1D/2D Transient Heat Conduction (Diff. Eqs)

[minger]

Homework Statement

My problem is not necessarily the 2D problem, it's getting the answer in one dimension (I have the y-direction). The problem is the boundary conditions...anyways.

Solve for the temperature of a 1D transient metal bar, with the following boundary conditions. Imposed temperature at the left boundary. Convection all else around.

Homework Equations

Heat Condution Eq.
[tex]
\newcommand{\pd}[3]{ \frac{ \partial^2{#1} }{ \partial {#2}^{#3} } }

\pd{\theta}{x^2}{} - m^2\theta= {\frac{\partial \theta}{\partial t}
[/tex]3. The Attempt at a Solution 1
The general solution is:
[tex]
\theta(x) = A sinh(mx) + B cosh(mx)
[/tex]

We know that the solution is the product of the individual directions.

[tex]\theta(x,y,t) = \theta(x,t) X \theta(y,t) [/tex]

We separate the variables and solve for X(x) and T(t) and Y(y) and T(t) separately:

Starting with theta(x,t)

By imposing convection at x=L [tex] \theta(x=0) = \theta_b [/tex]
And imposed temperature at x=0, we get the solution for the x-direction
[tex] \theta(x) = \theta_b cosh(mx) - \theta_b \frac{mk sinh(mL) + h cosh(mL)}{mk cosh(mL) + h sinh(mL)}sinh(mx)[/tex]

This is the steady 1-D fin solution for those boundary condition.

However, the solution in the time direction is:

[tex] \Gamma(t) = e^{-\alpha \lambda^2_n t} [/tex]

Where lambda_n are the characteristic values. I have no idea where to get these. They are somewhat easy in the y-direction. In the y-direction (homogeneous), we get:

[tex] \lambda_n tan(\lambda_n y) = \frac{h}{k} [/tex]

Giving a final solution in the y,t direction as:

[tex] \theta(y,t) = 2 \theta_i \sum_{n=0}^\infty e^{-\alpha \lambda^2 t} \frac{sin(\lambda_n y) cos(\lambda_n y)}{\lambda_n B + sin(\lambda_n B) cos(\lambda_n B)} [/tex]

But I was thinking that the y-direction is only homogeneous if theta = theta_infinity (the freestream temperature), which it will not be. So..I have an answer to one equation which is more than likely wrong. And the other equation I don't know how to get lambda values out of.

3. The Attempt at a Solution 2
Second attempt by using Laplace Transformations was suggested by the prof, since he couldn't figure it out either. In the x-direction only, we transform the equation into (where p's are like the normal s's):

[tex] \frac{\partial^2 \overline{\theta}}{\partial x^2} - m^2 \overline{\theta} = \frac{p}{\alpha} \overline{\theta}[/tex]

Grouping to get:

[tex] \frac{\partial \overline{\theta}}{\partial x^2} - (m^2 + \frac{p}{\alpha} \overline{\theta} = 0 [/tex]

Of which the general solution is:

[tex] \overline{\theta} = C_1 e^{\lambda x} [/tex]

Where lambda is the root, and imposing the boundary condition at x=0 giving:

[tex] \overline{\theta} = \frac{\theta_b e^{-x \sqrt{\frac{p}{\alpha}+m^2}}}{p}
[/tex]And...I cannot for the life of me, find the inverse laplace of that. Also, I never imposed the convection boundary condition...where was that supposed to go?

Basically, this one problem is stopping me from finishing this last project. If I cannot get this, then that stops me from getting:
1D Transient
2D Steady-State
2D Transient

All of which need the damn x solution. Thanks in advance, I greatly appreciate the help.

edit: I've tried both Matlab and Maple to get the inverse laplace of that function. Neither could do it (don't ask about Mathematica, our university just got rid of it).

 

[HallsofIvy]

What do you mean by "convection all around"? Exactly what is that boundary condition?

You say you are looking for the inverse Laplace transform of
[tex] \overline{\theta} = \frac{\theta_b e^{-x \sqrt{\frac{p}{\alpha}+m^2}}}{p}[/tex].

The inverse Laplace transfrom of [itex]e^{a x}[/itex] is [itex]1/(x- a)[/itex]. I'm surprised you couldn't find that!

 

[minger]

The boundary conditions are convection at y=-+B (assuming a height of 2B, and the x-axis running through the middle of the bar), and convection at x=L; imposed temperature at x=0.

Also, I do not believe that you can use the shifting theorem because in this case a, is a function of the transformed variable p. The inverse of:

[tex] \overline{\theta} = \theta_b \frac{e^{-x \sqrt{\frac{p}{\alpha}}}}{p} [/tex]

equals

[tex] erfc(\frac{x}{2 \sqrt{\alpha t}}) [/tex]

Which is close to what I would expect. But also, using Laplace transformations, where was I supposed to impose the x=L convection boundary condition?

 

[minger]

For the record here is the correct process. The problem is that neither direction had homoegeneous boundary conditions. So, what you do is solve the final answer as:

[tex] \theta = \theta_s + \theta_t [/tex]

Basically, the summation of a steady-state part and a transient part. Since we have already found the steady state part, we move to the transient. The equation becomes:

[tex] \frac{\partial^2 \theta_t}{\partial x^2} - m^2 \theta_t = \frac{\partial \theta_t}{\partial t} [/tex]

The reason to do this is that now, the boundary conditions must also be adjusted. Where before the imposed boundary condition was:

[tex] \theta(x=L) = \theta_b [/tex]

Is now:

[tex] \theta_t(x=L) = \theta_b - theta_s = 0 [/tex]

The unfortunate side-effect of this is that before the boundary condition in time was simply:

[tex] \theta(t=0) = \theta_i [/tex]

Is now:

[tex] \theta_t(t=0) = \theta_i - \theta_s [/tex]

Where the steady-state portion needs to be the entire answer. This makes the last constant very difficult to find, but at least you're farther.

Then the final answer is simply:

[tex] \theta = \theta_t + \theta_s [/tex]