For techies - Potential Flaw in Zs/ADS calculations?

[John D v2.0]

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Me: And I think you are absolutely right that the PFC measured as per the regs isn't the same as the actual PFC, even at the exact time of measuring. This is because firstly the calculation assumes all other loads will still take the same current even during a fault on your installation that would reduce the voltage at other loads ...
JohnW2: No - both my simulations and attempts at mathematical modelling take that into account (see my posted simulations).
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Sorry - "the calculation" I meant as per the regs, your model is clearly far more comprehensive! (although perhaps difficult for any electrician to work out on site.)

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Me: Since we don't know the voltage at the transformer and we don't know the effective impedance of all the other loads, we just have to assume the voltage is sufficient that given the impedence of the other loads, the voltage at the origin is 0.95 of nominal. And we can know that as we pull more current, less volt drop will be caused by other loads on the system so they will be less important. So call it a fudge factor!
JohnW2: That's not quite how I see it. I think the intended purpose of Cmin (currently 0.95) is simply to ensure that calculations remain valid even if supply voltage (without faults) is 0.95 of nominal (as I've said, that would much more logically be 0.94, since it would then correspond with the lowest permissible supply voltage!)
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Sorry I was purely making a statement of fact that as far as the regs are concerned, we effectively "just have to assume" that in all conditions when non fault origin voltage is 0.95 of nominal, the voltage under fault conditions is sufficient [to trip the protective device]. And then I went on to say that's it's not all as bad as we think due to the effect of other loads.

I wasn't saying it makes sense, or that the people who made the regs had the right thing in mind when they thought about it!

 

[John D v2.0]

I can't find V(h), is that V(s) really? And I(o) is already calculated, should that be I(h) ?

 

[John D v2.0]

your diagrams look spot on to me for TN-S-C, apart from the above possible typos. And good assumption for simplicity there. I haven't looked at your data tables at this stage, but I should have time later.

 

[JohnW2]

View attachment 108785I can't find V(h), is that V(s) really? And I(o) is already calculated, should that be I(h) ?

Yes, sorry, that's just a (multiple!) typo. It should read:
I(h) = V(s) / R(h)

Kind Regards, John

 

[JohnW2]

Sorry - "the calculation" I meant as per the regs, your model is clearly far more comprehensive! (although perhaps difficult for any electrician to work out on site.)

Yes, but my point was that even with my "more comprehensive model", the simulations still show a (albeit small) difference between fault current with the two methods of calculation, whenever there is a load (in the absence of any load {my first simulation}, the two methods give identical fault currents). I take that to mean that I (i.e. my model) is therefore overlooking some relevant factor.

Sorry I was purely making a statement of fact that as far as the regs are concerned, we effectively "just have to assume" that in all conditions when non fault origin voltage is 0.95 of nominal, the voltage under fault conditions is sufficient [to trip the protective device].

The simulations certainly seem to confirm that 'calculating' PFC by dividing supply voltage (with no fault) by Zs results in an almost correct answer. Pre-Amd3, the regs assumed that the "supply voltage with (no fault)" was 'nominal' (i.e. 230V), although it would be perfectly permissible for the supply to be as low as 216.2V. By introducing Cmin, with a value of 0.95, Amd3 ensured that the calculation would give an 'almost correct' answer if supply voltage (without fault) was as low as 218.5 - but that still leaves any people with supply voltages between 216.2V and 218.5V with a slight problem (which would not have existed with Cmin=0.94!).

Kind Regards, John

 

[John D v2.0]

I had a quick look at your tables and I don't see anything unexplained - you are comparing the regs Zs with the actual fault current, which we know the regs Zs uses a flawed (but conservative under "other load") model. That looks to be exactly what your tables have shown.

 

[JohnW2]

I had a quick look at your tables and I don't see anything unexplained - you are comparing the regs Zs with the actual fault current, which we know the regs Zs uses a flawed (but conservative under "other load") model. That looks to be exactly what your tables have shown.

In general terms, I agree. The thing which remains 'unexplained' (or, rather, not yet understood by me) is:

Why the regs' approach (dividing no-fault supply voltage by Zs) produces very-nearly-correct figures for fault current
AND/OR
Why that approach does not produce exactly correct figures for fault current

You may know/understand the answers to those question, but I don't .... yet!

Kind Regards, John

 

[JohnW2]

I've just noticed that the fifth (last) of the simulation tables I posted earlier was actually a repeat of the fourth one. The fifth one should have been this:

Kind Regards, John

 

[John D v2.0]

In general terms, I agree. The thing which remains 'unexplained' (or, rather, not yet understood by me) is:

Why the regs' approach (dividing no-fault supply voltage by Zs) produces very-nearly-correct figures for fault current
AND/OR
Why that approach does not produce exactly correct figures for fault current

You may know/understand the answers to those question, but I don't .... yet!

