Page 703 - Bird R.B. Transport phenomena
P. 703

§22.3  Correlation of Binary  Transfer  Coefficients  in One Phase  683

                           In computing x j, we  have assumed  ideal gas behavior, equilibrium  at the interface, and com-
                                       A
                           plete insolubility  of air in water. The mean mole fraction, x f, of the water vapor  is  sufficiently
                                                                          A
                           small  that it can be neglected  in evaluating  the physical  properties at the film conditions:
                                                с = 3.88  X 10~ 5  g-moles/cm 3
                                                p = 1.12xlO" g/cm  3
                                                           3
                                                           4
                                                fi  =  1.91  X 10"  g/cm  • s (from Table 1.1-1)
                                              ЯЬ  = 0.292 cm7s  (from  Eq. 17.2-1)
                                                АВ
                                                   ( _ M \  __L91X1O-  4
                                               S r  =                      =  0.58
                                                            (1.12  x  10" )(0.292)
                                                                    3
                                                                             3
                                                            (0.05X215X1.12 X  10" )
                                               Re =                            =  63
                                                                 1.91  X 10~ 4
                           When  these values  are used  in Eq. 22.3-21 we  get
                                                  Sh  w  = 2 + 0.60(63) (0.58) 1/3  =  5.96    (22.3-29)
                                                                 1/2
                           and the mean mass transfer  coefficient  is then
                                                                       5
                                               ,    сЯЬ АВ    (3.88 X  10~ )(0.292)
                                               k xm  = —  Sh w  =  —         (5.96)
                                                            3
                                                  =  1.35  X 10~  g-mol/s  • cm 2              (22.3-30)
                           Then from  Eq. 22.3-26 the evaporation rate is found  to be
                                                                         2
                                                                           02
                                                                3
                                                 W A0  = (1.35  X 10- )(тг)(0.05) ^ ^ 7  024 ^
                                                    = 2.70X10" g-mole/s                        (22.3-31)
                                                              7
                                                                    3
                           This result corresponds to a decrease  of  1.23  X  10  cm/s in the drop diameter and indicates
                           that a drop of this size will fall a considerable distance before  it evaporates  completely.
                               In  this  example,  for  simplicity,  the  velocity  and  surface  temperature  of  the drop  were
                           given. In general, these conditions must be calculated  from  momentum and energy  balances,
                           as discussed  in Problem 22B.1.

       EXAMPLE 22.3-2      We next turn to a problem for which the analogy between heat and mass transfer  leads to a sur-
                           prisingly  simple and useful,  if approximate result. The system, shown in Fig. 22.3-2, is a pair of
      The  Wet  and  Dry  Bulb  thermometers, one of  which  is covered  with  a cylindrical  wick kept saturated with  water. The
      Psychrometer         wick  will  cool  by  evaporation  into  the  moving  air  stream  and  for  steady  operation will ap-
                           proach  an asymptotic  value  known  as  the wet bulb temperature.  The bare thermometer, on the
                           other hand, will tend to approach the actual temperature of the approaching air, and this  value
                           is called the dry bulb temperature. Develop an expression  for determining the humidity  of the air
                           from the wet and dry bulb temperature readings neglecting radiation and assuming  that the re-
                           placement  of the evaporating  water  has no significant  effect  on the wet bulb temperature mea-
                           surement. In Problem 22B.2 we will see how radiation can be taken into account.

      SOLUTION             For simplicity,  we  assume  that the  fluid  velocity  is  high  enough  that the thermometer read-
                           ings  are  unaffected  by  radiation and  by  heat conduction along  the thermometer stems, but
                           not so high that viscous  dissipation  heating effects  become significant.  These assumptions are
                           usually  satisfactory  for  glass  thermometers and  for  gas  velocities  of  30 to  100  ft/s.  The  dry
                           bulb temperature is then the same as the temperature Т  of the approaching gas, and the wet
                                                                        ж
                           bulb temperature is the same as the temperature T  of the outside  of the wick.
                                                                   o
                               Let species  A  be water  and species  В be air. An energy  balance is made on a system  that
                           contains a length L of the wick (the distance between planes  1 and 2 in the figure).  The rate of
   698   699   700   701   702   703   704   705   706   707   708