# Thread: Electricity from ambient heat

1. ## Electricity from ambient heat

I was always interested in the idea that you could generate electricity from ambient heat. Such a thing would not violate either conservation of energy or momentum and thus should be possible. Please don't bring up the second law of thermodynamics - it is a *tendency* not a certainty. Please refer to http://en.wikipedia.org/wiki/Second_...thermodynamics if you don't believe me.

Anyways, I've drawn up an equation as a starting point in the thought process. The model in my head is a two-chamber device where a heat pump creates a gradient where one chamber is the hot one and the other is the cold one. The model assumes that the energy put into the pump is exhausted into the hot chamber (to increase that chambers temperature, and thus the heat gradient).

To keep things simple, I assume there is no outside environment. Only the two chambers exist and heat cannot be transfered out of both chambers.

The equation I came up with calculates at what efficiency the electric generation must occur in order for a net gain in useful energy to happen.

Eff = A/(2x + A)
where:
* A is the heat-energy added to the hot-chamber via powering the heat pump
* x is the heat-energy moved from the cold-chamber to the hot-chamber (2x is the difference in heat-energy), and
* Eff is the heat to electricity conversion efficiency needed for the system to break even.

This comes from the equation: gain = (2x+A)*Eff, where gain is the amount of electricity generated.

Now, since this kind of thing is generally regarded as impossible, I'm wondering what is wrong with my assumptions, or generally why this scheme doesn't work.

2. Please don't bring up the second law of thermodynamics - it is a *tendency* not a certainty. Please refer to http://en.wikipedia.org/wiki/Second_...thermodynamics if you don't believe me.
Hi Frencheneesz,
This might be a problem, because the answer to your problem lies in thermodynamics. Can you explain why you think that the 2nd law is not a certainty?

Anyway, to investigate your interesting idea properly, you will want to read up on thermodynamic cycles. You'll find that the maximum possible efficiency of a heat engine is less than 100%, and depends directly on the difference in temperature between the hold and cold reservoirs:

$Efficiency <= 1 - Cold/Hot$
(Cold and Hot are the respective reservoir temperatures in kelvins)

The performance of a heat pump is similarly limited:

$Q/W <= Hot/(Hot-Cold)$
Where Q is the heat transferred and W is the energy consumed.

I'm no expert on thermodynamics, so that's as far as I can go... but I think I've given you enough to be able to figure out whether or not you can get the required efficiency.

3. Also, with no 'environment', and heat not being transferred to/from it, you don't have an engine.

An engine can't be a closed system, or it can't do anything useful. How do you get heat into it, for one thing?
An engine inputs heat, and outputs work (so the heat has to be replaced from somewhere).

4. The system includes an energy source to drive the heat pump, and the generated electricity.

5. Vkothii: pete's got it right. The point of my question is that all the heat neccessary is inside the two chambers (which are at equal temperatures before the heat pump gets to work). The chambers are the environment, it is a closed system mind experiment.

"Can you explain why you think that the 2nd law is not a certainty?"

Of course. Did you read the wikipedia article I linked? Entropy is a very misunderstood measurement, and people end up saying things like "entropy always decreases". That statement, however, is false. Net entropy does not always increase. It simply has a *tendency* to increase in nature. It is an empiracle law.

"the entropy of an isolated system which is not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium."

On a microscopic scale, equilibrium really never happens. There are always small fluctuations throughout any ambient system. Thus the second law of thermodynamics is a macroscopic generalization, and not a law that holds at all times.

This link is related to that:
http://www.newscientist.com/article.ns?id=dn2572

Anyways. The law of thermodynamics can't be used to calculate how close my thought experiment can get to working - even if it could be used to calculate whether it can work or not. In other words, the law of thermodynamics is not a fundamental law, but is based on other laws - or based on simple observation.

I'll take a look at those canot efficiency and heat pump efficiency equations later.. they require some calculus to evaluate in this context...

6. Frencheneesz, you are right about the second law of thermodynamics. It's not a law at all, just a tendency. You are also "on to something" with your heat pump, and Maxwell came up with a very similar idea. Read about Maxwell's Demon, and the decades of debate that developed. The "bad" news is that it has been proven that you will always be required to put more energy in than you can pull out, but the "good" news is that you have concocted a thought experiment that is on par with a certified Physics genius...!

http://en.wikipedia.org/wiki/Maxwell%27s_demon

In a very similar setup, I've myself always wondered if you could generate energy by storing pressure from molecular movement with a "one-way gate" separating two tanks. This happens in biological processes such as osmosis...

