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    Default Re: Thermal iimits of solar power.

    Does anyone have the optical laws/explanation for this? Maybe that would help halfeye.
    My knowledge there ends at "image size is proportional to object size"

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    Quote Originally Posted by Chronos View Post
    Lasers are subject to the same limitations as any other light source. They don't have a thermal distribution of wavelengths, but a laser of any given wavelength will have an equivalent temperature, and no matter how many of those lasers you have, and no matter what optics you shine them through, you'll never be able to use them to heat a spot to higher than that equivalent temperature. Laser beams are also never truly perfectly unidirectional: You can make them pretty tight, but there will always be some spread, if nothing else from the diffraction limit imposed by the wavelength and the aperture.
    Thanks for that, it reads more or less as expected, so that looks likely to be the orthodoxy.

    Quote Originally Posted by Radar View Post
    Lenses and mirrors do not use energy to do what they do nor do they (in theory) create any losses to the directed and shaped beams of light. This is the key point of the whole problem and it is indeed connected to perpetuum mobile.

    If you were able to concentrate light (or anything really) without doing work (and that's what lenses and mirrors do) to obtain higher temperature than in your initial source, this creates temperature difference out of nothing lowering the universes entropy, which is not possible: you can obviously use that temperature difference to extract work out of the system using a standard heat engine. The residual heat from that engine could again be concentrated to the high temperature you need to keep the engine going. So in a closed cycle you convert thermal energy of a single source to useful work without resorting to using some lower temperature reservoir to dump the residual heat into.
    This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.

    Even with lasers there will be limitation, but what counts here is the entropy of the light from a given source. Lasers produce by construction light of extremely low entropy so it is easier to obtain high temperatures using those. Stars are pretty much black bodies (as weird as it sounds), so the entropy of emitted light is actually very high - if I remember correctly emission of light with that distribution produces maximum of entropy for a given energy emitted. So if you want to concentrate it you are quite limited unless you put some work into it.
    There are problems with lenses and mirrors, it is relatively easy to make spherical surfaces, so lenses and mirrors are often made with such, but perfect optical surfaces would be parabolic, and sometimes the effort is made to produce parabolic surfaces, but it is much harder to make those. I presume parabolic surfaces are being used in this discussion, but if not that would explain everything, spherical aberation would limit any lens to images bigger than a certain fraction of the lens' diameter.

    Quote Originally Posted by NichG View Post
    The issue isn't so much alignment as it is needing photons traveling on the same path to catch up with eachother. With reversible optics you'd have the problem that at some point on the reverse path, two photons at the same position and direction must spontaneously go in different directions. But with nonlinear optics or active optics (such as stimulated emission in a laser) you can have one point in space absorb two photons from different directions and times, and emit them in the same direction and time.
    That reads like that same explanation from the link above with the wonky lens and the infinitely small image. I agree that an infinitely small image with infinite temperature is impossible, it's not what I've at any point been trying to suggest.
    Last edited by halfeye; 2022-05-13 at 11:22 AM.
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    Quote Originally Posted by halfeye View Post

    That reads like that same explanation from the link above with the wonky lens and the infinitely small image. I agree that an infinitely small image with infinite temperature is impossible, it's not what I've at any point been trying to suggest.
    Eh, I don't think it has to do with infinity. Just, I was trying to imagine how one would even with arbitrary moving fiber optic cables make it so all of the light from two different point sources released at one moment in time would converge in the same location at the same time.

    In order for that to happen, I either had to have one photon 'catch up' with another on the same path, or have the ability to inject a photon into a unidirectional beam from two directions at once.

    So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.

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    Quote Originally Posted by NichG View Post
    So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.
    However, you can have sources with various densities? If the limit is the carrying capacity of the vacuum, how can you have different capacity sources?
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by halfeye View Post
    This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.
    You are mixing up the first principle of thermodynamics (conservation of energy, thus the impossibility to generate energy/heat from nothing) with the second one (entropy always increases in a closed system). Radar's point is relative to the latter, not the former.
    Let's say that the entire universe consists only of the star (body A), the lenses apparatus (body B) and the "point" you want to heat up (body C), with the star having a fixed temperature Ta (heat bath) and the other two bodies temperatures Tb, Tc < Ta. Now let's assume that the apparatus is somehow perfectly efficient and doesn't heat up at all in the process, so that only Tc actually changes: the moment Tc reaches Ta, body C can't heat up any further without employing active mechanisms, this is because you now can't physically produce a non-zero gradient between A and C without external work, as it would violate the second principle of thermodynamics, in fact, defining with dS the total variation of entropy and with Q the heat exchanged after equilibrium, we would have dS = Q/Tx - Q/Ta < 0 (where Tx is some value such that Ta < Tx < Tf, with Tf the final temperature you want to reach).

