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I have been taught that the heat resistance of a hollow cylinder is

R(hollow cylinder) is proportional to ln(outer radius/inner radius)

It was assumed in the derivation that the inner surface is at some temperature t and has some radius r. Steady State and no heat generation also follows. The outer surface is at some temperature T and radius R.

All I am trying to do here is take the limit as r goes to zero.

Why does a complete cylinder have infinite heat resistance?

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  • $\begingroup$ This question is too unprecise. Where is the heat source? where did you get this formula, what other resistance do you mean? $\endgroup$
    – trula
    Commented Aug 22, 2022 at 16:42
  • $\begingroup$ My answer here discusses why electrical resistances between "point electrodes"have infinite resistance. It may address your question. $\endgroup$ Commented Aug 22, 2022 at 17:24

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I have been taught that the heat resistance of a hollow sphere is: R(hollow sphere) is proportional to ln(outer radius/inner radius)

Well, this isn't correct; as derived in many textbooks (e.g., Incropera & DeWitt) and noted in many handbooks, the thermal resistance for steady-state conduction through a hollow sphere with uniform temperatures on the inside and outside surfaces (of radii $r_1$ and $r_2$, respectively) is

$$R_\text{sphere}=\frac{1}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r_2}\right),$$

where $k$ is the thermal conductivity.

(The thermal resistance of a cylinder includes a $\ln(r_2/r_1)$ term; perhaps you were thinking of that?)

In any case, we find the inner radius $r_1$ appearing in the denominator for both geometries. How is it that the thermal resistance of these objects (with radial symmetry) can increase to infinity as the internal cavity shrinks?

Because the area $A$ over which the temperature is applied drops to zero. Therefore, as mediated by Fourier's law of conduction $q=-kA\frac{dT}{dx}$, the temperature difference $dT$ required to achieve a given heat flux $q$ tends to infinity; this is equivalent to an infinite thermal resistance.

The situation is broadly similar to trying to heat up an object by heat conduction through a safety pin, say, touching the object's surface. The kinetics of the process are limited by the very small cross section of the pin, and you'd need an enormous temperature at the other end to push through appreciable heat.

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  • $\begingroup$ Mb. It was supposed be a cylinder. But anyways your argument explains it all. Thank you. $\endgroup$
    – Xp_Candy
    Commented Aug 24, 2022 at 8:02
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Heat resistance is very much like electrical resistance. Imagine a long thin electrical resistor. We know that the more we reduce the cross sectional area of a resistor the greater the resistance to electrical current becomes. If we pinch the middle of the resistor until the cross sectional area goes to zero, the resistance goes to infinite as we are effectively putting a break in the circuit.

In the case of heat resistance in the wall of a cylinder, the surface area of the inner wall goes to zero as the the radius goes to zero and so the heat resistance goes to infinite.

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