How to Find the location of a fault - Puzzler

[Colonel Hathi]

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

I confess I have no idea how to tackle it
...... 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

 

[JulianC]

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.

 

[Guest123]

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.

 

[Colonel Hathi]

Thanks
That makes perfect sense now that you point it out.

Seems a very useful method for locating faults.

CH

 

[chris gray]

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

 

[Colonel Hathi]

Thats interesting.

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

Thanks

CH

 

[Colonel Hathi]

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

CJ

 

[bugsy]

And there's me thinking I'd lost the plot,

 

[Guest123]

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.

 

[simonspark]

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!

 

[Colonel Hathi]

Aha!

I see what you are saying Lenny

I am still thinking though

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

 

[bugsy]

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 ,

 

[Guest123]

This is all based on assumption, dont read too much into it matey.

 

[Colonel Hathi]

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

 

[ian.settle1]

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.