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..
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.
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.
But the starting requires more power....
Can I know how much percentage of rated power is required at start up and for continuous running ?
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?
@ Meet Patel
The starting current is always much higher than the working current.
Say 1.5 times or more higher.
Mechanically I found out it to be less than 10 KWh.
But is there any formulation for the same ?
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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