Example 9.2-1: Finding the Voltage required for a Given Temperature Rise in a Wire Heated by an Electrical Current

See worksheet p268.mws for a derivation of the equations used in this 
example.
Start with equations 9.2-1 and 9.2-14 to get equation 9.2-17
>  restart;I^2; Since Maple thinks I is sqrt(-1), we don't want to 
use I for the current
>  Se:=i^2/ke; eq 9.2-1 but using i for the current density instead 
of I.
>  dTmax:=Se*R^2/(4*k); eq 9.2-14 now leads to eq. 9.2-17
>  i:=ke*E/L; eq. 9.2-18
>  dTmax; giving eq. 9.2-19
>  eq19:=dTmax=Tmax-T0;
>  s:=solve(eq19,E);
>  E:=s[1]; Giving us eq 9.2-20. Note the square brackets in s[1].
>  R:=2*mm; L:=5*m;T0:=293*K; Tmax:=T0+10*K; Given data with 
units
>  k:=2.23e-8*volt^2*K^(-2)*ke*T0; Given the value for the Lorenz 
number of copper
>  E; Close to eq. 9.2-21 except:
>  mm:=m/1000; We need the relation between m and mm.
>  E; And we need to tell Maple that ke and volt are positive real 
variables.
>  assume(ke>0);assume(volt>0);
>  simplify(E);  The text found 40 volts.

>