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

>  restart;
>  Nol:=3;  Number of concentric layers

>  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...Nol); T:=array(0...Nol);  Defining R and T as 
vectors with Nol+1 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...Nol+1);ks:=array(1...Nol); 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 Nol-1 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[Nol+1]:=T[Nol]-Tb=
Q0/(h3*A(R[Nol]));  Now we will add the equations for each position 
where h0 and h3 are heat transfer coefficients
>  eq:=sum(eqs[k],k=0...Nol+1);  Summing all Nol+1 equations
>  Q0:=solve(eq,Q0);




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