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5.5: Untitled Page 103

  • Page ID
    18236
  • Chapter 5

    humidity precisely defined by Eq. 5‐36. We start by examining the ratio of mass flow rates (water to dry air) given by

    v n dA

     H2O

    ( m

    )

    H O

    A

    1

    1

    2

    (7)

    ( m

    )

    air 1

    v n dA

    air

    A 1

    If the velocity, v n , is uniform across the entrance, or the species densities,

    and 

    , are uniform across the entrance, we can follow the H O

    air

    2

    development in Sec. 4.5.1 to conclude that

    v n dA

     H2O

    ( m

    )

    

    Q

    

    H O 1

    A

    H O 1

    1

    H O 1

    1

    2

    2

    2

    (8)

    ( m

    )

      Q

     

    air 1

    v n dA

    air 1

    1

    air 1

    air

    A 1

    In addition, if we accept the ratio of the area average species densities as our measure of the humidity we obtain

    

    ( m

    )

    H O 1

    H O

    2

    2

    1

     humidity

    (9)

    1

     

    ( m

    )

    air 1

    air 1

    This suggests a particular strategy for solving this mass balance problem.

    Dividing Eq. 5 by the mass flow rate of dry air in Stream #1 leads to ( m

    )

    ( m

    )

    ( m

    )

    H O 1

    H O 2

    H O

    2

    2

    2

    3

     0

    (10)

    ( m

    )

    ( m

    )

    ( m

    )

    air 1

    air 1

    air 1

    and on the basis of the mass balance for “dry air” given by Eq. 4, we can express this result in the form

    ( m

    )

    ( m

    )

    ( m

    )

    H O 1

    H O 2

    H O

    2

    2

    2

    3

     0

    (11)

    ( m

    )

    ( m

    )

    ( m

    )

    air 1

    air 2

    air 1

    Use of equations of the form of Eq. 9 leads to a “humidity balance”

    equation given by

    Two‐Phase Systems & Equilibrium Stages 187

    ( m

    )

    H

    humidity  humidity

    O

    2

    3

    (12)

    1

    2

    ( m

    )

    air 1

    Our objective in this example is to determine how much liquid water is removed from the air, and Eq. 12 provides this information in the form

    ( m

    )

     ( m )

    humidity

     humidity

    (13)

    H O 3

    air 1 

    2

    

    1

    2 

    Here it becomes clear that the solution to this problem requires that we determine the humidity in Streams #1 and #2. This motivates us to make use of Eq. 5‐37 that we list here as

    MW

    p

    H O

    H O

    2

    2

    point humidity =

    (14)

    MW

    p p

    air

    H2O 

    In addition, we are given information about the percent relative humidity defined by Eq. 5‐38 and listed here as

    p H O

    2

    % relative humidity =

    100

    (15)

    p H O, vap

    2

    In Stream #1 the percent relative humidity is 80% and this provides

    p H2O1

    % relative humidity =

    100  80

    (16)

    1

    p H O, vap

    2

    1

    which allows us to express the partial pressure of water vapor in Stream #1

    as

    p

    .

    p

    H

    0 80

    (17)

    O

    O vap

    2

     H ,2 

    1

    1

    The vapor pressure of water, in both Stream #1 and Stream #2, can be determined by Antoine’s equation (see Appendix A3) that provides

    p

    p

    H O ,

    H O ,

    ( T

    80F)

    3 484

    .

    kPa

    vap

    vap

    (18a)

    2

    2

    1

    p

    p

    ( T  15C)  1 705

    .

    kPa

    H O , vap

    ,vap

    (18b)

    2

    H2O

    2

    188