Page 417 - Modelling in Transport Phenomena A Conceptual Approach
P. 417

9.5.  MASS TRANSFER WITH CONVECTION                                397


          Analysis
           The Reynolds number is
                Re=- D(V%)P
                        CL
                   - (6 x 10-2)(1.5 x 10-3)(1000)  = 101   +    Laminar flaw   (1)
                   -
                              892 x
          Note that the tern (D/L) Re Sc  becomes
                          (+&=(                 ) (101)(737) = 2233



          Since the concentration at the surface of the pipe is constant, the we of Eq.  (9.5-73)
          gives
                             Sh = 3.66 +   0.668 [(DIL) Re%]
                                         1 + 0.04 [(D/L) ReScI2I3
                                           0.0668 (2233)
                                = 3.66 +  1 + 0.04 (2233)2/3  = 22.7           (3)
           -                                           - (CAb)OUt] - [cA~ - (CAb)iTZ]
           Considering  the  water  in the pipe  as a  system,  a  macroscopic  mass  balance  on
          benzoic acid gives

            (.D2/4)(v%) [(CAb)out - (CAb)in] = (@(kc)
                                                                -
                Q                          AM            ln  cA~ (CAb)Wt  3
                                                                - ( CAb )in
                                                                 ~
                                                   \                           /
                                                              (ACA)LM
                                                                               (4)
          Since  (cA~)~~ Eq.  (4) simplifies  to
                        0,
                      =




          Substitution of  numerical values into Eq.  (5) gives
                                     {
                      (cA~)~~ 1 - exp  -             4 (2)(22.7)   I>
                                3.412
                              =
                                             [  (6 x 10-2)(101)(737)
                              = 0.136kg/m3                                     (6)
          Comment:  One could  also use Eq.  (4.5-31) to calculate the Sherwood number,
          %.e.,

                                  Sh = 1.86 @~SC(D/L)]~/~
                                     = 1.86 (2233)'13  = 24.3

          which is not very much digerent jhm 22.7.
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