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BS7671 Max Zs and Maximum Measure Zs

BS7671 Max Zs and Maximum Measure Zs

I’ve asked 3 of my lectures, but have yet to get an answer to this question. I have my own theory, but would like to ask if I’m on the right track or not.

 The question is why is the maximum permitted measured Zs = BS7671 Max Zs x 0.8 (appendix 3 page 363). I know it is to do with colder conductors, its the 0.8 factor used, as the table in the On Site Guide do not seem to match is they are at 10⁰C

Guidance note 3 gives the 0.8 as the temperature correction. With the values of Zs in the On Site Guide and Guidance note 3 given for Zs at 10⁰C

The equation in the On Site Guide for temperature correction is given as
(1+0.004(New Temp-20)) as 0.004 is the coefficient correction when starting at 20⁰. As we are not converting from 20⁰C to 10⁰C, we do not need the lower part of the equation.

This would give us a (1+0.004(70-20))=1.2 therefore this would equate to 80% or the 0.8. The quoted 0.8 multiplier.

As the table in the On Site Guide are given at 10⁰C, the equation would look like this (1+0.004(70-20))/(1+0.004(10-20))= 1.25. therefore this would equate to a 75% or 0.75. This correction factor is given as 1.25 in the Note on page 126 of the On Site Guide.

However the figures in the On Site Guide are given as Zs(BS7671) x 0.8 = Zs(On Site Guide).

The true calculation from appendix 3 of BS7671 includes a correction for U0 taking into account Cmin. With Cmin at 0.95, perhaps this would account for the missing 5%?

My theory is that the On Site Guide figures have been corrected for 20⁰C and not 10⁰C, which would bring them in line with Table I1 of the On Site Guide.

Although this is academic, as in design if our calculated Zs figures were equal to or approaching the Max Zs figures, we are likely if the circuit will allow, used additional protection with and RCD. It just my OCD nagging at me.

Thank you for your help.


  • This would give us a (1+0.004(70-20))=1.2 therefore this would equate to 80% or the 0.8. The quoted 0.8 multiplier.



    Not quite  to get back from 1.2 Ohms at 70 degrees to 1 Ohm at 20 degrees you need a factor of 1/1.2 which isn't 0.8 but 0.83' - it's 1/1.25 that gives you 0.8.

      - Andy.
  • Hi Andy


    Thank you for your answer much appreciated. Back on track, 


    Regards

    Ian
  • It’s a rule of thumb estimate of the acceptable measurement allowing for all the derating factors, not just ambient temperature.


    We used to be told to use three quarters of the BS7672, being 75% of the tabulated figure rather than 80%, if you cannot meet that requirement with the test results then you need to check the figure in BS7671 and if it is less than that actually sharpen your pencil and do the full calculation.


    Stop trying to apply a rule of thumb assessment of the test results so rigidly.


    Andy Betteridge
  • Former Community Member
    0 Former Community Member
    Nothing to do then with the contribution made to Zs by the external Ze that could be assumed to remain constant?


    Regards


    BOD

  • perspicacious:

    Nothing to do then with the contribution made to Zs by the external Ze that could be assumed to remain constant?


    Regards


    BOD




    Not including Ze in your calculation Zs will be considered a major error.


    The English phrase rule of thumb refers to a principle with broad application that is not intended to be strictly accurate or reliable for every situation. It refers to an easily learned and easily applied procedure or standard, based on practical experience rather than theory. Wikipedia

  • More years ago than I care to mention I did the inspection and testing course and the lecturer passed comment that older electricians had issues using digital loop testers.


    When guys used an analogue meter the needle would wobble about and they would look at it and say to themselves that’s okay, but with digital meters they get a very precise test result and start reaching for a book to check the result in a way that they would never have done when using the analogue tester.


    You are checking that there is a satisfactory earth connection, if the test result is well within limits don’t get over involved.


    Andy B

  • Not including Ze in your calculation Zs will be considered a major error.



    I suspect Bod was referring to the temperature adjustment (0.8x) only applying to the conductors within the installation (effectively presuming the DNO's conductors were at working temperature) not ignoring Ze altogether.


    Curiously that (seemingly sensible) approach seems to have been phased out of the 18th Ed - implying we should now compare the whole Zs reading on the 80% values.


      - Andy.
  • Ze is not a precise figure, it varies from installation to installation and during the day at the intake of the installation.


    Loop testing is not a precise process, even the figures given in BS7671 for maximum Zs is only a guide, manufacturers data overrules them.


    Andy B.
  • Former Community Member
    0 Former Community Member
    I suspect Bod was referring to the temperature adjustment (0.8x) only applying to the conductors within the installation (effectively presuming the DNO's conductors were at working temperature) not ignoring Ze altogether.

    Curiously that (seemingly sensible) approach seems to have been phased out of the 18th Ed - implying we should now compare the whole Zs reading on the 80% values.



    Spot on Andy. I'd like to know the thought process behind its apparent exclusion.


    Regards


    BOD
  • Don’t try and tie me in knots, it will completely confuse a guy on a training course.


    What Ian needs to know is what to do if he does a loop test on a socket circuit protected by a B32 MCB and the test result of say 1.2 ohms, which is more than the OSG Table B6 figure of 1.1 ohms, but less than the 1.37 ohms given in BS7671 Tabs 41.3.


    A Zs measurement without including  Ze using a loop tester would not be a Zs measurement, it would be R1+R2 determined using the loop tester.


    How do you determine Ze could not be considered as part of the test results?


    Andy B