Rawlemon’s Spherical Solar Energy Generation From Moonlight

[EEVblog]

I predict about half a bdW (Bee's Dick Watt) of power from the moonlight.

http://inhabitat.com/rawlemon%E2%80%99s-spherical-solar-energy-generating-globes-can-even-harvest-energy-from-moonlight/

[2N3055]

I predict about half a bdW (Bee's Dick Watt) of power from the moonlight.

http://inhabitat.com/rawlemon%E2%80%99s-spherical-solar-energy-generating-globes-can-even-harvest-energy-from-moonlight/

bdW 

[RGB255_0_0]

It hurts the green movement when these lunatics are given the limelight (could have made a joke here but this is no joke). I wonder how long it'd take for the energy gathered from the "moon's" energy to recoup the cost to manufacture such monstrosity.

[rob77]

wow ! what a stupid design !    probably it's 35% better on concentrating sunlight/moonlight but it must weight a few tons .... who in his right mind would design a solar system which can't be installed on roofs because of it's sheer weight ? 
and i would also have a look at losses in that giant marble... probably the losses in the glass will be higher than the 35% gain

[T3sl4co1l]

https://what-if.xkcd.com/145/

[Someone]

https://what-if.xkcd.com/145/

Except thats still an open ended question without any consensus on an answer, etendue is not preventing it from being possible.

[Red Squirrel]

I jokingly mentioned here what if you could use a bunch of mirrors to concentrate moon light to a solar panel and get usable power. Someone ran the numbers and you need a ridiculous amount that may as well simply be said to be impossible.  This design would be better served as a solar thermal generator.

[Towz]

It's so ridiculous!! I doubt it would ever even match the energy wasted manufacturing it... but of course! We've got an elegant way to power a solar calculator at night! A bdW should suffice   

[TheWelly888]

If Wallace and Gromit can make a man and a rabbit nuts by concentrating moonlight then this is a perfectly viable idea.

 

[Kleinstein]

There is an even more ridiculous / confusing version: one could point a kind of telescope to a really dark part of sky. Here the radiation temperature is really low, well below ambient and thus very little IR light comes from there. So an IR sensitive photodiode would show a reverse voltage (other polarity than with normal light), as there is IR radiation send to the sky with no return. So you could generate power from sending out radiation to a cold place.

I have not done the numbers but I would expect less than a bee's ...

[T3sl4co1l]

There is an even more ridiculous / confusing version: one could point a kind of telescope to a really dark part of sky. Here the radiation temperature is really low, well below ambient and thus very little IR light comes from there. So an IR sensitive photodiode would show a reverse voltage (other polarity than with normal light), as there is IR radiation send to the sky with no return. So you could generate power from sending out radiation to a cold place.

I have not done the numbers but I would expect less than a bee's ...

This isn't as bad as it sounds!  But it will be thermal in nature.

You could build a moderately effective solar-thermal array by doing...

1. Build a radiator.  Assuming clear skies overall, then: during the day, it will absorb sunlight and heat up.  During the night, it will cool down.

2. During the day, pump the heat into a holding tank.  Preferably phase change (paraffin or molten salt) storage.

3. At night, run a heat engine from the tank, using the radiator as the cold side.

This boosts efficiency because the radiator goes down quite cool at night: perhaps 40 degrees below ambient.

The heat transfer rate is fairly low, so you will need quite a lot of radiator, of course.  But it will still perform better (higher Carnot efficiency) even if the radiator isn't much below ambient temperatures -- because, if you used convection cooling to ambient air instead, you can only ever have the heatsink somewhat hotter than ambient, never equal or below.

Probably, the solar array should be a concentrator type, like the trench or tower kind.  The Sun is highly directional, so it's a big net win to thermally isolate your hot side from anything that's not line-of-sight with the Sun.  That way you can get really hot fluid coming out.  (As opposed to a flat thermal field, like say, a blacktop parking lot with thermal pipes underneath, which loses heat as radiation in all directions, plus convection off the huge area of air it's in contact with.)

Conversely, the cooling radiator should be exactly the opposite kind: because, at night (cloudless), the whole sky is largely quite cool, and you need to maximize area because the temperature is low and radiation has crappy conductivity at low temperatures.

