The conveyor lifts 420 m3 of gravel to a height of 5 in 1.5 hours. Determine the work

| August 7, 2021 | education

The conveyor lifts 420 m3 of gravel to a height of 5 in 1.5 hours. Determine the work done for this, as well as the power developed by the engine, if the efficiency of the lifting device is 75%. Density of gravel 2400 kg / m3

These tasks: V (volume of gravel raised) = 420 m3; h (lifting height of gravel) = 5 m; t (duration of the conveyor) = 1.5 h = 5400 s; η (hoist efficiency) = 75% = 0.75.

Constants: g (acceleration of gravity) ≈ 9.8 m / s2; according to the condition ρg (gravel density) = 2400 kg / m3.

1) Perfect work of the conveyor: A = m * g * h = ρg * V * g * h = 2400 * 420 * 9.8 * 5 = 49392000 J ≈ 49.4 MJ.

2) Power of the conveyor motor: η = Np / Nd = (A / t) / Nd, whence we express: Nd = A / (η * t) = 49392000 / (0.75 * 5400) = 12196 W ≈ 12.2 kW …

Answer: The conveyor performed work at 49.4 MJ, its engine power is 12.2 kW.



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