Question

You own a mechanically ventilated swine facility. The dry bulb temperature inside the facility is 50 degrees F and outside is only 30 degrees F. Inside you have a relative humidity of 70% and outside the relative humidity is 20%. To keep the humidity from getting any higher you need to remove 12 gallons of water per hour from the facility with the ventilation air. At what rate (cfm) must the fans pull inside air out of the building?

Fan Rate =  [Num]cfm

Answer 1

Inside condition => T = 50F, RH = 70%

0.0055 lb moisture/lb dry air

Outside condition => T = 30F, RH = 20%

0.0008 lb moisture/lb dry air

So a ventilation of 1 lb air removes 0.0055 - 0.0008 = 0.0047 lb of water

amount of water to be removed = 12 gallon/hour = 12 * 8.34 lb per hour = 100.08 lb/hour = 0.0278 lb/s

So 0.0278 lb/s moisture can be removed by 0.0278/0.0047 = 5.915 lb/s of air = 354.89 lb/min

density of air = 1.225 kg/m³ = 0.0765 lb/ft³

rate (cfm) must the fans pull inside air out of the building = 354.89/0.0765 = 4639​ cfm

[Note: the data of relative humidity are taken from http://www.greenbuildingadvisor.com/sites/default/files/psychrometric-chart-quantities-carrier.jpg graph]