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7.32: Untitled Page 177

  • Page ID
    18310
  • Chapter 7

    Here we have marked with a double asterisk (**) the three molecular species that are contained in the recycle stream. In order to determine the portion of Stream #4 that is recycled in Stream #5, we would normally require macroscopic balances for only those components that appear in the recycle stream. However, in this particular problem, the molar flow rates of oxygen and nitrogen entering the mixer are connected by Eq. 5b, thus we require the macroscopic balances for all four species that enter the mixer. This means that only Eqs. 10a, 10b, 10c and 10d are required and we can move on to the remaining control volumes making use of only the mole balances associated with these four species.

    Control Volume II (Reactor)

    In this case the four mole balances include a term representing the net global rate of production owning to chemical reaction.

    ** C H :

    ( M

    )

     ( M

    )  R

    (11a)

    2

    4

    C2H4 3

    C2H4 2

    C2H4

    ** CO :

    ( M

    )

    ( M

    )  R

    (11b)

    2

    CO2 3

    CO2 2

    CO2

    ** N :

    ( M

    )

    ( M

    )

    (11c)

    2

    N2 3

    N2 2

    O :

    ( M

    )

    ( M

    )

     R

    (11d)

    2

    O2 3

    O2 2

    O2

    Control Volume III (Absorber)

    No chemical reaction occurs in this unit, thus the four balance equations are given by

    ** C H :

    ( M

    )

     ( M

    )

    (12a)

    2

    4

    C2H4 4

    C2H4 3

    ** CO :

    ( M

    )

    ( M

    )

    (12b)

    2

    CO2 4

    CO2 3

    ** N :

    ( M

    )

    ( M

    )

    (12c)

    2

    N2 4

    N2 3

    O :

    ( M

    )

    ( M

    )

     0

    (12d)

    2

    O2 4

    O2 3

    Here we have made use of the information that all of the oxygen in the feed reacts, and we have used the information that ethylene ( C H ) , 2

    4

    carbon dioxide ( CO ) , nitrogen ( N ) or oxygen, ( O ) do not appear in 2

    2

    2

    Stream #7 and Stream #8.

    Material Balances for Complex Systems

    331

    Control Volume IV (Splitter)

    This passive unit involves only three molecular species and the appropriate mole balances are given by

    ** C H :

    ( M

    )

     ( M

    )

     ( M

    )

    (13a)

    2

    4

    C2H4 5

    C2H4 6

    C2H4 4

    ** CO :

    ( M

    )

     ( M

    )

    ( M

    )

    (13b)

    2

    CO2 5

    CO2 6

    CO2 4

    ** N :

    ( M

    )

    ( M

    )

    ( M

    )

    (13c)

    2

    N

    5

    N

    6

    N

    2

    2

    2 4

    Sequential Analysis for O2

    At this point it is convenient to direct our attention to the molar balances for oxygen and carry out a sequential analysis to obtain C.V. 1:

    ( M

    )

     ( M )

    (14a)

    O2 2

    O2 1

    C.V. 2:

    5

    ( M

    )

    ( M

    )

     (3  Y)  Y

    (14b)

    O2 3

    O2 2

    2

    C.V. 3:

    ( M

    )

    0

    (14c)

    O2 3

    These results allow us to determine the oxygen flow rate entering the mixer as

    5

    ( M

    )

     (3  Y)  Y

    (15)

    O2 1

    2

    Here it is important to recall Eq. 5b and use that result to specify the molar flow rate of nitrogen ( N ) in Stream #1 as

    2

    5

    ( M

    )

     (3  Y)  Y

    (16)

    N2 1

    2

    We are now ready to direct our attention to the molar flow rates of ethylene ( C H ) ,carbon dioxide ( CO ) and nitrogen ( N ) in the recycle 2

    4

    2

    2

    stream and use these results to construct an iterative procedure that will allow us to calculate the fraction  .

    Algebra

    We begin by noting that the splitter conditions can be expressed as C H :

    ( M

    )

     (1  ) ( M

    )

    (17a)

    2

    4

    C2H4 5

    C2H4 4

    CO :

    ( M

    )

     (1  ) ( M

    )

    (17b)

    2

    CO2 5

    CO2 4

    332