Ze external earth loop impedance

[Chr!s]

Eh???? I think you need to have a think about that.

Why, whats wrong with it

 

[Silly Sausage]

Also, with a TN supply, if yar Ze is too high, an Earth fault won't take out the intake fuse. (depending on A rating)

 

[Rockingit]

Why, whats wrong with it

Just in case I'm misunderstanding you...walk through it once more.

 

[Engineer54]

Eng,
Do you mean 0.04 Ohms or even .004 ohms? (thats a lotta raisins)

I did indeed mean 0.04 ohms!! That decimal point can get you into some real problems if your not careful!! lol!!

 

[Chr!s]

Just in case I'm misunderstanding you...walk through it once more.

Well on a single phase system whats the mimimum voltage.

So for a 100 amp supply, assuming a current of 100amps 35 volts will be dropped in the supply.

So if at Tx terminals we have 250v then and 35 v dropped across supply then only 215 v is supplied at the consumers service head

 

[Plonker 3]

Unless I am being completely thick, I have no idea what Chr!s is on about.

 

[KAS1]

Nor me lol

 

[Silly Sausage]

I just can't be arsed!

Why am I even typing this?!?!

 

[Chr!s]

Unless I am being completely thick, I have no idea what Chr!s is on about.

Maybe its me who's thick

 

[HandySparks]

Maybe its me who's thick

No, I can see where you're coming from. On a TN-C-S supply (single phase), with a combined neutral and earth conductor, the line impedance will be identical to the earth fault loop impedance. So if the Ze is high, so will the Z (line), resulting in a high voltage drop between the local transformer and the consumer under maximum load conditions.

 

[Chr!s]

No, I can see where you're coming from. On a TN-C-S supply (single phase), with a combined neutral and earth conductor, the line impedance will be identical to the earth fault loop impedance. So if the Ze is high, so will the Z (line), resulting in a high voltage drop between the local transformer and the consumer under maximum load conditions.

You got it

 

[Rockingit]

Well on a single phase system whats the mimimum voltage.

So for a 100 amp supply, assuming a current of 100amps 35 volts will be dropped in the supply.

So if at Tx terminals we have 250v then and 35 v dropped across supply then only 215 v is supplied at the consumers service head

Okaaaayyy....... you sure? Have another think about voltage.

 

[Chr!s]

Okaaaayyy....... you sure? Have another think about voltage.

V= I x Z

What do you get?

 

[telectrix]

this is why the lights go dim when i fire up the tardis. volt drop, due to the high current. but seriously, when does the average house draw anywhere near 100A? once in a blue meerkat's --------.

 

[HandySparks]

Okaaaayyy....... you sure? Have another think about voltage.

OK, I'm not saying that VD is the only, or even the main reason, for setting a particular value of Ze on a TN-C-S supply but, with all due respect, where are Chr!s and myself going off track?

 

[Rockingit]

Your pd is always between your two reference points, A & B. So in order to be able to calculate ANY value, we need to know what they are. If we are saying that A is 245V (or whatever) then it can ONLY be 245V referenced against B, which is why B (N) is zero. B cannot ever change from that otherwise we can't calculate a load (or whole circuit load inc VD if you wish) as we have an unknown variable. Therefore, it matters not what is downstream of the neutral, therefore using Ze as an expression of Zl is just irrelevant for practical purposes.

I do kind of see where you're coming from, but.......

IMO, and I open the floor to debate.

 

[Chr!s]

Your pd is always between your two reference points, A & B. So in order to be able to calculate ANY value, we need to know what they are. If we are saying that A is 245V (or whatever) then it can ONLY be 245V referenced against B, which is why B (N) is zero. B cannot ever change from that otherwise we can't calculate a load (or whole circuit load inc VD if you wish) as we have an unknown variable. Therefore, it matters not what is downstream of the neutral, therefore using Ze as an expression of Zl is just irrelevant for practical purposes.

I do kind of see where you're coming from, but.......

IMO, and I open the floor to debate.

Will their be any current in your neutral?

 

[Rockingit]

I = V/R. Therefore I = 0/R, therefore no.

 

[Chr!s]

I = V/R. Therefore I = 0/R, therefore no.

I think you need to rethink that one

You have a current of 100 amps flowing in a loop.

You have a voltage 250

You have the impedance of half of that loop

So if you use Kirchoffs voltage law we know that the sum of the voltages in a loop add to zero.

 

[Rockingit]

I think you need to rethink that one

You have a current of 100 amps flowing in a loop.

You have a voltage 250

You have the impedance of half of that loop

So if you use Kirchoffs voltage law we know that the sum of the voltages in a loop add to zero.

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

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.

 

[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.