Unravelling the Mystery of Free Energy

[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

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

 

[ZapperZ]

:uhh: 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)

 

[Count Iblis]

Sam, I have a better idea. Find a way to genetically manipulate bacteria or plants so that they will produce electricity using photosynthesis.

 

[cesiumfrog]

Sam, I have a better idea. Find a way to genetically manipulate bacteria or plants so that they will produce electricity using photosynthesis.

Just burn plants. Or is that politically incorrect?

 

[Count Iblis]

Just burn plants. Or is that politically incorrect?

Anyway, if you collect solar radiation during some time t, then you know the Gibbs energy, so you can compute the maximum work that can be performed. You can set an upper limit to the temperature by equating the maximum efficiency that follows from this Gibbs energy to the efficiency of a Carnot engine.

Of course, the temperature can be made arbitrarily high as you can choose the time t as high as you want. But the maximum power does not depend on t.

The Carnot engine has to operate between a reservoir that is kept at the high temperature and the ambient temperature, so the power it extracts from the reservoir as to be supplied to the reservoir.

New Edit: Actually using the Gibbs energy doesn't work, as the radiation when it reaches Earth cannot be considered to be at constant temperature and pressure when we are extracting work from it.

But we can simply use the second law and demand that when the work is extracted the total entropy stays the same at best. The relation between entropy end energy of black body radiaton is well known, so this seems to be a trivial excercise to me.

 

[russ_watters]

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)

I'm curious so I may do that math myself, but the Stefan-Boltzman law doesn't say anything about area/intensity. Ie, it doesn't matter what the intensity is, if two objects facing each other are emitting the same spectrum of light (ie, they are at the same temperature), there will be no net energy transfer between them. You can't focus one's light to transfer its energy to the other.

 

[Dale]

the Stefan-Boltzman law doesn't say anything about area/intensity.

Hmm, the version of the http://en.wikipedia.org/wiki/Stefan-Boltzmann_law" I learned does. It talks about "total energy radiated per unit surface area of a black body in unit time". I got this from Wiki, so it could easily be incorrect.

 

[russ_watters]

Oops...yah, so I was way wong about that part (and I really knew that). The point I was trying to make (the next sentence) is that area doesn't matter in whether objects radiate toward each other. They won't exchange energy if they are the same temperature, regardless of how the energy is concentrated.

The equation is q = ε σ (Th4 - Tc4) Ac http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

I don't like the way they did their subscripts, but in any case, Ac is the area of the object. The area of the surroundings doesn't enter the equation at all and the result is that if there is no temperature difference, there is no heat transfer.

I would say it was too early, but it was 10:00 in the morning...

 

[Sam Lee]

Oops...yah, so I was way wong about that part (and I really knew that). The point I was trying to make (the next sentence) is that area doesn't matter in whether objects radiate toward each other. They won't exchange energy if they are the same temperature, regardless of how the energy is concentrated.

The equation is q = ε σ (Th4 - Tc4) Ac http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

I don't like the way they did their subscripts, but in any case, Ac is the area of the object. The area of the surroundings doesn't enter the equation at all and the result is that if there is no temperature difference, there is no heat transfer.

I would say it was too early, but it was 10:00 in the morning...

Something is funny here, although I cannot pinpoint where.

The case in point is that the body receiving the focused radiation from the sun has a hole for the focused radiation to enter, thereby acting like a black body in absorbing all the radiation that passes through the lens.

The rest of the body is shinny and has low emissivity. Ratio wise, only 1% of the body's surface area behaves like a black body while the remaining 99% of the body's surface area behaves as a white body.

Because of this set up, the 1% of surface area receives a lot of radiation due to the focusing of the parallel rays from the sun. The remaining 99% of the surface area emitts little radiation because of its shiny surface.

So mathematically, even if the body is at the same temperature as the sun, it will emit less radiation than it receives. Hence the temperature of the body will continue to rise.

 

[russ_watters]

Doesn't matter how many times you say it - it doesn't work that way. It will keep absorbing only until its temperature equals the temperature of the sun.

 

[Count Iblis]

Sam, the temperature is not relevant. You can attain temperatures of a billion K using energy from a source which is at 1 K. What matters is how much useful work you can extract from the system.

So, let's try to find the thermodynamic limit on the maximun amount of work you can extract for an amount E of solar energy. Naively you could think that this is given by Carnot's formula with the two temperatures of the reservoir the temperature of the photons as they escape from the sun and the temperature of the environment here on Earth. But this is wrong, basically because we are using the photons that the Sun emits which have a certain entropy, we are not in control over what happens at the solar end of the reservoir.

What is the entropy of a box containing a energy of E in the form of blackbody radiation? Consider filling the box with black body radiation by heating it from absolute zero. The internal energy is proportional to T^4, so we put E = c T^4. The entropy change from T to dT is the internal energy change divided by T, so

dS = 4 c T^2 dT ------>

S = 4/3 c T1^3 = 4/3 E/T1

T1 is the temperature of the black body.

