Question Bank • QB_APE_DRY_016 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Psychrometry – properties of air-water vapour mixture

Question Statement:

Ambient air at $25^\circ\text{C}$ dry bulb temperature with humidity ratio $W_1 = 0.0120\text{ kg water / kg dry air}$ is sensibly heated in a grain dryer air heater to $65^\circ\text{C}$. The hot air then passes through a moist grain bed, undergoing an adiabatic drying process and leaving the bed at $35^\circ\text{C}$. The specific enthalpy of moist air is given by $h = 1.005 T + W (2501 + 1.88 T)\text{ kJ/kg dry air}$. Assuming an adiabatic saturation path ($h_{\text{exit}} = h_{\text{heater}}$), the moisture removed from the grain per kg of dry air is ________ $\text{g water / kg dry air}$ (round off to two decimal places).

Answer Key:

12.01

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Enthalpy Balance Along Adiabatic Saturation Path** 1. At heater exit (State 2: $T_2 = 65^\circ\text{C}$, $W_2 = W_1 = 0.0120\text{ kg/kg}$): $h_2 = 1.005(65) + 0.0120 [2501 + 1.88(65)]$ $h_2 = 65.325 + 0.0120 [2501 + 122.2] = 65.325 + 0.0120(2623.2)$ $h_2 = 65.325 + 31.4784 = 96.8034\text{ kJ/kg dry air}$ 2. At grain bed exit (State 3: $T_3 = 35^\circ\text{C}$, adiabatic $h_3 = h_2 = 96.8034\text{ kJ/kg}$): $96.8034 = 1.005(35) + W_3 [2501 + 1.88(35)]$ $96.8034 = 35.175 + W_3 [2501 + 65.8] = 35.175 + W_3 (2566.8)$ $W_3 = \frac{96.8034 - 35.175}{2566.8} = \frac{61.6284}{2566.8} \approx 0.024010\text{ kg/kg dry air}$ 3. Moisture picked up per kg dry air: $\Delta W = W_3 - W_2 = 0.024010 - 0.012000 = 0.012010\text{ kg/kg dry air}$ In grams: $\Delta W = 0.012010 \times 1000 = 12.01\text{ g water / kg dry air}$ **Method 2: Approximate Humid Heat Formula** $\Delta W \approx \frac{c_{s} (T_2 - T_3)}{\lambda} \approx \frac{1.028 \times (65 - 35)}{2567} = \frac{30.84}{2567} \approx 0.01201\text{ kg/kg} = 12.01\text{ g/kg}$
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