Kind Regards, John

In both calculations you are using I=V/R

In the accurate calculations you are:
1 using transformer voltage rather than local voltage: effect=higher fault current than using regs standard
2 taking account of the exact "pull down" of other loads on the eventual fault current: effect=lower fault current than using regs standard

Now in the situation where local voltage is the transformer voltage, and there is no pull down from other loads, you got a zero difference. However as soon as they differ, you get a discrepancy because the "pull down" effect (that was automatically taken into account by point 1) gets smaller when you are causing a voltage drop for the other consumers with your fault.

Regarding the 0.94 or 0.95 thing, that is just covering the situation as the other load varies over time, but since that hasn't actually been taken into account with your models or the regs calculations detailed in your tables, we can consider that in a new thread or something!!

 

[JohnW2]

In both calculations you are using I=V/R

Ultimately, yes, but only directly, and straightforwardly, in the case of the "accurate" calculation. That is the simplest possible case of Ohm's Law. The supply voltage (during the fault) is simply connected across (R1+R2), and nothing else, so even a schoolchild could probably calculate the current through R1+R2.

The 'conventional' calculation is far less straightforward. One is estimating the fault current by dividing the supply voltage when there is no fault by Zs, which includes a component (Ze) which is not even within the installation which has that supply voltage. Mathematically, that approach will only give the exactly correct answer if the ratio of supply voltage during fault to supply voltage without fault is exactly the same as ratio of (R1+R2) to Ze, and will become increasingly inaccurate as the difference between those two ratios increases.

Now in the situation where local voltage is the transformer voltage, and there is no pull down from other loads, you got a zero difference. However as soon as they differ, you get a discrepancy because the "pull down" effect (that was automatically taken into account by point 1) gets smaller when you are causing a voltage drop for the other consumers with your fault.

Yes, conceptually/qualitatively, that is undoubtedly what is going on, but it is (the small extent of) the quantitative effect which has surprised me, and about which I have yet to convince myself. An ~1,500% increase in the load on the network only seems to increase the 'error' in PFC estimation from ~0.9% to ~8%. I'm going to look at bit further at these figures and make sure that they really do 'add up', and that it's only my intuition which is 'surprised'!

Regarding the 0.94 or 0.95 thing, that is just covering the situation as the other load varies over time, but since that hasn't actually been taken into account with your models or the regs calculations detailed in your tables, we can consider that in a new thread or something!!

The value of Cmin (whether 0.94, 0.95 or anything else) does not alter any of the calculations per se. All it alters (with the conventional approach) is the 'estimated' maximum Zs that will produce the required PFC for a guaranteed magnetic trip of the MCB. Since I was 'thinking pre-Amd3' in all of my tabulations I included (and highlighted) a row for exactly 230V supply voltage and exactly 160A fault current. However, I could just as easily have included and highlighted a row for 218.5V (230V x 0.95) and 160A - but my calculations would have been exactly the same.

Kind Regards, John

 

[JohnW2]

I think I'm now fairly convinced ....

Mathematically, that approach will only give the exactly correct answer if the ratio of supply voltage during fault to supply voltage without fault is exactly the same as ratio of (R1+R2) to Ze, and will become increasingly inaccurate as the difference between those two ratios increases. ........ it is (the small extent of) the quantitative effect which has surprised me, and about which I have yet to convince myself. An ~1,500% increase in the load on the network only seems to increase the 'error' in PFC estimation from ~0.9% to ~8%. I'm going to look at bit further at these figures and make sure that they really do 'add up', and that it's only my intuition which is 'surprised'!

See the simulation below. It's essentially the same as one you've already seen, but I've extended the range of the 'loads', and have added those two ratios I mentioned above - (R1/R2)/Zs and the (during fault)/(no fault) supply voltage ratios.

As you can see (and as we've seen and discussed before), when the total load on the network is extremely small (below about 3.3A) those two ratios are identical. (R1+R2)/Zs obviously remains constant (at 0.794), but as load increases the other ratio progressively increases a little, but even over a range of total loads from ~3.3A to ~836A, that ratio only increases (from 0.794) to 0.876, and over a more realistic range of loads, up to ~370A, it only rises to 0.830, at which point the error in underestimating true fault current is only ~4.3%.

It therefore seems that those ratios do, indeed, remain pretty close over a wide range of load currents, such that the 'conventional' method will give a pretty accurate estimate of true fault current (mainly with less error than likely measurement error of Zs) - and, as previously discussed (and explained), what small errors occur are all in the 'safe' direction (under-estimation of PFC).

