Supplementary Bonding Revisited

[studentspark]

To illustrate with some figures, if one assumes that the path to earth from ECP had the same impedance (0.175Ω) as your R2e, the two of those in parallel would have a combined impedance of 0.0875. Total fault loop impedance would then be 0.4975Ω (0.175Ω + 0.09Ω +0.145Ω +0.0875Ω), hence fault current about 462A (230/0.4975), and hence the 'touch voltage' (voltage across R2) would be about 67V (462 x 0.145) - which, co-incidentally, is close to the figure you got.

Thanks John I reduced the R2e as the parallel earth paths gave a combined impedance of 0.0875Ω.

Yes I am confusing and complicating things...

I will have a go at the Ze of 0.8

I think I am mixing up the job of the cpc and the bonding.
That fig 5.3 in guidance note 8 is causing me confusion as the external R2 is removed from the calculation when bonding is added

 

[JohnW2]

Thanks John I reduced the R2e as the parallel earth paths gave a combined impedance of 0.0875Ω.

Do you mean the R2e in parallel with the ECP's 'path to earth'? If so, as I said, unless the latter is very low, the 'parallel paths' will not reduce the total significantly below the 'actual R2e' figure.

...Yes I am confusing and complicating things... I will have a go at the Ze of 0.8

That will be interesting, and maybe 'educational'!

Kind Regards, John

 

[studentspark]

OK. This is a simplified version

So with bonding we have a touch voltage below 50V.
This is because we are not using the external R2 in out calculations.
Why the external R2 is left out confuses me, but I think this is why...

So we are not talking about EFLI and low impedances here, i.e getting a low resistance path to enable the OPD to operate in the correct time.

This is about trying to prevent as much as possible different potentials.

So the voltage on the Extraneous CP would be considered at earth potential (0V)
If a fault were to put a voltage on the Exposed CP . You would have a large PD between the two.

Now we have not connected that Extraneous CP to make it a more conductive path to clear a fault.
We have connected it because we want to try and equalise the difference between the Exposed CP and the Extraneous CP.

So a proportion of the fault current at the Exposed CP will ( resistance depending) appear at the MET.
Which will pass that voltage on to the Extraneous CP.

This will mean that the the Exposed and the Extraneous CP, although at a different potential,
that potential difference is not so large, and shock risk is reduced.

What I still need to clear in my mind is what is the Voltage at the fault.

This is again what I think is happening..

The voltage 'left' after passing through R1e and R1 ( I work out at 121.12v) appears on the Exposed CP

The voltage on R2 appears ( again resistance depending ) on the MET but it has it has dropped another 32.22v
putting 88.88v onto the MET, and Extraneous CP.

This rises the voltage on the Extraneous CP to 88v.

The potential difference is the 121.12v minus the 88.88v which works out at 32.222v

(Hurray) Possibly... But I have another question which Im not quite sure how to word yet...

 

[studentspark]

Yes. With no SB present, all of the extraneous-c-ps, and all exposed-c-ps other than those on the circuit which has a fault, should be at MET potential (even during a fault).

Oh dear Im back to Square 1

 

[JohnW2]

OK. This is a simplified version .... So with bonding we have a touch voltage below 50V.

Indeed - as I predicted

This is because we are not using the external R2 in out calculations. Why the external R2 is left out confuses me ...

I don't understand what you mean. Are you referring to the 'earth side' of the Ze, what you called R2e in previous diagrams? If so, then you are "using" that in your calculations, since it is part of the total loop impedance which determines the fault current - and, in turn, that fault current multiplied by the R2 (within the final circuit) gives the 'touch voltage'.

... So we are not talking about EFLI and low impedances here, i.e getting a low resistance path to enable the OPD to operate in the correct time. This is about trying to prevent as much as possible different potentials.

Indeed so - that's why, in terms of terminology, it is bonding, not 'earthing'.

As for the rest of what you write,I think you are probably making things unnecessarily complicated. It's really very simple, and I will try to help you a best as I can! ....

So the voltage on the Extraneous CP would be considered at earth potential (0V)
If a fault were to put a voltage on the Exposed CP . You would have a large PD between the two.

That would be true without bonding. With bonding, the potential of the Extraneous-c-p is not at earth potential but, rather, is raised to the potential of the MET (which will be considerably above earth potential during the fault). The potential difference between extraneous- and exposed-cps during the fault (i.e. the 'touch voltage') is therefore a lot lower than it would be if the extraneous-c-p were at earth potential (i.e. without bonding).

Now we have not connected that Extraneous CP to make it a more conductive path to clear a fault. We have connected it because we want to try and equalise the difference between the Exposed CP and the Extraneous CP.

As above, agreed - which is why we are talking about bonding, not 'earthing'.

So a proportion of the fault current at the Exposed CP will ( resistance depending) appear at the MET. Which will pass that voltage on to the Extraneous CP.

