Thread: Thermal iimits of solar power.

1. Re: Thermal iimits of solar power.

Originally Posted by halfeye
I mean optical infinity in the camera/glasses/telescope sense, the focus is at infinity, the rays are parallel. By the time you are 3 metres from an object, you are more or less at infinity, the angle the rays arrive from is almost exactly 90 degrees. This is a peculiarity of cameras, the sort of glasses that are prescribed by opticians, and telescopes, it's really a bit odd, but it's only in really close to a subject that the angle of the light coming through the lens is significantly different from parallel. I am not suggesting that this is particularly significant to this debate, that was just a point of interest.

The inverse square law on the other hand, is in my opinion, entiely relevant to the debate.
They're not independent things, the inverse square law derives from the spreading of the light rays. If you assume parallel rays, you're already at infinity and the flux per unit area can't further decrease.

Think of it this way - if I put a Dyson sphere of radius r around a star, or a Dyson sphere of radius 2r, the energy collected over the entire surface of the sphere must be the same. But the area of the latter is 4 times higher, so the flux per unit area must by 4 times lower. It's not that the photons are becoming dimmer or losing energy, they're just spreading out. A lens can reverse that spread only to the level of the most concentrated the light has ever been, not beyond that point.

2. Re: Thermal iimits of solar power.

Originally Posted by NichG
They're not independent things, the inverse square law derives from the spreading of the light rays. If you assume parallel rays, you're already at infinity and the flux per unit area can't further decrease.
To be fair, the parallel rays approximation can be used also when not at infinity (if it wasn't the case, it would be useless), you just need to be aware of the error you are introducing by doing so: in this case, the relative error would go as L/R, with L the characteristic dimension of your apparatus and R the distance from the centre of the sun; if the radius r of the sun is far greater than the apparatus' size L, then you can reasonably adopt the parallel rays approximation throughout the range between surface of the sun and infinity.

Of course, for systems of the size of a Dyson sphere that envelope the sun the approximation always fails.

3. Re: Thermal iimits of solar power.

Originally Posted by Lord Torath
(emphasis mine)

I... think you've got that backward? If you're much closer to the sun than it is wide, your object is going to be having light rays hitting it from all angles. If your object is very far from the sun, then the rays are closer to parallel.

If the sun takes up 30% of your field of view (ie you're very close to it), then its rays will span 54 degrees. If it takes up 1% of your field of view, it's rays are spread only across 1.8 degrees.
I was thinking about the net radiation flux (the quantity you actually apply the inverse-square law to) when I wrote the post, but yeah, the rays themselves are not parallel.

4. Re: Thermal iimits of solar power.

Originally Posted by NichG
Think of it this way - if I put a Dyson sphere of radius r around a star, or a Dyson sphere of radius 2r, the energy collected over the entire surface of the sphere must be the same. But the area of the latter is 4 times higher, so the flux per unit area must by 4 times lower. It's not that the photons are becoming dimmer or losing energy, they're just spreading out.
That's all true, but at this point I'm not talking about a Dyson sphere or anything like it.

A lens can reverse that spread only to the level of the most concentrated the light has ever been, not beyond that point.
That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?

We have a working solar furnace that gets up to 3,500 C / 3,200 K, if we replicate the relevant parts of that and put it half the distance to the sun of the original, it gets 4 times the number of photons. How does that not work out to four times the energy? I don't think the energy difference between temperatures is exponential, but I really don't remember, if it is we can move the replicated furnace even closer to the sun, until that exponential is reached.

5. Re: Thermal iimits of solar power.

Originally Posted by halfeye
We have a working solar furnace that gets up to 3,500 C / 3,200 K, if we replicate the relevant parts of that and put it half the distance to the sun of the original, it gets 4 times the number of photons. How does that not work out to four times the energy?
Just because you input 4 times the energy it doesn't mean the object retains 4 times the energy.

Originally Posted by halfeye
I don't think the energy difference between temperatures is exponential, but I really don't remember, if it is we can move the replicated furnace even closer to the sun, until that exponential is reached.
The thermal energy of a body is directly proportional to its temperature.

6. Re: Thermal iimits of solar power.

Originally Posted by halfeye
That's all true, but at this point I'm not talking about a Dyson sphere or anything like it.