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    Quote Originally Posted by halfeye View Post
    However, you can have sources with various densities? If the limit is the carrying capacity of the vacuum, how can you have different capacity sources?
    It has nothing to do with the carrying capacity of the vacuum though. It's that photons can't catch up with one another along the same path. You can of course irreversibly produce different numbers of photons at once, but that's not a passive optical process anymore.

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    Quote Originally Posted by halfeye View Post
    This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.
    This was not about getting heat for free - nothing in the process would create new energy. It is all about the entropy of the energy and your ability to convert thermal energy to useful work. Getting higher temperature with the same energy means lower entropy. To use some energy for work, you need some energy with low entropy and a place to dump the high entropy leftover energy afterwards. Typically, this means two reservoirs: one of high temperature to take energy from and one with low temperature to dump the residual heat into - think of a typical heat engine (steam or internal combustion for example).

    Now let's see, how increasing temperature without spending any extra energy would be a problem:
    1. We have just one reservoir of whatever temperature.
    2. By this hypothetical process, without putting in any extra work we take some energy from the reservoir and make a reservoir of higher energy with it (no new energy - just more "densely packed").
    3. Since now we have two reservoirs of different temperatures, we can use them to do useful work.
    4. If you continue this for long enough, you would fully convert the whole initial source of heat into useful work and that completely breaks the thermodynamics as it would allow you to effortlessly decrease entropy.

    And breaking thermodynamics basically means breaking statistics as a whole.
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    Quote Originally Posted by Captain Cap View Post
    You are mixing up the first principle of thermodynamics (conservation of energy, thus the impossibility to generate energy/heat from nothing) with the second one (entropy always increases in a closed system). Radar's point is relative to the latter, not the former.
    I believe you are mistaken about what I am thinking about.

    Quote Originally Posted by NichG View Post
    Eh, I don't think it has to do with infinity. Just, I was trying to imagine how one would even with arbitrary moving fiber optic cables make it so all of the light from two different point sources released at one moment in time would converge in the same location at the same time.
    What have point sources got to do with it? Photons start at a point, but any five or six start at five or six.

    We can shine two torches at the same wall with no problem.

    In order for that to happen, I either had to have one photon 'catch up' with another on the same path, or have the ability to inject a photon into a unidirectional beam from two directions at once.

    So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.
    I am sorry, I don't understand this at all.

    Quote Originally Posted by NichG View Post
    It's that photons can't catch up with one another along the same path.
    They travel at the same speed, so they wouldn't catch up? I'm definitely not understanding this.

    Quote Originally Posted by Radar View Post
    It is all about the entropy of the energy and your ability to convert thermal energy to useful work.
    Yeah, that's right. I am not in any way intentionally suggesting that that doesn't happen.

    Getting higher temperature with the same energy means lower entropy.
    Not necessarily. A higher temperature difference means more work can be done, a lower temperature difference means less work can be done, if the only energy you have is thermal. If everything you have is at a high temperature, you can do very little work with it. The heat death of the universe (if that's how it ends) is not at exactly zero Kelvin.

    Now let's see, how increasing temperature without spending any extra energy would be a problem:
    There is some energy put into the creation of lenses, trivial though it may be.

    1. We have just one reservoir of whatever temperature.
    2. By this hypothetical process, without putting in any extra work we take some energy from the reservoir and make a reservoir of higher energy with it (no new energy - just more "densely packed").
    No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.

    3. Since now we have two reservoirs of different temperatures, we can use them to do useful work.
    This needs thinking about, and might have something going for it, but I don't think I have so far read of a mechanism by which the universe might enforce this described.

