How much pressure is required by an exhaust fan when it exhausts straight up?

Actually, would it be the difference in air pressure between the room where the exhaust fan is taking air from and the roof where it is exhausting to? That comes out to be more reasonable, more like 0.3 inches but still seems high.
And if I do 70 °F intake and 0 ° outdoor, it comes out to 8 in w.c. so I'm back to being confused.

loss is velocity dependent, without knowing that this is all I can offer

Consider:
[ul]
[li]Differential pressure inside to outside[/li]
[li]Inlet loss[/li]
[li]Duct friction[/li]
[li]Regain from 4" round to 12x12 rectangular duct[/li]
[li]more duct friction[/li]
[li]Exit loss[/li]
[/ul]

Hi.. Density change in air with 30 feet height is insignificant and change in pressure due to this 30 feet altitude is also insignificant. You can ignore it in case of air. Just calculate the duct+fittings losses and safety margin of 15% and proceed further.

Calculating on the exhaust discharge side is barely any different than on the suction inlet side

You do your normal calculation and fitting loss lookups based on your exhaust airflow, and the duct size and velocity the air is traveling through.

You also may have to account for velocity pressure if you are narrowing/squeezing the duct after the fan exit you need to account for that exit velocity increase. The fan itself should be coming with enough pressure for it’s specific velocity pressure associated with its outlet size but not more.

30’ of height in a building with typical operating pressure wouldn’t even factor in very much. If you want to get into the details, this link will probably help.


And finally see if the fan selection performance has factored in a ducted discharge, sometimes fan manufacturers can account for that and it re-rates the performance you’ll get.

Thank you, I'll read through that.