Unravelling the Mystery of Free Energy

[colin9876]

free energy??

water disasotiates to H+ and 0H-
if you could take the dissasociated ions away, and combine them how much energy could you get out?
In the liquid remaining, more would dissosciate, and the process could be continued.

Where does this energy come from - it sounds like free energy which can't be the case?

 

[kateman]

it takes energy to break the bonds and gives energy out to make them. therefore it isn't "free energy".

dont have an energy table on me but if you look up exothermic reactions on google, you should find what you are after

 

[colin9876]

It takes energy but where does the energy for this dissociation come from.
Im wondering if it comes from the heat withing the liquid - if so possibly removing the ions would allow more to remain dissociated and possibly lower the temp of the liquid

 

[russ_watters]

It takes energy but where does the energy for this dissociation come from.

The usual way is to apply an electric current to the water to disassociate it to H2 and O2.

Im wondering if it comes from the heat withing the liquid - if so possibly removing the ions would allow more to remain dissociated and possibly lower the temp of the liquid

No, it has to be external. Conservation of energy applies.

 

[Andy Resnick]

water disasotiates to H+ and 0H-
if you could take the dissasociated ions away, and combine them how much energy could you get out?
In the liquid remaining, more would dissosciate, and the process could be continued.

Where does this energy come from - it sounds like free energy which can't be the case?

The energy you get out, from the ions re-combining, is equal (at best) to the energy required to separate them in the first place.

Water normally has 10^-7 mol H+ and 10^-7 mol OH-. (If that's the correct unit.. molar concentration is what I'm going for) But, those ions don't preferentially separate in space unless you (the experimenter) apply an electric field. Once you put some energy into the water to separate the charges, you can get that energy back by allowing the charges to move.

 

[Sam Lee]

I was just wondering, if we have an infrared to electric converter, that is it converts infrared energy to electric energy, then we can get "free energy". Everything around us radiates infrared radiation. We will never run out of infrared radiation. So we can keep converting the infrared energy to other forms of useful energy.

 

[kmarinas86]

I was just wondering, if we have an infrared to electric converter, that is it converts infrared energy to electric energy, then we can get "free energy". Everything around us radiates infrared radiation. We will never run out of infrared radiation. So we can keep converting the infrared energy to other forms of useful energy.

Waste of time.

Solar.

Wind.

Geothermal.

Wave.

Installation costs are not free.

Operating costs are not free.

Some human labor and oversight required.

 

[kateman]

I was just wondering, if we have an infrared to electric converter, that is it converts infrared energy to electric energy, then we can get "free energy". Everything around us radiates infrared radiation. We will never run out of infrared radiation. So we can keep converting the infrared energy to other forms of useful energy.

yes, and it is the future of solar cells
you could probably find somthing on google under "spray on solar cell"
it could be worn on clothes to recharge batteries, put on houses, cars, pretty much anywhere that is exposed to sunlight (this is theoretically speaking of course). It may mean hydrogen cars that never need to be recharged/refuled, etc.

it would be most useful for the army, who could use their equipment and recharge it anywhere during the day, even in the clowdy days.

one day - but it would be expensive. take 100 years or so to refine i am guessing.
may not be that much point in the end though, depends. but might as well move forward

 

[cesiumfrog]

I was just wondering, if we have an infrared to electric converter, that is it converts infrared energy to electric energy, then we can get "free energy". Everything around us radiates infrared radiation. We will never run out of infrared radiation. So we can keep converting the infrared energy to other forms of useful energy.

The laws of thermodynamics say no, which is the well known reason why we can't power ships by the abundant (local) thermal energy of the ocean (leaving wakes of ice). It is why refrigerators consume electricity. Your "infrared converter" will emit just as much black-body radiation as it absorbs.

Solar panels only work because most of the sky is dark (and hence the panel itself stays relatively cold); if the sun encompassed the whole sky then we would come into thermal equilibrium with the sunlight, and despite there being "more energy present" it would be entropically impossible to harness any of it to do work.

 

[kateman]

The laws of thermodynamics say no, which is the well known reason why we can't power ships by the abundant (local) thermal energy of the ocean (leaving wakes of ice). It is why refrigerators consume electricity. Your "infrared converter" will emit just as much black-body radiation as it absorbs.

Solar panels only work because most of the sky is dark (and hence the panel itself stays relatively cold); if the sun encompassed the whole sky then we would come into thermal equilibrium with the sunlight, and despite there being "more energy present" it would be entropically impossible to harness any of it to do work.

