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CHEMICAL ENGG. INSTRUMENTATION & PROCESS CONTROL GATE-2019

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C O N T E N T1. INTRODUCTION OF LAPLACE TRANSFORM ……………… 03-11

2. TRANSFER FUNCTION ………………………………………….. 12-18

3. PHYSICAL EXAMPLE OF FIRST ORDER SYSTEM…............. 19-24

4. RESPONSE OF FIRST ORDER SYSTEM IN SERIES ………… 25-31

5. HIGHER ORDER SYSTEM ………………………………………. 32-38

6. THE CONTROL SYSTEM …………………………………………….. 39-41

7. CONTROLLERS ………………………………………………………. 42-48

8. CLOSED LOOP TRANSFER FUNCTION ………………………….. 49-53

9. ERROR ANALYSIS STEADY STATE ERROR ………..………….. 54-59

10. STABILITY ANALYSIS …………………………………………. 60-68

11. FREQUENCY RESPONSE …………………………………………… 69-79

12. STABILITY …………………………………………………………….. 80-84

13. INTRODUCTION TO FREQUENCY RESPONSE ………………… 85-91

14. CONTROL SYSTEM DESIGN BY FREQUENCY RESPONSE …... 92-93

15. ADVANCE CONTROL STRATEGIES ………………………………. 94-96

16. MEASUREMENT OF PROCESS VARIABLES ……………………. 97-99

17. PRACTICE SET-I with Solutions …………………………………….. 100-116

18. PRACTICE SET-II ……………………………………………………. 117-121

19. PRACTICE SET-17 (Subjective) ……………………………………... 122-126

20. INSTRUMENTATION & PROCESS CONTROL- Topic Wise TEST 127-156

1.1 Block diagram and signal flow graph

1.2 Time response analysis

1.3 Routh-hurwitz stability criterion

1.4 Bode plot









7. CONTROLLERS ………………………………………………………. 42-48





12. STABILITY …………………………………………………………….. 80-84






18. PRACTICE SET-II ……………………………………………………. 117-121






1.4 Bode plot









7. CONTROLLERS ………………………………………………………. 42-48





12. STABILITY …………………………………………………………….. 80-84






18. PRACTICE SET-II ……………………………………………………. 117-121






1.4 Bode plot



CHAPTER-1INTRODUCTION OF LAPLACE TRANSFORM

Need For Process ControlWhy process control subject in Chemical Engineering? When we run a kinetics experiment, how do we maintain the temperature and level at desired values? How do we manufacture products with consistently high quality when raw material properties

change? How much time do I have to respond to a dangerous situation?There is a need for continuous monitoring of the operation of a chemical plant and external control toguarantee the satisfaction of operational objectivities.A control system must: Suppress the influence of external disturbances Ensure the stability of a chemical process. Optimize the performance of a chemical process.Every engineer needs basic knowledge about control. With this insight, we will be able to design plantsthat can be controlled safely and produce high quality products

1A Linear differential equation of order n1

1 1( ) ( ) ....... ( )

n n

n nn n

d y d yL y A t A t y

dt dt

Linear differential equations arise from mathematical modelling of chemical processes.Use of Laplace transformLaplace transform offers a very simple method of solving linear differential equations.Using Laplace transform, a linear differential equation is reduced to an algebra problem. (Which issimpler than solving differential equation directly).

DEFINITION OF LAPLACE TRANSFORM:The Laplace transform of a function f(t) is defined as f(s) which can be find according to the equation

-st

0f(s)= f(t)e dt

Notation of Laplace transform of f(t) is {f(t)}=f(s)Example 1.1 Laplace transform of function ( ) 4F t

-st-st

00

- 4e 4f(s)= 4e dt= =

s s

4

{4}=s







1 1( ) ( ) ....... ( )

n n

n nn n


dt dt



-st

0f(s)= f(t)e dt


-st-st

00

- 4e 4f(s)= 4e dt= =

s s

4

{4}=s







1 1( ) ( ) ....... ( )

n n

n nn n


dt dt



-st

0f(s)= f(t)e dt


-st-st

00

- 4e 4f(s)= 4e dt= =

s s

4

{4}=s



FACTS ABOUT LAPLACE TRANSFORM1. The Laplace transform is not defined for the function f (t), when the value of ‘t’ is less than zero .

