The guidance in Appendix 14 of BS 7671 are rules of thumb which will overstate the actual fault current.
The maximum prospective fault current on a 3-phase circuit will depend on a number of factors.
- The installation is “cold”. A loaded installation will increase the running temperatures of conductors which in turn will reduce fault current.
- The prospective fault current will be reduced by the current limiting effect of any upstream circuit protection from the fault.
- The prospective fault current will be increased by any stored energy in the installation such as power factor correction capacitors and the capacitive effect of the installation itself. In addition, there may be a contribution from motors running down. Also, additional parallel supplies.
- Calculations are usually carried out assuming “an infinite bus” i.e. the HV supply being able to deliver unlimited current under short circuit conditions on the LV side of the transformer. In reality the HV system cannot deliver unlimited current so this will reduce fault current on the LV side of the transformer.
- The highest prospective fault current will a simultaneous short circuit of negligible impedance between all 3 phases. In reality the fault will not be of negligible impedance.
- This short circuit condition between the 3 phases is known as a” bolted fault” which is a useful description that describes all 3 phases bolted together. The bolted fault would look like a balanced load to the installation so there would be no neutral current.
- The star point of the transformer would sit at 0V to the line conductors and so would the potential of the “bolt”.
Assuming the transformer has been tapped at 433V the phase voltage will be 433 / √3 = 250V.
The maximum prospective fault current can be calculated from the phase voltage (250V) divided by the impedance of the transformer winding added to the impedance of one line conductor.
So, putting some values in for the purposes of illustration, and accepting there will be a margin of error due to the rounding down the value of √3, we can do a calculation. The calculation does not include any of the factors that would impact on the actual values of fault current listed above.
If we have a phase winding impedance of 0.1Ω and each of the line conductors have an impedance of 0.2Ω. The prospective fault current for a bolted fault will be 250/ (0.1+ 0.2) = 833A (0.833kA).
If we adopted the guidance in Appendix 14 and measured the fault current with a loop tester between one line and neutral and doubled that reading the calculated equivalent of this would be. 250/ (0.1+ 0.2 +0.2) = 500A if this value was doubled the PFC would be 1000A. This would overstate the value of PFC by a significant margin.
To get a more realistic value of PFC from field measurement. Measure the value of fault current from line to line. Then divide the measured value by √3/2 = 0.87.
So, using the figures from the example above the calculated value of fault current from line to line would be 433/ (0.1 + 0.1 +0.2 +0.2) = 721.67A. Divide this figure by 0.87 = 830A (0.83kA) which is very close to the actual calculated figure.
What concerns me is the standard inspection and testing teaching is for a 3-phase installation measure line to neutral and double the reading. This is fine if the equipment and circuit breakers are rate at a higher level of PFC but not fine if the inspector does not consider that this measured figure considerably overstates the actual PFC and enters C1 or C2 on the EICR for underrated circuit protection devices.
Have a look at the latest Edition of GN3. The IET provides additional guidance of fault current measurement, advising measuring PFC line to Line and dividing by 0.87.