If you say its impossible, why then do I know of several universities that are trying to pursue it?

http://news.nationalgeographic.com/news/2005/01/0114_050114_solarplastic.html

 

[Creator]

I was just wondering, if we have an infrared to electric converter, that is it converts infrared energy to electric energy, then we can get "free energy". Everything around us radiates infrared radiation. We will never run out of infrared radiation. So we can keep converting the infrared energy to other forms of useful energy.

Even though IR is 'all around us', the intensity is extremely low. The issue is intensity when trying to usefully harness it. That is why , as kateman has pointed out, it is the solar IR that is being exploited...since its intensity is high. ...However, the main issue is still cost per watt.
Can the material be economically produced? If so it will be a great boost to solar 'cell' efficiency.

 

[cesiumfrog]

If you say its impossible, why then do I know of ...

It is impossible for ambient temperature devices to extract work from the black-body radiations of everything else that is at the same ambient temperature. This is not to say that it isn't possible for ambient temperature devices to extract work from any given components of the black-body radiations of something that is above ambient temperature (particularly the IR component of sunlight). I doubt you will find many scientific journal papers seriously advocating the extraction of work from the IR emissions of things that are just slightly different to ambient temperature (like people, or machines, or the night sky) because (since the available energy and Carnot efficiency are just so low) it is far more effective to harness sunlight. Learn thermodynamics and be critical of what you read.

 

[russ_watters]

Even if you could extract energy from a small delta-T via infrared radiation, the energy density is so low it wouldn't be worth it. Solar panels get probably a thousand times more energy from the sun's spectrum and their energy per unit area is still too low to be economical. Why buy a square meter solar panel that produces 1 watt of power?

 

[kateman]

very interesting, thank you

 

[russ_watters]

Oops, I forgot - the device you guys are describing already exists. http://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator

They are used to power space probes, which radiate against the black coldness of space (using nuclear decay to generate the heat). People are also trying to sell them to paste onto truck exhaust pipes. The efficiency is too low to be worth the money for even that application (where you do have a decent delta-T), though in that application, they capture the heat by conduction and dissipate it by convection.

The efficiency is still pretty low - perhaps if it gets better it could become viable for higher delta-T waste-heat generators.

 

[Sam Lee]

It is impossible for ambient temperature devices to extract work from the black-body radiations of everything else that is at the same ambient temperature. This is not to say that it isn't possible for ambient temperature devices to extract work from any given components of the black-body radiations of something that is above ambient temperature (particularly the IR component of sunlight). I doubt you will find many scientific journal papers seriously advocating the extraction of work from the IR emissions of things that are just slightly different to ambient temperature (like people, or machines, or the night sky) because (since the available energy and Carnot efficiency are just so low) it is far more effective to harness sunlight. Learn thermodynamics and be critical of what you read.

Pardon my ignorance,
Is a photovoltaic cell an ambient temperature device?
Is photoelectric effect an ambient temperature effect?
If they are not, then they maybe we can extract work from black-body radiations.

The problem with harnessing sunlight energy is that sometimes there is no sunlight!
Whereas infrared is in abundance and always there.
So even if the efficiency is restricted by Carnot efficiency, we can still achieve 1 to 2% efficiency. And this 1 to 2% of a large amount of infrared radiation can translate into a very high W.

 

[russ_watters]

Is a photovoltaic cell an ambient temperature device?

They run hotter than ambient due to the absorption of infrared.

Is photoelectric effect an ambient temperature effect? If they are not, then they maybe we can extract work from black-body radiations.

That question has no meaning that I can see. If you are asking if a photovotaic cell will be at ambient temperature if there is no sunlight, the answer is yes. As a result, there is very little incident radiation of any kind.

The problem with harnessing sunlight energy is that sometimes there is no sunlight!
Whereas infrared is in abundance and always there.

Ever-present, yes. Abundant, no. There is more energy to be had in capturing the ambient lighting in a room than there is in capturing the radiant heat from the people in it.

So even if the efficiency is restricted by Carnot efficiency, we can still achieve 1 to 2% efficiency. And this 1 to 2% of a large amount of infrared radiation can translate into a very high W.

There is not "a large amount of infrared radiation".

A human being dissipates about 70W of energy at rest. Of that, about 60% is in the form of radiation for a naked person in a dark room. Capturing those 42 watts would require wrapping your body in about 2 square meters of collector, for an energy density of 21 Watts per square meter. Incident radiation from the sun is over 1000 W/m^2.

