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5.4: Untitled Page 102

  • Page ID
    18235
  • Chapter 5

    relation given by Eq. 5‐64 will be applied at the gas‐liquid interface. We express this type of equilibrium condition as

    Equilibrium relation:

    y

    K

    x ,

    at the gas‐liquid interface

    (5‐65)

    A

    eq,A

    A

    When the form given by Eq. 5‐64 represents a valid approximation, the techniques studied in this section can be applied with confidence.

    In the preceding paragraphs we have illustrated liquid‐liquid and gas‐liquid systems that can be treated as equilibrium stages. Such systems are ubiquitous in the world of chemical engineering where purification of liquid and gas streams is a major activity. In addition to the contacting devices illustrated in Figures 5‐6

    and 5‐7, there are many other processes that can often be approximated as equilibrium stages and some of these are examined in the following paragraphs.

    EXAMPLE 5.7. Condensation of water in humid air

    On a warm spring day in Baton Rouge, LA, the atmospheric pressure is 755 mm Hg, the temperature is 80 F, and the relative humidity is 80%.

    A large industrial air conditioner operating at atmospheric pressure is illustrated in Figure 5.7. It treats 1000 kg/h of air (dry air basis) and lowers the air temperature from 80 F to 15 C. The cool air leaving the unit is assumed to be in equilibrium with the water that leaves the unit at 15 C.

    Thus the air conditioner is treated as an equilibrium stage.

    Figure 5.7. Air conditioning as a separation process

    Here we want to determine how much liquid water, in kg/h, is removed from the warm, humid air by the air conditioning system. In this case the obvious control volume cuts all three streams and encloses the air

    Two‐Phase Systems & Equilibrium Stages 185

    conditioner. If we assume that the system operates at steady state and we conclude that there are no chemical reactions, the appropriate form of the species mass balance is given by

    dA  0 ,

    A  1 , 2 , ...., N

    (1)

    v n

    A A

    A

    In problems of this type, it is plausible to neglect diffusion velocities, u

     v , so that Eq. 1 takes the form

    A

    A

    dA  0 ,

    A  1 , 2 , ...., N

    (2)

    v n

    A

    A

    Under these circumstances we can combine the equations for nitrogen and oxygen to obtain the following mass balance for dry air:

    v n dA

    v n dA

    v n dA  0

    (3)

    air

    air

    air

    A 1

    A 2

    A 3

    Once again we note that 

    indicates the density of oxygen and nitrogen

    air

    as indicated in Eq. 5‐36. Since significant amounts of nitrogen and oxygen do not leave the system in Stream #3, the mass balances for dry air and water take the form

    Dry Air:

     ( m )  ( m )

     0

    (4)

    air 1

    air 2

    Water:

     ( m

    )  ( m

    )  ( m

    )

     0

    (5)

    H2O 1

    H2O 2

    H2O 3

    in which we have used the nomenclature illustrated by

    dA  ( m )

    v n

    (6)

    air

    air 1

    A 1

    Equations 4 and 5 represent the fundamental balance equations associated with the system illustrated in Figure 5.7, and we need to use the information that is given to determine ( m

    ) .

    H2O 3

    Since information is given about the humidity of the incoming air stream, we need to think about how we can connect the mass flow rates in Eqs. 4 and 5 with the humidity vaguely described by Eq. 5‐35 or the point

    186