A horizontal screw conveyor of length 3.2 m conveys solid grain having bulk density of 725 kg m⁻³. The screw diameter, shaft diameter and pitch of the screw are 0.6 m, 0.24 m and 0.48 m, respectively. If the screw is operating with 80 % of its volumetric capacity at a speed of 50 rpm, the actual discharge of the screw is _________ ton h⁻¹.
Official Answer Key:
196.00 to 200.00
Step-by-Step Solution & Explanation
Volume conveyed per revolution = (π/4)(D² − d²) × pitch
= (π/4)(0.6² − 0.24²) × 0.48 = 0.7854 (0.36 − 0.0576)(0.48) = 0.7854 × 0.3024 × 0.48 = 0.11400 m³ rev⁻¹.
Theoretical volumetric capacity = 0.11400 × 50 × 60 = 342.0 m³ h⁻¹.
Actual volumetric capacity at 80 % fill = 0.80 × 342.0 = 273.6 m³ h⁻¹.
Mass capacity = 273.6 × 725 = 198 372 kg h⁻¹ = 198.4 ton h⁻¹ — within the official range 196.00–200.00.
The conveyor length (3.2 m) affects the power requirement, not the discharge rate.
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