Question

Moist air is treated as an ideal gas mixture of water vapour and dry air (molecular weight of air = 28. 84 and molecular weight of water = 18). At a location, the total pressure is 100 KPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapour per kg of dry air is ________ grams.

Answer 1

Concept:

Humidity ratio (w) is the mass of water associated with each kilogram of dry air. It is known as specific humidity or absolute humidity.

\(\omega = \frac{{{m_v}}}{{{m_a}}} = 0.622\left( {\frac{{{P_v}}}{{{P_t} - {P_v}}}} \right) \)

Relative humidity (ϕ):

It is defined as the ratio of partial pressure of water vapour (pv)in a mixture to the saturation pressure (ps) of pure water at the same temperature of the mixture.

\(\phi=\frac{p_v}{p_{vs}}\)

Calculation:

\(\phi = \frac{{{P_v}}}{{{P_v}_s}} \Rightarrow 0.55 = \frac{{{P_v}}}{{4.246}} \Rightarrow {P_v} = 2.3353\;kPa\)

\(w = 0.622\left( {\frac{{{P_v}}}{{{P_t} - {P_v}}}} \right) =0.622\left( {\frac{{2.3353}}{{100 - 2.3353}}} \right) = 0.0149\;kg/Kg\;dry\;air\)

= 14.9 gm/ Kg dry air