this post was submitted on 05 Oct 2026
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[–] monketman82@lemmy.world 60 points 1 day ago (9 children)

Agate LLC is reporting 52.65 megawatts of electricity during peak electrical demand and 13.299 megagallons of water between cooling towers, evaporative systems and site operations in the last year.

That totals out to 13 million gallons of water.

[–] fonix232@lemmy.world -2 points 1 day ago (4 children)

53 megawatts "during peak electrical demand" or "last year"?

Because timespans matter. 53 megawatts is the rough average annual power consumption of 5 households. So if that's the datacenter's annual power usage... they're not using that much power after all. On the other hand, if that's the monthly usage, that's more in line with compute/data center usage. And if it's daily use... 😬

Same with the water. 13 million gallons is about 260 thousand bathtubs worth of water. That's basically what a town of ~300k will use in one night for bathing only. So for annual usage it's not that bad. For monthly usage it's passable. For daily usage... again 😬

[–] StaticallyLazy@lemmy.blahaj.zone 3 points 1 day ago* (last edited 1 day ago)

You'll need to give me a source for those numbers. My 1,300 square foot house has a 200 ampere main breaker, so I can't draw more than about 0.024 megawatts. It would therefore take more than two thousand similar-sized houses to draw 53 megawatts—and that's only at full power draw, which I never come remotely close to.

Also it's worth mentioning that timespans don't actually matter here: monthly usage and annual usage are the same number when dealing with watts. When a source says "average annual power consumption", they likely mean that the power measurement is averaged throughout the year and that hot/cold seasons are accounted for.

If you wanted to do math involving time, you would need to use watt-hours (or kWh, MWh, etc.), for which time is a factor. For example, if you drew a constant rate of 1 kilowatt for a full day, you would have 24 hours * 1 kilowatt = 24 kilowatt-hours.

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