I'm not sure what you are saying here. Unless the impedance to earth of the extraneous-c-p is very low (in comparison with R2+R2e), nearly all of the fault current will flow through R2+R2e, via the MET (and relatively little will flow through the extraneous-c-p)

This will mean that the the Exposed and the Extraneous CP, although at a different potential,
that potential difference is not so large, and shock risk is reduced.

That is, indeed, the whole point of the bonding.

What I still need to clear in my mind is what is the Voltage at the fault. ... This is again what I think is happening.. The voltage 'left' after passing through R1e and R1 ( I work out at 121.12v) appears on the Exposed CP

Exactly - it's as simple as that. The potential of the exposed-c-p (relative to earth) will simply be the potential at the transformer (relative to earth) MINUS the voltage drop through R1+R1e.

With bonding present, the extraneous-c-p will be at MET potential, so the 'touch voltage' will simply be the difference between the potential (relative to earth) at the exposed-c-p (as above) MINUS the potential of the MET (relative to earth).

It really is pretty simple!

However, as you have illustrated with your latest calculations, this does not alter my point that it seems as if, with perfectly credible figures for R1 and R2 (within the final circuit), it is probably impossible to get touch voltage' below 50V unless the total Ze is above some figure (about 0.45Ω with the R1+R2 you've been using) - which I think might come as a surprise to many people..

(Hurray) Possibly... But I have another question which Im not quite sure how to word yet...

Fair enough. Let me know when you've worked out how to word it!

In the meantime, I hope the above helps a bit!

Kind Regards, John

 

[JohnW2]

Oh dear Im back to Square 1

You seem to have jumped back to the Supplementary Bonding situation. I would suggest that you read what I have just written and try to get a clear understanding about main bonding ("MPB") before we move back onto the year-old discussion about SB!

Once you fully understand MPB, moving from that to SB ought not to be too much of a challenge..

Kind Regards, John

 

[studentspark]

Thanks again for you help John.
Re. MPB I think I've made very hard work trying to understand something I basically understood, Well the concept of it anyway...

The better understanding of the actual volt drops will be very helpful going forward. I've never tried to calculate these things before, you just install the 10mm...So thank you for that. I have a better understanding of touch voltage, and MPB as a result.
But like you I found the 50v scenario difficult to understand as mentioned in this thread I posted some time ago..(I'll have to re read it.) But I didn't have the basics down, to understand.
https://www.diynot.com/diy/threads/touch-voltage.538485/

I put it down to not really being able to grasp Ohms law

The relationship between current, voltage and resistance is expressed by Ohm's Law. This states that the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit, provided the temperature remains constant.
I get it, but like a double negative, I struggle to visualise it.

Anyhow I'm waffling on


Oh I just found another thread of yours SP conclusions, will have a read through that

 

[JohnW2]

Yes. With no SB present, all of the extraneous-c-ps, and all exposed-c-ps other than those on the circuit which has a fault, should be at MET potential (even during a fault).

Oh dear Im back to Square 1

I'm not sure why you're having a problem with that - since, again, it's very simple ...

You presumably understand that if no current is flowing through a conductor, then there is no 'voltage drop' along it's length, so that the potential (relative to earth or anything else) at both ends will be the same?

All exposed-c-ps are connected to the MET via CPCs and all (almost all) extraneous-c-ps are connected to MET via bonding conductors. In the absence of any faults, no current flows through any of those 'protective conductors', hence, as above, all exposed- and extraneous-c-ps are at MET potential.

In the absence of SB, most of that remains true even during a fault - all extraneous-c-ps and most exposed CPCs in the installation remain at MET potential, for the above reason. The only exceptions are the exposed-c-p(s) of at least some items on the final circuit which has an L-CPC fault, since considerable (fault) current will then flow through the CPC to the MET, resulting in a voltage drop in that CPC between the fault and the MET, hence raising the potential of the affected exposed-c-p to appreciably above MET potential.

As I've said before, nothing other than 'local' SB can alter that situation.

Does that make any sense?

Kind Regards, John

 

[studentspark]

I'm not sure why you're having a problem with that - since, again, it's very simple ...

You presumably understand that if no current is flowing through a conductor, then there is no 'voltage drop' along it's length, so that the potential (relative to earth or anything else) at both ends will be the same?

Well no...and well yes kind of.

As in say a radial socket circuit. Nothing is flowing if no load attached. You calculate the volt drop based on a expected load. When unloaded, the conductors just sit there waiting to be called into action ( sorry for the not technical terms) There is a potential for them to do some work, if a resistance is put across live and neutral, or in a fault condition, earth. Then current will start to flow.

And I can see that a voltage may appear on the MET, but not on the connected CPC of another circuit, because there is no load, to call on that current to flow.

 

[JohnW2]

Well no...and well yes kind of. .... As in say a radial socket circuit. Nothing is flowing if no load attached. You calculate the volt drop based on a expected load. When unloaded, the conductors just sit there waiting to be called into action ....

