VFD Output Current is More Than Input Current

At this slow speed you are running the motor in a very inefficient manner...

https://av8rdas.wordpress.com/2010/12/18/variable-frequence-drive-system-efficiency/

I am concerned about the motor heating, not the efficiency sir.

This is what leads to heating and motor failure...the motor is designed to run at minimum 50% load...Why do you run the motors at such a low speed? If this is all the ventilation you require, your motors are grossly oversized, or not designed for this type application...

It is designed for running between 0-1500 rpm as per process requirement. We also have thermistor protection to cut off power supply from vfd to motor in case of winding temp is high

Well then, I don't see a problem....

https://en.wikipedia.org/wiki/Direct_torque_control#:~:text=Direct%20torque%20control%20(DTC)%20is,and%20current%20of%20the%20motor.

If you have your ACS800 programmed properly, the motor will not overheat, nor will the life of the equipment be shortened. This is especially applicable to variable torque loads like your fans. The winding excitation is adjusted to the speed of the rotor, so you do not generate more heat than the motor was designed to tolerate. If you have some poorly built motors, perhaps operation below 10% rated speed might be moderately risky, especially if you operate right at the motor rated ambient temperature, usually 40C.

Your motor is heating partly because of the current flowing through the windings, which is less than the full load current anyway, and the fact that the fan is running slowly.

Why on earth you would want to do what you’re doing is a mystery to me, Mildred.

The power drawn by a fan is proportional to something like the cube of the tip speed.

It appears from here that the inverter is doing a proper job. When a motor is taking little-to-no mechanical load it is effectively a coil of wire as far as the supply is concerned; one would expect a lot of current out-of-phase with the applied voltage in such circumstances, and out-of-phase doesn't appear on the motor shaft. When it is at full load a lot more current is drawn, this time in closer phase to the applied voltage.

Magnetizing current and no load current of induction motor is same? No load current is calculated at rated voltage and speed , i guess. In this case, current seems to be near no load current in low voltage, low speed.

No load current includes some real (mechanical) power components, bearing friction, I2R heat losses of the windings, windage of the rotor, and the motor cooling fan air movement power.

The bulk of the amperes observed will be due to the reactive power, the magnetizing current, when compared to total no load amperes.

That’s because at low speed the thing isn’t doing any work.

At <...92rpm...> the shaft would need only 6-10W to turn it. Try turning it by hand at this speed, obviously with the motor isolated from the supply first; it should go easily.

In any case, the motors are oversized for this job by a factor of between 2 and 3.

The actual real power required by the motor is very low, and the input power to the inverter will be very close to the real power, since many of these inverters present a 0.95 power factor to the supply.

The magnetizing current required by the motor can be supplied by the input capacitors in the inverter supply. The power factor of your lightly loaded motor is likely around 0.55, so you measure lots of amperes that, just like with power factor correction capacitors, can be sourced independent of the supply mains.

You could actually check the running power factor of the motor, assuming your input amperes are probably about 0.95.

I admit I’ve never actually checked the input amperes against output amperes, with some 70 or 80 PWM drives install & operate, so this was an illuminating discussion.

Reconfigure the intake and output fan manifolds.....cycle one motor at a time at higher load. This also gives you longevity and redundancy.

That assumes that reverse flow through parallel fans can be prevented somehow.

You must be getting some resonance that is inflating the voltage with ghost voltage...

In many VFD apps, the output voltage is proportional to the output frequency. Thus lower frequency ==> lower voltage ==> higher current.

Has there been any problem with overheated motors?

As well as modulating speed, the <...VFD [variable frequency drive]...> will modulate voltage. If at the running conditions the motor is drawing very little load, then the current in the supply cable to it could under some circumstances be higher than the current drawn by the <...VFD...>. It would appear that in this condition, the current is way out of phase with the voltage waveform, indicating magnetising current.

The <...fan application...> is lightly loaded. It might be worth turning it off.

Current on the motor side of the inverter is inclusive of the motor reactive current, i.e. the motor power factor is very low, probably as low as .2, so most of the current read by the drive output sensors is reactive. On the line side, the power factor remains at .95 irrespective of what is going on at the motor, so the current that you read is representative of the real power being consumed in by load and motor.