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5.10: Untitled Page 108

  • Page ID
    18241
  • Chapter 5

    Special Condition:

    ( x )

     0

    (5‐75)

    A o

    so that Eq. 5‐74 takes the form

    ( x ) M

     ( y ) M

     ( y ) M

    (5‐76)

    A 2

    A 1

    A 3

    At this point our objective is to determine ( y ) in terms of ( y ) , and we begin A 1

    A 3

    Figure 5‐12. Two‐unit extraction process

    by arranging this result in the form

    ( y )  ( x )

    M

    M

     ( y )

    (5‐77)

    A 1

    A 2 

     

    A 3

    Here we note that the process equilibrium relation is given by Process equilibrium relation:

    ( y )

    K

    ( x )

    (5‐78)

    A 2

    eq,A

    A 2

    and use of this result allows us to simplify the mole balance for species A to ( y )  ( y )

    M

    M

    K

     ( y )

    (5‐79)

    A 1

    A 2 

    eq,A

    A 3

    We now use Eq. 5‐73 to determine ( y ) so that Eq. 5‐79 provides the following A 2

    result

    ( y )  ( y )  A 1  A   ( y ) (5‐80)

    A 1

    A 1 

    

    A 3

    which can be solved for ( y ) to obtain

    A 1

    ( y )

    Two Equilibrium Stages:

    A 3

    ( y )

    (5‐81)

    A 1

    2

    1  A A

    index-206_1.png

    Two‐Phase Systems & Equilibrium Stages

    197

    We are now ready to determine ( y ) when three equilibrium stages are A 1

    employed as illustrated in Figure 5‐13.

    Figure 5‐13. Three‐unit extraction process

    In this case the molar balance for species A can be expressed as ( x ) M

     ( y ) M

    ( x ) M

     ( y ) M

    (5‐82)

    A 3

    A 1

    A o

    A

    

    

    4 

    molar flow of species A

    molar flow of species A

    out of the control volume

    into the control volume

    and we continue to impose the condition

    Special Condition:

    ( x )

     0

    (5‐83)

    A o

    so that Eq. 5‐82 takes the form

    ( x ) M

     ( y ) M

     ( y ) M

    (5‐84)

    A 3

    A 1

    A 4

    At this point our objective is to determine ( y ) as a function of ( y ) , and we A 1

    A 4

    begin by arranging this result in the form

    ( y )  ( x )

    M

    M

     ( y )

    (5‐85)

    A 1

    A 3 

     

    A 4

    Here we note that the process equilibrium relation is given by Process equilibrium relation:

    ( y )

    K

    ( x )

    (5‐86)

    A 3

    eq,A

    A 3

    and use of this result allows us to simplify the mole balance for species A to ( y )  ( y )

    M

    M

    K

     ( y )

    (5‐87)

    A 1

    A 3 

    eq,A

    A 4

    198