Subtopic: Centrifugal separation of solids, liquids and gases
Question Statement:
Air carrying dust particles enters a cyclone separator of 800 mm diameter at a tangential velocity of 30 m·s⁻¹ near the wall. The average diameter of the dust particles is 100 µm. The separation factor is nearest to (Take g = 9.81 m·s⁻²)
Options:
(A) 229.36
(B) 611.62
(C) 485.72
(D) 3.75
Official Answer Key:
A
Step-by-Step Solution & Explanation
The separation factor of a cyclone is the ratio of the centrifugal acceleration imparted to a particle to the acceleration due to gravity:
S = (v²/r)/g = v²/(r g)
Step 1 — cyclone radius:
r = D/2 = 0.800/2 = 0.40 m
Step 2 — substitute:
S = 30²/(0.40 × 9.81) = 900/3.924 = 229.36
Note: the particle diameter (100 µm) is redundant data — it governs the collection efficiency and cut-size, not the separation factor, which depends only on cyclone geometry and inlet velocity.
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