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PFC for Andy J

John Peckham
Andy rather than enter a thread drift post on the CT scanner topic I thought I would put my views on PFC on a new post. So here is my thinking on PFC.

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.
  1. 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. 

  • 0
    Yes but ...


    IIRC, some testing machines cannot do L-L, so the testing person doubles L-N as an approximation. Most of the time this will be within limits so all is well.


    However, I would expect a competent tester to know that the value is an approximation and what to do if it appears that there is a bare fail.
  • 0
    Chris


    That is correct some older instruments are not rated for 400V. That said if the instrument is not rated for 400V then it should not be used inside a 3 phase DB in the first place!
  • 0
    John, my opinion is that  testing pfc at intake is just a punt, erring on the safe side. Double the line to neutral or divide LL by 0.87 is really not worth the debate. Most installation testers are a bit of a hit and miss affair at low values of impedance and do not account for reactance. As I understand it, some do not use the actual voltage but calculate output at a readjusted 230v. There are so many variables, many identified in your post, that an accurate value is really only an aspiration. 

    Two brand new MFTs L to N 0.1 on one 0.15 on the other LL 0.07 and 0.1 respectively. Three phase + N installation. Ipf = ? I recorded it as 4.6 KA but I think that all we can say is that it is somewhere in that range give or take a thousand!
  • 0
    IMHO,

    In essence, we make quite a few "rules of thumb" in our industry. As such all such rules of thumb must err on the safe side. Any testers we use only give approximations. Our results therefore give a similar outcome of trying to convict a person of a crime but in reverse sense. We go to great lengths to try to avoid an innocent person being found guilty that the result is that many guilty get off with it. That is a safeguard we all enjoy. Any readings we take for verifications of electrical safety mean that any satisfactory conclusions probably are in fact satisfactory, whereas lots of unsatisfactory conclusions actually are quite satisfactory installations really. I don`t think we can really do much else.
  • 0
    HI John -

       Yes, I can see that treating the winding's impedance separately from the line conductor's can make quite a difference to the approximation - although the effect would be less pronounced where the winding impedance is small compared to the line conductor's.


       As a counter example, say we had an old system with a reduced N and comparatively small winding impedance - say 0.01Ω for each winding, 0.1Ω for each line and 0.15Ω for N. Then we'd be looking at a bolted fault current of 250/(0.01+0.1) = 2273A on each line. But a L-N loop would only be 250/(0.01+0.1+0.15) = 962A. Doubling that would only yield 1923A - which would seem to be rather un under-estimate.


       - Andy.
  • 0
    Yes another good reason for not measuring L-N and doubling it. 


  • Former Community Member
    0 Former Community Member
    To some extent, all of the above does tend to ignore the power factor of the fault - which tends to change depending on the relative ratio's between the winding impedance and the line conductor resistance etc. ie the fault power factor changes as the fault moves closer to, or further from the transformer winding.


    OMS
  • 0
    OMS:

    To some extent, all of the above does tend to ignore the power factor of the fault - which tends to change depending on the relative ratio's between the winding impedance and the line conductor resistance etc. ie the fault power factor changes as the fault moves closer to, or further from the transformer winding.


    OMS


    Indeed, which is why I don’t think it matters one jot how it is measured, especially close to the tx. But JP makes a good point which I am going to explore. I have an inspection looming for a client with a 11Kv intake via 2no transformers and LV circuits to feeder pillars around the site with very large swa cables in parallel. Some distribution circuits are relatively short some very long. However, I know the tx characteristics and the almost exact length and make up of the distribution cables. I will have the lads note the LN and LL readings at each pillar and compare with the expected based on the known data. Instruments calibrated in December 2019. Will gather from at least two. 


  • 0
    Do you also know the impedance of the HV side as well ? For very high fault currents this effect we normally neglect does start to bite.

    That said it will be interesting if you do not mind to publish the readings and their variation.
  • Former Community Member
    0 Former Community Member
    The DNOs will provide you with the fault level at the HV point of connection if you ask nicely,  normally within a week or two. The problem is finding clients who have accurate record information detailing the types and sizes of buried cables and drawings showing the cable routes for their private HV networks.

    As others have already mentioned measuring the LV fault current close to a Tx or large switchboard is likely to result in unrealistic values being recorded. There's also the safety of the inspectors and others on site to consider. (SSoW)


    Regards 

    Parsley

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