    4. If you continue this for long enough, you would fully convert the whole initial source of heat into useful work and that completely breaks the thermodynamics as it would allow you to effortlessly decrease entropy.
    We are talking about a utterly miniscule fraction of the energy that the sun emits, and the timescale is billions of years.

    The Sun's core fuses about 600 million tons of hydrogen into helium every second, converting 4 million tons of matter into energy every second as a result. This energy, which can take between 10,000 and 170,000 years to escape the core, is the source of the Sun's light and heat.
    Core
    Main article: Solar core

    The core of the Sun extends from the center to about 20–25% of the solar radius.[70] It has a density of up to 150 g/cm3[71][72] (about 150 times the density of water) and a temperature of close to 15.7 million kelvins (K).
    https://en.wikipedia.org/wiki/Sun

    Quote Originally Posted by Radar View Post
    And breaking thermodynamics basically means breaking statistics as a whole.
    Nope, breaking thermodynamics would break the universe, statistics can take its chances.
    Last edited by halfeye; 2022-05-14 at 12:49 AM.
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    Quote Originally Posted by halfeye View Post
    I believe you are mistaken about what I am thinking about.



    What have point sources got to do with it? Photons start at a point, but any five or six start at five or six.

    We can shine two torches at the same wall with no problem.
    A torch doesn't fill every direction, so two torches each occupy a different range of incoming angles. If you had a solid sphere of torchfire surrounding a point and added an extra torch outside of that sphere, it's light would have to pass through another physical emitter in order to be able to add to the rest.

    Now imagine rather than an emitter, we replace each torch with a fiberoptic cable carrying light from elsewhere. If that cable bends light into a particular direction there, then because of reversibility it cannot let light through from behind it in that same direction if it's a passive element.

    For a passive optical element, every outgoing direction must correspond to exactly one incoming direction.

    I am sorry, I don't understand this at all.

    They travel at the same speed, so they wouldn't catch up? I'm definitely not understanding this.
    Let's say you wanted to use the fiberoptics from the previous example to inject light on a path and then moved them out of the way so you could get light from behind as well. Because light travels at the same speed, you can't use this to increase the instantaneous intensity either. So merely allowing optical elements to move, even though it gives you something passively irreversible, isn't enough to let you stack light along a direction vector.

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    Quote Originally Posted by Bavarian itP View Post
    Maybe start here.
    But what if you had a bunch of lenses at different angles?

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    Quote Originally Posted by halfeye View Post
    Not necessarily. A higher temperature difference means more work can be done, a lower temperature difference means less work can be done, if the only energy you have is thermal. If everything you have is at a high temperature, you can do very little work with it. The heat death of the universe (if that's how it ends) is not at exactly zero Kelvin.
    You are right, higher temperatures don't imply lower entropy per se, higher temperature gradients do. The point is that if you want to heat up a certain object to a higher temperature than the star, you must reach its temperature first: once that is done, any further increase of temperature would decrease entropy.

    Quote Originally Posted by halfeye View Post
    There is some energy put into the creation of lenses, trivial though it may be.
    What matters is that our ideal lenses don't expend energy to do their job of redirecting light once there, they do it passively. The work you used to assemble the system is now irrelevant in the interaction between star and lenses; they may have very well been there since ever, for all it matters.

    Quote Originally Posted by halfeye View Post
    No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.
    The reservoir emits energy in "our" direction because there is a temperature gradient between the star and "us". If our apparatus had the same temperature as the star, it would irradiate as the star does and the net flux of photons would be zero.
    Last edited by Captain Cap; 2022-05-14 at 02:46 AM.

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    Quote Originally Posted by halfeye View Post
    We can shine two torches at the same wall with no problem.
    And yet, we would not be able to make the wall hotter than a torch regardless of how many torches we use. In fact, we can explain the initial problem this way: if we view a star from any point, it takes some part of the sky (in other words, a given part of the whole spherical angle). What we can do with lenses and mirrors is to create a situation that at given point we see the star taking a bigger part of the sky. The absolute limit would be if a given point looked as if it is completely surrounded by the surface of the star. Such problems can be solved directly and you simply get at such a point temperature equal to that of the surface of the star.

    You may wonder, why is that. Well, imagine a sphere as hot as the star surrounding some point. Each inside point of that sphere emits light in all directions - not just toward the center. This results in the radiation density being equal everywhere within that sphere and there is no particularly high heat concentration anywhere.

