Subtopic: Fluidization of granular solids – pneumatic, bucket, screw and belt conveying
Question Statement:
The average interstitial velocity in a packed bed is 2.5 m s⁻¹ and the superficial velocity under minimum fluidized condition based on the cross-section of the empty container is 1 m s⁻¹. Solid particles having density of 1200 kg m⁻³ and size of 0.15 mm are to be fluidized using air at 3 atm absolute pressure and 30 °C temperature having a density of 3.5 kg m⁻³. The empty bed cross section is 0.4 m² and the bed contains 400 kg of solid. Pressure drop at minimum fluidized condition in Pa is
Options:
(A) 3.5 × 10³
(B) 6.5 × 10³
(C) 9.7 × 10³
(D) 17.1 × 10³
Official Answer Key:
C
Step-by-Step Solution & Explanation
At minimum fluidization, the pressure drop across the bed is set by a force balance: the upward drag of the fluid supports the net (buoyed) weight of the particles: ΔP = (Weight of solids − buoyant force of air on solids)/(bed cross-section) = [Ms·g − Ms·(ρ_air/ρ_solid)·g] / A = Ms·g·(1 − ρ_air/ρ_solid)/A. Using Ms=400 kg, g=9.81 m/s², ρ_air=3.5 kg/m³, ρ_solid=1200 kg/m³, A=0.4 m²: ΔP = 400×9.81×(1−3.5/1200)/0.4 = 3924×0.99708/0.4 ≈ 9770 Pa ≈ 9.7×10³ Pa, matching option (C). (The simpler ΔP = Ms·g/A ≈ 9810 Pa without the buoyancy correction gives nearly the same value, confirming the answer.)
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