A large room contains moist air at 35 degree C 104 kPa The p

A large room contains moist air at 35 degree C, 104 kPa. The partial pressure of water vapor is 1.8 kPa. Determine the relative humidity. the humidity ratio. the dew point temperature, in degree C. the mass of dry air, in kg, if the mass of water vapor is 15 kg.

Solution

Formula used Pw =PWs* RH ( partial pressure = pressure of saturated * Rel Humidity)

PWs = A*10 ^( mT/(T+273.16))

A = 6.116441, m = 7.591386

calculate PWs= 6.1164*10^^( ) = 44.58 *100 Pa = 4.458 KPa

RH = partial pressure/Saturated pressure = 1.8/4.458 =40.4 %

Dew Point is given by the formula : 273.16/{ m/log10 (Pw/A) -1) = 273.16/(7.59139/.4688 -1) = 18 deg C

HUMIDITY RATIO: using ideal gas law can be expressed as .62198* Pw/(Pa-Pw) = .62198*1800/104000=0.108 kgwater/kgair

Mass of dry air in Kg can be found if mass of water vapour is 15 kg, as 15/Hum ratio = 15/.108 = 1393 Kg

(Note on line calculators may give some different values. These formulae are from Vaisala.