For techies - Potential Flaw in Zs/ADS calculations?

[JohnW2]

These are all red herrings. ... You are saying allowing for 218.5V is unnecessary because the transformer is always 240V (or whatever) but you may have the shower and oven on when a fault occurs.

I don't think that makes any difference. However, I clearly need to do some thinking, in the hope that I can convince myself that I'm right (or not, as the case may be!). Give me a bit of time and I'll try to look at some numerical examples. Watch this space.

Kind Regards, John

 

[John D v2.0]

These are all red herrings.

You are saying allowing for 218.5V is unnecessary because the transformer is always 240V (or whatever) but you may have the shower and oven on when a fault occurs.

I think you hit the nail on the head, the fault current is reduced when it's split amongst other consumers on the same lv circuit from the substation, including yourself. That split is evident in the volt drop you see at your origin. So if the whole street is having a shower and a roast dinner, your ze will be the same but your pfc will be lower (assume tn/c/s to avoid confusing the matter)
You just can't get the same current through the same zs if the voltage is lower.
My mistake was I thought the pfc was the reciprocal of the ze, but actually there's the voltage in there also.

 

[JohnW2]

I think you hit the nail on the head, the fault current is reduced when it's split amongst other consumers on the same lv circuit from the substation, including yourself. That split is evident in the volt drop you see at your origin.

I'm certainly coming to realise that the situation is a lot more complicated than I first thought, but I'm not sure that what you say above is actually the issue - at least, not quite as you have stated it.

One of the issues that I'd previously overlooked is that the 'supply voltage' as seen at the origin of one's installation will fall dramatically during the duration of the fault. For example (again assuming the simple case of TN-C-S), if one had a Ze of 0.25Ω and a PFC of, say, 400A, the voltage at the origin of one's installation (and, indeed, at the orgin of the installation of every other consumer downstream on the same LV circuit) will fall by 100V for the duration of the fault. However, that effect will be fairly trivial at the currents used for Zs measurement - with the '12A maximum' used by the meters that EFLI and I have, the supply voltage reduction would only be 3V ('maximum').

In an attempt to get my head around all this, I'm going to do dome sums and simulations, but that may take a little time. watch this space!

Kind Regards, John

 

[ericmark]

As I see it one would need to use Norton's theorem to work out the current at the MCB. When the fault happens there will be a greater volt at that consumers premises and less at other peoples premises, so one would have to know the load at every house to work out the fault voltage.

So since we don't have that information it is near impossible to work out.

I remember talking about this problem with a narrow boat supply. To ensure the hull is not eaten away an isolation transformer was used and there was no connection to the shore supply earth. This resulted in the maximum impedance being that of the isolation transformer, which was hardly good enough to cause a MCB to open on the magnetic part. But with a 230 volt supply it did just hit the mark.

However the volt drop was not only across the transformer but also the shore supply, the latter could change with every hook up, so one could test the system in socket and it passed, and next day plugged in somewhere else it could fail.

With the other system the question is if a MCB is fast enough? Or should it be fed with a semiconductor fuse? The whole idea of having diodes in the earth connection seems wrong, but that is how it is done.

I see that with a farm house with the transformer some distance away from the house and it supplying only the house the voltage would be that of the transformer not the voltage at the house, as the impedance is the whole system not just the impedance in the house. But once you have many houses on the same supply it becomes more complex.

 

[JohnW2]

As I see it one would need to use Norton's theorem to work out the current at the MCB.

It's certainly proving to be much more complicated than I had initial thought. I suppose one could say that one was using Norton's Thorem but with just a single power source, it really reduces to just "electrical common sense" (aka Kirchoff's and Ohm's Laws). 'Full-blown' application of Norton's Theorem is really only necessary if there are multiple (voltage and/or current) sources.

In any event, I don't think that Theorem tells one how to do anything - as I understand, it merely states that it is possible to replace any circuit containing (only) current sources, voltage sources and resistances with an equivalent circuit which is just one current source and one resistance.

When the fault happens there will be a greater volt at that consumers premises and less at other peoples premises, so one would have to know the load at every house to work out the fault voltage.

I'm not sure what you mean by "fault voltage" but, for the purpose of the present exercise, I think that one can (assuming a simple case) consider all the loads (in all the premises) at one point in time to be a single resistive load between L and N at/near the origin of the installation of interest.

