Unravelling the Mystery of Free Energy

[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)

 

[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!)

 

[Sam Lee]

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!)

Great effort.

My gut feel is that the 1 meter lens diameter (800W) will not provide enough energy to bring something all the way up to 6300K. The scale is off. It should be more fruitful to look at something much bigger, like 100 meters!

Working backwards, in order to receive 9000W, we need to collect the sun's rays from an area of 9 m2 (1kW/m2). Hence we need a lens diameter of 3.4m or more(not 11 times bigger but square root of 11 times bigger). It doesn't look too daunting after all.

Let's check whether such a lens is possible (diameter 3.4m, focal length 1.07m, refractive index 2.4). The power of a thin lens, P = (nlens - no)/no x (1/R1 -1/R2), and the focal length is the inverse of P.
A lens with R1 = 3m and R2 = -3m does give a focal length of 1.07m.

In any case, if lens is a problem, then we can use other means of focusing the sun's ray such as using mirrors or mirrors in combination with lens, (or maybe even gravity to bend and focus light).

Using mirrors, we can bring together square kilometers of sun's radiation to the body in question. This is being used in some solar projects.

 

[russ_watters]

If all we are doing is concentrating the sun's rays to send them to a solar panel, why do we care if we get 6300K anyway?

 

[Sam Lee]

Count Iblis, I was trying to follow your reply.
It went on quite well in the beginning but it was difficult to link the end and the beginning together.

While the total energy in a system is constant, the amount of energy that can be extracted from it to do useful work is dependent on the temperature, right?

 

[russ_watters]

The amount of energy being extracted depends on temperature, yes. But in a solar cell, efficiency goes down as temperature goes up. If you use a collector to drive a thermodynamic cycle (a steam cycle), efficiency goes up as temperature goes up. But there is a practical limit of perhaps 500 K.

 

[Dale]

Using mirrors, we can bring together square kilometers of sun's radiation to the body in question. This is being used in some solar projects.

The square kilometers of mirrors you refer to do not even attempt to focus the energy down to a 1 cm target. As russ mentions, there is no benefit in focusing the energy that tightly so it isn't done. I am not at all convinced that is possible, and your handwaving argument certainly doesn't demonstrate that.

In any case, solar power with mirrors is already being done, so there is nothing new about this.

 

[Count Iblis]

Count Iblis, I was trying to follow your reply.
It went on quite well in the beginning but it was difficult to link the end and the beginning together.

While the total energy in a system is constant, the amount of energy that can be extracted from it to do useful work is dependent on the temperature, right?

That's right. In this case the efficiency is worse than in case of an ideal Carnot process, basically because of entropy production when the emited photons by the blackbody are replenished.

 

[uart]

Great effort.

My gut feel is that the 1 meter lens diameter (800W) will not provide enough energy to bring something all the way up to 6300K. The scale is off. It should be more fruitful to look at something much bigger, like 100 meters!

Working backwards, in order to receive 9000W, we need to collect the sun's rays from an area of 9 m2 (1kW/m2). Hence we need a lens diameter of 3.4m or more(not 11 times bigger but square root of 11 times bigger). It doesn't look too daunting after all.

Yep I made a simple mistake and forgot to square-root it. You really think a focal length less then one third the lens diameter is easy?

Let's check whether such a lens is possible (diameter 3.4m, focal length 1.07m, refractive index 2.4). The power of a thin lens, P = (nlens - no)/no x (1/R1 -1/R2), and the focal length is the inverse of P.
A lens with R1 = 3m and R2 = -3m does give a focal length of 1.07m.

In any case, if lens is a problem, then we can use other means of focusing the sun's ray such as using mirrors or mirrors in combination with lens, (or maybe even gravity to bend and focus light).

Using mirrors, we can bring together square kilometers of sun's radiation to the body in question. This is being used in some solar projects.

Firstly a 3.4 meter diameter lens with a radius of curvature 3 meters is hardly a "thin lens", make a drawing and do the maths, it's over a meter thick!. You have to be joking, the critical angle (beyond which you get total internal reflection) is only 23.5 degrees with n=2.5!

