Question

A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial pressures of the dry air and water vapor, (b) the humidity ratio of the moist air, and (c) the specific volume of the moist air. Assume standard sea-level pressure.

Answer 1

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg