Question

A sample of moist air at a total pressure of 90 kPa ha a dry bulb temperature of ${30}^{\circ }C$. If the absolute humidity of air is $22.9gm/kg$ of dry air, then then relative humidity of air is closest to
$(P_{sat}at{30}^{\circ }C=4.24kPa)$

• A. 79%
• B. 75%
• C. 83%
• D. 71%

Answer 1

The correct option is B 75%

$\omega =\frac{0.622P_v}{P_t-P_v}22.9\times {10}^{-3}=\frac{0.622\times P_v}{90-P_v}{P}_v=3.196kPa\varphi =\frac{P_v}{P_{sat}}=\frac{3.196}{4.24}=0.754=75.4\%$