Supplementary Bonding Revisited

[EFLImpudence]

Right, John, I do understand what YOU are saying but it goes against everything I have thought and read (for ever?).
I shall be heartened by the fact that you have only just thought of it.

That you used the ratio between R1 and R2 (rather than the actual resistance values) for the calculations obviously indicates that it does not matter how long - or short - the cable is, albeit the shorter the cable the higher the fault current but the relative calculations would result in the same voltages.


Another thing I have now thought of is the supposed standard thinking that during a fault all earthed and bonded parts will become 'live'.
This cannot be so either, can it? Because - with no SB present - all other exposed-c-ps (except the one at the fault and others on the same circuit) and extraneous-c-ps will be in the same situation as the extraneous-c-p in our diagram.

So, if SB is required then it is always required no matter what 50/Ia results in.

Apart from 50V being considered 'safe' where did it come from, and to what does it apply, as far as the formula R<50/Ia is concerned when R in that situation is irrelevant?

I think.

 

[JohnW2]

Right, John, I do understand what YOU are saying but it goes against everything I have thought and read (for ever?). I shall be heartened by the fact that you have only just thought of it.

Fair enough, and I'm pleased to hear that you understand what I have been saying. If it's any consolation to you (as you seem to imply it is), as I said at the very start of this thread, I've been 'looking at' these things for many years, and have never felt the need to think about them particularly deeply until something I very recently wrote in another thread stimulated me to do that 'thinking'.

I presume that much of the reason that I have previously "never felt the need to think about them particularly deeply" is that I have probably always 'assumed' (always dangerous!) that, since practices had been established, and regulations written, over many years/decades, by a lot of people who were much cleverer than me (many of them 'experts'), and certainly more-specifically 'trained' in the discipline than me, that these practices/regs had to be correct, and to make sense.

However, please understand that I am probably little more comfortable than you are. Although I am as sure as I can be that everything I've thought and written is correct, in view of what I say about all the 'experts', I have to wonder whether I am, nevertheless, in some way 'making a fool of myself'! In that respect, it would be reassuring (or otherwise ) if some others provided some input to this discussion.

That you used the ratio between R1 and R2 (rather than the actual resistance values) for the calculations obviously indicates that it does not matter how long - or short - the cable is, albeit the shorter the cable the higher the fault current but the relative calculations would result in the same voltages.

That's true - one can work it all out with just resistances and currents (for a particular cable (I'll illustrate later) but it's simpler (although mathematically identical) to just work with the R2/R1 ratio - since that will apply for any length of any cable, without having to do the specific calculations.

Another thing I have now thought of is the supposed standard thinking that during a fault all earthed and bonded parts will become 'live'. .... This cannot be so either, can it? Because - with no SB present - all other exposed-c-ps (except the one at the fault and others on the same circuit) and extraneous-c-ps will be in the same situation as the extraneous-c-p in our diagram.

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).

If 'full SB' is present (connecting all extraneous- and exposed-c-ps together with 'bonds'), then it becomes true that all of those parts (extraneous and exposed) will become 'live' in the event of a single fault - but only 'live' in the sense that they will all be appreciably above MET potential. However, that "does not matter" - this is a good illustration of the purpose of bonding and creating an equipotential zone. Provided that all simultaneously touchable conductors are at the same potential, it is 'safe', regardless of whether that 'same potential' is MET potential or (for the purpose of illustration!) 1,000V above MET potential.

However, although it doesn't really 'help' (since, as above, the actual potential of an equipotential zone does not matter), another consequence of having 'full SB' is that, by creating multiple additional paths to the MET in parallel with the CPC of the circuit with a fault, the effect of instally SB is that the "R2" of the R2/R1 ratio will considerably decrease, whereas the "R1" (L-conductor of circuit with the fault) remains unchanged. The effective R2/R1 ratio, hence the magnitude of the pd between the exposed-c-p (with fault) and MET will therefore be appreciably decreased with 'full SB' as compared with the situation without any SB.

This does, however, remind me of another thing I have come to realise for the first time (and mentioned previously) - namely that if one connects ('bonds') one extraneous-c-p to one (or more) exposed-c-p(s), then I think one must extend that bonding to all extraneous-c-ps and exposed-c-ps in the room - otherwise, in the event of a fault, one of the extraneous-c-ps (the 'bonded' one) could come to be at considerably above MET potential, whereas the other (potentially simultaneously touchable) ones would still be AT MET potential.

