Ze external earth loop impedance

[Chr!s]

Indeed. But where in that loop are you taking your A point to support a % vd?

At the service Head, Line and Neutral, so A and B


You're taking it at the entry point of our practical load, which means it's being referenced to zero at the end of that load

Why is it being referenced to Zero, is that the potential you think the neutral will be at.So based on your assumption then no current flows int the neutral? Its simple ohms law, you have a current you have an impedance therefore you have a voltage.

- otherwise, we end up in a muddle of trying to define a %'ge as 245/245 which is meaningless. Hence, what follows after the connection to the neutral point is irrelevant.

No its not! Lets start with an open PEN, what voltage would you expect at the PME LINK assuming no extraneous conductive parts for current to flow?

Now lets add a impedance to the the neutral 100 ohms, whats the voltage now at the head, then do it at 10 ohms 1, 0.1 ... Whats happening to the Voltage on the neutral at the service head?

Besides, WPD around here will allow a PME connection to be as high as a TN-S at .85 So now work that one out at your 100A as before. 85VD at source? I think not.

WPD are my DNO, and if they are putting 100 amp supply's on a weak supply like that there in for a lot of trouble. Im not aware of 100 amp supply's being offered on weak systems like that!!

I know they offer PME on supply's slightly above 0.35, but you will get an 80/60 amp supply.

 

[Rockingit]

This is simple maths! PD - potential difference - the difference between two points.

Therefore A-B = C Therefore, if A = 250 B = 0 V = 250.

So, if we are stating that we have a V of 250 to start with (as in, before we decide to do anything else to it), then clearly it can only be 250 if B is zero, else we end up with a different value. Otherwise, we end up in ever decreasing circles. The %'ge will stay the same in direct proportion, as mathematics dictate, but the real world value will have no meaning as it spirals to infinity.

Otherwise, in order for a voltage meter to give you a reading it would need to know the full loop impedance first of the whole circuit, and it doesn't, it just references A > Bzero.

As for WPD, around here an 80A supply is all you'll get anyway but they'll let existing properties go as high as .8 whilst still on PME. I know that first hand after an argument I had with a grumpy engineer after I'd called them out last year to look at one which was .45ish!!

 

[Chr!s]

This is simple maths! PD - potential difference - the difference between two points.

Correct

Therefore A-B = C Therefore,

Correct


if A = 250 B = 0 V = 250.

This is where your going wrong, A, wont be at 250, that is at the Tx, so voltage is dropped along the Line, ohms law, we have impedance we have current therefore a voltage will be dropped along its length.

B will not equal zero, B will depend upon the loop impedance's.

So, if we are stating that we have a V of 250 to start with (as in, before we decide to do anything else to it), then clearly it can only be 250 if B is zero,

That holds true at the source assuming an ideal source, so yes at the terminals Tx


else we end up with a different value. Otherwise, we end up in ever decreasing circles. The %'ge will stay the same in direct proportion, as mathematics dictate, but the real world value will have no meaning as it spirals to infinity.

Its pretty straight forward

Otherwise, in order for a voltage meter to give you a reading it would need to know the full loop impedance first of the whole circuit, and it doesn't, it just references A > Bzero.

As for WPD, around here an 80A supply is all you'll get anyway but they'll let existing properties go as high as .8 whilst still on PME. I know that first hand after an argument I had with a grumpy engineer after I'd called them out last year to look at one which was .45ish!!

LOL the maths it obviously not that simple, you've still not grasped it.

 

[Rockingit]

:mad2: I officially give up. Nothing personal Chr!s, but we're just clearly on two different pages here!

 

[HandySparks]

I did myself a little sketch to help visualise where the various volt drops occur and what the voltage readings at several places would be. I find it easier to use example real-world values rather than symbols. Is this any help?

It assumes that the supply impedance at the local transformer is negligible. Other earthing points (for a PME supply) are not shown, but I don't think that they make a fundamental difference.

 

[Chr!s]

Not sure he'll accept it lol, he's adamant this Neutral is tied down to 0, he will see its the star point that is tied down 0(well close to) and not the neutral in the service head.

So in your example 40 V is dropped along the supply, so when he puts his meter across A and B his meter will show 210 V, im sure the penny will drop

 

[the pict]

Tell you what though this little debate has been put very intelligently with both sides stating their veiws professionally, very refreshing considering the sometimes purile responses from some members to queries, and on a personal level has been infomative on a subject I admit was somewhat unclear to me a times
Pict

 

[BillyTheSparks]

Why in your diagram do you have a VD of 20v on the N from the transformer to the meter?

If it's PME then any stray voltage along the N will be returning to E at each N-E connection. Therefore all along the length of the N conductor the voltage will always be 0v. So there would never be a VD on the N.

I see that you will have a VD on the phase (line) of 20v; and when measured at the meter will shown a voltage of 230v.

 

[BillyTheSparks]

Sorry re-reading that, what I'm trying to say is, if the N is always connected to E at each point, then there is little/no resistance on that conductor. Being that you only get VD with a resistance of cable then you won't have any VD on that N from the Transformer.

 

[Chr!s]

Why in your diagram do you have a VD of 20v on the N from the transformer to the meter?

That is correct, V = I x Z

If it's PME then any stray voltage along the N will be returning to E at each N-E connection. Therefore all along the length of the N conductor the voltage will always be 0v. So there would never be a VD on the N.

Yes your right in a way, there are multiple paths a long the PEN for current to return to the star point, the voltage will divide along the PEN. If you make the assumption that the 0.4(in the above diagram) is the total impedance of the supply including any parallel paths along the PEN and the total impedance on that PEN is 0.2 then as stated V = I x Z

I see that you will have a VD on the phase (line) of 20v; and when measured at the meter will shown a voltage of 230v.

The voltage measured at the service head between line and neutral will be 250 - 40 = 210 volts or to put it another way, you have 250V at the source, 20V is dropped along the line to the service head, So voltage at the service head is 230 V. We know 20 V is dropped along the PEN, its the potential difference between the two points.

Hope that makes it a little clearer

 

[HandySparks]

Sorry re-reading that, what I'm trying to say is, if the N is always connected to E at each point, then there is little/no resistance on that conductor. Being that you only get VD with a resistance of cable then you won't have any VD on that N from the Transformer.

With PME, yes there will be the 'earth' in parallel with the N/E core. However, you can't say that, just because a cable has multiple earth points, it will be at earth potential. That would only be the case if the earth rods had a Ra of zero. They won't, so there will be a small, but finite, resistance to 'earth'.

So, yes, the N supply conductor may not have the same volt drop as the L, but it will have some.

Actually, in reality, the N supply conductor may have very little current for some or all of its length if it's a three phase distribution cable, as is usual. This could be to actual three phase loads or multiple balanced single phase loads.