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7.28: Untitled Page 173

  • Page ID
    18306
  • Chapter 7

    Here we have used

    (o)

    ( M

    )

    to identify the first assumed value for the

    C2H4Cl2 5

    molar flow rate of dichloroethane in the tear stream. Moving on to Control Volume II, we obtain

    Control Volume II

    C H :

    ( M

    )

     ( M

    )  R

    (4a)

    2

    6

    C2H6 3

    C2H6 2

    C2H6

    C H Cl :

    ( M

    )

     ( M

    )  R

    (4b)

    2

    4

    2

    C2H4Cl2 3

    C2H4Cl2 2

    C2H4Cl2

    HCl :

    ( M

    )

     ( M

    )  R

    (4c)

    HCl 3

    HCl 2

    HCl

    C H Cl :

    ( M

    )

     ( M

    )  R

    (4d)

    2

    3

    C2H3Cl 3

    C2H3Cl 2

    C2H3Cl

    The global net rates of production are constrained by Axiom II that provides the relations

    Axiom II: R

     0 , R

      R

    , R

      R

    (5)

    C H

    HCl

    C H Cl

    C H Cl

    C H Cl

    2

    6

    2

    4

    2

    2

    3

    2

    4

    2

    which can be expressed in terms of the conversion (see Eq. 1) to obtain R

     0 ,

    R

    ( M

    ) ,

    R

    ( M

    C

    C

    ) (6)

    C2H6

    HCl

    C2H4Cl2 2

    C2H3Cl

    C2H4Cl2 2

    These representations for the global net rates of production can be used with Eqs. 4 to obtain

    Control Volume II

    C H :

    ( M

    )

     ( M

    )

    (7a)

    2

    6

    C2H6 3

    C2H6 2

    C H Cl :

    ( M

    )

     ( M

    ) 1

    (  C)

    (7b)

    2

    4

    2

    C2H4Cl2 3

    C2H4Cl2 2

    HCl :

    ( M

    )

     ( M

    ) 

    ( M

    C

    )

    (7c)

    HCl 3

    HCl 2

    C2H4Cl2 2

    C H Cl :

    ( M

    )

     ( M

    ) 

    ( M

    C

    )

    (7d)

    2

    3

    C2H3Cl 3

    C2H3Cl 2

    C2H4Cl2 2

    This completes the first two steps in the sequence, and we are ready to move on to Control Volume III shown in Figure 7.7. Application of Eq. 2

    to Control Volume III provides the following mole balances:

    Material Balances for Complex Systems

    323

    Control Volume III

    C H :

    ( M

    )

     ( M

    )

    (8a)

    2

    6

    C2H6 4

    C2H6 3

    (1)

    C H Cl :

    ( M

    )

     ( M

    )

    (8b)

    2

    4

    2

    C2H4Cl2

    C

    5

    2H4Cl2 3

    HCl :

    ( M

    )

     ( M

    )

    (8c)

    HCl 4

    HCl 3

    C H Cl :

    ( M

    )

     ( M

    )

    (8d)

    2

    3

    C2H3Cl 4

    C2H3Cl 3

    Here we have indicated that the molar flow rate of dichloroethane ( C H Cl ) leaving the separator in Stream #5 is the first approximation 2

    4

    2

    based on the assumed value entering the mixer. This assumed value is given in Eq. 3b as ( M

    )(o) . Because Stream #5 contains only a single

    C2H4Cl2 5

    component, this problem is especially simple, and we only need to make use of Eqs. 3b, 7b and 8b to obtain a relation between (1)

    ( M

    )

    and

    C2H4Cl2 5

    ( M

    )(o) . This relation is given by

    C2H4Cl2 5

    C H Cl :

    ( 1

    (

    ) )

    M

     1 

    C ( M

    )  ( M

     (13)

    C H Cl

    )(o)

    2

    4

    2

    2

    4

    2

    C

    5

    2H4Cl2 1

    C2H4Cl2 5 

    Neither of the two molar flow rates on the right hand side of this result are known; however, we can eliminate the molar flow rate of dichloroethane ( M

    ) by working in terms of a dimensionless molar flow rate C2H4Cl2 1

    defined by

    ( M

    )

    C H Cl

    2

    4

    2 5

    M

    5

    ( M

    (14)

    )

    C2H4Cl2 1

    This allows us to express Eq. 13 as

    (1)

    (o)

    M

    1 C1

     M

    ,

    C  0 . 30

    (15)

    5

    5

    (o)

    (1)

    and we can assume a value of M

    in order to compute a value of M

    .

    5

    5

    (1)

    On the basis of this computed value of M

    , the analysis can be repeated

    5

    (2)

    to determine M

    that is given by

    5

    (2)

    (1)

    M

    1 C 1

     M

    ,

    C  0 3

    . 0

    (16)

    5

    5

    324