In: Mechanical Engineering
1a. The relative humdity outdoor is 70% and the outdoor temperature is 90 degree F what is the actual partial pressure of water vapor in the air
1b. Humidty ratio (w) is the ratio of the mass of the water (Mw) to the mass of dry air (Mw) note that two partial pressures sum to the total pressure (P =Pa + Pv) derive the equation
w = 0.622Pv / (P-Pv). Use the ideal gas equation for mass
1c. Solve for humidity ratio if the partial pressure of water vapor is 0.1816psia
1d. What is the relative humidity of 70 degree F moist air in a classroom
1e. Estimate the mass (Ibm) of water vapor in the classroom
1a
Relative humidity phi = Pv / Pvsat
From psychrometric tables, At 90 deg F, we get Pvsat = 0.6988 psi
Therefore, partial pressure of water vapor Pv = 0.7 * 0.6988 = 0.4892 psi = 3.373 kPa
1b
w = mv / ma
= (MvPvV / RT) / (MaPaV / RT)........where M denotes molecular mass
= MvPv / MaPa
Molecular weight of water vapor (H2O) Mv = 18
Molecular weight of air Ma = 28.96
Also, P = Pa + Pv
Thus, w = (18 / 28.96) Pv / (P - Pv)
w = 0.622 Pv / (P - Pv)
1c
Pv = 0.1816 psia
Assuming atmospheric pressure as ambient, we have P = 14.5 psia
Thus, Humidity ratio w = 0.622 * 0.1816 / (14.5 - 0.1816)
= 0.00788
1d
From tables, At 70 deg F, Pvsat = 0.3632 psi
Rel Humidity RH = Pv / 0.3632......Pv can be found from Dew point temperature.