Subtopic: Centrifugal separation of solids, liquids and gases
Question Statement:
In a cyclone separator of radius $r = 0.20\text{ m}$, dust-laden air enters tangentially with an inlet velocity of $14\text{ m/s}$. Taking acceleration due to gravity $g = 9.81\text{ m/s}^2$, the separation factor ($S = v^2 / (r g)$) developed in the cyclone is ________ (round off to 1 decimal place).
Answer Key:
99.9
Step-by-Step Solution & Explanation
The separation factor $S$ is the ratio of centrifugal acceleration to gravitational acceleration:
$S = \frac{v^2}{r g} = \frac{14^2}{0.20 \times 9.81} = \frac{196}{1.962} = 99.898 \approx 99.9$