VFD removal

[rob194]

I need to remove a vfd for a motor that is currently running at 48.1 hz. The motor is connected to a chain drive. I want to keep the same running rpm (7.5 rpm). Is this a proportional relationship between hz and rpm? I am not experienced in this area and any help would be greatly appreciated. Thanks.

 

[electricpete]

Is this a proportional relationship between hz and rpm?

Yes, neglecting slip which tends to be small.

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[fangas]

To Calculate the speed of a induction motor, apply this formula: rpm = (120 x F)/P
rpm = synchronous revolutions per minute.
120 = constant
F = supply frequency (in cycles/sec)
P = number of motor winding poles

So, the synchronous speed of a 4-pole motor is 1800RPM, a 2-pole is 3600.

As Pete said that's ignoring slip. Slip is the difference between sync speed and developed speed. All squirrel cage motors have it to some degree. It's why a 4-pole motor has a name plate of 1775 say, and the formula says 1800.

Plug in the numbers and you can get the drive ratios you have with the VFD, you might get away changing a pulley or 2. I doubt it though.

BTW, your VFD might have something called "Slip compensation". It's a method of of making a commanded 60Hz speed go at sync speed, rather then nameplate.

Good luck

Ed

 

[rob194]

Thanks greatly.