Indeed, an all copper round loop 0f  100m of 10mm2 running hot will be about 200 milliohms.  One might hope the earth core is colder, but that does not make much difference.

The inductance per unit length of the internal CPC is largely cancelled  compared to a wire in free space we can estimate this from the dielectric constant of the insulation and the capacitance per unit length of the cores, which is something the better multi-meters can measure directly.

If we know the insulation is polyethylene  say, (and XPLE is just polyethylene abused a bit  by radiation or nasty chemicals to toughen it up ) the dielectric constant  is about 2.3 - so the capacitance between the wires is about 2.3 times what it would be if the wires were in the same layout in free space. This means the wave velocity is c*= c /√2.3 where c is the speed of EM waves (light, radio waves etc ) in vaccuum. and c* is the velocity in the dielectric. So c~ 300m/microsecond and  c* is ~ 200m/microsecond..

If you know the wave velocity and the capacitance, the inductance follows as c*= 1/ √(L.C)  where L and C are the inductance and capacitance  per unit length respectively.

So if you measure 1000pF core to core for a 10m length, that is 100pF per metre, and we can go to L= 1 / (c*)² *C
1/ ((200,000 000) ²  *100 pico) = 0.00000025 H or in units we can chew more usefully 250nH per metre..

A CPC on the outside does a lot worse than this, and will look like more like 1000 nH per metre.

However, even at that level, at 50Hz, a nano Henry is ~ 314 nano ohms reactive at 50 Hz, so we re looking at 1/3 of a milliohms  per metre of inductive reactance for the loop of live and the external CPC - the simple resistance still dominates with an external CPC but the inductance is not totally negligible - with the internal core, it pretty much is.

Mike.

Thanks for all the help.,thought I could use a 40A bs88 fuse because of the higher zs it allows on a submain and might be a cheaper option than a type S rcd.

                                                    Regards,

                                                        Hz

Maybe you can, Is the Zs really as high as 0.8 ohms at the origin ? - if you can get a PSCC of a couple of hundred A when cable is hot and supply is low you can make the 5 seconds needed for sub-main protection. Achieving the half a heartbeat of  0,4 seconds is not so likely, needing more like 300-400 A.

Fuse curves for BS88 part C

Mike

hertzal123 said:

Thanks for all the help.,thought I could use a 40A bs88 fuse because of the higher zs it allows on a submain and might be a cheaper option than a type S rcd.

You may well be able to "get away" with a BS 88-3 fuse at 45 A ... without using the armour

(but if you use that you're well in).

Disconnection time of sub-main is 5 s on TN-S.

Table in Fig 3A1 says we need 220 A for a 45 A cartridge fuse.

Ze of 0.8 ohms + (R1+R2) of 0.2196 = 1.0196 ohms.

230 V / 1.0196 ohms = 225 A.

(so, if you parallel the armour, you're well above that, but not at the 380 A required for 0.4 s required for a final circuit).

Check adiabatic for the copper means S > 4.4 sq mm

So, again, if you parallel the armour with the copper cpc (bond at both ends), you'll be well in for ADS and adiabatic for a sub-main. If necessary, you could use RCBOs for final circuits if ADS were considered an issue (or you didn't want to do more calculations).

The down-side is that you will have a calculated 8.8 V volt-drop at 40 A down the sub-main ... which means lighting circuits in the outbuilding can't comply with (i) of Table 4Ab. Not knowing your design currents and arrangement of final circuits, I don't know if that's a problem for you.

hertzal123 said:

Thanks for all the help.,thought I could use a 40A bs88 fuse because of the higher zs it allows on a submain and might be a cheaper option than a type S rcd.

You may well be able to "get away" with a BS 88-3 fuse at 45 A ... without using the armour

(but if you use that you're well in).

Disconnection time of sub-main is 5 s on TN-S.

Table in Fig 3A1 says we need 220 A for a 45 A cartridge fuse.

Ze of 0.8 ohms + (R1+R2) of 0.2196 = 1.0196 ohms.

230 V / 1.0196 ohms = 225 A.

(so, if you parallel the armour, you're well above that, but not at the 380 A required for 0.4 s required for a final circuit).

Check adiabatic for the copper means S > 4.4 sq mm

So, again, if you parallel the armour with the copper cpc (bond at both ends), you'll be well in for ADS and adiabatic for a sub-main. If necessary, you could use RCBOs for final circuits if ADS were considered an issue (or you didn't want to do more calculations).

The down-side is that you will have a calculated 8.8 V volt-drop at 40 A down the sub-main ... which means lighting circuits in the outbuilding can't comply with (i) of Table 4Ab. Not knowing your design currents and arrangement of final circuits, I don't know if that's a problem for you.