangent at t=0 is

A

Therforeslope

A

)e(A)t(Y t 1 tAeA)t(Y

)t(YAAe t

A

)t(YAe t

A

0.63 A

τ t

A

Y(t)slope

t


A

)t(YAlnt

Let A

)t(YAB

1

1

slope

tx

Blny

let

tBln

Y(t)

A

)t(YAB

Bln t

slope

slope

1

1

lnB

1slope

t


Physical Examples of first Order System

1.Liquid Level Tank

Consider the system shown in Figure which consists of:

1. A tank of uniform cross-sectionalarea A.

2. Valveattached to the output flow which resistance constant=R.

qo : The output volumetric flowrate (volume / time) through the resistance, is related

to the head h by the linear relationship.

ValveLinearhqo

R

hqo ……………..(1)

Where:

R is related as a linear resistance

if )n(hq no 1 Non linear valve

q(t) is a time varying volumetric flowrate with constant density ρ.

Find the T.F. that relates the head to the input flowrateq(t).

We can analyze this system by writing a transient mass balance around thetank:

Mass flow in - mass flow out = rate of accumulation of mass in the tank

dt

dvqq o

)Ahv(

dt

dhAqq o

dt

dhA

R

hq ………………………(2)

At staedy state

0dt

dhA

R

hq sss ………………………(3)

Substracting Eq(3) from Eq. (2)

dt

)hh(dA

R

hh)qq( sss

QR

H

dt

dHA

RQHdt

dHAR Taken laplace for both sides of equation

where)s(RQ)s(H)s( 1 AR


)s(

R

)s(Q

)s(H

1 Firstorder system equation………………………(4)

)s(Q)s(

R)s(H

1 ………………………(5)

Comparing the T.F. of the tank with thr T.F. of the thermometer we see that eq. (5)

contain the factor (R) which is relate H(t) to Q(t) at s.s. as ts 0

For this reason, a factor K in the transfer function1s

Kis called the steady state gain

To show that

Takes

)s(Q1

Therefores

.s

R)s(H

1

1

Final value theorm

Rs

Rlim

s.

s

Rslim)s(sHlim)t(Hlim

ssst

1

1

1 000s.s gain

This show that the ultimate change in H(t) for a unit change in Q(t) is R

?)s(Q

)s(Qo

R

hqo

R

hq sos

R

HQo

)s(RQ)s(HR

)s(H)s(Q oo

1

s

R

)s(Q

R)s(Qo

1

1

s)s(Q

)s(Qo


)e()t(H t 110

Example 6.1 A tank having a time constant of 1 min and a resistance of 1/9 ft/cfmis

operating at steady state with an inlet flow of 10 ft3 /min (or cfm). At time t =0, the

flow is suddenly increased to 100 ft3/min for 0.1 min by adding an additional 9 ft3 of

water to the tank uniformly over a period of 0.1 min. (See Fig. 6-2a for this input

disturbance) . Plot the response in tank level and compare with the impulse response.

Solution:

1

s

R

)s(Q

)s(H

1

1

9

1

s)s(Q

)s(H

The input may be expressed as the difference in step

function as was done in example

Pulsesecsec

sec.*.

secmin.a

Pulse.aIf

pulseIm.aIf

36

360050050

610

050

050

Pulse Function)].t(u)t(u[A)t(Q 10

90=10-100=A

)e(s

)s(Q s.10190

)e(ss

)s(H s.10190

1

1

9

1

)e)s(s)s(s

()s(H as

1

1

1

110

for t 0.1

Simplifying the expression for H(t) for t>0.1 gives

)eee()t(H tt. 1010

)1052.1(10)( tt eetH

tetH 052.1)( t>0.1


If the input function is assumed impulse function

If t


0dt

dyVFyFx sss

Subtracting the above both equations and introducing the deviation variables

dt

)yy(dV)yy(F)xx(F sss

s

s

yyY

xxX

FXFYdt

dYV

XYdt

dY

F

V

XYdt

dY

)s(X)s(Y)s( 1

)s()s(X

)s(Y

1

1

1st order system ,where

F

V

Example:Find the T.F for the system shown in figure

dt

dhA

R

hFF 21

dt

dhA

R

hFF ssss 21

dt

dHA

R

HQQ 21

RQRQHdt

dHAR 21

R)s(QR)s(Q)s(H)s( 211

)s(Q)s(

R)s(Q

)s(

R)s(H 21

11

When F1 constant 01

1

)s(Q)s(

R)s(Q

)s(

R)s(H 2

1

When F2 constant 01

2

)s(Q)s(

R)s(Q

)s(

R)s(H 1

1

h

F1F2

qo

=


3.Stirred Tank Heating System

Energy balance equation.

Q)TT(wCdt

dTCV i

Assumption : constant liquid holdup and

constant inflow(w is constant), a linear

model result.

If the process is at steady-state, 0dtdT

ssis Q)TT(wC 0

Subtract equations

)()]()[()(

ssisis QQTTTTwC

dt

TTdCV

Define some important new variables(Deviation variables).

sisiis QQ'Q,TT'T,TT'T

By substituting deviation variables for variables, the transfer functions are not related

o initial conditions.

'Q)'T'T(wCdt

'dTCV i

Let wV,wCk 1

Apply Laplace Transform.

)s('kQ)s('T)s('T)s( i 1

)s('Qs

k)s('T

s)s('T i

11

1

If 0)s('Ti1

1

s

k

)s('Q

)s('T)s(G

If 0)s('Q1

12

s)s('T

)s('T)s(G

i

Homework

1) Solve 6.1, 6.3, 6.5, 6.8

2) Drive an equation represent the response of a level H(t) in liquid level tank

system for the follwing inpput fuctions.

a) Step change Q(t)=10 at time>0

b) Ramp change Q(t)=0.5t at time>0

Figure Continuous stirred-tank heater.

)s('kQ))s('T)s('T()s('sT i


Response of 1st order systems in series

Many physical systems can be represented by several first-order processes connected

in series as shown in figure:-

Figure 7-1Two-tank liquid-level system: (a) Non-interacting; (b) interacting.

