Thanks all.

Will check the referenced regs when I get back to my Big Brown Book.

AJJewsbury Andy - surely if trying to figure out how much fault current would travel down the SWA vs a parallel/additional CPC then it's 'simply' a matter of using the resistances of each to figure out the current split - although the resistance of the SWA would be affected by how good the gland terminations are, and would possibly be more susceptible to corrosion - especially outdoors.

Jason.

surely if trying to figure out how much fault current would travel down the SWA vs a parallel/additional CPC then it's 'simply' a matter of using the resistances of each to figure out the current split.

Quite, except that the word 'magnetism' seems to strike fear into the hearts of those classically trained in electricity only, who presumably slept through the EM part of the lecture and woke up near the end, when all the things that go wrong were being discussed leading to irrational worries about eddy currents, additional inductance and confusion about if we are cutting grooves into gland plates to interrupt lines of magnetic field or eddy currents.
Now wires inside tubular magnetic cores do indeed experience substantially higher inductance, as the magnetic field around the wire is augmented by the elemental magnets in the tube wall lining up in rings nose to tail spinning around the wire and reversing 50 times a second. So steel conduits with outbound current up one tube and return down another are a very bad idea, while if the tubes were plastic or indeed any non magnetic metal the inductane would be largely unaffected.

Worse, the sort of steel used in conduits does not have a particular magnetic or electrical spec, so  the heating associated with that regular magnetising and demagnetising, as well as any eddy currents flowing parallel to the currents, are not easily quantified, except as 'here be dragons, avoid them' Single core wire armour is always aluminium or copper, never iron or steel,  just in case this is a problem.

But the steel wires of an SWA do not really form a continuous cylinder enclosing the wires  - really there are air gaps, slots  and bunching and the magnetic effect is more like split tube with a distributed 10% gap, rather than welded conduit.  As such the extra impedance in the fault loop for an external CPC versus one up the middle is only a small fraction of the case if closed rings or short tubes of steel had been threaded on the wire, and not that significant, at least at 50Hz,  and for a CPC we are not really pushing the current rating to its limit..

Oh and at really high currents once all the magnetic domains (elemental magnets in the iron or steel) have aligned, and there are no more to move, the inductor saturates anyway, and the in effect the inductance is falls to its free air value.

So 'here be dragons, but really not very big ones' The  final Zs should ideally be verified at AC not as a an R1/R2 at DC, or at least expect the AC case to have a lower PSSC, and allow slack in the ADS design for that.

Mike

jbrameld said:

'simply' a matter of using the resistances of each to figure out the current split

As mapj1 says, it's not a simple resistive split, but rather an 'impedance split'.

Just a reminder that, whilst we generally ignore impedances for conductors with csa up to 16 mm², with the exception of 2c 1.5 mm² SWA, the csa of the armour is >16 mm² and reactance is important in the calculations involving the armour (used as cpc or in parallel with a copper conductor as cpc) for 50 Hz mains circuits.

Of course if we think that 16mm2 limit is related to current handling capacity,  or more accurately heating while handing fault currents, then we should also allow for the fact that the resistance, and so the heating, of the steels in SWA can be that much higher. 

The resistivity of  steel typically varies with carbon content from 100 to 1000 x 10-9. while drawn copper has a much lower resistance than even the best steel -  17.2 x10-9 ohm.metre at 20C.

The resistivity of something is just the resistance between opposite faces of a 1m cube of it - but a 1m length of 1m2 cross section is a pretty inflexible sample converting m2 to mm2 and turning the nano-ohns to milli-ohms is more practical !

According to JPs authoritative text on the topic, IEE GN8 suggests using a factor of 8 for the ratio of the resistivity of copper to steel - though  this may assume more conductive steel is  used in SWA than for average construction steel.

In an case to get the equivalent earthing impedance to 16mm2 of copper needs  well over 200mm2 of steel. More dragons...

Mike

Definitely resistivity is important if we are looking at "copper equivalent".

However, I'm talking about the approx 5 % to 10 % difference in (Z₁+Z₂) over (R₁+R₂) on the "unsafe side" of max Z_s.

SWA is one case where actual Zs measurement would be preferable than using basic resistance measurements, as the measured resistances would have to be adjusted for temperature, and then put back through the PD CLC/TR 50480 formulas, to render a loop impedance.

Definitely resistivity is important if we are looking at "copper equivalent".

However, I'm talking about the approx 5 % to 10 % difference in (Z₁+Z₂) over (R₁+R₂) on the "unsafe side" of max Z_s.

SWA is one case where actual Zs measurement would be preferable than using basic resistance measurements, as the measured resistances would have to be adjusted for temperature, and then put back through the PD CLC/TR 50480 formulas, to render a loop impedance.