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What will happen if I operate normal circuit breakers designed to operate in lagging PF in leading PF? Will the breaker interrupt normally?
Very helpful for a student like me traversing and exploring HV field. Thank you Phil!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])
by Bob Odhiambo
by Seth Price