distance to a live to earth fault in an under ground cable

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I have a problem that I hope you will be able to help with. I have just bought an old property that has a bore hole for the water.

I have an armoured cable that runs to the pump. However I have an Live to earth fault on it. This cable runs several hundred feet and is buried. I don't at this stage want to run a replacement cable.

How can I find where on the cable the fault lies. I have disconnected both ends so I know it is a fault with the cable.

Any help appreciated.
 
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You may be able to hire a ranger/reflectometer to determine the distance to the fault.
 
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You could find it by passing a steady DC current along the Live conductor to create a voltage drop along it. Then by measuring the voltage on the Earth conductor and knowing the voltage drop along the cable you can tell where the Earth conductor is touching the Live conductor. Using a high impedance digital meter means the meter is drawing virtually no current through the fault so is no voltage across the fault.

A practical way to do this is, assuming the Neutral conductor is not faulty, is to join Live and Neutral at one end. Using a car battery and a 50 watt 12 volt lamp to run 4 amps through the looped Live and Neutral. Measure the voltage between Live and Neutral and that will give the voltage drop along the loop. This is twice the voltage drop along the Live Call the voltage drop along the Live conductor VdL . Measure the voltage between Live and Earth and that is the votage on the Live at the fault. call it VatF

If VatF = 1/2 VdL then the fault is at the mid point of the cable's length

If VatF = VdL then the fault is very close to the far end of the cable

If VatF = 0 then the fault is very close to the ends at which you are measuring

For other values the ratio of voltages is the same as ratio of cable length to distance to the fault

The amount of current in the cable is not important provided it produces a readable voltage drop along the cable and is constant during the measurements.

Using AC current will work but the capacitive affects will make the resultl less accurate.
 
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That will only work if the actual fault connection is, effectively, of zero impedance.

As would (with a lot less messing about) using a low-ohm meter to just measure the L-E resistance at one end and comparing the value to the published resistance per length for that size of conductor.

Actually I've just thought - you could use a low-ohm meter to get quite a good idea even if the fault resistance is not zero.

Measure L-E resistance at each end - that'll allow you to cancel out any fault resistance value.

If it's a 3-core cable with one core used as an earth you will need to remove any connection between that and the armour and test both L-core and L-armour to check you have the right path and remove parallel paths.
 
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That will only work if the actual fault connection is, effectively, of zero impedance.

Not so. The resistance of the fault connection between Live and Earth conductors is not part of the current flow creating the voltage drop along the Live conductor. Provided the volt meter does not pull current from the circuit it is measuring there is no significant current flowing from the Live to the Earth at the fault so there is no significant voltage drop across the fault.

There is only a significant current through the fault if the volt meter used in a low impedance meter, most digital meters are high impedance.
 
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Not so. The resistance of the fault connection between Live and Earth conductors is not part of the current flow creating the voltage drop along the Live conductor.
It is in series with the L-fault-E path whose comparative resistance you think you can calculate via measuring comparative voltage.

So if you were right about your test working then the resistance would be relevant - as it increases you would "see" the fault point moving further along the cable by the length of conductor with a resistance half that of the fault

But - it wasn't until I read your idea in more detail, to produce a worked example that I realised it wouldn't work anyway.

You've applied a voltage between the ends of a conductor (your L-E loop) which at some point is connected to another conductor that's not part of your potential zone - it's floating, there will be no potential difference between it and any part of your loop, any more than there would be between the loop and the garden gate.


A practical way to do this is, assuming the Neutral conductor is not faulty, is to join Live and Neutral at one end. Using a car battery and a 50 watt 12 volt lamp to run 4 amps through the looped Live and Neutral. Measure the voltage between Live and Neutral and that will give the voltage drop along the loop.
In this circuit, measuring the voltage between A & B will tell you what?

t262166.jpg
 

STI

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Also before you go any further are you sure its not the pump thats causing the fault ? Far more likely the fault is in the pump than the cable if properly installed .
 
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You've applied a voltage between the ends of a conductor (your L-E loop) which at some point is connected to another conductor that's not part of your potential zone - it's floating, there will be no potential difference between it and any part of your loop, any more than there would be between the loop and the garden gate.]

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.
 

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