Unravelling the Mystery of Free Energy

[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.

 

[uart]

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".

Good, this part is making sense. At a fixed location on Earth we can think of it in terms of the Sun glowing in the daytime and the Earth glowing (infra-red) at night. If we want to make full use of that night-time glow we could build a huge tower many km to the cold upper atmosphere, aim some IR solar panels down at a warm "glowing" ocean and (if we could get the efficiency of IR solar panels to about the same as present day solar panels) collect maybe 30 to 40 Watts per square meter. Maybe you could even collect enough power to run the aircraft navigation warning lights down the length of the tower. ;)

Realistically I think it would be much more productive to look at improved energy storage technologies and just collect the energy in the daytime when it's cheap and easy.

 

[Dale]

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.

OK, so rough estimates. Blackbody radiation peaks at 3 kT and you need at least 10 kT for good conduction, so your emitter needs to be at least 10/3 hotter than your photovoltaic. Is that correct?

With a room-temperature (300 K) photovoltaic the sun is a good source since 6000 K (6 kK?) is much greater than 1000 K (10/3 300 K). But to use the background IR radiation we would need to cool our photovoltaic down to less than 90 K. Liquid nitrogen would do the trick (77 K) and might allow you to collect IR energy down to 256 K (2ºF), but of course producing the liquid nitrogen is pretty expensive energy-wise.

 

[dhruvbhalla]

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?

When does water disasotiate like that?.
If you are referring to salts (ionicly bonded compounds) that disasotiate to there composite ions in water , such as common salt (NaCl) to [Na+] and [Cl-] is due to the highly polar nature of the water and the energy to perform this is derived from the electostatic forces of bonding and from the dipole moment in the water molecule.

 

[Sam Lee]

Great inputs from all of you. We are making good progress in harnessing energy from the surrounding.

If we take Earth as the radiating body (300K), and outer space as the cooler body (close to 0k), then we have a temperature difference!

Say we make use of the roof of a building.
We have a 1 m2 plate above the roof.
Say we start with the initial condition whereby the roof and the plate are both 300K.
At night, the roof will be radiating 400W/m2. The plate will also be radiating at 400W/m2.

The interesting bit is that the plate has two sides. So on the side facing the roof, it will be emitting 400W and at the same time absorbing 400W (from the roof). However for the side facing the sky, it will emit 400W but will absorb very little.

I wonder how much power we can extract from such a system.

 

[russ_watters]

Bad news - on a hunch, checked-out the atmosphere's transparency and the black-body curve of a 300 K object. They don't overlap. Just about all of the 300K black body curve is between 5 and 25 nm. The atmosphere cuts off at 200 nm. That would be the greenhouse effect in action...

300k Black Body: http://irweb.info/archives/28
Sun and atmosphere: http://science.nasa.gov/headlines/images/sunbathing/sunspectrum.htm

 

[mgb_phys]

Just about all of the 300K black body curve is between 5 and 25 nm.

I think you meant um

That would be the greenhouse effect in action...

Yep, the moon is the same distance from the sun, receives the same level of illumination, isn't such a nice place to live.

 

[russ_watters]

Oops, I forgot my prefixes. So the solar spectrum goes from about 0.2 to 20 um (peak at 0.5 um) and the 300 K black body is 5-25 um (peak at 10) . So then there is some overlap.

 

[Dale]

Great inputs from all of you. We are making good progress in harnessing energy from the surrounding.

If we take Earth as the radiating body (300K), and outer space as the cooler body (close to 0k), then we have a temperature difference!

Say we make use of the roof of a building.
We have a 1 m2 plate above the roof.
Say we start with the initial condition whereby the roof and the plate are both 300K.
At night, the roof will be radiating 400W/m2. The plate will also be radiating at 400W/m2.

The interesting bit is that the plate has two sides. So on the side facing the roof, it will be emitting 400W and at the same time absorbing 400W (from the roof). However for the side facing the sky, it will emit 400W but will absorb very little.

I wonder how much power we can extract from such a system.

I am really glad I don't live where you live. I would think it would be very inconvenient living in a place where it is around 300 K in the day and around 0 K at night. Especially with the atmosphere around your place liquifying each night, I would think that would be somewhat dangerous.

 

[cesiumfrog]

I am really glad I don't live where you live. I would think it would be very inconvenient living in a place where it is around 300 K in the day and around 0 K at night. Especially with the atmosphere around your place liquifying each night, I would think that would be somewhat dangerous.

