Ipf Measurement on a 3-phase system

Thinking about this multiply by 2 (a round-up of 1.732)

My understanding is that 2x is meant to be an exact doubling - not a round up of √3. As you say, the assumption is that L & N conductors have the same impedance - with a bolted 3-phase fault, the fault itself effectively forms an artificial N point - so as with any balanced load, the N currents cancel and so the impedance of the N conductor makes no contribution to the loop impedance. So looking at any one individual phase, compared with the single phase fault case (which the loop meter measured when connected L-N) we have the same driving voltage but half the conductor length, so all else being equal we'd have double the current. (For a fault between just two of the three lines, where the N currents wouldn't entirely cancel, we might indeed have a √3 factor to consider).


If you know the relative impedances of your L and N conductors you should, in principle at least, be able to adjust the 2x factor to better suit your situation. Although in practice, as others have already mentioned, loop meters are unlikely to give you an accurate starting figure so close to such a large supply, so might not be worth the effort.


 

as the meter only measures the impedance on the LV side of the transformer and the downstream cables i.e. assumes an ‘infinite source’ on the 11kV side

  As I understand it most loop meters work on the principle of drawing a known current and noting the corresponding drop in voltage to calculate the impedance - as such they do take into account the entire supply, including the HV side (within the meter's limitations at least). A milli-Ohm on the HV side does make considerably less difference than a milli-Ohm on the LV side and will be scaled down accordingly by the transformer's ratio within your reading. (BS 7671's description of Earth fault loop impedance isn't exactly helpful in that regard).


   - Andy.

Thinking about this multiply by 2 (a round-up of 1.732)

My understanding is that 2x is meant to be an exact doubling - not a round up of √3. As you say, the assumption is that L & N conductors have the same impedance - with a bolted 3-phase fault, the fault itself effectively forms an artificial N point - so as with any balanced load, the N currents cancel and so the impedance of the N conductor makes no contribution to the loop impedance. So looking at any one individual phase, compared with the single phase fault case (which the loop meter measured when connected L-N) we have the same driving voltage but half the conductor length, so all else being equal we'd have double the current. (For a fault between just two of the three lines, where the N currents wouldn't entirely cancel, we might indeed have a √3 factor to consider).


If you know the relative impedances of your L and N conductors you should, in principle at least, be able to adjust the 2x factor to better suit your situation. Although in practice, as others have already mentioned, loop meters are unlikely to give you an accurate starting figure so close to such a large supply, so might not be worth the effort.


 

as the meter only measures the impedance on the LV side of the transformer and the downstream cables i.e. assumes an ‘infinite source’ on the 11kV side

  As I understand it most loop meters work on the principle of drawing a known current and noting the corresponding drop in voltage to calculate the impedance - as such they do take into account the entire supply, including the HV side (within the meter's limitations at least). A milli-Ohm on the HV side does make considerably less difference than a milli-Ohm on the LV side and will be scaled down accordingly by the transformer's ratio within your reading. (BS 7671's description of Earth fault loop impedance isn't exactly helpful in that regard).


   - Andy.