Question

Air carrying dust particles enters a cyclone separator of 800 mm diameter at a tangential velocity of 30 \(\text{m.s}^{-1}\) near the wall. The average diameter of the dust particles is 100 \(\mu\text{m}\). The separation factor is nearest to (Take \(g = 9.81\) \(\text{m.s}^{-2}\))

• A. 229.36
• B. 611.62
• C. 485.72
• D. 3.75

Hint:

Use the separation factor formula S equals v_t squared over r g, with the cyclone radius, not diameter.

Answer 1

The Correct Option is A

Step 1: Recall the separation factor formula.
In a cyclone separator, the separation factor compares the centrifugal force pulling a particle outward to the gravitational force pulling it down, \(S = \dfrac{v_t^2}{r g}\).
Here \(v_t\) is the tangential (whirl) velocity of the air near the cyclone wall and \(r\) is the radius at which that velocity acts.

Step 2: Substitute the given values.
The cyclone diameter is 800 mm, so the radius is \(r = 0.4\) m, and the tangential velocity is \(v_t = 30\) \(\text{m.s}^{-1}\).
\[ S = \frac{(30)^2}{0.4 \times 9.81} = \frac{900}{3.924} \]
\[ S \approx 229.36 \]

Step 3: Note that particle size does not enter this formula.
The 100 \(\mu\text{m}\) particle diameter is extra information; the separation factor as defined here depends only on wall velocity and wall radius, not particle size.

Final Answer:
The separation factor is nearest to 229.36. \[ \boxed{S \approx 229.36} \]