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I have this problem involving conduction of heat:

Suppose that $z = 0$ represents the ground level on a street where an electric cable is buried at $x = 0$ and $z = −D$. The ground is kept at a fixed temperature $u = 0$ and the cable is releasing heat (Joule effect) at $Q$ units of energy per unit time and unit length of the cable. The cable lies inside a protecting pipe of radius $d$ ($0 < d < D$). Calculate the steady-state temperature $u=u_0$ of the protecting pipe. We suppose $u_0$ = constant, the conductivity of the pipe is supposed to be high. The result will depend on the thermal conductivity $k$ of the ground.

I've studied some 1-D case with a uniform exterior temperature, but in this case being a 3-D problem with a boundary (at $z=0$) where the temperature is fixed I do not think that it is the same way of procedure. Am I right? Does the steady-state temperature of the protecting pipe is constant all over their area?

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  • $\begingroup$ Did you make a sketch? I expect this is just an ill-phrased axi-symmetric (1D) situation. $\endgroup$
    – Bernhard
    Dec 29, 2020 at 8:42
  • $\begingroup$ I feel like if we sketch as a 1-D case, considering that the cylinder becomes the interval [-d,d] (diameter of the cylinder), we will have the boundary at D and we know there the temperature is 0 degrees, but it is not clear to me how to model then the temperature at d and -d, since the temperature is not uniform for all the other space (for example, in -D the temperature might be way different since there is not a source there.) $\endgroup$
    – Alejandro
    Dec 29, 2020 at 8:48
  • $\begingroup$ What is in the space between the cable and the protecting pipe? Does the protecting pipe have a high thermal conductivity? $\endgroup$ Dec 29, 2020 at 13:12
  • $\begingroup$ @Alejandro What does it mean that the cable is buried at $x=0$ and $z=-D$? What is the orientation of the cable? Is it pointing towards $x$, $y$ or $z$? Also, why is the radius called $d$, that is a bit uncommon too. $\endgroup$
    – Bernhard
    Dec 29, 2020 at 13:13
  • $\begingroup$ @Bernhard it means that the cable is buried at a distance -D, and I assume that the cable is parallel to the y axis, so it remains at x=0 all the time, and always at a distance -D. $\endgroup$
    – Alejandro
    Dec 29, 2020 at 15:06

1 Answer 1

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I have an approximate method of solving this, based on the method of images. Treat the cable as a line source of heat flow Q (per unit length) at z = -D, and place an image line sink of heat flow -Q at z = +D. (Of course, the medium would be extended upward to infinite z). These boundary conditions will guarantee a temperature u = 0 at z = 0. Then solve for the average temperature around the circle of radius d surrounding the line source. It won't be exact (because the temperature at the pipe won't be exactly constant), but you can see how much the temperature varies circumferentially, and judge whether it is not too excessive. Certainly, this method will be very accurate at large values of D/d.

Based on this analysis, I get $$u_0=\frac{Q}{2\pi k}\ln{(2D/d)}$$

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  • $\begingroup$ I think your answer is good! Do you have any document that provides the derivation of the formula? $\endgroup$
    – Alejandro
    Dec 30, 2020 at 14:47
  • $\begingroup$ Sorry. I know of no document that derives this result. I derived it on my own. We are not provide complete solutions to the questioner. We are only supposed to provide hints to help the questioner solve the problem on their own. I think I provided enough information to help you proceed with a solution. $\endgroup$ Dec 30, 2020 at 15:36

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