2. The Laplace transform is linear. Mathematically 1 2 1 2( ) ( ) ( ) ( )af t bf t a f t b f t Where a and b are constants.

3. Laplace transform of the function f(t) exists if the integral0

( ) stf t e dt

takes a finite value (i.e. remains

bounded)4. Laplace transform is a transformation of a function from time domain (where time is independentvariable) to s – domain (where s is independent variable)s is a variable defined in complex plane (i.e. s = a + jb)

LAPLACE TRANSFORMS OF SIMPLE FUNCTIONS1. The step function

00( )

0f t

A

t

t

tst

00

e [u(t)]= ( )e dt

s

s AA A

s

AA

s

[ Step function of size A] = A/s , when function is unit step.

i.e.0, 0

( )1, 0

tu t

t

1( )u t

s

2. The exponential function

f (t) = at

o t ot oe

-at -(s+a)t -(s+a)t

00

-1{e }= e dt= e

s+a

-at 1{e }=

s+a

Similarly,1ate

S a

3. The ramp function

f (t) = o t oat t o

-st

0

{at(t)}= ate dt

-st2 2

0

t 1 aat = a -e + =

s s s






( ) stf t e dt




00( )

0f t

A

t

t

tst

00

e [u(t)]= ( )e dt

s

s AA A

s

AA

s


i.e.0, 0

( )1, 0

tu t

t

1( )u t

s


f (t) = at

o t ot oe

-at -(s+a)t -(s+a)t

00

-1{e }= e dt= e

s+a

-at 1{e }=

s+a

Similarly,1ate

S a


f (t) = o t oat t o

-st

0

{at(t)}= ate dt

-st2 2

0

t 1 aat = a -e + =

s s s






( ) stf t e dt




00( )

0f t

A

t

t

tst

00

e [u(t)]= ( )e dt

s

s AA A

s

AA

s


i.e.0, 0

( )1, 0

tu t

t

1( )u t

s


f (t) = at

o t ot oe

-at -(s+a)t -(s+a)t

00

-1{e }= e dt= e

s+a

-at 1{e }=

s+a

Similarly,1ate

S a


f (t) = o t oat t o

-st

0

{at(t)}= ate dt

-st2 2

0

t 1 aat = a -e + =

s s s



4. The sine function

{ }o t osinkt t oF t

-st

0(sinkt)= sinkt e dt

-st

2 2

0

-e= (s.sin kt+kcos kt)

s +k

2 2

k{sin kt}=

s +k

Function Graph Transform

u(t)

1

s

t u(t)

2

1

s

tn u(t)

1

!n

n

s

e-at u(t) 1

s a

( )n att e u t

1

!

( )n

n

s a

sin k t u(t)

2 2

k

s k

cos ktu(t)

2 2

s

s k





-st


-st

2 2

0


s +k

2 2

k{sin kt}=

s +k


u(t)

1

s

t u(t)

2

1

s

tn u(t)

1

!n

n

s

e-at u(t) 1

s a

( )n att e u t

1

!

( )n

n

s a

sin k t u(t)

2 2

k

s k

cos ktu(t)

2 2

s

s k





-st


-st

2 2

0


s +k

2 2

k{sin kt}=

s +k


u(t)

1

s

t u(t)

2

1

s

tn u(t)

1

!n

n

s

e-at u(t) 1

s a

( )n att e u t

1

!

