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I would like to derive an expression for the thermal conduction resistance through a square pipe.

For example, consider a cylindrical pipe, the conduction resistance can easily be obtained from Fourier's law in polar coordinates:

$$q=-kA\frac{dT}{dr}$$ where we integrate from $r_{i}$ to $r_{o}$ and $T_{i}$ to $T_{o}$. After integration, we end up getting the expression: $$q=2\pi Lk\frac{T_i-T_o}{ln(\frac{r_o}{r_i})}$$ Where we take $\frac{ln(\frac{r_o}{r_i})}{2\pi Lk}$ as the conductivity resistance through a cylindrical pipe wall of thickness $L$.

So what about a pipe with a square cross section with wall thickness $L$? How can this be done using integration?

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  • $\begingroup$ have you googled? what did you find? $\endgroup$ Feb 22, 2017 at 2:23
  • $\begingroup$ nothing. I think what I am going to do is consider a single (trapezoidal) face, and where the area that heat passes through increases as you go further towards the outer surface. Integrate like that, then multiply by 4 (for each edge). Either that or find an "effective" radius of the square and use the above formula $\endgroup$ Feb 22, 2017 at 2:28

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Notice the symmetries of the square. Symmetry lines are shown dashed. Due to symmetry, there will be no heat transfer across these lines. Which means that you can just take the hatched region with adiabatic boundary condition applied along lines of symmetry. This should be easy to solve by integration. enter image description here

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  • $\begingroup$ @MikeJamesJohnson Could you find a closed form expression for temperature field? $\endgroup$
    – Deep
    Feb 24, 2017 at 4:03
  • $\begingroup$ what do you mean by closed form expression? $\endgroup$ Feb 25, 2017 at 17:37
  • $\begingroup$ @MikeJamesJohnson I mean an exact solution. $\endgroup$
    – Deep
    Feb 26, 2017 at 5:11

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