Question Bank • QB_APE_EXP_014 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Psychrometry – properties of air-water vapour mixture

Question Statement:

Air at standard atmospheric pressure ($101.325\text{ kPa}$) has a dry bulb temperature of $30^\circ\text{C}$ and partial pressure of water vapor $p_v = 2.50\text{ kPa}$. Using the humidity ratio formula $W = 0.622 \frac{p_v}{P - p_v}$, the humidity ratio in $\text{kg water / kg dry air}$ is ________ (round off to 4 decimal places).

Answer Key:

0.0157

Step-by-Step Solution & Explanation

Humidity ratio formula: $W = 0.622 \frac{p_v}{P - p_v}$ Given: • $P = 101.325\text{ kPa}$ • $p_v = 2.50\text{ kPa}$ $P - p_v = 101.325 - 2.50 = 98.825\text{ kPa}$ $W = 0.622 \times \frac{2.50}{98.825} = \frac{1.555}{98.825} \approx 0.015735\text{ kg/kg dry air} \approx 0.0157\text{ kg/kg}$
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