In fig (7-1 a) variation of h2 does not effect on q1then1

11

R

hq

In fig (7-1 b) variation of h2 does effect on q1then1

211

R

hhq

1-Non Interacting System

Material balance on tank 1 gives

dt

dhA

R

hqi

11

1

1

At s.s. 0111

1 dt

dhA

R

hq ssis

By substracting both equations

dt

)hh(dA

R

hh)qq( ssisi

111

1

11

11

11

1 R]dt

dHA

R

HQ[ i

dt

dHARHQR i

11111


11

11

s

R

)s(Q

)s(H

iwhere 111 RA

Material balance on tank 2 gives

dt

dhA

R

h

R

h 22

2

2

1

1

At s.s. 0222

2

1

1 dt

dhA

R

h

R

h sss


dt

)hh(dA

R

hh

R

hh sss 222

2

22

1

11

22

22

2

1

1 Rdt

dHA

R

H

R

H

11

22

222 H

R

RH

dt

dHRA

)s(HR

R)s(H)s(sH 1

1

2222 222 HR

)s(HR

R)s(H)s( 1

1

222 1

)s(H)s(

RR)s(H 1

2

122

1 By substituting the lapace transform of H1(s)

11 1

1

2

122

s

R)s(Q

)s(

RR)s(H i

)s(Q)s)(s(

R)s(H i

11 21

22

)s)(s(

R

)s(Q

)s(H

i 11 21

22

Non-interacting system

In the case of three non-interacting tanks in sereies the transfer function of the system

will be as below:-

)s)(s)(s(

R

)s(Q

)s(H

i 111 321

33


Example 7.1.

Two non-interacting tanks are connected in series as shown in Fig. 7-1 a. The time

constants are τ2 =1 and τ1 =0.5; R 2=1. Sketch the response of the level in tank 2 if a

unit-step change is made in the inlet flow rate to tank 1.

Solution:

The transfer function for this system is found directly from Equation above thus

)s(Q)s)(s(

R)s(H i

11 21

22

Substitutings

)s(Qi1

Unit step change in Qi

)s()s(s

s)s)(s(

R)s(H

o

11

1

11

2

2

1

1

21

22

)s(s)s(s)s)(s(R o 1111 1221212

20 Rslet o

221

1212

121

212

12

11

1

11

111R)(R)(R))()((slet

)(R12

21

21

222

2122

222

122

21

22

2

11

111R)(R)(R))()((slet

)(R21

22

22

)s()(R

)s()(R

s

R)s(H

1

1

1

1

221

22

2112

21

22

2

])s(

)()s(

)(s

[R)s(H1

1

1

11

21

2

21

21

12

1

12

2122

])/s(

)()/s(

)(s

[R)s(H2121

21

1221

2122

1

11

1

111

)ee)(((R)t(H /t/t 21

1221

2122

111

)e.

e)(.

.()t(H /t./t 1502

50

1

1

1

150

5011


)ee)(.

.()t(H tt

250

501 22

)ee)t(H tt 21 22

)s(Q.s

R)s(H i

11

11

s.

s

R)s(H

1

11

11

)e(R)t(H /t 1111

Let R1=1

)e()t(H ./t 501 11

te)t(H 21 1

Example:

Obtain the transfer function of the following system (no reaction):

Where:

F = volumetric flow rate, Fi = F1C = conc. of solute in stream.

V = liquid volume in tank.

Solution:

Mass balance on concentration; i.e.

In – out = accumulation

Tank 1:dt

dCV=CF-CF 111ii

1

iCCdt

dc 1

11 where τ1 = V1/F1

Laplace transform → τ1 s C1(s) + C1(s) = Ci(s)

1

1

1

1

s)s(C

)s(C

i…. (1) Ci(s) C1(s)

Tank 2:dt

dCV=CF-CF+CF 222iiii11

2

iiii

11

22 C

F

F+C

F

FC

dt

dC

F

V

22

2

2

iiCKCKCdt

dC2112

22

H1(t)

H2 (t)

1

1

1 s

(non-interacting system)

FiiCii

F2, C2V2, C2

F1, C1V1 c1

Fi , Ci


22

2

11

2

22

F

FK,

F

FK,

F

V ii

Laplace transform → τ2s C2(s) + C2(s) = K1 C1(s) + K2 Cii(s)

)s(Cs

K)s(C

s

K)s(C ii

11 2

21

2

12

Substitute C1(s) from Eq. (1)

)s(Cs

K)s(C

)s)(s(

K)s(C iii

111 2

2

21

12

2. Interacting System

Material balance on 1sttank

dt

dhAqqi

111

dt

dhA

R

hhqi

11

1

21

Steady state

0111

21

dt

dhA

R

hhq sssis


dt

)hh(dA

R

hh

R

hh)qq( sssisi

111

1

22

1

11

11

11

1

1

2 R]dt

dHA

R

H

R

HQ[ i

dt

dHRAHHRQi

111121

2111

1 HRQHdt

dHi

)s(H)s(QR)s(H)s( i 2111 1

)s(H)s(

)s(Q)s(

R)s(H i 2

11

11

1

1

1

…………..(1)

Material balance on second tank

dt

dhA

R

h

R

hh 22

2

2

1

21


0222

2

1

21

dt

dhA

R

h

R

hh ssss

22

22

2

1

2

1

1 R]dt

dHA

R

H

R

H

R

H[

)HH(R

RH

dt

dHRA 21

1

22

222

))s(H)s(H(R

R)s(H)s( 21

1

222 1 …………..(2)

Substituting for H1(s)from eq(1) in eq(2)

)]s(H)s(H)s(

)s(Q)s(

R[

R

R)s(H)s( i 22

11

1

1

222

1

1

11

)s()]s(HR

R

)s(

)s(H

R

R

)s(

)s(QR)s(H)s[( i 1

111 12

1

2

1

2

1

2

1

222

)]s(HR

R)s()s(H

R

R)s(QR)s(H)s)(s( i 2

1

212

1

22212

111

)s(QR)s(HR

sR)s(H)sss( i22

1

21221

221 1

Let 12211

211

1

21

RAR

RRA

R

R

)s(QR)s(H)s)(s( i2212212

21 1

)s(Q.s)(s

R)s(H i

112212

21

22

Interacting system

)s(Q.s)(s

R)s(H i

1212

21

22

Non- Interacting system

The difference between the transfer function for the non-interacting system, and that

for the interacting system, is the presence of the cross-productterm A1R2 in the

coefficient of s. 2112 RA

To understand the effect of interaction on the transient response of a system,

considera two-tank system for which the time constants are equal (τ1=τ2=τ).


