# Thermocouple-Heat pump-Power plant?

1. Jan 3, 2009

### Asok_Green

This isn’t what you think. I’m not out to ignore or challenge the laws of thermodynamics and I have no free energy fantasies. This is just a thought experiment for anyone willing to indulge a layperson. For those of you thinking, “Go educate yourself. Read a physics book and stop wasting my time,” you’re absolutely right; this is a selfish request. I suspect, though, that there are a few people who might occasionally enjoy explaining things like this to people like me. If that’s you and you’re still with me:

Suppose you’ve got an electrical power source and you use it to create heat (I’m under the impression that it doesn’t much matter how you use it to create heat; it will ultimately have the same efficiency. I don’t know this for a fact, though, so please tell me if I’m misinformed). Amount of electricity X yields amount of heat Y. This heat could, in turn, be harnessed in some way to create amount of electricity Z, but Z would always be less than X (in practice I’m guessing it would end up being a lot, lot less).

But how about if the power source is connected to a thermocouple? It would create a temperature gradient which, when averaged, should be about equal to Y, correct? (I’m really asking, here; I don’t know) But measuring the hot side alone should show an amount of heat greater than Y (I’m more confident about this). If this greater amount of heat were harnessed to create electricity, it should be greater than Z. Now here’s the question: Could it (even theoretically) be greater than X?

My instinct is no, it is not even theoretically possible, but I don’t have any knowledge of thermodynamics (my critical thinking isn’t so sharp, either), so I’m of no consolation to me. Obviously, if the concept would break conservation of energy, there’s not much point in giving it any further thought. But part of me, and again, what do I know? But part of me thinks that it needn’t break conservation of energy, because the extra thermal energy isn’t coming out of nowhere; it’s coming from the cold side of the thermocouple. Now, as soon as the cold side of the thermocouple had transferred most of its thermal energy to the hot side, the whole thing would have to stop working. But if you attached a large heat sink to the cold side of the thermocouple (possibly buried in the ground, depending on the scale of the design), perhaps it could passively gather enough ambient thermal energy to replace that which is being transferred to the hot side of the thermocouple? It would end up being a sort of low-temperature geothermal power plant, assuming the concept isn’t hopelessly flawed.

So, who’s up for explaining how the concept is hopelessly flawed?

Last edited: Jan 3, 2009
2. Jan 4, 2009

### Staff: Mentor

Correct.
A temperature gradient is not power, so no, it can't be equal to Y.

Energy/power in a heat pump is a little tricky because you're both using and generating heat (that's a simplification), which makes it seem like you've got more at the end than you had at the beginning. But you don't. Imagine as an analogy, pushing a rock up a hill. It takes a certain amount of energy to push a rock up a hill. If you double the size of the rock, the same amount of energy can get the rock only halfway up the hill. The hill is your temperature gradient and the rock is a bundle of heat energy you are trying to move.

So using a heat pump or other thermoelectric device, you can create as large a temperature gradient as you'd like, but that does not imply the ability to generate more power.

Last edited: Jan 4, 2009
3. Jan 4, 2009

### Asok_Green

Hmm. I may not know the right words to say it, but when I said that the temperature gradient averages out to something close to Y, what I meant was that the hot side of the thermocouple is made hotter by the same amount the cold side is made colder, so that seems to be a wash. And the energy used to move the heat also shows up as heat, so in the end it's just a heater, and (I believe) the amount of heat you end up with should be close to Y. That is, it should be no different to what you'd get if you used the same amount of energy and a resistor (Was I correct in this?).

So the hot side of the thermocouple should have a heat of Y, plus whatever ‘s been moved from the cold side (something less than Y, but nonzero). The amount of heat on the hot side of the thermocouple should, then, be greater than Y. What I’m curious about is, converted in to electricity, could it theoretically be greater than X?

I realize it bears a striking resemblance to an “overunity” (ugh) question, but I don’t believe it is one. I believe this is a “maximum theoretical efficiency of a thermocouple + maximum theoretical efficiency of a heat engine” question. If both were, for example, 70%, then I believe this would work. But if both were 60%, then it wouldn’t.

4. Jan 4, 2009

### Staff: Mentor

Perhaps the hot side gets 10F hotter and the cold side gets 10F colder. Is that what you're going for? Power is watts or BTU/h (energy is BTU or kWh). If you have 1 cup of boiling water and one gallon of boiling water, they are both at 212 F, but one took 16 times as much energy to bring to a boil - 16 times as much power if they took the same time.
Yes.
Actually, in a heat pump, the amount of heat energy (Y) output is greater than the amount of electrical energy input (X), which is where most people get confused. Again, you can use my example of rocks being carried up a hill - or better yet, a bunch of car batteries being carried up a hill. You can transport energy (car batteries) from one place to another for an output of more than the input without it being a violation of conservation of energy.
Ok, we're getting closer. Yes, a hypothetical thermoelectric (TEC) device might have a hot side that is 20 F above ambient and a cold side that is 10 F below ambient due to the fact that the electrical energy is also dissipated as heat.
And that's what I discussed before: no, it can't. That TEC that was able to generate a 30 F delta-T, say it took 10 W of power to do it. If it were exactly as efficient in power generation mode would produce 10 W of power.
You're on the right track. Because heat pump thermodynamic engines are basically the opposite of power generating thermodynamic engines, their efficiencies are the inverses of each other. Ie, a heat pump with a COP of 3:1 (3x as much heat out as in or 300%) can be used to power an engine with an efficiency of 1:3 (33%).

Going back to that boiling pots of water example, just because you have a certain temperature difference, that doesn't equate to a specific amount of power. Ie, a TEC that generates a 60F delta-T will not necessarily generate or consume twice as much power as one with a 30F delta-T. In reality, as the delta-T gets greater (for a heat pump), the efficiency goes down.

5. Jan 4, 2009

### Asok_Green

Ah, okay. I hadn't intended for a TEC to be used to generate the power. I was thinking more along the lines of a turbine, though maybe that's irrelevant, I don't know; I think it's clear we're moving beyond the limited scope of my understanding.

I'm curious, though: If the turbine were more efficient as a power generator than the TEC was as a heat pump, couldn't you have the situation of a TEC that uses 10 W of power to generate a 30 F delta-T, and a turbine that produces >10 W of power (assuming ambient heat is able to continuously replace the energy moved from the cold side of the TEC)?

Also, thanks a ton for your patience. I find this all very interesting.

6. Jan 4, 2009

### Staff: Mentor

It is irrelevant - some of the same laws that limit the efficiency of one limit the efficiency of the other.
Well that's why you're here, isn't it!?
With any kind of thermodynamic device, the efficiency depends on the delta-T. A steam engine works better if you use higher temperature steam (and therefore higher pressure). A TEC works better as a generator when it is given a higher delta-T. But with either, if you reverse it and use it as a heat pump, the efficiency drops when you increase the delta-T of the output. So what you end up with is that any attempt to increase the efficiency of the generator by increasing the delta-T reduces the efficiency of the heat pump. Those two principles cancel each other out, making it impossible to generate more electricity than you put in.

You might want to read up on heat pumps (particularly the part about efficiency and "lift"): http://en.wikipedia.org/wiki/Heat_pump

And Carnot's principle:
http://www.engineersedge.com/thermodynamics/carnots_principle.htm

I know the rules are similar for thermoelectric devices and their efficiencies, but I'm not very knowledgeable on the particulars of how such devices work. Perhaps someone else can chime in on how Carnot's principle relates to such devices.
You're welcome!