That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?
This all comes from reversibility, and basically corresponds to what that XKCD book called 'the law of etendue'. If I have some kind of propagating light wavefront, I can describe that as a distribution of energy over both position 'r' and direction 'k' that evolves with time. Passive optics apply reversible operations to that distribution, by which I mean that for the forward operation L, I can write down a unique inverse operation L^{-1}. That implies that I can never (with passive optics) perform any operation which would make it ambiguous where each parcel of energy came from originally - that would violate reversibility, and would correspond (thermodynamically) to performing work on the system - e.g. it would require energy or at least dissipation.

So imagine if I were to have an optical element which mapped a broad distribution in 'r' to a delta function in 'r'. That would in turn have to map a narrow distribution in 'k' to a broad distribution in 'k' for the operation to be reversible, because otherwise it would be ambiguous how much energy should go where on the reverse path. If you are 'out of space' to make the distribution in 'k' broader, then you can no longer make the distribution in 'r' more narrow. If you did, that would violate reversibility.

So its not like 'a new physical principle comes into play to stop you from making it coverge more', and rather that if you write down the equations for how optical elements interact with these distributions, you'll find that the full form of those equations always ends up having these diminishing returns built into them just because of the way that the optical elements behave *in detail*. The issue you're having is that you're combining a bunch of asymptotic approximations as if those approximations were the full story, but you're combining them in regimes where they no longer hold true, and as a result finding an outcome > 1.

Edit: Tried to plot it and this bit is wrong, I think I have a 1/x somewhere flipped. Will post again if I figure it out.
They have a formula for the 'maximum concentration': Cmax = n^2/sin^2(alpha), where n is the index of refraction of the medium in which the receiver is placed, and 2*alpha is the collection angle (the fraction of the incoming field of view along which light from the source is being captured). So if you set n=1 and consider something like a pinhole camera, 'alpha' is basically the angle subtended by the light source, which is proportional to 1/r with r being the distance, in which case you derive the 1/r^2 law in the limit that alpha << 1, because sin(alpha)~alpha when alpha is small. However, when you get close to the source, alpha approaches pi/2 and you're in a different limit entirely, so the 1/r^2 law cuts off.

7. Re: Thermal iimits of solar power.

Originally Posted by halfeye
I mean optical infinity in the camera/glasses/telescope sense, the focus is at infinity, the rays are parallel. By the time you are 3 metres from an object, you are more or less at infinity, the angle the rays arrive from is almost exactly 90 degrees.
This is exactly the sense that matters and that you cannot simplify away. The imperfection doesn't matter much for a camera, but it matters for solar furnaces.

Let's say you have a 1 km radius reflecting dish heating a 1 meter radius solar furnace. If the light travels to the side one meter on the thousand meter journey between outer edge of the reflector dish and the furnace, it misses the solar furnace.

One meter out of a thousand is 3.5 arcminutes. Sunlight varies from parallel by 16 arcminutes, so the vast majority of the light that hits the rim of the reflector will miss the solar furnace.

8. Re: Thermal iimits of solar power.

Originally Posted by halfeye
That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?
Geometric argument: because the incoming photons are not going in parallel directions. A perfect lens can in theory (and ignoring more involved effects here) focus a parallel beam into a mathematical point. The thing is, ideally parallel beams do not and cannot exist. This approximation is especially true for black body radiation as every point of the source spreads the light equally in all directions. So instead of focusing a coherent beam, you try to focus light coming from many different direction with the same lens or mirror and if you try to filter it somehow, you are just losing energy in the process, which will not help you at all. Where the light will be directed by the lens or a mirror will obviously depend on the direction of the incoming light, so for larger angular size of the source you get proportionally larger focal area you can obtain even with perfect equipment.

9. Re: Thermal iimits of solar power.

When you get twice as close to the Sun, the Sun is four times as bright, because it's taking up four times as much area in the sky. If you had a piece of paper held at arms' length, with a 30 arcsecond diameter hole cut in it, and looked at the Sun through that hole, you couldn't tell if you were at 1 AU and seeing the whole sun, or at 1/2 AU and seeing only a part of it. The "brightness density" of the Sun isn't changing, just its angular area.

And that's the same thing a collecting mirror does: It doesn't change the brightness density, just the effective area.

In either case, getting closer to the Sun or using a bigger mirror, the increase in total incoming energy is entirely due to the Sun (or equivalently, its image) taking up more angular area. And so, in both cases, the incoming energy maxes out when the angular area maxes out.

