Page 634 - Bird R.B. Transport phenomena
P. 634

614  Chapter 20  Concentration Distributions with More Than One Independent Variable

      SOLUTION             For this system the equation  of continuity for the mixture, given in Eq. 19.1-12, becomes
                                                              dv*
                                                             -^  = 0                            (20.1-1)

                           in which v* is the z-component of the molar average velocity. Integration with respect to z gives
                                                             v* = i;* (0                        (20.1-2)
                                                                  0
                           Here and elsewhere in this problem, the subscript "0" indicates a quantity evaluated at z =  0.
                           According to Eq. (M) of Table 17.8-1, this velocity can be written in terms of the molar  fluxes
                           of A and В as
                                                                    N B
                                                          vZ =                                  (20.1-3)

                           However, N  is zero because of the insolubility of species  В in liquid A. Then use of Eq.  (D)
                                     Bz0
                           of Table  17.8-2 gives  finally
                                                               V AB  dx fi
                                                        v- = - 1 - x  dz                        (20.1-4)
                                                                  A0
                           in  which x  is the interfacial gas-phase concentration, evaluated here on the assumption  of
                                    A0
                           interfacial  equilibrium. For an ideal gas mixture this is just the vapor pressure of pure A di-
                           vided by the total pressure.
                               The equation of continuity of Eq. 19.1-17 then becomes

                                                                              z
                                                                             d x
                                                             dX A                               (20.1-5)
                                                  dt     -  x A  \ dZ
                           This is to be solved with the initial and boundary conditions:
                           I.C.:                        at t = 0,  x  = 0                       (20.1-6)
                                                                    A
                           B.C.I:                       at z = 0,  x  = x A0                    (20.1-7)
                                                                    A
                           B.C. 2:                      at z =     x A  = 0                     (20.1-8)
                           We  can try the same kind  of combination  of variables used  in Example 4.1-1; namely, X =
                           X  X  a n
                            A/ AO  ^ 2 = z/\/4£b ABt.  However, since Eq. 20.1-5 contains the parameter x A0, we can an-
                           ticipate that X will depend not only on Z but also parametrically on x A0.
                               In terms of these dimensionless variables, Eq. 20.1-5 can be written as
                                                        2
                                                        d X
                                                                                                (20.1-9)
                                                                    dZ
                           Here the quantity
                                                                      dX
                                                                                               (20.1-10)
                                                                         z=o
                           is a dimensionless molar average velocity, <p = v*\/t/%b AB, as can be seen by comparing  Eqs.
                           20.1-10 and  20.1-4. The initial and boundary conditions in Eqs. 20.1-6 to 8 now become
                            B.C. 1:                  atZ  -o,   X = 1                          (20.1-11)
                            B.C. 2 and I.C.:         atZ =  00  X = 0                          (20.1-12)

                            Equation 20.1-9 can be attacked by first letting dX/dZ  = Y. This gives a first-order  differential
                            equation for У that can be solved to obtain

                                                       =  C x p [ - ( Z - 0 ] - ^              (20.1-13)
                                                                        2
                                                           i e
   629   630   631   632   633   634   635   636   637   638   639