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7.24: Untitled Page 169

  • Page ID
    18302
  • Chapter 7

    Control Volume I

    Since there are no chemical reactions in the mixer illustrated in Figure 7.6c, we can make use of Eqs. 29 along with Eqs. 13 and 15 to obtain the following mole balance equations:

    C H :

     ( M

    )  ( M

    )  ( M

    )

     0

    (30a)

    2

    4

    C2H4 1

    C2H4 5

    C2H4 2

    CO :

     ( M

    )  ( M

    )

     0

    (30b)

    2

    CO2 5

    CO2 2

    O :

     ( M )  ( M )

     0

    (30c)

    2

    O2 1

    O2 2

    N :

     ( M )  ( M )  ( M )

     0

    (30d)

    2

    N2 1

    N2 5

    N2 2

    The single constraint on these molar flow rates is given by Eq. 14 that we repeat here as

    ( M

    )

     ( M ) ,

     

    (31)

    O

    1

    N

    1

    21 79

    2

    2

    We cannot make any significant progress with Eqs. 30, thus we move on to the second control volume.

    Control Volume II

    In this case we make use of Eqs. 28 and 29 to obtain the six species mole balances. These six equations are simplified by Eq. 18, Eq. 19 and Eq. 25 along with the observation that there is no ethylene oxide (C H O) 2

    4

    or water (H O) in Stream #2. The six balance equations are given by 2

    C H :

     

    ( M

    ) 

    ( M

    )

     

    C ( M

    )

    (32a)

    2

    4

    C2H4 2

    C2H4 4

    C2H4 2

    C H O :

    ( M

    )

    Y C ( M

    )

    (32b)

    2

    4

    C2H4O 7

    C2H4 2

    H O :

     ( M

    )  ( M

    )

     2 ( 1  ) ( M

    C

    Y

    )

    (32c)

    2

    H2O 8

    H2O 7

    C2H4 2

    CO :

     ( M

    )  ( M

    )

     2 ( 1  ) ( M

    C

    Y

    )

    (32d)

    2

    CO2 2

    CO2 4

    C2H4 2

    O :

    5

     ( M )

    (

     3) ( M

    C

    Y

    )

    (32e)

    2

    O2 2

    C

    2

    2H

    2

    4

    N :

     ( M )  ( M )

     0

    (32f)

    2

    N2 2

    N2 4

    Equation 21 along with Eq. 22 and Eq. 23 provide the constraint

    Material Balances for Complex Systems

    315

    ( M

    )

      CY

    (33)

    C2H4 2

    indicating that the molar flow rate of ethylene entering the reactor is specified. In addition, we can use Eq. 32d along with Eq. 32f and Eq. 17 to determine ( M

    )  ( M

    ) . A summary of these results is given by

    CO2 4

    N

    4

    2

    ( M

    )

      CY

    (34a)

    C2H4 2

    ( M

    )

     (1  )

    C CY

    (34b)

    C2H4 4

    ( M

    )

     

    (34c)

    C2H4O 7

    5

    ( M

    )

     (3  Y) Y

    (34d)

    O2 2

    2

    ( M

    )

        2 ( 1  Y ) Y

    (34e)

    H

    2O 7

    ( M

    )

      

    (34f)

    H2O 8

     

    1

    ( M

    )

    ( M

    )

    1   ( 1  C  CY )

    (34g)

    CO2 4

    N2 4

    2

    CY

    At this point we need information about Stream #4 and this leads us to the splitter contained in Control Volume III.

    Control Volume III

    As indicated in Figure 7.6c, Stream #4 contains only ethylene ( C H ) , 2

    4

    carbon dioxide ( CO ) and nitrogen ( N ) , and the appropriate mole 2

    2

    balances are given by

    C H :

     ( M

    )  ( M

    )  ( M

    )

     0

    (35a)

    2

    4

    C2H4 4

    C2H4 5

    C2H4 6

    CO :

     ( M

    )  ( M

    )  ( M

    )

     0

    (35b)

    2

    CO2 4

    CO2 5

    CO2 6

    N :

     ( M )  ( M )  ( M )

     0

    (35c)

    2

    N2 4

    N2 5

    N2 6

    On the basis of Eq. 12, and the fact that the splitter streams contain only three species (see Figure 7.6c), we apply Eq. 7‐44 to obtain

    ( x

    )

     ( x

    )

    (36a)

    CO2 6

    CO2 4

    ( x

    )

     ( x )

    (36b)

    N2 6

    N2 4

    316