Question

Calculate the efficiency in percent (rounded off to 1 decimal place) of an oil expeller which yields 37 kg oil containing 5% solid impurities from 100 kg mustard seeds. The oil content of the mustard seed is 38%.

Hint:

Always subtract impurities to get the true oil yield, then divide by theoretical oil content for efficiency.

Answer 1

Step 1: Theoretical oil content.
Mustard seed mass = 100 kg.
Oil content = 38%.
Theoretical maximum oil = \(100 \times 0.38 = 38 \ \text{kg}\).
Step 2: Actual oil obtained.
Expeller yield = 37 kg oil (but contains 5% solid impurities).
So, actual pure oil = \(37 \times (1 - 0.05) = 37 \times 0.95 = 35.15 \ \text{kg}\).
Step 3: Efficiency calculation.
\[ \text{Efficiency (%)} = \frac{\text{Actual oil recovered}}{\text{Theoretical oil}} \times 100 \] \[ = \frac{35.15}{38} \times 100 = 92.5% \] \fbox{Efficiency = 92.5%}