10.6 Heat conduction through composite Walls: Addition of Resistances
         Q=H*W*hN*(T[N]-Tb)
           Tb
       _____________ x=X[3], T=T[3] for N=3
      /   ks[3]    /
     /------------/|x=X[2], T=T[2]
    /   ks[2]    /||
  x/------------/ ||x=X[1], T=T[1]
 _/___ks[1]____/| ||x=X[0], T=T[0]
 ^|            || ||
 ||            || |/
 H|            || /
 ||            ||/
 _|____________|/
  |<---W------>|
       Ta
     Q=H*W*h0*(Ta-T[0])
>  restart;
>  N:=3; Number of sections. This may be set to any positive integer.
>  A:=W*H;  The area perpendicular to the direction of heat conduction
>  Q:= (x,k)-> -k*A*D(t)(x); Both k and x determine the heat flow rate
>  deq:=limit((Q(x+dx,k)-Q(x,k))/(k*A*dx),dx=0)=0;  The heat into any element at x must equal the amount that leaves at x+dx
>  X:=array(0...N); T:=array(0...N);  Defining X and T as vectors with N+1 elements in each one
>  s:=dsolve({deq,t(X[m])=T[m],t(X[m+1])=T[m+1]},t(x));  Solving the DE with BC at X[m] and X[m+1].
>  assign(s); t:=unapply(t(x),x);  Making t a function of x 
>  eqs:=array(0...N+1);ks:=array(1...N); Making eqs and ks vectors
>  Qv(m):=Q(x,ks[m+1]);  The heat transfer rate in each interval is a constant independent of x 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 N-1 do  We will set our equations so the LHS equals the temperature difference in it.
>   eqs[j+1]:=simplify((Qv(j)=Q0)*(X[j]-X[j+1])/(ks[j+1]*A));
>  od;  This specifies the equations for each of the intervals in the solid
>  eqs[0]:=T[0]-Ta=-Q0/(h0*A); eqs[N+1]:=Tb-T[N]=-Q0/(hN*A);  Now we will add the equations for each end where h0 and hN are heat transfer coefficients
>  eq:=simplify(sum(eqs[k],k=0...N+1));  Summing all the equations
>  Q0:=solve(eq,Q0);
>  U:=simplify(Q0/((Ta-Tb)*A));
>  InvU:=expand(1/U);  This looks a lot like the books answer
>  U;Ubook:=1/((1/h0)+sum((X[i]-X[i-1])/ks[i],i=1...N)+(1/hN));difference:=simplify(U-Ubook);  Here is a comparison of the expression we found for U to the formula given as eq. 10.6-16 in the text.