That is the essence of my question. Is the 6mm connection also being used as a bonding connection.
See below - if extraneous-c-ps
entering the building are (as required) main-bonded
close to their entry into the building, then any protective conductor going to a flat will not be being used as a
main bonding conductor.
Where you find 16mm2 T&E as the supply from a switch fuse to a CU. On a EICR, what code would that be?
If, and only if, it
IS being used as a main bonding conductor, that surely depends upon the adequacy of its CPC (as determined by the requirements of BS7671 {for TN-C-S}). If it's not adequate, I presume that it would probably be a C2, wouldn't it? Any additional 'rules' that DNOs may have is irrelevant to an EICR - which is undertaken in relation to BS7671.
Well I was thinking the fuse would be a 1361 type 2 - But these are not listed in 7671 time current graphs. So I went with the BS 88-3, which is a 63 A fuse. Though in the old switch fuses they were 60A.
With a Zdb of 0.21Ω I calculated a fault current of 1095A As this is the Zs reading I am taking this as the PEFC
To cause a 63A BS 88-3 fuse to trip in 0.1 seconds takes 820A (600A for 0.4 seconds)
So (Square root) 1095 x 1095 x 0.1 / 115 = 3.01 mm So that would be 4mm2 Protective conductor required.
[I had forgotten that we're talking about fuses, but that does make it a little easier than with MCBs, because of the shape of the curves ]
What you've done will give a 'safe' estimate of the minimum required CSA, but it will probably be a considerable over-estimate of what is actually needed.
The BS7671 curve for a 63A BS88-3 fuse stops at 0.1s but, if one assumes that the curve continues down in the way it is going at that point, by extrapolation it looks as if the disconnection time at 1,095A would be around 0.01 sec, probably a bit less. If one repeats your calculation for 0.01s it would become:
[ √(1095 x 1095 x 0.01) ] /115 = 0.95 mm²
So I went with 0.1 for disconnection time rather than 0.4 as the PEFC suggested a fault current greater than the 820A value in Fig 3A1 7671
The required CSA of the protective conductor depends upon the actual I²t (given the actual PEFC and disconnection time at that current for the OPD concerned), not the 'regulatory maximum disconnection time' you are designing for. Hence, as above, a 'proper' calculation of required protective conductor size would use a figure of around 0.01s for your PEFC and fuse, not "0.1s" (or "0.4s").
This all stemmed from a post I was reading on the IET forum. ... The answer in part was...
Is it also a bonding conductor (or at least the main earthing conductor) - - If this is TNCS, your first requirement is to comply with Table 54.8 - then proceed to an adiabatic check.
1 - it is the main earthing conductor for the installation so needs to meet the requirements of the adiabatic expression or Table 54.7 if you don't want to calculate .... 2 - it is also a part of the bonding conductors so needs to meet the the requirements of 54.8. '
As I've been trying to explain in my responses to others (particularly flameport and plugwash), in the scenario you have described, everything really depends upon whether any extraneous-c-ps (e.g. gas pipes)
which enter the building are (as they should be) main-bonded to the DNO's earth terminal close to where the extraneous-c-p enters the building, then no other 'main bonding' is required, and no other protective conductors within the building are 'main bonding conductors'.
As I've said, if the gas pipework enters a flat, that will be extraneous to the flat, and therefore will need to be bonded to the flat's CPCs (at the 'local MET') in order to ensure that the flat is an equipotential zone. The 'earth conductor' from the DNO's earth terminal to the flat (necessary for fault protection) will then be in parallel with the actual main-bonding of that pipe, and therefore will carry a proportion of the current due to, say, a CNE supply-side fault.
In practice, the length of the true main bonding conductor is likely to be very much shorter (hence lower resistance) than the length of the 'earth' connection to the flat so, even if that were of the same CSA as the main bonding conductor, it's likely that only a small proportion of the fault current would flow through that conductor - as I said, rather ironically, the l
arger the CSA of the earth connection to the flat, the
greater will be the proportion of the fault current that it will carry.
Does any of that help?
Kind Regards, John