It may be of order 300K at night where the apparatus is located, but since the atmosphere is mostly transparent, the apparatus can be considered in "radiative contact" with the 3K microwave background. From the 300K perspective, direct thermal contact with a 3K bath would be a great sink of entropy (think 99% efficiencies).. unfortunately radiation specifically involves further inherent entropy than an ordinary heat engine (and here, low power densities).

Similarly devices are useful for obtaining water in arid climates (without any input they cool to the dew point, at least some of the time). In fact, by using special materials (that only interact with EM in the range at which the atmosphere does not interact with EM: thermally emitting radiation at wavelengths that pass the atmosphere but transparent or reflective at all wavelengths received from the atmosphere) it is possible to construct an apparatus that radiates to sub-ambient temperature even during the day (particularly with vacuum insulation).

 

[Dale]

It may be of order 300K at night where the apparatus is located, but since the atmosphere is mostly transparent, ... In fact, by using special materials (that only interact with EM in the range at which the atmosphere does not interact with EM: thermally emitting radiation at wavelengths that pass the atmosphere but transparent or reflective at all wavelengths received from the atmosphere) it is possible to construct an apparatus that radiates to sub-ambient temperature even during the day (particularly with vacuum insulation).

You are certainly correct here. Up until now we have been considering everything as simple black bodies, and in reality that is just an approximation, but for overall energy balance it is a good approximation.

Let's say that, using special "grey body" materials designed to enhance the radiation at just the right frequency, you are able to lower the temperature by 10 K without any energy input. And let's further say that you have 100% efficient conversion of the energy difference. You get a maximum of about 30 W/m².

By the way, if I were going to actually get energy this way I would try this:
1) build the ambient IR focusing lens
2) use it to heat some material to >1000 K
3) use a photovoltaic to capture the energy from the heated element
4) passively cool the photovoltaic to keep it at ambient temp (~300 K)

I doubt 1) is possible, and 2) would require some serious engineering, but 3) and 4) should be relatively straight-forward. Overall I doubt it could be efficient enough to get any net energy, but it seems more reasonable than any other suggestion so far.

 

[cesiumfrog]

1) build the ambient IR focusing lens
2) use it to heat some material to >1000 K
[..] it seems more reasonable [sic] than any other suggestion so far.

http://en.wikipedia.org/wiki/Second_law_of_thermodynamics"

 

[Dale]

http://en.wikipedia.org/wiki/Second_law_of_thermodynamics"

D'oh! OK, pardon the temporary insanity, but now I have my brain twisted. Say we have the following scenario:

A () B

A is a 1 m² black body maintained at 300 K, B is a .01 m² black body, and () is a lens arranged so that B receives the full 400 W radiated from the 1 m² surface area of A. Since B cannot be raised above 300 K then B will only radiate 4 W. What happens to the other 396 W? If B is a black body then it should absorb that energy, but then that violates the 2nd law of thermo.

 

[Moonbear]

There's some "iffy" content in this thread, so it is temporarily locked pending review by one of the physics mentors. Once they decide what, if anything, needs to be done, it may be reopened. Please stand by!

 

[Doc Al]

Be advised that discussion of recognized pseudoscience is not welcome here. (Review our posting rules, which are linked at the top of every page.) With that in mind, I am reopening this thread.

 

[cesiumfrog]

D'oh! OK, pardon the temporary insanity, but now I have my brain twisted. Say we have the following scenario:

A () B

A is a 1 m² black body maintained at 300 K, B is a .01 m² black body, and () is a lens arranged so that B receives the full 400 W radiated from the 1 m² surface area of A. Since B cannot be raised above 300 K then B will only radiate 4 W. What happens to the other 396 W? If B is a black body then it should absorb that energy, but then that violates the 2nd law of thermo.

OK, basic problem is your presumption that the radiation can be focussed on one spot. (Liouville theorem.) I suggest you draw the ray diagram yourself (easier if you put a curved mirror: "AB)" instead of the lens). Only 1% of the rays from any single point on A will go in the right direction to reach B, the rest will be focussed elsewhere (maybe back to A).

There's some "iffy" content in this thread, so it is temporarily locked pending review by one of the physics mentors. Once they decide what, if anything, needs to be done, it may be reopened. Please stand by!

lol

 

[Sam Lee]

From the 300K perspective, direct thermal contact with a 3K bath would be a great sink of entropy (think 99% efficiencies)..

Does that mean that 99% of the thermal radiation will flow to the 3K bath?

unfortunately radiation specifically involves further inherent entropy than an ordinary heat engine (and here, low power densities).

Can you explain more on this?
Is it good or no good in relation to what we are trying to do here?

