Rather than trying to explain mysef ,here is a working example
Why bother?
It is about making sure the CPC of a circuit will be able to carry the fault current produced during a short circuit or earth fault, without deterioration, or heat/ fire damage. Remember we are talking about fault currents possibly in 100’s if not 1000’s of amps, and even though the CPC will only carry the fault current for a short while, it must not cause fires or heat damage and remain intact so as to enable the protective device to clear the fault.
Isn’t there an easier way of checking this out?
Well actually, Yes! Look at table 54G (page 116 of BS7671: 2001 (2004). Here we have clear guidance; in a nutshell the table shows that for:
·[FONT="] [/FONT]all cable
up to 16mm2 the CPC must be the
same size as the phase conductor
·[FONT="] [/FONT]phase conductors
between 16mm2 and 35mm2 require a
CPC of 16mm2
·[FONT="] [/FONT]a phase conductor
above 16mm2 requires a CPC of
at least ½ it’s size
So where’s the problem and why all the mathematical jargon?
Think about it, every PVC/PVC sheathed cable your likely to install in a domestic situation (other than 1mm2) has a CPC between a third and two thirds the size of the phase conductor (not meeting the requirements of section 543 of BS7671).
So what are we to do?
Well, for certain circuits at least, if we select all our cables and all our circuit arrangements in accordance with the IEE On-Site Guide, then we are told no further calculations will be necessary. Remember compliance with the On-Site Guide does not ensure compliance with BS7671 (particularly section 513) and it only covers installations with a supply cut-out of up to 100Amps. When using the standard circuit arrangements described in the On-Site Guide (Appendix 8) it is still the responsibility of designers and installers to satisfy themselves of amongst other things the suitability of the size of protective conductors (Chapter 54).
What about other circuit arrangements? This is where we might have to apply the adiabatic equation, so let’s have a go with a standard radial socket circuit arrangement to see, if in fact pushed to the limit, the 1.5mm2 CPC is sufficient in size.
Circuit design
An existing double socket outlet wired in 25 metres of PVC/PVC 2.5/1.5mm2 cable and protected by a 20A BS 1361 cartridge fuse. The Zs measured at the socket is 1.40 Ω.
Adiabatic equation :
Ö [FONT="](I² t) [/FONT][FONT="][/FONT]
[FONT="] [/FONT]
[FONT="] [/FONT]
[FONT="] k[/FONT]
[FONT="] [/FONT]
S is the minimum protective conductor size (mm2)
I is the fault current (Amp)
t is the opening time of the protective device (sec)
k is a factor depending on the conductor material and insulation,
and the initial and maximum insulation temperatures
(typically 115 for copper conductors)
The problem is that when you are using the adiabatic equation you will normally be able to obtain a value for the fault current
I, but then you have to obtain a protective device operating time (at that fault current) from the graphs in Appendix 3 of BS7671.
However, to continue:
Fault current
I = Uo = 230 = 164 Amp
Zs 1.40
From the graph depicted in fig 3.1 (BS7671 Appendix 3), a fault current of 164 Amps would cause a 20 Amp BS1361 type fuse to operate in just less than 0.2 sec.
So the minimum size : [FONT="]S =[/FONT]
Ö [FONT="](164² x 0.2)[/FONT][FONT="] = [/FONT][FONT="]0.64[/FONT] mm2[FONT="][/FONT]
[FONT="] [/FONT]
[FONT="] 115[/FONT]
Therefore our CPC of 1.5mm2 meets the requirements of Section 543.