So, if SB is required then it is always required no matter what 50/Ia results in.

It depends on what you mean by "if SB is required" - are you perhaps talking about the regulatory requirement (i.e. 'always required', in a bathroom, unless the conditions for omission are satisfied)?

However, as I hope I have successfully illustrated (and unless there is some flaw in my reasoning), if one's aim is to avoid the possibility of dangerous potential differences arising between two touchable parts (extraneous- and/or exposed-c-ps), then it would seem that ('full') SB is always required (i.e., if that is the aim, then it's not a matter of "IF required"), and in any room (regardless of 50/Ia). As you have said, I'm not convinced that a bathroom is really all that different from any other room in this regard - the only increased risk being due to the possibility of wet skin in someone who comes into simultaneous contact with two parts which are at appreciably different potential.

Apart from 50V being considered 'safe' where did it come from, and to what does it apply, as far as the formula R<50/Ia is concerned when R in that situation is irrelevant?

Although the same idea is perpetuated in our relatively relaxed attitude to ELV, I'm not even sure that I understand where "50V being considered safe" came from! It's far from uncommon for the resistance across the 'shock path' of a human body to be ≤1kΩ, which means that 50V will often be enough to result in a shock current appreciably above 30mA (which we seem to consider 'unsafe').

As for what on earth 415.2.2 is all about, or 'where it came from', that is MY question, so I can't provide an answer It would seem that the "R<50/1a" is pretty ridiculous when there there IS SB (since a metre or 5 of 4mm² G/Y is surely never going to fail that test?), yet totally inappropriate when there is not SB (since it then relates to a path through which a fault current is virtually never going to flow).

Kind Regards, John

 

[JohnW2]

That you used the ratio between R1 and R2 (rather than the actual resistance values) for the calculations obviously indicates that it does not matter how long - or short - the cable is, albeit the shorter the cable the higher the fault current but the relative calculations would result in the same voltages.

That's true - one can work it all out with just resistances and currents (for a particular cable (I'll illustrate later) but it's simpler (although mathematically identical) to just work with the R2/R1 ratio - since that will apply for any length of any cable, without having to do the specific calculations.

OK - and so to the “illustrations later” . Firstly, a couple of disclaimers:

• Firstly, some of the figures are not totally precise. Neither the ‘rounded’ VD figures for cables in BS7671 nor the more precise conductor resistance figures in the OSG correspond exactly pro-rata to the CSA’s of different-sized conductors. I have, for simplicity, used the ‘rounded’ BS7671 70°C figures - namely 22.0, 14.5 and 9.0 mΩ/metre for 1.0, 1.5 and 2.5 mm² conductors respectively.
• 
  
• It has been assumed that the L-conductor and CPC will be at the same temperature and, since (as below) it is only the ratio of the resistances of those two conductors that matters for this discussion, it does not matter what temperature is assumed, since any changes in temperature will affect the two equally, hence leaving the ratio unchanged. The standard BS7671-tabulated 70C° figures have been used. However, it seems unlikely that, during 'normal operation' (e.g. immediately before appearance of a fault) the CPC will be as hot as the L-conductor - so the calculations below might be over-estimating the resistance of the CPC
• 
  
• Secondly, again for simplicity, I have essentially assumed that Ze is ‘zero’ - or, at least, that the supply voltage at the origin of the installation is 230V during a fault. In practice, of course, voltage at the origin would decrease during the fault (because of VD across Ze). However, for the purpose of this discussion/illustration, all that matters is the voltage at the origin during the fault, regardless of what is happening external to the installation

So now consider the simple situation shown in this first diagram, in which a supply voltage of 230 is maintained whilst there is an L-E short at the end of 10m of 2.5/1.5 mm² cable.

The L-conductor has a resistance of 0.090Ω and the CPC has a resistance of 0.145Ω so a total path resistance of 0.235Ω. That means that with the voltage of 230V, the current is 979 A. That current flowing through the CPC (resistance of 0.145Ω) therefore results in a ‘voltage drop’ of about 142V (0.145 x 979). The voltage between the MET at the location of the fault (i.e. the exposed-c-p, at the other end of the CPC) is therefore about 142V.

