Page 311 - Bird R.B. Transport phenomena
P. 311

§10.2  Heat Conduction with  an Electrical Heat Source  295

                               To  illustrate  further  problems  in  electrical  heating, we  give  two  examples  concern-
                           ing  the temperature rise in wires: the first indicates the order  of magnitude  of the heating
                           effect,  and  the second  shows  how  to handle  different  boundary  conditions.  In addition,
                           in  Problem  10C.2 we  show  how  to take  into account the temperature dependence  of  the
                           thermal and electrical  conductivities.




       EXAMPLE  10.2-1     A  copper  wire  has  a radius  of  2 mm and  a length  of  5 m.  For what  voltage drop  would  the
                           temperature rise at the wire axis be  10°C, if  the surface  temperature of the wire is 20°C?
      Voltage Required for  a
      Given Temperature Rise  QQJJJJIQJSI
      in  a Wire Heated  by  an
      Electric Current     Combining  Eq. 10.2-14 and  10.2-1 gives


                                                           max                                 (10.2-17)
                                                                °  4kk e
                           The current density  is related  to the voltage drop E over  a length L by

                                                                                               (10.2-18)
                                                             I  =  K
                                                                 L
                           Hence
                                                                  2
                                                            _ T  _  E R 2  (K                  (10.2-19)
                                                              °   4L  U
                                                                     2
                           from  which

                                                                                               (10.2-20)
                                                                               2
                                                                                  2
                                                                           8
                           For  copper, the Lorenz number  of  §9.5 is k/k T  = 2.23  X 10~  volt /K .  Therefore, the voltage
                                                                0
                                                              e
                           drop needed to cause a  10°C temperature rise is
                                             E = ( 5 0 0 0 m m \   -  vglt  93)(10)K
                                                2            V 2 2 3 x l 0  8  V(2
                                                  \  2 mm  /          К
                                                              4
                                               =  (5000X1.49 X 10~)(54.1)  = 40 volts          (10.2-21)
       EXAMPLE  10.2.2     Repeat the analysis  in  §10.2, assuming  that  T  is  not known, but  that instead  the heat flux at
                                                               o
                           the  wall is given by  Newton's  "law  of  cooling"  (Eq. 10.1-2). Assume that the heat transfer  co-
      Heated  Wire  with   efficient  h and the ambient air temperature T  are known.
                                                               air
      Specified Heat  Transfer
      Coefficient  and     SOLUTION  I
     Ambient  Air            e
      Temperature          ^   solution proceeds  as before  through Eq. 10.2-11, but the second integration constant is de-
                           termined  from  Eq.  10.1-2:

                           B.C. 2':                                                            (10.2-22)

                                                                                 2
                           Substituting  Eq. 10.2-11 into Eq. 10.2-22 gives C  = (S R/2h) + {S R /4k)  + T , and  the tem-
                                                                                e
                                                                                         air
                                                                 2
                                                                      e
                           perature profile  is then
                                                                                               (10.2-23)
                                                            4k
                           From this the surface  temperature of the wire is found  to be T  + S R/2h.
                                                                            air  e
   306   307   308   309   310   311   312   313   314   315   316