healyx
Electrical
- Apr 7, 2009
- 115
I know that if a circuit breaker is interrupting a known symetrical fault, you can find out the disconnection time by looking at the breaker's time-current curve (Given you have calcualted the prospective short circuit current (PSCC) using cable impedance etc.). Can you still calc disconnect time this way if a fault current limiting breaker is placed upstream of the breaker closest to the fault? The upstream breaker will/may reduce the peak current so that the downstream breaker will take longer to disconnect. Anyone know how to work this out? The peak limited current is quoted as Asymmetric (I think) where as the PSCC is a symetrical RMS value used to index the time-current curve for the breaker. I wouldn't have thought you could use the asymetric limited peak current with the breaker time-current curve to determine the "new" disconnection time??? I have noticed that for lower fault values the limited peak kA is higer than the PSCC kA value (obviously due to RMS calc). Anyone know how to find the actual disconnection time? (FYI, I am trying to work out if an upstream FCL breaker could cause a downstream breaker to move into the thermal (slow) region instead of magnetic (fast) region due to the FCL action and hence cause an earth fault to not be cleared fast enough).
Cheers.
Cheers.