# Efficiency of thermoelectric material and Carnot

Is Carnot efficiency also relevant for "open systems"?

Example - If we take a thermoelectric material with the hot side connected to a stable heat source and the cold side to an infinite heat sink (say the earth), this is not a closed system. Is Carnot efficiency still relevant? I thought It's basically relevant only for closed system but somehow the Carnot efficiency always pops back. Why?

• Why is it important here that the material be thermoelectric? Commented Jan 21, 2022 at 17:05
• Why do you say that the system is open? There is no net mass flowing in or out, is there? Commented Jan 21, 2022 at 20:02

A closed system in thermodynamics is one that does not exchange mass with its surroundings. If the thermoelectric material is the system and the surroundings are the heat source and heat sink, the thermoelectric material is a closed system since it does not exchange mass with its surroundings.

The efficiency $$\eta$$ of the thermoelectric device equals the energy provided to the electrical load divided by the heat absorbed at the hot junction. Since thermoelectric devices are heat engines, their maximum efficiency $$\eta$$ is theoretically limited to the Carnot efficiency of

$$\eta=\frac{T_{H}-T_{C}}{T_H}$$

For thermoelectric devices the actual maximum efficiency is a function is a fraction of the Carnot efficiency based on the devices "figure of merit". See devices efficiencies in the following: https://en.wikipedia.org/wiki/Thermoelectric_materials

Hope this helps.

• What you say is true, however their efficiency is even more limited than Carnot (see the thermoelectric figure of merit, where in the case of a perfect, albeit impossible to create material, zT equals infinity and only in that case does the efficiency reaches Carnot's). Commented Jan 22, 2022 at 15:09
• @AccidentalBismuthTransform Not sure what your point is. I never said how close to the Carnot limit the efficiency would be. Just that the theoretical limit for any heat engine is the Carnot limit and thus that would be the theoretical limit for the thermoelectric device. Commented Jan 22, 2022 at 15:11
• My point is that you're giving an upper bound (which is correct), but there is another lower upper bound for thermoelectrics. So, I wouldn't say "Just that the theoretical limit for any heat engine is the Carnot limit and thus that would be the theoretical limit for the thermoelectric device.". It is "a" limit, yes, but there are others, too, which are even more constrainable than Carnot's. Commented Jan 22, 2022 at 20:44
• @AccidentalBismuthTransform I am aware of that. Again, I never said how close it is, which as I understand it is based on the devices Figure of Merit, as described here: en.wikipedia.org/wiki/Thermoelectric_materials I have updated my answer to reflect that. I hope that satisfies you. Commented Jan 22, 2022 at 21:28
• It does. +1 from me. Commented Jan 23, 2022 at 11:36

The Carnot efficiency arises because entropy can't be destroyed and because work carries no entropy—but heat transfer does carry entropy. Thus, if we wish to extract work from a heat source (termed the hot reservoir), we have to dump the associated entropy somewhere, and we achieve this by heating something else (termed the cold reservoir). The energy we spend to do this reduces our efficiency below 100% even with the gentlest operation and the highest-quality engineering.

There's nothing incompatible with an open system here, and if our system is our heat engine, then we require an open system for these energy and entropy transfers from and to the hot and cold reservoir, respectively. Is this what you're asking about?

• Thank you Chemomechanics for a very lucid and helpful explanation. Commented Jan 22, 2022 at 19:31