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I have a hollow cylinder with a heat source inside. The generated heat flow is $\dot{Q}_g = 14\ 962 \ \text{W}$ , the cylinder has the radius $r = 20 \ \text{mm}$, the length $l = 50 \ \text{mm}$ and the thickness $t = 5 \ \text{mm}$. The inside temperature is $T_1 = 1300 \ \text{K}$. If the material is for example stainless steel with a conductivity of $\lambda = 30 \ \text{W/(m * K)} $ I get a reasonable outside wall temperature $T_2$:

$T_2 = T_1 - \dot{Q}_g \cdot \frac{\text{ln}(\frac{r + t}{r})}{2 \pi \cdot \lambda \cdot l} = 946 \ \text{K} $

But if I have a material with a low conductivity like $\lambda = 1 \ \text{W/(m * K)}$, the same equations gets me a negative outside wall temperature:

$T_2 = -9328 \ \text{K} $

That doesn't make sense to me. I think I made some stupid mistake, so is there another equation for this case or did I made a false assumption? Or something else? I already checked the units, and it looks all right. The unit for the conductivity is watt per meter kelvin, not milli kelvin. This looked wrong in the first version, sorry.

I would be really happy, if somebody could help me with this. Thanks!

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  • $\begingroup$ Can you show your calculations? I get $T_2$ of 1289 K for the low conductivity. Also, look out that the units relate properly, you sometimes have K, sometimes mK, also sometimes mm. Figuring out the units is a good way to verify results. $\endgroup$
    – OpticalResonator
    Aug 12, 2020 at 17:19
  • $\begingroup$ Ah, right, thermodynamics is not my field, but W/m*K makes more sense indeed :) $\endgroup$
    – OpticalResonator
    Aug 13, 2020 at 11:20
  • $\begingroup$ 15 kW is a very high energy output for such a small volume. Not quite nuclear reactor levels, but still high. I think the calculated temperature is "correct", at least mathematically. In other words, the ~10,000$K$ temperature difference is what would be required to dissipate this much power over such a small area. The fact that a negative temperature is non-physical is not part of the heat conduction equation. $\endgroup$
    – Carlton
    Aug 18, 2020 at 19:02

2 Answers 2

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The results are telling you that—according to this model—there is no outer temperature that you can apply to obtain an inner temperature as low as 1300 K at steady state for this power output, geometry, and material thermal conductivity (1 W/m-K). If you were to apply ~0 K, this model would predict an inner temperature of 9328+1300=10628 K (at which point you would have to evaluate the model's usefulness).

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Many years since I took heat transfer, but it seems to me you do not get to decide the inside temperature. It will depend on the heat generation. Also the outside pipe temperature will determine the convective heat transfer, for steady state they need to equal out.

The trick on the units is length is .o5 meters but stated as 50 mm. The radius and thickness are in a ratio so you can use mm or meters to get same result.

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