ends to ends

[kellogs64]

hi all. wats the crack with high earth reading. i get r1 40ohms. rn 40ohms. r2 is 80ohms. earth should not be higher than 1,67 times highest reading plus 0.05. is this correct. any imput would be helpfull does it make it a c2 code?

 

[kellogs64]

hi all. wats the crack with high earth reading. i get r1 40ohms. rn 40ohms. r2 is 80ohms. earth should not be higher than 1,67 times highest reading plus 0.05. is this correct. any imput would be helpfull does it make it a c2 code?

0.40,0.40 and 0.80

 

[telectrix]

with those readings the rfc must be a hell of a big one. about half a mile.

 

[telectrix]

just seen your edit.

either

1. high resistance connection on cpc.
2. cpc is 1.0mm instead of 1.5mm as found on some older installs.

 

[telectrix]

and it's no code if the adiabatic determines that the circuit will disconnect within the 0.4 secs.

 

[polo1]

Unusually restrained Tel:smilewinkgrin:

 

[kellogs64]

can anyone help me with adiabatic equation break it down for me cheers

 

[telectrix]

wait till i've had a few more beers. only on my 4th.

 

[Simonslimline]

can anyone help me with adiabatic equation break it down for me cheers

Look in your BGB!

 

[telectrix]

bgb. 543.1 says it all.

 

[Guest77]

hi all. wats the crack with high earth reading. i get r1 40ohms. rn 40ohms. r2 is 80ohms. earth should not be higher than 1,67 times highest reading plus 0.05. is this correct. any imput would be helpfull does it make it a c2 code?

Gasp, just reported you for foul and abusive language

 

[Richard Burns]

can anyone help me with adiabatic equation break it down for me cheers

Adiabatic calculation
basics




k for a single copper conductor is 143 (usually the case for MEC and MPB)
k for cpc in a cable is 115 (usual for final circuits)
S is minimum csa of the cpc
I is the fault current (PEFC) (as limited by the current limiting effects of the PD)
t is the disconnection time with fault current I

If you have the I[SUP]2 [/SUP]t for the device use that.
If not then use PEFC to get I and time / current graphs for the device to get t.
You can use minimum disconnection time and fault current at that time but this gives higher values.

 

[Baker1988]

and it's no code if the adiabatic determines that the circuit will disconnect within the 0.4 secs.

I may be getting things totally mixed up or just reading this wrong but I thought a adiabatic was to determin the minimum size of a conductor like a cpc or bonding and to make sure it is of adequate size, if you wanted to make sure the circuit would disconnect within 0.4 seconds wouldnt you use the Zs of the circuit? Sorry if I am totally off the mark as It is a bit late haha

 

[alanl]

but I thought a adiabatic was to determin the minimum size of a conductor like a cpc or bonding and to make sure it is of adequate size, if you wanted to make sure the circuit would disconnect within 0.4 seconds wouldnt you use the Zs of the circuit?

Not really. For standard circuits, using standard sized conductors, like the typical house in the UK, then the Zs will usually give the correct result.
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.

 

[telectrix]

sorry. i worded that badly. i meant to say that if the adiabatic showed conductor to be adequately size for the duration of the fault.. i blame the beer. lol.

 

[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.