For techies - Potential Flaw in Zs/ADS calculations?

[JohnW2]

Oh originally I was talking about simultaneous faults on multiple circuits in the same property, I'm not sure who mentioned faults in other properties first. Probably just a general misunderstanding.

Well, that's even 'worse'! Given that, if everything is right, the first fault to occur should be cleared with 0.4 secs, then (except, perhaps, if the building were in flames!) we really would be in the realms of "vanishingly improbable" (sorry EFLI!) for a second L-E fault to arise in a different circuit in the same property within that 0.4sec window

And even that I only mentioned as the only situation I could possibly fabricate where the circuits might not disconnect in time due to voltage dropping to 150v or whatever the quoted figures were.

I don't think we really need to consider situations in which (in the absence of faults) supply voltage drops to below the lowest permissible (216.2V), since it would then be impossible to guarantee that an MCB would operate (magnetically). During the presence of an uncleared fault in a high-current circuit (say 32A or 40A), the supply voltage may well fall to extremely low levels (<100V), but that's a different matter.

Kind Regards, John

 

[John D v2.0]

As far as I can tell we're all entirely and firmly in agreement! I forgot what the remaining point for discussion is now, if any?

 

[JohnW2]

As far as I can tell we're all entirely and firmly in agreement! I forgot what the remaining point for discussion is now, if any?

The mathematical/physics issues I raised in post #35. If I can discipline myself to stop reading/posting messages, I'm going to get onto that

Kind Regards, John

 

[John D v2.0]

There is no doubt (unless I'm going mad!) that if one just looks at the installation in question in isolation, then:

Current through fault = supply voltage (at origin of the installation, during fault) / (R1+R2) [to the point to the fault]

(that is just Ohm's Law). However, we, the regs and probably the way in which meters measure Zs have always presumed that that:

Current through fault = supply voltage (at origin of the installation without fault) / Zs [i.e. (R1+R2)+Ze, to the point to the fault]

If both of those equations are correct, the right-hand sides of both must be equal. In other words, supply voltage during fault divided by (R1+R2) must be equal to supply voltage without fault divided by (R1+R2+Ze). I will attempt to prove that mathematically and, if I can, then try to work out what it means in terms of the physics.

Kind Regards, John

You're right they should be the same. I had a go and ended up with an extra factor of ze², but I am sure it's my mistake as I haven't done algebra in ages. I was trying to be wary of dividing both sides by zero and proving 1=2 but I think there another error.

 

[JohnW2]

You're right they should be the same. I had a go and ended up with an extra factor of ze², but I am sure it's my mistake as I haven't done algebra in ages. I was trying to be wary of dividing both sides by zero and proving 1=2 but I think there another error.

Join the club! The algebra I'm trying to do (a bit more complex than yours, since I'm considering a more general situation) is proving very tedious and complex. I can but ask you to continue 'watching this space' whilst I do battle with it!

Kind Regards, John

 

[John D v2.0]

Actually I now think the reason I couldn't prove it is because it doesn't hold. In short the situation is "better" safer than the meter makes out.

This is because when the fault happens, (margianlly) less current flows through the "other loads" (Zo) due to the voltage drop they see due to our fault. Therefore we get more current through our fault.

Since Ze would be small and Zo would be relatively large (i.e. the voltage without fault is 0.9 or whatever of the transformer), we would only have a small effect, but positive nonetheless. So maybe that covers the difference between the minimum 0.94 and the specified 0.95 factor!

 

[JohnW2]

Actually I now think the reason I couldn't prove it is because it doesn't hold. In short the situation is "better" safer than the meter makes out.

You may be right - I can't tell you yet. I think I may have to give up my attempt to algebraically model as 'general a situation (I have variable loads within the premises and in other premises, am splitting the external fault loop impedance into the 'local service' and 'common' parts etc.) - since, quite frankly, I haven't got big enough sheets of paper to get the algebra tidily on!

