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

Answer:

 Load =  2.42 tons

Lifting speed = 24 RPM

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

$G=\dfrac{Speed\ of\ motor}{Speed\ of\ drum}$

$60=\dfrac{1440}{N}$

N=24 RPM

We know that

$P=\dfrac{2\pi NT}{60}$

Where T is the torque

$9000=\dfrac{2\pi \times 24\times T}{60}$

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