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• SOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTED

PROBLEMS INPROBLEMS INPROBLEMS INPROBLEMS IN

PROCESS SYSTEMS ANALYSIS AND

CONTROL

DONALD R. COUGHANOWR

COMPILED BY

M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)

CATCH ME AT gopinathchemical@gmail.com

Disclaimer: This work is just a compilation from various sources believed to be reliable and I am not responsible for any errors.

• CONTENTS

PART 1: SOLUTIONS FOR SELECTED PROBLEMS

PART2: LIST OF USEFUL BOOKS

PART3: USEFUL WEBSITES

• PART 1

1.1 Draw a block diagram for the control system generated when a human

being steers an automobile.

1.2 From the given figure specify the devices

• Solution:

• Inversion by partial fractions:

3.1(a) 0)0()0(1 '2

2

===++ xxxdt

dx

dt

dx

)0()0()( '22

2

xsxsXsdt

dxL =

)0()( xsXsdt

dxL =

L(x) = X(s)

• L{1} = 1/s

+ )0()0()( '2 xsxsXss

sXxsXs1

)()0()( =+

ssXss

1)()1( 2 =++=

)1(

1)(

2 ++=

ssssX

Now, applying partial fractions splitting, we get

)1(

11)(

2 +++

=ss

s

ssX

2222

2

3

2

1

2

3

3

2

2

1

2

3

2

1

11)(

+

+

+

+

+=

ss

s

ssX

tetCosesXLtt

2

3sin

3

1

2

31))(( 2

1

2

1

1 =

+

=

tSintCosetX

t

2

3

3

1

2

31)( 2

1

b) 0)0()0(12 '2

2

===++ xxxdt

dx

dt

dx

when the initial conditions are zero, the transformed equation is

s

sXss1

)()1( 2 =++

• )1(

1)(

2 ++=

ssssX

12)1(

122 ++

++=

++ ssCBs

s

A

sss

CsBsssA ++++= 22 )12(1

)(20

)(0 2

sofeffecientscotheequatingbyCA

sofeffecientcotheequatingbyBA

+=

+=

2,1,1

2

1

0

)(1

===

=

=

=+

=

CBA

AC

B

BA

constofeffecientscotheequatingbyA

12

21)(

2 +++

=ss

s

ssX

( )( )

+

++=

2

11

1

111)}({

s

s

sLsXL

( )

++

+=

2

1

1

1

1

11)}({

ssLtX

)1(1)}({ tetX t +=

3.1 C 0)0()0(13 '2

2

===++ xxxdt

dx

dt

dx

by Applying laplace transforms, we get

ssXss

1)()13( 2 =++=

)13(

1)(

2 ++=

ssssX

• 13

)(2 ++

++=

ss

CBs

s

AsX

CsBsssA ++++= 22 )13(1

)(30

)(0 2

sofeffecientscotheequatingbyCA

sofeffecientcotheequatingbyBA

+=

+=

3,1,1

33

1

0

)(1

===

==

=

=+

=

CBA

AC

B

BA

constofeffecientscotheequatingbyA

+++

= 13

31)}({

2

11

ss

s

sLsXL

+

+=

22

11

2

5

2

3

31)}({

s

s

sLsXL

+

+

+=

2222

11

2

5

2

3

2

5

5

2.

2

3

2

5

2

3

2

3

1)}({

ss

s

sLsXL

tt

CosetX

t

2

5sinh

5

3

2

5(1)( 2

3

+=

3.2(a)

1)0(

0)0()0()0(;

11

''''

3

3

4

4

=

====+

x

xxxtCosdt

xd

dt

dx

• Applying Laplace transforms, we get

1)0()0()0()()0()0()0()0()(

2

'''23'''''1234

+=+s

sxsxxssXsxsxxsxssXs

1)1()()(

2

34

+=++s

sssssX

34

2)1()1

1()( ssss

ssX +

+++

+=

= )1)(1(

12

)1)(1(

123

23

23

23

+++++

=++++++

sss

sss

sss

ssss

11)1)(1(

1223223

23

++

++

+++=+++++

s

FEs

s

D

s

C

s

B

s

A

sss

sss

)1()()1()1)(1()1)(1()1)(1(12 323222223 +++++++++++++=+++ ssFEssDssscssBsssAssss

A+B+E=0 equating the co-efficient of s5.

A+B+E+F=0 equating the co-efficient of s4.

A+B+C+D+F=0 equating the co-efficient of s3.

A+B+C=0 equating the co-efficient of s2.

B+C=2 equating the co-efficient of s.

A+B+E=0 equating the co-efficient of s2.

C=1equating the co-efficient constant.