( )n

n

s a

sin k t u(t)

2 2

k

s k

cos ktu(t)

2 2

s

s k



sinh ktu(t)

2 2

k

s k

sin ( )ate ktu t

2 2( )

k

s a k

cos ( )ate k tu t

2 2( )

s a

s a k

(t), unit impulse

1

A (t), unit pulse

1 (1 )sAe

A s

LAPLACE TRANSFORMS OF DERIVATIVESa) First order derivative

df(t)=sf(s)-f(o)

dt

b) Second order derivative2

t=02

d f d df df df(t)= =s -

dt dt dt dt dt

2 (1)=s f(s) -sf(0) - f (0)

c) nth order derivativen

n n-1 n-2 (1) n-2 n-1n

d f=s f(s)-s f(0)-s f (0)............sf (0)-f (0)

dt

Example 1.2 Find the Laplace transform of the function given below2

12

d x dx+ +x=1, x(0)=x (0)=0

dt dt

Solution :2

2 12

d x= s x(s) -sx(0) - x (0)

dt

dx=sx(s) -x(0)

dt



sinh ktu(t)

2 2

k

s k

sin ( )ate ktu t

2 2( )

k

s a k

cos ( )ate k tu t

2 2( )

s a

s a k

(t), unit impulse

1

A (t), unit pulse

1 (1 )sAe

A s


df(t)=sf(s)-f(o)

dt


t=02


dt dt dt dt dt

2 (1)=s f(s) -sf(0) - f (0)


n n-1 n-2 (1) n-2 n-1n

d f=s f(s)-s f(0)-s f (0)............sf (0)-f (0)

dt


12

d x dx+ +x=1, x(0)=x (0)=0

dt dt

Solution :2

2 12

d x= s x(s) -sx(0) - x (0)

dt

dx=sx(s) -x(0)

dt



sinh ktu(t)

2 2

k

s k

sin ( )ate ktu t

2 2( )

k

s a k

cos ( )ate k tu t

2 2( )

s a

s a k

(t), unit impulse

1

A (t), unit pulse

1 (1 )sAe

A s


df(t)=sf(s)-f(o)

dt


t=02


dt dt dt dt dt

2 (1)=s f(s) -sf(0) - f (0)


n n-1 n-2 (1) n-2 n-1n

d f=s f(s)-s f(0)-s f (0)............sf (0)-f (0)

dt


12

d x dx+ +x=1, x(0)=x (0)=0

dt dt

Solution :2

2 12

d x= s x(s) -sx(0) - x (0)

dt

dx=sx(s) -x(0)

dt



x = x(s)

11

s

We get 2 1 1s x(s)-sx(0) -x (0) +sx(s) - x(0) + x(s)=

s

2 1(s +s+1)x(s)=

s

2

1x(s)=

s(s +s+1)Laplace transform of an integral

if t

o

f(s)f(t) =f(s) then f(t)dt =

s

Inversion by partial fractionIn the following example, the technique of partial fraction inversion for solution of differential equation isrepresented.Example 1.3 solve the following equation for x(t)

t

o

dxx(t)dt t

dt

x 0 3

Solution: Taking Laplace transform of above equation

t

0

dx= x(t)dt - t

dt

2

x(s) 1sx(s)-x(0)= -

s s

2

x(s) 1sx(s)-3= -

s s23s -1

x(s)=s(s+1)(s-1)

Expanding it by partial fraction method23s -1 A B C

= + +s(s+1)(s-1) s (s+1) (s-1)

2A(s -1)+B{s(s-1)}+C{s(s+1)}=

s(s+1)(s-1)2 23s -1 s (A+B+C)+s(C-B)+(-A)

=s(s+1)(s-1) s(s+1)(s-1)

Comparing the coefficients on both sideA+B+C = 3, C – B = 0, -A = -1We get A = 1, B =1, C= 1

1 1 1X(s)= + +

s s+1 s-1-t tX(t)=1+e +e



x = x(s)

11

s


s

2 1(s +s+1)x(s)=

s

2

1x(s)=


if t

o


s


t

o

dxx(t)dt t

dt

x 0 3


t

0

dx= x(t)dt - t

dt

2

x(s) 1sx(s)-x(0)= -

s s

2

x(s) 1sx(s)-3= -

s s23s -1

x(s)=s(s+1)(s-1)


= + +s(s+1)(s-1) s (s+1) (s-1)