Example:

τ1 = τ2 = τ12=τ

Q2(t)=? Output flow rate

s)s(Qi

1

Non-interacting system

21

212

21

22

1

=

s)(s

R

)s(Q

)s(H

i

122222

ss

R

)s(Q

)s(H

i

but2

22

R

)s(H)s(Q

2

222

1

1

11

1

12

1)

s(

)s)(s(ss)s(Q

)s(Q

i

Ifs

)s(Qi1

11

1

1

1 22

122

s)s(ss.

)s()s(Q o

By multiplying both sides by 21)s(s and expanding, we get

111 212 )s(ss)s(o

112 22122 )ss(s)ss(o

12 21222 ooo )(s)(s

10 os

222

222 00os

11211 0202 os

11

122

s)s(s)s(Q

11

122

s)s(s)s(Q

/s)/s(s)s(Q

1

1

1

11122

/t/t e

te)t(Q 12 for non-interacting


Interacting system

If the tanks are interacting, the overall transfer function, according to Equation of

interacting system (assuming further that A1=A2)

s.

ss)s(Q

1

13

1222

By application of the quadratic formula, the denominator of this transfer function

canbe written as

)s.)(s.(s)s(Q

16221380

112

1622138021

2

s.s.s)s(Q o

10 oslet

06640

1380

1622

1380

380

11 .

).

(.

..

slet

6643

1622

1380

1622

622

11 .

).

(.

..

slet

1622

06643

1380

0664012

s.

.

s.

.

s)s(Q

6221

62206643

3801

3800664012

./s

./.

./s

./.

s)s(Q

62.2/1

17.1

38.0/1

17.01)(2

ssssQ

2623802 1711701./t./t e.e.)t(Q

Figure7–7Effect of interaction on step response of twotank system.


Homework1. The two-tank liquid-level system shown in Figure H-1 is operating at steady state

when a step change is made in the flow rate to tank 1. The transient response takes

1.0 min for the change in level of the second tank to reach 50 percent of the total

change.

If the ratio of the cross-sectional areas of the tanks is A1/A2 = 2, calculate the ratio

R1/R2. Calculate the time constant for each tank. How long does it take for the change

in level of the first tank to reach 90 percent of the total change? Note: 1 = 2

Figure H-1

2. The two tanks shown in Fig. H-2are connected in an interacting fashion. The

system is initially at steady state with q = 10 cfm. The following data apply to the

tanks: A1 = 1 ft2, A2 = 1.25 ft

2, R1 = 1 ft/cfm, and R2 = 0.8 ft/cfm.

(a) If the flow changes from 10 to 11 cfm according to a step change, determine

H2(s),i.e., the Laplace transform of H2 where is the deviation in h2.

(b)Determine H2(1), H2(4), and H2(∞).

Figure H-2

R1 R2

A1 A2

h1 h2

Q2

Q

R1

R2

A1

A2

h1

h2

Q1

Q2

Q0


Non-linear systems

To solve non-linear systems there are two methods:-

1-Linearization method

Making the non-linear function as linear using Taylar series and give approximate

results.

2-Non-linear solution

It is difficult and give exact solution

Linearization Technique

xty

ty 2Linear(all terms to power =1)

xlny

ty

ty

2

2

Non-Linear (power ≠1)

To make the non-linear function linear one use Tayler series.

...............)xx(dx

)x(fd

!)xx(

dx

)x(fd)x(f)x(f o

xx

oxx

o

oo

2

2

2

2

1

Neglacting the non linear terms because their value are very small.

Then

)xx(dx

)x(fd)x(f)x(f o

xxo

o

…….(1)


Example:The flow of water through a valve or other construction usually follow a

square-root law.

IfR

hqo Linear valve

hcqo Non-linear valve

c is a constant

dt

dhA

dt

)V(dqq oi

dt

dhAchq /i

21 ……..(1)

qo may be expanded around the s.s. valuehsusing linearization method

dt

dhA)hh(

hhcq s/

/si

s

21

21 1

2

1…….. (2)

021 dt

dhAhcq s/sis …….. (3)at s.s h=hs

dt

)hh(dA)hh(

hc)qq( ss/

s

isi

21

1

2

1

dt

dHAH

h

cQ

/s

i

21

1

2

Rh

cassume

s

1

2

dt

dHA

R

HQi

iRQHdt

dHRA

By taking laplace transform

)s(RQ)s(H)s( i 1

1

s

R

)s(Q

)s(H

i………….. 1st order system

Wherec

hR s

2 AR

1-Transfer function is similar to linear.