 

[Sam Lee]

I think infrared radiation is currently underestimated.
I read somewhere that half of the sun radiation is infrared radiation.

Furthermore, if a body dissipates 70W, then 10 bodies will dissipate 700W.
If we focus the energy, like using a lens, we can increase the Watts per square meter.
Using a fresnel lens, we can focus 1 m2 into 1 cm2. Thats 10000 times more powerful!

So even with 21 W/m2, it becomes 21 x 10000 W/m2!

 

[Dale]

Hi Sam,

Infrared radiation is not underestimated. You seem to think that the physical principles involved are somehow not well understood, but that is not the case this is very well understood.

Have a look at the http://en.wikipedia.org/wiki/Stefan-Boltzmann_law" . Even if you had a perfectly effecient IR converter that was somehow magically kept at absolute zero, the incident IR energy is less than 0.5 kW/m² even on a hot summer day. Now, let's say that instead of absolute zero we are only going to keep your IR absorber 10º cooler than ambient and let's say that you had a really good 50% efficiency, that brings you down to about 15 W/m² even on a hot summer day.

As any homeowner can tell you, it takes a lot more than 15 W/m² to cool something down 10º below ambient. There is just nothing to be gained this way, TANSTAAFL.

 

[russ_watters]

I think infrared radiation is currently underestimated.

It isn't.

I read somewhere that half of the sun radiation is infrared radiation.

That's true, but what you suggested was getting ir radiation from objects around us, not from the sun.

Furthermore, if a body dissipates 70W, then 10 bodies will dissipate 700W.
If we focus the energy, like using a lens, we can increase the Watts per square meter.
Using a fresnel lens, we can focus 1 m2 into 1 cm2. Thats 10000 times more powerful!

And all you have to do is line all the walls, ceiling, and floors of a building with lenses! That's just not realistic.

 

[cesiumfrog]

I think infrared radiation is currently underestimated. ... half of the sun radiation is infrared radiation. ... fresnel lens

Are you aware that many solar designs already harness infrared (eg. solar-thermal)?

As for your scheme to use lenses to concentrate biothermal radiation (to mitigate the power density problem) don't forget one could apply the same thing with solar; solar remains more effective by three orders of magnitude (after accounting for thermodynamic efficiency).

 

[Sam Lee]

Yes, we know that solar energy is significant, but not enough.
We need to tap many other energy sources.

Based on an ambient temperature of 25 deg C (300K), there is more than 400 W/m2 of ambient (or background) radiation. If we can focus this background radiation somehow, then it can become 4 000 W/m2 or 40 000 W/m2.

We should explore ways to tap the background radiation.

 

[Dale]

Is it even possible to focus a directionless "ambient" light source? I'm not an expert at optics, but I wonder if that is even theoretically possible. However, for the sake of argument let's say that it is possible.

A focusing factor of 100 still doesn't give you more energy. You still need a 1 m² lens to collect 400 W even if you can then use a small IR photovoltaic of .01 m² area. If your lens material is much cheaper than your photovoltaic then that makes design sense, but you still only have 400 W/m² of energy incident on your lens.

And again, the real problem remains the fact that you have to magically cool your photovoltaic down to absolute zero without using energy in order to get even that 400 W/m².

 

[Sam Lee]

I'm not sure whether we can focus a directionless radiation source.
Can someone assist in this area?

The price of photovoltaic is about $1000/m2 whereas fresnel lens cost about $150/m2. It's much cheaper to use lens to focus the source before using the more expensive photovoltaic to convert the radiation into electricity.

Another big question here. Why do we need to cool the photovoltaic down to absolute zero here?

 

[G01]

Yes, we know that solar energy is significant, but not enough.
We need to tap many other energy sources.

Based on an ambient temperature of 25 deg C (300K), there is more than 400 W/m2 of ambient (or background) radiation. If we can focus this background radiation somehow, then it can become 4 000 W/m2 or 40 000 W/m2.

We should explore ways to tap the background radiation.

The problem is not that solar energy is not enough, it is that current photovoltaics can't absorb enough of it. A solar cell works by absorbing sunlight and exciting electrons across a semiconductor's band gap. These released electrons produce the photo current from the cell. The problem is that band gaps are properties of a material and are hard to tune to the energy of incoming radiation. This means that certain solar cells will only absorb light with energy that is larger than its band gap. Thus, many parts of the solar spectrum may not be absorbed in a given cell. Along with energy loss to heat radiated from the device, this means that solar cells are far from 100% efficient.

