# Steady State conduction of cylindrical pipeline

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?

• Did you make a sketch? I expect this is just an ill-phrased axi-symmetric (1D) situation. Commented Dec 29, 2020 at 8:42
• 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.) Commented Dec 29, 2020 at 8:48
• What is in the space between the cable and the protecting pipe? Does the protecting pipe have a high thermal conductivity? Commented Dec 29, 2020 at 13:12
• @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. Commented Dec 29, 2020 at 13:13
• @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. Commented Dec 29, 2020 at 15:06

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