Energy let-through for Type 2 MCB's ?

You may need to change the reduced CPC cable with modern stuff because it has just been cooked with a bolted fault within a few metres of  the consumer unit. My point, perhaps not too well worded, was that there is really no need to change the cable on the off-chance of a fault, just because the adiabatic check is not met, until after the fault has occurred. The 1mm CPC racing a 32A breaker (and a 100A fuse as back up energy limit) will still operate the ADS and the 1mm2 copper will survive intact - it is the plastic near it that will not. The nightmare scenario that some folk imagine, of the CPC burning to open circuit and leaving a live fault, is not what happens. (though if the core area has been reduced by over tightening at a joint, maybe, but it has to be quite a noticeable core reduction)

I probably could have worded that better the first time.

Mike.

PS, it is perhaps amusing to ask how far down a 1mm cpc do you need to be on a 230V circuit for faults of various magnitudes to become impossible, because at that current the voltage drop along that section of the copper alone would have to exceed 230V.
Assuming 16 millioms per metre per mm2, i.e. cold cable. It is likely to be a touch higher for  cables that are already warmed by a load. This is the longest length of cable you may need to replace from this failure mode...

6000A 230V = 38milliohms  = 2.38m of 1mm2 core. (and assuming all  the rest of the fault path is much fatter cable and infinite source current available)
3000A 230V = 77milliohms  = 4.8m  of 1mm2 core. ( assuming as above)
1500A 230V = 153milliohms = 9.5m  of 1mm2 core. (assuming as above)
1000A 230V = 230milliohms =14.3m of 1mm2 core. ( assuming as above)

For other cable sizes than 1 square mm, the length for the same resistance scales by the cross-section, so for example to get the same current limiting resistance with 2.5mmmsq cable you need 2.5 times the distance etc.

Have cables been empirically tested and results obtained to show the results of transient high currents due to faults, by say the B.S.I or B.A.S.E.C? Or is this just theoretical stuff?

www.youtube.com/watch

Z.

Have cables been empirically tested and results obtained to show the results of transient high currents due to faults, by say the B.S.I or B.A.S.E.C? Or is this just theoretical stuff?

www.youtube.com/watch

Z.

no - its the adiabatic equation at work. What we calculate is how hot the copper gets for a given I2t, assuming no time for the heat to escape. 

for 1mm2 copper we should ask at the limit

I²t <= K²S² when k=143 what are we really doing ?

well as I have said many times, 1mm² copper wire is 16 to 19 milliohms per metre and amps² times seconds are really joules per ohm- so we are looking at the number of joules of energy dumped into a given resistance.
The second figure buried in there relates those joules to a temperature rise, and that is the heat capacity of copper, about 0.4 joules per gram.degree. or if you want to turn it over about 2.5 degrees rise per joule.gram. So how many grams in our 1m length of 1mm ² Note that is 1000 cubes of 1mm ³ laid out end to end - same mass and volume as one cm³. Copper density is 8.8 times that of water or 8,8 grams per cubic cm.

So,  we have 8.8 grams of copper, and regs say 143² joules per ohm is only just OK, and we have 16 milliohms, well that is  143² *0.016 = 327 joules - so in 8.8 grams a rise of about 90 degrees.

From a 70 degree start this would certainly stuff the PVC if it was that hot for any time but is a very long way from melting the copper, that needs nearly a 1000 degree rise - so ten times the energy, perhaps 3 and a bit times the fault current. And of course the energy taken to break the bonds in the PVC does actually cool the copper a bit ;-) - the heating of the copper is never truly adiabatic.

Mike.

J.P.J.  and the beer connection.....

https://www.simscale.com/docs/simwiki/heat-transfer-thermal-analysis/what-is-joule-heating/

Z.