2- R depends on the s.s. condition (at steady state the flow entering the tank equals to

the flow leaving the tank,thenqo=qos).

h

qi

qo

= √


Example : Mixing tank with chemical reaction

2

2

kcrRateactionRe

CBA

?)s(C

)s(C)s(G

i

ci,c : Composition of component (A)V

V: Constant=L

F: Constant=L/min

In-out-rate of reaction= accumulation

dt

dcVVkcFcFci

2 Un-steady state

)cc(ccc sss 222

dt

dcV)]cc(cc[VkFcFc sssi 2

2

02 dt

dcVVkcFcFc sssis Steady statec=cs

dt

)cc(dV)]cc(c[Vk)cc(F)cc(F ssssisi

2

)VkcF(FCC)VkcF(dt

dCV sis 22

iRCCdt

dC

1

s

R

)s(C

)s(C

i1st order system

Where isiis ccC,ccC

ss VkcF

FR,

VkcF

V

22

F, c

F, ci

CBA 2


Linearization of nth order nonlinear differential equation

Consider thethn order nonlinear differential equation

Expanding the nonlinear function in a Taylor series about an operating point that

satisfies the original differential equation and retaining only the linear

terms yields

)xx(dx

)x,...x,x(df)xx(

dx

)x,...x,x(df

)xx(dx

)x,...x,x(df)x,...,x,x(f)x,...,x,x(f

onn

xxn

no

xx

n

o

xx

nxxnn

iiii

ii

ii

00

0

0

2122

2

21

111

212121

Example:

Mixing tank

?)s(F

)s(C)s(G

c : Variable (kg/L)

F: Variable (L/min)

Ci: Constant

V: Constant

dt

dcVFcFc

termlinearnoni

)yy(y

f)xx(

x

f)y,x(f)y,x(f s

yyxxs

yyxxss

s

s

s

s

dt

dcV)]cc(F)FF(ccF[Fc ssssssi Un s.s

0dt

dcVcFcF sssis s.s

dt

)cc(dV)cc(F)FF(c)FF(c ssssssi

dt

dYVYFXcXc ssi

wheres

s

FFX

ccY

F, c

F, ci

)cc(F)FF(ccFFc ssssss

n...,,i,xoi 21

)t(u)x...,x,x(f n 21


ssis F]X)cc(YFdt

dYV[

XF

)cc(Y

dt

dY

F

V

s

si

s

RXYdt

dY

Where

s

si

s F

)cc(R,

F

V

1

s

R

)s(X

)s(Y1

storder system

The Time Delay

The most commonly used model to describe the dynamics of chemical process is

First-Order Plus Model Delay Model. By proper choice τd , this model can be

represent the dynamics of many industrial processes.

Time delay or dead time between inputs and outputs are very common

industrial procsses, engineering systems, economical, and biological systems.

Transportation and measurement lags, analysis times, computation and

communication lags.

Any delay in measuring, in controller action, in actuator operation, in computer

computation, and the like, is called transportation delay or dead time, and it always

reduces the stability of a system and limits the achievable time of the system.

The Transportation Lag

The transportation lag is the delay between the time an input signal is applied to a

system and the time the system reacts to that input signal. Transportation lags are

common in industrial applications. They are often called “dead time”.


Dead-Time Approximations:-

qi(t) = Input to dead-time element.

qo(t) = Output from dead-time element.

The simplest dead-time approximation can be obtained qraphically or by physical

representation.

s

i

o

sio

dio

d

d

e)s(Q

)s(Q

e)s(Q)s(Q

)t(q)t(q

The accuracy of this approxiamtion depends on the dead time being sufficicently

small relative to the rate of the change of the slope of qi(t). If qi(t) were a ramp

(constant slope), the approximation would be perfect for any value of τd. When the

slope of qi(t) varies rapidly, only smal τd’s will give a good approximation.

If the variation in x(t) were some arbitrary function, as shown in figurebelow, the

response y(t) at the end of the pipe would be identical with x(t) but again delayed by t

Figure Response of transportation lag to various inputs.

DeadTime

qi(t) qo(t)


Example:Thermal system

If measured at T1 this can be modelled as:

11

s

K

)s(V

)s(T

Due to thedelay time the temperature T2represented by:

12

s

eK

)s(V

)s(T sd

Example:Mixing tank with time delay.

F, V : Constants

Time delay

sde)s(C

)s(C 1

2

u

L

uA

AL

q

AL

rateflowVolumetric

tubeofVolumed

11

s

R

)s(C

)s(C

i

s

i

des

R

)s(C

)s(C

1

2

u

Ld

F, ci

F, c1F, c2

Vc1

L

)s(C

)s(C

)s(C

)s(C

)s(C

)s(C

ii

1

1

22

Process Control /Lec. 8 49

Fourth Class

Second order system

A linear second order system under dynamic condition is given by the differential

equation:-

kXYdt

dY

dt

Yd

nn

212

2

2

n

1

XkYdt

dY

dt

Yd 2

2

22

Where:-

k : Steady state gain

Y : Response value

X : Input disturbing variable

ωn: Natural frequency of oscillation of the system.

000 )(Y)(Y

ψ : Damping factor (damping coefficient)

By taking laplace for the above second order differential equation

)s(kX)s(Y)s(sY)s(Ys 222

)s(kX)s(Y)ss( 1222

1222

ss

k

)s(X

)s(Y)s(G T.F. of second ordersystem

If x is sudden force, such as,step change, inputs Y will oscillate depending on the

value of damping coefficient ψ .

ψ 1 Response will oscillate (Over damped)

ψ=1 Response critical oscillation (critical damped)


Fourth Class

Response of second order system

1) Step response

s

A

ss

k

s

A

ss

k)s(Y

s

A)s(X

2

2

2

22 1212……(1)

The quadratic term in this equation may be factored into two linear termsthat containthe roots

2

12

2

44

2

4222

2

22

2

2

2

21

)(

s ,

12

=Two real roots

11 2

2

2

1 sands ……(2)

Eq. (1) can now be re-written as

)ss)(ss(s

kA)s(Y

21

2

ψ > 1 Overdamped Two distinct real roots

ψ = 1 Critically Damped Two equal real roots

0 < ψ < 1 Underdamped Two complex roots

1212 22210

22

ss

s

ss

A

ss

k)s(Y

kAss)ss(o 22

122 12

kAs 00

kAs 202 2201

211

20

2 0 kAs

]ss

s

s[kA)s(Y

12

2122

2


Fourth Class

]

)s(

s

s[kA)s(Y

2

22 1

21

1) For ψ under damped system

]

)()s()()s(

s

s[kA

]

)()s()()s(

s

s[kA

]

)()s(

s

s[kA]

