OT - VFD Hz to Motor RPM?

[skyfox]

Hello all,

Quick question. If motor name plate says 1400 RPM @50 Hz and VFD speed is set to 20Hz does that mean the motor shaft is rotating at 560RPM?
Is the calculation below correct to determine the shaft speed of the motor as commanded by the VFD?

Name Plate RPM 1400/50Hz = 28RPMS per 1Hz
VFD SPEED 20Hz * 28 = 560RPM

Or is there some other variable involved with the calculation assuming there is no load on the shaft.

Cheers.

 

[parky]

In a word yes, however, induction motors do slip on loads.
If a motor has four poles it will rotate half a revolution per mains cycle. At 50 Hz this will be 50 x 60 / 2 = 1500 RPM. If at 25Hz then 25 x 60 / 2 = 750
Therefore 20 X 60 / 2 = 600
The only thing I find weird is that according to normal motor calculations is that
A 4 pole motor should be 1500 at 50hz, yours apparently is 1400, not sure if this speed takes into account of slip/ losses I suppose the only way to measure it at 50 Hz then do the calculation. Same at the lower frequency.
VFD's can display the speed as rpm I assume they know the actual rotation based on the same calculation.
But of course varying loads may affect the actual speed.

 

[lfe]

As parky says, there is a slip, almost no slip when there is no load but the loss of revolutions increases with torque.

It also depends on the motor, small motors usually have more slip.


What the plate indicates is with the motor running at rated power, without load that motor would rotate at almost 1500 rpm

 

[doomsword]

Hello all,

Quick question. If motor name plate says 1400 RPM @50 Hz and VFD speed is set to 20Hz does that mean the motor shaft is rotating at 560RPM?
Is the calculation below correct to determine the shaft speed of the motor as commanded by the VFD?

Name Plate RPM 1400/50Hz = 28RPMS per 1Hz
VFD SPEED 20Hz * 28 = 560RPM

Or is there some other variable involved with the calculation assuming there is no load on the shaft.

Cheers.

It means it's rotating with speed corresponding to 20Hz, and rpm speed depends on the load on the shaft. If it had 0 load it would rotate with synchronous speed for 20 Hz (formula is 60*Hz/pair_of_poles).



So for nominal 50 Hz your motor would rotate with 1500 rpm if it had no load, with nominal 100% load it would rotate at 1400 rpm, this difference is due to slip, to generate torque motor needs to rotate slower than synchronous speed, torque is coming from rotor magnetic field trying to catch stator magnetic field, so more load means more torque which means less speed.

With 20 Hz your motor would rotate with speed 600 rpm with no load and with 560 rpm at nominal (100%) load.


So if you look only for stable situation (no acceleration and deceleration, load is constant and doesn't change) you can say yes, rpm is proportional to Hz setpoint.