Why don' we use RCD trip times for adiabatic equation

When using adiabatic equation for calculating minimum size of CPC, every example I have seen uses 0.1 second or whatever the disconnect time of the mcb element of the RCBO  or MCB will be.

In a domestic sittuation most circuits are protected by RCD's with a trip time of 40mS with significant fault currents, in this sittuation why don't we use 40mS as T in the adiabatic equation?

  • Quite simply because adiabatic is about protection against overcurrent, and RCCBs, or the RCD element of a combination protective device, cannot provide protection against overcurrent.

    GN6 tells us (Section 1.5):

    While residual current devices (RCDs) can provide protection against electric shock by automatic disconnection of supply, they do not provide protection against overcurrent. Residual current circuit-breakers (RCCBs) must always be backed up by a separate overcurrent protective device to protect against fault current (and, if required for the particular circuit, overload current). Overcurrent protection may be included in the same device, for example, residual-current circuit-breaker (with overcurrent protection) (RCBO).

    This does bring into question how we approach the situation for TT systems. The important factor is that we can't assume the prospective earth fault current is determined by the earth electrode alone (in the way we do for ZS for ADS), because extraneous-conductive-parts, or fortuitous earthing, may well reduce the overall effective earth electrode resistance ... and increase prospective fault current.

    In the worst-case, prospective earth fault current could well be the same as L-N prospective fault current (see Section 6.4.3 of GN6), and therefore we ought to consider using the same approach for protection against overcurrent for earth faults in TT system earth faults, as TN system.

  • OK, so what sort of fault will exercise the insulation on the CPC to its thermal limit, without triggering the operation of any non-broken  RCD?

    I agree that an RCD is not much  good against L-N or L-L faults.

    Also  while an RCBO is built with contacts that can open the full 6kA, 10kA or 16kA or whatever, many RCD only devices have smaller arc chutes and may only be guaranteed to open a maximum Im of 1kA to 2kA without welding contacts or other internal damage.

    Devices to IEC 1008 typically have a spec simiilar to this

    Rated residual non-operating current (l△no)

    0.5 l△n

    Residual current off-time

    ≤0.1s

    Minimum value of rated making and breaking capacity (lm)

    1KA

    Rated conditional short-circuit current (lnc)

    ln=25,40A Inc=1500A

    ln=63A Inc=3000A

    Mike.

  • OK, so what sort of fault will exercise the insulation on the CPC to its thermal limit, without triggering the operation of any non-broken  RCD?

    Perhaps life is the other way round ... for faults of rather high-current, the OCPD operates before the RCD ... that's why RCDs are tested (to the product standard) with their backup protection and have a conditional short-circuit rating, because of the "grey area" in the middle?

    At the end of the day, regardless of what an RCD will or won't do, the standard for RCDs assumes (quite rightly) that there will be an OCPD as well. And the tests cover for the fact the RCD may operate first ... or not

    And for designers using BS 7671, RCDs don't have a "let-through energy" or equivalent time-current curve to use with the adiabatic criterion ... fuses and circuit-breakers (including RCBOs) do ... but with RCBO's (and CBRs) the let-through energy is that of the mcb (CB) element, not the RCD element.

  • To give more context.

    During my training my instructor said that houses with sockets wired using 2.5/1.0 T+E would probably fail on the adiabatic equation and need rewiring. In my opinion if the circuits are RCBO protected, due to the quick switching time of the RCD element they are ok. I have yet to do some calculations but wanted to explore the general principles first.

    Fault current discussed earlier is flowing through live and neutral, size of which is checked against appendix 4 and voltage drop calculations, therefore not a major concern. I wouldn't condone designing new circuits that fail on the standard adiabatic equation calculation method but don't believe a rewire with 2.5/1.0 T+E in good condition is justified.

  • well, even if it fails the adiabatic analysis, one may as well wait until the insulation has actually been cooked by a short circuit fault, before re-wiring - the effort is the same, and it may allow you to keep the installation running for several years longer, maybe pretty much for ever. You will be far enough away from melting the copper that there is no risk of that, just that the insulation around the CPC will be taken over the recommended temp during a 'silver spanner' short circuit.

    Typical figures (Hagar)


    The let-though of an MCB varies with available PSSC, as unlike a fuse it does not get faster and faster with increasing fault current but we can say something fairly intelligent.
    Mike.

  • IIRC, my tutor said that we needn't worry about T&E because somebody cleverer than us had done the calculations, although I do take the point about 2.5/1.0.

    I think that you only really need to worry about it for substantial circuits in singles.

    Let's not forget that a particular device may go faster than 0.1 s and that in a domestic installation, PFC may be relatively modest. In any event, if the most likely place for a fault is at the plug or the appliance's flex, the actual fault current will be smaller still.

  • In a domestic sittuation most circuits are protected by RCD's with a trip time of 40mS with significant fault currents, in this sittuation why don't we use 40mS as T in the adiabatic equation?

    You could do - but you'd have to work it with the maximum fault current (which as others have mentioned could be relatively high under some circumstances) - and that then rapidly gives you some quite large numbers - e.g. 1kA fault yield I²t 40,000 A²s, 3kA gives 360,000 A²s, and 6kA 1,440,000 A²s - compared with a withstand (k²S²) of say a 1.5mm² c.p.c (k=115) of less than 30,000. Generally using the energy let-through figures for the MCB gives a more favourable answer (because at those sort of currents, the MCB is faster - which is what you'd want anyway to protect the RCD's contacts from having to break massive fault currents).

    Thanks Mike - that one's actually a fair bit more favourable than the generic values from the MCB standard (I think that quotes 18k, 45k and 90k for a B32 at 3, 6 & 10kA).

    my tutor said that we needn't worry about T&E because somebody cleverer than us had done the calculations

    Even the cleverest people will need to have made certain assumptions to do calculations (or to put it another way, there will be limits beyond which their results might no longer hold true). From the above a 1.5 mm² c.p.c. in 2.5mm² (withstand just under 30k A²s) might well be OK in domestic situations where the PFC is relatively low, but you might be on soft ground to assume that T&E will be good in other situations - e.g. some commercial and industrial settings where the PFC could be significantly higher.

       - Andy.

  • T&E has no place in commercial or industrial settings. Discuss.

  • Fault current discussed earlier is flowing through live and neutral

    But also in a fault to an exposed-conductive-part, the cpc.

    The issue with 1.0 sq mm is likely the use of circuit-breakers instead of fuses (and one potential issue with "upgrading to modern devices").

  • In my opinion if the circuits are RCBO protected, due to the quick switching time of the RCD element they are ok.

    But for higher fault current, the magnetic trip of the mcb part of the RCBO operates in < 0.01 s (in fact, < 0.001 s is also not unheard of) - far quicker than the RCD 0.04 s. Same, though, happens with cartridge fuses and RCDs as I've seen many times - BS 88-3 fuse blown on earth fault, RCD still "closed".