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I'm in a vineyard. There is a 300m buried SWA from the ccu direct to a 2 gang socket.
I'm getting 247V across N and L at the ccu.... But at the socket I'm getting 100V across E and N and 147 across L and N.
Continuity as a radial tested ok
IR All over 500 M
Please list possible issues here. Thank you.
 
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B

Bobster

Are there any loads attached at the time of measuring?

What were the continuity results and how did you measure them?

300m is a long way, for what I presume if connecting directly into a socket, is going to be <= 2.5mm SWA.
 

Pete999

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I'm in a vineyard. There is a 300m buried SWA from the ccu direct to a 2 gang socket.
I'm getting 247V across N and L at the ccu.... But at the socket I'm getting 100V across E and N and 147 across L and N.
Continuity as a radial tested ok
IR All over 500 M
Please list possible issues here. Thank you.
What size SWA?
 

marconi

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How did you check end-to-end continuity? Is the SWA two core or three core?

First thought is a broken neutral or SWA/CPC somewhere along the length of the cable so am surprised your radial continuity checks are ok.
 
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  • Thread Starter Thread Starter
  • #9
Are there any loads attached at the time of measuring?

What were the continuity results and how did you measure them?

300m is a long way, for what I presume if connecting directly into a socket, is going to be <= 2.5mm SWA.
It's 6mm SWA
WAS GETTING 2 ohm continuity r1+r2
Measured with L and N cables removed from socket.
 
  • Thread Starter Thread Starter
  • #10
How did you check end-to-end continuity? Is the SWA two core or three core?

First thought is a broken neutral or SWA/CPC somewhere along the length of the cable so am surprised your radial continuity checks are ok.
It's 2core, using armour as cpc
6mm
 
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What size SWA?
6mm 2 core, using armour to carry earth
 

marconi

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Did you measure 2Ohms for Rn + R2? ie: the same as R1 + R2?

Assuming 2.9mV/A/m volt drop for 2 core 6mm2 core SWA, R1 + Rn is circa 0.0029 x 300 x 2 = 1.74 Ohms.

Please could you describe more the wrong N bus bar wiring?

:)
 
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richy3333

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2 ohms. What’s the protective device?
 

Marvo

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As above, if you put any load above a few Amps on that length and size of of cable you're going to get an unacceptably high N-E voltage....
 

happyhippydad

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As above, if you put any load above a few Amps on that length and size of of cable you're going to get an unacceptably high N-E voltage....
Could you explain that a little please Marvo? I understand why approx 4A is the most you should draw due to VD, but how would increasing the load cause there to be voltage between N-E?
 
In a TN system the potential difference at source should be 0v in an ideal world. A circuit designed to have no volt drop should register 0v between the conductors at its point of utilisation. Once you start getting volt drop you must remember in normal circumstances there is no drop in the cpc but there is in the neutral and this drop will measure a potential difference between the conductors, the higher the drop the higher the pd between neutral and earth.
 
... how would increasing the load cause there to be voltage between N-E?
The cpc should have no current in it (unless a fault condition exists) so it can act as a reference point.
since Line conductors are the same csa they should share the pd evenly.
 

Lucien Nunes

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As above, the maximum N-CPC voltage that should be measured at the far end of the circuit, is half the permissible volt drop. 5% x 230V /2 = 5.75V. That is the drop along the length of the neutral conductor (the other 5.75V is in the line). If N & E are at the same potential at the origin, they must be 5.75V apart at the end of the circuit.

But returning to the OP's problem, all the evidence points to the neutral being disconnected at the source end, therefore floating for the whole length of the cable. Because of its length, there is quite a high capacitance between L & N, and between N & E, which can pass enough current to give a steady indication on a voltmeter or MFT. It's an extreme case of 'ghost voltage' if you like.

The big clue is that the voltage L-N and N-CPC add up to 'about 230V' so it is working like a voltage divider, with 147V of the 247V supply across L-N and the remaining 100V across N-CPC. The difference is accounted for by the difference in capacitance. Because the CPC (armour) has a greater surface area facing the floating N conductor than the line, its stray capacitance is higher, its capacitive reactance lower and therefore the neutral will float to a voltage somewhat less than half of the supply, in this case 100V.
 

marconi

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Is this final circuit safe? What is the touch voltage of earthed metalwork in the event of a direct fault connection between the line and earthed metalwork at the socket end?
 
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