Question Bank • QB_APE_EXP_087 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Psychrometry – properties of air-water vapour mixture

Question Statement:

Air at a dry bulb temperature of $25^\circ\text{C}$ has a humidity ratio of $W = 0.010\text{ kg water / kg dry air}$. Using the approximate enthalpy equation $h = 1.006 t + W (2501 + 1.86 t)$, calculate the specific enthalpy of the humid air in $\text{kJ/kg dry air}$.

Answer Key:

50.62

Step-by-Step Solution & Explanation

$h = 1.006(25) + 0.010 [2501 + 1.86(25)]$ $1.006 \times 25 = 25.15\text{ kJ/kg}$ $2501 + (1.86 \times 25) = 2501 + 46.5 = 2547.5\text{ kJ/kg}$ $0.010 \times 2547.5 = 25.475\text{ kJ/kg}$ $h = 25.15 + 25.475 = 50.625\text{ kJ/kg dry air}$
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