Mining room cooling - How much airflow for 20kW?

iceTony

copper member

Activity: 30

Merit: 0

If this calculation for one m3 for 20kw ?

Quote from: NiklasFalk on July 28, 2018, 01:32:58 PM

https://www.engineeringtoolbox.com/cooling-heating-equations-d_747.html

hs = cp ρ q dt

hs= 20 kW
cp = specific heat of air (1.006 kJ/kg C)
ρ = density of air (1.202 kg/m3)
q = 2000/3600=0.555555
dt = temperature difference (C)

So with 20kW that airflow will see a delta T of 20/(1.006*1.202*0.55555)=29.8 C

Lets take a real cool ambient temp of 20C, that would mean that the air exiting will be 60C, and you cards will then most probably will be running at 80+C.

Double the Air flow and the deltaT will be reduced to half, with exit T of 45C, a much better situation.
The math is actually really simple when you reduce the situation to total air flow and total power inserted to the system.

With very controlled airflow (so all cards get the ambient air and no exhaust from another card) the cards will see a lower peak. Water cooling would be even better to control where the heat goes. But the heat transfer by air is till the same, 20kW transferred to 2000 m3/h air gives deltaT of 30C.

philipma1957

legendary

Activity: 4116

Merit: 7849

'The right to privacy matters'

Quote from: onegiantcock on July 28, 2018, 09:10:18 AM

I’d use the whole room for 80% cold/intake air - and isolate the hot side at the exhaust fan

this is the way to go.

Andartis

jr. member

Activity: 306

Merit: 7

No way I invest in an AC for that room Grin

dagarair

sr. member

Activity: 847

Merit: 383

I am a AC guy.

Here is the calc if you want to cool the place. (I would not, but some do.)
Total watts x 3.14 BTU / 12,000 = Total Tonnage

Example
20K watts x 3.14 = 62,800 / 12,000 = 5.23 tons of air will do it without ventilation. So 6 Tons counting room etc would suffice to keep that beast about 72 F

Andartis

jr. member

Activity: 306

Merit: 7

I have to say, that I have to spend some more time understanding the physics of heat.
But would you say, I will have to change the ventilator for my room i described in the first post?
Its something I have, so i mighr give it a try. Since my rigs are splitted at 3 locations. 2 of the locations are just ventilated by open windows Tongue

TheRealCashen

newbie

Activity: 56

Merit: 0

My next task to do when it starts to cool off is to have both fans work off the thermostat and have a simple 1 way flap above it so when one is on it will not draw from the other one.

I'll most likely set them at 80F and 100F or something like that.

https://quietcoolsystems.com/ makes an extremely efficient variable fan which would be totally awesome.

https://www.amazon.com/Quietcool-Attic-Gable-Ventilation-Exhaust/dp/B073ZNGR6S

TheRealCashen

newbie

Activity: 56

Merit: 0

Quote from: NiklasFalk on July 28, 2018, 02:17:25 PM

Quote from: TheRealCashen on July 28, 2018, 02:10:18 PM

23 / (1.006 * 1.202 * (10194.064774 / 3600) ) = 6.7 C = 44F

Keep track of your imperial units.
T(°F) = T(°C) × 9/5 + 32

So a delta T of 6.7C is the same as a delta T of 12F, the 32 is just an offset for when the two scales are zero.

Thanks,

It works perfect then!! Exactly what I see. I'm guessing I really dont get all 6000 cfm out of the fans. I but its more like 5000 after advertising and having to pull through the filters.

Thanks again.

NiklasFalk

jr. member

Activity: 94

Merit: 1

Quote from: TheRealCashen on July 28, 2018, 02:10:18 PM

23 / (1.006 * 1.202 * (10194.064774 / 3600) ) = 6.7 C = 44F

Keep track of your imperial units.
T(°F) = T(°C) × 9/5 + 32

So a delta T of 6.7C is the same as a delta T of 12F, the 32 is just an offset for when the two scales are zero.

