Answer to Question #210944 in Mechanical Engineering for ali

Question

A hoisting system has a drum of 750 mm with 25 mm wire rope diameter, mounted on the same shaft with a worm gear having 90 teeth. A triple threaded worm is to drive the wheel and this system is to raise a 25 kN load a a speed of 24m/min. If the system has an efficiency of 70%, how much power should be supplied to the worm’s shaft? What is the torque on the worm’s shaft?

Efficiency = (Power out) / (Power in) 

Answer 1

F = 25kN = 25000N

speed = 24m/min = 24m/60s = 0.4m/s

Power out = 25000 × 0.4 = 10,000W

Efficiency = $\dfrac{\textsf{ Power out }}{\textsf{ Power in }}× 100\%$

70% = 10000/P_in × 100%

0.7 = 10000/P_in

P_in = 10000/0.7 = 14285.7W