σ is such a small number in Stefan-Boltzman that it makes no difference at all until your radiators get hot enough to start melting.

You not only need absolute huge radiators for a space data centre, you need an active cooling/pumping system to make sure the heat is evenly distributed across them.

I'm fairly sure no one has built a kilometer-sized fridge radiator before, especially not in space.

You can't just stick some big metal fins on a box and call it a day.

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torginus 10 hours ago | parent [-]

Out of curiosity, I plugged in the numbers - I have solar at home, and a 2 m2 panel makes about 500w - i assume the one in orbit will be a bit more efficient without atmosphere and a bit more fancy, making it generate 750w.

If we run the radiators at 80C (a reasonable temp for silicon), that's about 350K, assuming the outside is 0K which makes the radiator be able to radiate away about 1500W, so roughly double.

Depending on what percentage of time we spend in sunlight (depends on orbit, but the number's between 50%-100%, with a 66% a good estimate for LEO), we can reduce the radiator surface area by that amount.

So a LEO satellite in a decaying orbit (designed to crash back onto the Earth after 3 years, or one GPU generation) could work technically with 33% of the solar panel area dedicated to cooling.

Realistically, I'd say solar panels are so cheap, that it'd make more sense to create a huge solar park in Africa and accept the much lower efficiency (33% of LEO assuming 8 hours of sunlight, with a 66% efficiency of LEO), as the rest of the infrastructure is insanely more trivial.

But it's fun to think about these things.

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jdhwosnhw 9 hours ago | parent | next [-]

This argument assumes that you only need to radiate away the energy that the solar actively turns into electricity, but you also need to dissipate all the excess heat that wasn’t converted. The solar bolometric flux at the earth is 1300 w/m2, or 2600 for 2 sq m. That works out to an efficiency of ~20% for your home solar, and your assumed value of 750 w yields an efficiency of ~30%, which is reasonable for space-rated solar. But assuming an overall albedo of ~5% that means that you were only accounting for a third of the total energy that needs to be radiated.

Put another way, 2 sq m intercepts 2600 w of solar power but only radiates ~1700 w at 350 k, which means it needs to be run at a higher temperature of nearly 125 celsius to achieve equilibrium.

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yetihehe 10 hours ago | parent | prev | next [-]

> 2 m2 panel makes about 500w

It receives around 2.5kW[0] of energy (in orbit), of which it converts 500W to electric energy, some small amount is reflected and the rest ends up as heat, so use 1kW/m^2 as your input value.

> If we run the radiators at 80C (a reasonable temp for silicon), that's about 350K, assuming the outside is 0K which makes the radiator be able to radiate away about 1500W, so roughly double.

1500W for 2m^2 is less than 2000kW, so your panel will heat up.

[0] https://www.sciencedirect.com/topics/engineering/solar-radia...

	▲	fc417fc802 3 hours ago \| parent [-]
		You can't just omit the 500 W of electric. That ultimately ends up as heat too.

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uplifter 9 hours ago | parent | prev | next [-]

>Depending on what percentage of time we spend in sunlight (depends on orbit, but the number's between 50%-100%, with a 66% a good estimate for LEO), we can reduce the radiator surface area by that amount.

You need enough radiators for peak capacity, not just for the average. It's analogous to how you can't put a smaller heat sink on your home PC just because you only run it 66% of the time.

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two_handfuls 10 hours ago | parent | prev [-]

Yes it's fun. One small note, for the outside temp you can use 3K, the cosmic microwave background radiation temperature. Not that it would meaningfully change your conclusion.