Discuss How to Find the location of a fault - Puzzler in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

C

Colonel Hathi

Hi
I came across this question on the "learning lounge".

I confess I have no idea how to tackle it:confused:
...... any ideas/explanations very welcome


"A radial circuit wired in 2.5mm2 PVC singles fails an insulation resistance test between live conductors. A continuity test reveals a reading of 0.67 Ohm, how far along the length of the circuit is the fault?"

Thanks

CH
 
Right, off the top of my head with no books to hand, you will need to know the resistance per metre of 2.5mm cable. Assuming the continuity test is carried out between Line and Neutral and the IR fault is a dead short (has to be for the question to make sense) you have a R1+RN value of 0.67ohm. Work out the length of 2.5mm cable that gives you a resistance of 0.67ohm, divide it by 2 and that is your distance from test point to fault.
 
Well the resistance per meter of a 2.5mm single is 7.41milliohm per meter (OSG page 166).

As all conductors are the same size at 2.5mm technically it is the same calculation you would use for R1+R2 of a circuit, only this is R1+RN as the question states "live conductors".

So the R1+RN for 1 meter of 2.5+2.5 is 7.41milliohm x 2 = 14.82milliohm per meter.

Knowing this you can now work out at what length the "short" is at.

So with the reading at 0.67ohms and the resistance for 1 meter at 0.014ohms

0.67/0.014 = 47.8M/2 = 23.9M.
 
Last edited by a moderator:
was going to say 'quite a distance from point of test' and offer no technical reasoning.
ive done a fair bit of work on european installations using triple insulated cables, their t+e conductors are of equal csa and they only use radial socket circuits.
i was able to tell by the actual reading how far down the circuit the fault was likely to occur
 
Last edited:
Thanks for the answers chaps.

Having thought it through, I think the answer is ...
.67ohms divided by 14.82=0.0452 times 1000 = 45.2metres

I doubled up the joint resistances of the 2.5 mm R1 and RN at 7.41 m ohms each per metre.

I think that you can either double up the resistance (to 14.82) ...
or take it as 7.41 and then divide by 2 for the distance - but not both:)

I think this is correct ?

Thanks again for straightening this out:D

CJ
 
I'm confused now......the 0.67 reading is the loop reading i.e out on R1, through the short and back on RN.

Obviously the short is at the half way point of the loop i.e 0.335.

Thats why I divided my result of 47.8M to give the midpoint of the loop, i.e the point of fault.
 
Last edited by a moderator:
All very useful if you can actually see the run of cable. But absolutely useless if the singles go into a conduit and then into a concrete slab via a conduit. And there are tees. And more than one fault!
 
Aha!

I see what you are saying Lenny:D

I am still thinking though:D

The actual question came up as part of a random online test - so I don't actually have the "correct" answer myself.

I take other posters' points about the practical application onboard though.

CH
 
I'm confused now......the 0.67 reading is the loop reading i.e out on R1, through the short and back on RN.

Obviously the short is at the half way point of the loop i.e 0.335.

Thats why I divided my result of 47.8M to give the midpoint of the loop, i.e the point of fault.


Now you've got me at it ,
 
Hi
I came across the question again on the exam simulator.:)
The answer given is 45m.
I assume they have just rounded down from 45.2.

Thanks for the input.

CH
 
Well the resistance per meter of a 2.5mm single is 7.41milliohm per meter (OSG page 166).

As all conductors are the same size at 2.5mm technically it is the same calculation you would use for R1+R2 of a circuit, only this is R1+RN as the question states "live conductors".

So the R1+RN for 1 meter of 2.5+2.5 is 7.41milliohm x 2 = 14.82milliohm per meter.

Knowing this you can now work out at what length the "short" is at.

So with the reading at 0.67ohms and the resistance for 1 meter at 0.014ohms

0.67/0.014 = 47.8M/2 = 23.9M.


Not if it is T & E it's not.;)
 
Thats interesting.

I was going to ask whether anybody actually uses this in fault finding.

Thanks

CH

Yep, I've used it just the once doing initial verification on a nut house wing. Was doing my tests at the board, got 0 on insulation resistance, put it to continuity and got an extremely low reading. Guessed that it was a dead short as a continuity will only pick up a reading when conductors are touching, took the reading, looked up the table for R1+R2 for 2.5 t+e, and divided my reading by it. About 4 metres it worked out as, so measured about 4 metres from the MCB and lo and behold there amongst about 20 other cables in basket was a cable that had been severely twisted to the point that internally it was short to earth. It seemed like a common sense approach at the time, but I still was chuffed when the method worked perfectly!! Now if only the same worked for insulation resistance reading, life would be so much simpler!!
 
This reminds me of many years ago whilst looking for a fault on a Co-axial underground cable.
We used an ocilloscope thing that put out a very narrow pulse and measured the time it took to reflect from the fault. So having got 35 metres or so we set out with the digger to have a look.
We found no fault at all within about 15 metres both sides of the distance. So we checked again with the kit and got the same result. ( 35 metres ). Then we lifted the examination cover and found 20 metres of cable coiled up in the hole..:eek::eek::eek:
 

Reply to How to Find the location of a fault - Puzzler in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

Similar Threads

- I want to know what these -unknown- resistors Wattage are !? First of all, they are scrapped. Then, we really dont know in what part of the...
Replies
25
Views
2K
I am not a qualified electrician, I was employed as an electrician after my MoD electronics apprenticeship but that was 30 odd years ago. So I...
Replies
9
Views
3K
Hi, Can anyone help please. I am doing an assignment for college level 3. I am trying too understand how to find a fault on a ring main circuit. I...
Replies
14
Views
36K
Hi guys, Just wanted to get your advice on answering a 2391 question. If i was to get a question "Describe how to carry out an IR test on a...
Replies
34
Views
4K
wayn3
W
W
Hi all, would really appreciate some feedback to my mid life crisis, lol. i currently work as a technician on 11kV and Lv main circuits. My job...
Replies
3
Views
1K
lauriehurman
L

OFFICIAL SPONSORS

Electrical Goods - Electrical Tools - Brand Names Electrician Courses Green Electrical Goods PCB Way Electrical Goods - Electrical Tools - Brand Names Pushfit Wire Connectors Electric Underfloor Heating Electrician Courses
These Official Forum Sponsors May Provide Discounts to Regular Forum Members - If you would like to sponsor us then CLICK HERE and post a thread with who you are, and we'll send you some stats etc
This website was designed, optimised and is hosted by Untold Media. Operating under the name Untold Media since 2001.
Back
Top
AdBlock Detected

We get it, advertisements are annoying!

Sure, ad-blocking software does a great job at blocking ads, but it also blocks useful features of our website. For the best site experience please disable your AdBlocker.

I've Disabled AdBlock