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2D Heat Conductivity

  1. May 9, 2010 #1
    Please Help,

    I am not an mechanical engineer but am trying to learn Finite Element Methods to solve a 2D Steady State Heat Conductivity problem.

    Can anyone point me to an exact solution to the 2D Heat conductivity problem? Please be gentle, I have not had a physics class since 8th grade although I have taken calculus. My idea is to solve a specific 2D model analytically and then compare this exact solution to a solution that I will generate using FEM approximation equation.

    Thanks.:smile:
     
  2. jcsd
  3. May 9, 2010 #2
    Given your limited background, I would highly suggest going through a 1D problem. If that it something you are interested in, I would be happy to help. Let me know. :smile:
     
  4. May 9, 2010 #3

    Mapes

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    There is no single solution to the steady-state heat equation (which, for uniform thermal conductivity and no heat generation, is just Laplace's equation [itex]\nabla^2T=0[/itex]). Various boundary conditions will result in different solutions.

    As Saladsamurai suggested, 1D is a good way to get your feet wet. The equation is identical, but the solution is always a line (or, if heat generation exists, a parabola).
     
  5. May 10, 2010 #4
    Thanks for the reply's.

    I took your suggestions and did the 1D Heat Conduction problem. I finished it and it looks correct. Of course there is a lot of information on the web on the 1D problem, but the 2D is another issue.:confused:
     
  6. May 10, 2010 #5

    Mapes

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    I was thinking about this a little bit more, and I think the first nontrivial 2D configuration to try would be a square; hold two adjacent sides at a temperature gradient from 0 to 0.5 (i.e., each side goes linearly from 0 at the common corner to 0.5 at the ends) and the other two sides held from 0.5 to 1 (so that the corner opposite from 0 is held at 1). Then you can verify that one diagonal is a uniform 0.5 and the other is a linear gradient from 0 to 1. The temperature profile is simply T(x,y) = x + y, which can be verified to solve Laplace's equation. Know what I mean?
     
  7. May 10, 2010 #6
    Thank you.

    I think I understand what you are saying - the temperature profile T(x, y) = x + y is "plugged-into" the Laplace equation and solved? To solve the Laplace equation I will need to take the second order partial derivatives of T(x, y)?

    Is this correct?
     
  8. May 10, 2010 #7

    Mapes

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    Yes, exactly. And the solution also needs to solve all the boundary conditions: [itex]T(x,0)[/itex], [itex]T(x,L)[/itex], [itex]T(0,y)[/itex], and [itex]T(L,y)[/itex] if the temperatures are specified at the edges, [itex]\partial T(x,0)/\partial y[/itex] and so on if the heat fluxes are specified at the edges, a combination of these, or conditions set within the domain. That's why no single, simple solution exists.
     
  9. May 10, 2010 #8
    Cool, I am glad I understood something.

    The 2D model that I am trying to solve consists of the a square structure with the temperature T(0) = 1000 and T(L) = 100 where L = 100. I think this should be relatively simple but I am a bit rusty on differential equations.

    Thank you again.
     
  10. May 10, 2010 #9

    Mapes

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    If the sides aren't constrained and there's negligible convection in the third dimension and no heat generation, this is a 1D problem: [itex]T(x,y)=100x/L[/itex].

    EDIT: Nope, please see below.
     
    Last edited: May 10, 2010
  11. May 10, 2010 #10
    Sorry but I should have specified the boundaries better.

    Temperature at Node 1 (x=0, y=0) is 1000, Temperature at Node 4 (x=0, y=100) is 1000, Temperature at Node 2 (x=100, y=0) is 100, Temperature at Node 3 (x=100, y=100) is 100.

    Does this make any difference?

    Also can you please show me how you arrived at the equation T(x, y) = 100x/L?

    Thank you for all your help.:smile:
     
  12. May 10, 2010 #11

    Mapes

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    Whoops, I must have looked at your earlier post too fast; make that [itex]T(x,y)=9(100-x)+100[/itex]. Does that make more sense?
     
  13. May 10, 2010 #12
    The results make more sense, thank you:smile:. I just seems strange that the y coordinates have no influence on the final temperature at a given node.

    Can you provide any links to explain this kind of equation? I am really trying to learn this stuff.
     
  14. May 10, 2010 #13

    Mapes

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    Have you googled "Laplace's equation"? That's where I'd start.
     
  15. May 11, 2010 #14
    Thanks Mapes.

    You have been very helpful, I really appreciate it.
     
  16. May 11, 2010 #15

    Mapes

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    Any time!
     
  17. May 11, 2010 #16
    Mapes,

    Once again, thank you. I derived the linear equation from the original T(x, y) = x + y and got T(x, y) = -9(100 - x) + 1000. This equation gives the same results as the equation you derived but I must have arrived at it in a different fashion.

    Please excuse my ignorance with Laplace equations:confused:, but as I said I am trying to learn. Can you please give me the details at how you arrived at your equation?

    Thank you.
     
  18. May 11, 2010 #17

    Mapes

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    Actually, I just did it by inspection. I knew y wouldn't appear, as there was nothing in the boundary conditions to cause heat flux in the y direction. (Any temperature change in a certain direction implies heat transfer in that direction; that's the physical interpretation of the conduction equation.) Then I just played around until I had the equation for a line with [itex]T(x=0)=1000[/itex] and [itex]T(x=100)=100[/itex].

    But to do it systematically, you'd integrate Laplace's equation in 1D, [itex]\nabla^2 T=d^2T/dx^2=0[/itex], to get [itex]T(x)=C_1x+C_2[/itex], then find [itex]C_1[/itex] and [itex]C_2[/itex] by plugging in the boundary conditions. If this is what you did, you're in great shape.
     
  19. May 13, 2010 #18
    Mapes,

    Yes, that is exactly what I did. I just wanted to be sure I was correct.

    Again, a million thanks for all your help.
     
  20. May 13, 2010 #19

    Mapes

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    You're welcome!
     
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