)()s(

s

s[kA)s(Y

22

2

2

2

22

2

22

222

2

22

222

2

1

1

1

1

1

11

1

1

1

1

21

]tsinetcose[kA)t(Y t)/(t)/(

2

2

2 1

1

11

21w

]wtsinwt(cose[kA)t(Y t)/(21

1

2

2

2

22

1

1

11

)(qpr

21

2

11 1

1

1tantan

q

ptan

))]wtsin(r(e[kA)t(Y t)/( 1

]

)ss(

s

s[kA]

)ss(

s

s[kA)s(Y

2

2

22

22

2

2

2

2

2

2 12

21

12

21


Fourth Class

2) For ψ>1 Overdamped system

]wtsinhwt(coshe[kA)t(Y t)/(

11

2

where

12w

]

)()s()()s(

s

s[kA

]

)()s()()s(

s

s[kA

]

)s(

s

s[kA]

)s(

s

s[kA)s(Y

22

2

2

2

22

2

22

222

2

2

22

2

22

1

1

1

1

1

11

1

1

1

1

21

ψ

ψ


Fourth Class

Terms Used to Describe an Underdamped SystemSecond order system response for a step change

Figure (8-3) Terms used to describe an underdamped second-order response.

1. Overshoot (OS)

Overshoot is a measure of how much the response exceeds the ultimatevalue (new

steady-state value)following a step change and is expressed as the ratio

in the Fig.

(8-3).

OS%OS

)expOS

100

1 2

2. Decay ratio (DR)

The decay ratio is defined as the ratio of the sizes of successive peaksand is given by

in Fig. (8-3).where c is the height of thesecond peak

2

21

2)OS(expDR

3. Rise time(tr)

This is the time required for the response to first reach its ultimate valueand is labeledin Fig. (8-3).

w

1tan

t

21

r

1.05 b

0.95 b

tp


Fourth Class

4. Response time

This is the time required for the response to come within ±5percent of its ultimatevalue and remain there. The response time is indicated inFig. (8-3).

5. Period of oscillation (T)

The radian frequency (radians/time) is thecoefficient of t in the sine term; thus,

21

2

T

6. Natural period of oscillation

If the damping is eliminated (ψ=0), thesystem oscillates continuously without

attenuation in amplitude. Underthese “natural” or undamped condition, the radian

frequency is . This frequency is referred to as the natural frequency wn.

1nw

The corresponding natural cyclical frequency fnand period Tnare related bytheexpression:-

2

11

nn

Tf Thus, τ has the significance of the undamped period.

7- Time to First Peak(tp ) :Is the time required for the output to reach its first maximum value.

21

wt p

Figure(8-4) Characteristics of a step response of underdamped second-order system.

1

߰


Fourth Class

1-Over shoot

nwt

w

nt

max or min n=1, 2, 3 ……

If n=0,2,4, 6, ……………. min

If n=1, 3, 5, 7, ……………. max

1st max when n=1

ww

nt

)]w

w(sine[kA)t(y w

21

11

)]sin(e[kAymax

21

21

11

For Underdampded system

22 11 tan,sin,cos

)](e[kAymax21

21

1

11

2

]e[kAymax211

]kA

kA]e[kAOvershoot

211

)expOvershoot

21

B

Bmax

B

AOvershoot

)]wt((sinhe[kA)t(Yt

21

11


Fourth Class

where

21

2 11tan,w

Overshoot could be optained by getting 0dt

dy

))]wtsin(r(e[kA)t(Y t)/( 1

)](e)wt(sinw)wt(cose[dt

)t(dYtt

21

100

)]wt(sin)wt(cos[et

210

)]wt(sin)wt(cos

21

21

)wt(cos

)wt(sin

21)wttan(

21 1tan)wt(

2-Decay Ratio

A

CratioDecay

A

CratioDecay (The ratio of amount above the ultimate value of two sucessive

peaks).

w

nt

for n=3 then

wt

3

First peak at n=1 ]e[kAymax211

Second peak at n=3 ]e[kAymax21

3

1


Fourth Class

2

2

2

2

2

1

2

1

1

3

1

1

3

]1[

]1[

e

e

e

kAekA

kAekARatioDecay

21

2exp

RatioDecay

3. Rise time (tr)

It is the time required for the response to first tauch the ultimate line.

)]tw(sine[kA)t(yt

21

11

At tr y(t)=kA

)]wt(sine[kAkA r

t

21

11

)]wt(sin r 0

w

)(sinrt

01

w

tann

w

nrt

21 1

w

tan

rt

2

1 1

for n=1

4-Period of oscillation (T)

fTalsofw

frequencyRadianw

12

1 2

2

1 2f

21

2

T


Fourth Class

5. Natural period of oscillation (Tn).

The system free of any damping forψ=0

011 2

forwfrequencyofradianw, n

2

11nnnn ff2f2w

21

2

nn

w

2w

2nT for ψ=021

nf

f

6-Response time(ts)

The time required for the response to reach (±5%) of its ultimate value and remainthere.

7- Time to First Peak(tp )Is the time required for the output to reach its first maximum value.

w

nt

First peak is reached when n=1

21

ww

nt p

2- Impulse Response

If impulse δ(t) is applied to second order system then transfer response can be written as.

)s(Xss

k)s(Y

1222

AArea)s(X

A.ss

k)s(Y

1222

22

2

2

2

2

2

2

1212)()(ss

/kA

ss

/kA)s(Y


Fourth Class

2

22

2

22

22

2

112

)s(

/kA

)()(ss

/kA

i) ψ>1

22

2

2

22

22

2

2

2

22

2

1

1

1

11

)s(

kA

)s(

/kA

)s(

/kA)s(Y

wtsinhekA

)t(Yt

12

12w

ii)ψ


Fourth Class

Homework

1) For each of the second-order systems that follow, find ψ, τ, Tp, Tr ,OSand %OS.

Sol

040020

040

163

16

2

2

.s.s

.)s(Y)b

ss)s(Y)a

2) The transfer function for a thermometer in a CSTR reactor is given by.

Sol)s)(s()s(T

)s(T

actorRe

rthermomete

11013

16

Estimate the following.a) find ψ, τ b) The rise time.c) The peak time.d) The 2% settling time.e) The percentage overshoot.