    Quote Originally Posted by halfeye View Post
    There is some energy put into the creation of lenses, trivial though it may be.
    Irrelevant as lenses and mirrors in principle are not used up - they can work arbitrarily long.

    Quote Originally Posted by halfeye View Post
    No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.
    If some portion of energy is emitted, it is no longer at the star. Stars gradually lose energy due to radiation, so we actually do take the energy, if we try to use it for something.

    Quote Originally Posted by halfeye View Post
    This needs thinking about, and might have something going for it, but I don't think I have so far read of a mechanism by which the universe might enforce this described.
    It is not about enforcing - it is just about allowing something. If you have two reservoirs of different temperature, you can use the thermal energy to perform useful work.

    Quote Originally Posted by halfeye View Post
    We are talking about a utterly miniscule fraction of the energy that the sun emits, and the timescale is billions of years.
    Timescale does not matter, if we are talking about principles of physics. If something should cannot work in large scale due to fundamental reasons, it cannot work in a small scale for the same reasons.

    Quote Originally Posted by halfeye View Post
    Nope, breaking thermodynamics would break the universe, statistics can take its chances.
    What I meant is that thermodynamics is just applied statistics - breaking it means that statistics is not working in our universe. And yes, it would break the universe as we know it.
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    Quote Originally Posted by Bohandas View Post
    But what if you had a bunch of lenses at different angles?
    The best you can get is a bunch of mirrors and lenses that make it so that where ever you look, all you see is the sun. Which means you are effectively surrounded by the sun, and thus you will take the ambient temperature of the sun.

    Again, see Randall Munroe's What If article:
    Quote Originally Posted by Fire from Moonlight
    There's another way to think about this property of lenses: They only make light sources take up more of the sky; they can't make the light from any single spot brighter,[7] because it can be shown[8] that making the light from a given direction brighter would violate the rules of étendue.[9] In other words, all a lens system can do is make every line of sight end on the surface of a light source, which is equivalent to making the light source surround the target.


    If you're "surrounded" by the Sun's surface material, then you're effectively floating within the Sun, and will quickly reach the temperature of your surroundings.
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    I'm not entirely sure I understand why it would necessarily match the temperature. After all, it's not transmitting temperature, it's transmitting energy that excites the molecules of the target and thus raises its temperature

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    Quote Originally Posted by Bohandas View Post
    I'm not entirely sure I understand why it would necessarily match the temperature. After all, it's not transmitting temperature, it's transmitting energy that excites the molecules of the target and thus raises its temperature
    Thermal energy travels along temperature gradients.

    That this holds true so consistently even for energy transfer through radiation is somewhat unintuitive; after all, we should be able to just get a bunch of mirrors and lenses and pile all these photons into a denser area, right?

    But for a variety of reasons explained up thread, it doesn't actually work. Its like thinking you can stick a bunch of heat pipes into the surface of the ocean and expecting to boil tea with it: the energy is definitely there, but without doing some work yourself you have no way of moving the energy around beyond the point of thermal equilibrium.
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    To put it a different way, light (and heat) can travel both ways along the mirrors and through the lenses, right? In the event that you managed to heat your target up hotter than the source of heat, light and heat would instantly start flowing the other direction, away from your target and back toward the source until their temperatures are in equilibrium.
    Last edited by Lord Torath; 2022-05-17 at 08:34 AM.
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    It might be helpful to think of things in terms of distributions of photon frequencies. Even if you collect all the photons from the surface of the Sun somehow, those photons will still have a frequency distribution governed by the Planck function for a 5800 K blackbody. Concentrating the radiant output of a blackbody source means an increase in brightness, not temperature.

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    Relating this back to the Cooking a chicken by slapping it thread, could the chicken be heated to arbitrary temperatures by increasing the rate of the slaps without increasing their individual strength? I bring this up because it seems to be ultimately the same question
    Last edited by Bohandas; 2022-05-17 at 11:43 AM.