If we change the length of the cable, say to 100m, then the resistances of L and CPC become 0.90Ω and 1.45Ω respectively, hence the total resistance 2.35Ω, the current with 230V is 97.9 A (230/2.35), and hence the voltage across the CPC (hence voltage from MET to exposed-c-p is, again, about 142V (1.45 x 97.9). We could similar do the same for various lengths of different-sized cables (using different R1:R2 ratio).

However, we don’t really need to calculate the current. If the fault current is If, then the voltage across the CPC (i.e. from MET to exposed-c-p) is (If x R2), but we know that (at 230V) If is equal to 230/(R1+R2) - so, taking those two together, we get:

Voltage = 230 x R2 / (R1+R2)​

… so the voltage depends just on the ratios of resistances, which remains the same for any length of a particular cable, without needing calculation of the current.

So, now turning to the real-world situation of interest, we get the following diagram. At first sight it looks complicated, but it’s really just the same as the above ‘simple one’ with the addition of a ‘shock victim’ (assumed to have a body resistance of 1kΩ), an extraneous-c-p to connect him/her to the MET (whilst also touching the ‘live’ exposed-c-p) and a fair bit of ‘annotation’.

It is hopefully fairly self-explanatory. The relatively very small current flowing through the victim (142 mA with my figures, if resistance of extraneous-c-p to MET is negligible) also flows through the L-conductor (R1), hence fractionally increases the VD across that conductor (hence fractionally reduces the potential of the exposed-c-p {above MET potential}) - but that is totally negligible - in my 10 metres of 2.5mm² cable example, it would result in the current increasing from 979 to 979.142 A.

The voltage across the victim is the potential of the exposed-c-p (above MET potential) minus the voltage drop across the extraneous-c-p, that VD being the current through the victim (~142mA) multiplied by the resistance (to MET) of the extraneous-c-p. If, as will usually be the case, that resistance is very low (a handful of ohms), then that VD will be ‘negligible’, hence the voltage across the victim will be more-or-less the ‘full’ 142V.

This is where we get to the important bits in terms of our discussion, when we start considering the resistance (to MET) of the extraneous-c-p. It should be apparent from the above that the higher the resistance from extraneous-c-p to MET, the lower will be both the voltage across the victim and the current through the victim.

If, for example, the resistance from extraneous-c-p to MET rises to just high enough to ‘fail’ the 701.415.2 / 415.2.2 test (i.e. 1,677Ω with an RCD), then the total resistance of the path through the victim becomes 2,677Ω, hence (compared with the situation in which extraneous-c-p has negligible resistance to MET) the current through the victim falls from 142 mA (142V / 1,000Ω) to only about 53 mA (142V / 2,677Ω), and the voltage across the victim hence falls from 142V to about 53V (53 mA through 1,000Ω).

So, that’s the anomaly I’ve been trying to highlight. When used as suggested in 701.415.2 (to determine whether SB can be omitted), the 415.2.2 test wants the resistance from extraneous-c-p to MET to be below a certain value (by implication, “the lower the better”), yet (in the fault scenario I am considering) the lower that resistance, the higher will be the voltage across a victim and the higher the current through that victim. Conversely, the higher the resistance from extraneous-c-p to MET, the ‘less severe’ will be the shock.

As I’ve said, I don’t know what, if anything, I’m missing and/or getting wrong - but if I’m getting it roughly right, then the regulations seemingly make no sense at all (suggesting that SB can be omitted when the magnitude of the possible shock is, if anything, the greatest) - and, as we have both therefore said, the only way to minimise the ‘touch voltage’ between extraneous-c-ps and exposed-c-ps during an L-E fault would seem to be to always install supplementary bonding, regardless of any measurements.

If anyone can find a flaw in my reasoning, please let me know - since I am far from comfortable with 'disagreeing with the regs' to this extent

I’ve quite probably ‘lost you’, quite possibly because my diagram looks much more complicated than it actually is, so please do ask me for whatever clarification or further explanations you may need!

Kind Regards, John

 

[JohnW2]

Damnit - this again You'll have to be patient waiting to see the rest of this fairly lengthy/detailed post! ...

