Subtopic: Centrifugal separation of solids, liquids and gases
Question Statement:
Air carrying particles of density 700 kg m⁻³ and average diameter of 25 μm enters a cyclone of 0.7 m diameter at a tangential velocity of 30 m s⁻¹ at 0.35 m. The density and viscosity of air are 1.1614 kg m⁻³ and 1.85 × 10⁻⁵ Pa s respectively. The terminal radial velocity of the particle is
Options:
(A) 0.17 m s⁻¹
(B) 1.69 m s⁻¹
(C) 3.37 m s⁻¹
(D) 16.52 m s⁻¹
Official Answer Key:
C
Step-by-Step Solution & Explanation
In a cyclone, gravity is replaced by centrifugal acceleration (v_t²/r) in Stokes' law: v_radial = d_p²(ρ_p−ρ_a)·(v_t²/r) / (18μ). = (25×10⁻⁶)²×(700−1.1614)×(30²/0.35) / (18×1.85×10⁻⁵) = (6.25×10⁻¹⁰×698.8×2571.4)/(3.33×10⁻⁴) = 1.1234×10⁻³/3.33×10⁻⁴ = ≈3.37 m/s.
Section 7: Dairy and Food Engineering
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.