Page 486 - Modelling in Transport Phenomena A Conceptual Approach
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466     CHAPTER io.  UNSTEADY mmoscoprc BAL. WITHOUT GEN.

               Introduction of  the dimensionless quantities


                                                                            (10.3-43)

                                                 z
                                             t=-                            (10.3-44)
                                                 L
                                                                            (10.3-45)

            reduces Eqs.  (10.3-39)-(10.3-42) to

                                                                            (1 0.3-46)



                                         at  r=O     6=1                    (10.3-47)
                                                     ae
                                         at  (=O     --                     (10.3-48)
                                                     at -O
                                         at  (=1     @=O                    (10.3-49)

            The transformation
                                                 U
                                                                            (10.3-50)
            converts the spherical geometry into the rectangular geometry. Substitution of En.
            (10.3-50) into E!q.  (10.3-46) leads to
                                            azc
                                                 8%
                                           --  --                           (10.3-51)
                                            ar   at2
            which is identical with Eq. (10.2-69). Therefore, the solution is

                                   = e-  A%  [A sin(X) + B cos(A<)]         (10.3-52)

            or,
                                e=e-AaT  [.- sin( At)
                                               6  +B=l 6                    (10.3-53)
            The boundary condition defined by Eq.  (10.3-48) indicates that B = 0. Application
            of  Eq. (10.3-49) yields

                       sinA=O       +     A,,=nr     n=l,2,3, ...           (10.3-54)

            Therefore, the general solution is

                                                                            (10.3-55)
                                       n=l              h
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