3) A second-order system is described by the differential equation:

)t(u)t(Ydt

)t(dY

dt

)t(Yd25255

2

2

a) Write down the transfer function Y(s)/U(s) of the system, where U(s) and Y(s) arethe Laplace transforms of u(t) and y(t), respectively.b) Obtain the damping ratio ψ and the natural frequency Wn of the system.c) Calculate the rise time and percent overshoot of the system.d) Evaluate y(t) for a unit-step input u(t).


The Control System

The control system

A liquid stream at a temperatureTi,enters an insulated, well-stirred tank at a constant

flow rate w (mass/time). It is desired to maintain (or control) the temperature in the

tank at TRby meansof the controller. If the indicated (measured) tank temperature

Tmdiffers from the desiredtemperature TR,the controller senses the difference or

error,E = TR- Tm

Figure(9-1) Control system for a stirred-tankheater.

There are two types of the control system:-

1) Negative feedback control system

Negative feedback ensures that the difference between TRand Tm is used to

adjust the control element so that tendency is to reduce the error.

E=TR-Tm

2) Positve feedback control system

If the signal to the compartos were obtained by adding TRand Tm we would

have a positive feedback systems which is inherently unstable. To see that this

is true, again assume that be system is at steady state and that T=TR=Ti.

If Ti were to increase, T and Tm would increas which would cause the signal

from the compartor to increase, with the result that the heat to the system

would increse.

At s.s. T=TR=TinE=TR+Tm


Servo Problem versus Requlator Problem

Servo Problem

There is no change in load Ti, and that we are interested in changing the bath

temperature (change in the desired value (set point) with no disturbance load).

Requlating problem

The desired value TR is to remain fixed and the purpose of the control system is

to maintain the controlled variable TR in spite of change in load if there is a change in

the input variable (disturbance load).

Control system elements

Control system elements are:-

1) Process

2) Measuring element

3) Controller

4) Final Control Element

EP Q

Closed Loop Feedback control

Gm

GL(s)

T or YFinal controlelement

Process

Measuringdevice

Controller

Load

GPGVGC

TR or YR

Ti(s)

Comparator

Set point

Tmor Ym


Development of block Diagram

ProcessThe procedure for developing the transfer function remain the same.

An unsteady-state energy balance around the heating tank gives.

dt

dTVCpTTCpWqTTCpW ooi )()(

Where To is the reference temperature

At steady state, 0dtdT

0dt

dTVCp)TT(CpWq)TT(CpW ossois


dt

)TT(dVCpqq))TT()TT((CpW sssisi

Note that the refernece temperature To cancels in the subtraction. If we introducethedeviation variables.

s

s

isii

qqQTTT

TTT

Taking the laplace transform gives

CpWTVsCpsQsTsTCpW i )())()((

)()(

)( sTWCp

sQsTTs

W

Vi

The last expression can be written as

1

)()(

)1(

1)(

s

sT

WCp

sQ

ssT i

Where

iablevaredisturbanc)s(dor)s(Tiablevardmanipulate)s(mor)s(Q

iablevarcontrolled)s(Yor)s(TW

V

i

If there is a change in Q(t) only then 0)t(Ti and the transfer function relating

QtoTi is

WCpssQ

sT 1

)1(

1

)(

)(

dt

TdVCpQ)TT(CpW i


)s(T

)s(Y

If there is a change in )(sTi only thenQ(t)=0and the transfer function relating iTtoT

is

)s()s(T

)s(T

i 1

1

Block Diagram for process

)()()( sdGsmGsY dp

Measuring Element

The T.F. of the temperature-measuring element is a first order system

1)(

)(

s

k

sT

sT

m

mm)()( sTGsT mm

1

s

kG

m

mm

Where T and mT are deviation variables defined as

smmm

s

TTT

TTT

input

OutputgainstatesteadyKm

τm=time lag (time constant)=(1-9) sec

1s

KG

m

mm

Figure Block diagram of measuring element

+

+

1

1

s

1

1

s

WCp/

)s(T

)s(d

i

)s(Q

)s(m

i

Gd

Gp

)s(T )(sTm


Controller and final control elementThe relationship for proportional controller is

)()(

)(sG

sG

sPc

Where

)()( sEKsQ C

mR

s

TTE

PPP

G(s) for propertional controller Cc KsG )(

TTT mR at steady state

)(sE Kc )(sQ

GcTsp(or TR)+ E P(s)

-Tm

Controller


Controllers and Final Control Element

Final control Elements:

Control valve, Heater, Variec, Motor, pump, damper, louver, …. etc.

Control valveControl valve that can control the rate of flow of a fluid in proportion to the

amplitude of a pressure (electrical) signal from the controller. From experiments

conducted on pneumatic valves, the relationship between flow and valve-top pressure

for a linear valve can often be represented by a first-order transfer function:

Air supply Air supply

Control valve (Air to close) Control valve (Air to open)

Figure Pneumatic control valve (air-to-close).

0rateflow

rateflowmax

psig)153(

closeopenclose)(air to

P

rateflowmax

0rateflow

psig)153(

opencloseopen)(air to

P

Transfer Functionof Control Valve

1)(

)(

)(

)()(

s

K

sP

sQ

sp

smsGv

v

V

Kv=steady state gainS.S12

12

PP

QQ

Input

Output

(Good)sec01

lagTime

v

v

Where:


Kv: steady-state gain i.e., the constant of proportionality between steady-state flow

rate and valve-top pressure.

τv: time constant of the valve and is very small compared with the time

constants of other components of the control system.A typical pneumatic valve has a

time constant of the order of 1 sec. Many industrial processes behave as first-order

systems or as a series of first-order systems having time constants that may range

from a minute to an hour.So the lag of the valve is negligible and the T. F. of the

valve sometimes is approximated by:

VK)s(P

)s(Q

The time constant of lag valve depends on the size of valve, air supply characteristics,whether a valve positioner is used, etc.

Control Action

It is the manner, in which the automatic controller compares the actual value of the

process output with the actual desired value, determines the deviations and produce a

control signal which will reduce the deviation to zero or to small value.

