9.5 Heat Conduction with Chemical Heat Source: Part b

Solving the Dimensionless equations

>  restart;
>  eq1:=D(D(theta1))(Z)=B*D(theta1)(Z); eq. 9.5-23
>  eq2:=-(1/B)*D(D(theta2))(Z)+D(theta2)(Z)=N*theta2(Z);
eq. 9.5-24
>  eq3:=D(D(theta3))(Z)=B*D(theta3)(Z);; eq. 9.5-25
>  s:=dsolve(eq1,theta1(Z));
>  assign(s);
>  T1:=unapply(theta1(Z),Z);
>  _C1:=1;_C2:=c2;  Using BC 9.5-13 to get _C1
>  s:=dsolve(eq2,theta2(Z));
>  assign(s);
>  T2:=unapply(theta2(Z),Z);
>  _C3:=c4; _C4:=c3;
>  s:=dsolve(eq3,theta3(Z));
>  assign(s);
>  T3:=unapply(theta3(Z),Z);
>  _C6:=0;_C5:=c5;  Using BC 9.5-18
>  T1(Z); As seen in eq. 9.5-26 with c1 = 1 to satisfy BC 9.5-13
>  T2(Z); As in 9.5-27 with m3 and m4 as in 9.5-29 and 30.
>  T3(Z); As in eq. 9.5-28 with BC 9.5-18 used to eliminate c6.
>  eq1:=T1(0)=T2(0);  BC eq. 9.5-14
>  eq2:=D(T1)(0)=D(T2)(0); BC eq. 9.5-15
>  eq3:=T3(1)=T2(1); BC eq. 9.5-16
>  eq4:=D(T2)(1)=D(T3)(1); BC eq. 9.5-17
>  s:=solve({eq1,eq2,eq3,eq4},{c2,c3,c4,c5}); A rather long 
answer
>  assign(s);
>  B:=8;N:=2; Try one set of parameters: B is the same as in 
Fig. 9.5-2 and N=2 should give the top curve.
>  T1(z);
>  c2;
>  N:=1.99; Cut back on N a little
>  T2(z);
>  p1:=plot(T1(z),z=-.5....0): The colon keeps this plot from 
being shown.
>  p2:=plot(T2(z),z=0....1):
>  p3:=plot(T3(z),z=1...1.5):
>  with('plots'):
>  display([p1,p2,p3]); This agrees with the top curve (for N=2) 
in figure 9.5-2

>