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Minimum size of CPC

bakey1959

When using the Adiabatic equation, only to calculate the minimum size CPC, I am thinking that using the fault currents given in Appendix 3 of BS7671 rather than that calculated (UO/ZS) would give a more realistic outcome.

For example: A D6 circuit breaker is protecting a circuit where the calculated fault current is 328A. The table associated with Fig 3A6 in Appendix 3 of BS7671 shows that an overcurrent of 120A will operate the 6A breaker in 0.4s. Using these values and a K value of 115 results in a minimum CPC of 0.66mm2 (1.0mm2), whereas if you use the 328A the calculation results in a minimum CPC of 1.80mm2 (2.5mm2).

I cannot see a problem with using the lower value, but would appreciate confirmation and guidance from others.

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    The 0.4 seconds figure is too long - a real breaker will not be that slow once you are above 20times the rated current - hence the 120A - the highest possible level at which the internals of the MCB changes from being slow and thermal to being magnetic and fast.
    In practice you will struggle to damage a 1mm cable with a D6 breaker at any fault current that the breaker can break

    Mike

  • 0 in reply to mapj1
    mapj1 said:
    In practice you will struggle to damage a 1mm cable with a D6 breaker at any fault current that the breaker can break

    The data in the EIDG (Tables 8.7 and 8.8) doesn't support that assertion for all manufacturers' breaker (using manufacturer provided data)- even with Type C at 3 kA and 6 kA.

    But there's probably some "engineering overhead" in "cable fault survival" assumptions.

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  • 0 in reply to mapj1
    mapj1 said:
    In practice you will struggle to damage a 1mm cable with a D6 breaker at any fault current that the breaker can break

    The data in the EIDG (Tables 8.7 and 8.8) doesn't support that assertion for all manufacturers' breaker (using manufacturer provided data)- even with Type C at 3 kA and 6 kA.

    But there's probably some "engineering overhead" in "cable fault survival" assumptions.

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