10. Re: Thermal iimits of solar power.

Originally Posted by Captain Cap
I was thinking about the net radiation flux (the quantity you actually apply the inverse-square law to) when I wrote the post, but yeah, the rays themselves are not parallel.
I misread your example, and realized you were correct. Then I deleted my post in embarrassment, hoping no one had seen it.

So much for hoping!

11. Re: Thermal iimits of solar power.

Originally Posted by halfeye
That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?.
It is called the diffraction limit. As you concentrate light it starts to interfere with itself. Any telescope (i.e. lens system) has a limit where it cannot make images smaller and still have them be images. At that point everything starts to blur together. I know it falls out of Maxwell's equations but I cannot remember exactly how to derive it.

12. Re: Thermal iimits of solar power.

Originally Posted by Rockphed
It is called the diffraction limit. As you concentrate light it starts to interfere with itself. Any telescope (i.e. lens system) has a limit where it cannot make images smaller and still have them be images. At that point everything starts to blur together. I know it falls out of Maxwell's equations but I cannot remember exactly how to derive it.
In the case of concentrating Sun's energy things brake down long before the diffraction limit and the problem at hand is grounded in the optical geometry - there is no need to involve the wave equations here.

13. Re: Thermal iimits of solar power.

And for what it's worth, the diffraction limit "falls out of Maxwell's equations" only in the sense that Maxwell's equations show that light is a wave. Diffraction works the same way for all waves, light, sound, ripples on the surface of water, whatever.

14. Re: Thermal iimits of solar power.

Originally Posted by halfeye
First off, how about transformation of electricity? The voltage can go up or down, the amperage can go up or down, and the only necessarily moving parts are switches to start and stop the process. This clearly doesn't break entropy.
Wattage=Voltage*Current

The total energy in a system is identical on both sides of a spherical transformer in a vacuum (practically, the output is slightly lower because some energy is wasted as heat within the transformer). For a given power source, it is impossible to get an output higher than the input, although you can add waste loads if you want to reduce the output for some reason. Wattage is constant. Thus the transformation of electricity does not have any unusual relationship to entropy.

Similarly, a star can only put out temperatures equal to what it itself is. You can get lower temperatures by adding waste load (all the areas around the star that aren't being illuminated), and you can shuffle some of the output around in the same conceptual way that you can shift voltages and currents around (in this case, you prevent solar energy from striking some areas in order to strike a different area more), but you can never, ever, ever get an output higher than the input.

There's a very simple (albeit potentially expensive) experiment that can be done to prove this, requiring a laser and a lens. If you take an industrial laser and shine it against different materials, using a lens will get a much thinner line on those materials that it is powerful enough to mark or cut, and it will produce those results significantly faster. What it will not do is mark or cut a material that the un-lensed beam cannot. This is because the final determiner of what effect the laser has is how hot it can make the target, and that is identical for both beams because they have the same source. The sole exception to this is if you use significantly different (but same cuttable/markable material) targets between the two beams - it is possible that if you try to use a larger sample for the wider beam it will transfer energy away faster than it can be applied, even if a thinner example would be affected.

I know this because I've done it - I spent most of a week back in high school trying to figure out why our lens didn't make the beam exponentially more powerful the way our simple area calculations suggested.

15. Re: Thermal iimits of solar power.

Originally Posted by Gnoman
Wattage=Voltage*Current

The total energy in a system is identical on both sides of a spherical transformer in a vacuum (practically, the output is slightly lower because some energy is wasted as heat within the transformer). For a given power source, it is impossible to get an output higher than the input, although you can add waste loads if you want to reduce the output for some reason. Wattage is constant. Thus the transformation of electricity does not have any unusual relationship to entropy.
That is exactly what I was saying. My point is that you can take either voltage or current to be equivalent to temperature, and overall wattage is exactly equivalent to overall wattage, but in the case of electricity there is not a hard limit on the output voltage or amperage provided you take the hit in the other aspect, so that overall output wattage falls due to the system not being 100% efficient..

Similarly, a star can only put out temperatures equal to what it itself is.
In the core of the sun that's millions of degrees Kelvin. The temperature at the surface is less because the heat is spread out and emitted.