 

[studentspark]

I've been reading and re reading this.. and the Guidance note 8 post a couple of weeks ago I posted, and the touch voltage post I posted as well.

I'm still confused.

Re the above. I have used you figures, but don't you need Ze as well, to get the fault current, to work out the volt drops.
How I read it is, you are using just R1 and R2.

Also should you include the voltage that will appear on the MET, which would also appear on the extraneous CP?
And this would reduce the potential difference between exposed CP and extraneous CP.
But still not to 50V

This is how MPB can reduce the touch voltage example 1 is with out bonding, 2 with bonding.
I haven't go on to supplementary bonding yet, as Im trying to find out if I have the maths correct for MPB.
As Voltage drop is also causing some confusion.

I have a diagram....

 

[JohnW2]

I've been reading and re reading this.. and the Guidance note 8 post a couple of weeks ago I posted, and the touch voltage post I posted as well. I'm still confused.

OK. I'll have a proper look later and see if I can help (but I'll also need to revisit what I wrote a year ago!) - but I am currently about to (almost literally) 'make hay whilst the sunshines' in my garden! However, in the meantime, for starters ...

Re the above. I have used you figures, but don't you need Ze as well, to get the fault current, to work out the volt drops. How I read it is, you are using just R1 and R2.

I think you are probably overlooking the pre-amble to my diagrams, which read ...

.... So now consider the simple situation shown in this first diagram, in which a supply voltage of 230 is maintained whilst there is an L-E short at the end of 10m of 2.5/1.5 mm² cable....

That was, as I said, a similification for the purpose of illustration. The actual 'supply voltage' (at origin of installation) will obviously depend upon the voltage at the DNO transformer and the voltage drop in the distribution cable due to the fault current flowing through the Ze. For simplicity, I was considering the situation in which the voltage at transformer, minus that VD during the fault, was 230V during the fault - which would obviously mean that the voltage at the transformer was appreciably higher than that.

So, regardless of what is going on external to the installation, IF the supply voltage to the installation were 230V during the fault then, per Ohm's Law, the fault current would simply be 230/(R1+R2).

In fact, with your figures, if one assumes a Ze of, say, 0.35Ω then a fault current of 979A would mean a VD in the distribution cable of about 343V, so if there were still 230V at the installation, the voltage at transformer would have to be about 573V - which is clearly ridiculous!

More later.

Kind Regards, John

 

[JohnW2]

I've been reading and re reading this.. and the Guidance note 8 post a couple of weeks ago I posted, and the touch voltage post I posted as well. .... I'm still confused. .... Re the above. I have used you figures, but don't you need Ze as well, to get the fault current, to work out the volt drops. ... How I read it is, you are using just R1 and R2.

I have already addressed that - by saying that I was talking about a simplified situation in which the supply voltage to installation was 230V during the fault, I was effectively taking Ze (and the VD across it) into account. In fact, I think this same issue goes a fair way to addressing the remainder of your points/questions....

Also should you include the voltage that will appear on the MET, which would also appear on the extraneous CP? And this would reduce the potential difference between exposed CP and extraneous CP. But still not to 50V .... This is how MPB can reduce the touch voltage example 1 is with out bonding, 2 with bonding. .... I have a diagram....

Your Diagram 1 (without MPB) and calculations are essentially correct (albeit I get fractionally different answers to my calculations), but probably not very realistic.

You appear to be assuming that the voltage at the transformer is 230V. I presume that your R1e and R2e are the two 'legs' of Ze (which will only be equal, as you have assumed, with TN-C-S) which means that, as you have said, there will be a VD of about 69V in each of those legs (hence about 138V VD in total), so that the 'supply voltage' (at the origin of the installation) would fall to about 92V during the fault - very different from my ('simplified') assumption of a supply voltage of 230V during the fault! However, that aside, your calculations are roughly correct. Since the ECP is assumed to be at true earth potential (since little/no current flowing through it), the 'touch voltage' will simply be the sum of the voltages across your R2 and R2e - which, as you say, would be around 125V (slightly higher by my calcs).

However, when one moves onto your Diagram 2 (with MPB), things start going a bit wrong, because you have assumed that the ECP has a zero impedance/resistance to earth.

