Question

Taking the acceleration due to gravity as \(10\text{ m/s}^2\) and with a cyclone \(0.5\text{ m}\) in diameter and having a tangential velocity of \(20\text{ m/s}\) near the wall, then the separation factor is

• A. \(180\)
• B. \(160\)
• C. \(240\)
• D. \(350\)

Hint:

For cyclone, separation factor \(=\frac{v^2}{rg}\). Use radius, not diameter, in the formula.

Answer 1

The Correct Option is B

Separation factor in a cyclone is the ratio of centrifugal acceleration to gravitational acceleration. It is given by: \[ S=\frac{v^2}{rg} \] where, \[ v=\text{tangential velocity} \] \[ r=\text{radius of cyclone} \] \[ g=\text{acceleration due to gravity} \] Given: \[ v=20\text{ m/s} \] Diameter of cyclone: \[ D=0.5\text{ m} \] So radius is: \[ r=\frac{D}{2}=\frac{0.5}{2}=0.25\text{ m} \] Also: \[ g=10\text{ m/s}^2 \] Now: \[ S=\frac{20^2}{0.25\times10} \] \[ S=\frac{400}{2.5} \] \[ S=160 \] Therefore, the separation factor is: \[ 160 \]