2A(s -1)+B{s(s-1)}+C{s(s+1)}=

s(s+1)(s-1)2 23s -1 s (A+B+C)+s(C-B)+(-A)

=s(s+1)(s-1) s(s+1)(s-1)


1 1 1X(s)= + +

s s+1 s-1-t tX(t)=1+e +e



x = x(s)

11

s


s

2 1(s +s+1)x(s)=

s

2

1x(s)=


if t

o


s


t

o

dxx(t)dt t

dt

x 0 3


t

0

dx= x(t)dt - t

dt

2

x(s) 1sx(s)-x(0)= -

s s

2

x(s) 1sx(s)-3= -

s s23s -1

x(s)=s(s+1)(s-1)


= + +s(s+1)(s-1) s (s+1) (s-1)

2A(s -1)+B{s(s-1)}+C{s(s+1)}=

s(s+1)(s-1)2 23s -1 s (A+B+C)+s(C-B)+(-A)

=s(s+1)(s-1) s(s+1)(s-1)


1 1 1X(s)= + +

s s+1 s-1-t tX(t)=1+e +e



PROPERTIES OF TRANSFORMSFinal value theoremIf f(s) is the Laplace transform of f(t), then

t s 0lim f t =lim sf s

Provided that sf(s) does not become infinity for any value of s satisfying Re (s) 0. the limit of f(t) isfound to be correct only if f(t) is bounded as t approaches infinity

The final value theorem allows us to compute the value that a function approaches as t when itslaplace transform is known.Example 1.4 Find the final value of the function x(t) for which the Laplace transform is

3 2

1x(s)=

s(s +3s +6s+8)Solution : Applying final value theorem

t s 0

1Lim lim sf s[x(t =)]

8

t

1[x(t)Lim ]=

8

The conditions of the theorem satisfied unless s=–2 or (s+2)≠ 0Initial value theorem

t 0 s lim f t = lim sf s

Translation of transform

If f t = f s then

( )

0

{e f(t)}=f(s+a)= ( )at s a tf t e dt

Translation of function

if f t = f s then

00f t-t =e f sst for t > 0

Unit Pulse FunctionUnit pulse function of duration A is defined by.

0, 0

1( ) , 0

0,

A

t

t t AA

t A

Unit pulse function is the difference of two step functions of equal size 1/A.First step function occurs at t = 0

1

0, 0( ) 1

, 0

tf t

tA







3 2

1x(s)=


t s 0


8

t

1[x(t)Lim ]=

8




If f t = f s then

( )

0



if f t = f s then



0, 0

1( ) , 0

0,

A

t

t t AA

t A


1

0, 0( ) 1

, 0

tf t

tA







3 2

1x(s)=


t s 0


8

t

1[x(t)Lim ]=

8




If f t = f s then

( )

0



if f t = f s then



0, 0

1( ) , 0

0,

A

t

t t AA

t A


1

0, 0( ) 1

, 0

tf t

tA



Second step function occurs at t = A

2

0,( ) 1

,

t Af t

t AA

1 2

1 1

1 1

( ) ( ) ( )

( ) ( ) ( )

1 1( ) ( ) ( )

1 (1 )( )

A

A

sA sAA

sA

A

t f t f t

t f t f t A

L t L f t e L f t eAs As

eL t

A s

Unit Impulse FunctionAs A 0, unit pulse function becomes unit impulse or Dirac function. Represented by (t)

0

( ) 1

( ) lim ( )AA

t dt

t t

0 00

1 1lim ( ) lim

sAst

AA A

et e dt

A s

Using L’ Hospital rule

Example 1.5 Solve the following equation for y (t)

t

o

d y ty t dt= ,y o =1

dtSolution taking Laplace transforms

t

o

d(y(t))y(t)dt =

dt

1 y s = sy s – y 0s

2( )

1

sy s

s

1 12

sy(t)= [y(s)]= =cosht

s -1

0 0

1 (1 )lim lim 1

( ) 1

sA sA

A A

e se

A s st




2

0,( ) 1

,

t Af t

t AA

1 2

1 1

1 1

( ) ( ) ( )

