Example4.8 rev 1.mw

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)));

| (3)

The governing equation is further simplified here:

> | eq2:=expand(eq2*t);

| (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]);

> |