Does exceeding the tested X/R ratio of a piece of equipment place it in excess of it equipment duty even though the short circuit at that point is lower than what the equipment is rated for?

Any in put would be helpful.

Thanks
Mike.

Maybe. Look at the ANSI standards to determine the proper derating.

There are multiplying factors for this. Depending on how much above the tested X/R the calculated current is, the asymmetry of the short circuit current from the larger amount of X in the circuit could cause the peak current to be greater than the device saw during testing. The test X/R is defined by ANSI or UL depending on the breaker type - 6.6 for ANSI type breakers. The multipliers are used to adjust the calculated current for a more direct comparison to the tested current at it's X/R. Most software can account for this and you may see a note about multiplying factors in the study results.

Equipment and overcurrent devices have ratings that are related to RMS and "peak" values of current. The X/R ratio that defines a fault condition defines the peack current associated with the RMS fault current value. As Jim states in his reply different type of devices, based on defining standard and manufacturer's disgression, are tested at several different X/R ratios. They vary by standard and by fault current they are being tested for.

Unless dealing with relatively low values for UL 489 MCCB devices, and, I assume, associated equipment such as switchboards the X/R ratio is 4.899 which yields a fault power factor of 20%, this in turns yields a peak to RMS ratio of 2.183. The table below shows these ratios for PFs of 20-10%.

PF% X/R Peak/RMS
10 9.9501 2.455
11 9.0354 2.424
12 8.2733 2.394
13 7.6271 2.364
14 7.0721 2.336
15 6.5912 2.309
16 6.1695 2.282
17 5.7967 2.256
18 5.4649 2.231
19 5.1672 2.207
20 4.8990 2.183

If the fault current you calculate has an X/R ~7 then the peak to RMS ratio is ~2.33. Let us say the current you calculate has an RMS value of 45kA. As far as the equipment is concerned, from the peal current perspective it feels like a current of (2.33/2.183) x 45kA = ~ 48kA. For the purposes of comparing the available fault current to equipment it is common to "derate" the equipment by the ratio of the tested current/available current. But it can also be done as I descrived above, by calculating the equivalent current at the tested power factor and then comparing to the equipment rated current. Generally with switchboards you are using a starting X/R of 4.9, for switchgear, as Jim stated in his answer, you start from a X/R of 6.6.

Thanks you for the information it has come in very handy.