Example 10.2.2 Heat Conduction in a Cylinder with an Electrical Heat Source and Newton's Law of Cooling
Following the same steps as BS&L used.
>  restart;
>  A:=r-> 2*Pi*r*L;  area of conduction 
>  Q:=r-> A(r)*qr(r);  Heat conducted across cylinder of radius r in the direction of increasing r: qr is the flux of energy.
>  Q(r);  the rate of thermal energy into cylindrical shell at r: eq. 10.2-2 
>  Q(r + dr);  rate of thermal energy out across cyllindrical surface at r+dr: eq. 10.2-3
>  V:=r->A(r)*dr;  Volume of shell
>  Epro:=(r)->V(r)*Se;  Thermal production rate by electrical current: eq. 10.2-4
>  Epro(r);
>  LHS:=limit((Q(r+dr)-Q(r))/dr,dr=0);  Left hand side of eq. 10.2-5
>  de:=simplify(r*(LHS=Epro(r)/dr)/A(r)); energy balance, out - in = energy production by electrical dissipation giving 10.2-6
>  sol:=dsolve({de,qr(0)=finite},qr(r));  Solving 10.2-6 with BC 10.2-8 to get 10.2-9
>  assign(sol);  Giving qr(r) a value
>  qr:=unapply(qr(r),r);  Making qr a function of r
>  deFL:=qr(r)=-k*D(T)(r);  Fourier's Law
>  sol:=dsolve({deFL,D(T)(R)=-h/k*(T(R)-Tair)},T(r)); trying to solve the de with the BC 10.2-22
>  sol:=dsolve(deFL,T(r)); leaving out the BC.
>  assign(sol);T:=unapply(T(r),r);
>  bc:=-k*D(T)(R)=h*(T(R)-Tair);
>  _C1:=solve(bc,_C1);
>  dT:=simplify(T(r)-Tair); Eq. 10.2-23
>  dT:=unapply(dT,r);

>  T0:=simplify(T(R));
>  dT(R); Checking eq. 10.2-24