Solving 2D Heat Conductivity Problem - Where to Start?

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Discussion Overview

The discussion revolves around solving a 2D steady-state heat conductivity problem using finite element methods (FEM). Participants explore various approaches to find exact solutions, boundary conditions, and the implications of different configurations in the context of heat conduction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks guidance on finding an exact solution to a 2D heat conductivity problem, expressing a desire to compare it with FEM approximations.
  • Another participant suggests starting with a 1D problem due to the inquirer's limited background in physics.
  • It is noted that the steady-state heat equation does not have a single solution; various boundary conditions lead to different solutions.
  • A specific 2D configuration is proposed involving a square with varying temperature gradients along its sides, leading to a temperature profile T(x,y) = x + y.
  • Participants discuss the need to verify that the proposed temperature profile satisfies Laplace's equation and the associated boundary conditions.
  • One participant describes their own attempt at solving a 1D problem and expresses confusion about extending that to 2D.
  • Clarifications are made regarding the boundary conditions of the problem, with a focus on how they influence the temperature distribution.
  • Participants explore the implications of boundary conditions on the derived equations, with one participant expressing surprise that the y-coordinate does not influence the final temperature at certain nodes.
  • There is a discussion about deriving the temperature equation systematically versus through inspection, with one participant sharing their approach to integrating Laplace's equation.

Areas of Agreement / Disagreement

Participants generally agree on the complexity of the problem and the importance of boundary conditions, but multiple competing views and approaches remain regarding how to solve the 2D heat conductivity problem and the implications of different configurations.

Contextual Notes

Participants express varying levels of familiarity with differential equations and Laplace's equation, indicating a potential gap in foundational knowledge that may affect their understanding of the problem.

Who May Find This Useful

Individuals interested in finite element methods, heat conduction problems, and those looking to understand the application of Laplace's equation in 2D contexts may find this discussion beneficial.

dinaharchery
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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:
 
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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:
 
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 \nabla^2T=0). 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).
 
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:
 
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?
 
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?
 
dinaharchery said:
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?

Yes, exactly. And the solution also needs to solve all the boundary conditions: T(x,0), T(x,L), T(0,y), and T(L,y) if the temperatures are specified at the edges, \partial T(x,0)/\partial y 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.
 
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.
 
If the sides aren't constrained and there's negligible convection in the third dimension and no heat generation, this is a 1D problem: T(x,y)=100x/L.

EDIT: Nope, please see below.
 
Last edited:
  • #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:
 
  • #11
Whoops, I must have looked at your earlier post too fast; make that T(x,y)=9(100-x)+100. Does that make more sense?
 
  • #12
Whoops, I must have looked at your earlier post too fast; make that T(x,y)=9(100-x)+100. Does that make more sense?

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.
 
  • #13
Have you googled "Laplace's equation"? That's where I'd start.
 
  • #14
Thanks Mapes.

You have been very helpful, I really appreciate it.
 
  • #15
Any time!
 
  • #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.
 
  • #17
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 T(x=0)=1000 and T(x=100)=100.

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

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