Example 15.5-1 Heating of a Liquid in an Agitated Tank

>  restart;
>  eq1:=diff(Utot(t),t)=w*H1+Q(t); eq. 15.5-1
>  H1:=0;Utot:=t->rho*Cp*V(t)*(T(t)-T1); By use of simple 
reference conditions.
>  Q:=t->U0*A(t)*(Ts-T(t));
>  eq1; Identical to eq. 15.5-2 except the derivative of a product 
has been expanded.
>  V:=t->w*t/rho;A:=t->V(t)*A0/V0; Eq. 15.5-3,4
>  eq1; eq. 15.5-5 should be here.
>  tr:={T=T1+(Ts-T1)*Theta,t=rho*V0*tau/w};
>  U0:=w*Cp*N/A0;
>  with(PDEtools[dchange]);
>  eq2:=dchange(tr,eq1,[Theta(tau),tau]);
[Maple Math]
>  eq2:=simplify(eq2/(Ts-T1)/Cp/w); This checks with 15.5-9
[Maple Math]
>  s:=dsolve({eq2,Theta(0)=0},Theta(tau)):
>  assign(s); Theta:=unapply(Theta(tau),tau); This agrees 
with 15.5-13
[Maple Math]
>  limit(Theta(tau),tau=0);
[Maple Math]
>  N:=5;
>  plot(Theta(tau),tau=1e-5...2); This looks like Fig. 15.5-2, 
but note that eta is N*tau. In out case this is 5*tau, so as tau goes 
from 0 to 2 eta goes from 0 to 10.
[Maple Plot]
>  V0:=1000*ft^3;Lcoil:=10*4*ft*Pi;A0:=Lcoil*Pi*(1/12)*ft;
w:=20*60*lb/hr;U0:=100*Btu/(hr*ft^2*F);Cp:=Btu/(lb*F);
>  T1:=20*C;Ts:=105*C;
>  N:=evalf(U0*A0/w/Cp); This gives the value of N and agrees with 
the value stated at the end of the example.
[Maple Math]
>  T:=tau->T1+(Ts-T1)*Theta(tau);
>  T(1.);
[Maple Math]