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Comment by skrebbel | Hacker News Reader

> A bigger issue is the lack surface area for cooling. Not even The Expanse gets that one right.

Care to elaborate? What's "lack surface area" and how does the Expanse get it wrong?

Scifi ships tend have engines with stupendous power ratings (Gigawatts and higher). Even with 99% of efficiency they'd have to shed megawatts of heat. Since there is no convective or conductive cooling in space your steady-state option is radiative cooling which is constrained by surface area, temperature difference and emissivity. You don't have much flexibility on the latter two. So you need to increase surface area. The ISS' radiators take up about 200m² and can reject 70kW. There are non-steady-state options such as sacrificial coolants or plain thermal mass but that would require extra mass, resupplies or additional downtime. In The Expanse the ships just don't have enough area considering they're powered by fusion reactors burning constantly.

What about lasers for emitting waste heat? Obviously there are practical issues with having powerful-enough lasers essentially powered by a heat engine, but are there thermodynamic reasons it couldn't work?

This question is not 'Expanse'-related but inspired by David Brin's 'Sundiver'.

I think you're going to run into problems turning waste heat into usable energy, because it implies there's a cold sink already that you can use to convert the temperature differential into electrical energy to drive your laser. High end laser efficacy ranges from 30% to 50%?[1] But regardless, you're asking to get low-entropy energy out of high-entropy energy, and that's one of those "you can't win or break-even at thermodynamics in a closed system" kind of thing. Since spaceships come conveniently jacketed in a vacuum-insulating outer layer (i.e. space :) you're still stuck with some variant of radiative heat transfer, and "a big piece of metal being used as a infrared heat radiator because we pushed heat into it using a refrigerator loop) is still the cheapest option.

The last option techie that I've ever heard for rejecting heat already onboard on a spaceship (and this is pure technobabble from proposals about Star Wars ships massive power generation needs) would be some magical way to convert heat into neutrinos, which, since neutrinos can pass though most matter, could be inside the ship and still function as a heat rejection mechanism. [2]

[1] https://www.laserfocusworld.com/lasers-sources/article/16547...

[2] https://www.theforce.net/swtc/power.html

There's a writeup here that you may be interested in, if you haven't seen it already:

https://toughsf.blogspot.com/2019/10/the-expanses-epstein-dr...

The author suggests creating a fusion reaction well behind the ship, but channeling the useful parts of it in an electromagnetic field tunnel to create thrust while dissipating the heat more effectively. Vaguely like Project Orion, but with a more-or-less continuous external fusion reaction instead of a series of nuclear explosions. I am not a physicist, but they seem to have done their homework.

Incredibly detailed analysis. That site is what the Internet was supposed to be all along, dammit.

At least on the TV show the reactors are internal though, so this (quite neat!) speculation doesn't work there.

Whatever you do with internally generated energy, you either get it out or it accumulates, overheating you.

You need 3-10 m^2 of radiators for every kW of power you use. If your ship is powered (I mean internal power, not engine exhaust) by a 1MW reactor, then you need radiators close to the scale of a football field pushing out waste heat, no matter what tech that reactor uses.

Delightfully counterintuitive given the low temperatures of space - would it help said ships to hide on the dark side of the moon and avoid solar radiation?

Where does that 3-10 m^2 per kW estimate come from? The best I could guess based on my shaky grasp of radiation is to equate internally generated power to total radiated power using the Stefan-Boltzmann Law:

https://www.wolframalpha.com/input/?i=1+kW%2F%28Stefan-Boltz...

Based on that, you need 3-10 m^2 of radiative surface area per kilowatt, but that's assuming your equilibrium temperature is 203 to 275 K (-70 to 2 °C). Assuming I haven't made some basic mistake, couldn't you decide to heat part of your surface to some much higher temperature and radiate most of your internal power out of that part?

This is what our current spacecraft radiators manage. You also have to take into account that radiators also interact with the sun and with the rest of the spacecraft.

Of course, you could have a heat pump pushing heat to a higher temperature surface, some current radiators do that, but it has its limitations requires more energy for the pump the higher temperature difference you want to sustain.

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