kartracer087
Electrical
- Apr 18, 2020
- 61
Hello,
Just was curious if anyone knows approximately what LADWP's RS substation transformer impedances are. I know their breakers at their substations were rated 25,000A for many years and the IEEE standard rates those at a test X/R of 17. Based on that the minimum impedance considering infinite primary would be :
25,000A interrupting at X/R = 17 considering a system X/R ration of 30 ---> maximum allowable current that the breaker could safely interrupt would be somewhere in the range of 21,000A symmetrical given IEEE test X/R for breakers is 17 which is less than estimated system X/R of 30, so my de-rate to 21,000A is suitable. X/R for transformers with 100MVA base is about 30-35, roughly. So it would need to interrupt roughly 21,000A given the X/R of 30 to not exceed the tested interrupting capability of the breaker.
My figuring on 100MVA base (100/130/160MVA transformer): 230kV-34.5GY/19.92
21,000A = 1,255MVA -----> 100MVA/1255MVA = 7.97% minimum impedance.
Of course, if anyone here knows roughly what their bus current fault levels are you could back-calculate in the same way. I do know that they use S&C Vista switches on their distribution system so I'd imagine fault currents would be limited to approximately 25,000A in a non-bus tied situation (off of one transformer).
LADWP's stations are usually double bus-double breaker or breaker and half and their new station breakers are rated 40kA symmetrical. So in a tied situation you would still have half the current interrupted by one breaker and the other half by the other. The only exception to this would be if you had a situation per my attached sketch. In this case the full 40kA would be required to be interrupted by the breaker in question, so in this case if the interrupting ratings of the breakers are 40kA symmetrical, you would realistically need to reduce fault currents down more so that when two transformers are tied and one breaker would be required to interrupt the fault on a line bay, you would be no greater than say 37kA. In that situation, maximum fault current from each transformer would be limited to about 18,500A. Doing the same analysis as above, on 100MVA base, impedance would be 1105MVA short circuit or 9.05% minimum impedance.
Anyone have any ideas on what their impedance values actually are per spec on 100MVA base?
Just was curious if anyone knows approximately what LADWP's RS substation transformer impedances are. I know their breakers at their substations were rated 25,000A for many years and the IEEE standard rates those at a test X/R of 17. Based on that the minimum impedance considering infinite primary would be :
25,000A interrupting at X/R = 17 considering a system X/R ration of 30 ---> maximum allowable current that the breaker could safely interrupt would be somewhere in the range of 21,000A symmetrical given IEEE test X/R for breakers is 17 which is less than estimated system X/R of 30, so my de-rate to 21,000A is suitable. X/R for transformers with 100MVA base is about 30-35, roughly. So it would need to interrupt roughly 21,000A given the X/R of 30 to not exceed the tested interrupting capability of the breaker.
My figuring on 100MVA base (100/130/160MVA transformer): 230kV-34.5GY/19.92
21,000A = 1,255MVA -----> 100MVA/1255MVA = 7.97% minimum impedance.
Of course, if anyone here knows roughly what their bus current fault levels are you could back-calculate in the same way. I do know that they use S&C Vista switches on their distribution system so I'd imagine fault currents would be limited to approximately 25,000A in a non-bus tied situation (off of one transformer).
LADWP's stations are usually double bus-double breaker or breaker and half and their new station breakers are rated 40kA symmetrical. So in a tied situation you would still have half the current interrupted by one breaker and the other half by the other. The only exception to this would be if you had a situation per my attached sketch. In this case the full 40kA would be required to be interrupted by the breaker in question, so in this case if the interrupting ratings of the breakers are 40kA symmetrical, you would realistically need to reduce fault currents down more so that when two transformers are tied and one breaker would be required to interrupt the fault on a line bay, you would be no greater than say 37kA. In that situation, maximum fault current from each transformer would be limited to about 18,500A. Doing the same analysis as above, on 100MVA base, impedance would be 1105MVA short circuit or 9.05% minimum impedance.
Anyone have any ideas on what their impedance values actually are per spec on 100MVA base?