3 ms·
Let us, per the article's title, do the math... * Total energy consumption = 10^26 watts (hope the caret comes through HN formatting) * Assume some magical ge
by mgsouth 4y ago
Let us, per the article's title, do the math...
* Total energy consumption = 10^26 watts (hope the caret comes through HN formatting)
* Assume some magical generation and transport mechanisms which don't involve energy gradiants (avoiding pesky thermodynamic realities), and are 99.9% efficient.
* Total waste energy is 10^23 watts.
* Radiators are on Earth, and we dedicate 1/2 of the entire surface area (not just land) to radiators. Earth has about 5.1 x 10^8 km^2, so say 2.5 x 10^14 sq. meters.
* Radiators for the waste heat would be dumping 10^23 W / 2.5 x 10^14 m^2 = 4 x 10^8 watts per square meter. That's 400 megawatts. Per square meter.
* That's a black-body temperature of 9,165 K (the Sun's surface is 5,772 K). Peak wavelength is 316 nm. So half the Earth's surface is a glowing UV light pointed at the sky. I don't think we'd have to worry about the ozone layer disappearing. (A very bright, inefficient UV lamp--plenty of infrared, hard UV, and soft X-rays to go around!)
* OK, so put the power plants in geostationary orbit. Say 1 million power stations, each with radiators 1 square kilometer.
* 1e6 stations * 1e9 square meters = 1e15 sq meters.
* So, that's only 1e8 watts/meter! Hmm. Still 6,480 K, but peak is blue-violet rather than UV. Still lots of UV and infrared, but it's space, who cares. Just don't cross the beams (ha ha, meaning no deep-space traffic). And mind the Moon! (Wouldn't want to give anbody on earth a reflected sunburn).
* The radiation pressure from the waste engery for one power stations would be about 730 kN (160,000 pounds). Would need counter-balanced beams.
* Geostationary orbit is about 26,000 miles. The surface area of a sphere that big is 2e16 square meters. So about 1/20th of the sky would be covered with powerplant radiators.
And this is just for the hypothetical 0.1% waste energy. Multiply all of the above by 1,000 for the actual used energy. So you'd have half the Earth's surface as power receivers slurping up 400 gigawatts/square meter of X-rays and hard UV. Each power station would see a force of 739 mega-newtons; you'd have to have a counter-balancing beam pointing away from Earth. And then you'd need to radiate away all that energy after it was used.
Edit: Oh, and while 1/2 of the Earth is receiving 400 GW/m^2, the other half is using that amount. Which would be the real limit. I don't think anybody would carry around a 10 GW iPhone 237+.
- whatshisface 4y agoWhy not have one square meter of radiators that emits gamma rays?
- lumost 4y agoAs I recall, gamma rays cannot be generated by thermal processes such as electron orbit decay - so we would need a nuclear radiator which converted waste heat into atoms which would rapidly decay into gamma rays. Assuming we’ve done all of that… you would still need to deal with the rocket engine you’ve created. Sending out a tight beam of a suns worth of energy is equivalent to a rocket engine with 1x10^18 newtons, creating challenges of its own - including an appreciable acceleration of the planet earth. A more plausible escape hatch for the energy may be a black hole heat sink. In theory, such a device could perfectly “recycle” waste heat into usable energy.
- mgsouth 4y agoOops, I miscalculated the total satellite area; (meant 1e6 x 1e6 = 1e12). To keep everything else the same, replace "each with radiators 1 square kilometer" with "each with radiators of 1,000 square km."