Fault in SWA cable

[JohnW2]

the distance of a fault of whatever resistance can be determined from both ends in a simple swa run

It can, provided that (as you said before) one uses tables (or measurement of a known intact conductor and armour) to determine how much of the measured resistance is attributable to the conductors and armour. To do as EFLI suggested, and simply take measurements from both ends (without injection of the additional information) will not work unless the fault resistance is zero.

Kind Regards, John.

 

[EFLImpudence]

You don't need the tabulated resistance. If the measurement was 3Ω from one end and 2Ω from the other then the fault would be 3/5ths of the distance from the first end.

I think you're assuming a fault of zero resistance. As 17thman implied, things get more complicated when the fault has appreciable resistance. Without any other information, what would you say about the location of the fault if your measurements were, say 10,002Ω from one end and 10,003Ω from the other end (assuming you had a 5-digit DVM - and therein lies another potential problem)? ... 'your method' would presumably say that the fault was virtually in the middle - yet, if the fault had a resistance of 10,000Ω, it would actually again (as in your example), be 3/5ths from one end.

Kind Regards, John.

I did say in my first post on this subject "if there was direct contact between the conductors".

However in your scenario it would be necessary to measure the resistance of the two conductors concerned from end to end and subtract from the sum of both loop measurements and divide by two - this would be the resistance of the fault.
The remaining resistance would be the measurements as if the fault were zero Ω and proportional to the distance.

 

[JohnW2]

I did say in my first post on this subject "if there was direct contact between the conductors".

You did indeed ... but you then responded to 17thman's post suggesting how it could be done in the presence of a 'fault of unknown resistance' with a method that would only work if the fault resistance were zero.

IHowever in your scenario it would be necessary to measure the resistance of the two conductors concerned from end to end and subtract from the sum of both loop measurements and divide by two - this would be the resistance of the fault. The remaining resistance would be the measurements as if the fault were zero Ω and proportional to the distance.

Exactly. As I've just written in another message, one either has to measure the resistance of the conductors (assuming them to be intact) or calculate from tabulated resistivity, and then use that to facilitate the calculation. There may be some problems with doing it by measurement (as you describe) if the coinductor has been seriously damaged in the vicinity of the fault. Also, as I mentioned before, there could be a problem of meter precision if the fault resistance was very high (I don't know if the 0.01M Ω figure quoted was simply the lowest the IR meter could display, or whether it really meant 10,000&#937.

Have a good weekend!

Kind Regards, John

 

[Paul_C]

This is where techniques such as the Varley bridge I mentioned last night become useful. The Varley method arranges the connections in such a way that the resistances of the conductor from the point of the fault to each end become "arms" of the bridge, but the resistance of the fault itself to earth is not part of the bridge.

Edit: Here's an old article which describes the technique, along with others:
http://www.alcatel-lucent.com/bstj/vol10-1931/articles/bstj10-3-382.pdf

The basic Varley method is shown on the sixth page (page 387 of the original journal).

 

[JohnW2]

This is where techniques such as the Varley bridge I mentioned last night become useful. The Varley method arranges the connections in such a way that the resistances of the conductor from the point of the fault to each end become "arms" of the bridge, but the resistance of the fault itself to earth is not part of the bridge.

Of course,as I hinted last night, none of these resistance-based mathods will give a correct answer if (as is far from impossible) damage resulting in a fault brings about increased resistance in the conductor itself, unless that increased resistance is exactly equal on both sides of the point at which the main fault (connection to another conductor) occurs.

Kind Regards, John.

 

[EFLImpudence]

Even if the fault resistance were unequal on each side this could be determined by measuring from one end of one wire to the other end of the second wire and vice versa.
With all the measurements it should be possible to obtain a rough estimate as to where the fault was.

After all, had the IR test been done from the other end as well the reading may have been 0.00 or 0.02 which may have given you a very rough idea that it was to one end or the other - not conclusive, I know.

It just seemed a drastic first step to dig up and cut the cable.

I suppose the moral is - Check the IR before you bury the cable and stay on friendly terms with the neighbour.
I know from personal experience that sometimes the latter is not an option.



Here is an exercise in logic and random theory.

If you decide to find a fault by cutting the cable, (or just a fault in a Ring Final) is the centre the best place to make the first cut?
Would it be better to first cut at a quarter or a third the distance. You may be lucky and find the fault on the short side. If not - have you lost anything?

Which method would result in the fewest number of cuts needed to find the fault?

