Surely it is not a 69kva TX - as far as I know, no such beast exists - this is a virtual transformer at the point of arrival of the supply to force fit the PSCC measured - in reality  more likely to be a 1/8 or 1/4 megawatt lump and perhaps a hundred metres of street main cable . Actually if that is the software calculation method, then that is misleading, as there is a lot of difference in being close to a small transformer, where the impedance is largely inductive, and  far from a big one (I see shades of the father Ted small / far away cow sketch here ) where the impedance is more resistive.

In a real fault this alters the degree of overshoot of the waveforms compared to a clean slice of sine-wave quite considerably and so ratio of the peak fault current to the RMS is higher nearer the TX.

But Lyle's method is fine in any case, though as I like easy rules of thumb, I'm a 'just double it' sort of person  at least when sizing things, more forgiving reviewing designs of others that are already in place.

The lazier argument for double it goes like this. IF you have a 3 wire bolted fault, then at the point of fault, the voltage is zero - identical to that at the transformer star point, assuming all 3 fault currents equal there is no neutral current to speak of. . So the current is set by the line conductor resistances only. In the P_N loop test it was set by the series resistance of line and neutral conductors, likely equal. So for the 3 wire bolted fault, the current is double the LN loop test result. . Tada - no hard sums. Just wrong if the supply cable is reduced neutral of course. Of if you are close to the transformer and the supply is not just 3 identical resistors made out of the line cores of the supply cable.

Users of approximations, know your limits, as you might say. With a nod to one of Harry Enfield's historic characters..

Mike.

Surely it is not a 69kva TX - as far as I know, no such beast exists - this is a virtual transformer at the point of arrival of the supply to force fit the PSCC measured - in reality  more likely to be a 1/8 or 1/4 megawatt lump and perhaps a hundred metres of street main cable . Actually if that is the software calculation method, then that is misleading, as there is a lot of difference in being close to a small transformer, where the impedance is largely inductive, and  far from a big one (I see shades of the father Ted small / far away cow sketch here ) where the impedance is more resistive.

In a real fault this alters the degree of overshoot of the waveforms compared to a clean slice of sine-wave quite considerably and so ratio of the peak fault current to the RMS is higher nearer the TX.

But Lyle's method is fine in any case, though as I like easy rules of thumb, I'm a 'just double it' sort of person  at least when sizing things, more forgiving reviewing designs of others that are already in place.

The lazier argument for double it goes like this. IF you have a 3 wire bolted fault, then at the point of fault, the voltage is zero - identical to that at the transformer star point, assuming all 3 fault currents equal there is no neutral current to speak of. . So the current is set by the line conductor resistances only. In the P_N loop test it was set by the series resistance of line and neutral conductors, likely equal. So for the 3 wire bolted fault, the current is double the LN loop test result. . Tada - no hard sums. Just wrong if the supply cable is reduced neutral of course. Of if you are close to the transformer and the supply is not just 3 identical resistors made out of the line cores of the supply cable.

Users of approximations, know your limits, as you might say. With a nod to one of Harry Enfield's historic characters..

Mike.