Question Bank • QB_APE_SIZE_010 MCQ • 1 Mark

Section 6: Agricultural Process Engineering

Size Reduction and Material Handling

Subtopic: Centrifugal separation of solids, liquids and gases

Question Statement:

In a tubular bowl centrifugal separator of bowl radius $r = 0.25\text{ m}$ rotating at an angular speed of $\omega = 200\text{ rad/s}$, the centrifugal separation factor ($C = \frac{r \omega^2}{g}$, with $g = 9.81\text{ m/s}^2$) is closest to:

Options:

  • (A) 1020
  • (B) 510
  • (C) 2040
  • (D) 100

Answer Key:

A

Step-by-Step Solution & Explanation

The centrifugal separation factor represents the acceleration generated relative to Earth's gravity: $C = \frac{r \omega^2}{g} = \frac{0.25\text{ m} \times (200\text{ rad/s})^2}{9.81\text{ m/s}^2} = \frac{0.25 \times 40000}{9.81} = \frac{10000}{9.81} \approx 1019.37 \approx 1020$
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