C=1

-B=-C+2=1

A=1-B-C=-1

D+F=0

E+F=0D+E=1

D-E=0

2D=1

A=-1; B=1; C=1

D=1/2; E=1/2; F =-1/2

• { }

+

++

+++

= 1

)1(2/1

1

2/1111)(

232

11

s

s

ssssLsL

{ }

+

++

+++

= 1

)1(2/1

1

2/1111)(

232

11

s

s

ssssLsXL

{ } int2

1

2

1

2

1

21)(

2

StCoset

ttX t ++++=

2)0(;4)0(2 122

2

==+=+ qqttdt

dq

dt

qd

applying laplace transforms,we get

23

'2 22)0()(()0()0()(ss

qssQqsqsQs +=+

+=++ 112

424))((2

2

ssssssQ

)(

)24()1(2

)(2

3

ss

ss

s

sQ+

+++

=

= )1(

24224

34

++++

ss

sss

)1(

3*2

)1(

2

1

14)(

4 ++

++

+

=sssss

sQ

31

3

1)1(24)())(( teetqsQL tt ++==

therefore tet

tq ++= 23

2)(3

• 3.3 a)

+

+=

++ 41

1

1

3

3

)4)(1(

32222 ss

s

ss

s

+

+=

2222 2

1

1

1

ss

tCosCostss

L 22

1

1

12222

1 =

+

+

b) [ ] 522)1(1

)52(

12222 +

++=

+=

+ ssCB

s

A

sssss

A+B=0

-2A+C=0

5A=1

A=1/5 ;B=-1/5;C=2/5

We get

+

+=

52

21

5

1)(

2 ss

s

ssX

Inverting,we get

=

+ tCosetSine tt 222

11

5

1

=

+ tCostSinet 222

11

5

1

c) 2222

22

)1(1)1(

233

+

++=

+

s

D

s

C

s

B

s

A

ss

sss

• 233)1()1()1( 232222 +=+++ sssDssCssBsAs

233)()12()2( 2322323 +=+++++ sssDsssCssBsssA

A+C=3

-2A+B-C+D=-1

A-2B=-3

B=2;

A=2(2)-3=1

C=3-1=2

D=2(1)-2+2-1=1

We get 22 )1(

1

1

221)(

+

+++=

sssssX

By inverse L.T

[ ] tt teettXL +++= 221)(1 [ ] )2(21)(1 tettXL t +++=

3.4 Expand the following function by partial fraction expansion. Do not evaluate

co-efficient or invert expressions

)3()1)(1(

2)(

22 +++=

ssssX

3)1(11)(

222 ++

++

+++

++

=s

F

s

EDs

s

CBs

s

AsX

22222 )1)(1()3)(1)(()1)(3)(1)(()3()1( ++++++++++++++= ssFssEDssssCBsssA

)14)(1()34)(()1)(34)(()3)(12( 222224 ++++++++++++++++= sssFssEDssssCBssssA

• 2333)4

34()346()342()43()( 2345

=++++++

+++++++++++++++++=

FEACAFE

BCAsCBCAsBCBAsFBCAsFBAs

A+B+F=0

-3A+C+4B+F=0

2A+B+4C+3B=0

6A+C+4B+3C=0

A+4C+3B+3D+4E+F=0

3A+3C+3E+F=2

by solving above 6 equations, we can get the values of A,B,C,D,E and

33 )3()1)(1(

1)(

+++=

sssssX .

3232 )3()3(321)(

++

++

++

++

++++=

s

H

s

G

s

F

s

E

s

D

s

C

s

B

s

AsX

by comparing powers of s we can evaluate A,B,C,D,E,F,G and H.

c))4()3)(2(

1)(

+++=

sssssX

4321)(

++

++

++

+=

s

D

s

C

s

B

s

AsX

by comparing powers of s we can evaluate A,B,C,D

3.5 a) )15.0)(1(

1)(

++=

ssssX

)15.0(1)15.0)(1(

1

++

++=

++ sC

s

B

s

A

sssLet

1)(2

12

3

2

222

=++

++

++= ssCs

sB

ssA

A=1

2

1

20

22=+==++ C

BC

BA

2

30

2

3=+==++ CBCB

A

• B/2=1/2 *-3/2=-1;

B=-2;

C= -3/2+2=1/2

+

++

=15.0

1

2

1

1

21)(

ssssX

tt eetxsXL 21 21)())(( +===

b) 0)0(;22 ==+ xxdt

dx

Applying laplace trafsorms

ssXxssX /2)(2)0()( =+

)2(

2))((1

+=

sssXL

+

= )2(

22))(( 11

ssLsXL

=

+

= 2

2/12/12))(( 11

ssLsXL

=te 21

3.6 a) 52

1)(

2 +++

=ss

ssY