thermoelectricity

n the backyard,

Is Peltier and thermopile the same thing?

backyard,

It's the same effect run backwards. Peltier is current->heat flow, Seebeck is heat flow->current. You can use a Peltier device as a really amazingly high output voltage thermocouple, for instance.

Cheers

Phil Hobbs

in the backyard,

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I had this idea once (yet to be tried) of using a Peltier as a heat flow sensor. As opposed to an other method which would be two temperature sensors and a known thermal resistance.

George H.

the backyard,

That sounds good.

I wonder if the thermal resistance would go down if you short the leads.

John

the backyard,

It's a pretty good method in principle, but bismuth telluride's figure of merit has a fairly strong temperature dependence, iirc, so your calibration table would need to have at least two dimensions and possibly three: I vs (T_hot, T_cold, Qdot). If you came up with a decent strategy, this might work fine. It might be something as simple as two reservoirs, one well insulated and the other driven by a second TE cooler--ramping the TE cooler up and down at various rates while looking at the insulated reservoir's temperature would get you I, Qdot, Thot, and Tcold. That would probably take overnight to calibrate, but then it would work pretty well--and it would be faster if Qdot is known to be small, so that Thot tracks Tcold closely, so you only need two dimensions. (Stuff the insulated reservoir fairly full of fine brass wool before putting in the water--that's almost as good as stirring and much much easier.)

The Peltier's many series-connected thermocouples would average out the thermal flux over the end plates, which would help accuracy in principle. You'd probably have trouble getting close to a Peltier's pancake geometry with the classical method, unless you did something like soldering copper and constantan disks face-to-face and using that as the heat transfer surface. (Copper and constantan both solder well with ordinary radio solder.)

Cheers

Phil Hobbs

out in the backyard,

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Ahhh.. not by much I wouldn't think. But it's not clear to me if one would want to measure the 'short circuit' current from the Peltier or the open circuit voltage.

George H.

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Thanks Phil, You're much quicker than I am. I was still thinking about what to measure and you're already on to a calibration setup. But having to measure the hot and cold side temperatures sorta defeats 'my' purpose of using the peltier. (which was to eliminate the two temperature sensors) (Hmm you may be able to get away with measuring T on only one side... the other could be infered from the other parameters.)

I like the copper/constantan/copper sandwich idea! (you might also put the copper in the middle.) I assume you were proposing to use the TC voltage between the two materials to determine the temperature on each side.

George H.

That would make sense. But as I understand the physics, the seeback effect is absolute -- constant voltage (= energy) at each junction, so current is limited by bulk resistivity only (Zo = R). Fortunately, resistivity is high enough that heat energy is conserved.

As a bidirectional effect, it should work just the same as a synchronous DC-DC converter, except it's DC-heat. So as you load it, you're making a back EMF that opposes the heat loss. And if you supply power to it instead, you're driving your own EMF that forces heat backwards.

It could very well be that, because efficiency is always so low, no one really bothered to determine the overall thermodynamic effect. Or it could be that I've never read about it. The best test would be a superconducting thermocouple: pure voltage, zero resistance. Is a superconducting bimetal loop also a super thermal conductor: whatever happens at one junction is followed exactly at the other?

Tim

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It gets complicated; at EQUILIBRIUM one gets a Seebeck voltage, but if you actually draw current, the charges are not kept in equilibrium but are pulled out through the external circuit, and you have to generate new charges before the Seebeck voltage recovers.

So, in semiconductors, there's a limit (because the charges are regenerated by the impurities at very few sites), and in superconductors there's a blockage (because a superconductor's charge carriers do not interact with the temperature of the material AT ALL). The electrons in a superconductor do not cause heat conduction, and will not reach the kind of equilibrium that a thermocouple requires in any reasonable amount of time. Superconductors are super-insulators of heat.

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So my picture of this is that the effect is different for different materials. In metals the heat energy is carried 'about' equally between electrons and phonons. (at least above the Debye temperature) For semiconductors the heat energy is carried mostly by the phonons.. and the charge carriers (holes or electrons) are carried along for the ride... which is what produces the heat generated electrical current or open circuit voltage. Electrically shorting a semiconductor Peltier device will allow the charge carriers to carry some of the heat load... but there are just not very many of them.... when compared to the number of phonons. (However I'm far from an expert and would love to be corrected.)

George H.