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7.18: Untitled Page 163

  • Page ID
    18296
  • Chapter 7

    We can utilize Eqs. 2 of Example 7.4 to express the molar flow rates according to

    At Entrances & Exits:

    M

    x M ,

    A  1 , 2 , 3 , 4

    A

    A

    (9a)

    At Entrances & Exits:

    M

    M  M  M  M

    (9b)

    A

    B

    C

    D

    and in many cases it will be convenient to use Eq. 9a in the form

    M

    At Entrances & Exits:

    x

    A ,

    A

    A

    1 , 2 , 3 , 4

    (9c)

    M

    Use of Eq. 8 allows us to express the species mole balances for Control Volume I in the form

    Control Volume I

    C H :

     ( M

    )  ( M

    )

     R

    (10a)

    2

    6

    C2H6 1

    C2H6 4

    C2H6

    C H Cl :

     ( M

    )  ( M

    )

     R

    (10b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2 4

    C2H4Cl2

    :

     ( M

    )  ( M

    HCl

    )

     R

    (10c)

    HCl 1

    HCl 4

    HCl

    C H Cl :

     ( M

    )  ( M

    )

     R

    (10d)

    2

    3

    C2H3Cl 1

    C2H3Cl 4

    C2H3Cl

    Since there are no chemical reactions taking place in the mixer illustrated in Figure 7.5c, the mole balances for Control Volume II take the simple forms given by

    Control Volume II

    C H :

     ( M

    )  ( M

    )  ( M

    )

     0

    (11a)

    2

    6

    C2H6 1

    C2H6 5

    C2H6 2

    C H Cl :  ( M

    )  ( M

    )  ( M

    )

     0

    (11b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2 5

    C2H4Cl2 2

    HCl :

     ( M

    )  ( M

    )  ( M

    )

     0

    (11c)

    HCl 1

    HCl 5

    HCl 2

    C H Cl :

     ( M

    )  ( M

    )  ( M

    )

     0

    (11d)

    2

    3

    C2H3Cl 1

    C2H3Cl 5

    C2H3Cl 2

    We begin our analysis with Control Volume I and simplify the mole balances in terms of conditions on the molar flow rates. Some of those molar flow rates are zero and when we make use of this information the mole balances for Control Volume I take the form

    Material Balances for Complex Systems

    303

    Control Volume I ( constraints on molar flow rates)

    C H :

     ( M

    )  ( M

    )

     R

    (12a)

    2

    6

    C2H6 1

    C2H6 4

    C2H6

    C H Cl :

     ( M

    )

     R

    (12b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2

    :

    ( M

    HCl

    )

     R

    (12c)

    HCl 4

    HCl

    C H Cl :

    ( M

    )

     R

    (12d)

    2

    3

    C2H3Cl 4

    C2H3Cl

    We now make use of the results from Axiom II given by Eqs. 7 and impose those constraints on global rates of production to obtain Control Volume I ( constraints on global net rates of production) C H :

     ( M

    )  ( M

    )

     0

    (13a)

    2

    6

    C2H6 1

    C2H6 4

    C H Cl :

     ( M

    )

     R

    (13b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2

    HCl :

    ( M

    )

      R

    (13c)

    HCl 4

    C2H4Cl2

    C H Cl :

    ( M

    )

      R

    (13d)

    2

    3

    C2H3Cl 4

    C2H4Cl2

    At this point we can use Eq. 1 to express the global net rate of production in terms of the conversion leading to

    Control Volume I ( global net rates of production in terms of the conversion) C H :

     ( M

    )  ( M

    )

     0

    (14a)

    2

    6

    C2H6 1

    C2H6 4

    C H Cl :

    ( M

    )

    ( M

    C

    )

    (14b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2 2

    HCl :

    ( M

    )

    ( M

    C

    )

    (14c)

    HCl 4

    C2H4Cl2 2

    C H Cl :

    ( M

    )

    ( M

    C

    )

    (14d)

    2

    3

    C2H3Cl 4

    C2H4Cl2 2

    Moving on to Control Volume II, we note that all global rates of production are zero and that only ethane and dichloroethane are present in this control volume. This leads to the two non‐trivial species mole balances given by

    Control Volume II ( constraints on molar flow rates)

    C H :

     ( M

    )  ( M

    )

     0

    (15a)

    2

    6

    C2H6 1

    C2H6 2

    304