 

[sparkwright]

If you suspect sabotage you could make the cut where the neighbour would have had easiest access.

Who needs neighbours like that?

 

[JohnW2]

Even if the fault resistance were unequal on each side this could be determined by measuring from one end of one wire to the other end of the second wire and vice versa. With all the measurements it should be possible to obtain a rough estimate as to where the fault was.

You're probably right, but I need time to think and scribble, since it goes against my initial (not very carefully thought-through) intuition!

Here is an exercise in logic and random theory. If you decide to find a fault by cutting the cable, (or just a fault in a Ring Final) is the centre the best place to make the first cut?
Would it be better to first cut at a quarter or a third the distance. You may be lucky and find the fault on the short side. If not - have you lost anything? Which method would result in the fewest number of cuts needed to find the fault?

I remember, an large number of moons ago, spending many days attempting to derive a comprehsive general solution to this problem, but I can't remember the answer. Intuition 'obviously' suggests that, in the absence of any other information, 'halving' has got to be the best general solution. However, as I think you're implying, one has to bring the concept of 'utility' into it - since, if one is a gambler, starting away from the centre will give you a quicker/cheaper answer if your initail guess was right (but the opposite effect if your initial guess was wrong). The answer also depends upon how precise an answer one is looking for (i.e. what length of cable you wish to 'narrow down' to).

In practice, this may be further modulated by practicalities. Excavating and making additional cuts costs time/money. Repairing cuts costs appreciable time/money, and there may also well be a maximum number of cuts/repairs that one is prepared to contemplate before deciding to replace the whole cable. When one brings those sort of considerations into the equation (which I was required to do for my 'comprehensive general solution') I think that starting somewhere other than the middle may sometimes make sense - since it can, on average, speed (and/or reduce the cost of) the process of getting to either the 'success' point (if lucky) or 'give up' point (if unlucky).

I'll try to locate (don't hold you breath) my efforts from all those decades ago!

Kind Regards, John.

 

[JohnW2]

If you suspect sabotage you could make the cut where the neighbour would have had easiest access.

Indeed. It hardly needs saying but, whether one suspects accidental or deliberate damage, the first stage is to consider (and investigate) where the damage is most likely to occurred - and that could often totally change the mathematical approach (being discussed with EFLI) which is based on 'no other information being available'.

Indeed, in the case of suspected deliberate damage with a known suspect, torture offers one possibly means of locating the fault

Kind Regards, John.

 

[Jackthebiscuit]

Yes the machine is a TDR (Time Domain Reflectometer) we've been using them for years.

http://uk.rs-online.com/web/c/test-measurement/electrical-test/tdr-cable-fault-locators/

However they are not cheap (the one I use cost £3500 some years ago but there are far less expensive versions) Even if you get one you will need specialist training and will need to use it on a regular basis to get "good" at interpreting the results and justify the cost.
Though, we have a much advanced HV version that could be used on LV which will give you a distance to the fault - trouble is they are in the £13,000 range.

As for the cut and test method, it works and has been done for years.

Or,, make you own Wheatstone Bridge,,? Dearest component is a bread board to mount the bits on

 

[JohnW2]

Or,, make you own Wheatstone Bridge,,? Dearest component is a bread board to mount the bits on

Indeed, and that's effectively what we've been discusing here. However, it has it's limitations - primarily that it can only be used to locate faults which are connections between conductors. To detect a broken conductor, you either need a shovel and wirecutters or something like TDR technology.

I do recall some pre-TDR (and much cheaper than TDR) methods for attempting to locate broken conductors - probably the most 'obvious' being to measure capacitance between conductors. Given an estimate of the approximate expected capacitance/metre, that approach can (sometimes!) give a reasonable estimate of the distance to a break. In the case of reasonably accessible cables, one can also send a signal down the conductor and use non-contact detectors to determine where there is a sudden change in signal strength.

Kind Regards, John.

 

[Paul_C]

probably the most 'obvious' being to measure capacitance between conductors. Given an estimate of the approximate expected capacitance/metre, that approach can (sometimes!) give a reasonable estimate of the distance to a break. In the case of reasonably accessible cables, one can also send a signal down the conductor and use non-contact detectors to determine where there is a sudden change in signal strength.

Again, all methods which have been employed for years with telecommunication cables. The Ohmmeter 18 provided an option for injecting an a.f. signal and then using a capacitance bridge arrangement to determine the location of an open - Earlier versions relied upon an earphone as the null indicator, while later versions incorporated a transistorized amplifier to drive the same center-zero meter which was used for d.c. Wheatstone/Varley measurements.

