Question Bank • QB_APE_EXP_079 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Drying kinetics

Question Statement:

A batch of $500\text{ kg}$ of grain is dried from an initial moisture content of $25\%$ wet basis to $15\%$ wet basis. Calculate the total mass of water removed during the drying process in $\text{kg}$.

Answer Key:

58.82

Step-by-Step Solution & Explanation

Initial total mass $W_1 = 500\text{ kg}$ at $M_1 = 0.25$ (w.b.). Bone-dry solid matter: $W_d = W_1 (1 - M_1) = 500 \times (1 - 0.25) = 375\text{ kg}$ Final total mass at $M_2 = 0.15$ (w.b.): $W_2 = \frac{W_d}{1 - M_2} = \frac{375}{1 - 0.15} = \frac{375}{0.85} \approx 441.176\text{ kg}$ Water removed: $\Delta W = W_1 - W_2 = 500 - 441.176 = 58.824\text{ kg}$
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