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Sizing of protective conductors - what am I missing here

Scrambled

Evening all. As ever I'm wrestling with a conundrum that I've hit a bit of a brick wall with. I could use a few pointers as to what I am missing/fill in some blanks.

Sizing of protective conductors:

If we take say a circuit wired in 1mm cable, protected by a suitable device, lets say a Type C 6A 60898. By selection from 54.7 we can use a 1mm cpc. From the design guide it is stated that if the current carrying capacity of the protective conductor is the same as the line conductor then the adiabatic equation will be satisfied (so long as the circuit has been designed for protection against overload) but if it is of reduced current capacity then we have to either select (54.7) or calculate (adiabatic/verify energy let through) to check the conductor is suitably protected. As a 1mm cpc is the same as the line conductor apparently no issues...

But...

Table B7 of the onsite guide states that the minimum CPC size for a class 3 energy limiting device Type C 16A or less, at a sub 3kA fault level is 1.5mm2. Presumably because the A2S according to BSEN60898 is permitted to exceed the K2S2 of a 1mm conductor for this device. I've checked through a number of manufacturers energy let through charts and indeed for a 6A Type C breaker at certain fault current levels (some around 2.0kA) - the I2T does exceed the 13225 K2S2 for a 1mm conductor. Surely though this would apply to a line-neutral fault as well?

So...

Why is it ok that potentially I could have a conductor with a K2S2 less than the I2T of the protective device in the case of a 1mm2/1mm2 cable just because it has been selected rather than calculated?

Furthermore... if that is truly ok - why (if fault current levels are at the aforementioned) does that 1mm conductor potentially become unsuitable if it is partnered with a 1.5mm2 line conductor instead?

I have a suspicion that the impedance/resistance of the line conductor (being the same size) is coming into play here and this will be the difference effecting the thermal stress on the protective conductor. But I could do with a bigger brain to point me in the right direction!

I hope that made sense! Thanks as ever for tolerating my questions!

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    Further to Andy's comments above /below sideways in this forums odd format.. 

    It all hinges on the fact that as fault currents rise, fuses and cable damage, both being essentially thermal, both get faster more or less pro-rata together.

    The rather unhelpful unit of amps squared seconds can also be considered to be units of joules per ohm, and maybe that makes it more intuitive what is happening  during the instant of flash crackle and pop - each ohm ( more likely small fraction of an ohm for real cables ) of metal in the fault path gets heated by that many joules, but the volume of metal to absorb those joules determines how hot it will get. (less metal volume or more joules, then it is pro-rata hotter...)

    For a fuse once you get into the fast part of the blowing curve, at least until onset of catastrophic rupture, the joules per ohm are more or less constant, and set by the no of joules to melt the (resistive )fuse element.

    A breaker on the other hand cannot get much faster than supersonically parting  contacts, as the arc has to be in free air - in an HR fuse it is normally in sand, which breaks it up as it forms.

    But we are not anywhere near burning a 1mm CPC open circuit with a 16A breaker on a 16kA  PSSC - though the PVC around it will look very unhappy afterwards.

    But as others have noted,  1m down the 1mm cable is 30-40 milliohms of extra fault loop assuming L and E are both 1mm2, so our PSSC starts to fall back from that theoretical 16kA quite fast . (1kA is 230millohms, 10kA is 23 milliohms, so 1m of 1mm  T and E limits you to about 6-8kA on a really bad day, even if the origin PSSC was infinite, and it never is. ( - and that is even if you can manage a silver nail lossless fault, and I bet you can't, a truly lossless fault has no flash and is silent, real ones go bang and do some mechanical work as well)

    Now I suppose you could turn a screw into the lighting cable as you fixed the consumer unit to the wall, so the cable is a few inches long, but that is about the only way you will see the full  fault current. - and if you did, then the fact that cable insulation gets singed is a 'so what ?' you'd  strip back and replace at least the last foot or so  anyway.

    Mike.

Reply
  • 0

    Further to Andy's comments above /below sideways in this forums odd format.. 

    It all hinges on the fact that as fault currents rise, fuses and cable damage, both being essentially thermal, both get faster more or less pro-rata together.

    The rather unhelpful unit of amps squared seconds can also be considered to be units of joules per ohm, and maybe that makes it more intuitive what is happening  during the instant of flash crackle and pop - each ohm ( more likely small fraction of an ohm for real cables ) of metal in the fault path gets heated by that many joules, but the volume of metal to absorb those joules determines how hot it will get. (less metal volume or more joules, then it is pro-rata hotter...)

    For a fuse once you get into the fast part of the blowing curve, at least until onset of catastrophic rupture, the joules per ohm are more or less constant, and set by the no of joules to melt the (resistive )fuse element.

    A breaker on the other hand cannot get much faster than supersonically parting  contacts, as the arc has to be in free air - in an HR fuse it is normally in sand, which breaks it up as it forms.

    But we are not anywhere near burning a 1mm CPC open circuit with a 16A breaker on a 16kA  PSSC - though the PVC around it will look very unhappy afterwards.

    But as others have noted,  1m down the 1mm cable is 30-40 milliohms of extra fault loop assuming L and E are both 1mm2, so our PSSC starts to fall back from that theoretical 16kA quite fast . (1kA is 230millohms, 10kA is 23 milliohms, so 1m of 1mm  T and E limits you to about 6-8kA on a really bad day, even if the origin PSSC was infinite, and it never is. ( - and that is even if you can manage a silver nail lossless fault, and I bet you can't, a truly lossless fault has no flash and is silent, real ones go bang and do some mechanical work as well)

    Now I suppose you could turn a screw into the lighting cable as you fixed the consumer unit to the wall, so the cable is a few inches long, but that is about the only way you will see the full  fault current. - and if you did, then the fact that cable insulation gets singed is a 'so what ?' you'd  strip back and replace at least the last foot or so  anyway.

    Mike.

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