Kiryel712
Electrical
- Feb 18, 2016
- 4
Hello All,
I'm new to Eng-Tips (and to forums in general), but I find that my "googling" skills and personal resources have reached their limit. So...onto my question - which is more based on curiosity rather than needing an actual solution: Upon the loss of Utility Power, is it possible for the "leftover capacitance" (and I use this term loosely to define the voltage still in the conductor, after the supply has been de-energized) to be strong enough to power the main breaker's trip coil unit to activate within say one (1) second of the power loss? Some general background: We have separate dual utility 15kv feeds to our site. Two (2) MAIN breakers (each fed from it's own Utility source), with a set of TIE breakers (one side's TIE is always CLOSED, and one side's TIE is always OPEN). In the event that we have a single Utility feed outage, say from "Side A", our Control Power (as commanded by our breaker relays) is supposed to TRIP OPEN the MAIN breaker on that Side (Side A) and then CLOSE the TIE breaker that is normally OPEN, all so that Side B's MAIN breaker can supply power to the side that lost power. All of our breaker control power is being supplied by a 120V source, stemming off of the actual incoming feed. Normally, we have backup power to keep control power energized, but in THIS case, the reason we had an actual problem is that upon the loss of that one (1) feed, we lost all control power (due to a problem that we are correcting)...so we definitely lost all control power. We know this because the TIE failed to CLOSE, even though our relays (powered by backup power) had commanded the breaker to close. We also know this because all of our indicator lights (which are powered off the same lines as our control power) went dark. So.....the conundrum I have is that our MAIN breaker tripped OPEN as it should have...how did this happen if control power was gone? Hence, my question on whether or not leftover capacitance could power a trip coil unit long enough to activate as commanded...
Thanks ahead of time for any time put towards my question and cheers!
I'm new to Eng-Tips (and to forums in general), but I find that my "googling" skills and personal resources have reached their limit. So...onto my question - which is more based on curiosity rather than needing an actual solution: Upon the loss of Utility Power, is it possible for the "leftover capacitance" (and I use this term loosely to define the voltage still in the conductor, after the supply has been de-energized) to be strong enough to power the main breaker's trip coil unit to activate within say one (1) second of the power loss? Some general background: We have separate dual utility 15kv feeds to our site. Two (2) MAIN breakers (each fed from it's own Utility source), with a set of TIE breakers (one side's TIE is always CLOSED, and one side's TIE is always OPEN). In the event that we have a single Utility feed outage, say from "Side A", our Control Power (as commanded by our breaker relays) is supposed to TRIP OPEN the MAIN breaker on that Side (Side A) and then CLOSE the TIE breaker that is normally OPEN, all so that Side B's MAIN breaker can supply power to the side that lost power. All of our breaker control power is being supplied by a 120V source, stemming off of the actual incoming feed. Normally, we have backup power to keep control power energized, but in THIS case, the reason we had an actual problem is that upon the loss of that one (1) feed, we lost all control power (due to a problem that we are correcting)...so we definitely lost all control power. We know this because the TIE failed to CLOSE, even though our relays (powered by backup power) had commanded the breaker to close. We also know this because all of our indicator lights (which are powered off the same lines as our control power) went dark. So.....the conundrum I have is that our MAIN breaker tripped OPEN as it should have...how did this happen if control power was gone? Hence, my question on whether or not leftover capacitance could power a trip coil unit long enough to activate as commanded...
Thanks ahead of time for any time put towards my question and cheers!