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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