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7.33: Untitled Page 178

  • Page ID
    18311
  • Chapter 7

    N :

    ( M

    )

     (1  )

     ( M )

    (17c)

    2

    N2 5

    N2 4

    and we will use these conditions in our analysis of ethylene (C H ) , 2

    4

    carbon dioxide ( CO ) , nitrogen ( N ) . Directing our attention to the 2

    2

    ethylene flow rate leaving the reactor, we note that the use of Eq. 4 in Eq. 11a leads to

    ( M

    )

      1  C CY

    (18)

    C H

    3

    2

    4

    Moving from the reactor to the absorber, we use Eq. 12a with Eq. 18 to obtain

    ( M

    )

      1  C CY

    (19)

    C H

    4

    2

    4

    and when this result is used in the first splitter condition given by Eq. 17a we obtain

    ( M

    )

     (1  )

      1  C CY

    (20)

    C H

    5

    2

    4

    Here we must remember that  is the parameter that we want to determine by means of an iterative process. At this point we move on to the mole balances for nitrogen ( N ) and make use of Eqs. 10c, 11c, 12c, 16

    2

    and 17c to obtain

    1   

    5

    3 

    Y

    2

    ( M

    )

    (21)

    N2 5

     Y

    Finally we consider the mole balances for carbon dioxide ( CO ) and 2

    make use of Eqs. 7b, 10b, 11b, and 17b to obtain the following result for the molar flow rate of carbon dioxide in the recycle stream.

    1   2 1 

    ( M

    Y

    )

    (22)

    CO2 5

    Y

    At this point we consider a total molar balance for the mixer M

    M

    M

    (23)

    1

    5

    2

    and note that the total molar flow rate of Stream #2 was given earlier by Eq. 6b. This leads to the mole balance around the mixer given by Mixer:

    M

      

    M

    C Y

    (24)

    1

    5

    Material Balances for Complex Systems

    333

    and we are ready to develop an iterative solution for the recycle flow and the parameter  .

    At this point we begin our analysis of the tear stream, i.e., Stream #5

    that can be expressed as

    M

     ( M

    )  ( M

    )  ( M

    )

    (25)

    5

    C2H4 5

    CO2 5

    N2 5

    Given these values, we make use of Eq. 24 to express the molar flow rate entering the mixer in the form

    M

      CY   

    ( M

    ) 

    ( M

    ) 

    ( M

    )

    (26)

    1

    C2H4 5

    CO2 5

    N2 5 

    Representing the total molar flow rate of Stream #1 in terms of the three species molar flow rates leads to

    M

    ( M

    ) 

    ( M

    ) 

    ( M

    ) )

    1

    C2H4 1

    O2 1

    N2 1

    in which the molar flow rates of oxygen ( O ) and nitrogen ( N ) are 2

    2

    specified by Eqs. 15 and 16. Use of those results provides

    5

    M

     ( M

    )  (3  Y)  1  

    Y

    (27)

    1

    C H

    1

    2

    4

    2

    and substitution of this result into Eq. 26 yields

    5

    ( M

    )

      CY  (3  Y)  1   Y

    C H

    1

    2

    2

    4

    (28)

     ( M

    )  ( M

    )  ( M

    ) 

    C

    2H4 5

    CO2 5

    N2 5 

    Here we can use Eqs. 4b and 10a to obtain

     ( M

    )  ( M

    CY

    )

    (29)

    C2H4 1

    C2H4 5

    which allows us to express Eq. 28 as

    5

      

     (3 

    ) 1     ( M

    )  ( M

    CY

    CY

    Y

    Y

    )  (30)

    CO

    2

    2 5

    N2 5 

    At this point we return to Eqs. 20, 21 and 22 in order to express the molar flow rates of carbon dioxide ( CO ) and nitrogen ( N ) in the tear stream 2

    2

    as

    334