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09/17/2018 5:32 PM

How does a VFD protect Induction motors at lower speeds? Example: if the full load amps for a 30KW motor is 50Amps at 50Hz, at 25Hz it will use 10Amps (fan). How does the VFD now protect this motor from oversaturation?

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#1

09/17/2018 5:46 PM

Are you measuring the amperage coming in to the VFD, or going out of the VFD? ...That is the question....you must ask yourself...

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#2

09/17/2018 7:46 PM

The impedance of the motor is proportional to frequency.

Below the rated frequency, the VFD voltage is proportional to frequency and current (torque) is constant. At the rated frequency, voltage x current = rated power of the motor. Above the rated frequency, the VFD voltage is held constant and the current (torque) decreases as the motor impedance increases with frequency. Above the rated frequency motor power (torque x rpm) is constant.

Zmotor = 2*pi*f*L

http://lhp.co.in/index_without_right.php?file=vfd_school

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#7

09/18/2018 8:48 AM

Indeed. Torque is what the load presents to the motor, and varies with speed according to the load's characteristic curve. A motor that is able to sustain the relationship between speed and torque at the operating point on the load characteristic curve is not in <...overload...>.

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#3

09/17/2018 8:23 PM

"Overload" in an AC induction motor is about exceeding the motor's designed thermal damage limits. Heat in the motor is a function of the current flow (I2R losses or "copper losses"), magnetic hysterisis (eddy current or "magnetic") losses, bearing friction and windage. But the vast majority (>70%) of the heat is coming from the copper losses based on current flow. So if your motor is designed to handle 50A, but it is only pulling 10A, there is no "overload" taking place.

As to "saturation", that is why you use a VFD. It is controlling BOTH the frequency AND the voltage, so that the motor always gets the proper ratio of V and hz that it was designed for. So in your example the motor was designed for 380V 50Hz, a ratio of 7.6 V/hz. At 25Hz then, it will reduce the voltage to 190V*. No saturation.

*In reality, the required torque of a centrifugal fan is much lower than 50% at 50% speed, it will be 50% cubed, so 12.5%. So that means at that speed, the motor torque could be much lower and the VFD will, if told to, reduce the voltage even more, essentially UNDER fluxing the motor as a way to reduce those magnetic losses. It's not much, but over time it does add up.

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#4

09/17/2018 11:02 PM

Thanks for understanding the question. At 25Hz the Voltage should be around 200V but I noticed the drive in question is closer to 260V at 25Hz. Nameplate is 400V at 50Hz. The PF is about 0.5.

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#5

09/18/2018 3:15 AM

At those figures, there is no <...overload...>.

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#9

09/18/2018 12:33 PM

What type of meter are you using?

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#8

09/18/2018 12:15 PM

"In reality, the required torque of a centrifugal fan is much lower than 50% at 50% speed, it will be 50% cubed, so 12.5%."

No, the torque varies as speed squared, the power as speed cubed.

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#10

09/18/2018 12:35 PM

<...the torque varies as speed squared, the power as speed cubed...>

For a <...fan...>, close enough.

If the motor were connected to anything else, a conveyor belt for example, then it would be different.

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#11

09/18/2018 2:06 PM

Yes I know, but this thread is about a fan. see original post.

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#12

09/18/2018 2:21 PM

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#13

09/18/2018 2:54 PM

Not mine it won't

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#14

09/19/2018 1:38 PM

"No, the torque varies as speed squared, the power as speed cubed."

Good catch... my mistake.

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#6

09/18/2018 8:16 AM

It will modulate the power delivered to the motor to regulate the speed of it.

At <...10Amps...>, a motor with a full load current of <...50Amps...> is not in <...Overload...>.

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