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248   CHAPTER 8.  STEADY MICROSCOPIC BALANCES  WITHOUT GEN.


            Example 8.1  Consider a solid wne of  circular cross-section  as shown in Figure
            8.6.  The diameter at z = 0 is 8cm and  the diameter at z = L is 10cm.  Calculate
            the steady rate of  heat transfer if the lateral surface is well insulated and the thermal
            conductivity  of  the solid material  as a function of  temperature is given by

                                       k(T) = 400 - 0.07T

            where  k  is an  W1m.K and  T is in degrees Celsius.


                                            L = 40 cmd












                           Figure 8.6  Conduction through a solid cone.

            Solution

            The diameter increases  linearly in the z-direction,  i.e.,

                                       D(z) = 0.05 z + 0.08

            Therefore,  the  cross-sectional  area perpendicular  to  the  direction  of  heat flu  is
            given as a function of  position in the form

                                        aD2    T
                                 A(z) = - - (0.05 z + 0.08)2
                                             =
                                         4     4
            The use of  Eq.  (A) in Table 8.1 with To = 80"C, TL = 35 "C and  L = 0.4m gives
            the heat transfer rate as


                                     /80  (400 - 0.07 T) dT
                              Q= J35                     = 280W
                                    r0.4       dz
                                   I,,  40.05 z + 0.08)2/4


            Example 8.2  Consider the problem given in Example  2.2.  Determine  the tem-
            perature distribution within the slab.
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