The tower crane evenly lifts a load weighing 0.5 tons. to a height of 30m in 2 min. the current in the

The tower crane evenly lifts a load weighing 0.5 tons. to a height of 30m in 2 min. the current in the electric motor is 16.5 A. at a voltage of 220 V. Determine the efficiency of the crane electric motor.

Given:

m = 0.5 tons = 500 kilograms – the mass of the cargo;

h = 30 meters – the height to which the load is lifted;

t = 2 minutes = 120 seconds – the time during which the load is lifted;

g = 10 m / s ^ 2 – acceleration of gravity;

I = 16.5 Amperes – current in the electric motor;

U = 220 Volts – voltage in the electric motor.

It is required to determine the efficiency of the electric motor N.

Let’s find the useful power that is needed to lift the load:

Wuseful = A / t = m * g * h / t = 500 * 10 * 30/120 =

= 150,000 / 120 = 1250 watts.

The total power of the electric motor is:

W full = I * U = 16.5 * 220 = 3630 watts.

Then the efficiency of the electric motor will be equal to:

N = W useful / W full = 1250/3630 = 0.34 = 34%.

Answer: The efficiency of the electric motor is 34%.