Example 10.2-1: Finding the Voltage required for a Given Temperature Rise in a Wire Heated by an Electrical Current
See worksheet sec102.mws for a derivation of the equations used in this example.
Start with equations 10.2-1 and 10.2-14 to get equation 10.2-17
>  restart;I^2; Since Maple thinks I is sqrt(-1), we don't want to use I for the current
>  Se:=i->i^2/ke; eq 10.2-1 but using i for the current density instead of I.
>  dTmax:=Se(i)*R^2/(4*k); eq 10.2-14 now leads to eq. 10.2-17
>  i:=ke*E/L; eq. 10.2-18
>  dTmax; giving eq. 10.2-19
>  eq19:=dTmax=Tmax-T0;
>  s:=solve(eq19,E);
>  E:=unapply(s[1],R,L,T0,Tmax,k,ke); Giving us eq 10.2-20. Note the square brackets in s[1]. We used unaaply to make E a function of its parameters.
>  E1:=T0->E(2*mm,5*m,T0,T0+10*K,k,ke(k,T0)); Specifying R, L, and noting that the result should be just a function of T0 if ke/k is given by the Lorenz ratio.
>  ke:=(k,T0)->k/T0/2.23e-8/volt^2*K^2; The Lorenz number given in the problem.
>  Eallow:=E1(293*K); Now our allowed E is given by:
>  mm:=m/1000; We need the relation between m and mm.
>  Eallow; And we need to tell Maple that k, K and volt are positive real variables.
>  simplify(Eallow,assume=positive);  The text found 40 volts.