Draw a ray diagram and show how all the rays from the entire surface of that lens can be focused to one cm without any rays exiting at more than 23.5 degree from the normal. It's impossible, your proposed lens absolutely will NOT work.

 

[Dale]

Firstly a 3.4 meter diameter lens with a radius of curvature 3 meters is hardly a "thin lens",

Not only that, but the 1m focal distance would be inside the lens!

You will be better off with mirrors since you don't need to worry about refractive index, but I think you will ultimately hit similar geometric limitations.

 

[Sam Lee]

Not only that, but the 1m focal distance would be inside the lens!

You will be better off with mirrors since you don't need to worry about refractive index, but I think you will ultimately hit similar geometric limitations.

You are absolutely right.
Forget about lens. No wonder big solar projects use mirrors to heat up a body.

Now we can really scale things up.
We can have a 100m parabolic dish and focus all the energy to a spot.

Anyone knows how to work out the maths?

 

[Sam Lee]

The amount of energy being extracted depends on temperature, yes. But in a solar cell, efficiency goes down as temperature goes up. If you use a collector to drive a thermodynamic cycle (a steam cycle), efficiency goes up as temperature goes up. But there is a practical limit of perhaps 500 K.

This body (6300K) that we are talking about could be a thermal stroage body.
It is not a solar cell.

Solar cells are more efficient if they receive radiation from bodies with higher temperatures.

 

[russ_watters]

You want to use thermal storage to hold energy to later pass to a solar cell? Why?

 

[Dale]

I agree with russ. Even assuming that you can achieve sufficient focusing (I am still skeptical) this is a really bad way to store energy. The material engineering challenges of handling and insulating a 6kK object are not even worth considering.

The discussion about lenses and 6kK objects has merely been to demonstrate the applicability of the 2nd law to optical systems, i.e. to demonstrate that you cannot focus sufficient energy to cause heat to radiate from a colder body to a hotter one.

 

[uart]

My understanding was that the question of focusing the sun to produce a temperature hotter than the sun came about as a result of Sams desire to focus normal background IR radiation to get hotter than ambient temperature suitable for useful energy conversion. People were saying it was impossible so Sam brought up the example of using a lens to focus the Sun to such high temperatures as an example (now debunked, at least for the lens) of why it could work.

I'd like to see some argument of why a parabolic mirror would also suffer from some similar limitaion to the lens so we could finally put this to rest.

 

[russ_watters]

I'd like to see some argument of why a parabolic mirror would also suffer from some similar limitaion to the lens so we could finally put this to rest.

Again, it's the second law of thermodynamics:

Heat cannot spontaneously flow from a material at lower temperature to a material at higher temperature.

It's similar to the principle of pressure, where you have a large vessel and a small vessel sitting next to each other, both partially with water to the same level. The vessels are connected at the bottom. The downward force of the water in the larger vessel is much, higher than the downward force of the water in the smaller vessel. Does the water flow from the larger vessel to the smaller one?

 

[uart]

Again, it's the second law of thermodynamics:

Yes obviously I know that, and that's why I know it would be impossible. I'm saying that it would be interesting to see a specific physical or technological limitation in the case of the mirror.

 

[Mapes]

Yes obviously I know that, and that's why I know it would be impossible. I'm saying that it would be interesting to see a specific physical or technological limitation in the case of the mirror.

The issue was discussed in J Phys D in 1992-1993. S. Panse ("Non-spontaneous radiative heat transfer," J. Phys. D: Appl. Phys. 25 (1992) 28-31) argued that exceptions to the Second Law could occur in some optical systems similar to the one Sam Lee is proposing. In a rebuttal article, K.M. Browne ("Focused radiation, the second law of thermodynamics and temperature measurements," J. Phys. D: Appl. Phys. 26 (1993) 16-19) showed that in fact the Second Law was not violated in these systems because the finite size of the radiating body prevented focusing its rays to a point. (DaleSpam was right to request a ray diagram for these types of focusing proposals!) Browne concluded that the Second Law could also be expressed as: "No system may focus from one object, through an aperture onto a second identical object, more radiation than may be emitted by the second object through the same aperture."

