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

1. Dec 13, 2007

### minger

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Heat Condution Eq.
$$\newcommand{\pd}[3]{ \frac{ \partial^2{#1} }{ \partial {#2}^{#3} } } \pd{\theta}{x^2}{} - m^2\theta= {\frac{\partial \theta}{\partial t}$$

3. The attempt at a solution 1
The general solution is:
$$\theta(x) = A sinh(mx) + B cosh(mx)$$

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

$$\theta(x,y,t) = \theta(x,t) X \theta(y,t)$$

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 $$\theta(x=0) = \theta_b$$
And imposed temperature at x=0, we get the solution for the x-direction
$$\theta(x) = \theta_b cosh(mx) - \theta_b \frac{mk sinh(mL) + h cosh(mL)}{mk cosh(mL) + h sinh(mL)}sinh(mx)$$

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

However, the solution in the time direction is:

$$\Gamma(t) = e^{-\alpha \lambda^2_n t}$$

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:

$$\lambda_n tan(\lambda_n y) = \frac{h}{k}$$

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

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

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):

$$\frac{\partial^2 \overline{\theta}}{\partial x^2} - m^2 \overline{\theta} = \frac{p}{\alpha} \overline{\theta}$$

Grouping to get:

$$\frac{\partial \overline{\theta}}{\partial x^2} - (m^2 + \frac{p}{\alpha} \overline{\theta} = 0$$

Of which the general solution is:

$$\overline{\theta} = C_1 e^{\lambda x}$$

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

$$\overline{\theta} = \frac{\theta_b e^{-x \sqrt{\frac{p}{\alpha}+m^2}}}{p}$$

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 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).

Last edited: Dec 13, 2007
2. Dec 14, 2007

### HallsofIvy

Staff Emeritus
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
$$\overline{\theta} = \frac{\theta_b e^{-x \sqrt{\frac{p}{\alpha}+m^2}}}{p}$$.

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

Last edited: Dec 14, 2007
3. Dec 14, 2007

### 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:

$$\overline{\theta} = \theta_b \frac{e^{-x \sqrt{\frac{p}{\alpha}}}}{p}$$

equals

$$erfc(\frac{x}{2 \sqrt{\alpha t}})$$

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?

Last edited: Dec 14, 2007
4. Dec 18, 2007

### 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:

$$\theta = \theta_s + \theta_t$$

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:

$$\frac{\partial^2 \theta_t}{\partial x^2} - m^2 \theta_t = \frac{\partial \theta_t}{\partial t}$$

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

$$\theta(x=L) = \theta_b$$

Is now:

$$\theta_t(x=L) = \theta_b - theta_s = 0$$

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

$$\theta(t=0) = \theta_i$$

Is now:

$$\theta_t(t=0) = \theta_i - \theta_s$$

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:

$$\theta = \theta_t + \theta_s$$