Skip to main content
Engineering LibreTexts

7.19: Untitled Page 164

  • Page ID
    18297
  • Chapter 7

    C H Cl :

     ( M

    )  ( M

    )  ( M

    )

     0 (15b)

    2

    4

    2

    C2H4Cl2 1

    C2H4Cl2 5

    C2H4Cl2 2

    At this point we use Eq. 14b to obtain

    1  C

    ( M

    )

    ( M

    )

    (16)

    C2H4Cl2 5

    C

    C

    2H4Cl2 1

    Since the molar flow rate of dichloroethane entering the system is not specified, it is convenient to work in terms of dimensionless molar flow rates, and this leads to

    ( M

    )

    C H Cl

    5

    1  C

    C H Cl :

    2

    4

    2

    (M

    )

    (17a)

    2

    4

    2

    C2H4Cl2 5

    ( M

    )

    C

    C2H4Cl2 1

    ( M

    )

    C H Cl 4

    C H Cl :

    2

    3

    (M

    )

    .

    (17b)

    2

    3

    C2H3Cl 4

    ( M

    1 0

    )

    C2H4Cl2 1

    ( M

    )

    C H

    4

    0 . 02

    C H :

    2

    6

    (M

    )

    2

    6

    C2H6 4

    ( M

    (17c)

    )

    0 9

    . 8

    C2H4Cl2 1

    ( M

    )

    HCl :

    HCl 4

    (M

    )

     1 .

    HCl 4

    ( M

    00

    (17d)

    )

    C2H4Cl2 1

    In addition, the molar flow rates of the four species can be expressed in terms of the total molar flow rate entering the system, and this leads to C H Cl :

    ( M

    )

     2 2867

    .

    M

    (18a)

    2

    4

    2

    C2H4Cl2 5

    1

    C H Cl :

    (M

    )

     0 9

    . 8 M

    (18b)

    2

    3

    C H Cl 4

    1

    2

    3

    C H :

    (M

    )

     0 0

    . 2 M

    (18c)

    2

    6

    C H

    4

    1

    2

    6

    HCl :

    (M

    )

     0 9

    . 8 M

    (18d)

    HCl 4

    1

    These results can be used to determine the total molar flow rates in Streams #4 and #5

    M

     2 2867

    .

    M

    ,

    M

     1 98

    .

    M

    (19)

    5

    1

    4

    1

    along with the composition in Stream #4 that is given by

    ( x

    )  0 . 495 ,

    ( x

    )

     0 01

    .

    ,

    ( x

    )

     0 . 495

    (20)

    C2H3Cl 4

    C2H6 4

    HCl 4

    index-314_1.png

    index-314_2.png

    index-314_3.png

    index-314_4.png

    Material Balances for Complex Systems

    305

    A direct solution of the recycle problem given in Example 7.5 was possible because of the simplicity of the process, both in terms of the chemical reaction and in terms of the structure of the process. When purge streams are required, as illustrated in Figure 7‐5, the analysis becomes more complex.

    EXAMPLE 7.6. Production of ethylene oxide with recycle and purge In Figure 7.6a we have illustrated a process in which ethylene oxide ( C H O ) is produced by the oxidation of ethylene ( C H ) over a 2

    4

    2

    4

    catalyst containing silver. In a side reaction, ethylene is oxidized to carbon dioxide ( CO ) and water ( H O ) . The feed stream, Stream #1, 2

    2

    consists of

    Figure 7.6a. Production of ethylene oxide

    ethylene ( C H ) and air ( N and O ) which is combined with Stream #5

    2

    4

    2

    2

    that contains the unreacted ethylene ( C H ) , carbon dioxide ( CO ) and 2

    4

    2

    nitrogen ( N ) . The mole fraction of ethylene in Stream #2 entering the 2

    reactor must be maintained at 0 . 05 for satisfactory catalyst operation. The conversion of ethylene in the reactor is optimized to be 70% ( C  0 . 70 )

    306