When someone proposes a violation of the Second Law, it's easiest to infer the Law as a rebuttal. It's harder (but perhaps more satisfying) to show specifically why the idea won't work. Browne does an admirable job in the article mentioned above.

 

[Sam Lee]

The issue was discussed in J Phys D in 1992-1993. S. Panse ("Non-spontaneous radiative heat transfer," J. Phys. D: Appl. Phys. 25 (1992) 28-31) argued that exceptions to the Second Law could occur in some optical systems similar to the one Sam Lee is proposing. In a rebuttal article, K.M. Browne ("Focused radiation, the second law of thermodynamics and temperature measurements," J. Phys. D: Appl. Phys. 26 (1993) 16-19) showed that in fact the Second Law was not violated in these systems because the finite size of the radiating body prevented focusing its rays to a point.
...

Where can we find the articles online?
Any web sites?

 

[uart]

K.M. Browne ("Focused radiation, the Second Law of Thermodynamics and temperature measurements," J. Phys. D: Appl. Phys. 26 (1993) 16-19) showed that in fact the Second Law was not violated in these systems because the finite size of the radiating body prevented focusing its rays to a point.

Yes that is exactly along the lines I was thinking (similar to the situation with the lens) except I'm not so familiar with the detail of a parabolic mirror when used with diverging light rays so I wasn't sure how to prove it. Perhaps some time I'll be able to look up that reference, thanks Mapes.

 

[Habanabasa]

I have read this thread with interest because I face exactly the problem being discussed.

I have devised a device to focus RANDOM ambient Radiant Energy (RE) to an area smaller (19.63 sq mm) than the collection area (153.94 sq mm).

The first part of the device redirects the input RE that falls on a circle (180̊) and outputs it in the format of a cone with an internal angle of 30̊. This part has been built and tested

I still don’t know what will happen when I bombard the final target at 300K with photons also at 300K (in a black body distribution curve).

Using the Stephan Boltzmann formula we can calculate the optimal energy radiated into the collection area as 7.07 joules but in the real world we may have an emitter with emissivity of 0.9 and a net efficiency of 80% giving us 5.09 joules.

Provided the emitter emits the calculated amount of energy and provided the target will absorb all the photons, if we concentrate all this energy and have it absorbed by the target then the temperature of the target should rise to 482K.

It has been suggested to me that the emitter will not emit because it is at ambient. I don’t buy this. My understanding is that all surfaces above 0K will output RE - the amount dictated by the temperature and emissivity of the surface.

Remembering that a radiation distribution at a particular temperature will have some photons at higher and some at lower temperatures, I have difficulty understanding the absorption of the photons. I am lead to believe by some that photons that do not have sufficient energy to bump an electron up a level will simply pass through the object. I have difficulty with this because if this were the case then all the lower temperature photons would never be absorbed and we would end up in a sea of lower and lower temperature photons. This would cause the ambient temperature to decline as fewer photons of sufficient temperature would be emitted in the system. Furthermore, these low temperature photons would simply pass through all the matter and exit the system.

It makes more sense to me that the photons are absorbed. If they do not have enough energy to bump the electron up to the next level then they presumably will add to the energy content of the electron in some way. Another subsequent photon might cause the electron to have enough energy to bump it up a level.

My calculations may be flawed and please correct me if I am wrong but to me the amounts of energy involved are HUGE and not as small as some have indicated in this thread. Remember that the amount of energy is calculated from the temperature rise above absolute zero. For example: Energy emitted from 1 sq m of black body at 300K = 45,927 joules or 62 horse power. This figure will decline based on efficiencies of the system used to collect the energy.

This is nearly 46 times as much energy than that which arrives from the sun at noon at the equator on a clear day which is generally accepted as being 1000 joules.

