Question Bank • QB_APE_EXP_041 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Processing of Agriculture Produce

Subtopic: Processing of seeds, spices, fruits and vegetables

Question Statement:

Soybean seeds containing $20.0\%$ oil (wet basis) are processed in a solvent extraction plant using hexane. The extracted de-oiled cake retains $1.0\%$ residual oil on dry basis. If the initial moisture content of seeds is $10.0\%$ (w.b.) and extraction is solvent-free in cake, the percentage recovery of oil from the seeds is ________ (round off to 1 decimal place).

Answer Key:

95.5

Step-by-Step Solution & Explanation

Basis: $100\text{ kg}$ soybean seeds. • Initial oil = $20.0\text{ kg}$ • Moisture = $10.0\text{ kg}$ • Non-oil solids (defatted dry matter) = $100 - (20 + 10) = 70.0\text{ kg}$ Residual oil in de-oiled cake: $\text{Residual oil} = 70.0 \times 0.01 = 0.70\text{ kg}$ Oil extracted: $\text{Extracted oil} = 20.0 - 0.70 = 19.30\text{ kg}$ Oil recovery: $\text{Recovery} = \frac{19.30}{20.0} \times 100 = 96.5\% \implies \text{wait: } \frac{19.3}{20} \times 100 = 96.5\%$ (setting correct_answer: "96.5", range min: 95.5, max: 97.5).
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