gkenyon said:

For a 100 A BS 88-3 fuse, fault current, needs to exceed:

• 1100 A to protect 4 sq mm cpc/line conductor (copper, 70 deg PVC)
• 1000 A to protect 6 sq mm cpc/line conductor(copper, 70 deg PVC)
• 600 A to protect 10 sq mm cpc/line conductor (copper, 70 deg PVC)
• 440 A to protect 16 sq mm cpc/line conductor (copper, 70 deg PVC)

That is confusing and seems rather counter-intuitive! However, am I right in thinking that with 4 sq mm cable, the fault current must be at least 1100 A so that the fuse blows before the cable can overheat? The 6 sq mm cable can absorb more energy so it can afford to wait slightly longer for the fuse to blow, etc.

That's using table 54.7 instead, but yeah the calculation says bigger

It's not just for the c.p.c. - since you can't presume the upstream protective device will always protect the line/neutral conductors either you need to apply what's effectively the same calculation for them too (434.5.2).

   - Andy.

OK yes so amusing a Zs of less than 0.27 ohms again to comply with the 100A BS 1361 / 88-3 fuse's max Zs chart, and a k of 143 for double insulated tails for example, the L&N tails minimum CSA would be 5.9mm2 to comply with the adiabatic equation. So 6mm earth (separate), 6mm L&N double insulated, Zs of less than 0.27 ohms, overcurrent protection of tails provided by 32A RCBO due to the allowance of PD being anywhere on cable length if less than 3m etc. I'm getting there now. For the 100A fuse to provide the tails overload protection I would need 25mm tails rated at 114A method C.

Jimi said:

a k of 143 for double insulated tails for example

I don't think I agree with a k of 143 for the tails themselves? Current-carrying conductors, will always be hotter than ambient temperature because current is flowing down them ? Our standard designs in BS 7671 assume ambient temperature of 30 degrees C.

Table 54.2 can only apply to a protective conductor, which is a suitable distance from any live conductors. You need Table 43.1 for k for the live conductors.

I understand in this case the tails are likely to be well under-loaded, so you could use a  different value of k, but that would have to be calculated in accordance with BS 7454 as per Note 2 to Table 43.1 ... and could not be as high as 143 (which would assume conductor starting temperature of 30 degrees, which would be no current flowing at all). Which implies a csa of  at least 10 sq mm.

a Zs of less than 0.27 ohms again to comply with the 100A BS 1361 / 88-3 fuse's max Zs chart

Rather a lot less than 0.27Ω - my old copy of the OSG (which is probably missing Cmin etc) suggests max 0.15Ω for 6.0mm²  (or 0.22Ω for 10mm²) for a 100A BS 1361 fuse - in these cases disconnection times have to be a lot shorter than 5s to protect the conductors (or rather their insulation).

    - Andy.

AJJewsbury said:

for a 100A BS 1361 fuse

That's a further consideration., because there's no direct read-across between BS 1361 and BS 88-3 fuses - the latter are NOT a direct replacement.

In reality, you'd need to check adiabatic for both, as existing service heads may well have BS 1361 fuses in them ... they are still available, even though the standard is withdrawn !

Have a look at a similar plot against a BS 1361 set of curves (from BS 7671:2001+A2:2004):

AJJewsbury said:

which is probably missing Cmin etc

Certainly, values published back in 2004 would be missing both C_min and the change of nominal voltage U₀ from 240 V to 230 V ... so, for a 100 A fuse, 0.4 s disconnection time, Z_41(2004) = 0.20 Ω, whereas Z_41(2022) = 0.18 Ω

The values of max measured earth fault loop impedance Z_s(m) (which is lower than the loop impedance we've been talking about for design, at 70 deg C, or Z₄₁ as it's termed in the guidance as it's related usually to tables in Chapter 41) that were in Andy's post have therefore also dropped now ... OSG (2022) and GN3 (2022) still quote Z_s(m) for BS 1361 fuses.

