# Earth Loop Impedance

In practice, unless we were talking about a very short circuit (length!) wired in high CSA cable, I think that the risk of a CPC melting (as a result of a fault) would probably be almost non-existant, even if the Ze (hypothetically) fell to zero, since the R1+R2 of the circuit would be a significant portion of the total Zs.
A fault can occur anywhere on a circuit, for example cables chafing on a rough edge of the cu. That would be a very short circuit indeed and be almost the worst case
For adiabatic the pefc is used rather than the zs because pefc is the worst case.
For disconnection time zs gives the worse case as it's at the end of the circuit

A fault can occur anywhere on a circuit, for example cables chafing on a rough edge of the cu. That would be a very short circuit indeed and be almost the worst case
Agreed.
For adiabatic the pefc is used rather than the zs because pefc is the worst case.
One obviously cannot use a current 'rather than' an impedance I presume that you mean that one uses the PEFC at the origin of the circuit (or origin of the installation?), 'rather than' a PEFC calculated from Zs at the most distant part of the circuit? If so, that sounds horribly logical, but is it what people actually do (or are expected to do)? ...

... given that final circuits in domestic installations usually start more-or-less at the origin of the installation, does that not mean that an adiabatic calculation of required CPC size for any final circuit should be based on a PEFC of Uo/Ze? If so, that would be at least 657A for a TN-C-S installation, which would surely lead to some enormous 'required CPC CSAs', even for lighting circuits. It also means that the required 'minimum CPC size' would be the same regardless of the CSA of the L & N conductors of the cable. I think that I must be missing something pretty basic!

For disconnection time zs gives the worse case as it's at the end of the circuit
Very true, but that's obviously a different matter.

Kind Regards, John

does that not mean that an adiabatic calculation of required CPC size for any final circuit should be based on a PEFC of Uo/Ze? If so, that would be at least 657A for a TN-C-S installation, which would surely lead to some enormous 'required CPC CSAs', even for lighting circuits
Yes the current you quote is correct, but i2t also depends on the disconnection time. For a B6 mcb you could only go down to a disconnection time of 0.1s, 30A x 0.1s which would require a very small CSA. One you get to 657A you would read the i2t directly off the graph for energy limiting class 3 and a 100A upstream fuse, as applicable and where appropriate. Then you could find that 1.0mm would be fine for a lighting circuit in all cases up to 10kA.

Yes the current you quote is correct, but i2t also depends on the disconnection time. For a B6 mcb you could only go down to 30A x 0.1s which would require a very small CSA.
Are you not making the mistake that so many do of assuming that a protective device (be it a fuse, MCB or RCD) somehow 'limits' the magnitude of the current (as well as its duration)? Whilst it is true that a current ≥30A will cause a B6 to trip magnetically, if the loop circuit's impedance is such that the fault current is, say, 600A, the current which would flow through the B6 would be 600A, not 30A, until it operated and cleared the fault.

Kind Regards, John

I'm saying that the higher the current the quicker the disconnection time, keeping i2t below a limit of 35000 for an mcb under 16A.
Furthermore when using back up protection from a bs88 fuse, that would limit the I2t even further, as at very high fault currents, the fuse would interrupt the current before the mcb.

I'm saying that the higher the current the quicker the disconnection time, keeping i2t below a limit of 35000 for an mcb under 16A.
I've never fully understood this, so perhaps you can help ....

I would imagine that we can probably agree that 'the fault current is the fault current' (determined only by supply voltage and impedance of the fault path loop) - so that if that is 600A, then 600A is what would flow though, say, a B6 until the fault was cleared.

That means that, in that situation, to get a 'small' I²t (and particularly given that the "I" is squared) one has to have a 'very short' disconnection time. However, given that there are physical/mechanical factors and Laws of Physics (e.g. in relation to inertia, friction, acceleration etc .) in the melting pot, I can but assume that there must be a finite lower bound to how short a disconnection time can be achieved, no matter how high the current. In other words, if the fault current (over which the MCB has no control) increased by, say, a factor of 10, disconnection time would have to decrease by a factor of 100 to maintain the same I²t - and I would have thought that would be impossible because of the 'lower bound'.

