Example 4.8 Heat Conduction in a Rectangular Slab
Example 4.1 is solved here with the boundary and initial conditions switched.[4] The governing equation is
=α
u(x,0) = 0 (4.1.10)
(0,t) = 1 and u(∞,t) is defined
The following transformation is used to combined the variable:[7]
η=
The variable n in the new coordinate η is represented by U. The governing equation (ordinary differential equation) for U is obtained by converting the time and special derivative in equation (4.1.10) (partial differential equation) to derivatives in the η coordinate. The boundary conditions for U are:
U(0) = 1 (4.1.12)
U(∞) = 0 (4.1.13)
The governing equation for U is then solved with the above boundary conditions to obtain the final solution. Example 4.8 is solved in Maple below:
| > | restart: |
The with(student) package is called to facilitate variable transformations:
| > | with(student): |
The governing equation is entered here:
| > | eq:=diff(u(x,t),t)-alpha*diff(u(x,t),x$2); |
| (1) |
First, u(x,t) is transformed to U(η(x,t)):
| > | eq1:=changevar(u(x,t)=U(eta(x,t)),eq); |
| (2) |
The transformation for η is substituted here:
| > | eq2:=expand(simplify(subs(eta(x,t)=x/2/(alpha*t)^(1/2),eq1))); |
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(3) |
The governing equation is further simplified here:
| > | eq2:=expand(eq2*t); |
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(4) |
| > | eq2:=subs(x=eta*2*(alpha*t)^(1/2),eq2); |
| (5) |
| > | eq2:=convert(eq2,diff); |
| (6) |
The final form for the governing equation is:
| > | eq2:=expand(-2*eq2); |
| (7) |
Enter the boundary condition here:
| > | bc1:=U(0)=1; |
| (8) |
| > | bc2:=U(infinity)=0; |
| (9) |
U is solved as:
| > | U:=rhs(dsolve({eq2,bc1,bc2},U(eta))); |
| (10) |
| > | U:=convert(U,erfc); |
| (11) |
Next, us as a function of x and 1 is obtained as:
| > | u:=subs(eta=x/2/(alpha*t)^(1/2),U); |
| (12) |
The solution obtained can be plotted:
| > | plot3d(subs(alpha=0.001,u),x=1..0,t=500..0,axes=boxed,title="Figure Exp. 4.1.17.",labels=[x,t,"u"],orientation=[-60,60]); |
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| > |