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power factor correction

power factor correction

power factor correction

(OP)
I'm in Melbourne, OZ, ie 50Hz.  Can anyone point me to orthodox guidance on FUSE sizing for motors with PF correction capacitors connected in parallel.  I will soon have a 25 kW 0.56 PF 415 volt 16-pole motor.  I want to bring the 77 amp motor FLC back to a line current of only about 50 amps using about 37.5 kVAR of capacitor.  Trouble is I don't have much in the way of fuseholder capacity, I'm limited to a 63M100 amp fuse and am not sure if I'm courting disaster.

RE: power factor correction

Hi, how did you calculate that you need 37.5kvars of correction?

RE: power factor correction

(OP)
The 37.5 kVA is derived from a number of sources:  Siemens and the OZ agent for Sprecher and Schuh do a very good algorithm which I used, aiming for correction to 0.95 PF.  An old GE diary also has the same data in the form of a table.  My actual load will be about 22 kW of the 25 rating.  The precise value of capacitor doesn't matter much, rather the inrush currents, even when using capacitor-switching contactors.

RE: power factor correction


37500    vars
415    volts
1.732    root3
52.2    /ΓΈ bank amperes
1.35    US code multiplier
70.4    A overcurrent device

Subject to interpretation and applicability to regional variations and local customs.

RE: power factor correction

Suggestion: Please, would you post the motor efficiency and indicate the motor horsepower or whether kW is input or shaft output in order to get the 77A in correct relationships.

RE: power factor correction

Purplepete;

Can you please post this "algorithm" used to work out the PFC kVAr.  It seems as though your correction may be a little on the high side (possibly creating a leading PF).

RE: power factor correction

Interesting, I calculate the correction required as 35.7KVAR for a power factor of 0.95 so the figure is about right!!
Best regards,
marke

Mark Empson
http://www.lmphotonics.com

RE: power factor correction

Suggestion: A clarification is needed for the following relationships (Hi Mark!)
25kW = ? = sqrt3 x 0.415kV x 77A x 0.56PF = 30.99kW
Mark, is anything wrong in the posting?

RE: power factor correction

Hello jbartos.
Dont forget that the 25KW is shaft power. If you take into account efficiency, then the KW input will be higher.In this case an efficiency of 81% does not seem untoward. 16 pole motors tend to have a low power factor and a lowish efficiency, but still often better that the combination of a high speed motor and gearbox.

Calculate another way, KVA = 77 x sqrt3 x 0.415 = 55.3
=> KVAR = 55.3 x sin (acos(0.56)) = 45.9

Critical correction would be 45.9KVAR

Best regards,
Mark.

Mark Empson
http://www.lmphotonics.com

RE: power factor correction

(OP)
Gents,
thanks for the guidance.  I'm seriously not concerned with the accuracy of my correction.  I will have capacitors with taps toplay with anyway.  The big concern is that I can only put a 63M100 fuse in as a MAXIMUM. I simply don't know if this will cope with the starting current of the 16-pole motor (about 250 amps) as well as the inrush of the capacitor..

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