Thanks all 

I think I have it now as in the first 2 pics

In the 3rd I have tried to calculate touch voltage with supplementary bonding in place.
Im I in the right area?

Thanks very much

I used the 126.5V as a constant voltage?

JohnE said:

Im I in the right area?

More or less, except that -

the Zs will be reduced by the SB being fitted, and 

not quite all of the fault current will flow through the SB; a (small) part of it will flow through R2.

So Ut = If (in SB) x 0.05

Also, with reference to the formula R≤50/Ia as stated in the regulations:

It is not the resistance between exposed and extraneous-c-ps which is the relevant value for touch voltage but, as in the above examples, solely the resistance of the SB. The resistance of the extraneous-c-p between touch point and SB connection is irrelevant.

I used the 126.5V as a constant voltage?

Probably safer to use 230V (x1.1 if you like = 253V) across the entire loop - 126.5V is based on the assumption that the L and PE sides have the same impedance - which may be true if you have equal sized L and PE or PEN conductors - but having a supplementary bond undermines that assumption (as does using some common cable types, e.g. T&E which have a reduced c.p.c.s).

   - Andy.

This is really helpful thank you. I promise not to keep bothering you.


Just on the Supplementary though..

geoffsd said:

So Ut = If (in SB) x 0.05

AJJewsbury said:

the supplementary bond is in effect parallel with the c.p.c - which reduces the overall earth fault loop impedance, and so increases the fault current

So calculation of the  new Zs is needed

 253V

0.05Ω SB and 0.5Ω R2 in parallel = 0.05Ω

plus R1 0.5

+ Ze 0.35Ω

Zs 0.5+0.05+0.35 =0.9

253 / 0.9 = 281.1A

281 x 0.05 = 14.05V

So without bonding we had  Ut = 126.5V
With bonding Ut = 93.7V

With Supplementary  = 14.05V?

Thank you

Nearly -

With a fault current of 281.1A flowing through the CPC of 0.5Ω and SB/pipe/MEB of  0.05Ω (say negligible for pipe and MEB) then 90% of the fault current will flow through the SB; i.e. 253A.

So Ut = 253 x 0.05 = 12.65V

JohnE said:

Just on the Supplementary though..

geoffsd said:

So Ut = If (in SB) x 0.05

AJJewsbury said:

the supplementary bond is in effect parallel with the c.p.c - which reduces the overall earth fault loop impedance, and so increases the fault current

No, I disagree with the 'supplementary bond is in effect parallel with the cpc' in all cases. The resistance cpc's of the circuits alone (R₂ of circuit 1 + R₂ of circuit 2) may be of sufficiently low resistance to a common DB earth terminal (or in a smaller installation, all the way back to MET) to meet the requirements of Regulation 415.2, especially if RCDs are used. And, see 544.2.1, protective conductors in sheathed cables, or in conduit or trunking, mean the minimum csa can be less than 4 sq mm; the only issue is if there's a disparity in cpc csa between the circuits ... you'd have to reinforce the smaller one with a conductor of csa not less than the largest of the cpc's.

Also, highlights the fact that 415.2 does not necessarily reduce touch-voltage. It 'reduces touch-voltage to 50 V AC or 120 V DC or less for currents that will not operate the relevant overcurrent protective device in 5 s, or alternatively the relevant RCD in either 0.3 s (for non time-delayed devices) or 0.5 s (for S time-delay devices).'

If the fault current is more than I_a, the touch-voltage may well be > (or even >>) 50 V AC or 120 V DC, but will be disconnected in a known time.

Look closely at the formulae in 415.2.2:

415.2.2 The resistance R between simultaneously accessible exposed-conductive-parts and extraneous-conductive-parts shall fulfil the following condition:

R ≤ 50 V/I_a in AC systems

R ≤ 120 V/I_a in DC systems

where I_a is the operating current in amperes (A) of the protective device or:

(i) for RCDs, I_Δn

(ii) for overcurrent devices, the 5 s operating current.

Nothing in BS 7671 limits touch-voltages absolutely in all possible fault conditions, except SELV (and even then, caution if you choose certain modern electronic sources, because they don't all conform to 414.1.1 with respect to voltage limits in certain Sections in Part 7).

If this exposed to extraneous-c-p resistance happens to be 1.5Ω on a 6A circuit where R<50/Ia is 1.67Ω,

how does that have any bearing on touch voltage?