Guidance note 8 help...

[JohnW2]

That would mean a person had to touch two points on a conductor between which the impedance was 0.55Ω. That is not possible in the diagram as part of that 0.55Ω is the supply cabling.

No, it would require that they simultaneously touched the unbonded extraneous-c-p (which would be AT true earth potential) and the exposed-c-p with the fault - which, if a fault current of 100A flowing would, with the GN8 figures, be at a potential 55V above true earth potential.

Kind Regards, John

 

[EFLImpudence]

Ok. I'm getting a bit confused then - because you did not mention fault current in your previous calculations - because as you have explained (ignoring the 0.19Ω) - whatever it is, as long as R1 = R2 then the touch voltage is 115V.

 

[JohnW2]

Ok. I'm getting a bit confused then - because you did not mention fault current in your previous calculations - because as you have explained (ignoring the 0.19Ω) - whatever it is, as long as R1 = R2 then the touch voltage is 115V.

What I think you're missing is that the 115V assumes that only the impedance in the fault loop is due to the impedance of the conductors in the fault loop (i.e. Zs in total), the impedance of the fault being 'negligible'.

To get the fault current down to 100A would require an additional impedance (the 'non-negligible' impedance of the fault) in the loop, and that would change the answer from 115V to 55v in the case we are considering (the 'missing' 60V being 'across the fault').

Put another way, as compared with the assumed "R1=R2", we now have "R1=R2+Rx", Rx being the (now non-zero) impedance of the fault.

Does that make sense?

Kind Regards, John

 

[EFLImpudence]

That makes sense but I don't see what it has to do with the diagram in GN8 nor why you have chosen to have that happen.

 

[JohnW2]

That makes sense ...

I'm glad you agree.

... but I don't see what it has to do with the diagram in GN8 nor why you have chosen to have that happen.

I don't think it has got much to do with the GN8 diagram and I'm not sure what you mean by my having "chosen it to happen".

I was merely responding to your comment that the touch voltage would persist for only "a very short time" before an OPD cleared the fault, pointing out that, as an example, a 100A fault current (due to a fauklt of non-negligible impedance) would (with GN8 figures) result in a touch voltage of 55 V which could persist for an 'appreciable' period of time.

Kind Regards, John

 

[studentspark]

Thanks so much for your time and effort on this. I am struggling a bit with Rx . I do feel the mist is clearing, but every-time I think I get a break through, another question arrises.
I hope to have some sensible input shortly.

 

[JohnW2]

Thanks so much for your time and effort on this. I am struggling a bit with Rx . I do feel the mist is clearing, but every-time I think I get a break through, another question arrises.

Rx is simply enough, it's just the effective resistance of (Rpipe + Rbonding in series) in parallel with R3.

The resistance/impedance of (Rpipe + Rbonding in series) is obviously just Rpipe + Rbonding - let's call that Ry.

We therefore have Ry and R3 in parallel. The 'long' way to calculated that is:

... but I think it simpler to use

Does that help?

Kind Regards, John

 

[studentspark]

Thanks for your help with this John. I really do struggle with simple algebra, so following your excellent posts does take me some time.
After reading my following post, you will probably want to start banging your head against the computer screen. I completely understand if you want to bail out

But I have 5 scenarios.

1 is just the very basic 115V at the point of a fault as all the resistances are the same. Edit 230/1.1 = 202.09A (not 202.09V)

2 is the calculation using half the Zs ( As we would do in the real world?), and also the values from GN8

3 is you values with some values for Rx

4 is with bonding added - Im trying to understand why the 0.25 external R2 is removed from the equation - Im was thinking it is a lower resistance path for the fault
But then you would have a high resistance return path to the transformer via the mass of earth.

5. But all of these calculations have the Extraneous CP at 0 V, when it would actually (with bonding attached) rise to the potential of the MET,
So the touch voltage would be as shown.

 

[JohnW2]

Thanks for your help with this John. I really do struggle with simple algebra, so following your excellent posts does take me some time.

No problem.

After reading my following post, you will probably want to start banging your head against the computer screen. I completely understand if you want to bail out ... But I have 5 scenarios. ....

I'm certainly not going to 'bail out', but I'm pretty busy with other things at the moment, so it may be a little time before I can digest your "5 scenarios" and hopefully have something useful to say in response - so watch this space!!

Kind Regards, John

 

[JohnW2]

But I have 5 scenarios.

OK, I'm going to take this is easy stages, first with your two simple ones ...

1 is just the very basic 115V at the point of a fault as all the resistances are the same. Edit 230/1.1 = 202.09A (not 202.09V)

Yep, once the typo is corrected, that's all correct. You don't really need to work out the current and multiply that current by the resistance between fault and earth - since the resistances on both sides of the fault (0.55) are the same, the voltage at the fault will obviously be half the total supply voltage.

2 is the calculation using half the Zs ( As we would do in the real world?), and also the values from GN8

You have, perhaps unwittingly, raised an interesting point, which I (and probably many others) haven't really thought about much in the past - in that I am starting to understand the relevance of GN8's "0.19Ω".

Zs (or Ze) is the total impedance of the relevant 'loop' and we (at least, I !!) do not often think about the fact that the two 'arms' of that loop' (relative to where they join, at the 'truth earth' point) may not (indeed, usually will not) both have the same impedance (to earth).