You can get lower temperatures by adding waste load (all the areas around the star that aren't being illuminated), and you can shuffle some of the output around in the same conceptual way that you can shift voltages and currents around (in this case, you prevent solar energy from striking some areas in order to strike a different area more), but you can never, ever, ever get an output higher than the input.
I'm definitely not talking about getting more Watts out than were originally present.

There's a very simple (albeit potentially expensive) experiment that can be done to prove this, requiring a laser and a lens. If you take an industrial laser and shine it against different materials, using a lens will get a much thinner line on those materials that it is powerful enough to mark or cut, and it will produce those results significantly faster. What it will not do is mark or cut a material that the un-lensed beam cannot. This is because the final determiner of what effect the laser has is how hot it can make the target, and that is identical for both beams because they have the same source. The sole exception to this is if you use significantly different (but same cuttable/markable material) targets between the two beams - it is possible that if you try to use a larger sample for the wider beam it will transfer energy away faster than it can be applied, even if a thinner example would be affected.

I know this because I've done it - I spent most of a week back in high school trying to figure out why our lens didn't make the beam exponentially more powerful the way our simple area calculations suggested.
That is much better. Do you have any sources for that that aren't purely anecdotal? Is there any theory as to why that should be that isn't vague hand-waving?

16. Re: Thermal iimits of solar power.

Originally Posted by halfeye
In the core of the sun that's millions of degrees Kelvin. The temperature at the surface is less because the heat is spread out and emitted.
Radiation wise, the sun interacts with the rest of the universe only through the energy irradiated by its surface, so its temperature anywhere else is irrelevant.

17. Re: Thermal iimits of solar power.

Originally Posted by halfeye
That is much better. Do you have any sources for that that aren't purely anecdotal? Is there any theory as to why that should be that isn't vague hand-waving?
I mean there's a What-If on the topic, concerning starting a fire with moonlight. Give it a read for a Munroesian Explanation.

18. Re: Thermal iimits of solar power.

Originally Posted by halfeye
That is exactly what I was saying. My point is that you can take either voltage or current to be equivalent to temperature, and overall wattage is exactly equivalent to overall wattage, but in the case of electricity there is not a hard limit on the output voltage or amperage provided you take the hit in the other aspect, so that overall output wattage falls due to the system not being 100% efficient..
The point is neither voltage nor current can be equivalent to temperature, since electrical current is the coherent motion of electrons, so its entropy is low and is not connected to the voltage or current values in any significant way - this is why you can transform the voltage almost however you like. With temperature and any kind of heat transfer (be it by radiation, convection or whatever else) it is completely different.

19. Re: Thermal iimits of solar power.

Also relevant is Maxwell's demon, and I'm surprised it hasn't been mentioned yet with regards to materials. The idea is that any sort of one way door for energy is prohibited, so that materials that absorb light at a particular frequency also emit it efficiently, and these two things are perfectly correlated. From that it might be easiest to consider each path individually. Lenses and mirrors can align beams quite closely, but they are always reversible. There is no possible way for two paths to be identical over some of their distance, and not over some other part. They can be close, but if they were aligned with mirrors they can be separated with mirrors.
If we consider the light entering and leaving a single path from a single point, at a single frequency, then we can say that either the material is perfectly reflecting or transmitting the light, or it is a good emitter at that frequency. If we follow the path back in time, and hit the sun, we can work out how much energy the material will be absorbing. If we know the temperature of the material, we can work out how much energy is going out at that frequency too*. If the material is hotter than the surface of the sun then we are guaranteed to be radiating more than we absorbing. While it might seem like we can do interesting things with manipulating the paths, this is true along every single individual paths. Managing to heat something that was already hotter than the surface of the sun would require adding up the effects of lots of paths, all of which are emitting more energy than they are absorbing, and coming up with a total where we are absorbing more than we are emitting**.

* You might then wonder if there is some way to trap the photons, such as by moving things so that our target will absorb light from the sun, but photons from it cannot escape and are redirected towards it. You can, but Maxwell's demon tells you that when those photons arrive back at our target they are on a path that could have lead back to the sun, so they are instead of energy from the sun, not in addition to it. In order to avoid energy escaping you need to close the door that lets energy get in.

**We cannot say where the energy will end up. Things move, light emitted back along a path that came from the sun is following a whole new path, but we can say that if any energy is getting in, then energy can get out.

