1st thought should be to Google 'convert watts to BTU'. Why does no one seem to know how to use search engines....
FWIW, BTU/hr=Watts x 3.4129
The CFM, fair question. Don't think there is any data on that. I believe the fans move 250CFM free-air but under load, pure guesswork. I'd ballpark it as around 150CFM
I didn't know figuring out BTU was as simple as plugging in the number of watts, I thought it was different for every machine. Some machines are less efficient so they put out more heat per watt, some are designed better than others so they put out less heat per watt, etc. was my line of logic.
The miner efficiency is irrelevant and only compares hash rate to power used, has nothing to do with heat load.
For any electric power heat source they are
all 100% efficient. Power in = power out as heat.
Cooling is of course another matter as it involves work
moving heat from one location to another, not creating heat. In that case efficiency
does matter but only in the sense that it takes x amount of power turned into y amount of mechanical energy (with losses turned into heat) to move z amount of heat energy.
Hi, I just wanted to double check this and clarify that it is still correct to say that 100% of power in = heat out for the antminer.
See:
https://physics.stackexchange.com/questions/132026/is-it-possible-for-an-electrical-device-to-generate-more-heat-than-another-usingAccording to that, a device that is less efficient at converting input power into light, motion, work etc... will generate more heat than a device that is better at converting power into motion, etc.
Given this, is it still correct to say in the case of the antminer that 100% power in = heat out? (or something close like 95% power in = heat out since i would guess the device would not be 100% efficient at converting power into heat).
My best guess is given that there are no moving parts (except for low wattage fans) on the antminer, that probably close to 100% power in would equal heat out...?
Thanks for your thoughts!