Short circuit protection - let through value of fuses

Good afternoon, I am looking for some clarification on regulation 434.5.2 for a short duration fault k^2 x s^2 > I^2 x t 

Im looking at the calculation for a very small run of cable 2 x 10mm conductors over 3 metres. Due to this the values will be high and I can not guarantee the cable integrity.

In rspects to the above calculation, I have total brain fog around how I determine I^2 x t 

Can any one help or provide a link to some literature that would aid me here? 

Many thanks 

Parents
  • Ultimately it depends on the device, but depending on tripping time you either look up at the I-t curve (generally the worst case is at the prospective fault current), or if the fault is large enough to trip in <100ms, consult the manufacturer's datasheet where there should be a value in A2s. Note that you want the energy let through for fault clearance, not pre-arcing or other intermediate time!

    BS7671 has charts for BS88 gG fuses in the appendices, as well as several MCBs. Other devices you'll need to get data from the manufacturer either way.

    PS - Beaten to it!

Reply
  • Ultimately it depends on the device, but depending on tripping time you either look up at the I-t curve (generally the worst case is at the prospective fault current), or if the fault is large enough to trip in <100ms, consult the manufacturer's datasheet where there should be a value in A2s. Note that you want the energy let through for fault clearance, not pre-arcing or other intermediate time!

    BS7671 has charts for BS88 gG fuses in the appendices, as well as several MCBs. Other devices you'll need to get data from the manufacturer either way.

    PS - Beaten to it!

Children
  • Thaks for the replies, greatly appreciated. I had referenced the graphs etc but due to the size of fault the tripping time was well below 0.1 sec, in this case I will need to consult with the MI. Unless im missing something! This would be the below calculation I am working to? 

    Does this look correct in terms of time?  So in terms of t i need my circuit to disconnect within this time before the cable is damaged beyond use? Cables I am using are 10mm live and 10mm cpc double insulated.

    (5) - Short circuit impedance

    mOhm/m x length x 1.2 / 1000 = r1 +rN

    3.66 x 3 x 1.2 / 1000 = 0.013

    Ipscc = 230 / 0.013 = 17.7 ka    - app 14 – 16Ka

     

    K = 115

    S = 10

    I = 16 ka

     

    T = k² x S² ÷ I² = 0.005 seconds

    So from K^2xS^2 > I^2 x t   How would this calculation look for a BS88-3 63 Amp or an 80 Amp? Using the above?

    Apologies for this...

  • For a typical makers data, look here Similar date is available from Lawson and MEM

    note that the thinner fuses on the first page let through a slightlu higher I2t
     For 22mm diameter BS88-3

    vs the  30mm diameter  a few pages later.

    treat the pre-arc as a 'never blow' limit for co-ordination with other things downstream, and the '240V total' being the 'always blow' most energy that will ever be let through prior to disconnection on a phase to ground or phase to neutral fault. If you think a phase to phase fault is credible or you are using 400-690 3 phase, read the 'total at 415' column instead.
    In any case take the makers I2t figure, and work back from your max I to an equivalent let go time 't', or forward to the cable damage energy. directly, as that is in the same  units..
    If this  all feels a bit "measure with micrometer cut with axe",  it is, as no real fault is zero impedance, and no two fuses are identical, and as the curevs in the linked data show, the idea that a single figure covers the I2t is not really right - this is just a safe upper limit to the problem.

     A line of constant I2t does not fit this curve especially well, anywhere.

    Come back if this makes no sense.

    Mike.

  • Thanks for this Mike, really helpful information. 

  • mOhm/m x length x 1.2 / 1000 = r1 +rN

    3.66 x 3 x 1.2 / 1000 = 0.013

    Ipscc = 230 / 0.013 = 17.7 ka    - app 14 – 16Ka

    I suspect you're missing the contribution of the supply to the loop impedance (it's all part of the circuit)... you won't have a infinite current available at the start of your circuit (pretty large perhaps, but not infinite).

    For example, if the supplier quotes (as is typical for a domestic PME/TN-C-S supply) Max PFC 16kA, max Zs 0.35Ω then you can deduce that the supply impedance is somewhere between 230V/16,000A= 0.014Ω and 0.35Ω (L-N and L-PE are the same at the origin for TN-C-S). You then add any further installation cable impedances onto that.

    Sometimes it's not obvious which is going to be the worst case - low fault currents with longer disconnection times (e.g. for a fault at the far end of a circuit) or large fault currents and hopefully faster disconnection times (e.g. for faults soon after your protective device) - after all faults could occur anywhere in practice. MCBs and fuses differ quite a bit in that respect. If in doubt it's good to work through both possibilities and compare the results. 

       - Andy.

  • That is exactly ewhat I hacve done wrong here I cant believe I forgot that external impedence.

    Thanks very much Andy.