TheRealCashen

newbie

Activity: 56

Merit: 0

23 / (1.006 * 1.202 * (10194.064774 / 3600) ) = 6.7 C = 44F

Something doesn't seam right. That is about 3 times what I see. Is it because not all the kW is converted to heat?

Those are my calculations and I don't see anywhere near 44F delta.

Example - Heating Air
An air flow of one cfm is heated from 32 to 52oF. Using (1) the sensible heat added to the air can be expressed as:

hs = 1.08 (1 cfm) ((52 oF) - (32 oF))

= 21.6 (Btu/hr)

this one works for me.

1.08 * 5000 * (100-94)

crairezx20

legendary

Activity: 1638

Merit: 1046

The important is you must take out the hot air on the room, adding an exhaust fan or seperate the miners with duct host in the exhaust fan to push the air out from the room.

This is just a sample image below
Sample image.

NiklasFalk

jr. member

Activity: 94

Merit: 1

https://www.engineeringtoolbox.com/cooling-heating-equations-d_747.html

hs = cp ρ q dt

hs= 20 kW
cp = specific heat of air (1.006 kJ/kg C)
ρ = density of air (1.202 kg/m3)
q = 2000/3600=0.555555
dt = temperature difference (C)

So with 20kW that airflow will see a delta T of 20/(1.006*1.202*0.55555)=29.8 C

Lets take a real cool ambient temp of 20C, that would mean that the air exiting will be 60C, and you cards will then most probably will be running at 80+C.

Double the Air flow and the deltaT will be reduced to half, with exit T of 45C, a much better situation.
The math is actually really simple when you reduce the situation to total air flow and total power inserted to the system.

With very controlled airflow (so all cards get the ambient air and no exhaust from another card) the cards will see a lower peak. Water cooling would be even better to control where the heat goes. But the heat transfer by air is till the same, 20kW transferred to 2000 m3/h air gives deltaT of 30C.

Lunga Chung

member

Activity: 277

Merit: 23

If you wanna use the entire room you will need at least 40 air change per hour, if you can add another fan for intake on the other side of room that should do it....

But i like TheRealCashen design, I might try this for couple of ASIC

TheRealCashen

newbie

Activity: 56

Merit: 0

Quote from: Andartis on July 28, 2018, 08:52:51 AM

Hi guys,

I have a cold basement room. It has a window, in which a ventilator is installed, that pushes 2000m3 per hour out through a short pipe and gets fresh air through an opening from outside.

The room has 90m3. But I can downsize it to 30m3.

Will 20kW of miners and some ASICs work with that ventilator?
Should I downsize the room to exchange the hotter air faster or leave it big, so it can distribute?

Thank you in advance!

I love this topic.

I use 2 x ~3Kcfm fans for my 23kW cabinet. Works awesome. You do not need cold air. You need a lot of flow and that flow needs to actually transfer the heat and move the heat away from the equipment.

I'll NEVER use anything again besides this cabinet design. Its amazing. No noise, no heat, all contained.

I have a large screened window open in my garage which pulls outside air into my garage, filters it through the box, exhausts the hot air into the attic and then vents back to outside. I have never seen the exhaust over 110 degrees. Which we have had plenty of 100 degree days here in Missouri.

Good luck

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onegiantcock

copper member

Activity: 96

Merit: 1

I’d use the whole room for 80% cold/intake air - and isolate the hot side at the exhaust fan

Andartis

jr. member

Activity: 306

Merit: 7

Hi guys,

I have a cold basement room. It has a window, in which a ventilator is installed, that pushes 2000m3 per hour out through a short pipe and gets fresh air through an opening from outside.

The room has 90m3. But I can downsize it to 30m3.

Will 20kW of miners and some ASICs work with that ventilator?
Should I downsize the room to exchange the hotter air faster or leave it big, so it can distribute?

Thank you in advance!

Topic: Mining room cooling - How much airflow for 20kW? (Read 384 times)