Injecting an audio signal from an oscillator and then using probes and induction coils along the route would then typically be used to pinpoint the location of the break after getting an approximate location from the bridge measurement.

 

[JohnW2]

Here is an exercise in logic and random theory. If you decide to find a fault by cutting the cable, (or just a fault in a Ring Final) is the centre the best place to make the first cut?
Would it be better to first cut at a quarter or a third the distance. You may be lucky and find the fault on the short side. If not - have you lost anything? Which method would result in the fewest number of cuts needed to find the fault?

OK, it's all starting to come back to me now! Back in the 70's, of course, I had to tackle the problem on the basis of theory and manual analysis. Today, I have the luxury of being able to do simulations (something I would have given a king’s ransom for back then!) ....

I have very quickly undertaken some simulations (and therefore can't yet guarantee that I've got it all right!). One of the things it reminds me (as previously mentioned) is how important is the degree of precision one seeks (i.e. how short a length of cable one wants to ‘narrow down to’).

Anyway, I’ve simulated 100,000 instances of what I think is a realistic situation - a fault occurring at a random point along the length of a 100m cable and, for each of those 100,000 instances, have determined how many cuts one would have to make in order to locate the segment of cable within which the fault was no more than 2m from either end. I have repeated this exercise for a range of positions for the first cut, ranging from 0.1 to 0.9 times the length of the cable from one end (given the symmetry of the situation, I needn't have gone beyond 0.5, but I didn't twig that until after I'd done it ). The results are summarised here:

Untitled

• JohnW2
• 26 Nov 2011
• 1

In keeping with my vague recollections from all those decades ago, the answer is ‘complicated’, but what you implied seems to be qualitatively not far off, in that (probably contrary to most people’s intuition) it does not make ‘all that much difference’ where one makes one’s first cut. As you’ll see, in my simulation, the average number of cuts required did get progressively less as the starting point got closer to the middle, but the difference was only between an average of 5.72 cuts if one started at the “50% point” (i.e.middle) and an average of 6.26 cuts if one started at the “10% point” (i.e. near one end). Correspondingly, if one starts close to the middle, one never needs more than 6 cuts, but the need for a 7th becomes progressively more likely as the starting point moves away from the mid-point – getting up to a 52% probability that the 7th will be needed if one starts at the 10% point. It appears that (at least with the parameters I’m using) one would never need more than 7 cuts if one’s first cut is anywhere between the 10% and 50% points.

In the other direction, if one starts near the middle, one will never get the answer (with required precision) with less than 5 cuts, whereas one stands a chance of getting an answer with only 3 or 4 cuts if one’s initial cut is only 10% from one end – so it’s really ‘horses for courses’, depending upon one’s attitude to gambling!

I am ‘surprised’ by my simulations appearing to indicate that that, if one starts 10% from one end, one will never need 5 cuts – one may need 3, 4, 6 or 7 cuts, but never 5! Despite searching, I haven’t yet found any flaw in my simulation to explain that, so maybe it’s just a quirk of the numbers – I’ll be investigating that (and re-checking my simulation code!) further!

Hope that's of some interest, and I hope I haven't got it all wrong (wouldn't be the first time!).

Kind Regards, John

 

[EFLImpudence]

Wow - that deserves several thanks.

Very interesting.

 

[JohnW2]

Wow - that deserves several thanks. Very interesting.

You're very welcome - I just hope it's right (and I'm not yet convinced, because it isn't quite what I was expecting)!

Needless to say, I did it more for myself than for you - remembering all that slaving I did over it 'way back' (it was probably 2 or 3 weeks, rather than just 'days'), you made me realise that I had never really 're-visited' the question since much better 'tools' were available to us!

I wonder what you were 'expecting'- that it made absolutely difference where one started, perhaps? As I realised as I was doing this exercise, it's really all about how quickly one can get down to the segment size of interest (4m or less in my simulation) by successive halving - and that is going to be different in the two different directions if one starts 'off centre'.

For the acadmically inclined, there is, of course, a far more complicated variant of this question (and I'm very thankful that 'they' didn't think of this one back when I was the victim!) - in addition to varying the location of the first cut, one can also vary how one 'divides' the segments - i.e. one doesn't necessarily have to do 'halving' - one could, for example, do 'thirding'! I'm not sure I'm interested enough to go exploring all the possibilities that throws up!

Kind Regards, John.