Ze external earth loop impedance

[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.