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# Why don't they build really high poles to mount solar panels?

 Why don't they build really high poles to mount solar panels?21-04-2019 21:13 Tai Hai Chen★★★★☆(1013) Considering the atmosphere absorbs more than 98% of incoming solar energy, making solar panels useless to use at ground level as compared to in space where solar panels are used on spacecraft like artificial satellites, why don't they build poles tens of thousands of meters high reaching above the clouds to mount solar panels where at such heights solar panels would work much better than on ground level? 21-04-2019 21:40 James___★★★★☆(1172) Tai Hai Chen wrote:Considering the atmosphere absorbs more than 98% of incoming solar energy, making solar panels useless to use at ground level as compared to in space where solar panels are used on spacecraft like artificial satellites, why don't they build poles tens of thousands of meters high reaching above the clouds to mount solar panels where at such heights solar panels would work much better than on ground level? This is where you're getting into atmospheric physics. I have a "pet" theory that there is actually less solar radiation at such elevations. Solar panels are passive. Basically what they are absorbing is background electromagnetic radiation. There's simply more of it at ground level. Their efficiency might actually be related to the Boltzmann constant and the definition of heat. Heat is the flow of energy. If it's is 70º F. or 21.11º C. it is 343.15 kelvins outside. I know, can you believe it? Some people still use Fahrenheit and Celsius. If that is the average KE and if we assumed for the purpose of discussion that the average KE is 3/2kT or 3/2k343.15, then if we said 3/2k333.15 was the low and that 3/2k353.15 was the high, then heat would be KE = 3/2k20. And it would be the KE = 3/2k20 that solar panels are absorbing and because of the crystals they use, would be converting it into electrons. 22-04-2019 01:54 Into the Night★★★★★(7663) Tai Hai Chen wrote:Considering the atmosphere absorbs more than 98% of incoming solar energy, making solar panels useless to use at ground level as compared to in space where solar panels are used on spacecraft like artificial satellites, why don't they build poles tens of thousands of meters high reaching above the clouds to mount solar panels where at such heights solar panels would work much better than on ground level?The atmosphere does not absorb 98% of the incoming solar energy.The Parrot Killer Edited on 22-04-2019 01:54 22-04-2019 21:53 Wake★★★★★(4021) Tai Hai Chen wrote:Considering the atmosphere absorbs more than 98% of incoming solar energy, making solar panels useless to use at ground level as compared to in space where solar panels are used on spacecraft like artificial satellites, why don't they build poles tens of thousands of meters high reaching above the clouds to mount solar panels where at such heights solar panels would work much better than on ground level?Huh? 51%, on the average, solar radiation gets to ground levels. The absorption begins in the upper stratosphere and in the Troposphere it is a linear reduction with altitude. The Troposphere height is about 8 miles. There is NO materials that has the structural strength to support solar cells for as much as a quarter of a mile high and hence there would be virtually no improvement placing these cells on the tops of poles. Also the solar energy gathered would result in large scale shadowing of the ground beneath such a thing and cause large scare environmental damage. 22-04-2019 21:59 IBdaMann★★★★★(3531) Tai Hai Chen wrote: Considering the atmosphere absorbs more than 98% of incoming solar energy, making solar panels useless to use at ground level as compared to in space where solar panels are used on spacecraft like artificial satellites, why don't they build poles tens of thousands of meters high reaching above the clouds to mount solar panels where at such heights solar panels would work much better than on ground level?The atmosphere absorbs more than 98% of only the UV-C band. Most of the remainder plows right on through.Global Warming: The preferred religion of the scientifically illiterate.Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafnYou are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank :*sigh* Not the "raw data" crap. - Leafsdude IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist 22-04-2019 22:02 IBdaMann★★★★★(3531) James___ wrote: I have a "pet" theory that there is actually less solar radiation at such elevations. Technically there is greater power received the closer the panel is to the sun. This is a physics thing called the Inverse Square law. If you successfully place panels thousands of meters closer to the sun then you will achieve a gain in power. Having said that, incident photons arriving at earth are essentially on parallel paths so the increase in power over a few thousand meters would be negligible, ... yet it will be non-zero. How much will you gain by elevating them on towers? Let AFTM = a few thousand meters, D = distance from the sun to earth's surfacePower(surface solar cells) = Power(incident) / D^2Power(elevate solar cells) = Power(incident) / (D-AFTM)^2At this point you might try to salvage your theory by asking Into the Night about the electrical resistance over the conductors (power cables) used to deliver the new solar-cell-acquired electricity those few thousand meters down to the earth's surface. Maybe the loss induced by those power cables would exceed the negligible gain from having the solar cells a few thousand meters high. Maybe just letting the solar energy just continue down on its own to the surface renders a greater yield, i.e. Maybe Power(incident) / D^2 > Power(incident) / (D-AFTM)^2 - cable lossWho knows? I bet Into the Night can tell us. I bet he knows the answer. Did you ever think that Into the Night might be the salvation of one of your "theories"? Yes, I was pretty shocked at the idea as well. I wonder how this will work out for you. James___ wrote: Solar panels are passive. Basically what they are absorbing is background electromagnetic radiation. ... yeah, when it's the sun that's in the background!James___ wrote: There's simply more of it at ground level.No, I assure you that as you get closer to it, there is more of it. You don't have to take my word for it. You don't have to accept the Inverse Square law. If you can just find a way to get yourself a lot closer to the sun, you will have all the convincing you need. James___ wrote: Their efficiency might actually be related to the Boltzmann constant and the definition of heat.Aaaah, I should have guessed. You high-tailed it to Wikipedia and have obsessed over the misinformation therein. You eagerly confused the idea of the kinetic energy of a gas particle at a particular temperature and volume with the completely separate idea of the thermal energy of a particle determining it's temperature. You passionately accepted the mistaken notion that the Boltzmann Constant is a measure of kinetic energy rather than correctly identifying it as an energy scalar for temperature. James___ wrote: Heat is the flow of energy. ... as long as the energy is specifically thermal energy.James___ wrote: If it's is 70º F. or 21.11º C. it is 343.15 kelvins outside. I know, can you believe it? Some people still use Fahrenheit and Celsius. If that is the average KE ... Except here you seem quite content to shift goalposts on yourself and declare thermal energy to be kinetic energy. Either you are aware that kinetic energy and thermal energy are separate and distinct forms of energy, or you are not. If you are aware then you are blissfully deluding yourself. If you are unaware then you are obviously scientifically illiterate. There is one way to find out which case is true:Are you aware that kinetic energy and thermal energy are separate and distinct forms of energy? Edited on 22-04-2019 22:11 22-04-2019 23:50 Into the Night★★★★★(7663) IBdaMann wrote:James___ wrote: I have a "pet" theory that there is actually less solar radiation at such elevations. Technically there is greater power received the closer the panel is to the sun. This is a physics thing called the Inverse Square law. If you successfully place panels thousands of meters closer to the sun then you will achieve a gain in power. Having said that, incident photons arriving at earth are essentially on parallel paths so the increase in power over a few thousand meters would be negligible, ... yet it will be non-zero. How much will you gain by elevating them on towers? Let AFTM = a few thousand meters, D = distance from the sun to earth's surfacePower(surface solar cells) = Power(incident) / D^2Power(elevate solar cells) = Power(incident) / (D-AFTM)^2At this point you might try to salvage your theory by asking Into the Night about the electrical resistance over the conductors (power cables) used to deliver the new solar-cell-acquired electricity those few thousand meters down to the earth's surface. Maybe the loss induced by those power cables would exceed the negligible gain from having the solar cells a few thousand meters high. Maybe just letting the solar energy just continue down on its own to the surface renders a greater yield, i.e. Maybe Power(incident) / D^2 > Power(incident) / (D-AFTM)^2 - cable lossWho knows? I bet Into the Night can tell us. I bet he knows the answer. Did you ever think that Into the Night might be the salvation of one of your "theories"? Yes, I was pretty shocked at the idea as well. I wonder how this will work out for you. There are two things to consider here. One is the length of the conductors, their size, the voltage you put on them. With that you can calculate loss. You can also calculate the weight of the conductors, which must be added to the overall weight on the pole.The second thing to consider is the pole itself. I'll start with this.Since you used 'thousands of meters', I will assume the pole to be at least 2000 meters high, or approx. 