Page 544 - Bird R.B. Transport phenomena
P. 544

524   Chapter 17  Diffusivity  and  the Mechanisms  of Mass  Transport

                              This problem  can also  be  solved  by  method  (iii) without  an experimental  value  of сЯЬ *.
                                                                                                   АА
                           Equation  17.4-2 gives  directly
                                                              eM      i_Y /2    ,2/3
                                                  \  = 2.96  X  1(Г  +
                                                               ^44.01  46^  (304.2)  1
                                                              6
                                                    =  4.20  X  10~  g-mole/cm  •  s           (17.2-7)
                           The  resulting  predicted  value  of  сЭ)^* is  5.1  X 10"  g-mole/cm  • s.
                                                                  6


      EXAMPLE   17.2-3     Estimate  c9) AB  for  a  mixture  of  80 mole% CH  and  20 mole% C H  6  at  136 atm and  313K.  It is
                                                                             2
                                                               4
                                                                                 5
                           known  that, at  1 atm  and  293K, the molar  density  is  с  =  4.17  X  10~  g-mole/cm  3  and  9) AB  =
     Estimation  of  Binary  0.163  cm7s.
     Diffusivity  at  High
     Density
                           SOLUTION
                           Figure  17.2-1  is used, with  method  (i). The reduced  conditions for  the known  data  are
                                                       T           293
                                                                          =  1.22              (17.2-8)
                                                             V(190.7)(305.4)
                                                                 1.0
                                                 Pr =                  = =  0.021              (17.2-9)
                                                      /p p   V(45.8)(48.2)
                                                       cA  cB
                           From  Fig.  17.2-1  at  these  conditions  we  obtain  (c%b )  =  1.21.  The  critical  value  (c$) )  is
                                                                      AB r                      AB c
                           therefore
                                                                         5
                                                         c4b AB  _  (4.17  X  l(T X0.163)
                                                      ~~ (c4b )  ~    1.21
                                                           AB r
                                                                 6
                                                      =  5.62  X  10" g-mol/cm-s              (17.2-10)
                           Next we  calculate the reduced  conditions for  the prediction  (T  =  1.30,  p  = 2.90) and read  the
                                                                            r       r
                           value  (c£b AB) r =  1.31  from  Fig.  17.2-1. The predicted value of  сЯЬ  is  therefore
                                                                             АВ
                                                                                6
                                                               ) c  =  (1.31)(5.62  X  10~ )
                                                             6
                                                    =  7.4X  10~  g-mole/cm  -s               (17.2-11)
                                                                         6
                           Experimental  measurements  8  give  сЯЬ АВ  =  6.0  X  10 ,  so  that  the  predicted  value  is
                           23%  high.  Deviations  of  this  magnitude  are  not unusual  in  the  estimation  of  c£b AB at  high
                           densities.
                               An  alternative  solution  may  be  obtained  by  method  (iii).  Substitution  into  Eq.  17.4-3
                           gives
                                                                     V  / 2  (45.8 X  48.2) 1/3
                                                         6
                                         (c% AB) c  =  2.96 X 10~ |
                                                         6
                                               =  4.78  X 10"  g-mole/cm  • s                 (17.2-12)
                           Multiplication by  (сЯЬ ) . at the desired  condition gives
                                             АВ }
                                                                  6
                                                   сЯЬ  =  (4.78  X  10~ )(1.31)
                                                      АВ
                                                                 6
                                                       =  6.26  X  10~  g-mole/cm  •  s       (17.2-13)
                           This is in closer  agreement with  the measured  value.  8



                               s  V.  J. Berry, Jr., and  R. C. Koeller, AIChE  Journal, 6,274-280  (1960).
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