7. I have an old Apple //e program, in which it does an education animation of that Maxwell's demon idea, a little guy with a tennis racket, holding it up to deflect a slow molecule (or vice versa) and lowering it to let pass a fast molecule. After a while one tank naturally becomes hot, and the other cold, setting up the needed temperature differential to power up a turbine to do work. All you need then, is some reverse refrigerant, with the right temperature properties, so that it boils on the hot side, and condenses on the cold side. Then the resultant gas pressure can power up a turbine. Problem is, supposedly, Maxwell's Demon uses energy just to make the decision and move his tennis racket, about the energy of a refrigerator, so no "free lunch" or "perpetual motion machine" here.

I agree that the 2nd Law of Thermodynamics, is probably just a general observation, and not really an absolute law, so someday, I may yet see my imagine car with the "cold" radiator, sucking up FREE waste heat out the ambient air, to provide the power to zip along down the freeway. Frosting up problems, I imagine could be solved, by dividing the radiator into thirds, cycling which third is in defrost mode, and designing it to avoid too much rain blowing into it?

Thermal batteries would self-recharge, tending to "sweat" or attract condensate, any time too much power is withdrawn from them, as by "conservation of energy," higher power draw would necessitate an increased cooling effect upon the batteries.

Thermal batteries, or nuclear batteries, would be way cool. Imagine buying a digital watch, that never will need another battery. Imagine calculators, that still power up, 50 years after you bought it. Imagine cars, never needing refueling. Imagine a cellphone, that never runs dead, nor needs recharging, nor a charger. Imagine power lawn mowers, tools, computers, cordless phone bases, with no power cord and no gasoline tank. Don't touch that part of the lawn mower. It could be very cold? That frost there, might be colder than a freezer?

But I have no idea how to produce materials with such absolute thermal absorbing properties. But then, people didn't know so much about superconductors, until what? Maybe decades ago?

8. Originally Posted by Pete
The system includes an energy source to drive the heat pump, and the generated electricity.
If it includes an energy source, it's "self-contained" then? How do you get energy out of a closed system?

That's not a problem??

9. I always find these threads highly amusing.

What none of these people seem to realize is that others have spent tremendous amounts of time and effort attempting to find a 'loophole' in the 2nd law. Some have even dedicated the major part of their lives and resources in the effort - all to NO avail.

On a macroscopic scale - which is all that matters for any practical purpose - it simply cannot be done. Entropy will always win in the end.

A major flaw in all their plans is that they always assume no external environment and completely disregard the ordinary losses in ANY energy conversion process AND simple friction. Those things cannot be avoided, yet the dreamers dream on and every new generation produces yet another batch of foolish people who will try yet again.

10. Originally Posted by Pete
Hi Frencheneesz,
This might be a problem, because the answer to your problem lies in thermodynamics. Can you explain why you think that the 2nd law is not a certainty?

Anyway, to investigate your interesting idea properly, you will want to read up on thermodynamic cycles. You'll find that the maximum possible efficiency of a heat engine is less than 100%, and depends directly on the difference in temperature between the hold and cold reservoirs:

$Efficiency <= 1 - Cold/Hot$
(Cold and Hot are the respective reservoir temperatures in kelvins)

The performance of a heat pump is similarly limited:

$Q/W <= Hot/(Hot-Cold)$
Where Q is the heat transferred and W is the energy consumed.

I'm no expert on thermodynamics, so that's as far as I can go... but I think I've given you enough to be able to figure out whether or not you can get the required efficiency.

Not hard at all considering the two efficiencies are best case reciprocal so the total Eff <= 1.

11. Originally Posted by Frencheneesz
Did you read the wikipedia article I linked? Entropy is a very misunderstood measurement, and people end up saying things like "entropy always decreases". That statement, however, is false. Net entropy does not always increase. It simply has a *tendency* to increase in nature. It is an empiracle law.

"the entropy of an isolated system which is not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium."
Yes, I read the article, but I don't think you did... or if you did, you didn't understand it. You're reading way to much into the word "tend" in the first sentence.

Read a little further into the article, and you'll find:

The second law can be stated in various succinct ways, including:
• It is impossible to produce work in the surroundings using a cyclic process connected to a single heat reservoir (Kelvin, 1851).
• It is impossible to carry out a cyclic process using an engine connected to two heat reservoirs that will have as its only effect the transfer of a quantity of heat from the low-temperature reservoir to the high-temperature reservoir (Clausius, 1854).
• If thermodynamic work is to be done at a finite rate, free energy must be expended.[5]

Immediately after that, you'll find the mathematical formulations.

On a microscopic scale, equilibrium really never happens. There are always small fluctuations throughout any ambient system. Thus the second law of thermodynamics is a macroscopic generalization, and not a law that holds at all times.
This link is related to that:
http://www.newscientist.com/article.ns?id=dn2572
From the article:
...the second law can also be consistently broken at micron scale, over time periods of up to two seconds.

