Question

The hoist of a crane consists of a 10 kW electric motor running at 1440 rpm drivine 300 mm diameter drunn through a 60:1 gear reduction unit. If the efficiency is 90% . calculate the load, in tonnes that can be lifted at the rated motor capacity, and the lifting speed.

Answer 1

Load = 2.42 tons

Lifting speed = 24 RPM

Step-by-step explanation:

Given that

Power=10 KW

Efficiency = 90%

So the actual power,P=0.9 x 10 =9 KW

Speed of motor = 1440 RPM

Diameter of drum = 300 mm

radius =150 mm

G=60:1

Lets take speed of drum =N

We know that

`G=(Speed\ of\ motor)/(Speed\ of\ drum)`

`60=(1440)/(N)`

N=24 RPM

We know that

`P=(2\pi NT)/(60)`

Where T is the torque

`9000=(2\pi * 24* T)/(60)`

T=3580.98 N.m

Lets take load =F

So T= F x r

3580.98 = F x 0.15

F=23.87 KN

We know that

1 KN=0.109 tons

So 23.87 KN= 2.42 tons

So the load = 2.42 tons

Lifting speed = 24 RPM