PM motor current

[thinker]

Please advise: for VFD system (unity PF at AFE) with PM synchronous motor - how to estimate total power/current consumed from the grid when the motor is spinning up to max speed with no shaft load.

 

[itsmoked]

Not easy to do. I'd tell the VFD to display it as that would be the final word anyway. That's to motor only. You'd need to measure the power going INTO the VFD to include the VFD losses.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[thinker]

Thanks for response, Keith. A bit more details. This is a test stand for 4WD vehicle transmission testing, with 4 PMSM wheel dynamometers(absorbers) and a common DC bus VFD system. We selected input drive transformer kVA rating (reduced comparing with rated for full scale system size)based on specific test vehicle prime mover (ICE) power capacity. As customer also requested to spin all 4 motors to the max speed (with no shaft load, but at the same time) we have a problem to evaluate if this transformer is still adequate to cover for 4 motors no load current. It would much easier for us to estimate that for induction motors, but we have very limited experience with PM motors. So, some advice would be helpful.

 

[waross]

Calculate the stored energy for the spinning masses. That is the energy used to spin up. Add some for losses.
But with VFDs, why not just set the drives to ramp up at a current limit that the transformer will support.
The question now is, how does supply current correlate with motor current at lower speeds?
But, how are you going to dissipate the absorbed energy?
If you are going to regenerate into the grid, the full load is what you should size your transformer for, not the no load acceleration.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

Regardless of whether they are PMAC or SCIM motors, current is still basically directly related to torque, then add a few percent for efficiencies. A PMAC motor is typically MORE efficient than an equivalent (in terms of torque) SCIM motor, most are in the >90% efficiency range. So if you can determine your peak torque requirement for accelerating the unloaded test system, you can estimate that the peak current will be roughly 10% higher than that value.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[thinker]

Thanks for your responses, gentlemen. We found out the source of our confusion - the fact that above the base speed VFD supplies to the motor additional reactive current component to reduce the flux (field weakening simulation) in order to limit raise of the motor generated back EMF at higher speeds. In reality, the front end has to cover for losses in the system and shaft acceleration even with no shaft load. Thanks again for prompting us to think better.