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

50.5

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38.5

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56.5

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68.5

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Solution

The correct option is B 38.5
Relative humidity,

$ϕ = \frac{p_{v}}{p_{v s}}$

Specific humidity,

$ω = 0.622 (\frac{p_{v}}{p - p_{v}})$

$11.5 \times 10^{- 3} = 0.622 (\frac{p_{v}}{90 - p_{v}})$

$p_{v} = 1.63378 KPa$

Relative humidity,

$ϕ = \frac{p_{v}}{p_{v s}} = \frac{1.63378}{4.28} = 0.385$

= 38.5%

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