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?

  • $\begingroup$ have you googled? what did you find? $\endgroup$ – aaaaa says reinstate Monica Feb 22 '17 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 $\endgroup$ – Mike James Johnson Feb 22 '17 at 2:28

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

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

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