Quite so. When there is no load on, say, a radial circuit, there is no current flowing, hence no voltage drop in any of the conductors - so, no mater how long the circuit (it could be 'miles') the potential of the L conductor at the far end of the circuits will be at exactly the same potentiakl as that where it is connected to the CU.

When you undertake a 'VD calculation' for such a circuit, you are not calculating what the VD actually "is" (since, as above, it could be zero if there is no load) but, rather, what the VD would be if the circuit were fully loaded (i.e. carrying the maximum design current).

.... And I can see that a voltage may appear on the MET, but not on the connected CPC of another circuit, because there is no load, to call on that current to flow.

You've got that the wrong way around. Since there will be no current flowing in the CPC of the 'other circuit' (without a fault), there will be no VD across the length of that other CPC, so that the potential at the end of it will be exactly the same as the potential of the MET.

Of course, that ceases to be true if one does install SB, since the SB will then connect together all of the CPCs (even of different circuits) in the room/location in question.

Kind Regards, John

 

[studentspark]

Excellent, some good learning going on here. Really appreciate your patience, and time.

 

[JohnW2]

Excellent, some good learning going on here. Really appreciate your patience, and time.

You're welcome. I'll carry on trying to help as much as I can - just ask the questions! ... and don't ever worry about 'feeling stupid' - we've all been there and realise how 'stupid we can look when we haven't grasped something that someone else may think is 'obvious'!

Kind Regards, John

 

[EFLImpudence]

John: Is this what is confusing Studentspark - or am I the confused one?

In this scenario:

Isn't the situation without bonding that the 121.11V is the product of 230x0.545/1.035 (rather than 222.22x0.545) with the person merely receiving a shock of 121.11V while providing a path to earth via the extr-c-p and not actually dependent on the fault current?
This would be similar to the person simply touching the expo-c-p while standing on a well earthed floor.
That the product (121.11) is the same as 222.22x0.545 being an inevitable coincidence of the arithmetic.

Whereas with bonding the person is only subject to the voltage drop of 32.22V between expo-c-p and MET because of the current and impedance between them (222.22x0.145) - the extr-c-p and bond merely being (regarded as) a negligible impedance method of touching the MET.

 

[JohnW2]

John: Is this what is confusing Studentspark - or am I the confused one? ... In this scenario:
... Isn't the situation without bonding that the 121.11V is the product of 230x0.545/1.035 ....

Indeed so, but ...

... (rather than 222.22x0.545) ....

I'm not sure what you mesn by "rather than" since, as you go on to admit, it's exactly the same. Below you call it an "inevitable consequence", but it's really just a matter of very simple arithmetic - nothing to do with 'coincidences'! There are several ways one can look at it (and get the same answer) but probably the simplest is...

as you say ... touch voltage = voltage across (R2+R2e) = fault current x (R2+R2e) = fault current x 0.545 ....equation 1

However, fault current = 230 / (R2+R2e+R1+R1e) = 230 / 1.035 = 222.22 V

Hence, putting that value for fault current into equation 1, you get ....

touch voltage = 222.22 x 0.545

... so, the two things are identical not due to any 'coincidence' but because of very simple arithmetic.

... with the person merely receiving a shock of 121.11V while providing a path to earth via the extr-c-p and not actually dependent on the fault current?

I don't understand what you are suggesting. The magnitude of the shock is totally dependent upon the fault current [since its magnitude is fault current multiplied by (R2+Re)]. Edit: for example, if If were to change, due to a change in R1 and/or R1e, then the touch voltage would become something other than ~121 V.

This would be similar to the person simply touching the expo-c-p while standing on a well earthed floor.

Well yes, of course, since in the absence of bonding there's no difference between an extraneous-c-p and a 'well-earthed floor' In fact, strictly speaking, I suppose that a 'well-earthed floor' actually IS an 'extraneous-c-p',

That the product (121.11) is the same as 222.22x0.545 being an inevitable coincidence of the arithmetic.

As above, it is, indeed, inevitably the same - but as a consequence of simple (Ohms Law-based) arithmetic, not any 'coincidence'.

Whereas with bonding the person is only subject to the voltage drop of 32.22V between expo-c-p and MET because of the current and impedance between them (222.22x0.145) - the extr-c-p and bond merely being (regarded as) a negligible impedance method of touching the MET.

Quite so. Are you implying that anyone has suggested otherwise?

Perhaps the simplest way of thinking about is is probably that without bonding, the touch voltage is If x (R2+R2e), whereas with bonding it is just If x R2 - since, in the latter case (with bonding), the potential of the extraneous-c-p is essentially that of the MET (regardless of what potential that may be above true earth, which is determined by If flowing through R2e).

Kinde Regards, John

 

[EFLImpudence]

What I am trying to do is help with where I think Studentspark's confusion lies.

I am trying to distinguish between just getting a shock when a person is only earthed by the ground/floor and the situation when a person touches a live expo-c-p and another item which is also connected to the MET where bonding might help.