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)
GATE AG - 2026,
Answer: A
The separation factor is about 229.36.
The separation factor of a cyclone is the ratio of the centrifugal acceleration at the wall to gravity:
\[ \begin{aligned} S &= \frac{v^{2}}{r g} = \frac{30^{2}}{0.4 \times 9.81} \\ &= \frac{900}{3.924} = 229.36 \end{aligned} \]
The radius is half the 800 mm diameter, 0.4 m, and the particle diameter does not enter the separation factor at all. Using the full 0.8 m diameter instead of the radius halves the result to about 115, and leaving the velocity unsquared gives a value below 8; 611.62, 485.72 and 3.75 do not follow from \(v^{2}/(rg)\) with the cyclone's radius.
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