Okay so I lied, it's not as exciting as when I started typing...  The reason the collector and radiator must be asymmetrical is because the radiation source and sink are asymmetrical in nature.  So, unfortunately, there's no easy two-for-one sale on that.

Tim

[amspire]

https://what-if.xkcd.com/145/

Total nonsense! Is this a new tactic - fighting bad science with even worse science?

The site is right in saying that lenses and mirrors are bi-directional and so that an equilibrium is reached, but it mixed up power and temperature. It claims that as the Suns temperature is 5000 deg, then if all the light from the Sun was directed to, say, a 1M square black body, it would only reach 5000 degrees maximum. As radiated power is approximately proportional to the fourth power of the temperature for a Sun and the Sun's surface area is over 6 x 10¹⁸ square meters, the What-If page is possibly incorrect by a factor of over 100000000000000000000000000000000000000000000000000000000000000000000000000000000%.

That is what I call the error of all errors!

I have to admit I cannot say precisely how a 1m² black body behaves when receiving all the Sun's energy - I haven't done the test myself.

[T3sl4co1l]

Well, I can assure you with great confidence that the 1 m^2 area would, very quickly at that, heat up to 5000 K.  Assuming the material somehow remains within that 1 m^2 region at such temperatures, it would simply stay there in equilibrium.

Global (so to speak) equilibrium would certainly be off by a bit, though -- all the power being radiated by the Sun would be reflected back into itself, creating the heaven's grandest greenhouse of all.  It is, of course, identical to coating the inside of your Dyson sphere with a mirror finish -- why you put the bling on the inside, I have no idea.  You must not be a very smart Dyson-sphere-constructor.  Which kind of begs the question...

But anyway, the Sun's surface temperature would rise, and the photosphere, puff up.

The temperature, of our somehow-still-stationary 1 m^2 plasma cloud, would track nearly perfectly with the Sun's surface temperature (give or take propagation delay).

Pretty soon, the temperature will rise to that which is normally seen at much deeper layers, and hard UV and even some soft x-ray radiation will dominate.  Those better be some damned good mirrors!

In any case, focusing that much radiation, from an astronomically sized body, doesn't sound like a good idea...

Tim

[amspire]

Well, I can assure you with great confidence that the 1 m^2 area would, very quickly at that, heat up to 5000 K. 
...
But anyway, the Sun's surface temperature would rise, and the photosphere, puff up.
...

In any case, focusing that much radiation, from an astronomically sized body, doesn't sound like a good idea...

Tim

Tim, have you read what you have written? A 1m² black body at 5000 degrees would cause the Sun to heat? You realise that to heat the Sun significantly, the power has to be radiated from the 1m² black body. A 1m² black body at 5000 deg may radiate enough to warm a decent sized building, but not the Sun.

To radiate the same power back to the Sun that the whole Sun is emitting, a 1m² black body would have to be at a nonsensical temperature - something like 3 x 10⁶⁹ degrees.

[T3sl4co1l]

So what you're saying is, a cup of water, at room temperature, can't possibly be in thermal, radiative equilibrium with the Earth, because the Earth has a surface area of (whatever) and the cup doesn't, and so the total radiative powers of both can't possibly be reconciled?

Tim

[amspire]

So what you're saying is, a cup of water, at room temperature, can't possibly be in thermal, radiative equilibrium with the Earth, because the Earth has a surface area of (whatever) and the cup doesn't, and so the total radiative powers of both can't possibly be reconciled?

Tim

No I am not saying that.

The subject here is two objects of different sizes connected by lenses so there are two completely different power densities on each side of the lens, and I am saying that in this case they have to reach a radiated power equilibrium regardless of the temperature that that implies. If you put a cup of water in space with a huge lens focusing the Earth's emitted light, the water in the cup would boil. It has to because if you are putting megawatts of power into a cup, it has to get very hot. It cannot re-emit energy to return it back to the Earth without getting very hot. You seem to think that somehow megawatts of power can be emitted from the cup back to the Earth via the huge lens without the cup getting hot. That is not physically possible. Emitted power is driven by temperature. If there are megawatts of emitted power, then the cup must be very, very, very hot.

