Subtopic: Thermodynamic principles of I.C. engines
Question Statement:
A fat globule of 1.5 µm diameter is rising up in a stagnant skim milk medium of 1005 kg m⁻³ density and 1.5 cP viscosity. If the density of the fat globule is 915 kg m⁻³, the steady rising velocity of the globule in µm s⁻¹ is _____.
Official Answer Key:
0.072 to 0.076
Step-by-Step Solution & Explanation
Since the fat globule is less dense than the surrounding medium, it rises under a buoyant (Stokes') settling force. Stokes' law: V = d²(ρ_medium−ρ_particle)g/(18μ).
d = 1.5×10⁻⁶ m, Δρ = 1005−915 = 90 kg/m³, μ = 1.5 cP = 1.5×10⁻³ Pa·s, g = 9.81 m/s².
V = (1.5×10⁻⁶)²×90×9.81/(18×1.5×10⁻³) = 1.987×10⁻⁹/0.027 = 7.358×10⁻⁸ m/s.
Converting to µm/s: V = 0.0736 µm/s. This lies within the accepted range 0.072–0.076.
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