9.6 Heat conduction through composite Cylindrical Walls: Example 9.6-1

> restart;
> A:=r->2*Pi*r*L; The area perpendicular to the direction of heat 
conduction
> Q:= (r,k)-> -k*A(r)*D(t)(r); Both k and r determine the heat 
flow rate
> deq:=limit((Q(r+dr,k)-Q(r,k))/(k*dr),dr=0)=0;
 The heat into any element at r must equal the amount 
that leaves ar r+dr
> R:=array(0...3); T:=array(0...3);
 Defining R and T as vectors with 4 elements in each one
> s:=dsolve({deq,t(R[m])=T[m],t(R[m+1])=T[m+1]},t(r)); 
Solving the DE with BC at R[m] and R[m+1].
> assign(s); t:=unapply(t(r),r); Making t a function of r
> eqs:=array(0...4);ks:=array(1...3); Making eqs and ks vectors
> Qv(m):=Q(r,ks[m+1]);
 The heat transfer rate in each interval is a constant independent 
of r and must also be equal to some common value in all intervals 
since heat is not lost or gained at the interfaces. Making Qv a 
function of m
> Qv:=unapply(Qv(m),m);
If Q0 is the common value of Q in the intervals then, we should 
have the following hold for the first interval:
> m:=1;
> eqs[m+1]:=Qv(m)=Q0;
> for j from 0 to 2 do
 We will set our equations so the LHS equals the temperature 
difference in it.
>    eqs[j+1]:=simplify((Qv(j)=Q0)*(log(R[j+1])-log(R[j]))/(2*Pi*L*ks[j+1]));
> od;
 This specifies the equations for each of the intervals in the solid
> eqs[0]:=Ta-T[0]=Q0/(h0*A(R[0])); 
eqs[4]:=T[3]-Tb=Q0/(h3*A(R[3]));
 Now we will add the equations for each position where h0 and h3 are 
heat transfer coefficients
> eq:=sum(eqs[k],k=0...4); Summing all five equations
> Q0:=solve(eq,Q0);












> Q0book:=2*Pi*L*(Ta-Tb)/(1/(R[0]*h0)+(log(R[1])-log(R[0]))/
ks[1]+(log(R[2])-log(R[1]))/ks[2]+(log(R[3])-log(R[2]))/ks[3]+
1/(R[3]*h3));

> simplify(Q0-Q0book,assume=positive);

>