Quote from: philipma1957 on October 13, 2017, 09:57:28 PM

Quote from: cupfish on October 13, 2017, 09:24:41 PM

Quote from: NotFuzzyWarm on May 05, 2017, 02:48:49 PM

Quote from: GMPoison on May 05, 2017, 02:29:27 PM

Quote from: NotFuzzyWarm on May 05, 2017, 02:01:26 PM

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-using

According 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!

almost all the energy converts to heat . 99.0% ? most of that stays in the room where the psu and the miner is located. unless you vent it.

the small amount that leaves the rooms is in the signal on the ethernet cable. and if the cable goes to a switch in the mining room I would argue that it is 100%.

and the switch is sending the energy out of the room not the miner

the fans produce heat and the move the heat they produce and the heat the miner produces. maybe 10 watts in the mother of each fan. if you measure the motor of the fans they are warm but the fan blows the heat away.

Information does contain some energy but for all practical reasons 99.9999999999999999999999999999999999% of the energy is converted to heat. That is rounded up alot so please don't use it as perfect.