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?
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.
