finding an earth break on a ring if in a JB?

[HappyHippyDad]

Morning all..

I have a bugger of a ring problem to solve on monday. All the following faults are on the one ring.

It has a N-E IR fault which I have almost found.
It has an intermittant tripping 32A MCB which looks like an overloaded circuit (one ring for whole house) and...
It has a broken CPC.

This house is absolutely jam packed with clutter.

I can find the broken CPC by doing the following:

1. Split CPC at any socket
2. Test at CU which side has broken CPC (wander lead or R1+R2)
3. Connect line and CPC of good side of ring in CU.
4. R1+R2 all sockets - those with continuity are ok and eliminated.
5. Reconnect at broken socket and break CPC at one of the suspect sockets.
6. Go to 2 above and repeat.

However, I have a feeling from spending a little time there that the fault will be in JB. The above method will find the fault if in a socket but not if its in an unreachable JB. How can I narrow it down to between 2 sockets? (i.e the ones either side of the faulty JB). I have to say I dont like the EFLI method, I dont find it accurate enough and it leaves you even more confused.

Is it just a case of slowly working out the layout of the ring and getting an idea of where the fault might be during all the testing or is there a better way?

 

[Deleted member 26818]

You narrow it down to the two sockets either side, then lift floorboards or dig out walls.

 

[cliffed]

Sounds so easy,these kinda jobs make you wanna finish earlier for Christmas.

 

[Dave OCD]

If you can find the link with missing CPC continuity and it's inaccessible without big upheaval HHD [especially with clutter everywhere which is a real pain] I'd consider abandoning the dodgy bit and converting to 2 x 20a radials at the CU if space permits, of course this isn't ideal for the kitchen load requirements though and if the break is in a hidden JB with a spur off you'd lose that point as well. I always seem to get jobs like this near Christmas too.

 

[HappyHippyDad]

I have to admit I have thought of changing to 2 radials, but I'd have to find the 2 sockets either side of the JB first! Also, like you say it could be feeding something. Plus if I've got as far as narrowing it down to 2 sockets then I like to fix the fault. But... how to narrow it to between 2 sockets?

However, if I did do this and one radial was much longer than the other AND this was the side that had the overload on it then its going to trip even more so not an option! Arghhh...

I don't want this job! I like fault finding but not in a house like this where you can hardly move. I even pricked myself on a needle last time (not drugs, just an elderly lady's sowing stuff literally everywhere).

 

[HappyHippyDad]

As a point of interest.

The adiabatic gives the following for the socket circuit:

I = 237A
t = 0.1
K = 115

therefore S = 0.65mm

The CPC cable is 3/029 which I think is approx 1.5mm.

I'm really not going to use that as an excuse if I cant find it, but I would be very interested in peoples thoughts? The adaibatic shows that the CPC needs to be at least 0.65mm which it is.

 

[Deleted member 26818]

So you’re saying that it doesn’t comply but is still safe for continued use?

 

[Lucien Nunes]

3/.029 is 1.3mm²

It sounds like the line conductor is continuous and not faulted to anything. I'd remove all six ends in the CU and link an L & CPC. Then I'd wander round the house with a 13A plug on the MFT, plugging into each socket in turn, writing the R1+R2 on a post-it note and sticking it on the socket along with which CU leg it connects to, e.g. 0.36Ω - A.

Then repeat with the other L & CPC linked at the CU. The fault is probably between the 'A' and 'B' with the highest resistance, which I would open. It should then be possible to prove that the L is continuous between them but the CPC isn't. Of course if the fault is in a socket it will be one of them.

If there are nearby sockets with similar resistance, or the two sockets either side of the break are not adjacent on what seems to be a straight cable run, variability of contact resistance may have obscured the exact order of connection from the CU. I'd open these and break the L to prove they are not actually the nearer to the break.

