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5.6: Untitled Page 104

  • Page ID
    18237
  • Chapter 5

    Returning to Eq. 17 and making use of Eq. 18a leads to

    p

    .

      p

     

    H O

    0 80

    H O ,

    ( T

    80F)

    2 , 787 Pa

    vap

    2

    2

    1

    (19)

    and we are now ready to determine the humidity in Stream #1.

    We are given that the total pressure in the system is equivalent to 755

    mm Hg and this leads to a pressure given by

    3

    755 mm Hg

    101.3  10 Pa

    p

     100 , 634 Pa

    (20)

    760 mm Hg/atm

    atm

    Use of this result in Eq. 14 leads to the humidity in Stream #1 given by MW

    p

    H O

    H O

    2

     2 1

    ( humidity)

    =

    1

    p

    MW

    p

    air

     H 

    2O 

    1 

    18.015 g H O/mol

    2,795 Pa

    2

    (21)

    28.84 g dry air/mol 100 , 634 Pa  2795 Pa

     0 . 0178 g H O g dryair

    2

    We are given that Stream #2 is in equilibrium with Stream #3, thus the relative humidity in Stream #2 is 100% and the partial pressure of water vapor can be determined in terms of a process equilibrium relation as ( p

    p

    p

    H O )

    (

    vap )

    1 , 705 Pa

    (22)

    2

    H O ,

    2

    2

    3

    H

    , vap

    2O

    T 15 C

    This allows us to repeat the type of calculation illustrated by Eq. 21 to obtain

    MW

    p

    H O

    H O

    2

     2 2

    (humidity)

    =

    2

    p

    MW

    p

    air

     H 

    2O 

    2 

    (23)

     0 . 01076g H O g dry air

    2

    Finally we return to Eq. 13 to determine the mass flow rate of water leaving the air conditioning unit according to

    index-198_1.png

    index-198_2.png

    Two‐Phase Systems & Equilibrium Stages

    189

    m

    H2O  3

    kg H O

    2

     1000 kg dryair/h 0 01783

    .

     0 01076

    .

    (24)

    kg dry air

     7 0

    . 7 kg H O h

    2

    EXAMPLE 5.8. Use of air to dry wet solids

    In Figure 5.8a we have illustrated a co‐current air dryer. The solids entering the dryer contain 20% water on a mass basis and the mass flow rate of the wet solids entering the dryer is 1000 lbm/h. The dried solids contain 5% water on a mass basis, and the temperature of the solid stream leaving the dryer is 65 C. The complete design of the dryer is a complex process that requires a knowledge of the flow rate of the dry air entering the dryer. This can be determined by a macroscopic mass balance analysis.

    Figure 5.8a. Air dryer

    As the air and the wet solids pass through the dryer an equilibrium condition is approached and mass transfer of water from the wet solids to the air stream diminishes. As an example, we assume that the air leaving the dryer is in equilibrium with the solids leaving the dryer, and this allows us to determine the maximum amount of water that can be removed from the wet solids. For this type of approximation the dryer becomes an equilibrium stage.

    To construct a control volume for the analysis of this system, we need only make cuts where information is given and required and these lead to the control volume shown in Figure 5.8b. We begin this problem with the species macroscopic mole balance for a steady‐state process in the absence

    index-199_1.png

    index-199_2.png

    index-199_3.png

    190