Classification of industrial automatic controller:

They are classified according to their control action as:

1) On-off controller

2) Proportional controller (P)

3) Integral controller (I)

4) Proportional plus Integral controller (PI)

5) Proportional plus Derivative controller (PD)

6) Proportional plus Integral plus Derivative controller (PID)

The automatic controller may be classified according to the kind of power

employed in the operation, such as pneumatic controller, hydraulic controller or

electronic controller.

Self operated controller: In thiscontroller the measuring element (sensor) and theactuator in one unit.It is widely used for the water and gas pressure control.

Figure Closed loop block diagram of first order system

y(s)

T(s)

௩ܩ =ݒܭ

௩߬ݏ+ 1 ܩ =ܴ

+߬ݏ 1

ܩ =݉ܭ

߬ +ݏ 1

GC=Kc

ௗܩ =ܴ

+߬ݏ 1

ProcessFinal controlelement

Controller

TR

d=Ti(s)

Comparator

Set pointOr ysp

Tm(s) T(s)

E P m

Q


)1

1(s

KI

C

P(s)

Types of Feedback Controllers

1) Proportional controller (P):

For a controller with a proportional control action, the relationship between the

output of the controller, p(t), and the actuating error signal (input to controller) is

)()(

).()(

)(

)()(

)()(

)()(

sEKsP

FTsE

sPKG

ptEKtp

tEKptp

tEKtP

C

CC

sC

Cs

C

s.sC )E

p(K

Proportional Band (Band Width)

Is defined as the error (expressed as a percentage of the range of measured variable)

required to make the valve from fully close to fully open.

%100K

1P.B

C

On-Off Control

On-Off control is a special case of proportional control.

If the gain KCis made very high, the valve will move from one extreme position to the

other if the set point is slightly changed.So the valve is either fully open or fully

closed (The valve acts like a switch).

The P.B. of the on-off controller reaches a zero because the gain is very high0P.B

2) Propertional-Integral controller (PI):

This mode of control is described by the relationship

])(1

)([)(0t

ICs dttEtEKptp

KC : Steady state gain

I : Integral time constant

t

I

CCs dttE

KtEKtPptp

0

)()()())((

Taking L.T

)()1

1()(

)(sG

sK

sE

sPc

IC

Output signal

KCset point P(s)

Output signal


Prob(10-1):PI controller with step change in errors

A)s(E

s

A)

s

11(K)s(P

IC

tAK

AKtPI

CC

)(

Y=c+mX

3) Proportional-derivative control (PD):

sDC p])(

+E(t)[K=p(t) dt

tdE

dt

tdE )(K+E(t)KP(t)=)p-(p(t) DCCs

sK+K=E(s)

P(s)DCC

cDC G)s+(1K=E(s)

P(s)

KC: gainτD :Derivative time (rate time)

2s

A=E(s)(Ramp)tA=E(t)ErrorRampFor

DCC AKAtK=P(t)

4) Proportional-Integral-Derivative (PID) controller

sd

t

I

CC p

dt

tdEdttE

KtEKtp ]

)()()([)(

0

)1

1()(

)(s

sK

sE

sPD

IC

t E(t) P(t)

set point P(s)

P(s)P(t)P

E(t) p(t)KC.A

A KCA τi

0 ps0 0

t t

Response of a PI controller (lineaer)

KC(1+τDs)set point P(s)

)1

1( ss

K DI

C


Motivation for Addition of Integral and Derivative Control Modes

The value of the controlled variable is seen to rise at time zero owing to the

disturbance. With no control, this variable continues to rise to a new steady-state

value.

With control, after some time the control system begins to take action to try to

maintain the controlled variable close to the value that existed before the

disturbance occurred.

With proportional action only, the control system is able to arrest the rise of the

controlled variable and ultimately bring it to rest at a new steady-state value.

The difference between this new steady-state value and the original value (the

set point, in this case) is called offset.

The addition of integral action eliminates the offset; the controlled variable

ultimately returns to the original value. This advantage of integral action is

balanced by the disadvantage of a more oscillatory behavior.

The addition of derivative action to the PI action gives a definite improvement

in the response. The rise of the controlled variable is arrested more quickly,

and it is returned rapidly to the original value with little or no oscillation.

Figure: Response of a typical control system showing the effects

of various modes of control


Ex: A unit-step change in error is introduced into a PID controller. If Kc = 10, τI = 1,

and τD = 0.5, plot the response of the controller, m(t).

Solution:

The equation of PID controller is

)ss

11(K

)s(E

)s(PD

IC

m(t)

s

1)s(E slop = 10

)s5.0s

11(

s

10)s(P 5

5s

10

s

10)s(P

2 10

(t)5+10t+10=m(t) 0 t →

Ex: Consider the 1st order T. F. of the process with control valve

Valve processP(s) Y(s)

If we assume no interaction;

The T. F. from P(s) to Y(s) is

)1)(1()(

)(

ss

KK

sP

sY

v

pV

For a unit step input in P

Y(s) =)1)(1(

1

ss

KK

s v

pV

y(t) = KvKp

// 111 tvt eevv

v

τ>>τvthe T. F. is)1()(

)(

s

KK

sP

sY pv

If

For a unit step input in py(t) = KvKp (1 – e

-t/τ)

1s

KP1s

K

V

V


Ex: Consider the two tank reactor control system shown in figure below. The

volumetric flow rate ( F = 100 ft3/min) contains reactant A ( oc = 0.1 lbmol/ft3) which

decomposes with 1st order reaction A → B by rate (-rA = kc) where k is rate constant,

a function of temperature. k1 = 1/6 min-1&k2 = 2/3 min

-1 where T2> T1. The purpose

of the control system is to maintain c2 constant by adding pure A to tank 1 through

valve. The sample from tank 2 is withdrawn at a rate of 0.1 ft3/min. Another data are:

Mw of A = 100 lb/lbmol, ρA = 0.8 lbmol/ft3, m = 1 lbmol/min, V1 = V2 = 300 ft

3.

