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7.31: Untitled Page 176

  • Page ID
    18309
  • Chapter 7

    R

    O

    Y  Yield of C H O /C H

     0 . 50

    (2)

    2

    4

    2

    4 

    C H

    2 4

     RC H

    2 4

    We are given that 100 mol/h of ethylene oxide are produced by the reactor and we express this condition as

    Reactor:

    R

      ,

      100 mol/h

    (3)

    C2H4O

    The results expressed by Eqs. 1, 2 and 3 can be used to immediately deduce that

    R

       Y

    (4a)

    C2H4

    ( M

    )

     CY

    (4b)

    C2H4 2

    In addition, we are given the following relations:

    Streams #7 and #8:

    ( M

    )

      ( M

    ) ,

      100

    (5a)

    H2O 8

    C2H4O 7

    Steam #1:

    ( M

    )

     ( M ) ,

     

    (5b)

    O

    1

    N

    1

    21 79

    2

    2

    Stream #2:

    ( x

    )

      ,

      0 . 05

    (5c)

    C2H4 2

    Stream #3:

    ( M

    )

     0

    (5d)

    O2 3

    We can use Eq. 4b and Eq. 5c to provide

    ( M

    )

    C H

    2

     CY

    2

    4

    ( x

    )

     

    (6a)

    C2H4 2

    M

    M

    2

    2

    which indicates that the total molar flow rate entering the reactor is given by

    M

    (6b)

    2

     C Y

    In Example 7.6 we carefully constructed control volumes that would minimize redundant cuts, while in Figure 7.8 we have simply enclosed each unit in a control volume and this creates four redundant cuts. As in Example 7.7, we will solve the material balances for each control volume in a sequential manner, and we will assume a value for any variable that is unknown.

    Material Balances for Complex Systems

    329

    Stoichiometry

    The details associated with Axiom II (see Eqs. 7 through 12 of Example 7.6) can be used along with the definitions of the conversion and yield to express the global rates of production in terms of the conversion and the yield. These results are given by

    R

       Y

    (7a)

    C2H4

    R

     2 ( 1  Y )  Y

    (7b)

    CO2

    R

     0

    (7c)

    N2

    5

    R

      (3  Y)  Y

    (7d)

    O2

    2

    R

     

    (7e)

    C2H4O

    R

     2(1  Y) Y

    (7f)

    H O

    2

    Mole Balances

    We begin our analysis of this process with Axiom I for steady processes and fixed control volumes

    Axiom I:

    c

    dA  R ,

    A  1 , 2 , 3 , ..., 6

    (8)

    v n

    A A

    A

    A

    and we note that each mole balance can be expressed as

    ( M

     )

     ( M )  R ,

    A  1 , 2 , 3 , ..., 6

    (9)

    A out

    A in

    A

    Application of Axiom I to the first control volume leads to

    Control Volume I (Mixer)

    ** C H :

    ( M

    )

     ( M

    )  ( M

    )

    (10a)

    2

    4

    C2H4 2

    C2H4 1

    C2H4 5

    ** CO :

    ( M

    )

    0

     ( M

    )

    (10b)

    2

    CO2 2

    CO2 5

    ** N :

    ( M

    )

    ( M

    )

     ( M )

    (10c)

    2

    N

    2

    N

    1

    N

    2

    2

    2 5

    O :

    ( M

    )

     ( M )

    (10d)

    2

    O2 2

    O2 1

    C H O :

    ( M

    )

     ( M

    )

     0

    (10e)

    2

    4

    C2H4O 2

    C2H4O 1

    H O :

    ( M

    )

     ( M

    )

     0

    (10f)

    2

    H2O 2

    H2O 1

    330