What will happen if I operate normal circuit breakers designed to operate in lagging PF in leading PF? Will the breaker interrupt normally?

 

Responding to Anonymous' Aug 29, 10:45pm query.... an interesting, but wide-open question! I'm unaware of specific standards. but some "engineering practices" may be helpful:

1) In the USA pf is ignored, even for short-circuit. However, X/R (lagging pf) ratio's are used to determine interrupting and momentary
duty.

2) Most USA manufacturers advise "factory contact' if circuits are capacitive (leading pf.)

3) Some European manufacturers list duty limits, either as pf values or L/R ratios.

4) Such limits would require contacting the manufacturer with regard to capacitive circuits.

5) For predominantly capacitive (leading pf) circuits, like capacitor banks, are closely scrutinized to preclude the phenomenon known as
"re-striking."

6) Close attention is paid to circuits prone to ferroresonance, especially unground power system!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Can you be more specific about "normal" circuit breakers? 415 VAC, 4160 VAC, 11.5 KV, 13.8 KV, 120 VAC? Are these air-break circuit breakers or oil-filled or vacuum bottled? Are these used in generator output or utility tie? Are they single-phase circuit breakers or three-phase circuit breakers? Are they used on synchronous condensers or large synchronous AC motors?

And what about them makes them "designed to operate in lagging PF"?

What is the normal lagging PF experienced by these normal circuit breakers? What is the expected leading PF to be experienced by these normal circuit breakers?

markvguy

 

Further to my Aug 30, response on the subject... I want to apologize to Anonymous for not really addressing his/her question, "How.....!"

Markvguy listed some of the factors, i.e., kV level, type, and arc medium. But there are more than 30 factors that must be evaluated. Therefore I will address your specific question. It's actually simpler than often thought.

The breaker's primary duty is to safely handle the transitory current (arc) that flows when a circuit is opened, or closed. Interruption occurs when the current wave is near zero. At that time the "'recovery' voltage that appears across the contacts should be low so that the arc
can't be re-established. It is that 'recovery' voltage that determines success. Then, paraphrasing Anonymous... what role does pf play?
Following are three cases to consider:

Firstly, the unity pf case... a pure resistor. Voltage and current are in phase, so when the contacts open at 'zero' the 'recovery' voltage is
zero too.

Secondly, Now the lagging pf case... an inductive circuit. When the current wave is passing through zero, the 'recovery' voltage is
higher... max if the circuit is a pure inductor.

Lastly, the leading pf case... a capacitive circuit. Ignoring "inrush" concerns, then, in theory, the breaker's duty severity is about equal to the lagging case. But, in the real world, it's more onerous because in the presence of inductance the transient current is oscillatory. Hence, more difficult to control.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Further to my Aug 30, response on the subject... I want to apologize to Anonymous for not really addressing his/her question, "How.....!"

Markvguy listed some of the factors, i.e., kV level, type, and arc medium. But there are more than 30 factors that must be evaluated. Therefore I will address your specific question. It's actually simpler than often thought.

The breaker's primary duty is to safely handle the transitory current (arc) that flows when a circuit is opened, or closed. Interruption occurs when the current wave is near zero. At that time the "'recovery' voltage that appears across the contacts should be low so that the arc
can't be re-established. It is that 'recovery' voltage that determines success. Then, paraphrasing Anonymous... what role does pf play?
Following are three cases to consider:

Firstly, the unity pf case... a pure resistor. Voltage and current are in phase, so when the contacts open at 'zero' the 'recovery' voltage is
zero too.

Secondly, Now the lagging pf case... an inductive circuit. When the current wave is passing through zero, the 'recovery' voltage is
higher... max if the circuit is a pure inductor.

Lastly, the leading pf case... a capacitive circuit. Ignoring "inrush" concerns, then, in theory, the breaker's duty severity is about equal to the lagging case. But, in the real world, it's more onerous because in the presence of inductance the transient current is oscillatory. Hence, more difficult to control.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

Very helpful for a student like me traversing and exploring HV field. Thank you Phil!