That XKCD irks me somewhat, because it doesn't mention that it is possible to light a fire with a somewhat more complicated setup (though not with a magnifying glass), using the fact that we can filter by frequency. A material that does not absorb in the infrared spectrum but does absorb blue light can get extremely hot, even in moonlight. It only stops heating when it is emitting as much blue light as it is absorbing, and that means several hundred degrees! We can't heat a moon rock above 100C, but that doesn't actually tell us that we can't get a different material hotter than that. This is how greenhouses work, or in reverse, that cooling paint.

20. Re: Thermal iimits of solar power.

Originally Posted by Fat Rooster
That XKCD irks me somewhat, because it doesn't mention that it is possible to light a fire with a somewhat more complicated setup (though not with a magnifying glass), using the fact that we can filter by frequency. A material that does not absorb in the infrared spectrum but does absorb blue light can get extremely hot, even in moonlight. It only stops heating when it is emitting as much blue light as it is absorbing, and that means several hundred degrees! We can't heat a moon rock above 100C, but that doesn't actually tell us that we can't get a different material hotter than that. This is how greenhouses work, or in reverse, that cooling paint.
Careful here, the math doesn't quite work out the way you're saying...

Constants aside, the power emitted per surface area and angle is ~ integral dw w^3/(exp(w/T)-1) where w is the frequency of light. Equilibrium occurs when the total power in = total power out. So now lets say you put a filtering function 0<=g(w)<=1 on that for a given target body, which applies equally for power absorbed and power emitted.

Total power in = integral dw g(w) input_spectrum(w)
Total power out = integral dw g(w) w^3/(exp(w/T)-1)

Lets say the input spectrum comes from a black body at temperature T'. That means that equilibrium occurs when:

integral dw g(w) w^3/(exp(w/T')-1) = integral dw g(w) w^3/(exp(w/T)-1)

That still occurs only when T=T'.

So regardless of a material's spectrum of absorbance and emittance (call this a transfer function so I don't have to keep writing that), it has the same equilibrium temperature with respect to an external black body source. Now, if the source has it's own filtering function 0<=h(w)<=0, it should be clear that it cannot increase the left side of that equation, only decrease it. So even if your emitter isn't a blackbody source, as long as its thermally induced emission there is no pair of transfer functions that can cause it to heat something else up to above its own temperature, only to a lower temperature at best.

21. Re: Thermal iimits of solar power.

Originally Posted by NichG
Careful here, the math doesn't quite work out the way you're saying...

Constants aside, the power emitted per surface area and angle is ~ integral dw w^3/(exp(w/T)-1) where w is the frequency of light. Equilibrium occurs when the total power in = total power out. So now lets say you put a filtering function 0<=g(w)<=1 on that for a given target body, which applies equally for power absorbed and power emitted.

Total power in = integral dw g(w) input_spectrum(w)
Total power out = integral dw g(w) w^3/(exp(w/T)-1)

Lets say the input spectrum comes from a black body at temperature T'. That means that equilibrium occurs when:

integral dw g(w) w^3/(exp(w/T')-1) = integral dw g(w) w^3/(exp(w/T)-1)

That still occurs only when T=T'.

So regardless of a material's spectrum of absorbance and emittance (call this a transfer function so I don't have to keep writing that), it has the same equilibrium temperature with respect to an external black body source. Now, if the source has it's own filtering function 0<=h(w)<=0, it should be clear that it cannot increase the left side of that equation, only decrease it. So even if your emitter isn't a blackbody source, as long as its thermally induced emission there is no pair of transfer functions that can cause it to heat something else up to above its own temperature, only to a lower temperature at best.
Oh, sure. You absolutely cannot get hotter than the source of the light, but the moon is not the source of the light, and it irks me how he dismisses that and simplifies in a way that causes confusion because spectral effects are extremely important. You also could not use lenses to focus moonlight more effectively than if you just had a lump of your material sitting on the moon. You absolutely could craft a material that got hotter sitting on the moon than the moon rocks that surround it, even if it was not directly in sunlight. The factor of 400,000 that is lost to the whole reflecting off the moon part is more than made up for by the exponential in the equation and the fact we don't need to get that hot (250C is enough for paper). If our transfer function is 0 below the visible spectrum then the question becomes whether an object at 250C emits or absorbs more light in the visible spectrum when exposed to moonlight, or equivalently, whether we could see better looking into an oven at 250C with no other light sources, or a moonlit scene. Things don't even start to visibly glow until ~500C, and if we wanted to we could restrict ourselves further to the blue end of the spectrum.