 

[Deleted account]

on a property like this I always add an extra £150 per day so I can get an extra body in to do the furniture and junk moving. That way I can get on with finding the fault.
Quite often customer doesn't want to pay this so I don't get job either way I'm happy.

Interesting thought on adiabatic, but i would be concerned with a floating lose cpc.

 

[Lucien Nunes]

Re the adiabatic suitability of one leg of 3/.029, do you mean to protect the existing ring? If so, it would fail on 543.2.9, which specifically requires the CPC to be in the form of a ring if the L & N are, except where metallic containment provides the CPC.

 

[Ian1981]

As a point of interest.

The adiabatic gives the following for the socket circuit:

I = 237A
t = 0.1
K = 115

therefore S = 0.65mm

The CPC cable is 3/029 which I think is approx 1.5mm.

I'm really not going to use that as an excuse if I cant find it, but I would be very interested in peoples thoughts? The adaibatic shows that the CPC needs to be at least 0.65mm which it is.

If it’s a mcb protecting the circuit then t is likely to be less than 0.1 seconds.
Let through energy of the mcb will need to be obtained to confirm required CSA of the cpc.
1.5 mm will be in most cases adequate for a 20 amp mcb

 

[HappyHippyDad]

Cheers everyone, well almost everyone

on a property like this I always add an extra £150 per day so I can get an extra body in to do the furniture and junk moving. That way I can get on with finding the fault.
Quite often customer doesn't want to pay this so I don't get job either way I'm happy.

Interesting thought on adiabatic, but i would be concerned with a floating lose cpc.

Paignton isn't too far Pete, see you at 8am monday morning and thanks for the offer

Re the adiabatic suitability of one leg of 3/.029, do you mean to protect the existing ring? If so, it would fail on 543.2.9, which specifically requires the CPC to be in the form of a ring if the L & N are, except where metallic containment provides the CPC.

Excellent find Lucien!
Yes, I did mean to protect the existing ring. Your reg shows that it would not meet the regulations but would you class it as safe seeing as though it meets the adiabatic. Under fault conditions 1.3mm would handle the fault current for the duration of the fault until the MCB trips.

 

[HappyHippyDad]

3/.029 is 1.3mm²

It sounds like the line conductor is continuous and not faulted to anything. I'd remove all six ends in the CU and link an L & CPC. Then I'd wander round the house with a 13A plug on the MFT, plugging into each socket in turn, writing the R1+R2 on a post-it note and sticking it on the socket along with which CU leg it connects to, e.g. 0.36Ω - A.

Then repeat with the other L & CPC linked at the CU. The fault is probably between the 'A' and 'B' with the highest resistance, which I would open. It should then be possible to prove that the L is continuous between them but the CPC isn't. Of course if the fault is in a socket it will be one of them.

If there are nearby sockets with similar resistance, or the two sockets either side of the break are not adjacent on what seems to be a straight cable run, variability of contact resistance may have obscured the exact order of connection from the CU. I'd open these and break the L to prove they are not actually the nearer to the break.

Sounds good Lucien, but there are almost as many spurs as there are sockets on the ring, the spurs will give a higher R1+R2 so I think going by those results I could be led on a bit of a wild goose chase?

 

[Lucien Nunes]

When the ring is split at the CU, the resistance is directly proportional to cable length from CU to point (allowing for minor discrepancies where 2.5mm² additions have been made to a 7/.029 RFC). Sure, some spurs might be higher than the sockets either side of the break, but only if they had a greater overall length. If the 'A' leg was spurred at a point 3m short of the 'last' socket before the break, the spur cable itself would have to be greater than 3m long to override the result. It would be unlikely to run past the faulty point or accessory, or it would have been spurred there instead, so it is likely to be further away from the 'last' socket on the 'B' leg. A combination of visual and resistance clues is likely to give a better picture than either one alone.

My point was really to home in on the affected area before taking any sockets off the wall.

 

[Octopus]

Why not give a high quote and see if the customer wants to go for it ....

sounds like you don't want it!