Allied & ICAR Exam Bank • ALLIED_1286 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Psychrometry and Grain Drying

Subtopic: Enthalpy of Moist Air

Question Statement:

The approximate specific enthalpy of moist air is given by $h = 1.006 T + w(2501 + 1.86 T)\text{ kJ/kg dry air}$. For dry bulb temperature $T = 30^\circ\text{C}$ and humidity ratio $w = 0.015\text{ kg water/kg dry air}$, the enthalpy $h$ in kJ/kg dry air is:

Answer Key:

68.5

Step-by-Step Solution & Explanation

$h = 1.006(30) + 0.015(2501 + 1.86 \times 30) = 30.18 + 0.015(2501 + 55.8) = 30.18 + 0.015(2556.8) = 30.18 + 38.35 = 68.53\text{ kJ/kg}$.
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