Your system is neither microscopic nor short term. It is quite definitely bound by the 2nd law of thermodynamics.
The Law of Large Numbers is relevant.

Anyways. The law of thermodynamics can't be used to calculate how close my thought experiment can get to working - even if it could be used to calculate whether it can work or not. In other words, the law of thermodynamics is not a fundamental law, but is based on other laws - or based on simple observation.
If you take the time to learn a little more about thermodynamics than whoever has been misleading you knows, then you'll find that it contains everything you need to address your experiment.

Good luck!

12. Entropy can decrease, tho it is statistically unlikely. Do you agree pete?

Ok, so instead of doing calculus I'll assume the machine is held constant (at constant temperature).

I'll chose two temperatures and try the calculations for them:

* hot = 308 K
* cold = 273 K

My original equation:
minEff = A/(2x + A)
where:
* A is the heat-energy added to the hot-chamber via powering the heat pump
* x is the heat-energy moved from the cold-chamber to the hot-chamber (2x is the difference in heat-energy), and
* minEff is the minimum heat to electricity conversion efficiency needed for the system to break even.

A = x/Cp
where
* Cp is the coefficient of performance for the heat pump

This leads to a slightly modified equation:
minEff = 1/(2 Cp + 1)

The maximum efficiency for the carnot cycle is 1 - cold/hot. This is 11.36% for this example.
The maximum coefficient of performance for a heat pump is hot/(hot - cold). This is 8.8 for this example.

If we use the maximum coefficient of performance the equation is:
1/(2*8.8 + 1) = .0537634 = 5.37634% minimum break-even efficiency.

This gives almost a 6% leeway between the theoretical maximum efficiency, and the minimum neccessary break even efficiency.

These equations indicate that what I have proposed is possible. Does anyone see a flaw in my logic or my math?

13. Interestingly, using these equations, the leeway between the theoretical maximum efficiency and the minimum neccessary break even *increases* as the cold temperature decreases *and* as the temperature difference increases. However, if you look at the minimum efficiency as a percentage of the maximum theoretical efficiency, the opposite is true: the percentage *decreases* in the above conditions. This indicates that as heat is taken from the system, it becomes harder to continue taking energy from the system. This is exactly as one would expect. Comments?

14. Originally Posted by Frencheneesz
Entropy can decrease, tho it is statistically unlikely. Do you agree pete?

Ok, so instead of doing calculus I'll assume the machine is held constant (at constant temperature).

I'll chose two temperatures and try the calculations for them:

* hot = 308 K
* cold = 273 K

My original equation:
minEff = A/(2x + A)
where:
* A is the heat-energy added to the hot-chamber via powering the heat pump
* x is the heat-energy moved from the cold-chamber to the hot-chamber (2x is the difference in heat-energy), and
* minEff is the minimum heat to electricity conversion efficiency needed for the system to break even.

A = x/Cp
where
* Cp is the coefficient of performance for the heat pump

This leads to a slightly modified equation:
minEff = 1/(2 Cp + 1)

The maximum efficiency for the carnot cycle is 1 - cold/hot. This is 11.36% for this example.
The maximum coefficient of performance for a heat pump is hot/(hot - cold). This is 8.8 for this example.

If we use the maximum coefficient of performance the equation is:
1/(2*8.8 + 1) = .0537634 = 5.37634% minimum break-even efficiency.

This gives almost a 6% leeway between the theoretical maximum efficiency, and the minimum neccessary break even efficiency.

These equations indicate that what I have proposed is possible. Does anyone see a flaw in my logic or my math?
You are STILL making the same fatal assumptions as all the "perpetual motion crowd" by ignoring the very real considerations of friction and power-loss in energy conversions.

If anyone is foolish enough to exclude those factors, there are MANY devices that could produce the same results you ar getting.

It's a total waste of time and imagination.

15. "[I]by ignoring the very real considerations of friction and power-loss[\I]"

I really don't understand why you say that. I am explicitly talking about efficiencies that have to do with power loss. Let me break it down for you Read-Only:

Heat-pump pumps heat from cold to hot - energy loss contributes to the heat on the hot side as a result of a non-infinite Cp (coefficient of performance).

The heat gradient is then used to create electricity - the efficiency of this process also creates extremely significant heat loss.

The point of my equations is to not only show that this is possible, but determine how easy it is. In fact, the "leeway" I calculated is how far away from ideal this machine can be and still produce net energy. If you say I ignored power-loss considerations, you are mistaken.

