Page 308 - Bird R.B. Transport phenomena
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292  Chapter  10  Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow
                               d.  At a solid-fluid  interface,  the normal  heat flux component  may be related  to  the
                                 difference  between  the solid surface  temperature  T o and  the "bulk"  fluid  temper-
                                 ature T b:
                                                                q  = h(T 0 -  T b)             (10.1-2)

                                 This  relation  is referred  to as Newton's  law of cooling.  It  is not  really  a  "law"  but
                                 rather  the  defining  equation  for  h,  which  is  called  the  heat transfer  coefficient.
                                 Chapter  14 deals with methods  for estimating heat-transfer  coefficients.

                           All four  types  of boundary  conditions  are encountered  in  this chapter.  Still other  kinds
                           of boundary conditions are possible, and they will be introduced  as needed.



      §10.2  HEAT CONDUCTION WITH AN
             ELECTRICAL HEAT SOURCE
                           The first system we consider is an electric wire of circular cross section with radius  JR and
                                                     1   1
                           electrical conductivity  k e ohm"  cm" . Through  this wire there is an electric current  with
                                                  2
                           current  density  /  amp/cm .  The  transmission  of  an  electric  current  is  an  irreversible
                           process, and  some  electrical  energy  is converted  into heat  (thermal  energy). The rate  of
                           heat production per unit volume is given by the expression

                                                              S e = £                          (10.2-1)

                           The quantity  S e is the heat  source resulting  from  electrical dissipation.  We assume  here
                           that  the temperature  rise in the wire  is not  so large that the temperature dependence  of
                           either  the thermal  or electrical conductivity  need be considered.  The surface  of the  wire
                           is maintained  at temperature  T o. We now show how to find the radial temperature distri-
                           bution within the wire.
                               For  the energy  balance  we  take  the  system  to be  a  cylindrical  shell  of  thickness  Ar
                           and  length L (see Fig. 10.2-1). Since v  =  0 in this system, the only contributions to the en-
                           ergy balance are
                           Rate of heat in
                           across cylindrical          (27rrL)q r\,)  = (2irrLqX                 (10.2-2)
                           surface at r
                           Rate of heat out
                           across cylindrical     (2тг(г + kr)L){q\ )  = (2irrLq)\ Ar          (10.2-3)
                                                                          r
                                                              r r+Ar
                                                                           r+
                           surface at r + Ar
                           Rate of thermal
                           energy production by             (2irr^rL)S e                       (10.2-4)
                           electrical dissipation
                                                                                в
                           The  notation  q  means  "heat  flux  in the r direction/'  and  (• • )|,+д  means  "evaluated at
                                                                                    г
                                        r
                           r  + Ar."  Note that we  take "in" and "out" to be in the positive r direction.
                               We now substitute these quantities  into the energy balance  of  Eq. 9.1-1. Division by
                           2vrLAr and taking the limit as  Ar goes to zero gives

                                                               Ar
                           The  expression on the left  side  is  the first  derivative  of  rq  with  respect  to r, so  that Eq.
                                                                            r
                           10.2-5 becomes
                                                           j{rq )  = S r                       (10.2-6)
                                                              r     e
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