This is because when the fault happens, (margianlly) less current flows through the "other loads" (Zo) due to the voltage drop they see due to our fault. Therefore we get more current through our fault.

My (yet to be solved!) model takes that into account, and it's not as 'marginal' an effect as you may think. Looking at one of my simulations (with loads and Ze chosen to give supply voltage of 230V and fault current of exactly 5In for a B32; loads assumed to be passive - i.e. of fixed impedance -and set for this example at 10Ω in target premises and 3.55Ω for the other premises):

Transformer output fixed at 250 V
Without a fault:
Load within premises: 23.00 A
Load in other premises: 65.49 A
Supply voltage at premises: 230.00
During fault:
Load within premises: 18.50 A
Load in other premises: 57.19 A
Supply voltage at premises: 185.00
Fault Current: 160.0 A

Hence during the 160A fault there is a fall of about 13% in load current in other premises, about an 20% fall in load current in the 'target' premises (and the supply voltage at the 'target' premises'). If one looks at a much higher fault current (say 10In), then the effects are even more dramatic.....

Transformer output fixed at 250 V
Without a fault:
Load within premises: 23.00 A
Load in other premises: 65.49 A
Supply voltage at premises: 230.00 V
During fault:
Load within premises: 14.00 A
Load in other premises: 48.89 A
Supply voltage at premises: 140.00 V
Fault Current: 320.0 A

In this case, during the 320A fault the load current in other premises falls by about 24%, and the load current in the target premises (and 'supply voltage' there) falls by about 39%.

Kind Regards, John

 

[John D v2.0]

I'm looking forward to the result! But perhaps stick to modelling two situations rather then everything!

 

[JohnW2]

I'm looking forward to the result! But perhaps stick to modelling two situations rather then everything!

The way things are going, I might well be forced to. The problem is that it some of the 'everything' (and I don't necessarily what!) may well affect the results. To take the most obvious, trivial, case, if I eliminate the loads in all premises, then the two expressions (no-fault supply voltage/Zs and during-fault supply voltage/(R1+R2) obviously do give the same result - but that's little more than Ohm's Law. I press on!

Kind Regards, John

 

[SimonH2]

Discussion in another thread has made me think of something that has never occurred to me before.

I'm glad it's not just me then, more or less the same thing had been niggling me for some time - I'd just been a bit too timid (and lately, too busy) to stick my head above the parapet with it

I was thinking about it from a different (pratical) direction, namely : when specifying the maximum value of R1+R2 to compare your actual readings with, do you take into account Zs, Ze, PFC etc; or do you just use fixed values for the type of fuse/breaker on the basis that Zs & Ze are << R1+R2 ? I can see that in some cases you need to take into account the supply where Zs&Ze are not so small (as in the boat shore supply example), but mostly I can't say I've seen examples of sparks doing this calculation.

 

[JohnW2]

I'm glad it's not just me then, more or less the same thing had been niggling me for some time - I'd just been a bit too timid (and lately, too busy) to stick my head above the parapet with it

Thanks. It's reassuring to know that I am not alone!

I was thinking about it from a different (pratical) direction, namely : when specifying the maximum value of R1+R2 to compare your actual readings with, do you take into account Zs, Ze, PFC etc; or do you just use fixed values for the type of fuse/breaker on the basis that Zs & Ze are << R1+R2 ?

The short answer is 'Yes' (i.e. all those things are, implicitly or explicitly, 'taken into account').

What one needs to ensure is that a L-CPC fault of negligible impedance at the furthest point of a circuit will result in a fault current (PFC) high enough to cause the OPD to operate within the required disconnection time (0.4sec for TN systems) - which, for a Type B MCB, means that the PFC must be at least 5 times the rating (In) of the MCB.

The fault current PFC will obviously be related to the total loop impedance of the fault (Zs, or EFLI) which, in turn is obviously equal to the sum of the internal and external parts of that loop - (R1+R2) and Ze respectively. Since we work with the tabulations of values of 'maximum Zs' (to achieve the required PFC), to calculate the maximum permissible (R1+R2) of a circuit, we therefore have to subtract Ze (measured or 'estimated'!) from the maximum permissible total Zs.