Furthermore considering that this energy is emitted 24/7 and not only during sunny daylight hours, there is an immense storage of energy waiting to be tapped. The capture and storage of this energy is a tantalizing target.

 

[Mapes]

It looks like you're claiming that an arrangement of mirrors near two 300K objects will heat one of them to 482K. This is not going to happen; it would violate the Second Law. Kip Browne's paper (which I cite above) discusses these types of focusing systems and shows how to correctly quantify what's irradiating what. I recommend taking a look at it.

For example: Energy emitted from 1 sq m of black body at 300K = 45,927 joules or 62 horse power.

Check that math again!

 

[Habanabasa]

A impressively quick reply!

I have read K. Browne’s paper and in a nut shell he shows that for the two geometries studied, the target had a larger surface area than the sending object. Consequently the targets emitted all that they received from the emitter without raising their temperatures. My design is different in that the target is smaller than the emitter and so will receive a concentration of photons as envisaged by Sudhir Panse. Browne does not cover this eventuality even though he says that other geometries may be found, This is why I am still pursing it. Unless I learn something new that explains why it shouldn’t work, I will continue making the device. I expect to complete the device in the next 2 months when all my questions will be answered. If it doesn't work I shall still be looking for answers LOL

I have often heard the answer that the second law of thermodynamics says it can’t be done. People who say this have been unable to tell me why it won’t work or have some spurious illogical answer. There may well be a reason deep down in the realm of quantum mechanics why it won’t work but I have not heard it yet.

I suggest that the second law of thermodynamics was created from 300 or so years of empirical observation and is not tied to any mathematical proof. This leaves it open to future findings.

Your suggestions are welcome and may save me a great deal of work and expense.

As I said, my math may be wrong but here goes :
I used this site to compute the energy collected at the emitter :
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/stefan.html#c3
As I only want the energy emitted, not the net difference between two bodies, I used a T cold of 0K and a T hot of 300K with a collector area of 153.94 sq mm

(oops found my error - I put in 153.94 sq cm not 153.94 sq mm)

Collector collects 0.0707037 watts

I would need 1,000,000 / 153.94 collectors to get 1 sq m = 6,496.12 collectors

# of collectors * watts per collector = 0.0707037 * 6,496.12 = 459.27 watts or 0.62 HP

That certainly looks more reasonable, thanks for the pointer.

At this (now corrected ) rate it looks like the emitted energy is just under half that of the solar energy per unit area. The difference however is that radiant energy runs 24/7 and is not reliant on latitude or cloud cover and so is still a very attractive goal.

 

[russ_watters]

It is trivially easy to prove that this idea of yours is false - anyone who owns a telescope could point it at a wall and expect to be able to burn a hole in a piece of paper!

 

[Habanabasa]

You are correct. The telescope won’t burn a hole in a piece of paper. That is simply because the telescope would not focus ALL the radiant energy from the wall, only those minuscule small number of rays that travel in a parallel manner towards the telescope.

I like your site and am particularly impressed with the Hubble image.

 

[gmax137]

As I only want the energy emitted, not the net difference between two bodies, I used a T cold of 0K and a T hot of 300K with a collector area of 153.94 sq mm

Isn't this a problem? I have to admit I don't follow all of the arguments above, but think about what is keeping your emitter at the temperature of 300K. Isn't it just all of the 'random' radiant energy bouncing around between the emitter and its surroundings?

 

[russ_watters]

You are correct. The telescope won’t burn a hole in a piece of paper. That is simply because the telescope would not focus ALL the radiant energy from the wall, only those minuscule small number of rays that travel in a parallel manner towards the telescope.

Unless the telescope is right up against the wall, the amount of radiation that is not parallel will be miniscule! Besides - even if it misses a quarter or half the radiation, it can still focus light to thousands of times the intensity with which it was emitted: looking at a wall with a telescope as big as mine would be painful to your eyes.

I like your site and am particularly impressed with the Hubble image.

Thanks!

 

[russ_watters]

Anyway, sorry, but this discussion is explicitly against forum guidelines. We deal only in established science here.