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Author Topic: Re: pulling heat out of room with asic miner  (Read 122 times)
jackhandsome5858 (OP)
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July 25, 2023, 08:08:27 PM
 #1

So I have an Antminer E9 Pro in a 2400cubic sq ft (8Wx25Lx8H) room. If I wanted to pull the heat from that room is there a formula that will tell me which cfm fan I should put in there? There is 1 ac vent and also 2 doors that I can partially open if need be for fresh air coming in.
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July 25, 2023, 11:43:28 PM
Merited by Flexystar (2), vapourminer (1)
 #2

Yes, there is a formula that you can use to calculate the cfm that you need for that room.

Here's the sample
(Feet wide*feet high*feet long=cubic feet)
6*6*10=360

Multiply it by 2
2x360=720

And then divide it by 60 to get cfm the result should be 12cfm.

Why not just exhaust the hot air directly from the miner using duct host if it's not many?

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jackhandsome5858 (OP)
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July 26, 2023, 01:13:37 AM
Last edit: July 26, 2023, 04:02:50 PM by jackhandsome5858
Merited by vapourminer (1)
 #3

That's interesting you say that as it gives me a new idea. If that calculation is true then that means I could get an 80cfm bathroom type of exhaust. I could then put said exhaust in the ceiling tile (It is a drop ceiling in a warehouse) and vent the heat into the ceiling. Would that work? Does that trap the heat in the ceiling and not affect the rest of the room? It does have sheets of thick insulation in the ceiling. If that would work it would be perfect. I was initially going to vent it outside as there is a part of the room that has a wood wall that I could put an exhaust in but those types of exhaust are generally like 700cfm which would seemingly be too powerful. Thoughts?
*edit* Actually now that I think of it.  The best way to vent it out I am going to assume is to just vent it outside.
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August 11, 2023, 09:21:41 AM
Merited by vapourminer (1)
 #4

Yes, there is a formula that you can use to calculate the cfm that you need for that room.

Here's the sample
(Feet wide*feet high*feet long=cubic feet)
6*6*10=360

Multiply it by 2
2x360=720

And then divide it by 60 to get cfm the result should be 12cfm.

Why not just exhaust the hot air directly from the miner using duct host if it's not many?

That is cool to know that we have formula for even calculating the power needed to vent a room. Makes the life more easy. Though I am not into mining yet, I am always getting engaged in the conversation like this.
Recently I also posted about the cooling effect of the immersed hardware in BitCool coolant. This makes the venting issue go away since you are taking care of the heat with the help of Coolant that is circulating through the hardware chamber and taking away the heat on continuous basis.

However, in your case since it is just single unit with no complex farm, I think a simple vent will do it's job rightly.

Moreover you can also position a normal fan on one side that will point out to miner in such way that it will push the air towards Vent and rest will be taken care by the exhaust. If operations start to go above mid scale then you can think about the BitCool Coolant tanks.
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