Best way to find an electrical "leak"?

[HDRW]

John,
Thanks for the advice...

Turn off all the power at the main switch as well as the MCB, and remove all other loads from the circuit. If you measure the resistance with a simple multimeter, you will find that the resistance between L and N is so high you can't measure it, if there is no load on the circuit.

OK, I was just going to turn off the MCB - is there a reason this isn't enough? When the CU was changed (2004, of course!) I put in a couple of sockets on the downstairs ring near it, one on each leg of the ring, so I can work from there rather than having to open up and delve into the CU, so I was thinking that just turning that circuit off would be enough. It's going to be tricky reaching and working on the sockets as it is, without doing it in the dark!

I can see that if the leak is a complex device (switch-mode power supply for example) rather than just a resistance then the multimeter won't read it properly, or at all. But if it is a resistance such as a leaky cable due to decaying insulation I'd have thought 82 ohms should show up nicely? I realise I'm pre-judging what the problem is, though...

if the unexplained usage is on a ring, try to identify and draw the route that the ring takes. Remove a socket on each of the clockwise and the anticlockwise legs of the ring, so you interrupt the ring and any sockets between the interruptions will have no power going to them. Put the cores safely into chockblock.

Unfortunately the route of the ring doesn't seem to be obvious - a couple of sockets are on spurs, chased into the wall behind cabinets, and without lifting floorboards (and the stuff on top of them!) it will be tricky to work out where it all runs. If I assume most sparks will just run the shortest route around the outside walls I can guess the order, but can I rely on that?

If the load now disappears, you know it is somewhere between the two interuptions.

Interrupt the remaining part of the ring to narrow down the place where the hidden load is. And again. Eventially you will find it, or at least identify a length of hidden cable within which the fault exists. You might end up isolating that section and replacing with new cable.

OK, I was going to do a "BInary chop" with what I think is the sequence - starting with dividing it in two, then successively divide the faulty half each time until only one part is left. I think this results in the fewest disconnections needed?

Is there a good way to replace a cable in-situ? Again, lifting floorboards will be a major undertaking. I know the existing cable isn't clipped (none of them are), so I suppose the obvious way is to pull it out with the new one attached and hope it doesn't get stuck on the way, and/or get disconnected. But would it be feasible to use rods with a ring on the end around the old cable, pushing it along like a curtain on a rail? At least if this goes wrong there's nothing to stop me trying again, or trying another method. Pulling through the old cable seems such a risky one-shot deal!

700W is as much as a small electric heater, so I will be surprised if you can't find it once you know where to look, because it will be hot, or possibly scorched and melted.

Yes this had occurred to me - despite my signature, there is no smoke, smell or any warm spot apparent. I suppose that it could be a length of cable that's leaking all along, so the warmth will be well spread and not noticeable - but it will be interesting to find what was happening regardless of the fault-fixing!

I wonder if I'll find that one of the cables feels warm when I go to disconnnect it from a socket?

Cheers,
Howard

 

[HDRW]

Chris,

Wouldn't have an electric towel rad would you?

QuIte right, I wouldn't, but thank you for asking!

 

[Spark123]

OK, I was going to do a "BInary chop" with what I think is the sequence - starting with dividing it in two, then successively divide the faulty half each time until only one part is left. I think this results in the fewest disconnections needed?

Yep, also known as the half split method. Wether or not the 82 ohms will show with your fluke is anyones guess, only one way to find out!

 

[JohnD]

your "binary chop" is easier on a radial circuit. On a ring, it is fed from both ends, so you have to isolate both legs.

If easier, you could isolate one leg at the first outlet from the CU (IMO it is easier and safer than isolating in the CU if you don't often do it - some CUs are real rats' nests, and the supply terminals are permanently live) and then you can test the rest as a radial.

My reason for suggesting turning off the main switch is that there might be some accidental connection between two circuits. This is more common on upstairs/downstairs lighting, when there are usually two-way switches and top and bottom of the stairs, and not infrequently a borrowed neutral.

 

[viewer]

Going back to the original post; just checking; may help before pulling apart the house.
The 700W
....this is per hour isn't it?
... does it vary as you watch? If so what are max/mins?
....is there a diurnal or other pattern?
I can't see how you could lose this amount of heat directly from cabling without a really obvious result - it's like having an electric iron on all the time.
You don't have an (faulty) active damp course do you?

 

[JohnD]

700W is a load. It is not time dependent.

A load of 700W will use 0.7kWh per hour.

 

[securespark]

I can't wait to see what he finds....

 

[bernardgreen]

You could trace the route of the 3 amp "leak" by measuring the voltage drop between sockets that the "leak" curret creates

Temporarily convert the ring into two radials by breaking at a socket or a single radial by removing Live and Neutral of one end in the consumer unit. Protect the ends with terminal blocks. Leave the CPC ( earth ) connected

1 amp over 1 metre of cable produces a voltage drop of approx 13 millivolts. The resistance of the wander lead can be ignored as it is very small compared the the meter's input resistance when measuring volts.

Restore power and measure the voltage between neutral at the consumer unit and the neutral at the first socket on the radial using a wander lead.

If you have made two radialsand if there is almost zero voltage between a socket's neutral and the CU neutral then the "leak" is not on that radial or it is very close to the CU

The resistance of the wander lead can be ignored as it is very small compared to the meter's input resistance when measuring volts and virtually no current is flowing in the wander lead.

Note the reading for that socket and then move to the next socket, the reading should be higher at each socket as the cable distance between it the CU increases untill you go past the "leak". The readings for sockets beyond the leak will be the same as there will be almost zero current flowing in the cable between them ( capacity between Live and Neutral in the cable will produce a minute current ).

At 3 amp the drop of approc 40 millivolts per metre may provide enough accuracy to calculate the length of cable from socket to the point where the "leak" current is leaving the cable.

 

[viewer]

700W is a load. It is not time dependent.

A load of 700W will use 0.7kWh per hour.

My point exactly (if badly phrased) - is the OP using a measuring device which shows 700W or a consumption of 0.7kWh? Most of the "energy measuring devices" show consumption and will be using a time frame so that after 1 hour it will show that 0.7kW has been used, but may not show that there was a varying load during that hour. However some allow instantaneous measurements which equate to current passing at that moment. So, my question was really which measurements were being reported. Did the OP watch the instantaneous readings or leave the machine attached for a period. If the former then was a changing value seen, and, if so was there a pattern to the change?

 

[ban-all-sheds]

Also, how accurate is the device at low currents?

 

[JohnD]

that was my immediate thought too, as they are often inaccurate at low loads, but he said

"I've timed the meter's flashing and confirmed that it really is being consumed, not some phantom in the monitor. "

 

[martinxxxxxx]

Get a "fuse finder"off of ebay and sell it again once you have sorted it - it will be a bit quicker that faffing about with all those sockets.
I have one that works on dead circuits or live circuits, and it is well useful for completing Insulation resistance testing.

You would be surprised that most people don't even know where all their sockets are! But normally the hidden ones have nothing plugged into them apart from antenna amplifiers, air fresheners, doorbell recievers or mouse deterrents devices. But none of them take the whooping .7KW you are looking for. Could it be going next door?

 

[JohnD]

how does the fuse finder work?

 

[martinxxxxxx]

Well, with mine you can put the reciever in-line with the supply and then follow to the load with a detector. Great for finding low impedances such as nails through cables, I can even use it inplace of the fuse, if the fuse blows every time the circuit is energised for instance

 

[RF Lighting]

Which one have you got? Mine isn't as fancy as that. It just plugs onto a live circuit and tells you which fuse is supplying it.