Dust-laden air enters an 800 mm diameter cyclone at tangential velocity 30 m/s near the wall. Mean particle diameter is $100\,\mu\mathrm m$. With $g=9.81\,\mathrm{m/s^2}$,find the nearest separation factor.
The separation factor is the centrifugal acceleration divided by gravitational acceleration:$G=v_t^2/(rg)$. Near the wall,$r=0.800/2=0.400$ m. Hence $G=30^2/[0.4(9.81)]=229.36$. Particle mass cancels in the acceleration ratio, so the particle diameter is not needed. **A**.
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