I'm still playing with the maths and simulations, and will attempt to report back fairly soon.

Kind Regards, John

 

[John D v2.0]

Assume tn-c for extreme simplicity
This means If you measure your zs with the oven/kettle on and off you'll get different answers. Anyone got a loop impedence tester handy to check?

 

[JohnW2]

Assume tn-c for extreme simplicity .... This means If you measure your zs with the oven/kettle on and off you'll get different answers. Anyone got a loop impedence tester handy to check?

OK - watch this space. However, if the answers are different (which I'm inclined to agree they might well be), then what is being measured is surely not "Zs" in the normal sense - an impedance ('loop impedance' or any other) is an impedance, regardless of what currents are flowing.

In other words, true loop impedance (Zs) in your diagram is R1+R2+Ze, which is clearly not dependent on anything else.

Kind Regards, John

 

[John D v2.0]

Sorry sorry I'm confusing zs and pfc again!

If you measure PFC in both cases it will be differenct. Zs by definition must be corrected for voltage, so should be the same in all cases. But the point is it's your fault current that trips the MCB so the zs isn't quite enough to know it'll trip.

 

[JohnW2]

temporarily deletecd (posted too soon). Will be back shortly!

 

[JohnW2]

Sorry sorry I'm confusing zs and pfc again! If you measure PFC in both cases it will be differenct. Zs by definition must be corrected for voltage, so should be the same in all cases. But the point is it's your fault current that trips the MCB so the zs isn't quite enough to know it'll trip.

Yes, my tests indicate that my meter is measuring true (unchanging) Zs:

No load. V=248.6V. Zs = 0.68Ω. Calculated PFC = 365.6A.
2kw fan heater. V=247.4V. Zs = 0.68Ω. Calculated PFC = 363.8A
2 x 2kw fan heaters. V = 246.2V. Zs = 0.68Ω. Calculated PFC = 362.1A

[all of the following again assuming TN-C-S for simplicity]As I've said before, one problem which arises is that, during the duration of a negligible impedance fault, my 'supply voltage' will fall to a very low level. My current Ze is about 0.32Ω. With the Zs of 0.68Ω, that means that R1+R2 is about 0.36Ω. That means that with a negligle impedance fault, and assuming a no-load voltage of 248.6V, the 'supply voltage' (across R1+R2, and the fault) should fall to about 131.6V, giving a calculated PFC of 193.5A - albeit still enough to magnetically trip a B32. However, if the Ze was just a bit higher, it would not be enough toguarantee magnetic tripping the B32, even though the Zs was well below the specified maximum for a B32. For example, if the Ze were, say, 0.5Ω, making Zs 0.86Ω (still well 'within limits'), the during-fault voltage ought to be about 104V, leading to a calculated PFC of about 121A - far too low to guarantee a magnetic trip of a B32.

Given the answers that the meter gives, I can but presume that it uses the 'before fault' (not 'during fault') voltage to calculate PFC. However, as above, as far as I can make out, the actual current during a fault would be much lower than that figure (quite possibly too low for a guaranteed magnetic trip), since the voltage would be so much lower. Yet again, I suspect I'm missing something, or making some silly error!

Kind Regards, John

 

[John D v2.0]

yes it would be the voltage before the fault to give the PFC in that situation, as you're trying to ignore the effect of the meter on the circuit.

However this whole debate is admittedly about what voltage we should use in the equation - and it's been decreed to use 0.95 of nominal effectively, for the purposes of checking breakers will open.

edit:
"""as far as I can make out, the actual current during a fault would be much lower than that figure (quite possibly too low for a guaranteed magnetic trip), since the voltage would be so much lower. Yet again, I suspect I'm missing something, or making some silly error!"""
actually not, the fault in your circuit can do what it likes to the voltage, that's the thing that you're trying to measure the resistance of, so it's all taken into account.

What you are getting as, is if you have a fault on your socket circuit and your oven circuit at the same moment, the volt drop at the origin our your install would be to a lower voltage than anticipated by the zs calculations, therefore the fault current would be lower.

In an extreme case, where all your circuits fault at the same time, maybe none of the MCBs would trip and it would be down to the cutout to blow.