Taking the example of the GN8 figures, as you have said, the total Zs is 1.29Ω and the total Ze is 0.69Ω. However, because the 0.19Ω (the 'internal impedance' of the source) is entirely in the leg of the loop which goes via the L conductor (which is not really apparent from the way you have drawn your diagram), that leg contributes 0.74Ω to the total Zs and 0.44Ω to the total Ze, whereas the other ('earth') leg contributes only 0.55Ω to Zs and 0.25Ω to Ze.

Thinking only of the total Zs or Ze is fine for calculating fault currents etc. but, as GN8 and you are saying, it is not enough in relation to calculation of touch voltages.

So, as you (and GN8) have discovered, assuming that the two arms of the loop have equal impedance, hence calculating touch voltages on the basis of "half of Zs" is not correct (but your/their calculation is). In other words, using the GN8 figures, the ~98V touch voltage would be correct.

[ of course, in practice, the two arms of the loop will each be closer to "half the Zs" than the GN8 figures suggest since, although the 0.19Ω is only in the L side, other than for circuits wired in 1.0mm² T+E, the within-installation "R2" (CPC) (in the other, 'earth' side of the loop) will be greater than the within-installation "R1" (L conductor) - so that the "half the Zs" approach to calculating touch voltages may, in practice, not be too far off. ]

Having got that straight in my mind, and thereby having de-mystified the 0.19", I will subsequently (hopefully soon!) move on to the rest of your scenarios!

Kind Regards, John

 

[JohnW2]

Having got that straight in my mind, and thereby having de-mystified the 0.19", I will subsequently (hopefully soon!) move on to the rest of your scenarios!

3 is you values with some values for Rx
4 is with bonding added - Im trying to understand why the 0.25 external R2 is removed from the equation - Im was thinking it is a lower resistance path for the fault
5. But all of these calculations have the Extraneous CP at 0 V, when it would actually (with bonding attached) rise to the potential of the MET,
So the touch voltage would be as shown.

I'm afraid I have confused things, for which I apologise.

As I said, because I did not at the time understand what the "0.19Ω" was all about, I simply ignored it - but, as I recently wrote, I now understand it's importance/relevance. That means that your diagrams (hence calculations) for scenarios 3, 4 and 5 are all incorrect, because none of them include the 0.19Ω.

I'm afraid that means that you need to re-do 3, 4 & 5, with a 0.19Ω resistance/impedance included in the 'left-hand'("L", brown) arm of the fault loop in every case. Zs will then be 1.29Ω, and Ze 0.69Ω in all cases.

Again, my apologies.

Kind Regards, John

 

[studentspark]

No need to apologise John. I really appreciate your help.

 

[JohnW2]

No need to apologise John. I really appreciate your help.

Thanks

I think the problem (for me and, I suspect, many/most others) is that GN8 is almost being 'too clever' by introducing theoretical issues which are probably virtually never usable in practical real-world situations (other than by guesswork).

I think that the reason why I (and, I suspect, most others!) haven't really given much thought to the "0.19Ω" component of Ze that GN8 uses in its diagrams and calculations is that, in practice, we will never know what it is, and so will never be able to do calculations like theirs which involves that component of Ze.

In reality, all we ever know is the total Ze (or Zs), so all we can really do is to assume (actually incorrectly) that the impedances of the two 'legs' of it are equal (even though we know that the 'earth' side will actually have a lower impedance than the 'L' side, to the tune of GN8's 0.19Ω, or whatever) .

In other words, as you said at one point, 'the best we can usually do' is to assume (actually incorrectly) that the impedance of the 'earth side' of the Zs is "half of Zs". That will take us back to the situation in which we estimate touch voltage as 115V if R1=R2 within the installation (only true for 1.0mm² T+E) or, with any T+E larger than 1.0mm², >115V (e.g. 143.55V with 2.5/1.5mm² cable).

The only way one could do it strictly correctly would be if we could measure the loop resistance actually at the DNO's transformer (thereby measuring that 0.19Ω, or whatever) - but that is never going to be feasible.

Therefore, unless I'm still missing something, I would suggest that it's probably not very worthwhile (only of theoretical, but no real practical, value) for you to spend too much time worrying about calculations which involve a component of Ze like GN8's 0.19Ω, since you're never going to be able to do such calculations in real-world situations (unless you simply guess a value for that component of Ze !).

Kind Regards, John

 

[EFLImpudence]

Are you saying that although this 0.19Ω should be included for Ze and Zs, it should not be included in R1 for the Volt-drop calculations?

 

[JohnW2]

Are you saying that although this 0.19Ω should be included for Ze and Zs, it should not be included in R1 for the Volt-drop calculations?

More-or-less. As I've said, it is part of Ze and Zs (and therefore is implicitly used for any calculations involving Ze or Zs), but not a part that one can measure separately (one can only measure the total).

'Voltage drop calculations' as usually done, relate only to voltage drop within an installation (i.e. from the 'origin' of the installation), and therefore would certainly not involve the "0.19Ω" part of Ze, any more than it would involve the part of Ze (0.50Ω in the GN8 example) attributable to the supply cables upstream of the installation. VD within an installation (which is what one usually measures/estimates" is entirely due to the "R1+R2" within the installation, and does not involve any components of Ze.

Indeed, even if one wanted to determine the voltage drop from "the terminals of the DNO's transformer" one would still not use the "0.19Ω"m, since that is within the transformer and therefore does not result in any voltage drop external to the transformer. The "0.19Ω" would only be relevant if one wanted to determine the reduction in voltage (due to VD within the transformer) at the transformer terminals under load.

I'll be responding fairly soon to studentspark's ongoing questions!

Kind Regards, John