[amspire]

Just another example, places like the Livermore Labs have been using lasers to initiate Nuclear fusion. Now fusion needs temperatures like 14 billion degrees, and the lasers are not at 14 billion degrees. They use Neodymium glass lasers, so there is no way temperatures in the lasers can be very high at all.

If you focus the power of simultaneous pulses from 192 x 2.5 Terrawatt Lasers onto a 2.3mm diameter target, you get 14 billion degrees along with an inward pressure on the core of the target of 300 Billion atmospheres. It is conservation of Energy not conservation of Temperature.

[T3sl4co1l]

The subject here is two objects of different sizes connected by lenses so there are two completely different power densities on each side of the lens

The power densities are equal -- as you stipulated, they are equal temperatures!

and I am saying that in this case they have to reach a radiated power equilibrium regardless of the temperature that that implies. If you put a cup of water in space with a huge lens focusing the Earth's emitted light, the water in the cup would boil. It has to because if you are putting megawatts of power into a cup, it has to get very hot.

Fascinating...

So, how far does the lens have to be from the Earth?  Can it be on the Earth?  Is the atmosphere refractive enough (given suitable conditions) to meet your requirement?

So why does my cup not boil, sitting here?

Perhaps another riddle will prove fruitful?  Suppose you point an antenna at the Sun.  (It should be a large enough parabolic dish type, so the beam is reasonably pointy, and can be said to be fully receiving solar noise.)  For as-yet poorly explained reasons, the corona is microwave-active, with a temperature in the millions K in that frequency range.  This is reflected at the antenna terminals as a few nV/rtHz noise (the same temperature, expressed into 50 ohms, or whatever impedance the antenna is matched to).

You agree that this is all consistent, right?  There's a resistance, there are two degrees of freedom (Gaussian noise on a resistor), and there's a temperature, which can be no other temperature than that of the plasma which is radio-opaque in the relevant range, correct?

Tim

[T3sl4co1l]

Just another example, places like the Livermore Labs have been using lasers to initiate Nuclear fusion. Now fusion needs temperatures like 14 billion degrees, and the lasers are not at 14 billion degrees. They use Neodymium glass lasers, so there is no way temperatures in the lasers can be very high at all.

If you focus the power of simultaneous pulses from 192 x 2.5 Terrawatt Lasers onto a 2.3mm diameter target, you get 14 billion degrees along with an inward pressure on the core of the target of 300 Billion atmospheres. It is conservation of Energy not conservation of Temperature.

Nonequilibrium, and lasers are superluminous; in a certain sense, they have an exceedingly high temperature, over a very narrow passband.  (But a mere thermal bandpass wouldn't exhibit the spacial and spectral coherency that is characteristic.)

Tim

[IanB]

Total nonsense! Is this a new tactic - fighting bad science with even worse science?

This is why science is a useful tool, and why we can use science to make fun of people who believe in free energy and other impossible schemes.

I have to believe the xkcd argument because of science. There is a law of thermodynamics that says you cannot transfer heat from a colder body to a hotter body without doing work on the system. This law has been tested to exhaustion and is effectively immutable. It is as impossible to violate this law as it is to construct a perpetual motion machine. Under no circumstances can heat flow spontaneously from a colder body to a hotter body. You could spend your whole life trying and it would be as vain as the work of alchemists past trying to turn lead into gold.

[Someone]

Total nonsense! Is this a new tactic - fighting bad science with even worse science?

This is why science is a useful tool, and why we can use science to make fun of people who believe in free energy and other impossible schemes.

I have to believe the xkcd argument because of science. There is a law of thermodynamics that says you cannot transfer heat from a colder body to a hotter body without doing work on the system. This law has been tested to exhaustion and is effectively immutable. It is as impossible to violate this law as it is to construct a perpetual motion machine. Under no circumstances can heat flow spontaneously from a colder body to a hotter body. You could spend your whole life trying and it would be as vain as the work of alchemists past trying to turn lead into gold.