I'm not saying any of this is practical (at best you could manage ~4mW/m2, so even if you could get it to work perfectly it would take a while), but it does not violate thermodynamics to use the visible part of moonlight to heat something above the flash point of paper despite the fact that moon rocks don't get that hot. It is impossible to use even the near infrared part, because then you would cool faster than you could possibly heat, but things need to get extremely hot before they start to emit any significant energy in the visible spectrum. Thermodynamics just means you cannot exceed the initial temperature, it doesn't say anything like "if you only get 1/4 of the light you can only heat to 1/4 of the temperature", or any function at all. Even if we get a millionth of the light, our limit is still the same temperature, because the exponential means that for any constant c and temperatures T > T' there is a frequency above which a body at temperature T will emit c times more energy at those frequencies than one at T'.

22. Re: Thermal iimits of solar power.

Originally Posted by Fat Rooster
Oh, sure. You absolutely cannot get hotter than the source of the light, but the moon is not the source of the light, and it irks me how he dismisses that and simplifies in a way that causes confusion because spectral effects are extremely important. You also could not use lenses to focus moonlight more effectively than if you just had a lump of your material sitting on the moon. You absolutely could craft a material that got hotter sitting on the moon than the moon rocks that surround it, even if it was not directly in sunlight.
If it were in sunlight, you could do this. But you absolutely could not craft such a material that would get hotter in equilibrium than the moon rocks surrounding it when outside of sunlight. The equations in my last post are the proof of why that doesn't work.

23. Re: Thermal iimits of solar power.

Originally Posted by NichG
If it were in sunlight, you could do this. But you absolutely could not craft such a material that would get hotter in equilibrium than the moon rocks surrounding it when outside of sunlight. The equations in my last post are the proof of why that doesn't work.
They are really not. Run the numbers and see. Set a g(w)=0 for low values of w, and g(w)=1 for high values. For low T the w/T is big before g(w) becomes 1, and that's an exponential term keeping power down. The emissions are small everywhere on the spectrum, either because of low transfer function, or low temperature. Power in meanwhile can be significant because the badly reflected sunlight can interact with the high part of the transfer function. Even the 20 fold reduction we eat from reflecting off moon rock is nothing on the power of that exponential term, meaning temperature has to be surprisingly high in order to be in equilibrium.
The point is that we are just using the moon rocks as (really bad) mirrors. Their temperature doesn't matter when used like that. The equilibrium temperature of a moon rock in that situation is going to be the same as any other moon rock, but that doesn't actually tell us anything about the equilibrium temperature of any other materials. It always depends on the transfer function. Normal materials are reasonably similar in having some IR response, but that doesn't tell us anything about the materials that can be produced, and definitely nothing about the thermodynamic limits of what a material can do. With enough layers of glass effective IR response can be brought down arbitrarily low, so it doesn't even need nanomaterial tricks. A good greenhouse design can do it. You can see that you can get to 250C, because objects at 250C are less bright in the visible spectrum than objects illuminated by moonlight. If you isolate a 250C object in all except the visible spectrum it will receive more light than it emits, even from moonlight. The power is miniscule, so your isolation needs to be extremely good, but it will be getting hotter if you manage that.

Maybe I'm misunderstanding, and you are just objecting to the practicality. I'm under no illusions that this is practical, just that thermodynamics has no beef with it.

Also, black body radiation for power in isn't really what you are looking for, because that implies you are completely surrounded by it. A <1 scalar multiple of black body is more appropriate, and when you do that it is rapidly pretty clear that power in and power out are rarely the same for particular values of w. In anything other than the completely surrounded case the specifics of g make a big difference.

24. Re: Thermal iimits of solar power.

Originally Posted by Fat Rooster
They are really not. Run the numbers and see. Set a g(w)=0 for low values of w, and g(w)=1 for high values.
Sure. So lets take the source as a blackbody at T=90K (temperature of the dark side of the moon) and a threshold value of w at 10^15 Hz, well into the UV. Lets assume your material is completely buried in moon rock, so no losses to space or anything, just exchange between the sample and the moon rock.

