Subtopic: Particle size analysis for comminuted solids
Question Statement:
Spherical dust particles of 50 μm are settling under gravity in air at 21 °C and normal atmospheric pressure. Density of particles is 1250 kg m⁻³ and density of air is 1.2 kg m⁻³. Considering viscosity of air as 1.81×10⁻⁵ Pa s, the settling velocity of dust in mm s⁻¹ will be __________.
Official Answer Key:
93.9 to 94.1
Step-by-Step Solution & Explanation
• For a small spherical particle settling under Stokes’ law, Vt = g d²(ρp–ρf)/(18μ).
• d = 50×10⁻⁶ m, d² = 2.5×10⁻⁹ m², ρp–ρf = 1250–1.2 = 1248.8 kg/m³.
• Vt = 9.81 × 2.5×10⁻⁹ × 1248.8 / (18 × 1.81×10⁻⁵) = 3.063×10⁻⁵/3.258×10⁻⁴ ≈ 0.0940 m/s = 94.0 mm/s, inside the accepted range 93.9–94.1.
• (Check: particle Reynolds number ≈ 0.31, small enough that the Stokes’ law assumption is acceptable for this GATE-level calculation.)
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