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On the one hand, it is commonly said that thermoelectric effects are reversible. For Wikipedia they are thermodynamically reversible because as the factor of merit ZT approaches infinity, the efficiency of an engine working with thermoelectric effects approaches Carnot efficiency, i.e. a fully reversible engine. In practice though, ZT barely reach a value of 2 under reasonable conditions, with a lot of efforts.

On the other hand, thermoelectricity falls into the realm of non equilibrium irreversible thermodynamics, as Callen wrote in 1948 or Domenicali in 1954. There is indeed an entropy flux, as the charge carriers do transport entropy.

When I personally focus on time reversibility, I think any system displaying thermoelectric effects cannot be time reversible, because as soon as there is a temperature gradient, we know that heat flows from hot to cold. While if time is reversed, we would see the heat flux going the other way around, which is not physical.

How can we reconciliate the two points of view? Which one is correct?

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One cannot. First of all the figure of merit is never high because materials with desirable high electrical conductivity (metals) also have high thermal conductivity. And as you say, any solid will have irreversibilities like its thermal conductivity. There is no reversibility in a thermodynamic sense.

But in a course lab, it is easier to reverse an electrical current than to reverse the pump direction of the cooling fluid in a refrigerator or in a heat pump. Or diesel engines or steam turbines. Thermoelectrical devices are more easily analyzed, as it is possible to measure the parameters independently. So thermoelectrical devices are reversible, just by connecting the leads the other way around.

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  • $\begingroup$ First of all, thanks a lot for confirming what I thought, in your 1st paragraph. In your second paragraph, are you really sure the TE devices are reversible by connecting the leads the other way around? Say the n and p-legs are non homogeneous materials, that are doped differently according to the distance between the hot or cold plate. Then there is a Thomson effect solely due to the variation of S w.r.t. x (as well as S w.r.t. T too), the so called extrinsic Thomson effect. Then it seems to me that the TE device does not operate exactly the same if the current is reversed. $\endgroup$ – thermomagnetic condensed boson Nov 24 '18 at 14:16
  • $\begingroup$ I fail to see the relevance. Are you talking about the Joule-Thomson effect in ideal gases (en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect)? This has nothing to do with the thermoelectric Thomson effect. $\endgroup$ – thermomagnetic condensed boson Nov 24 '18 at 14:36
  • $\begingroup$ @Kentucker_Filled_Turkey I deleted my comment, would need to study more to answer that with confidence. A good answer is probably somewhere in this article and/or in the reference: mdpi.com/1099-4300/13/8/1481/htm $\endgroup$ – Pieter Nov 24 '18 at 15:15
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The thermoelectric effects are indeed thermodynamically reversible. You are absolutely right that they are always accompanied by irreversible effects like Fourier's conduction law and the Joule effect. However one can focus solely on the reversible part, which encompasses all the thermoelectric effects, and this part is thermodynamically reversible. More precisely, as you say there is a Thomson heat that is generated or absorbed. However this heat is equivalent to the heat one should supply along the sample for it to stay with an unperturbed temperature profile when both a current and a temperature gradient are simultaneously present in the sample. This heat exchange between the sample and its surrounding does not produce entropy, it is merely a transfer of entropy. So yes, as you say, there are entropy fluxes due to the thermoelectric effects, but these do not increase the entropy of the system "sample + surroundings". They can either increase or reduce (or keep constant) the entropy of the sample itself, but they always maintain the whole system's entropy constant. It is in this way that the thermoelectric effects are fully thermodynamically reversible.

I find the Wikipedia sentence involving the ZT factor misleading, for it does not help to get to know why TE effects are thermodynamically reversible. Your references are more valuable for that purpose.

More precisely the following explanation is what they (Wikipedia) have in mind: when $ZT = \sigma S^2 T / \kappa$ is worth infinity, what they have in mind is that both $\kappa$ and $\rho = 1/\sigma$ are worth $0$ (note that this is mathematically not a necessity). In that case there is no heat conduction via Fourier's law and there is no Joule resistance. Then they assume that such a system can still produce a finite current (different from superconductivity which has $S=0$) solely due to the reversible TE effects. When all of these impossible conditions are happening all at once, the TE material can be used to make a engine having the efficiency of a Carnot engine, i.e. the engine based on the TE effects is thermodynamically reversible, hence the TE effects are thermodynamically reversible. But as pointed out, logically it does not fully make sense, and not only because finding a material with $ZT$ equal to infinity is unreachable.

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