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We have a motor on the bench that draws 0.6 amps (no load) in each phase with 460V applied. When voltage increases to 473V the current increases to 0.81 amps and when 493V current is 0.93 amps that is above FLA of motor.

The applied voltage is within the +/- 10% of the motor nameplate. Would this increase in current be due to voltage saturation in the stator or maybe the air gap between stator and rotor is not correct?

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General rule of thumb is the NLA will be half that of the rated FLA for a two pole. If it's a 4 pole it will be smaller still. But why are you ramping up voltage beyond what it is designed for? I can't really answer your saturation question. From experience, people like to say that motors are over saturated when the preliminary test results are inconclusive. Maybe somebody else can answer that one!.

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HI.

460 V is its nameplate voltage? What is its temperature after 1 hour of work on 460V and without load? (and what has been ambient temp by that?)

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Well, its not the same if you applied 460 V and 500 V, imagine a situation like this: you have a resistor with ratings 2 ohms and you applied voltage about 4 volts, the current draw is 2 amps, if you increase voltage to 8 volts, current ramps up to 4 amps, the same thing is with your motor, if you applied 460 volts it vill draw 0.6 amps, if you increase voltage, the resistance of motor windings stay same and the current ramps up. BTW dont increase voltage of motor, you will burn out the windings because they are designed for .6 amps not for .93
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Hello NEMotorTech

There are two major components of the current drawn by a motor.

1. The magnetizing current

2. The work current.

The magnetizing current is the current that is required to develop the flux in the gap and is determined by stator turns, the stator design and the voltage applied.

Small motors are typically designed to operate with a much higher flux density than large motors.

With the higher flux density, the ratio of magnetizing current to rated current reduces.

Large three phase motors will typically have a magnetizing current in the order of 20 - 25% of the rated current, but small motors will commonly be as high as 60%.

If the flux density is high, a small change in applied voltage will result in a large change in magnetizing current. This is due to the saturation of the iron in the stator. This results in a dramatic increase in the iron losses and can cause the motor to overheat and fail.

As the voltage is increased, the magnetizing current also increases.

As the voltage is increased, the work current reduces.

The motor that you describe is operating at saturation.

Best regards,

Mark.

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Hello Mark!

The original poster has made the point that the motor operated within the allowable voltage range and by that it overpassed the FLA. That is, nominal voltage being 460, allowable +/- 10 %, there was applied + 8% (if I remember good), by that, FLA was overpassed - without any load.

Do I understood you right that the nameplate states active current(s) ? Thus, there measured 0.9 apparent but, let's say, there was only 0.7A active current, which still was less than FLA?

The best regards.

Yuri

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Hello Yuri

I would suggest that the +/- 10% on voltage is optimistic on a motor this size. Small motors are overfluxed by design and so are very sensitive to an increase in voltage.

Larger motors operate at a lower flux density and so are more tolerant of an increase in voltage to a certain point.

If the magnetising current was say 70% of the rated current, then an increase in load current from zero would have no influence on the measured current until the load current was significant relative to the magnetizing current. The two currents are at phase quadrature so it is vector adition rather than magnitude addition.

The current rating is the total current as you would read with an ameter, not the individual work or magnetizing currents.

Best regards,

Mark.

Thank you, Mark.

My best regards,

Yuri.

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