Question

A humidifier is used to introduce moisture into air supplied to an office building during winter days. Outside air at atmospheric pressure and 5 C is introduced into the heating system at a rate of 100 m^3/min, on a dry air basis. The relative humidity of the outside air is 95% and the heating system delivers warm air into the building at 20 C. How much water must be introduced into the warm air, in kg/min, to keep the relative humidity inside the building at 75%?

Answer 1

Pair(ext) = 1 atm; Tair = 5°C = 278 K; Vg = 100m3/min; HR = 95%

T(h) = 20°C = 293 K

PV = nRT

nH2O= 1 atm x 100 m3/min / (8.2057E-5 m3.atm/K.mol) x 278 K

nH2O= 4386.73 mol/min

Vw = ?? to keep HR = 75%

PH2O = Patm x n (h20)
PH2O = 1atm x 4368.73 = 4368.73

HR = PH2O/P*H2O x 100

Where P*H2O = Saturation pressure

P*H2O = PH2O/HR
P*H2O = 4368.73/0.95
P*H2O = 4598.66

If PH20 - 95%
PH20 - 75%

PH20 = 3448.99

We can see from this that it will not reach the P*H2O

PH2O = nH2O

nH2O = molar mass/

PV = nRT
V = nrt/P

V = 3448.99 mol/min x (0.082 L.atm/K.mol) x 293 K / 1atm

V = 82865.61 L/min = 82865.61 Kg/min