ends to ends

[Geoffsd]

However, say you test a circuit, on a 32A CB, it comes in as 1.15 ohms. That would pass the Zs test (80% of 1.44).
Then you delve deeper, and find it has been run in singles, and the Line conductors are 6mm, but the CPC is 1mm.
Simple testing shows this to pass, Inspection would be likely to fail it, as the CPC would not be able to withstand the thermal requirements - it would fail when calculated using the adiabatic.
Basically, the Zs is good, but if there was a fault current, the CPC would melt before the fuse/CB tripped.

Would it?

 

[alanl]

No idea, just an example that doing a Zs test does not always satisfy the thermal requirements of the CPC.

 

[7029 dave]

Would it?

Doubt it.

 

[alanl]

Sigh.
The point is, Zs is not always indicative of the conformity of the CPC size.

Here's an example.
6mm Line, 1.5mm CPC, fed from a 32A CB with a 30 metre length.
Ze assumed to be 0.35,
Zs = Ze + (R1 +R2)

R1 = 3.08m ohm/m x 30, R2 = 12.1 mohm/m
R1 + R2 = 0.45 ohms, + Ze of 0.35
gives a Zs of 0.90, - well under the limit of 1.44 (non-corrected).

Anyone who hadnt inspected it properly, and just did a live Zs test, would assume that to be a good reading, and tick it off as a pass.

Now we have been called back as there has been a short circuit at the switch, and the cable has burnt through.

We go back, see the 1.5mm CPC, then do the PEFC test, and find there is a 500A earth fault current.
Looking at the tables, we find the CB will trip in 0.1 seconds.
So why did it burn?
Do the adiabatic, and we see why:
Fault current 500A, disconnection time 0.1s so doing the calculation 500x500 x0.1 square root answer, divide by 115, gives an answer of 1.37 sq mm.
Thus showing the CPC is too small for the circuit thermally , but as above, Zs would pass easily.

OK, the CPC would probably not melt, but, it is going to get hotter than recommended, so should be avoided, hence why it is not good enough to just rely on the tested Zs result to pass a circuit as acceptable.

 

[Geoffsd]

We go back, see the 1.5mm CPC, then do the PEFC test, and find there is a 500A earth fault current.
Looking at the tables, we find the CB will trip in 0.1 seconds.
So why did it burn?
Do the adiabatic, and we see why:
Fault current 500A, disconnection time 0.1s so doing the calculation 500x500 x0.1 square root answer, divide by 115, gives an answer of 1.37 sq mm.
Thus showing the CPC is too small for the circuit thermally , but as above, Zs would pass easily.

Assuming you meant a 1 sq.mm. cpc as you originally said.

However, the 32A MCB will trip in 0.1s @ 160A so will be quicker at 500A.


Doing the calculation for 160A gives 0.44 sq.mm.

Without more complicated calculations this gives an indication that the cpc will survive.