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?
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.
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.