Control valve: flow of A through the valve varies 0 – 2 ft3/min as the valve top

pressure varies 3 – 15 psig.

Neglect τv.

Measuring element: the measuring element converts concentration of Ato pneumatic

signal. The output of the measuring element varies from 3-15 psig as the

concentration of Avaries from 0.01 to 0.05 lb mole/ft3. Neglect lag (τm = 0). The

controller is proportional type.

Figure : Control of a stirred-tank chemical reactor.

Solution:1. The process transfer function

Component M.B on A in tank 1:

1111 )( cVkc

mFmFc

dt

dcV

Ao

…. (1)

A

m


At s.s

0)( 111 dt

dcVcVkFmFc sssos …. (3)

By subtracting equation (3) from equation (2)

MFCC)VkF(dt

dCV o11

1

Where:

s

s111

osoo

mmM

ccC

ccC

)VkF(

MC

)VkF(

FC

dt

dC

)VkF(

V

1

o

1

11

1

)VkF(

1R

)VkF(

FR

)VkF(

V

12

11

11

MRCRCdt

dC2o11

11 …. (4)

By taking laplace transform for equation (4)

)s(MR)s(CR)s(C)1s( 2o111

)s(M)1s(

R)s(C

)1s(

R)s(C

1

2o

1

11

Transfer function of firsttank

iablevardmanipulate)s(M

edisturbanc)s(Co

1s

R

1

1

12

2

s

R

M(s)

Co(s)

C1(s)


150

1

)3006

1100(

1

)VkF(

1R

3

2

)3006

1100(

100

)VkF(

FR

2

)3006

1100(

300

)VkF(

V

12

1

1

1

1

Component A M.B on tank 2:

dt

dcVcVkFcFc 22221

022221 dt

dcVcVkFcFc sss

1222 )( FCCVkF

dt

dCV

12

22

2

C)VkF(

FC

dt

dC

)VkF(

V

12

22

2 C)VkF(

FC

dt

dC

1322 )()1( CRsCs

)()1(

)( 12

32 sC

s

RsC

Transfer function of second tank

)s(M

1s2

150/1)s(C

1s2

3/2

)1s(

R)s(C o

2

32

3

1

3

2300100

100

VkF

FR

23

1

3

2300100

300

)VkF(

V

2

2

)s(M

)1s2(

150/1

)1s(

3/1)s(C

1s2

3/2

)1s(

3/1)s(C o2

Ms=1 lbmol/min

0c)kVF(mFc s1sos at s.s

0c)50100(lbmol11.0100 s1

3s1 ft/molelb0733.0

150

11c


0cVkFcFc s22s2s1

0c300150

11100 s2

3s2 ft/mollb024.0

450

11

150300

10011c

2-The measuring element

For each element the transfer function (T.F) must be known and the s.s conditions

must be known too.

0)1(

mm

mm

s

KG

2

3

s.s

m

c

b

ft/mollb10001.005.0

420

input

outputK

024.0? 2 scbs

linear024.005.0

bs4100

01.0024.0

0bs

inbs 4.1026.01004

3- Controller proportion

E

pKG CC

Must be given or calculated

bs=1.4 ps=10.5 psig

4- Control valve0v

psi

cfm

6

1

315

02

essurePr

Flowk v

Since ms/ ρA= 1.25 cfm, the normal operating pressure on the valve is

)1s)(1s2(

9/2

)1s)(1s2(

450/1

M(s)

Co(s)

C2(s)


)315(2

25.13ps

psig5.10ps

psi

min/lbmol133.0

ft

lbmol8.0

psi

min/ft

6

1K

)s(P

)s(M3

3

AV

5-Transportation lagA portion of the liquid leaving tank 2 is continuously withdrawn through a sample

line,containing a concentration measuring element, at a rate of 0.1 cfm. The

measuring elementmust be remotely located from the process, because rigid ambient

conditions mustbe maintained for accurate concentration measurements. The sample

line has a lengthof 50 ft, and the cross-sectional area of the line is 0.001 ft2 .

The sample line can be represented by a transportation lag with parameter

min5.0min/ft1.0

ft001.0ft5

FlowrateVolumetric

Volume

u

1t

3

2

d

Block diagram

C2m(s)

C2sp(s) P(s)

)1s)(1s2(

9/2

)1s)(1s2(

450/1

+

+

se 5.0100

KC2

15-

+Km

Co(s)

C2(s)M(s)E(s)

controller Controlvalve


Ex: a pneumatic PI controller has an output pressure of 10 psi when the set point and

pen point are togather. The set point is suddenly displaced by 1.0 in (i.e a step change

in error is introduced) and the follwing data are obtained.

Time (s) 0- 0+ 20 60 80

Psi 10 8 7 5 3.5

Determine the actual gain (psi/inch displacement) and the integral time

E1.0 in

10 psi

cK

8psi

I

cK

For PI control

sI

cc pdtE

KKtp

)(

For E=1

sI

cc pt

KKtp

)(

From the above figure

40

2

2060

572

I

c

c

KK

sec40)2(2020 cI K

Ex: (A) a unit-step change in error is introduced into a pid controller, if Kc=10 , τI=1

and τD=0.5 plot the response of the controller P(t).

(B) if the error changed with a ratio of 0.5 in/min plot the response of p(t).

Solution:

(A) For a PID control

sDcI

cc p

dt

dEKdtE

KtKtp

)(

For a unit step change in error E(t)=1

At t=0 sc pKp )0(

t>0 sI

cc pt

KKtp

)(

tppP s 1010


I

(B) E=0.5 t 5.0dt

dEand dtdEdt 5.0

spdttttp 5.05.0105.0105.010)(

5.25.25)( 2 ttptp s2

s t5.2t55.2p)t(p)t(P t P(t)0 2.51 102 22.53 404 62.55 90

1E

0

0

10P

0

cK

0

E 0.5

0

0 t

P(t)

2.5

0 t


Dynamic Behaviour of Feedback Controlled Process

Overall transfer function of a cl

lect 1 introduction to process control€¦ · process control /lec. 1 3 fourth class heater...

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