It's obviously not a realistic real-world situation, but IF that impedance were zero, then all of the fault current would go through the ECP (totally 'bypassing' R2e), but the potential of the ECP (hence also potential of MET, if the resistance of the bonding conductor was 'negligible') would still remain at earth potential. However, the total fault loop impedance would then fall to 0.41Ω (0.175Ω + 0.09Ω +0.145Ω), so that the fault current (with 230V at the transformer, per your diagram) would become about 561A. The 'touch voltage' would then be simply the voltage across R2, namely about 81V (561 x 0.145).

In a real-world situation, the impedance/resistance of ECP to earth would obviously not be zero, in which case the fault current would be shared between the ECP and R2e. The potential of the ECP, hence also MET, would therefore rise to above earth potential. However, that would again result in some reduction in fault loop impedance (since R2e would now be in parallel with another path to earth, hence higher fault current, hence more VD across R2.

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.

More generally, as the impedance to earth of the ECP increases upwards from zero, the 'touch voltage' progressively decreases (from a maximum of about 81Ω 81V, when the ECP impedance is zero), and, with all your figures (including an assumption or 230V at the transformer) the touch voltage' approaches a minimum value of about 57Ω 57V when that impedance gets very high - per the graph below.

Assuming I've got it right, does any of the above make sense to you (or anyone else)??

Kind Regards, John
Edit: typos corrected

 

[JohnW2]

... More generally, as the impedance to earth of the ECP increases upwards from zero, the 'touch voltage' progressively decreases (from a maximum of about 81Ω, when the ECP impedance is zero), and, with all your figures (including an assumption or 230V at the transformer) the touch voltage' approaches a minimum value of about 57Ω when that impedance gets very high - per the graph below....

.... I could/should perhaps have elaborated .... the minimum 'touch voltage' of about 57V was a consequence of your particular figures - it would be different for different values of your R1 and/or R2 and/or Ze (your R1e + R2e). This graph shows (with your values of R1& R2, and 230V at the DNO transformer) how 'touch voltage' varies with the impedance of ECP to earth for various values of Ze (assumed equally split between both legs, as with TN-C-S) ...
[ Edit: I might have observed that, if I've got it right, using your figures/assumptions, touch voltages could only be <50V if Ze were above about 0.45Ω (and impedance to earth of ECP >1Ω or so). Having said that, 'your assumptions' clearly are not realistic, since I don't think that voltage at the transformer would ever be as low as 230V - probably usually at least 252V, I would guess ]...

Kind Regards, John

 

[boringoldcodger]

Deleted as no longer relevant

 

[JohnW2]

Apologies John, I've got my pedants hat on (yet again).

Not at all, it's good to have 'proof readers' around

You mean 81V and 57V (rather than Ω).

I did indeed- now corrected. As you presumably realise, I don't usually bother inserting the symbols whilst I'm typing, but go through and do them all at the end - but got a bit too trigger happy on this occasion. Thanks for noticing!

Kind Regards, John

 

[studentspark]

Thanks very much for this John...I Think i am starting to get somewhere, then I lose it all again.
Here are a couple more diagrams.

So in the first diagram, Its the figures you used.
The Black type is the resistance
The Orange is the volt drop
The red is the voltage appearing at each point, deducting as I go around I should end up with 0 volts (I got 0.15v, I presume thats just then maths)

So we have a touch voltage of 67v
Which is the potential difference between the exposed CP and Extraneous CP / MET



This second diagram

I have increased the voltage of the transformer to compensate for the volt drop at the load.
So we have 230v at the load in a healthy circuit and 235.85 at the transformer.

I have a very low External R2 which is caused by Pipe.
Now We are using this figure to get the EFLI, to get the PFC.

But the Pipe apparently being so low in resistance would take the most of the fault current. But that is presuming it has a return to the transformer of 0.05Ω, which it clearly will not have, as the mass of earth is the return path. A figure of 200Ω might be more reasonable.
(Which would give us a ZS of 0.82Ω and a fault current of 287.62A and a Touch voltage of around 86v)

But it does show how a low Ω return path increases the touch voltage.

Which I have at 101.74v

Again in orange I have indicated what I think the voltage will be at...

The service fuse 176.9 v
At the fault 116.26V
At the Met 14.72v
and back at the Transformer - 2.32 ( so some slightly out maths there !!)

