# Material selection of heating element

I am interested in heating coils which are heated using Ohmic/Joule/resistive heating - specifically, how the material for the heating coil is selected.

When I read about heating coil materials - nichrome, carbon fibre etc, the material quality that is mentioned most is electrical resistance (along with a high melting point and stability with changes in temperature). It is my understanding that a high resistance material is typically selected to make heating coils, but how does the thermal conductivity of a material impact the suitability of a material for use as a heating coil?

If two materials had comparable resistance but very different values for thermal conductivity - which material would be the better choice, assuming the goal was a heating coil that heated up quickly using the least amount of electrical power?

## 3 Answers

Consider that ohmic heaters are usually made of thin wire, and that the majority of the current flow going through any wire is close to the surface of the wire. Both these things mean that even if the wire itself might be a poor conductor of heat, that heat has only a very short distance to travel before it reaches the wire surface and gets carried away by conduction, convection and radiation. Because of this we wouldn't expect the wire's thermal conductivity to make much difference.

Even if it did, consider the following case where we assume the hot wire is quite nonconductive thermally. This means that the heat generated in the wire can't get out easily. This means the wire gets hotter and hotter with time, until it begins to glow with heat. At that point, the primary heat transfer mode becomes radiative and the heat escapes anyway.

1. For maximum power efficiency, you want to match the load impedance (i.e., the heating element resistance) to the source impedance (i.e., the resistance of everything else in the circuit, including the power supply). This is the maximum power transfer theorem. (Volumetric heat generation is $$J^2\rho$$, where $$J$$ is the current density and $$\rho$$ is the resistivity. With a lower load resistance, you're not taking advantage of the circuit current. With a higher load resistance you're dragging down the circuit current.) Choose your material (and its electrical resistivity) and its geometry (cross-sectional area $$A$$ and length $$L$$) accordingly, considering that the linear resistance is $$R=\rho L/A$$.

2. The presence of thermal conduction means that the temperature profile of the heating element will be parabola- or catenary-shaped to some extent. To suppress this profile and obtain a fairly uniform temperature, choose a material with a high thermal conductivity, maximize the cross-sectional area, and/or minimize the heating element length. In addition, choose connectors with poor thermal conductivity to avoid bleeding away heat.

• The maximum power transfer theorem is often irrelevant here. For example, the impedance of a house 240V mains supply is probably of the order of 0.1 ohms, and the max power theorem would only be relevant if the output current from the supply was thousands of amps - but that the breakers would trip long before the current reached that value.. – alephzero Aug 14 '20 at 2:44
• What is "here"? The question asks specifically about power transfer maximization, so my answer is a general one. I don't disagree with you if one is designing a household toaster, but the context might be a microfabricated element, a spacecraft, military gear, a drone, a deep-ocean submersible, etc. – Chemomechanics Aug 14 '20 at 6:18

Not really much to choose from : platinum , nichrome ( with modest variations like 0.5 % Al) and for some high temperatures ,silicon carbide . Platinum tends to disappear on weekends when a facility is shut down . So the practical choice is nichrome whatever the conductivity. The decision is wire diameter and coil configuration.