Question

A moist air sample has dry bulb temperature of 30ºC and specific humidity of 11.5g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at 30ºC is 4.24 kPa and the total pressure is 90kPa, then the relative humidity (in %) of air sample is

1. 50.5
2. 38.5
3. 56.5
4. 68.5

Answer 1

Correct Answer - Option 2 : 38.5

Concept:

Specific Humidity: 

\(ω = 0.622\left( {\frac{{{p_\nu }}}{{p - {p_\nu }}}} \right)\)

Relative Humidity:

\(\phi = \frac{{{p_\nu }}}{{{p_s}}}\)

Calculation:

Given:

t_DBT = 30°C, ω = 11.5g water-vapor/kg dry air.

p_sat = 4.24 kPa, p = 90 kPa

\(ω = 0.622 \times \left( {\frac{{{{\bf{P}}_{\bf{v}}}}}{{{{\bf{P}}_{\bf{t}}} - {{\bf{P}}_{\bf{v}}}}}} \right)\frac{{{\bf{gm}}\left( {\bf{v}} \right)}}{{{\bf{kg}}\left( {{\bf{d}}.{\bf{a}}} \right)}}\)

\(11.5 \times {10^{ - 3}} = 0.622\left( {\frac{{{p_\nu }}}{{90 - {p_\nu }}}} \right){\rm{\;}}\)

\(18.489 \times {10^{ - 3}} = \frac{{{p_\nu }}}{{90 - {p_\nu }}}\)

\(1.664 - 0.01849{p_\nu } = {p_\nu }\)

\({p_\nu } = 1.634\;kPa\)

\(\phi = \frac{{{p_\nu }}}{{{p_s}}} = \frac{{1.1634}}{{4.24}}\)

\(\phi = 0.3853 = 38.53\% \approx 38.5\% \)