short circuit 2

[eman30]

I've got myself confused. I've got a utility with 3phase Isc=68.1MVA X/R=3.80218 going to a 3000kVA, 13.8kv/480V, 3phase, z=5.69%. Am I to calculate asymmetrical or symmetrical for sc?

 

[tommom]

Symmetrical, unless you're applying a circuit breaker that was manufactured prior to I think 1969. The standard basis from ANSI C57 something has used symmetrical as the rating basis for equipment, assuming a maximum X/R ratio. That ratio is 6.6 for 600 volts and under, and 15 for anything above that (I'm not sure if there is an upper voltage to that X/R).

 

[nightfox1925]

The first cycle (or 1/2cycle)symmetrical S.C. current is calculated and applied with appropriate multiplying factors (MF) to obtain the asymmetrical value based from system X/R ratio at the location of fault. Generally this asymmetrical value is used to compare or specify the bus momentary ratings of both bus (bracing) and circuit breaker interrupting ratings. The bus is required by ANSI standards to have a momentary rating equivalent to the momentary rating of the circuit breaker, switch or fuse applied to the switchgear equipment.

The ANSI Standards has established Multiplying Factors for the calculated symmetrical fault currents. For MF applied to HV circuit breakers you may refer to ANSI/IEEE C37.010-1979 and for LV circuit breakers ANSI/IEEE C37.13-1990.

Bear in mind that PCBs are tested with X/R ratios specified by industry standard with a given symmetrical fault current and breaker contact parting time...then this establishes the allowed asymmetrical S.C. current the breaker is allowed to interrupt. The actual calculated X/R ratio is too high, then the asymmetrical fault current may exceed the PCB interrupting capability.

You may also want to check out IEEE Std. 141-1993 pages 129 up to 138 (Red Book).

You may also want ot check out the latest revision of the ANSI Std C37.04 and ANSI Std C37.09 (for HV PCBs)using the %Idc in establishing the PCB interrupting current at a specified contact parting time.

 

[eman30]

Thanks for the info, I guess my next question would be then do I even care about the X/R? It seems as if M multiplier for symmetrical = the asym factor, or am I way off?

 

[mpparent]

You still need to be aware of it, as tommom mentioned for instance, low voltage power circuit breakers have a test X/R of 6.6. If you X/R is greater than this, then you need to calculate the fault duty of the breaker, which will include the DC portion.

Mike

 

[nightfox1925]

If your actual calculated (at fault location) system X/R is greater than X/R test of the circuit breaker, then you may have a problem of exceeding your circuit breaker fault duty. The X/R is quite high for power systems which are highly inductive like in the case of industrial plants. A high X/R ratio yields slower DC component decay which is combined to the AC component of the S.C. further yielding the asymmetrical value of the short circuit. Your circuit breaker may have sufficient symmetrical current rating over the calculated symmetrical S.C. bu defficient on the asymmetrical side due to a high system X/R.

GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata