Naked Science Forum
Non Life Sciences => Technology => Topic started by: Airthumbs on 29/01/2012 23:10:39
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Working with the help of CliffordK we came up with the idea of a black solar panel with a thermoelectric heatsink on the back.
A good place to use this would be in space. Daylight surface temp on the Moon (http://lunar.gsfc.nasa.gov/moonfacts.html) is 1340C and Shade or night -1530C. I understand that thermoelectric (http://en.wikipedia.org/wiki/Thermoelectricity)
devices require a heat gradient to produce electricity.
How efficient would a black solar/thermo electric panel be given the much lower temp gradient on Earth?
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It's a heat engine, and, according to a French geezer called Carnot, the highest efficiency you could ever hope for can be found from -
"The efficiency η is defined to be:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fen%2Fmath%2F6%2Fb%2F4%2F6b479c964c05142f9376aa6de5f14e07.png&hash=df61762b786c36f81e2fa4299de3b216)
where
W is the work done by the system (energy exiting the system as work),
QH is the heat put into the system (heat energy entering the system),
TC is the absolute temperature of the cold reservoir,and
TH is the absolute temperature of the hot reservoir.
From http://en.wikipedia.org/wiki/Carnot_cycle (http://en.wikipedia.org/wiki/Carnot_cycle)
If you do the sums, you should get about 0.7 (which ain't too shabby).
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I spent two hours last night trying to work through the equation and got stuck on Entropy. However the values look pretty good for a temp gradient of 200oC
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If you play with the numbers you will find that a 200 degree difference at low temperatures allows far greater Carnot efficiency than a 200 degree difference at high temperatures.
I use a thermodynamic calculator at www.thermofluids.net If you are serious about this stuff, you might find it a great help.
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Putting asside the Carnot efficiency for a momnet.
How are you going to maintain the temperature gradient? (i.e. how would you cool the heat sink?)
It may be a better way of recovering waste heat (there is a lot of it about).
I do not know if you have seen these
http://www.ecofan.co.uk/woodstove-ecofans.html (http://www.ecofan.co.uk/woodstove-ecofans.html)
They rely on the thermoelectric effect to drive a motorised fan that helps distribute heat around a room, but more critically draws cool(er) air across the fins thus maintaining the temperature gradient.
(having experimented by isolated the fins from the air current, the fan speed drops off quite quickly)
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With solar power, it is easy enough to magnify the input, as long as one has a solar tracker.
I have my old 8 foot satellite dish. If I line it with tinfoil, it should do quite nicely at generating some heat on a hot day. [;)]
The efficiency is apparently written in Kelvin. So, the efficiency is a bit better for low vs high temps, but it isn't too bad.
I'm looking up on the internet. There are numerous thermoelectric cooling modules for sale (and a few thermoelectric generators). I assume they are essentially the same, just wired oppositely.
I'm seeing some rated with a TMax of 180℃.
For the Cooling modules, the Watts seem to vary a lot, with the highest rating of the cooling modules being:
TEC1-28716T125:
34V, 16A, 345W, 68℃ temp difference, 50mm×50mm×3.95mm
Hmmm,
So, if it is all based on about 20% efficiency,
Does that mean I need to give that little tiny patch, 1725W of Heat to expect to get the 345W back. Whew...
Maybe I do need a big concentrator.
The trick might be to take a concentrated solar panel such as this one.
http://www.ebay.com/itm/Clearance-New-20W-20-WATT-Portable-Solar-Panel-Concentrator-PV-/280815442965?pt=LH_DefaultDomain_0&hash=item4161e79c15
Which requires 2 axis tracking (although it could be done with 1axis tracking).
Add a triple junction solar cell (http://en.wikipedia.org/wiki/Multijunction_photovoltaic_cell) (about 40% efficient with concentrated sunlight) + the thermoelectric generator.
And, perhaps one could bump it up to a range of 50% to 60% efficiency.
Of course, high efficiency is only a measurement of watts per unit area. The plain flat silicon cells are relatively cheap compared to other methods, if the application area permits.
In space, though, it may make a difference, as long as it doesn't add too much weight, and doesn't cost longevity of the solar cells (or, even improves long-term power generation).
A couple of percent might also help an automobile application, but designing a good mobile solar concentrator is a bit of a pain.
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What about taking your basic Toyota Prius.