Let me use more realistic numbers in those equations for you:
* hot = 35 degrees C = 308 K
* cold = 10 degrees C = 283 K

* Cp = 7.2
:source GSHP ground at 10°C (http://en.wikipedia.org/wiki/Heat_pump#Efficiency)

* minEff = 1/(2 Cp + 1) = 0.0649351 ~= 6.5%

Instead of using the carnot efficiency, I'll use the less ideal endoreversible process measure which is:
* maxEff = 1-sqrt(cold/hot) = 0.0566352 ~= 5.6%

The leeway then is -0.829984% . So these equations using more or less real numbers indicate that producing usable electricity from this system would be very difficult. However, the ideal maximums still suggest it may be possible.

To all those who would rather yell at someone pursuing sound solid science - think about helping them discover reasons they might be wrong on their own and perhaps expanding your own mind to discover things you've never known.

For example, did you know that an ambient-heat electricity generator was theoretically possible before this thread?

16. Originally Posted by Frencheneesz
"[I]by ignoring the very real considerations of friction and power-loss[\I]"

I really don't understand why you say that. I am explicitly talking about efficiencies that have to do with power loss. Let me break it down for you Read-Only:

Heat-pump pumps heat from cold to hot - energy loss contributes to the heat on the hot side as a result of a non-infinite Cp (coefficient of performance).

The heat gradient is then used to create electricity - the efficiency of this process also creates extremely significant heat loss.

The point of my equations is to not only show that this is possible, but determine how easy it is. In fact, the "leeway" I calculated is how far away from ideal this machine can be and still produce net energy. If you say I ignored power-loss considerations, you are mistaken.

Let me use more realistic numbers in those equations for you:
* hot = 35 degrees C = 308 K
* cold = 10 degrees C = 283 K

* Cp = 7.2
:source GSHP ground at 10°C (http://en.wikipedia.org/wiki/Heat_pump#Efficiency)

* minEff = 1/(2 Cp + 1) = 0.0649351 ~= 6.5%

Instead of using the carnot efficiency, I'll use the less ideal endoreversible process measure which is:
* maxEff = 1-sqrt(cold/hot) = 0.0566352 ~= 5.6%

The leeway then is -0.829984% . So these equations using more or less real numbers indicate that producing usable electricity from this system would be very difficult. However, the ideal maximums still suggest it may be possible.

To all those who would rather yell at someone pursuing sound solid science - think about helping them discover reasons they might be wrong on their own and perhaps expanding your own mind to discover things you've never known.

For example, did you know that an ambient-heat electricity generator was theoretically possible before this thread?
And let me break it down for you. Your assumption that you can ignore/isolate such a system from the environment is also silly and foolish.

No, I didn't know it was possible - and I STILL know it it isn't possible.

I'm not "yelling" at you and I have no problem with you wasting your life and resources chasing after this Maxwell's Demon just as so many thousands before you have done. So go right ahead - build your silly device and go insane trying to make it work. That will be nothing new, either. The 2nd law will defeat you just as it has done the others.

17. Originally Posted by Frencheneesz
Entropy can decrease, tho it is statistically unlikely. Do you agree pete?
Entropy can decrease over (very) short time scales. The larger the system, the shorter the time scale. It will always increase in the long term.

The likelihood of a decrease in entropy for even a millionth of a second for an engine of any appreciable size is extraordinarily low. Do you really want to design an engine that has maybe a 1 in several trillion chance of working for a tiny fraction of a second?

Ok, so instead of doing calculus I'll assume the machine is held constant (at constant temperature).
I don't think that will work... that would mean that the total energy added to the system is the same as the total energy extracted, but you want to get more energy out than you put in, don't you?
If this is to work, then the average temperature of the system must decrease over time.

You are STILL making the same fatal assumptions as all the "perpetual motion crowd" by ignoring the very real considerations of friction and power-loss in energy conversions.

If anyone is foolish enough to exclude those factors, there are MANY devices that could produce the same results you are getting.

It's a total waste of time and imagination.
No, what Frencheneesz is talking about is not practical limitations, but theoretical ones.

The laws of thermodynamics clearly say that there is no device that could extract net useful energy from ambient heat, even if did you have such things as perfect frictionless generators and perfect insulators.

19. Originally Posted by Pete
No, what Frencheneesz is talking about is not practical limitations, but theoretical ones.

The laws of thermodynamics clearly say that there is no device that could extract net useful energy from ambient heat, even if did you have such things as perfect frictionless generators and perfect insulators.
Theory is fine, but there are occasions - like this one - where "theoretical" is just a synonym for "nonsense."

Entropy always wins, fools always loose. (That should be inscribed in stone somewhere.)

20. Well, I think that "theoretical" is not a synonym for "nonsense" in this case... what Frencheneesz wants to do is not even theoretically possible. But the discussion of why it is not possible is worthwhile and interesting (to me, at least!)

Page 1 of 10 12345678910 Last

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•