4 ms·

By a factor of a million in my typo of “125M km2”, right? The available solar power calculation is still correct, I believe.

• ben_w 3y ago
  > The atmosphere attenuates our 1300W/m2 of solar energy down to 300W/m2 so for a 125M km2 planet you only get 40PW of available power, or 5PW after 80% loss in solar panels and distribution. I think you've got a mistake and a bad assumption here, too. I think the atmosphere attenuates by not to 300W/m^2, but you might also mean that the earth being spherical rather than flat geometrically reduces the solar irradiation to an annual average of 340W/m^2? But then you'd be double-counting the effect of night, and ignoring weather. The currently installed PV systems produce about 10% of their nameplate capacity, so I tend to just use that in my approximations of all the different effects together, so a 20% efficient cell[0] would produce 1kW/m^2 * 10% (capacity factor) * 20% (efficiency) gets 20 W/m^2 in practice. 1% of land with those percentages gets us 29.79 TW[1] which is indeed only about 1 order of magnitude away from current use (and we definitely want to use more energy than we do as most people don't have the luxury of energy abundance found in developed nations), but that already accounts for it being "neither sunny nor midday all the time either". But also, 1% of Earth's land area is already built up[2], so I think this may be too tight a constraint (after all, PV can be used as a surface covering for buildings and even vehicles). With sufficient political will (yes yes, I know that's wishful thinking), it's not unreasonable to build a thick enough set of conductors to make even a global power grid have negligible resistance — you'd need a 1m^2 cross section for an equatorial ring of aluminium[3] to have 1Ω resistance, and while this would be a big project today, it's not so large as to be absurd for a decade-long project. [0] The record for efficiency is 47.6%, though this will only matter if we ever ran out of land: https://en.wikipedia.org/wiki/Solar-cell_efficiency https://en.wikipedia.org/wiki/Solar-cell_efficiency [1] https://www.wolframalpha.com/input?i=land+area+on+earth+*+1%25+*+20W%2Fm%5E2 https://www.wolframalpha.com/input?i=land+area+on+earth+*+1%... [2] https://ourworldindata.org/land-use https://ourworldindata.org/land-use [3] Don't put 20 TW through a single ring. If I did the maths right, the magnetic field at the surface will be in the order of 1 tesla depending on how you choose to distribute current and voltage. Also, I have no idea how much energy such a thing would absorb from a CME, but I think the big CMEs can be in the range of 5e25 J, and if that's efficiently absorbed it would vaporise even a ring that big.
• jacquesm 3y ago
  Indeed. As for the available solar power insolence is about 1000 W / square meter at solar noon with panels aimed at the sun, depending on tracking you either have zero, one or two axis that are going to be 'off' and then there is panel efficiency to account for, so in practice a 100 square meter area of panels will provide out of the 50 KW available only about 17 KW peak, and usually less than that.