Kind Regards, John

 

[JohnW2]

OK, I think I can now at least partially summarise:

The “true” fault current (PFC) is equal to the supply voltage at the premises divided by the (R1+R2) at the point of the fault (worst case = furthest point in circuit).

With TN-C-S and no loads (in target house or any other) on the network, then the ‘conventional (per regs) method of calculating/estimating PFC will give exactly the “true” answer. That is because, in this very simple situation, the only current which ever flows is the fault current. Hence, the no-fault supply voltage is equal to the transformer voltage (say Vt), and the during-fault supply voltage is equal to:

Vt x (R1+R2)/Zs

The “true” fault current is therefore

[ Vt x (R1+R2)/Zs ] / (R1+R2)

= Vt/Zs

Since the no-fault supply voltage is equal to Vt, the ‘conventional’ calculation of fault current will also give PFC as Vt/Vs - which is the “true” value.

When there is any load on the network the supply voltage (with or without fault) will be lower than Vt, and that results in the estimate of PFC obtained by ‘conventional’ calculation being slightly lower than the true figure (i.e. error is in ‘safe’ direction), increasingly so as the load increases.

I need to think and play a bit more to satisfy myself as to why these errors are always pretty small, even at very high loads, but the fact is that such is the case.

With TN-S, if there are no loads, the situation is identical to that with TN-C-S. No current ever flows in the neutral feed, so it can be ignored, and Zs relates to the only loop which carries current. Conventionally-estimated PFC is therefore again exactly equal to the true figure. If there is some load, then the conventional calculation will again result in an underestimate of true PFC. However, since the loop determining the supply voltage is now different from (and usually of lower impedance than) the loop carrying the fault current, the situation is more complicated than in the TN-C-S case - and, without checking, I’m not sure whether the ‘errors’ would be greater of smaller (for a given load) than in the TN-C-S case (I suspect greater).

With “pure TT” (i.e. the TT electrode represents the only path from MET to true earth), there is really no hope of the PFC being high enough for ADS (even with a B6 MCB), so the question is moot (but mathematically analogous with TN-S).

I would add that I make no apology for having raised this issue because the reason why the conventional method of estimating PFC gives ‘almost correct’ answers is, I would say, far from obvious when one starts thinking about it. Indeed, I strongly suspect that nearly everyone ‘thinks’ (because they have not thought deeply enough!) that it is just a simply matter of “Ohm’s Law”, and that the figures obtained are exactly equal to the true PFC.

Kind Regards, John

 

[ericmark]

Thank you for your work. I would assume the 95% was to cater for lower voltage at the supply transformer. Before the feed in if the transformer is set to highest permitted voltage then one has quite a leeway before volt drop reaches the lowest permitted voltage. However this would cause the inverters to lock out on over voltage, so today the transformer needs to be set to a lower voltage to allow for the raise when inverters are feeding in. I note at home even after the official change in voltage from 240 to 230 the volts were above 240 most of the time. Then we got a load of solar panels fitted and my voltage dropped, in the main now 230 volt. Also noted fluorescent light slower to fire up.

So although it was stated that the 95% was to allow for volt drop, really I think it was to allow lower tapping to be used so power could be fed into the grid.

 

[JohnW2]

Thank you for your work. I would assume the 95% was to cater for lower voltage at the supply transformer. ... So although it was stated that the 95% was to allow for volt drop, really I think it was to allow lower tapping to be used so power could be fed into the grid.

You're welcome (it was more for myself than anyone else!) - but I personally very much doubt what you are suggesting. Rather, I think it was probably done to correct ('at long last'!) an almost ridiculous anomaly that had existed for decades....

If one ignores the very slight underestimation of PFC which has been discussed in this thread, prior to Amd3 a circuit which, in good faith (per BS7671), had be designed to have just a low enough Zs for ADS to 'work' would actually have failed to provide a magnetic trip (hence the required disconnection time) for any consumer who, for whatever reason, had a supply voltage between 216.2V and 230V.

With Amd3 calculations, at least the required disconnection times are now guaranteed down to at least 218.5V supply voltage, and with the 'slight underestimation of PFC' discussed in this thread, very probably all the way down to 216.2V.

Kind Regards, John

 

[John D v2.0]

2 points, one, I think with tn-s the situation would be better, as only the L has a load on it, the E is in the no load situation. Of course the possibly higher impedence of the E than the N is taken into account by the calculations.

The other point about why the error is so small, i think it's because (or maybe the reason why) the permitted voltage drop is small. If it helps, consider a situation where the transformer voltage is 460v and the supply voltage is 230v due to other loads.