Question

Determine the horse power required to drive a lift in raising a load of 2000 Kgf at a speed of 2m/sec ,if the efficiency is 70%.

Answer 1

Load (F)= 2000 kgf = 2000×9.8 N = 19600N = 19.6 kN
Speed = 2m/s
Power = Force × velocity = 19.6×2 = 39.2 kW
But efficiency = 70%
Let power required = P

⇒$\frac{39.2}{P}$ = $\frac{70}{100}$

⇒P = $\frac{100*39.2}{70}$

⇒P = 56 kW
1kW = 1.341hp
Thus power required = 56×1.341 = 75.1hp