The IET is carrying out some important updates between 17-30 April and all of our websites will be view only. For more information, read this Announcement

Rules and Guidelines
Want to participate in the discussions? Please read our Community Rules and Guidelines. Need some help? Visit Support and Answers.

  • State Not Answered
  • Locked Locked
  • Replies 39 replies
  • Subscribers 31 subscribers
  • Views 62916 views
  • Users 0 members are here
Options
You may also be interested in
This discussion is locked.
You cannot post a reply to this discussion. If you have a question start a new discussion

5 second disconnection times

Sparkymania
Hi all


Something that I have always wondered about since I started doing electrical work.


The 0.4 and 5 second disconnection times. 0.4 makes sense as it is quick.

However 5 seconds still seems a long time for exposed conductive parts to remain live. When I first started, lighting circuits had a 5s time.

Now it's 0.4 for all circuits feeding socket outlets up to 63A but only for fixed equipment up to 32A. So any equipment over 32A can be 5s.

The reason given in collage was that it was portable equipment that can be picked up and gripped but fixed equipment can be pulled away from.

Previously, in 16th ed regs, the 0.4 was for socket outlets and circuits supplying equipment that can be hand held.

However, 5 seconds still seems a long time for exposed metalwork to be live. I know with a low impedance earth the voltage will be lower, but still.


The other thing is that even a distribution circuit that can have 5s dis time, on an earth fault, say in an armoured cable, all earthed metalwork can be live for the full 5 seconds, even hand held equipment on circuits with a 0.4s dis time. I realise that if the fault was on the actual item of equipment itself the voltage would be higher.


Any equipment, though, above 32A can still have a 5s dis time. I come across fixed equipment all the time that is above 32A. This equipment quite often has parts of it that can actually be gripped. When the body has electricity passing though it the muscles contract so it may be hard to pull away.

I've seen a video of three men pushing a tower hitting an overhead HV line. all three dropped down but their hands still gripped the scaffold poles.

I know were dealing with LV but the muscles still react the same.

Even showers could once have a 5s dis time and the only thing that has changed that is the regs for RCDs in rooms containing a bath or shower. It's still on a circuit that, without the RCD, allows 5s.


The fact that the regs have tightened up of what circuits can have 5s dis times shows that there is still a danger on 5s. Otherwise, why change them to 0.4s.


Any thoughts?



  • 0
    Fire protection rather than personal protection.


    Andy Betteridge
  • 0

    Sparkingchip:

    Fire protection rather than personal protection.


    Andy Betteridge 




    Not fire protection, this is disconnection for protection against electric shock in a TN system - Chapter 41, Regulation 411.3.2.3


    Similarly, in a TT system, 1 s instead of 0.2 s is permitted by Regulation 411.3.2.4.

     

  • 0
    I stand corrected, though I think I would still rely on good insulation, earthing and workmanship rather than a 5 second disconnection time.


    Asan aside should that be good workpersonship or good craftspersonship rather than good workmanship?


    Andy Betteridge
  • 0
    First, 5 s is not the longest disconnection time in a TN system with U0 of 230 V ... disconnection times even longer than 5 s are permitted in cases where even that is not feasible ... provided supplementary protective bonding is provided to limit touch potentials within the installation to 50 V AC or 120 V DC in accordance with Regulation 419.3 and 415.2. Not forgetting of course, that this is no longer "supplementary local equipotential bonding" that we used to talk about - it may well apply across the whole installation because of the effect of transfer potentials from one room / area to another in a fault, see Note 3 to 415.2.


    Having said that, protection against fault current (i.e. against overheating / fire) might still be a limiting factor.


    Another example might be the use of 419.2, which again is aimed at limiting touch voltage under fault conditions within a specific time, where electronic converters are used. Again, there may be limiting factors relating to fault current protection though.



    The 5 s disconnection time is a sort of compromise and doesn't align fully with the touch voltage for time graphs in IEC/TR 60479-5, unless you make some assumptions about touch voltages being lowered by earthing and bonding - which is why TT systems have a 1 s disconnection time not 5 for the same kinds of circuit for the most part.
  • 0

    The other thing is that even a distribution circuit that can have 5s dis time, on an earth fault, say in an armoured cable, all earthed metalwork can be live for the full 5 seconds, even hand held equipment on circuits with a 0.4s dis time.



    Indeed - it's a well known problem which used to be known as the 'mixed disconnection times' problem. In times gone by there used to be a requirement to repeat main bonding at each DB that served 0.4s disconnection time circuits which could be subject to 5s faults. But it does get rather messy and the actual benefits of main bonding are hard to quantify (as BS 7671 doesn't have any requirements for its impedance for example).