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    Quote Originally Posted by McGarnagle View Post
    It might be helpful to think of things in terms of distributions of photon frequencies. Even if you collect all the photons from the surface of the Sun somehow, those photons will still have a frequency distribution governed by the Planck function for a 5800 K blackbody. Concentrating the radiant output of a blackbody source means an increase in brightness, not temperature.
    That's not exactly accurate since temperature of a black body governs both the frequency distribution and radiation intensity from the surface of a given object. Besides, frequency filters could be used to shape the distribution but it would not help you in heating things beyond the temperature of the original source since the key point is how well can you concentrate the energy. To heat something up, means to push in more energy than it can radiate out. Form of the energy or in particular frequency of the incoming radiation does not matter.
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    Quote Originally Posted by Bohandas View Post
    Relating this back to the Cooking a chicken by slapping it thread, could the chicken be heated to arbitrary temperatures by increasing the rate of the slaps without increasing their individual strength? I bring this up because it seems to be ultimately the same question
    There is a difference there. In the chicken slapping, you are doing work. You are converting electrical energy into kinetic energy, and then transferring that kinetic energy into the chicken, where it is converted into thermal energy.

    In the lenses-and-mirrors question, you are not doing work. The entire process relies on the natural flow of heat from a high-temperature object to a low temperature object. And as soon as the low temperature object is the same temperature as the high-temperature object, the flow of heat stops.
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    I want to be very clear that I am not arguing against entropy, I am arguing that maybe entropy doesn't apply in the way some people think it does in this particular situation, however entropy always applies.

    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.

    Mirrors and lenses are not 100% efficient, there is loss of photons by reflection and absorbtion from lenses and absorbtion and transmission though mirrors.

    Some people seem to be hung up on absolute temperature, I have a query, I would expect people not to go for this, but it seems worth asking because it's an odd one if people do, and the "why not" should clarify some things if as expected people don't. Suppose the target for the heating is cooled, does that change the amount of energy that can fall onto it?

    This thing apparently gets up to 3.5k C, in Earth's atmosphere:

    https://en.wikipedia.org/wiki/Odeillo_solar_furnace



    I suggest something in space without Earth's atmosphere interfering should be able to get much hotter, particularly if the object was much nearer the sun. Though materials would be a problem, and we probably don't have a use for a temperature that high.
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    Quote Originally Posted by halfeye View Post
    Mirrors and lenses are not 100% efficient, there is loss of photons by reflection and absorbtion from lenses and absorbtion and transmission though mirrors.
    Which would mean you can heat your object to the temperature of the sun minus some factor depending on the efficiency of your mirrors.

    Some people seem to be hung up on absolute temperature, I have a query, I would expect people not to go for this, but it seems worth asking because it's an odd one if people do, and the "why not" should clarify some things if as expected people don't. Suppose the target for the heating is cooled, does that change the amount of energy that can fall onto it?
    The same amount of energy falls into the object no matter what. Say your focusing array can dump x joules/s into your object. If you don't cool the object, it's temperature will increase until some temperature T where it radiates x joules/s as a blackbody. If you cool the object, then it will not reach T because you are pumping some of the heat (hence energy) out of the object. This makes no difference to the amount of energy being put into the object.

    This thing apparently gets up to 3.5k C, in Earth's atmosphere:

    https://en.wikipedia.org/wiki/Odeillo_solar_furnace
    The surface of the sun is about 5000 Kelvin. Nobody is saying you can't use the sun to get things very hot, they're saying you can't get it hotter than the sun.
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by warty goblin View Post
    The surface of the sun is about 5000 Kelvin. Nobody is saying you can't use the sun to get things very hot, they're saying you can't get it hotter than the sun.
    I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by halfeye View Post
    I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?
    You are considering only the energy input from the sun, but the higher the object's temperature becomes, the more energy it irradiates itself, and so the higher is the output. The maximum temperature you can reach without additional work is the equilibrium temperature for which input and output are the same.

    Let's see this setup explicitly: we know that the power W irradiated (output) by a black body at temperature T is proportional to T^4, and so, adopting the appropriate unites of measures, we can write W = T^4 and Ws = Ts^4 (where s indicates the sun); if R is the distance from the sun and r its radius, the input power is Wi = Ws * (r/R)^2, therefore at the equilibrium we have T^4 = Ws * (r/R)^2 = Ts^4 * (r/R)^2 <= Ts^4.
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by halfeye View Post
    I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?
    Lets say you have a lens system that makes the sun take up 1/2 of the sphere of view around a target point, and that gets you to, say, 2500K (for a 5000K source). If you move it closer, then depending on the optics, the sun may take up more or less of that sphere of view. If you can make it take up the full sphere of view, you could get it to 5000K. It's not physically possible for it to take up more than 100% of the sphere of view.