The falling down of that explanation of the XKCD comic is assuming that the light coming from the moon is emitted by its temperature, when we all know that you can look at a fire with a mirror and the mirror is not too hot to hold in your hand. Sun -> Moon (inefficient mirror) -> Super Lens -> Deathray

[amspire]

Nonequilibrium, and lasers are superluminous; in a certain sense, they have an exceedingly high temperature, over a very narrow passband.  (But a mere thermal bandpass wouldn't exhibit the spacial and spectral coherency that is characteristic.)

Tim

The output of these lasers is just UV photons. Photons have no temperature, so once the photons leave the laser, temperatures are totally irrelevant. The only thing that matters is the amount energy in the photon stream.

If the photons hit a black body, all the photon energy will be absorbed by the black body raising the temperature. As the temperature rises, the black body will radiate energy that can go back to the source but the amount of energy radiated is totally a function of temperature. For a small object to radiate a huge energy, it has to be much hotter then the large emitter on the other side of the lens.

To have a Black body release a large amount of UV, you would be looking at 10,000 deg to 20,000 deg but that is a very long way to the 14 billion degrees that can cause fusion in the case of the Livermore lasers.

All radiation is photons and so it has no temperature. Temperature is a statistical measurement of particle motion  and the old form of the 2nd law of thermodynamics was "Heat cannot of itself pass from a colder to a hotter body". The moment you add external heat pumps, lenses, lasers to push away hot atoms (in cooling helium down to near absolute zero), you can get heat passing from a cooler body to a hotter body as it is no longer a closed system. The way the 2nd law is worded nowadays, if you close the system (ie include the heat pumps, lenses, etc, power sources), then the total entropy of the closed system cannot decrease. That does not say that photons from the Moon cannot heat a small target in a vacuum to 10,000 degrees or more. As long as the total entropy is increasing, it is fine. It is merely a matter of how many photons you can get from the Moon and how small the target is. You could calculate the number of photons needed to get 10,000 degrees in a specified object.

[IanB]

Photons have no temperature

All radiation is photons and so it has no temperature.

I rather think photons do have a temperature characteristic. That is how astronomers can measure the temperature of the cosmic microwave background radiation and compare it with predictions from the big bang.

The important distinction that you mention in your post is the difference between an active system and a passive system. With an active system you can put work into it and can achieve any temperature you wish. A passive system with no external inputs cannot concentrate the photons from the moon to achieve any temperature greater than the surface of the moon. If it were possible it would violate the second law, no matter which way you look at it.

[amspire]

Photons have no temperature

All radiation is photons and so it has no temperature.

I rather think photons do have a temperature characteristic. That is how astronomers can measure the temperature of the cosmic microwave background radiation and compare it with predictions from the big bang.

The important distinction that you mention in your post is the difference between an active system and a passive system. With an active system you can put work into it and can achieve any temperature you wish. A passive system with no external inputs cannot concentrate the photons from the moon to achieve any temperature greater than the surface of the moon. If it were possible it would violate the second law, no matter which way you look at it.

You could say a lens is passive but it definitely allows a colder body to heat a hotter body.  As long as you look at the total entropy in a closed system, the 2nd Law is correct. If you get hung up on the colder body-hotter body thing, you can get yourself into trouble.

[amspire]

Just did some quick calculations. Based on an emission of about 26 lux, the amount of light radiated by a full moon would be about 10¹³ W. This excludes IR radiation which would be much higher - this is just visible light.

The amount of this power the Earth would receive during a full moon would be about about 1000th of this or about 10 Gigawatts.

So to get a useful amount of power from Moonlight - say 1Kw would keep life's barest essentials going - light, TV. modem and computer - we would need a sphere that had a diameter that was about 3200 times smaller then the earth.

That would be 8km in diameter and weigh 8 x 10¹⁶ kg.

A 1m diameter sphere would probably produce something like 16 uW during a full moon on a clear night. Over the month with clouds, you would be lucky to get 4uW.  A 1m glass sphere would weigh about 1300 kg and if you assume a 25% efficient solar cell, you would get about 1uW at night on average.

If you got a sphere the day you were born, and accumulated the moonlight energy every night, then on your 90th birthday, you could have the 1kW of power for a full 4 hours!