    I realise that if the fault was on the actual item of equipment itself the voltage would be higher.



    Not necessarily - it all depends on the relative impedances of the line and earth to the point of the fault. If a submain had a reduced c.p.c. (SWA armour only say) and a final circuit had full sized or better c.p.c.s. (steel conduit perhaps) a fault on the submain may well pose a higher voltage than one further downstream.



    Basically it's a compromise. If we had 0.4s disconnection time for everything you could have no discrimination - so would have to risk blacking out an entire installation for the smallest fault (which can also cause dangers) - and in some cases would risk a lot of nuisance tripping (things with large starting currents) if overcurrent devices had to disconnect within 0.4s.


    On the plus side, faults are far more common on appliances than final circuits and less common again on distribution circuits - so under the current approach the vast majority of actual earth faults do come under 0.4s (or 0.2s for TT) disconnection times.


       - Andy.
  • 0
    The 'technical report'  IEC 1200-413 (1996) gives some explanation and rationale for, as hinted by the second part of its number, the shock-protection principles in IEC 60364. This includes a few diagrams about the assumptions about transferred potential, bonding etc.  This report appears never to have been updated, but to have been withdrawn. Perhaps it was felt that giving details of reasons isn't a good idea? I'd love to share it, but sadly it's yet another document that's locked away unless one pays a ludicrous amount of 'CHF' to the IEC, even if it's not considered current.


    I gather this list has had a habit of occasional quizzes.

    Glancing through the above report, I see a couple of candidate questions, in case anyone wants to take the bait.  I've tried to avoid opportunities for pedantic humour (with interpretation) ... I almost neglected to specify the voltage in Q1, since it "should be obvious", but perhaps there's still a loop-hole: a challenge!


    Q1) Consider a 32 A single-phase (230 V) socket, with its own 20 m run of flat twin-and-earth (FTE) cable from a board where the prospective earth-fault current is 2 kA, with the nearest bonding being at that board or upstream.

    The required disconnection time nowadays would be 0.4 s.  However, an assumption on which this time was chosen for the standards is not well fulfilled in the above situation ... what's that assumption?

    Hint: FTE.


    Q2) Why is the letter 'I' the usual symbol for electric current?

    Hint: I was reminded of this when glancing at a French page from the report.


  • 0
    Thanks for your replies everyone.
  • 0

    Nathaniel:


    Q2) Why is the letter 'I' the usual symbol for electric current?

    Hint: I was reminded of this when glancing at a French page from the report.


     



     Intensité du courant

     

  • 0
    A1 to Q1.

    In UK T&E, (not Irish BTW)  the reduced CPC relative to L and N means that the touch voltage at the hot end of a faulty appliance could be rather more than half way up the local mains voltage, actually 2.5mm with 1.5mm CPC is pretty good  - try a 4mm socket radial or perhaps a 40A  electric shower on a 20m run of 10mm2 T & E with a CPC of 4mm2

    If you are in good contact with terra-firma earthed objects, like plumbing, this extra voltage may matter greatly.

    For this reason we had bathroom bonding - the assumption is that in other rooms like kitchens you are less likely to be wet and naked. Now we aim to achieve the same effect by fast ADS with an RCD, rather than EEBADS.


    In respect of the standards while there is a problem of quoting too much  with copyright for the doc itself, the concepts that underpin are not secret, and are not even original to the IEC actually,  and  as such can be discussed in a suitably tutorial  sort of way .


    V fault = Vsupply *(Rlive /(Rcpc +Rlive))



    230V supply.
    • 1.5 sqmm cable has 1.0 sqmm CPC   touch voltage on fault  becomes 138

    • 2.5 sqmm cable has 1.5 sqmm CPC   touch voltage on fault  becomes  144V

    • 4 sqmm cable has 1.5 sqmm CPC      touch voltage on fault  becomes  167V

    • 6 sqmm cable has 2.5 sqmm CPC  touch voltage on fault  becomes 162V

    • 10 sqmm cable has 4 sqmm CPC    touch voltage on fault  becomes  164V

    • 16 sqmm cable has 6 sqmm CPC  touch voltage on fault  becomes 167V


    All more than 120V...


    TT is not really much worse.




  • 0
    Errata

    Equation error -that way up for core cross-sectional areas (CSA)


    V touch = V supply *(L_CSA/(L_CSA+CPC_CSA))


    But with resistance


    V touch = V supply *( R_CPC)/(R_CPC+R_LIVE)


    But the voltages show the  trend OK.


    sorry.

Join us to get the best from IET EngX.

Joining EngX lets you personalise your experience so you stay up to date on the topics that interest you, plus you’ll be able to make connections who are looking to collaborate, exchange ideas and more.

Join us Not now

Community Rules & Guidelines