( ) ( ) ( )

1 1( ) ( ) ( )

1 (1 )( )

A

A

sA sAA

sA

A

t f t f t

t f t f t A


eL t

A s


0

( ) 1

( ) lim ( )AA

t dt

t t

0 00

1 1lim ( ) lim

sAst

AA A

et e dt

A s



t

o



t

o

d(y(t))y(t)dt =

dt

1 y s = sy s – y 0s

2( )

1

sy s

s

1 12


s -1

0 0

1 (1 )lim lim 1

( ) 1

sA sA

A A

e se

A s st




2

0,( ) 1

,

t Af t

t AA

1 2

1 1

1 1

( ) ( ) ( )

( ) ( ) ( )

1 1( ) ( ) ( )

1 (1 )( )

A

A

sA sAA

sA

A

t f t f t

t f t f t A


eL t

A s


0

( ) 1

( ) lim ( )AA

t dt

t t

0 00

1 1lim ( ) lim

sAst

AA A

et e dt

A s



t

o



t

o

d(y(t))y(t)dt =

dt

1 y s = sy s – y 0s

2( )

1

sy s

s

1 12


s -1



KEY POINTS

1. Laplace transform of a function f (t)

-st

of(s)= f(t)e td

2. t < 0 ; Laplace transform is not defined.

3. 1 2 1 2{af (t)=bf (t)}=a {f (t)}+b {f (t)}

4.df(t)

=sf(s) -f(0)dt

5.2

22

d f(t)=s f(s) -sf(0) - f (0)

dt

6. For nth order

nn (n-1) (n-2) 1

n

d f(t)= s f(s) -s f(0) -s f (0)................

dt

(n-2) (n-1)sf (0)-f (0)

7. t

0

f sf t dt =

s

8. Final value theorem t s 0lim f t = lim s f s

9. Initial value theorem 0

lim ( ) lim ( )t S

f t sf s

10. If f t = f s

0

(-at) (

0

)

-sto

Then, {e f(t)}= f(s+a) = f(t)e dt

[f(t-t )]= e f(s)

s a t



KEY POINTS


-st

of(s)= f(t)e td


3. 1 2 1 2{af (t)=bf (t)}=a {f (t)}+b {f (t)}

4.df(t)

=sf(s) -f(0)dt

5.2

22

d f(t)=s f(s) -sf(0) - f (0)

dt

6. For nth order

nn (n-1) (n-2) 1

n

d f(t)= s f(s) -s f(0) -s f (0)................

dt

(n-2) (n-1)sf (0)-f (0)

7. t

0

f sf t dt =

s



lim ( ) lim ( )t S

f t sf s

10. If f t = f s

0

(-at) (

0

)

-sto


[f(t-t )]= e f(s)

s a t



KEY POINTS


-st

of(s)= f(t)e td


3. 1 2 1 2{af (t)=bf (t)}=a {f (t)}+b {f (t)}

4.df(t)

=sf(s) -f(0)dt

5.2

22

d f(t)=s f(s) -sf(0) - f (0)

dt

6. For nth order

nn (n-1) (n-2) 1

n

d f(t)= s f(s) -s f(0) -s f (0)................

dt

(n-2) (n-1)sf (0)-f (0)

7. t

0

f sf t dt =

s



lim ( ) lim ( )t S

f t sf s

10. If f t = f s

0

(-at) (

0

)

-sto


[f(t-t )]= e f(s)

s a t



Table of Laplace Transforms

1) (1) = 1/s

2)1

( )ates a

3)1

!( ) when t > 0 & n Nn

n

nt

s

4)1

1( ) where n N n is functionn

n

nt

s




1) (1) = 1/s

2)1

( )ates a

3)1

!( ) when t > 0 & n Nn

n

nt

s

4)1


n

nt

s




1) (1) = 1/s

2)1

( )ates a

3)1

!( ) when t > 0 & n Nn

n

nt

s

4)1


n

nt

s

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