6,651 feet high. This puts the pole well into an air navigation hazard, but let's assume you somehow get the FAA to go along with it.However, a 6,651 foot pole will not get the solar array above the clouds in many cases. Cloud tops can go up to 35,000 ft in many cases. If you build a pole that high, you will need a pole that is some 11,000m high. For this calculation, I'll use this figure.Wind pressure to wind speed is a simple relationship: P=0.6 * W^2 where W is wind speed in m/s, and P is pressure in N/m^2.As of this moment, upper winds in Washington happen to be from 300 at 44 kts. This translates to approximately 23 meters per second. The resulting pressure on a 1 m^2 array is therefore 1161 N. All of this load is on a lever 11000m long and acting against it's base as the fulcrum. Let's assume a base of 100m (a 600 ft hole).From here, it's the same as a weight and balance calculation. 1161N at 11000m acting on a fulcrum at 600m will place a load of 12771000 N on the base. This is about 1852 pounds per square inch at the base. Since the base is 10ft in diameter, that means about 1,532,520,000 lbs of pressure will be applied to the base attempting to bend the pole.One of the strongest materials for this type of load is carbon fiber. Assuming a 10 ft diameter pole, the resulting load at the base is this modest pressure at the top with such a long moment arm will easily collapse the pole. Carbon fiber simply can't take that kind of bending stress. I'll let you picture the scene of an 11000 meter pole crashing to the ground. The weight of the pole made of that material will itself be approx 1.5 million lbs. Remember this is a 10 ft carbon fiber pole crashing from 30,000 ft.Remember, this is just the modest winds of 44 knots. In a hurricane or even a modest storm, loads are correspondingly higher. Remember it is wind speed squared.Guy wires help with tall structures like this, but remember, they are aluminum or steel cables, and subject to some elasticity. Over this distance they essentially produce no useful stabilization. The forces remain the same.Now, the wiring.Two conductors are required. Assuming you use a self-powered inverter to increase the voltage on the wire to 800kv, you will also need to add the weight of insulation on the wire, since they are next to a grounded pole. Bare aluminum wire with no insulation will weigh about 122 lbs per 1000 ft. There will need to be 30000ft * 2 or 60000ft of wire. This will weigh a total of 7320 lbs. No matter how you mount it, it will stretch like taffy. I'll ignore this and use the figures for new wire. Stretching thins the wire and only makes things worse.The resistance of such a wire is about 0.130 ohms per 1000 ft at DC. For AC at 60hz, the additional impedance added will be 7.8k ohms. At a one amp flow at 800kv, that will result in a loss of 6,240 megawatts.Note that this does not include the losses caused by the inductive or capacitive effects of such a cable at 60Hz.You can use copper to get a lower resistance, but it weighs twice as much as aluminum, corrodes a lot easier, and stretches more easily than aluminum does.IBdaMann has already shown that the additional power gained by placing the array at 30,000 ft is fairly negligible.In my opinion, constructing such a tower would be extraordinarily dangerous and foolish, and would give you negligible improvement in power, which would be more than eaten up by the cables you would have to run to it.The Parrot Killer Edited on 22-04-2019 23:52 23-04-2019 01:05 IBdaMann★★★★★(3531) Into the Night wrote: Wind pressure to wind speed is a simple relationship: P=0.6 * W^2 where W is wind speed in m/s, and P is pressure in N/m^2.Well, that was a bonus. I didn't expect you to delve into torque but I sure did get a kick out of imagining high winds on an 11,000-meter pole. I got a kick just out of the idea of planting an 11,000-meter pole for a solar-cell. Imagine trying to cut costs and using one 11,000-meter pole to hang a forest of solar cells! Maybe you could drop a hook-chain from a winch on a "stabilizer" satellite for added support from above! Thank you for "going there" ... and with a straight face no less. I had to catch my breath I was laughing so hard. Into the Night wrote: The resistance of such a wire is about 0.130 ohms per 1000 ft at DC. For AC at 60hz, the additional impedance added will be 7.8k ohms. At a one amp flow at 800kv, that will result in a loss of 6,240 megawatts.Note that this does not include the losses caused by the inductive or capacitive effects of such a cable at 60Hz.[ ... content cut ...]In my opinion, constructing such a tower would be extraordinarily dangerous and foolish, and would give you negligible improvement in power, which would be more than eaten up by the cables you would have to run to it.Well, it looks like James__ owes you a big "thank you" for saving him from snatching defeat from the jaws of victory. James__, was it your idea to run Hillary?Global Warming: The preferred religion of the scientifically illiterate.Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafnYou are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank :*sigh* Not the "raw data" crap. - Leafsdude IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist

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