Question
Asked 7th Aug, 2013

What is the power consumed by a 50 hp motor at no load condition ?

Many people say the following:
:"With a motor with an output of 50 HP, to convert the HP output into Watts is:
50 HP x 746 Watts/ HP = 37,300 Watts or converting to kiloWatts (kW) = 37.3 kW
A motor of this size probably typically has an efficiency of 90% to 93% (0.90 to 0.93).
For our purposes, assume the motor has an efficiency of 92% ( or 0.92).
So divide the output (50 HP) by the efficiency (0.92), to get the energy required to be input into the motor, to get 50 HP.
So --- 37,300 Watts / 0.92 = ~ 40,500 Watts is required to be input to the motor to get the 50 HP output.
In terms of kiloWatts (kW), --- the terms that your electric company charges you --- this is approximately 40.5 kW.
To run this motor for 1 hour, the energy is 40.5 kW x 1 hour = 40.5 kWh""
Its not true I guess... Its not for No load condition..

Most recent answer

5th Sep, 2013
Dimitrios Triantafyllidis
International Hellenic University
If you are interrested in a rough estimation, it should be in the range of a few hundred watts up to 1 kW, from my experience.
@David Johnson: A capacitive PF under no load? That is very peculiar! I have measured thousands of motors in my life and never found this phaenomenon before! Could it be that the measured motor has a local compensator connected?

All Answers (8)

7th Aug, 2013
David Johnson
Umxhumanisi Engineering
If you have physical access to the machine, you can measure the input currents, voltages and power factor and calculate the power consumed.. For most induction machines I've measured, the power factor is negative under no-load, i.e. capacitive. At no-load the machine only needs to overcome windage loss.
Otherwise, on the manufacturer's machine datasheet, there should be a graph where one can quickly look up the factory test result, which should give one a good approximation.
For calcutating approximations, one need to know the no-load-, locked rotor data and the winding resistances to calculate the parameters of the equivalent circuit. From this, one can estimate the machine's performance
Once you have the no load data, however, your question is answered.
1 Recommendation
7th Aug, 2013
Meet Patel
Nirma University
Yes that is right,but the rating of the motor is decided to its full load condition.
As the load reduces it will draw less current,that is obvious.
And for our use voltage remains constant and power factor as well as efficiency is also previously decided.
hence if current drawn by the motor is less then the input power to be applied to it , will also be less.
so for operating motor in no load condition less power will require as current drawn is reduced.
u can calculate exact amount of current at no load condition,and then the input power.
1 Recommendation
7th Aug, 2013
Geen Paul V
Daimler
But the starting requires more power....
Can I know how much percentage of rated power is required at start up and for continuous running ?
7th Aug, 2013
Tomasz Kraszewski
Silesian University of Technology
If you can just measure the current at start up with no load and then it is simple calculation. So, how often do you want to start the no-load-motor and the other thing is for what?
8th Aug, 2013
Meet Patel
Nirma University
Hmm u can measure starting current.and measure input power accordingly.
21st Aug, 2013
Geen Paul V
Daimler
@ Meet Patel
The starting current is always much higher than the working current.
Say 1.5 times or more higher.
4th Sep, 2013
Geen Paul V
Daimler
Mechanically I found out it to be less than 10 KWh.
But is there any formulation for the same ?
5th Sep, 2013
Dimitrios Triantafyllidis
International Hellenic University
If you are interrested in a rough estimation, it should be in the range of a few hundred watts up to 1 kW, from my experience.
@David Johnson: A capacitive PF under no load? That is very peculiar! I have measured thousands of motors in my life and never found this phaenomenon before! Could it be that the measured motor has a local compensator connected?

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