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Conceptionally there are things I'm not sure of in heat transfer, which I hope you can help me with.

Say for instance we have a long electrical wire with a uniform temperature distribution (higher temperature than ambient air). Because of the temperature difference, there will be a constant heat loss per unit length (W/m). Now in case, you insulate this wire, what do you actually do? The same heat has to escape? So as I understand, the only thing you do is slowing down the heat initially (depending on the thermal conduction in the material) and increasing the surface area so the heat loss per unit area (W/m^2) will decrease, but in steady state will the heat loss per unit length not be the same?

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The wire will heat up and reach a constant higher temperature.

Assuming that heat is not escaping from the ends of the wire, then all the heat generated inside the wire must escape; thus, the temperature of the wire will rise until the heat generated by the wire is equal to the rate at which heat is escaping. The temperature of the wire will then be constant but with a larger value than before.

However, assuming that the wire is a metal, the rise in temperature will increase the resistance of the wire. If the current through the wire is kept constant, then the rate at which heat is generated (i.e., $I^2R$) will be larger than before. If the voltage across the wire is kept constant, then the rate at which heat is generated (i.e., $\frac{V^2}{R}$) will be smaller than before.

So the actual rise in temperature of the wire will be determined by a number of factors, e.g., the insulation thickness, radius of the wire, etc., but also the characteristics of the power supply.

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  • $\begingroup$ Also worth noting that if you're using the cylinder to actually transport heat (like say moving hot water through a building), the heat insulation is extremely helpful, because you want to maximize the temperature in the pipe and minimize the temperature on the outside of the insulation surface. $\endgroup$ – JMac Oct 29 '17 at 15:28

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