neutral-earth voltage

[One_of_them]

guys,

head is battered its been a bad day and cant think straight.

Replaced a 3kW water heater element and everything went well. Circuit tested out fine so turned everything back on but when the element is pulling im getting L-N 238 L-N 234 N-E 4V!!! This was tested at element but is the same at the supply side of the 13A DP FCU but at the board (3phase) N-E is nothing (well mV anyway) When element or FCU is off everything is normal

I have checked all relative connections and even checked another heater on site and it is the same.

Is this just the way elements are?
Why would the supply side of FCU have 4V but the board not?

cheers all have a good weekend and some lovely cold beers

 

[telectrix]

you could use the 4V to charge your mobile. other than that, don't fret. could be induced voltage. try measuring with an analogue meter.

 

[Knobhead]

Welcome to the world of volt drop. You get VD* in the neutral as well as the line. Hence the 4V between N and E at the load.

(Deleted a load of waffle)

*(Sounds like a trip to the clap clinic is due).

But the following figures and the drawing are right

 

[Guitarist]

Listen to tel. In the scheme of induced voltage 4v isn't even worth mentioning

 

[Marvo]

The 4v is probably just voltdrop on the neutral as Tony mentioned.It might be worth doing a visual earthing check and an earth impedance test, if the earthing is tn c s it's a little higher than I'd expect to see.

 

[DurhamSparky]

unless your a BG engineer!! 4v is deadly and they need to upgrade all your earth and bonds and remove your split load box and install a dual rcd.... not to mention replace your boiler and pipework!!

all because his volt stick LIT up...!!

 

[Moses]

The 4v is probably just voltdrop on the neutral as Tony mentioned.It might be worth doing a visual earthing check and an earth impedance test, if the earthing is tn c s it's a little higher than I'd expect to see.

Its not Induced voltage. Its VD.

Doesn't matter what earthing system it is, If you'v got VD, you'v got VD mate!!!

It'l be same on any system.

You get VD under load conditions. Its normal and nothing to worry about. It shouldn't be more than 5% for an immersion heater.

 

[telectrix]

i might be having a blonde moment here. understand there is probably a bit of a VD under load, say there's a VD of 4V L/N. why is that not 4V L/E as both N and E are assumed to be 0V. so therefore 0V N/E?

 

[Knobhead]

There’s volt drop along the neutral. At the source (the CU) N and E should be more or less equal. At the point of load there will be a slight rise between N and E. 4V will give a total of 8V for the entire L/N loop. Which equals 3.47% volt drop, perfectly reasonable.

 

[Marvo]

i might be having a blonde moment here. understand there is probably a bit of a VD under load, say there's a VD of 4V L/N. why is that not 4V L/E as both N and E are assumed to be 0V. so therefore 0V N/E?

If the current is high enough that the l and n wires become part of the load and an equal voltage would be dropped across each. At the immersion heater you could get a 4 volt reading between n and e but the closer you get to the n-e bonding point on tncs the n-e voltage should get less. I actually misread part of the OP where he says he measured 4v at the FCU. I thought he was talking about the CU. All the terms like FCU and CU are not used here so occasionally I translate incorrectly so the blonde moment is at his end not yours

 

[telectrix]

There’s volt drop along the neutral. At the source (the CU) N and E should be more or less equal. At the point of load there will be a slight rise between N and E. 4V will give a total of 8V for the entire L/N loop. Which equals 3.47% volt drop, perfectly reasonable.

thanks for that, tony. i'm just trying to get my head round it. i see why, but not how, if yo know what i mean. :28:

 

[Moses]

guys,

head is battered its been a bad day and cant think straight.

Replaced a 3kW water heater element and everything went well. Circuit tested out fine so turned everything back on but when the element is pulling im getting L-N 238 L-N 234 N-E 4V!!! This was tested at element but is the same at the supply side of the 13A DP FCU but at the board (3phase) N-E is nothing (well mV anyway) When element or FCU is off everything is normal

I have checked all relative connections and even checked another heater on site and it is the same.

Is this just the way elements are?
Why would the supply side of FCU have 4V but the board not?

cheers all have a good weekend and some lovely cold beers

I think what you mean is that

L-N voltage is 234V
L-E Voltage is 238V and
N-E Voltage is 4V

Quite normal.

telectrix:-

Volt drop in the neutral wire when under load, therefore Neutral voltage at load end relative to MET = 4V
No voltdrop in earth wire as no current flow in it, Therefore CPC voltage at load relative to MET = 0V,
therefore N-E = 4V

 

[telectrix]

i can see the logic in that. just VD in the neutral is hard to grasp. a few more beers and all will become clear.

 

[Knobhead]

Try this

View attachment 12548

 

[telectrix]

i'm getting there, tony. can see it clearly from your diagram. many thanks.

 

[Knobhead]

I could be rotten and add a duff N/E in to the mix

View attachment 12549

 

[telectrix]

now that would put the cat amongst the pigeons. TC, don't you dare! LOL.

 

[Moses]

Try this

View attachment 12548

Thats marvellous, but I still don't quite understand!!!!

 

[LineProducts]

I have been having a quite night but after trying to get my mind round this I think it's time to reach for a bottle.

The way I understood this phenomenon was simple physics:
A is a measure of the number of electrons passing a point per second.
V is the potential difference so the force pushing the electrons along.

In plumbing speak the A is the volume of water and V is the pressure or head.

In a circuit the electrons will take the easiest route to ground. For example if you short E and N a RCD will trip. Back to plumbing speak equivalent of puncturing the pipe.

The electrons are all being squeezed through a small section of copper and therefore encounter resistance - V=I R. The back up due to resistance is the V measured between E and N. It is 0 at the consumer unit as this is near where E and N are merged tnc (not tt) so there is no relief of V by shorting N and E.

This leaves vd. Is this not caused by the resistance in the cable v=ir. E.g. If 10a of current flows through cable with 1 ohm of resistance the vd is 10v?

I am now going to pour a stiff one. Would be grateful for some clarity on this.

.

 

[Moses]

Try this

View attachment 12548

You are a magician!!!!

The OP does not tell us the 242V figure.

Where did you get it from????????

You sure that's r8????