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7.23: Untitled Page 168

  • Page ID
    18301
  • Chapter 7

    Stream #7:

    ( M

    )  ( M

    )  ( M

    )  ( M

    )

     0

    (18)

    C2H4 7

    O2 7

    N2 7

    CO2 7

    ( M

    )

     ( M

    )

     ( M )

    C H O 8

    C H

    8

    O

    8

    Stream #8:

    2

    4

    2

    4

    2

    (19)

     ( M )

     ( M

    )

     0

    N2 8

    CO2 8

    In addition, we have a condition involving both Stream #7 and Stream #8

    that was given earlier by Eq. 6 and repeated here as

    Streams #7 and #8:

    ( M

    )

      ( M

    ) ,

      100

    (20)

    H2O 8

    C2H4O 7

    The requirement that 100 mol/h of ethylene oxide are produced by the reactor can be stated as

    Reactor:

    R

      ,

      100 mol/h

    (21)

    C2H4O

    while the information concerning the conversion and yield are repeated here as

     RC H

    Reactor:

    2

    4

    C  Conversion of ethylene 

     0 7

    . 0

    (22)

    ( M

    )

    C2H4 2

    R

    O

    Reactor:

    Y  Yield of C H O /C H

     0 . 50

    (23)

    2

    4

    2

    4 

    C H

    2 4

     RC H

    2 4

    Here we conclude that there are 14 particular constraints associated with Control Volume II and we list this result as

    Control Volume II:

    particular constraints  14

    (24)

    Our third control volume encloses the splitter as indicated in Figure 7.6c. The problem statement indicates that “All of the oxygen in the feed reacts, ….all the ethylene oxide is absorbed,…..all the water leaves the absorber” thus it is clear Stream #4 is constrained by Stream #4:

    ( M

    )

     ( M

    )

     ( M )

     0

    (25)

    C2H4O 4

    H2O 4

    O2 4

    However, the splitter condition given by Eq. 12 along with the particular constraint given by Eq. 15 makes these three conditions redundant. Thus there are no new constraints associated with Control Volume III.

    Control Volume III:

    particular constraints  0

    (26)

    Material Balances for Complex Systems

    313

    The total number of particular constraints is 7  14  21 , thus the total number of constraints is given by

    Total constraints: 18  27  48

    (27)

    This means that we have zero degrees of freedom and the problem has a solution.

    Solution Procedure

    We can use Eqs. 3, 4 and 11 to express all the non‐zero global rates of production in terms of the conversion ( C ), the yield ( Y ), and the molar flow rate of ethylene into the reactor, 

    ( M

    ) . These global net rates of

    C H

    2

    4 2

    production are given by

    R

     

    ( M

    C

    )

    (28a)

    C2H4

    C2H4 2

    R

    ( M

    Y C

    )

    (28b)

    C2H4O

    C2H4 2

    R

     2 ( 1  ) ( M

    C

    Y

    )

    (28c)

    H2O

    C2H4 2

    R

     2 ( 1  ) ( M

    C

    Y

    )

    (28d)

    CO2

    C2H4 2

    5

    R

    (

     3) ( M

    C

    Y

    )

    (28e)

    O2

    2

    C2H4 2

    R

     0

    (28f)

    N2

    and this completes our application of Axiom II. Directing our attention to Axiom I, we begin our analysis of the control volumes illustrated in Figure 7.6c with the steady form of Eq. 7‐1 given by

    Axiom I:

    c

    dA  R ,

    A  1 , 2 ,..., 6

    (29)

    v n

    A

    A

    A

    A

    We will apply this result to Control Volumes I, II and III, and then begin the algebraic process of determining the fraction of M

     that needs to be

    4

    purged in order that 100 mol/h of ethylene oxide are produced by the reactor.

    314