The power absorbed by the sample per unit surface area is: 2h/c^2 * integral from 10^15 Hz to infinity [ w^3 / (exp(h w / k T) - 1)] dw. We can substitute u=(h/kT) w, and that becomes:

Power absorbed / m^2 ~= 2.78 * 10^-9 kg/(s^3 K^4) * T^4 * integral from 500 to infinity u^3/(exp(u)-1)

Now, at those values of u, the integral is basically u^3 exp(-u), which we can integrate in closed form, and it turns out to be 9 * 10^-210. So the total power absorbed per unit surface area is 1.64e-210 kg/s^3. A ridiculously small number because this material only emits or absorbs radiation in the high UV.

What's the temperature at which that is equal to the power emitted? Well, it's the same integral, but now with a temperature T'. Closed form here, approximating exp(big number)-1 ~ exp(big number), ends up being:

Power emitted / m^2 ~= T'^4 * 2.78 * 10^-9 * 9 * 10^-210 kg/(s^3 K^4). Notice something? It's the exact same equation, but with T' instead of T. So power emitted = power absorbed only when T' = T.

Now, what if we didn't consider the lunar rock to be a blackbody emitter? Well, that would strictly decrease the power absorbed at a given temperature, while leaving the power emitted to be the same. So in that case, you would have strictly T' < T.

25. Re: Thermal iimits of solar power.

What temperature, if any, would something have to be for a significant percentage of its blackbody radiation to emit mainly in the 300mhz-300-ghz range

EDIT:
And in particular, could frequencies like this dominate at some temperature or another that is less than or equal to 1000C?

Edit:
Because I read a Wired article about using radiation in this range to heat objects to 1000C

EDIT:
Tangential question. If you had unlimited time and resources, could you theoretically use the cosmic microwave background to cook chicken? Like if you had a parabolic reflector the size of a galaxy set up in intergalactic space or something like that

26. Re: Thermal iimits of solar power.

Originally Posted by Bohandas
What temperature, if any, would something have to be for a significant percentage of its blackbody radiation to emit mainly in the 300mhz-300-ghz range

EDIT:
And in particular, could frequencies like this dominate at some temperature or another that is less than or equal to 1000C?
So one thing to keep in mind is that there is no frequency at which the total energy at that frequency decreases with increasing temperature. As you increase the temperature, the higher frequencies increase faster than the lower ones, but everything increases. That means that even if you have a frequency range that dominates at low temperature, the object is still emitting more energy in those frequencies at a higher temperature in which that range is not dominant anymore.

The temperature at which 300mhz is the peak frequency is ~0.003 Kelvin. The temperature at which 300ghz is the peak frequency is ~3 Kelvin. A 30 Kelvin object will emit more radiation at 300ghz than a 3 Kelvin one.

Edit: There's evidently a subtle thing about what 'peak' means for a density function, so what I did here (c/peak wavelength) isn't strictly correct, because dw and dlambda are different measures...

27. Re: Thermal iimits of solar power.

In any case, the point is that you can use electromagnetic radiation in that range to heat things to at least 100°C, and I've heard of it being used to heat objects to 1000°C. You just need to have enough of it.

28. Re: Thermal iimits of solar power.

Originally Posted by Bohandas
In any case, the point is that you can use electromagnetic radiation in that range to heat things to at least 100°C, and I've heard of it being used to heat objects to 1000°C. You just need to have enough of it.
Yes, there's nothing about a certain frequency that inherently limits the temperature energy in that frequency can be used to heat something to. It's the balance of total power in versus total power out.

29. Re: Thermal iimits of solar power.

Originally Posted by Bohandas
EDIT:
Tangential question. If you had unlimited time and resources, could you theoretically use the cosmic microwave background to cook chicken? Like if you had a parabolic reflector the size of a galaxy set up in intergalactic space or something like that
The short answer is no. The longer answer involves the effective temperature of the CMB (a few kelvin), its extent (4 pi steradians), and essentially everything else in this thread.

30. Re: Thermal iimits of solar power.

Originally Posted by NichG
Yes, there's nothing about a certain frequency that inherently limits the temperature energy in that frequency can be used to heat something to. It's the balance of total power in versus total power out.
That and the ability to focus the energy well enough. If the source of radiation is directional, the light/microwaves/whatever can be focused far better than blackbody radiation.

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