Finding length of cable

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Hi,

I wonder if it would be possible to calculate the approximate length of a cable from a few know values? I work at a water treatment plant as a controls and instrumentation engineer (not an electrician) and have a panel that i cant for the life of me find out where it is fed from. The incoming cable is 4mm2 SWA presumably a mix of buried and in conduit underground, and is live. I have measure 253.5V to earth and 253V to neutral. This is ok as the panel has a local isolator so i can do the modifications that i need to do inside the panel, however for future reference i really need to know where the power is fed from. I'm hoping we can work out the approximate length of the cable to give me some more ideas of where to look. Currently there is only 1 flow meter in the panel and is pulling a load of appox 0.7A. I'm not hoping for too much, just a rough length.

TL;dr
Is it possible to calculate approx length of 4mm2 SWA with a measured voltage of 253V and of load of about 0.7A
 
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This is a diy forum and all we know is light switches.






Only joking.

I'm sure someone on here might help but I'm out as I have no idea.
 
I suppose i should have started with "i have a garage supplied with 4mm2 SWA but i dont know where its fed from".

[/joke]
 
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Not sure if this helps for electrical cables(*) but in computing circles, we use Time/Domain Reflectometers to tell us the distance to an open or short circuit on a cable;

http://en.wikipedia.org/wiki/Time-domain_reflectometer

(* the caveat is that we tend to do this on known-faulty / dead network cables - I don't know whether TDR works with live mains cables)
 
The length cannot be determined from a single voltage and load.

Whatever you do will not be able to distinguish the swa from the rest of the supply circuit.
 
If the supplies are RCD protected i connect a load L /E ! that narrows the search down :eek:

Regards,

DS
 
I built this http://gw7mgw.co.nf/Electric-Cal.html
How good it is to you not sure but with a loop impedance meter knowing impedance into site in example shown 0.35Ω and size of cable in example 2x2.5mm² and impedance at end of cable in this example 0.94Ω one can work out cable length in example 106 meters (58 meters really as working it out for ring).
 
... have a panel that i cant for the life of me find out where it is fed from. The incoming cable is 4mm2 SWA presumably a mix of buried and in conduit underground, and is live. I have measure 253.5V to earth and 253V to neutral. ... Currently there is only 1 flow meter in the panel and is pulling a load of appox 0.7A. I'm not hoping for too much, just a rough length. Is it possible to calculate approx length of 4mm2 SWA with a measured voltage of 253V and of load of about 0.7A
If you increased the load on that circuit substantially and measured the resultant fall in voltage at the end of it, that would enable you to determine the impedance of the entire L-N loop back to the supply transformer/substation. With certain assumptions about the magnitude of the 'external' part of that (between origin of cable and supply transformer/substation), you could thereby estimate the L-N impedance of the SWA (plus any other cable in the circuit, which could be the problem!), hence calculate an estimated length.

Kind Regards, John
 
The length cannot be determined from a single voltage and load.

Whatever you do will not be able to distinguish the swa from the rest of the supply circuit.

Oh well i figured this might be the case. I was hoping we could use volt drop calcs to work backwards. I suppose would need the voltage at the other end of the cable to work out the drop.



If the supplies are RCD protected i connect a load L /E ! that narrows the search down :eek:

Regards,

DS

I'm sure there are no RCD's here. This circuit (going from the installation date written on the old flow meter) was installed 1988.

As an aside, are industrial sites required to have RCD protected circuits for single supply's to instrumentation panels? What about the small lighting and power circuits?
 
Oh well i figured this might be the case. I was hoping we could use volt drop calcs to work backwards. ...
As I've just written, if you increased the load you could do that, but you would only be able to determine the impedance of the entire path back to the transformer/substation, and would then have to make some assumptions to translate that into an estimate of the length of your cable.

Kind Regards, John
 
0.011 V/A/m (both line and neutral) and at 0.5 volt difference earth to neutral and 0.7A I make it 130 meters. But if 0.8 not 0.7 it would be 113 meters and if 0.6 volt not 0.5 volt then 156 meters.

So 90 - 181 meters from point where earth and neutral are bonded together. As all ready said the more current drawn and the more accurate the readings are then also the more accurate the distance calculation.

We are assuming no current on the earth and the neutral is 0.0055 ohms per meter or volt drop 0.0055 volts per meter.

So 0.0055 * 0.7 = volts per meter = 0.00385 so 0.5/0.00385 = 129.8701 meters. But to find accuracy one has to do it with 0.6A and 0.8A and 0.4V and 0.6V which means 90 - 181 meters to the TN-C-S DNO head. If supply is TN-S then this is completely out. Same if there is a board some where between with other loads.
 
0.011 V/A/m (both line and neutral) and at 0.5 volt difference earth to neutral and 0.7A I make it 130 meters. But if 0.8 not 0.7 it would be 113 meters and if 0.6 volt not 0.5 volt then 156 meters. ... So 90 - 181 meters from point where earth and neutral are bonded together. As all ready said the more current drawn and the more accurate the readings are then also the more accurate the distance calculation.
I hesitated to suggest such a calculation, since the magnitude of the current seems fairly 'vague' and, in particular, the crucial difference between 253 and 253.5 volt likely to be not all that accurate (it would probably be better to directly measure the N-E voltage, using a more sensitive/precise meter range, than to look at the difference between L-E and L-N voltages). As we've both said, much higher load currents would lead to much more reliable calculations.

Kind Regards, John
 

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