Question Bank • QB_APE_EXP_058 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Psychrometry – properties of air-water vapour mixture

Question Statement:

Atmospheric air at a barometric pressure of $101.325\text{ kPa}$ has a dry bulb temperature of $30^\circ\text{C}$. The saturation vapor pressure of water at $30^\circ\text{C}$ is $4.246\text{ kPa}$. If the relative humidity of the air is $60\%$, calculate the humidity ratio (humidity ratio $W$) in $\text{kg water vapor / kg dry air}$.

Answer Key:

0.01605

Step-by-Step Solution & Explanation

Actual partial pressure of water vapor $p_v$: $p_v = \phi \times p_{vs} = 0.60 \times 4.246 = 2.5476\text{ kPa}$ Humidity ratio $W$: $W = 0.622 \times \frac{p_v}{P_b - p_v} = 0.622 \times \frac{2.5476}{101.325 - 2.5476} = 0.622 \times \frac{2.5476}{98.7774} \approx 0.01605\text{ kg/kg dry air}$
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