Unfused spur.

AJJewsbury:

The 1.5mm2 6242Y is no longer than 300mm and runs through a brick wall. It starts from a deeply sunk double metal box and terminates at a surface mounted all insulated 13 Amp. I.P. 64 rated single 13 Amp socket. Any physical damage is very unlikely.

It's not just about physical damage - faults can occur from many causes. I'm sure we've all come across back boxes where a wire has come loose and touched something it shouldn't or been nicked by one of the faceplate fixing screws (possibly after Mr DIYer has been loosening faceplates to redecorate or otherwise messing with things).


 

The metal box of the double socket in the house, on the ring,  is earthed by 2 X 1.5 mm2 C.P.C.s. The outdoor single socket is all insulated. So we have a very well earthed metal box in the house. Then the 300mm long  1.5mm2 6242Y cable runs through a brick wall to an all insulated outside socket.


Z.

davezawadi (David Stone):

Don't worry Zoomup, it wasn't really you! The case we are considering is as you have probably noticed is really quite complicated, and therefore not in the "normal" range of installation designs. If you had used a bit of 2.5mm, no one would have said a word, but some quite important stuff would not have got an airing here. The discussion has covered two different scenarios, the TT one where the Earth fault current is limited to a low value by Re, and the TN-S one where the fault current is limited by the source impedance and the wiring resistance. You have now told us that each leg of the ring is probably about 15 metres, and so we can calculate the maximum fault current. We also know the ability of a 1mm cable to withstand this current for the time necessary to trip the MCB, and can therefore decide that the installation is satisfactory.


Kind regards

David

Thanks. Would the swift operation of the R.C.D. limit the time the fault current flows, on a L to E fault, thus reducing the heating effect due to energy let through on the 1.0mm2 C.P.C? The R.C.D. should open within 2 cycles shouldn't it?


Z.

davezawadi (David Stone):

No Andy, it is not that you must assume the energy let through is the maximum, it cannot be unless the fault current is very high. The B32 for instance characteristics in the Regs shows a vertical line on the graph at 600A fault current, which according to the graph is between 10ms and 5 seconds. This is the instantaneous region, and the worst case is your energy let through. You are assuming the worst case can occur anywhere in the characteristic, but this is not the case. You will see the note on the graph, which says the worst case is at the high end of fault current. You do not and cannot have this, so the let-through must be less. Instantaneous operation is usually considered 10ms, not you see an unreasonably quick figure. Therefore the energy let through is I2t Watts, so the current is the important factor.

I read the graph at 160A down to 100ms - but that's an aside. I agree the energy let-though will be less than 18,000A²s, but my point is that in order to show compliance we need to actually quantify how much less - specifically that it'll be no higher than 13,225 A²s to suit 1mm² Cu in a PVC cable. I still think we don't have the MCB data to do that.


If from Z's latest posting we can deduce that the fault current will be below 575A (but above 250mA) then we can use the 40ms disconnection time of the 30mA RCD and that should be fine.


   - Andy.

Would the swift operation of the R.C.D. limit the time the fault current flows, on a L to E fault, thus reducing the heating effect due to energy let through on the 1.0mm2 C.P.C? The R.C.D. should open within 2 cycles shouldn't it?

At this current the breaker opens in less than 10ms

mapj1:
I am unclear about the energy let-through subject. I squared t, or A squared t. (Amp2 seconds).


I understand current and I understand time. Where did the current squared originate?


I suspect you are not alone, the choice of units is not the most descriptive.  Really it is a measure of the total energy dissipated in  every small resistance in the fault loop, in the time between start of fault, and the disconnection.  At the risk of a long post, and not one for those short of sleep, it may help to play some algebra games.

If need be print this out and take your time to digest it. (There is no exam at the end, except safe designs !)


V*I = watts. (1)  (by definition) 

V*I*t = joules (2)  (heat energy)

V= I*R           (3) (Mr Ohm)

so I*R*I*t = joules (merging statements 2 and 3)

regrouping


I*I*t*R = Joules

divide both sides by R, this remains a true statement

I²t = joules/ohms (5)

  Ta-da really these units are telling you that if you have say a switch contact of so many milliohms in series with a fuse of that let-through, it will be heated by that many joules per ohm  during the blowing time. If you know the contact heat capacity (so many degrees rise per so many joules of heat per certain mass or volume to be heated - watch you are using figures where the base  units are the same), you can then deduce a temperature rise, and decide if it will be damaged or not.


Take the example of a simple copper wire 1mm² in cross-section, and 1m long.

total resistance call it 16 milliohms per metre of length (or 19 if it is hot)

Total volume 1000mm³  

total mass

m=9 grams per metre of length

(it is 9 tonnes  per cubic metre, but rescaled... )

heat capacity per unit mass (S or Q in some texts), rescaled gives ~

S=  0.38 joules will raise one gram by 1 degree C   (in vacuum with no heat escape allowed)

temperature rise is more with more electrical power, but reduced by more mass of stuff to heat or higher heat capacity


dT=I²R*t/(S*m)          (6)


you can think of dT= difference in temperature


So for an example say a 100 degree temperature difference between before and after


100=I²R*t/(S*m)       (7)


re order to get


I²t=(S*m*dT)/R    (s)


= (0.38*9*100)/0.016

(note the per metre of length cancel as there is one on top one below.. )


 ~22 000 joules per ohm

or 22,000 amps² seconds will give a 100 degree rise in 1mm² copper cable.


for a 1000 degree rise, more like what it would take for copper to melt


more like 220,000 amps² seconds would be needed.


In reality there is always some cooling, so these figures will not quite match tabulated ones from real experiments, the faster it blows the less time there is for heat to escape, the better this approximation fits.

But it shows the technique.


Come back if you need walking through it in finer steps.



Mike.







 

Wow and thanks Mike. Thanks very much. I have never had it explained in the clear understandable terms that you have used. I am off to work now but will re-read your post later and retain it for future reference.


Z.