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Technical and scientific discussion of amps, cables and other topics.

Who's arguing? I'm being bashed for trying to explain.

I have tried, apparently with little success, to explain it..

I'll try again..

""In 1821, Estonian-German physicist Seebeck demonstrated the electrical potential in the juncture-points of two dissimilar metals when there is a heat difference between the joints. This was the thermoelectric effect and is known as the Seebeck Effect in Physics.""
(copied shamelessly from:
http://chem.ch.huji.ac.il/~eugeniik/history/seebeck.html

As can be seen from this link, the efficiency of the device is a function between the electrical conductivity of the junction and the thermal conductivity of the materials.
http://www.uni-konstanz.de/physik/Jaeckle/papers/thermopower/node3.html

This link provides a slightly more technical version, and explains carnot efficiency and the such..keep in mind...a dissimilar metal junction with small currents will have low temp gradient, hence the carnot efficiency will be practically zero..meaning almost no (but not zero) heat transfer. And little chance of recovering the energy that is lost from the peltier effect.

http://jchemed.chem.wisc.edu/Journal/Issues/1996/Oct/abs940.html

Here is the thermoelectric series:NOTE how far away nickel is from the other typicals copper, silver, gold..

Now, from this link:
http://www.bartleby.com/65/th/thermoel.html

The text states:Joule heating.....I R squared...while peltier..rate of heat transfer proportional to current ONLY..NOTE: it is not dependent on the voltage across the junction, so does not obey joule heating, hence not a linear resistor.

http://www.xyroth-enterprises.co.uk/thermser.htm

OK after reading that stuff....here goes.

Take a copper terminal, and plate it with nickel..

If you run current into it, there will be heat flow across the metal interface as a direct result of the peltier effect. It will be small, so it will not produce a large heat flow..But, according to Peltier, it will exist.

There will be a small heat gradient across the junction. Because it's small, the carnot efficiency will be very small..

Now, drop that current...(like an audio signal does twice frequency rate)..

The already established heat gradient will now produce electricity, (well, at least according to Seebeck...but what does he know??))

And again, the carnot efficiency is low..so it will return the heat energy very poorly.

In the meantime, while the current is going to zero, the thermal conductivity of the metals on each side of the junction are trying to re-establish thermal equilibrium, by moving heat opposite to what the peltier effect did.

Now...Look at what just happened..the peltier effect took energy from the current and setup a temp gradient..the seebeck effect took some back, but both are low efficiency, and thermal conductivity prevented all the heat gradient energy from being returned to the system as electricity. (just like two buckets of water with a small pipe connecting each at the bottom. If you pump the water into one bucket from the other, you are putting energy into the system. with no leak, you can use the water levels to regenerate the energy (assuming a reversible, 100% eff. pump) but, the leak reduces your efficiency, and actually sets up a time constant, where high speed back and forth regains the most energy back, while slow back and forth loses more.)

The discussion of thermal diffusivity involves the rate at which heat moves in a metal..high conductivity makes it faster, high heat capacity slows it down..Aluminum has high diffusivity because of it's low heat capacity.

For junctions with metals, the high diffusivity is quite bad for efficiency, the heat causes a thermal short circuit..that is one reason thermoelectric heaters use semiconductors, to reduce the thermal conduction between sides.

Nowhere have I said that there is a DC/thermal gradient effect going on..I have continued to state that the metal to metal junctions will be non-linear in their I/V curve..IE..not a linear resistor.

""If we can establish a hypothetical link between the Seeback/Peltier effects and what we are seeing on a spectrum analyzer, THEN it's worth pursuing this further and testing it. If not, what's the point of this discussion?""peter

I have known of one for a while now...

""so perhaps there is an opportunity for thermoelectric effects. You just haven't described any yet.""peter

I have done so three times here...perhaps this time sucessfully?

Cheers, John



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