Time is better spent on figuring out to absorb more of the given solar spectrum in a single solar cell, than there would be figuring out how to absorb infrared from our bodies. The possible energy yield is just that much greater.

 

[Dale]

Another big question here. Why do we need to cool the photovoltaic down to absolute zero here?

Look back at my post #19, in particular the link I posted. If the photovoltaic is the same temperature as the surroundings then it will radiate the same amount of energy that it receives. The only way to have it absorb more energy than it radiates is for it to be cooler than the surroundings. Any energy that it radiates comes right off of that meager 400 W/m² that you are hoping to get.

 

[russ_watters]

Based on an ambient temperature of 25 deg C (300K), there is more than 400 W/m2 of ambient (or background) radiation. If we can focus this background radiation somehow, then it can become 4 000 W/m2 or 40 000 W/m2.

We should explore ways to tap the background radiation.

Sam, you're not listening. In order for an object to radiate energy away, it must be warmer than its surroundings. Converseley, in order for an object to absorb radiant energy from its surroundings, it must be cooler than its surroundings. So unless you plan to cool the collector to near absolute zero (which, of course, requires energy), this energy you speak of does not exist.

Think about it logically: if the walls of your house were really radiating 400w/m^2 of heat at you, you'd be cooked in a matter of seconds.

 

[uart]

Think about it logically: if the walls of your house were really radiating 400w/m^2 of heat at you, you'd be cooked in a matter of seconds.

Hi Russ, it's a thermal equilibrium situation. In a way the walls are emitting that power (or some reasonable fraction of it since they're not ideal black bodies) but they are in thermal equilibrium with their surroundings and hence re-absorbing the same amount (from the ambient radiation of other walls and objects in the room).

The reason why we’re not burning up is that we’re absorbing lots of radiation from our surroundings but radiating away even more than we absorb (as we’re generally slightly warmer than our surroundings).

If you could somehow instantaneously "teleport" that wall into the dark of deep space then it actually would radiate that energy for a while as it cooled.

 

[mgb_phys]

If we can focus this background radiation somehow, then it can become 4 000 W/m2 or 40 000 W/m2.

Remember that you can't focus a thermal source so the target is hotter than the source - otherwise it would just radiate back the other way.

 

[uart]

Look back at my post #19, in particular the link I posted. If the photovoltaic is the same temperature as the surroundings then it will radiate the same amount of energy that it receives. The only way to have it absorb more energy than it radiates is for it to be cooler than the surroundings. Any energy that it radiates comes right off of that meager 400 W/m² that you are hoping to get.

There are also other fundamental reasons why it can't work Dale. Sam could possibly try to argue around your objections using the "lens concentration" idea (though I too am unconvinced that this is even possible for non directional background thermal radiation). Anyway here's some more reasons to consider why a photo-voltaic cell couldn't function under these conditions.

First remember that the peak energy of a the blackbody radiation curve occurs at a frequency of several kT/h (about 3kT/h in fact). So no matter how much we try to concentrate them (with a lens or otherwise) we are stuck with the fact that our photons have an energy of at most a few kT (BTW the lens can only increases the number of photons per second per m^2, it can do nothing to increase the energy of each photon!).

Now a solar (photovoltaic) cell is fundamentally a semiconductor junction and as such it simply will not function unless the band gap is much larger than kT. (Effectively the junction will "short circuit" itself with thermal leakage current if E_gap is not much larger than kT).

So let's summarize the situation. Our photons must have enough energy to kick an electron across the band gap or no photo-current will be produced, but our band-gap must be much bigger than kT or the junction can't work, but the blackbody radiation law tells us that our photons can have an energy of at most a few kT. What else can you say but snookered!

Edit. I hadn't read your response before posting this mgb_phys. Anyway the "snooker" situation with the photovoltaics would still occur even if the focusing of background thermal radiation were somehow possible. I guess that's a double snooker for Sam.

 

[Dale]

Now a solar (photovoltaic) cell is fundamentally a semiconductor junction and as such it simply will not function unless the band gap is much larger than kT. (Effectively the junction will "short circuit" itself with thermal leakage current if E_gap is not much larger than kT).

Thanks for the interesting info. When you say "much larger than kT" how much larger do you mean?

 

[russ_watters]

Hi Russ, it's a thermal equilibrium situation.

You are, of course, right. I didn't want to confuse the situation by explaining that. In the same way, water at ambient temperature in fog will continuously evaporate and condense in equilbrium. There is no net transfer (perhaps I should have used that word in the other post).