So far, so good - so there is no practical problem. My problem was theoretical. The implication of what we do (and the tabulated maximum Zs figures we use) is that the PFC is equal to the supply voltage at an installation (in the absence of a fault) divided by the total fault loop impedance (Zs). Whilst, at first sight, that sounds like no more than Ohm's Law, if one 'makes the mistake' of thinking too deeply, it becomes far less obvious why that calculation actually works. I'm not (any longer) suggesting that it doesn't work, but the reason it does is far from 'obvious' - given that one is dividing the voltage across one part of a circuit (when there is no fault) by the impedance of the whole circuit in order to determine the current in the whole circuit when there is a fault.

I think I'm getting quite close to 'working it out' - so, as I keep saying, 'watch this space'!

Kind Regards, John

 

[John D v2.0]

Excellent summary of the question there. 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 (this would actually cause the fault current to be bigger than calculated). And secondly because the loads at their installs might vary depending on time of day and the weather etc (hence the 0.95 figure being introduced)

If we really wanted to do it properly, as your first equation stated, the information we need to calculate the actual fault current would be the voltage during fault conditions at your incomer, and the R1+R2. To calculate the voltage during fault conditions we need to know the Ze, the voltage at the transformer, the effective impedance of all the other loads aggregated together*, and the R1+R2.

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!

* I guess this effective impedence is what you were trying to model, but I feel it could be modelled sufficiently by a simple voltage divider system?

 

[JohnW2]

I'm looking forward to the result! But perhaps stick to modelling two situations rather then everything!

OK, to hell with the algebra for the time being - for now, I’ll present some of my simulations!

In fact, if my simulations are correct (which I think they are - but will check again), then I think there must be some (very small) relevant factor missing from my attempts at algebraic modelling, since it appears from the simulations that there is a (very small) difference between PFC calculated in the two ways we have discussed. It is therefore probably the case that, unless I can refine my model, my algebra is never going to be able to prove that the two methods are exactly equivalent.

I hope the attached tabulations are readable, and also hope that they are essentially self-explanatory. Each table explores a different aspect (as described in the title) and seeks to compare the “actual” PFC (supply voltage during fault divided by R1+R2) with the PFC as normally calculated (Supply voltage without fault divided by Zs).

As you can see, no matter what scenario I consider, the convention calculation of PFC (using Zs) very slightly under-estimates the PFC (i.e. is very slightly ‘on the safe side’), the mall 'error' increasing with increasing load on the network. However, the differences are very small (generally not over 1%, except at very high load currents) and are well within the likely range of measurement error when measuring Zs.

Any thoughts?

Edit: I've just noticed that the fifth (last) of those tabulations is actually a copy of the fourth one. I have posted the proper fifth table below.
Kind Regards, John

 

[JohnW2]

Excellent summary of the question there. 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.

The simulations I've just posted seem consistent with that.

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 ...

No - both my simulations and attempts at mathematical modelling take that into account (see my posted simulations).

If we really wanted to do it properly, as your first equation stated, the information we need to calculate the actual fault current would be the voltage during fault conditions at your incomer, and the R1+R2.

Exactly - that's what I've called the "actual" fault current vin what I've just posted.

To calculate the voltage during fault conditions we need to know the Ze, the voltage at the transformer, the effective impedance of all the other loads aggregated together*, and the R1+R2.

Indeed - and both my simulations and modelling know all those things, and taken them all into account.

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!

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!)

I continue thinking about the reason for the very small difference between the two methods, and look forward to hearing your comments!

Kind Regards, John

 

[JohnW2]

So that you can see the basis of what I've simulated (as per tabulations posted above) and have attempted to model, the following couple of diagrams may be of assistance. Please do indicate if you can see anything wrong with it!!

Kind Regards, John