power factor correction 1

[purplepete]

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.

 

[cbarn24050]

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

 

[purplepete]

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.

 

[busbar]

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

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

 

[jbartos]

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.

 

[TheDOG]

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).

 

[Marke]

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

 

[jbartos]

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?

 

[Marke]

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

 

[purplepete]

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..