Solid food particles having nominal size of 0.2 mm with shape factor of 0.8 and density of 1040 kg m⁻³ are to be fluidized using air at 28 °C. The density and pressure of air at the above-mentioned condition are 1.175 kg m⁻³ and 1.013×10⁵ Pa, respectively. The voidage at minimum fluidizing condition is 0.48. Use the value of 'g' as 9.81 m s⁻². If cross section of the empty bed is 0.5 m² and contains 520 kg of solids, the pressure drop at minimum fluidization condition is _______ kPa.
Official Answer Key:
9.80 to 10.40
Step-by-Step Solution & Explanation
At minimum fluidization the drag force on the bed equals the net (buoyant) weight of the solids, so the pressure drop is simply the bed weight per unit area:
ΔP = M g / A = 520 × 9.81 / 0.5 = 5101.2/0.5 = 10 202 Pa = 10.20 kPa — within the official range 9.80–10.40.
Cross-check with ΔP = (1 − ε_mf)(ρ_s − ρ_f) g L: the buoyancy correction (1.175 vs 1040 kg m⁻³) is ~0.11 %, i.e. 10.19 kPa — the same within the accepted band.
Particle size, shape factor and air pressure are needed only if the minimum fluidization velocity were asked, not the pressure drop.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.