For techies - Potential Flaw in Zs/ADS calculations?

[John D v2.0]

Yes I think we're actually violently in agreement we just don't realise!
I was referring to the decrease in supply voltage due to other coincidental faults (or otherwise heavy load on the supply), as the decrease in supply voltage due to the fault in question is already built into the measurements.

 

[ericmark]

The PC with the software to simulate is under repair, so I can only look at likely results.
With a single short circuit and no other loads the voltage which is important is that at the supply transformer that's the easy bit. Voltage at the users consumer unit does not matter as the loop impedance used is total not just the bit within the users installation.

Where it gets harder is where other loads reduce the power which can be delivered to the short circuit. But since those other loads would be rather insignificant when compared with a short circuit the more I consider it, the more I am inclined to think voltage at the DNO head is not really that important, it is the voltage at the transformer which is the really important factor.

 

[John D v2.0]

The other loads should indeed be relatively insignificant (depending on your definition), their effect on the fault current are seen at the head as a decrease in voltage. In the regs they have defined the expected load will cause an voltage drop to 0.95 of nominal in the worst case.

In a highly extreme case of multiple faults, the 0.95 allowance wouldn't be enough, so you wouldn't have guaranteed disconnection in a compliant installation.

 

[JohnW2]

Yes I think we're actually violently in agreement we just don't realise!

You're probably right. My simulations have certainly confirmed that variations in local supply voltage, due to the load on the LV network, (without a change in voltage at the transformer) do change the fault current. However, simulations are all very well in indicating what happens, but they don't directly tell one why it is happening - so I'm going to get back to the algebra shortly!

I was referring to the decrease in supply voltage due to other coincidental faults (or otherwise heavy load on the supply) ....

I'm only really looking at changes in supply voltage due to loads on the LV network - coincidental multiple full-blown faults is such an incredibly unlikley scenario that I don't think it's worth considering as such. However, as far as the physics and maths are concerned, a 'fault' is (with TN-C-S supplies) not any from a very high (but not 'fault') load.

I haven't even thought about supplies other than TN-C-S yet. Things get appreciably more complicated with TN-S or TT, since the path taken by "L-E" fault current and the path taken by 'high loads' are different (as are the corresponding loop impedances), and partially physically separate.

Kind Regards, John

 

[JohnW2]

With a single short circuit and no other loads the voltage which is important is that at the supply transformer that's the easy bit. Voltage at the users consumer unit does not matter as the loop impedance used is total not just the bit within the users installation.

Indeed so - that was what started me thinking.

Where it gets harder is where other loads reduce the power which can be delivered to the short circuit. But since those other loads would be rather insignificant when compared with a short circuit the more I consider it, the more I am inclined to think voltage at the DNO head is not really that important, it is the voltage at the transformer which is the really important factor.

Again, that was, of course, my initial thinking, but it's starting to look as if the effect of other loads is more significant. One point is that, contrary to what you have said (and what I was thinking!) the current of 'other loads' is not necessarily 'insignificant' in comparison with the current due to a fault, particularly a fault in a 'low current' circuit. A fault ('short circuit') at the end of a long 6A lighting circuit might theoretically result in a fault current of only, say, 40A or 50A (30A needed for guaranteed trip of a B6), yet the genuine load from other consumers could presumably easily we 'a hundred amps or three' (particularly on Christmas day!).

The issue I am trying to get to grips with is essentially a mathematical one, to which I am seeking an explanation in terms of the physics. 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

 

[John D v2.0]

Well, you get the same voltage drop due to other loads L-N in all earthing systems, but remember that in general the voltage changes at your incomer is half due to the L dropping and half due to the N moving upwards (ignoring earth currents, which would be impossible to generalise about)

So the simple answer is the L-E fault current would only drop half as much due to load. You would use the (higher) L-E voltage to determine the fault current, not the L-N. However remember your Ze is determined by the resistance of the earth loop which would be higher than the l-n loop, could be 0.8 at the head rather than 0.35. Presumably the reason being that the earth wouldn't be lifted by normal currents so it's less risky, although in that case the 0.95 should only apply to TN-C-S supplies, and we should use 0.975 for TN-S supplies?

 

[JohnW2]

The other loads should indeed be relatively insignificant (depending on your definition), their effect on the fault current are seen at the head as a decrease in voltage.

See what I've just written to eric. I'm coming to realise that they can actually be quite 'significant' - the fault current (PSC) in a 6A lighting circuit might only be a little over 30A, yet genuine loads on the LV network could probably be at least 10 times that figure.

In a highly extreme case of multiple faults, the 0.95 allowance wouldn't be enough, so you wouldn't have guaranteed disconnection in a compliant installation.