 

[Sam Lee]

Harnesing energy from the surrounding is a challenging problem with non-obvious solutions. We have to keep an open mind and continue to exploring.

I suppose we can all agree that for black bodies, the radiation is more than 400 W/m2 at 300k. If we take the emissivity of our surrounding to be 0.5, then the surrounding radiation is more than 200 W/m2. This is not insignificant.

Of course at equilibrium in a closed system, the walls and objects radiate as much as they absorb.

In reality, we do not have a closed system. Our planet Earth receives radiation from the sun in the day and heats up. It cools down in the night. This is an open system, or open thermodynamics system. This gives us leeway to harness energy without breaking the law of thermodynamics or "creating energy from nothing".

Now from previous post, the problem with photovoltaic cells is that they will not function "unless the bandgap is larger than kT." I'm no expert in bandgap, but I think it is a property of material. That is to say, different materials will have differnt bandgaps. This gives us hope as we know a lot but too little about materials. Perhaps new materials (using nanotechnology) can have a small bandgap that can do the job.

Now this part about photovoltaic needing to be of a lower temperature than the surrounding temperature to operate is hard to understand. Is it because only then the bandgap required will be smaller than the radiation? Or is it because otherwise after abosrbing the radiation and the electron is excited, the electron will quickly give the energy back to the surrounding as radiation? Can we get the electron to do work before it returns the energy to the surrounding?

 

[russ_watters]

Harnesing energy from the surrounding is a challenging problem with non-obvious solutions. We have to keep an open mind and continue to exploring.

No, we have to understand the laws of thermodynamics and that what you suggest is a clear-cut violation of the second law.

In reality, we do not have a closed system. Our planet Earth receives radiation from the sun in the day and heats up. It cools down in the night. This is an open system, or open thermodynamics system. This gives us leeway to harness energy without breaking the law of thermodynamics or "creating energy from nothing".

You're changing the scenario again. Now you're talking about an outdoor radiator, radiating energy back into space. That would work - not well, but it it would work. Using a good radiator that is insulated against convection, you can cool something to a few degrees below ambient over a night (if it is clear). You'll get a small handfull of watts per square meter. Why? Unfortunately, the atmosphere still radiates some thermal energy back at you.

Or is it because otherwise after abosrbing the radiation and the electron is excited, the electron will quickly give the energy back to the surrounding as radiation?

Sort of, but I would alter that slightly to say that at the same time it is absorbing a photon, another atom next to it is emitting a photon. So the net energy level (temperature) of the object never changes.

Can we get the electron to do work before it returns the energy to the surrounding?

No. That's that second law of thermodynamics again. An object cannot spontaneously cool itself below the ambient temperature.

 

[uart]

Thanks for the interesting info. When you say "much larger than kT" how much larger do you mean?

I'd don't have an exact number but I think it would need to be at least 10 times larger to get anything close to a functional semiconductor junction. Essentially there can no P-type and no N-type, just a bland conductor unless E_g>>kT.

The following is a rough description of the physics involved, enough to give an overview. I’ll give the example of an n-type semiconductor but of course similar limitations apply to making a p-type as well.

Say we wanted to make a n-type semiconductor (Si for example), we’d have to dope it with a band-IV impurity at a level which is much greater than the intrinsic (undoped) electron concentration of the Si. There are limits however on just how large of an impurity concentration (dopant) we can use or else it will ruin the crystal structure of the primary semiconductor material. Typical dopant concentrations used in Si for example are in the range 10^13 to 10^18 atoms/cm3.

Now in addition to the free electrons created by the dopant there are also thermally generated electron-hole pairs, so unless the dopant level is much larger than this “intrinsic” carrier concentration we’ll never get a functional n-type semiconductor.

It turns out the number of thermally generated electrons is equal to $N_c e^{-(E_c-E_f) / 2kT}$, where $N_c$ is the effective density of states in the conduction band and E_f is the Fermi Level. In practice the Fermi Level is typically about mid-way in the band-gap hence $(E_c - E_f) \simeq E_g / 2$.

$N_c$, the density of states, is given by some complicated quantum mechanics that I don’t fully understand, but the simple upshot is that it’s a very large number, only a few orders of magnitude smaller the actual density of the atoms themselves (N_c is about 3x10^19/cm^3 in Si). So unless $e^{-E_g/2kT}$ is small (and I mean very small) we don’t have any chance of getting a functional doped semiconductor.

In Si for example, E_g=1.12 eV so at room temperature (300 Kelvin) e^{-E_g/2kT} is about 10^{-10}.

Hope that helps.