9.6 Heat conduction through composite Walls: Addition of Resistances

>  restart;
>  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...3); T:=array(0...3);  Defining X and T as 
vectors with 4 elements in each one
>  X[1];T[2];  This is the way to access specific elements
>  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...4);ks:=array(1...3); 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 2 do  We will set our equations so the LHS 
equals the temperature difference in it.
>   eqs[j+1]:=(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[4]:=Tb-T[3]=-Q0/(h3*A);  
Now we will add the equations for each end where h0 and h3 are heat 
transfer coefficients
>  eq:=sum(eqs[k],k=0...4);  Summing all five 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...3)+(1/h3));
difference:=simplify(U-Ubook);  Here is a comparison of the 
expression we found for U to the formula given as eq. 9.6-17 in the 
text.