As I've said, I'm inclined to ignore the unbelievably improbable situation of multiple simultaneous true faults - but your comment remains true in terms of genuine (not 'fault') high loads.

As I pointed out to them during the consultation on Amd3 of BS7671, if they had chosen 0.94, rather than 0.95, then that would have at least meant that any 'permitted' supply voltage (i.e. 216.2V or above) would give guaranteed magnetic disconnection. They appeared to agree with me, but were seemingly constrained by some 'international consdierations' to use the 0.95 figure!

Pleas also see what I've recently written to eric.

Kind Regards, John

 

[John D v2.0]

""""genuine loads on the LV network could probably be at least 10 times that figure.""""
yes true, but they wouldn't be over the same lighting circuit, they will be only over a lower impedence part of the circuit you are analysing and only affecting the voltage on that part of the network (whose influence stops before the "head" of your own install, and that influence is seen purely as a voltage drop from the transformer voltage)

 

[JohnW2]

Well, you get the same voltage drop due to other loads L-N in all earthing systems, but remember that in general the voltage changes at your incomer is half due to the L dropping and half due to the N moving upwards (ignoring earth currents, which would be impossible to generalise about)

That's my very point - that the impedance of the external path for non-fault loads (which causes those loads to reduce supply voltage) is Z(L)+Z(N), whereas for "L-E" fault currents is Z(L)+Z(E), and that (in a system other than TN-C-S) the voltage at the MET (relkative to 'true earth') will not be influenced by genuine loads but will be influenced by fault currents. That makes the calculations more complicated.

... Presumably the reason being that the earth wouldn't be lifted by normal currents so it's less risky, although in that case the 0.95 should only apply to TN-C-S supplies, and we should use 0.975 for TN-S supplies?

In the sense you mean, I suppose one might think that - but, as I've just written, even 0.95 (let alone 0.975) is really 'too high' (with any earthing system), given that the supply voltage in the absence of faults is allowed to be as low as 0.94Uo.

Kind Regards, John

 

[JohnW2]

... genuine loads on the LV network could probably be at least 10 times that figure.

yes true, but they wouldn't be over the same lighting circuit, they will be only over a lower impedence part of the circuit you are analysing and only affecting the voltage on that part of the network (whose influence stops before the "head" of your own install, and that influence is seen purely as a voltage drop from the transformer voltage)

I thought eric's point was about the relative effects of faults and 'high loads' on the supply voltage as seen at the origin of one's installation. If so, then, say, a 350A 'genuine load' on the network will result in a fall in 'supply voltage' which is almost (**) 10 times greater than that due to a 35A fault current in a 6A local lighting circuit.

** I say 'almost' because there will be a short run from the LV main to the premises which is not common to other consumers, and will probably be in lower CSA cable.

Kind Regards, John

 

[John D v2.0]

"""even 0.95 (let alone 0.975) is really 'too high' (with any earthing system), given that the supply voltage in the absence of faults is allowed to be as low as 0.94Uo."""
True, it is, but probably not a big worry. And my point is the 0.94 voltage drop is L-N, so if the voltage is at that level due to heavy load, that would leave approx 0.97 L-E voltage drop.

 

[John D v2.0]

"""350A 'genuine load' on the network will result in a fall in 'supply voltage' which is almost (**) 10 times greater than that due to a 35A fault current in a 6A local lighting circuit."""
Oh yeah I see what you mean, if someone else's lighting circuit shorts at the end of the cable, the dip in voltage would not be significant. Same as them turning their shower on.

 

[JohnW2]

True, it is, but probably not a big worry. And my point is the 0.94 voltage drop is L-N, so if the voltage is at that level due to heavy load, that would leave approx 0.97 L-E voltage drop.

Yes, but you could only vthink about taking that into account if the regs distinguished between different earthing systems. As things stand, the same value of Cmin applies to all earthing systems, and for TN-C-S, 'N' and 'E' are obviously the same thing.

Kind Regards, John

 

[JohnW2]

Oh yeah I see what you mean, if someone else's lighting circuit shorts at the end of the cable, the dip in voltage would not be significant. Same as them turning their shower on.

Indeed, but I wish you wouldn't keep talking about these incredibly unlikely scenarios of co-incidental simultaneous L-E faults in different properties I would say that someone turning on a shower somewhere on the LV network would.could have much the same effect on voltages (everywhere) as would someone else's light circuit developing an L-E 'short'!

I'm probably going to go quiet for a bit. All this writing is distracting me from actually exploring vthe maths/physics!

Kind Regards, John

 

[John D v2.0]

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.
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.