Moist air enters an airconditioning unit at the drybulb and

Moist air enters an air-conditioning unit at the dry-bulb and wet-bulb temperatures T_ = 35 degree C and T_wb = 30 degree C, respectively, and 1 atm of pressure. The flow rate of dry air in the mixture a 0.2 kg/s. Liquid water is drawn from the moist air at the rate of 0.003 kg/s. The air-vapor mixture leaves the unit at 25 degree C and 1 atm. Determine the humidity ratio (omega_1) and the mass flow rate of water vapor (m_n) in the moot air mixture at the inlet. Determine the humidity ratio (omega_2) and relative humidity (phi_2) of the moist air at the outlet.

Solution

a)

From psychrometric chart, at WBT = 30 deg C and DBT = 35 deg C, we get

w1 = 0.025 kg water / kg dry air and RH1 = 70%

w1 = mv1 / ma1

0.025 = mv1 / 0.2

mv1 = 0.005 kg/s

b)

mv2 = 0.005 - 0.003 = 0.002 kg/s

w2 = mv2 / ma

= 0.002 / 0.2

w2 = 0.0001 kg water / kg dry air

w2 = 0.622*Pv2 / (P - Pv2)

0.0001 = 0.622*Pv2 / (1 - Pv2)

Pv2 = 0.000160746 atm

At T2 = 25 deg C, we have sat vapor pressure Pvap = 0.0313 atm

Phi2 = Pv2 / Pvap

= 0.000160746 / 0.0313

= 0.00514 = 0.514%