Fault in SWA cable

[bernardgreen]

There is another way to approximately locate the short if the conductor shorting to ground or armour is still intact.

The conductor acts as the resistance track on a potentiometer and the voltage from the "wiper" at the fault is proportional to the distance along the cable to the fault.

Untitled

• bernardgreen
• 8 Feb 2011
• 3

Think of a potentiometer where the fault is the wiper mving along the resistance track formed by the Live conductor

The voltage drop along the Live conductor is linear between x and z and is equal to half of V loop as measured across Live and Neutral

The voltage on the Live at the location of the fault will via the fault affect the Earth conductor. If no current flows through the fault ( in purple ) then the two end of the fault resistance will be the same. Therefore the Earth conductor will be at the same voltage as the voltage on the Live at the lcoation of the fault.

Provided no current flows along the Earth conductor there can be no voltage drop along it. ( the meter take virtually no current ). Therefor the voltage at the end of the Earth conductor is the same as the voltage on the Live conductor at the location of the fault.

Read more: //www.diynot.com/forums/viewtopic.php?p=1894003#1894003#ixzz1ekURMvc1

 

[Taylortwocities]

Doesn't that distance to fault stuff only work if the fault (connection between live and earth) is a proper zero ohms short.

If the fault has any resistance it will skew the distance calculation.

 

[bernardgreen]

Provided the source of the test current has no connection to earth and the volt meter is high impedance and takes virtually no current then the current through the short is very small and therefor the voltage drop across the fault is insignificant.

 

[JohnW2]

Doesn't that distance to fault stuff only work if the fault (connection between live and earth) is a proper zero ohms short. If the fault has any resistance it will skew the distance calculation.

As Bernard explained, provided the voltmeter has a sufficiently high input resistance, current flow through the fault resistance (whatever its magnitude) will be negligible (hence voltage across fault resistance negligible) and therefore will not appreciably affect the accuracy of the method.

However, as Bernard said, the method as described will only work if the L conductor (and earth conductor or armour) remains intact but has a connection to earth conductor (or armour). The same method can be adapted for L-N or N-E faults - but there is always a requirement that the conductors are intact, even though 'shorting'. The apparent answer will be affected if, say, the L conductor is shorting to E but also is damaged to the extent that there is a high resistance segment within the L conductor on one side of the fault.

Kind Regards, John.

 

[ColJack]

what's wrong with the good old "keep resetting the breaker while you walk the cable run" method?
you feel the bump under your feet when you're over the fault

 

[westie101]

what's wrong with the good old "keep resetting the breaker while you walk the cable run" method?
you feel the bump under your feet when you're over the fault

Or as does happen you lift a the pavement up by a couple of cms.

 

[JohnW2]

The apparent answer will be affected if, say, the L conductor is shorting to E but also is damaged to the extent that there is a high resistance segment within the L conductor on one side of the fault.

I should have added that, although such a situation means that Bernard's method cannot be used, one can easily check to make sure that one is not dealing with that situation, since it will result in an unxpectedly high (end-end) resistance of the damaged conductor. If one doesn't want to play around with trailing test leads, one can test that from one end by joining pairs of conductors together, in turn, at the far end and then measuring the resistance of each 'loop'.

Kind Regards, John.

 

[Paul_C]

And that basic method above is the way that faults were located for many years on telephone cables using bridge instruments such as the PO/BT Ohmmeter 18, before TDR equipment came down in price enough to be made widely available in the field.

 

[17thman]

sorry wrong post

 

[17thman]

is it possible to measure the continuity of the fault from both ends and determine the distance proportionately using the tabulated resistance of the conductors? an unknown resistance fault would become known from both ends, unless it was intermittent

 

[EFLImpudence]

is it possible to measure the continuity of the fault from both ends and determine the distance proportionately using the tabulated resistance of the conductors? an unknown resistance fault would become known from both ends, unless it was intermittent

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.

 

[17thman]

true. saves 13k.

 

[JohnW2]

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.

 

[JohnW2]

..or, using EFLI's example rather than mine ...

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.

If the resistance of the fault were actually 1.5Ω, such the that 'conductor components' of those measurements were actually 1.5Ω and 0.5Ω respectively, then the fault would actually be one quarter of the way from the end, not as predicted by EFLI.

Kind Regards, John.

 

[17thman]

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