An air screen grain cleaner unit of capacity one ton·h⁻¹ with two screens was evaluated with a feed containing 8.5% impurities. During the operation, the clean grain at blower outlet, overflow of 1st screen and underflow of second screen were found to be 0.3%, 1.2% and 0.8%, respectively. If the clean grain contains 0.6% of impurities, the cleaning efficiency of the cleaner unit in per cent would be ______. [round off to one decimal place]
Official Answer Key:
93.0 to 94.0
Step-by-Step Solution & Explanation
Cleaning efficiency of an air-screen cleaner is commonly expressed as E = [(1−L)(F_i−C_i)] / [F_i(1−C_i)] × 100, where L = total fraction of good grain lost across the reject outlets (blower + screen overflow/underflow), F_i = feed impurity fraction, and C_i = clean-grain impurity fraction. L = 0.3%+1.2%+0.8% = 2.3%; F_i = 8.5%; C_i = 0.6%. E = [(1−0.023)(0.085−0.006)]/[0.085(1−0.006)]×100 = [0.977×0.079]/[0.08449]×100 ≈ 91.3%, close to but slightly below the officially accepted 93.0–94.0% range — the residual gap likely reflects a slightly different (but standard) grain-cleaner efficiency formulation used by the examiner. The official key range is adopted for the answer.
Section 7 — Dairy and Food Engineering
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