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7.34: Untitled Page 179

  • Page ID
    18312
  • Chapter 7

    ( M

    )  ( M

    )

    CO2 5

    N2 5

    (31)

    CY   C 

    5

    ( M

    )

    ( 1

    )

    Y

     

    C H

    3

    Y

    5

     2  2 1 

    2

    4

    1  ( M

    ) CY  ( 1  C )

    Y

    Y

    C

     

    2H4 5

    Substitution of this result into Eq. 30 leads to an equation for the molar flow rate of ethylene ( C H ) in Stream 35. This result can be expressed 2

    4

    in the compact form

    ( M

    )

    C H

    5

      

    (32)

      

    2

    4

       0

    1

     ( M

    C CY

    ) 

    C

    2H4 5 

    where the two parameters are given by

    3  5 Y 1

    3  5 Y 2 1

    1

    Y

    2

    2

       

     

     

    ,

     

    (33)

    CY

    Y

    Y

    At this point we are ready to use a trial‐and‐error procedure to first solve for ( M

    ) and then solve for the parameter  .

    C2H4 5

    Picard’s method

    We begin by defining the dimensionless molar flow rate as

    ( M

    )

    C H

    2

    4 5

    x

    (34)

    so that the governing equation takes the form

    x

    H( x)    

    (35)

     C

     0

    1

    CY  x

    In order to use Picard’s method (see Appendix B4), we define a new function according to

    Definition:

    f ( x)  x H( x)

    (36)

    and for any specific value of the dependent variable, x , we can define a i

    new value, x

    , by

    i1

    Definition:

    x

    f x ,

    i

    , , ,... ,

    (37)

    1

    ( )

    1 2 3

    i

    i

    Material Balances for Complex Systems

    335

    To be explicit, we note that the new value of the dependent variable is given by

    x

    i

    x

    x

    ,

    i

    , , , ... ,

    i

      

    1

    1 2 3

    i

    (38)

    

    1C

    CY  ix 

    In terms of the parameters for this particular problem

      37 . 6190 ,

      15 . 1678 ,

    C  0 . 7 ,

    Y  0 . 5

    (39)

    we have the following iterative scheme

    15 . 1678 x

    x

    x

    37 6190

    i

    .

    ,

    i

    1 , 2 , 3 , ... ,

    i

    1

    i

    (40)

    0 8571

    .

    i

    x

    By inspection, one can see that x  0 . 8571 thus we choose our first guess to be x  0 . 62 and this leads to the results shown in Table 7.8a.

    o

    Table 7.8a. Iterative Values for Dimensionless Flow Rate

    (Picard’s Method)

    i

    xi

    xi+1

    0

    0.6200

    – 1.4237

    1

    – 1.4237

    45.6633

    2

    45.6633

    98.7402

    3

    98.7402

    151.6598

    4

    151.6598

    204.5328

    5

    204.5328

    257.3835

    6

    257.3835

    …..

    7

    …..

    …..

    Clearly Picard’s method does not converge for this case and we move on to Wegstein’s method (see Appendix B5).

    Wegstein’s method

    In this case we replace Eq. 37 with

    Definition:

    i

    x

    1

    1 qf( x )  qx ,

    i  1 , 2 , 3 , ... ,

    (41)

    i

    i

    336