It's actually only about 3x.

As you intimated, the radiated heat Energy output of an object is described by the Stefan-Boltzmann Law, which is E = [Object Temp ]^4 * [Stefan-Boltzmann Constant]

However, Temp must be in units of an absolute temperature scale, typically Kelvin.

So the relative heat output of a 90C vs 20C objects will be (translating to K):

383^4 / 293^4 = 2.919x

Plugging in the constant (5.67 * 10^-8 W/(m^2*K^4)) The actual values for heat radiation energy output for objects at 90C and 20C objects is 1220 W/m^2 and 417 W/m^2

The incidence of solar flux must also be taken into account, and satellites at LEO and not in the shade will have one side bathing in 1361 W/m^2 of sunlight, which will be absorbed by the satellite with some fractional efficiency -- the article estimates 0.92 -- and that will also need to be dissipated.

The computer's waste heat needs to be shed, for reference[0] a G200 generates up to 700W, but the computer is presumably powered by the incident solar radiation hitting the satellite, so we don't need to add its energy separately, we can just model the satellite as needing to shed 1361 W/m^2 * 0.92 = 1252 W/m^2 for each square meter of its surface facing the sun.

We've already established that objects at 20C and 90C only radiate 1220 W/m^2 and 417 W/m^2, respectively, so to radiate 1252 W per square meter coming in from the sun facing side we'll need 1252/1220 = 1.026 times that area of shaded radiator maintained at a uniform 90C. If we wanted the radiator to run cooler, at 20C, we'd need 2.919x as much as at 90C, or 3.078 square meters of shaded radiator for every square meter of sun facing material.

[0] Nvidia G200 specifications: https://www.nvidia.com/en-us/data-center/h200/

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merman 14 hours ago | parent | next [-]

You use arbitrary temps to prove at some temps it’s not as efficient. Ok? What about at the actual temps it will be operating in? We’re talking about space here. Why use 20 degC as the temperature for space?

	▲	icegreentea2 12 hours ago \| parent [-]
		He didn't use 20C as the temperature of space. He used the OP's example of comparing the radiative cooling effectiveness of a heat SOURCE at 90C (chosen to characterize a data center environment) and 20C (chosen to characterize the ISS/human habitable space craft).

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terminalshort 11 hours ago | parent | prev [-]

You forgot about the background. The background temp at Earths distance from the sun is around 283K. Room temperature is around 293K, and a computer can operate at 363K. So for an object at 283K the radiation will be (293^4 - 283^4) = , and a computer will be (363^4 - 283^4)

(293^4 - 283^4) = 9.55e8

(363^4 - 283^4) = 1.09e10

So about 10x

I have no problem with your other numbers which I left out as I was just making a very rough estimate.

	▲	uplifter 8 hours ago \| parent [-]
		The background temp at Earth's orbit is due to the incidence of solar flux, which I took account of. I'm assuming the radiators are shaded from that flux by the rest of the satellite, for efficiency reasons, so we don't need to account for solar flux directly heating up the radiators themselves and reducing their efficiency. In the shade, the radiators emission is relative to the background temp of empty space, which is only 2.7 K[0]. I did neglect to account for that temperature, that's true, but it should be negligible in its effects (for our rough estimate purposes). [0] https://sciencenotes.org/how-cold-is-space-what-is-its-tempe...