Run the radiator COLD.
Insulate the engine and the exhaust system.
Put in one set of thermal electric generators between the engine coolant and the radiator.
Put a second set of thermal electric generators between an exhaust heat exchanger and an secondary radiator.
Even with only 20% thermal energy recovery, it should give a significant boost in system electrical energy.
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Your 8 foot dish should collect at least 4Kw but you should take care not to exceed 180°C on your hot junction whilst I see that cooling of the cold junction might be a problem.
Perhaps you could divert some of the beam to a Stirling engine to pump coolant via a mini cooling tower
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What about taking your basic Toyota Prius.
Run the radiator COLD.
Insulate the engine and the exhaust system.
Put in one set of thermal electric generators between the engine coolant and the radiator.
Put a second set of thermal electric generators between an exhaust heat exchanger and an secondary radiator.
Even with only 20% thermal energy recovery, it should give a significant boost in system electrical energy.
Sounds like a reasonable proposal although I am a bit confused as to where the solar element comes from...
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For anyone that (has some time) and has not seen it take a look at http://www.cockeyed.com/incredible/solardish/dish01.shtml (http://www.cockeyed.com/incredible/solardish/dish01.shtml)
Which is the blog about the construction and testing of the "light sharpener" a 12foot diameter dish covered with mirrors.
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Your 8 foot dish should collect at least 4Kw but you should take care not to exceed 180°C on your hot junction whilst I see that cooling of the cold junction might be a problem.
Perhaps you could divert some of the beam to a Stirling engine to pump coolant via a mini cooling tower
True,
Temperature regulation isn't too bad if one uses liquid filled heat pipes, with a liquid that has a boiling point near the target temperature (plus some kind of a condenser if one wants).
I have some trough panels that would likely be better than the satellite dish. I don't need to burn everything that I get near it. Although, I may have to keep my eyes peeled for some cheap mirrored bathroom tiles.
Cooling to ambient temperatures is easy enough.
I've got too many trees, but one of the places I've thought about putting up some solar panels is near a natural spring. Not a lot of flow, but perhaps enough.
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Even with only 20% thermal energy recovery, it should give a significant boost in system electrical energy.
"Only" 20% would be pretty amazing!
You're lucky if you can continuously convert 20% of the thermal energy in the gasoline into work.
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Run the radiator COLD.
Okkkk. You're in space. Space is a vacuum. Vacuums are thermal insulators.
You can't shift more power than the Stefan Boltzmann law will let you:
http://en.wikipedia.org/wiki/Stefan-Boltzmann_law
In other words the area of the cold plate needs to radiate the heat away, and the hotter it is, the better it does this, according to the 4th power of its temperature, so you need it to be as hot as possible (and/or as big as possible).
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Ok, I was reading a bit more of the thermoelectric generators.
Most seem to have efficiency rates in the couple of percent (3% or so) range. There are some predictions of high Carnot efficiency devices, but they seem to be few and far between.
I assume one could calculate the emissivity of the back (cool side) of the system in space. It is written in square meters, but it is unclear how it is affected by a convoluted surface in a vacuum, as one would increase surface area, but also increase reabsorption.
If j=σT4
And, the sun's TSI is about 1400W/m2 (at Earth).
I assume one can set j=TSI, and solve for T.
So, one gets the 4th root of 1400/5.7x10-8
Hmmm, I come up with about 396K... That just seems too hot. [xx(]
Even dropping it down to 700W (assuming 50% system efficiency, including solar cells), one still gets 332K.
Of course, one could increase the area with a large perpendicular heat sink.
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I agree with CliffordK. And as post started with the temp diff of surface of the moon, i.e. 400K to 120K, I suggest to look deep into it.
Moon has very large radius thus large area to absorb solar radiation and radiate heat energy. Now its exposure to sun is also large. Moon surface is poor conductor of heat so conduction is slow. Temp grad over the moon surface is very small and this leads to this high temp difference between two surface. For guys who wonder why something like this doesn't happen on earth? Well, it is just because of our atmosphere. Due to convection that temp grad is reduced drastically.
Now if you take a system in space to generate electricity using temp grad, you will be needing very large amount of area to go for. A small surface is good for nothing. Solar panels are almost at same temp on both sides. So in space you can have high temp grad in small place only in one condition: in spacecraft when electric components are generating heat. Everything else have very small grad.