Now, the photons that arrive at Earth from the Sun are, strictly speaking, not in thermal equilibrium anymore. The energy density is lower, also the distribution in momentum space is not uniform (the direction they move in is not random as they all come from the Sun). Nevertheless, the change is 100% reversible. In principle you can let photons escape from a black body and then reflect them back into the black body.

This means that for any process in which we extract work from solar photons at Earth we can construct an equivalent hypothetical process in which we use photons right at the solar surface. In that hypothetical process, the second law cannot be violated. This then implies that we can assign an entropy of S = 4/3 E/T1 to an amount of E of solar energy at Earth and assume the validity of the Second Law.

Now, if we destroy the solar energy photons and extract an amount of work W, the entropy change is:

-4/3 E/T1 + (E -W)/T2

T2 is the temperature at the Earth. So, the limit on W is:

W = (1 - 4/3 T2/T1) E

If we could extract solar energy using an ideal Carnot process, the limit would have been:

W = (1 - T2/T1) E

So, what is the source of the lower efficiency? Let's look at what goes wrong when we attempt to find an equivalent quasistatic process. Suppose we bring a small box of volume Vbox to the Sun, which we pretend to be a big volume filled with blackbody radiation.


When the box is filled, the internal energy that goes in the box, Ebox, is extracted from the system. Suppose that the box is open when it is moved into the Sun and then closed so that whatever radiation happens to be caught in it, stays in it. This changes nothing to the system as a whole, it is merely a partitioning of the volume of the system into a part Vbox and the the total volume minus Vbox.

When the box is removed from the system and brought to Earth, a volume Vbox in the system will refill with radiation. Now, the fundamental thermodynamic relation is:

dE = T dS - P dV

In the refilling process, dE = Ebox, dS is the same as the entropy that moved into the box, and dV is Vbox. So:

Ebox = 4/3 Ebox - P Vbox --------->

P = 1/3 Ebox/Vbox


So, we see that the blackbody radiation has a pressure of 1/3 times the energy density. We can now understand why the process is not quasistatic. When the box is removed, the volume re-fills with radiation, which is an irreversible process. One can say that an amount of work of P Vbox which could have been extracted in a quasistatic process, get's dissipated.

 

[uart]

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.

Ok Sam, let's do some rough calculations based on these figures.

When we use a lens to focus light from an object then the best focus we can obtain gives an image of the object with a size ratio of Image Size / Object Size = Image Distance / Object Distance, with distances measured from the lens. This is becasue an object at any finite distance has rays that are not parallel but diverging (I'm pretty sure you were already taking this into account).

So if we want for example a one cm diameter image of the Sun then the required focal length can be calculated from the thin lens formula, 1/I + 1/O = 1/F, where I and O are the image and object distances and F the focal length. This formula can be re-arranged to give the image linear scale factor of I/O = F/(O-F), which for the case of the Sun we may assume O>>F and hence,

$$\frac{Image Linear Size}{Oject Linear Size} \simeq \frac{F}{O}$$

So returning to the desired one cm diameter image we see that the required focal length (maximum) is F = Sun Distance * 1cm / Sun Diameter, which using numerical values gives F is about 1.07 meters. This is the maximum focal length to achieve a 1 cm image, with a shorter focal length we can certainly focus an even smaller image of the Sun.

Heres the problem though. If we stick with a one meter focal length then we can make the lens a fairly large diameter without requiring the light to be bent more than physically possible by the lens (there is a limit because as we go for higher refractive index in order to bend the light more we also run up against the problem of total internal reflection in the lens). If however we go for a short focal length then we can make the image smaller but the diameter of the lens will also have to be smaller to avoid needing to bend the light more than physically possible. In other words we can't really gain anything by going for a smaller image and a shorter focal length.

Ok so let's stick with a 1cm image and a 1m focal length for now. Let's assume we can achieve a focal length about the same as the lens diameter. This will give the lens an area of Pi/4 or about 0.8 m^2. By your figures we can collect about 800 Watts of power over this area. Let's assume 100% of that 800 Watts makes it to our 1cm^2 cavity and is 100% absorbed and the cavity has no thermal conduction losses and the "shiny" part has zero emissivity, that's about as favourable assumptions as we can make ok.

If the cavity did reach a temperature of 6300k then the radiation loss from the cavity would be sigma * T^4 * A = 5.67E-8 * 6300^4 * 1E-4 which is about 9000 Watts leaving our cavity. But by our most optimistic calculations the inflowing energy is only 800 Watts, it doesn't take much effort to workout which way the energy is flowing!

Now you could argue that we could overcome this problem by having a lens with a diameter very much bigger than it's focal length (diameter about 11 times the focal length) but I'm almost certain an optics guru (not my area) could prove that impossible. It certainly seems unlikely to me, I'm pretty sure that total internal reflection in the lens would ruin it even if a material with refractive index high enough to bend the light that much were available. (You'd be looking at bending the light through about 85 degrees!)