 

[JohnW2]

However, if the Ze was just a bit higher, it would not be enough toguarantee magnetic tripping the B32, even though the Zs was well below the specified maximum for a B32. For example, if the Ze were, say, 0.5Ω, making Zs 0.86Ω (still well 'within limits'), the during-fault voltage ought to be about 104V, leading to a calculated PFC of about 121A - far too low to guarantee a magnetic trip of a B32 ..... Yet again, I suspect I'm missing something, or making some silly error!

... one 'silly error' detected, I think. In fact, I wasn't paying attention to what I wrote at the start of this thread!

As I wrote back then, the current through the fault (and through R1+R2 and MCB) can be determined by dividing the ('during fault') 'supply voltage' by the R1+R2 (not Zs, which is what I used above). Hence, in the above example, a during-fault supply voltage of 104V across an R1+R2 of 0.36Ω would result in a current (through the fault circuit, including the MCB) of 288A, plenty to trip the MCB - but, of course, in practice one does not know (without measuring R1+R2 separately) what that 'during fault supply voltage' will be.

Going back to where this discussion started, if one makes a wild guess that the transformer output voltage is about 250V, then (as I anticipated) one gets almost the same PFC as above (actually 290.7A) if one divides the voltage at the transformer by the total Zs. Hence, at least for this example:

PFC = transformer voltage / Zs = supply voltage (during fault) / (R1+R2)

Kind Regards, John

 

[JohnW2]

yes it would be the voltage before the fault to give the PFC in that situation, as you're trying to ignore the effect of the meter on the circuit.

I wouldn't say that. What one wants to know is what current will flow through the MCB during a real fault, so the “effect of the meter on the circuit” (in artificially creating a fault current) is surely crucial? In any event (and contrary to what I implied before!), the meter creates only a very modest 'fault current' (far less than a real one), so the effect on supply voltage will obviously be far less than one would see with a real ('negligible impedance') fault - so it doesn't make that much difference whether it uses the 'before test' or 'during test' supply voltage to calculate PFC.

What you are getting as, is if you have a fault on your socket circuit and your oven circuit at the same moment, the volt drop at the origin our your install would be to a lower voltage than anticipated by the zs calculations, therefore the fault current would be lower.

No, I’m not saying that. I’m talking about just one fault, on one circuit, and the effect that has on the ‘supply voltage’ for the duration of the fault.

However, see what I’ve recently written about my ‘silly mistake’.

Kind Regards, John

 

[John D v2.0]

I think your fundamental confusion is on how the ze affects the voltage at the local origin in normal circumstances and how that's factored in to the ze and PFC or otherwise. The answer is the local origin voltage, ie the load on the lv network, affects the pfc but not the ze. The meter would measure those correctly. The means by which the meter measures is not relevant.

if the load on the lv network took the origin voltage below 0.95x, your MCB might not trip fast enough. My point, maybe confusingly made, was that if that voltage drop was caused by simultaneous faults on several of your final circuits, none of your MCBs might trip.

 

[JohnW2]

I think your fundamental confusion is on how the ze affects the voltage at the local origin in normal circumstances and how that's factored in to the ze and PFC or otherwise. The answer is the local origin voltage, ie the load on the lv network, affects the pfc but not the ze.

That much is self-evident. As I said before, an impedance (such as Ze) is an impedance - and as such is merely a property of the cables concerned. Nothing short of mechanical interference with the DNO's cable can change the Ze.

if the load on the lv network took the origin voltage below 0.95x, your MCB might not trip fast enough. My point, maybe confusingly made, was that if that voltage drop was caused by simultaneous faults on several of your final circuits, none of your MCBs might trip.

I think it may be you who is getting confused now. You seem to be talking about a reduction in 'local supply voltage' due to the fault itself (until the fault has cleared) - but, as I have illustrated, that voltage will almost certainly be very low during the fault, usually far lower than 0.95Uo - yet, as I have also illustrated, that may well be enough voltage to send enough current through R1+R2 (and the MCB) to magnetically trip the MCB. As I said, the important thing to understand that that 'low voltage' is driving current only through R1+R2 (and the MCB), not through the entire Zs (i.e. R1+R2+Ze).

I'm still plodding my way through the maths and simulations, and hope that it won't be too long before I have some 'chapter and verse' to show you.

Kind Regards, John