 

[Sam Lee]

OK, basic problem is your presumption that the radiation can be focussed on one spot. (Liouville theorem.) I suggest you draw the ray diagram yourself (easier if you put a curved mirror: "AB)" instead of the lens). Only 1% of the rays from any single point on A will go in the right direction to reach B, the rest will be focussed elsewhere (maybe back to A).

lol

Actually we can focus parallel rays on one spot.
Like focusing the sun's ray onto a spot.

Electromagnetic (EM) radiation is very different from other types of energy.
It can actually be focused!
Most, if not all, other forms of energy cannot be focused.
And EM rays can flow from low energy density areas to high energy areas!
In fact, the EM rays just travel at the speed of light forward.

So, we can use the sun's rays and make an object hotter than the sun!

 

[thomasxc]

all the free energy ideas are getting old.

 

[ZapperZ]

Actually we can focus parallel rays on one spot.
Like focusing the sun's ray onto a spot.

Electromagnetic (EM) radiation is very different from other types of energy.
It can actually be focused! Most, if not all, other forms of energy cannot be focused.

I can focus beams of electrons, i.e. focus their kinetic energy (not their "EM" radiation) using a series of quadrupole magnets.

And EM rays can flow from low energy density areas to high energy areas!
In fact, the EM rays just travel at the speed of light forward.

So, we can use the sun's rays and make an object hotter than the sun!

Can you prove this? Here's a lens.

Zz.

 

[Topher925]

I suppose it is possible, assuming you have a way to contain the matter that you are heating which practically you can't.

I am surprised that Tesla's idea of free energy hasn't been brought up yet. After all we do live in one enormous capacitor charged by solar radiation.

 

[Sam Lee]

Can you prove this? Here's a lens.

This can be proven theoretically, at least for a start.

We know the sun is about 6000 deg C (about 6300K).
We also know that the sun radiation is about 1kW/m2.
We also know that we can focus the sun's rays using fresnel lens to about 1 cm2 diameter.

Now, using a thermal storage cylinder with a small opening at the top, we can focus the sun's ray into the thermal storage cylinder through its opening.
The thermal storage cylinder receives about 90% of the radiation that hits the fresnel lens.

From here, we know the energy input into the thermal storage cylinder.

We can design the thermal storage such that the heat loss is less than the energy input for temperatures less than 6300K. (Example, the emissitivity of aluminium foil is 0.03).

So the internal temperature of the thermal storage will keep rising as energy input is greater than energy loss.

 

[ZapperZ]

This can be proven theoretically, at least for a start.

We know the sun is about 6000 deg C (about 6300K).
We also know that the sun radiation is about 1kW/m2.
We also know that we can focus the sun's rays using fresnel lens to about 1 cm2 diameter.

Now, using a thermal storage cylinder with a small opening at the top, we can focus the sun's ray into the thermal storage cylinder through its opening.
The thermal storage cylinder receives about 90% of the radiation that hits the fresnel lens.

From here, we know the energy input into the thermal storage cylinder.

We can design the thermal storage such that the heat loss is less than the energy input for temperatures less than 6300K. (Example, the emissitivity of aluminium foil is 0.03).

So the internal temperature of the thermal storage will keep rising as energy input is greater than energy loss.

Er.. that doesn't sound right. For example, the temperature of the sun is very much determined by purely the radiation that it is emitting, no? That's how you determine the temperature of a blackbody source, which we can approximate the sun to be. So essentially, all of the radiant energy that we get IS due to the actual temperature of the sun, to a first approximation.

Now, I think you are arguing that the energy density, if you focus this to a volume smaller than the sun, could raise its temperature higher than the sun. If that's what we are talking about here, then sure, that could happen since the energy density would be higher. But "focusing" an isotropic source of radiation, and THAT much energy, is definitely a major issue. That's why I gave you a lens and asked you to start from there.

Zz.

 

[cesiumfrog]

So, we can use the sun's rays and make an object hotter than the sun!

..if you focus this to a volume smaller than the sun, could raise its temperature higher than the sun. If that's what we are talking about here, then sure, that could happen

Seriously, you two. Don't make me repeat the second law of thermodynamics again in this thread (do follow post #46)!

 

[ZapperZ]

Seriously, you two. Don't make me repeat the second law of thermodynamics again in this thread (do follow post #46)!

Yeah, you're right. I should have been more careful.

Zz.

 

[Dale]

I think any further posts claiming that a lens can focus radiation in order to make the target hotter than the source should include a ray diagram. I won't be able to take the time any time soon, but I suspect that the geometry will side with cesiumfrog and the 2nd law of thermo. (although I would like to see the diagram anyway if someone has the time)