Question

Ambient air is at a pressure of 100 kPa, dry bulb temperature of 30°C and 60% relative humidity. The saturation pressure of water at 30°C is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is ________ (correct to two decimal places).

Answer 1

Calculation:

Let ‘P’ be the atmospheric pressure in (kPa)

‘ϕ’ be the relative humidity, ‘P_v’ be the partial pressure of the air in (KPa), ‘P_vs’ be the saturation pressure of water in (KPa), ‘ω’ be the specific humidity of air in (kg/kg of dry air)

\(ϕ = \frac{{{P_v}}}{{{P_{vs}}}} \)

\(\omega = \frac{{0.622{P_v}}}{{\left( {P - {P_v}} \right)}} \)

Calculation:

Given, ϕ = 60%, P = 100 kPa, P_vs = 4.24 kPa

\(\Rightarrow 0.6 = \frac{{{P_v}}}{{4.24}};{P_v} = 2.54\;kPa\)

\(\omega = \frac{{0.622{P_v}}}{{\left( {P - {P_v}} \right)}} = \frac{{0.622 \times 2.54}}{{\left( {100 - 2.54} \right)}}\)

ω = 0.01623 kg/kg of dry air = 16.23 g/kg of dry air

Mistake: By applying the wrong formula.

\(ϕ = \frac{{{P_{vs}}}}{{{P_v}}} \Rightarrow 0.6 = \frac{{4.24}}{{{P_v}}}\)

P_v = 7.06 KPa

\(\omega = \frac{{0.622 \times {P_v}}}{{\left( {P - {P_v}} \right)}} = \frac{{0.622 \times 7.06}}{{\left( {100 - 7.06} \right)}} = \;0.04724\;kg/kg\;of\;dry\;air\)

ω = 47.24 kg/kg of dry air

This is wrong as the partial pressure is always less than the saturation pressure. This is due to the wrong formula.