    So the curve of temperature vs distance from the sun will depend on that optics calculation, and you may indeed find that its easy to get from something like 1% of field of view to 80% of field of view by moving it closer, but you will at some point start to get diminishing returns.

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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by halfeye View Post
    I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?
    If you move that whole apparatus twice as close to the sun, the first set of mirrors will receive 4 times as much light. However, the direction of travel for that light will very twice as much and less than 4 times the amount of light will following the complicated mirror and lenses path to the furnace core.
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by NichG View Post
    Lets say you have a lens system that makes the sun take up 1/2 of the sphere of view around a target point, and that gets you to, say, 2500K (for a 5000K source). If you move it closer, then depending on the optics, the sun may take up more or less of that sphere of view. If you can make it take up the full sphere of view, you could get it to 5000K. It's not physically possible for it to take up more than 100% of the sphere of view.

    So the curve of temperature vs distance from the sun will depend on that optics calculation, and you may indeed find that its easy to get from something like 1% of field of view to 80% of field of view by moving it closer, but you will at some point start to get diminishing returns.
    At this point I disagree. What I suggested wasn't making the sun a bigger area of the object's sky, though coincidentally it would also do that to a small extent, but moving the object twice as close to the sun, where the radiation from the sun would be stronger by a factor of 4:

    https://en.wikipedia.org/wiki/Inverse-square_law

    I'm not sure how much energy is required for a given rise in temperature, but it seems to me that increasing the brightness of the source, ought to increase the temperature of the target.

    We've discussed the permiability of space, and apparently it's not that which is the issue.

    At all the ranges I am discussing, the sun would optically be at infinity.
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    Default Re: Thermal iimits of solar power.

    You can't apply the inverse square law and an optical infinity simplification at the same time. The inverse square law arises from the fact that your light source is not at optical infinity; that changing your distance from the light source increases or decreases the area of its image. If you're at optical infinity i.e. all your light rays are parallel, then you don't have that. It doesn't matter what your position is, or whether you move towards or away from the light source: the image of the light source has the same area, and thus the same wattage per area.

    If you're not at optical infinity, then the surface area of the image grows quadratically as you move away, and so the wattage per area must decrease inverse to that because total power output must remain invariant.

    If you're in a situation where moving towards or away from the light source substantially changes the power you can capture, you're in a position where the incident angle of the light rays change, and so you have to deal with the size of the image reflected off your solar furnace changing. Moving towards the light source increases the wattage you capture, but it also increases the area of the image you produce.
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by crayzz View Post
    You can't apply the inverse square law and an optical infinity simplification at the same time. The inverse square law arises from the fact that your light source is not at optical infinity; that changing your distance from the light source increases or decreases the area of its image. If you're at optical infinity i.e. all your light rays are parallel, then you don't have that. It doesn't matter what your position is, or whether you move towards or away from the light source: the image of the light source has the same area, and thus the same wattage per area.

    If you're not at optical infinity, then the surface area of the image grows quadratically as you move away, and so the wattage per area must decrease inverse to that because total power output must remain invariant.

    If you're in a situation where moving towards or away from the light source substantially changes the power you can capture, you're in a position where the incident angle of the light rays change, and so you have to deal with the size of the image reflected off your solar furnace changing. Moving towards the light source increases the wattage you capture, but it also increases the area of the image you produce.
    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.
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    Default Re: Thermal iimits of solar power.

    Quote Originally Posted by halfeye View Post
    I'm not sure how much energy is required for a given rise in temperature, but it seems to me that increasing the brightness of the source, ought to increase the temperature of the target.
    Technically, the brightness of the source is always the same, but I get what you mean: the higher is the power inputted by the sun (which in the setup considered can be described by an inverse-square law), the higher is the temperature.
    That's correct, but a quantity that increases can have an upper bound, and in this case the upper bound is the superficial temperature of the sun, which you reach when your furnace system and the sun are at contact with each other.

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