Question

if a motor have 2800 kW running 1 hour then how much unit will consume ​

Answer 1

If the motor is fully loaded, it will provide output power of 3HP = (3 X 0.746) kW = 2.24 kW.

Electrical input power = output power / (Efficiency /100).

Assuming 90 % efficiency, input power = 2.24 kW / (90/100)

= 2.49 kW

Unit consumed = Power in kW X numbers of hours of operation.

= 2.49 X 24 units

= 59.76 units.

Hence units burnt (consumed) by 3 HP motor in 24 hours will be 59.76, if the motor efficiency is 90% and motor is fully loaded