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6.8 Stationary heat flow in a sphere ∗          133




                                                    dr
                                                 r
                                                   q(r)  q(r + dr)









            Figure 6.13. Heat flow through a spherical shell.

            Energy conservation (6.118) in a spherical geometry can also be written as

                                       1 d  
  2
                                            r q(r) = S(r)                     (6.119)
                                       2
                                      r dr
            which is verified by carrying out the differentiation. The radial heat flow q is related to the
            radial temperature gradient by Fourier’s law
                                                  dT
                                         q =−λ(r)    ,                        (6.120)
                                                   dr
            where the heat conductivity is assumed to be a function of r. Inserting Fourier’s law
            into expression (6.119) for energy conservation gives the temperature equation for radial
            heat flow
                                    1 d     2  dT
                                          r λ(r)    =−S(r).                   (6.121)
                                     2
                                    r dr        dr
            The temperature equation (6.121) is straightforward to integrate when both the heat con-
            ductivity λ and the heat production per unit volume S are constants. Two integrations then
            yield

                                           1   S   2  c 1
                                   T (r) =−       r −    + c 2                (6.122)
                                           6   λ      r
            where c 1 and c 2 are integration constants that will be found from boundary conditions.
            One boundary condition is that dT/dr = 0 at the center of the sphere (r = 0). We see
            that the only way to have a finite temperature gradient at the center is to let c 1 = 0.
            The other boundary condition is to let T = T 0 be the temperature at the surface, r = r 0 .
            The temperature solution is then

                                          1    
  
  2  2
                                             S
                                   T (r) =       r − r   + T 0                (6.123)
                                                  0
                                          6  λ
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