Page 383 - Modelling in Transport Phenomena A Conceptual Approach
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9.3.  HEAT  TRANSFER WTH CONVECTION                                 363


           where ria is the mass flow rate given by
                                        ria = p(vZ)rR2                     (9.3-14)

              On the other hand, since ET/& = 0 as a result of  the symmetry condition at
           the center of the tube, the integral on the right-side of Eq. (9.3-11) takes the form

                                                                           (9.3-15)


           Substitution of  Eqs.  (9.3-13) and  (9.3-15) into Eq. (9.3-11) gives the governing
           equation for the bulk temperature in the form


                                                                           (9.3-16)

              The solution of  Eq.  (9.3-16) requires the boundary conditions associated with
           the problem to be known. The two most commonly used boundary conditions are
           the constant wall temperature and constant wall heat flux,

           Constant wall temperature
           Constant wall temperature occurs in evaporators and condensers in which phase
           change takes place on one side of  the surface.  The heat flux at the wall can be
          represented either by Fourier’s law of heat conduction  or by Newton’s law of cooling,
           1.e..
                                                                           (9.3-17)

          It is implicitly implied in writing Eq.  (9.3-17) that the temperature increases in the
          radial direction. Substitution of Eq.  (9.3-17) into EQ. (9.3-16) and rearrangement
          yields

                                                                           (9.3-18)

          Since the wall temperature, T,,  is constant, integration of  Eq.  (9.3-18) yields

                               mcp In ( Tw  -      = rD(h),z               (9.3-19)
                                         Tw - Tb
          in which (h)” is the average heat transfer coefficient from the entrance to the point
          z defined by
                                       (h)”  =  J”hdz                      (9.3-20)
                                                0
          If Eq. (9.3-19) is solved for Tb, the result is


                                                                           (9.3-21)
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