Bitcoin Forum

How many video cards can cool 3-4 Kilowatt air conditioner?  ???

I think it is incorrect to calculate like this: "60 video cards = 60 x 200 watts = 12 kilowatts of heat, so you need 4 x 3 KW air conditioners to cool it".

Video cards produce heat, so they can't cool anything.  ???

You need to look at the BTU output of the air conditioner not it's power consumption.  Then look at the BTU output of the cards.

Since computers convert power to heat almost perfectly take the power use of the miners into this formula:

http://www.rapidtables.com/convert/power/Watt_to_BTU.htm   (17060 BTU)

So if you have a 5000W setup, you need:

http://www.newegg.com/Product/Product.aspx?Item=96-865-002&IsVirtualParent=1  (21600 BTU)

You need extra cooling capacity, I am not sure how much.

agree. you can't cool with video cards.

The amount of heat generated by 1 kilowatt worth of computer (about four 200W video cards + system + 900W power supply inefficiency):

1kWh = 3,412 BTU

So we need 3400 BTU of A/C to remove that much heat generation.

The the next question is how much power will that cooling require. We must look at the the energy efficiency of air conditioning, which is typically given as SEER (seasonal energy efficiency ratio), which we must translate into an actual efficiency measure.

COP = 3.41214/EER

As a low-water benchmark, lets look at a 5000 BTU window AC unit ($110 on lowes.com website)
Specs: EER=9.7, Watts=515 (calculated efficiency multiplier: 2.84)

So this particular model moves 1462W of heat with 515W. A central AC will be more efficient, so 3000kW * 2.84+ = at least 8000 watts (but nothing left to actually cool your house against summer temperatures).

Real miners don't use AC though, why reduce profitability? Put the units in a closed room with a window fan, or in the garage.

Thank you for your response  :)

Will one window fan be enough to cool for example 13-15 kWt of heat?
What happens if it starts raining outside? Will the water get inside the room through the fan?

Water getting trough the AC?

Are you trolling?

First realize that AC is a heat pump and can remove ~3X - ~4X heat for the power it consumes from the main. (Coefficient of Performance)

A good rule of thumb is
Power consumption of AC in Watts * 3 = Maximum heats in watts removed.
So lets say you have 1850Watt AC (~18.5K BTU, 1.5 Ton) It will remove 5.4KW Heat. If each rig consumes 500 Watts, you can remove heats from ~10 rigs.  

depends on a few factors:

1.  sq. ft. of room
2.  insulation of room
3.   volume of room
4. max wattage per circuit in room  (15a, 20a, 30a, 40a)
5.    btu of air conditioner.
6.  if its window, duct or vent hoze (portable).
7.  1st story, basement, second story, etc.


For 1 bedroom about 500sq feet you need at least 8,000 to 12,000 btu of cooling for a huge mining farm.

Offc there are several factors. But the main ones are wattage/COP.

If the outside unit is in the sun or in the shadow, do windows have curtains and weather the airflow is restricted or not at both ends, is more important than the size of the room etc.