Question Bank • QB_APE_EXP_012 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Drying and milling of cereals, pulses and oilseeds

Question Statement:

One tonne ($1000\text{ kg}$) of paddy at $22\%$ moisture content (wet basis) is dried to $14\%$ moisture content (wet basis). The amount of water removed in kilograms during the drying process is ________ (round off to 1 decimal place).

Answer Key:

93

Step-by-Step Solution & Explanation

1. Dry matter conservation: $W_1 (1 - M_1) = W_2 (1 - M_2)$ Given: • $W_1 = 1000\text{ kg}$ • $M_1 = 0.22 \implies 1 - M_1 = 0.78$ Dry matter weight $W_{dm} = 1000 \times 0.78 = 780\text{ kg}$. 2. Final total weight $W_2$: $W_2 = \frac{W_{dm}}{1 - M_2} = \frac{780}{1 - 0.14} = \frac{780}{0.86} \approx 906.977\text{ kg}$ 3. Water removed $\Delta W$: $\Delta W = W_1 - W_2 = 1000 - 906.977 = 93.023\text{ kg} \approx 93.0\text{ kg}$
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