Conduction - Steady State Triangular Element

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SUMMARY

The forum discussion focuses on solving the temperature distribution T(x,y) for a right-isosceles triangular cross-section under steady-state heat conduction with constant thermal conductivity, k. The bottom boundary is maintained at temperature T2, while the left and diagonal sides are held at temperature T1. The challenge arises from the diagonal side not being insulated, complicating the application of symmetry and superposition methods previously used in similar problems.

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minger
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Homework Statement


Obtain the temperature distribution T(x,y) of the triangular cross section, assuming constant thermal conductivity, k.

The triangle is a right-isosceles triangle with the right angle at (0,0). The triangle goes "up" to a, and to the right to a, then diagonally across. Imagine a square split from top left to bottom right.

Most importantly the bottom is held at T2, and the left and diagonal sides are held at T1.


Homework Equations


The relevant equation would be the general heat conduction equation.


The Attempt at a Solution


We had a similar problem where the diagonal was insulated. This let you actually pretend you had a square where the top was at T1 also, and the right at T2. Then you used a symmetry line and used the principle of super-position to solve the problem. You would first define a variable theta = T - T2. Then you have homogenous boundary conditions on the bottom and right. Then superimposed solutions for the top theta = T1 - T2 and the left theta = T1 - T2.

The problem with this one is that you cannot do this, since the diagonal is not insulated, so it's not a symmetry line, and I have no idea how to start.
 
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I am working on the exact problem you discribe where the diagonal is insulated. I under stand the method I need to use is superposition by modeling the triangle as a square. I am confussed on what the symmetry boundary condition is down the center? How do I model that there is no delta T on the diagonal line that is in terms of x and y?
 

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