1a. The relative humdity outdoor is 70% and the outdoor temperature is 90 degree

Question

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

Explanation / Answer

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.

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