Question Bank • QB_APE_EXP_068 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Size Reduction and Material Handling

Subtopic: Centrifugal separation of solids, liquids and gases

Question Statement:

In a cyclone separator of radius $R = 0.40\text{ m}$, the tangential gas inlet velocity is $18\text{ m/s}$. Calculate the centrifugal acceleration factor (separation factor $G = v_t^2 / (R \cdot g)$). (Take $g = 9.81\text{ m/s}^2$)

Answer Key:

82.57

Step-by-Step Solution & Explanation

Centrifugal acceleration: $a_c = \frac{v_t^2}{R} = \frac{18^2}{0.40} = \frac{324}{0.40} = 810\text{ m/s}^2$ Separation factor: $G = \frac{a_c}{g} = \frac{810}{9.81} \approx 82.569$
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