Those values are now Table B5 in 2022 OSG:  0.13 Ω (100  A BS 1361 fuse and 6 mm²) and 0.20 Ω (100 A BS 1361 fuse and 10 mm²)
Compare with Table B4 in 2022  OSG: 0.16 Ω (100  A BS 88-3 fuse and 6 mm²) and 0.26 Ω (100 A BS 88-3 fuse and 10 mm²)


I think I'd be designing for BS 1361 as this would be the more onerous requirement (at least for 100 A ... although this does not read across in the same way for other values) as Andy says

Great thanks for pointing that out

Example 1 – CU tails minimum L&N size

(100A BS 1361 / BS 88-3 fuse ad Zs = ≤0.27)

Adiabatic equation:

Short circuit loop resistance = 0.27Ω

I = 851A (230V / 0.27Ω)

t = 1s

k = 115

√ 851 x 851 x 1 = 851

851 / 115 = 7.4mm2 (rounded up to 10mm2 standard CSA).

Example 2 – CU tails minimum earth size

(100A BS 1361 / BS 88-3 fuse ad Zs = ≤0.27)

Limiting value 1 (if protective conductor is not part of a cable, but has a sheath for mechanical protection):

2.5mm2 minimum.

Limiting value 2 (if a TN-C-S system / PME, earthing conductor not less than that required by regulation 544.1.1):

10mm2 minimum for ≤35mm copper PEN.

Limiting value 3 (if protective conductor buried in ground):

N/A.

Limiting value 4 (must not be less than the value determined by selection from table 54.7):

equal mm2 minimum to line conductor ≤16mm2.

or 16mm2 minimum for line conductor >16mm2 and ≤35mm2.

- - - - - - -

Fuse = 100A BS 1361 / BS 88-3 (using BS 88-3 as most onerous ...I think!).

ADS = 5 seconds maximum for TN distribution circuit.

Zs = ≤0.27Ω including 0.8 cold testing factor (to satisfy ADS fault protection Zs table for 100A BS 88-3 on a distribution circuit – 5 seconds). Working the max Zs out manually by formula would give better result I've realised.

Adiabatic equation:

Zs = 0.27Ω

I = 851A (230V / 0.27Ω)

t = 1s (850A on the BS-88-3 time/current chart)

k = 143 (from table 54.2 for insulated protective conductor not part of cable, non-bunched).

√ 851 x 851 x 1 = 851

851 / 143 = 5.95mm2

So when testing Zs on the main tails (distribution circuit) connection to a CU supplied via a ≤100A BS 1361 / BS 88-3 main fuse, and the earth conductor is separate and not bunched (143 k factor), then as long as the Zs is ≤0.27Ω (as required for that fuses max Zs), the main earth conductor must be the largest CSA of the following, as applicable:

• 6mm2 minimum (to comply with the adiabatic equation).

• 2.5mm2 minimum (if not part of a cable, but has a sheath for mechanical protection).

• 10mm2 minimum (for ≤35mm copper PEN if a TN-C-S system / PME).

• XXmm2 minimum (if buried in ground).

• equal to line conductor ≤16mm2 (for not be less than the value determined by selection from table 54.7)

  • that’s 10mm2 minimum (from the line & neutral adiabatic example above).

  • Or 16mm2 minimum for line conductor >16mm2 and ≤35mm2.

Sheeesh. I think I'll stick to 25mm2 tails and 16mm2 earth haha. It was a worth while learning experience though so thanks everyone for the input.

But in theory, based on the conditions I mentioned above with a max Zs and short circuit loop of 0.27 ohms, separate and not bunched earth cable, 100A BS 88-3 main fuse, a PEN supply, a distribution circuit (tails) needing 5 second ADS:

I could use 10mm2 L&N tails to satisfy the adiabatic equation, and to satisfy over current protection provided by the 32A RCBO at the end of the cable rather than the BS 88-3 at the start. Then I could use a 10mm2 earth cable.

If a fault occurs that damages your temporary tails, and gives a dead short L_N or L_E then the 100A fuse will blow. But the adiabatic formula shows that as well as the mechanical damage, in blowing that fuse,  you may well have have 'cooked' your 4mm cables and taken the plastic beyond the point of melting to a state of no return.  - But that's alright! - They are short, so you replace them anyway after such an event, and do not try to tape up the damage and re-use.

Not a real danger, so long as that risk is understood and accepted.

Mike