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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).
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