But its that R1 section
0.302 Ω 101.74 Volts dropped. Which is the touch voltage.

So the potential difference is between the 116.26V at the Exposed conductive part
and the 14.72V that appears on the MET. ?
So if I put a voltage tester on those two points I would get 101.74V

But this can't be correct can it ??
As touch voltage has to be below 50V.

I don't see how a touch voltage of less than 50V can be achieved.

Thanks...

 

[Swwils]

Isn't it all derived from iec 479-1?

 

[EFLImpudence]

Just a small point if I may:

Now We are using this figure to get the EFLI, to get the PFC.

That will be the PEFC.

 

[JohnW2]

Thanks very much for this John...I Think i am starting to get somewhere, then I lose it all again. Here are a couple more diagrams.

OK. I'll take it in stages - first in relation to your first diagram ....

So in the first diagram, Its the figures you used. ....

Sort of. You're still assuming that there is 230V at the transformer but you have, for some reason, halved the 'earth side' of the Ze (what you previously called R2e), from 0.175Ω to 0.0875Ω. That doesn't matter, so I'll go with your new figure.

The Black type is the resistance ... The Orange is the volt drop ... The red is the voltage appearing at each point, deducting as I go around ....

OK, I understand, although I do find your diagrams a little confusing. The red figures are, of course, voltages (potential differences) relative to earth.

... I should end up with 0 volts (I got 0.15v, I presume thats just then maths)

Indeed, that small discrepancy is just because you've been rounding some of the figures. The fault current (with your figures) is actually 462.311558..... V (rather than 462 V), and if you used that figure, and did your calculations to a reasonable number of decimal places, you'd end up with a figure (after all your subtractions) very close to zero.

So we have a touch voltage of 67v ... Which is the potential difference between the exposed CP and Extraneous CP / MET

Agreed, but you've made life a bit unnecessarily difficult for yourself, since the 'touch voltage' is simply the voltage across your "R2" (0.145Ω) - so, once you've calculated the fault current, you just need to multiply it by that 0.145Ω .... 0.145 x 462.311558...= 67.035... V.

I'll move to your second diagram in another post.

Kind Regards, John

 

[JohnW2]

This second diagram ... I have increased the voltage of the transformer to compensate for the volt drop at the load. So we have 230v at the load in a healthy circuit and 235.85 at the transformer.

Fair enough, but that will obviously still result in the 'supply' voltage (at the origin of the installation) being dramatically lower in the presence of the fault.

I think that your diagram, and thinking, is all a bit confused/confusing. I also think that, at least at this stage of the discussion, you are unnecessarily complicating things by introducing the 3kW 'healthy load'. Apart from anything else, under fault conditions there will be a very much reduced voltage across that load, so (assuming it is a 'simple', e.g. resistive, load), it will be drawing a lot less than 3kW during the fault.....

I have a very low External R2 which is caused by Pipe. ...

No - because, as you go on to say ...

But the Pipe apparently being so low in resistance would take the most of the fault current. But that is presuming it has a return to the transformer of 0.05Ω, which it clearly will not have, as the mass of earth is the return path. A figure of 200Ω might be more reasonable.

Exactly - although I think that 200Ω is pretty 'pessimistic' - a lot of the concerns about ECPs (particular in the face of TN-C-S 'lost neutrals') is in relation to ones which have 'very low' impedances to earth.

As you saw from my graphs, the touch voltage does alter significantly once the impedance to earth of the ECP gets much above 1Ω or 2Ω (let alone your 200Ω), since very little of the fault current then flows through the ECP.

Hence, the bottom line is therefore that if the path to earth from the ECP is more than "a few ohms", then one is essentially back to your Diagram 1 (albeit you reduced "R2e" from 0.175Ω to 0.0875Ω for that one, for some reason!).

.... But this can't be correct can it ?? ... As touch voltage has to be below 50V.

As I said, if my calculations and graphs are correct, then it's impossible (with your values of "R1" and "R2") to get 'touch voltage' below 50V unless total Ze (assumed equal in both legs) is above about 0.45Ω - why don't you try with a total Ze of, say 0.8Ω (0.4Ω + 0.4Ω) (which would be credible for TN-S)?

Kind Regards, John
Edit: clarification added (in red)