A cyclone receives gas at an inlet velocity of 18 m/s through a rectangular inlet 0.20 m wide, with 5 effective turns in the body. Gas viscosity is Pa·s and gas density is 1.2 kg/m³; particle density is 1,800 kg/m³. The cut diameter for 50% collection efficiency is most nearly:
- (A)5.66 µm
- (B)8.00 µm
- (C)12.7 µm
- (D)17.9 µm
Show worked solution
Answer: (A)
Centrifugal drift working against Stokes drag across the 0.20 m inlet width yields a cut diameter of about 5.7 µm.
Cut diameter follows from the particle that just traverses the inlet width under centrifugal drift opposed by Stokes drag during the effective number of turns.
Air viscosity in SI is Pa·s rather than the numerically similar poise value, and the gas density is only a 0.07 percent correction to the particle density.
Numerator and denominator combine to a cut size in the 5 to 10 µm band that defines high-efficiency cyclone duty and explains their use as precleaners.
FE Reference Handbook — Environmental Engineering: Air Pollution Control, Cyclone 50% Collection Efficiency for Particle Diameter
Why the other choices appear
- (B)Omits the factor 2 in the denominator, doubling the radicand to 6.405e-11 m^2 and giving 8.00 µm.
- (C)Leaves the number of effective turns out of the denominator, raising the radicand fivefold to 1.601e-10 m^2 and giving 12.65 µm.
- (D)Uses 1.81e-4 Pa·s for viscosity, the poise value read as a Pa·s value, raising the radicand tenfold and giving 17.9 µm.