Short Circuit Current contribution.

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Besoeker

Besoeker

Senior Member
Location
UK
Phil Corso said:
Besoeker...

Thus, a key was insertion of the VFD... But I'd still like to know about the increase of restored power, especially the reactive component, above conventional design!

Also, does your company participate in the American market?

Phil
Click to expand...

There is no VFD. The loads were centrifugal. As the speed drops, the rotor current drops and the rotor voltage increases. This results in maximum recovery of just under 15%.
And yes, it was an American corporation. We started as a privately owned UK company but were acquired by a US giant.
 
W

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
My _guess_ is that the rotor recovery side of things will not add much more than its normal maximum output in the event of a short. We are essentially talking about a semiconductor inverter, and much beyond its normal ratings I'd expect it to simply fail.

My further guess is that the short circuit contribution of the entire system will look like that of an ordinary induction motor operating at the same slip with the same effective rotor resistance.

-Jon
 
Wire-Smith

Wire-Smith

Senior Member
Location
United States
Besoeker said:
Could you expand on that please?
Click to expand...

crude motor contribution formula, i didn't see anything special in description. i don't know for sure, hence the ?


FLA = KVA / 1.732 x L-L Volts


FC = FLA / X”


Reactance Values for Induction and Synchronous Machine X” Subtransient

Salient pole Gen 12 pole 0.16, 14 pole 0.21

Synchronous motor 6 pole 0.15, 8-14 pole 0.20

Induction motor above 600V 0.17

Induction motor below 600V 0.25
 
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