Have I said enough to make the nature of my difficulty clear?

Kind Regards, John

I agree, in fact with these sub cycle disconnection times we don't even know whether it's at the peak that it happens, therefore it could be higher.
The mechanical side of the mcb is backed up by the bs88 fuse at high currents, but i agree even that is limited by the time for the fuse wiire to lose contact.
My assumption is that with a transformer of up to 16kA they have tested or calculated that even if you shorted at the output at the peak the fault current doesn't instantly rise to the maximum, and if it's not at the peak it will never get there.

I agree, in fact with these sub cycle disconnection times we don't even know whether it's at the peak that it happens, therefore it could be higher.
The mechanical side of the mcb is backed up by the bs88 fuse at high currents, but i agree even that is limited by the time for the fuse wiire to lose contact.
My assumption is that with a transformer of up to 16kA they have tested or calculated that even if you shorted at the output at the peak the fault current doesn't instantly rise to the maximum, and if it's not at the peak it will never get there.
I have never quite understood how the let through value is calculated, I have only once had the need, a tower crane with a 600A supply had a charging point for a radio for banks-man, this had a b16 RBCO rated 4.7 kA and I was measuring PFC in excess, so I fitted a fuse which it seemed had a let through value of under 4.7 kA think from memory 60A however the resistance of the fuse brought the PFC to under 4.7 kA so even if not using let through valve it would still comply.

I have not gone into the let through (i²t) valve of a BS88 fuse, but think this will allow the use of a MCB of 4.7 kA even if the PFC exceeds that valve however since I have never needed to work it out, I don't really know reading the data sheets if I have it right with a 100A BS 88 fuse the let though valve is between 1 kA and 2 kA at 0.01 seconds which is bottom of chart, so in domestic or any other where we have a maximum of 100A i.e. a consumer unit is used which is type tested for ordinary persons to use, we don't need to worry at the moment about PFC if the MCB/RCBO is 4.7 kA or more.

I say at the moment, I am not sure what the situation will be as smart meters are rolled out? From what I have read in some countries in the EU the charge for power has always been dependent on supply size, so there may be a 45A breaker to limit your supply to 9 kW with the smart meter they can measure the use more accurately and switch off the supply closer to the 9 kW your paying for, so it does seem in France they use a form of MCB not a fuse so it is questionable as to let through value. I was surprised to see a video of a smart meter taken apart and the relay is normally closed not normally open, so even if the supply is cut using the smart meter, it is not safe to work on supply. To me that defeats whole idea, in the case of a fire switching off smart meter you could not safely send in fire fighters as supply could become live again.

However the question is if we need to allow for a change in supply characteristics, not let through valves or smart meters.

I agree, in fact with these sub cycle disconnection times we don't even know whether it's at the peak that it happens, therefore it could be higher.
Yes, that's another issue/factor (which I almost mentioned). In fact, if one is talking very short disconnection times, it's nearly always going to be "slightly varying DC', rather than AC in any meaningful sense. However, as always, I presume we have to consider a 'worst-case scenario' - which I presume is a fault occurring at, or just before, the peak voltage point of the cycle.
The mechanical side of the mcb is backed up by the bs88 fuse at high currents, but i agree even that is limited by the time for the fuse wire to lose contact.
There certainly would seem to be fewer physical 'limiting factors' in the operation of a fuse, so I can imagine that, at very high currents, they are likely to operate quicker than an MCB (with all it's 'mechanical' issues). There probably will still be a physics-imposed 'lower bound' to disconnection time for a fuse, but I would expect it to be 'lower' than for a corresponding MCB. However, I think that considering 'secondary protection' by a fuse is likely to complicate our discussion about MCBs, isn't it?

I suppose one of my main problems is the fact that people seem to talking about I²t as an 'entity' ('let through') that can somehow be affected by external influences - and yet, in mathematical/computational terms it is simply "I squared times t" (I being the current which flows and t being the duration of that flow). If, for the purpose of a relatively extreme (but far from impossible with TN-C-S) case of Ze=0.1Ω, that obviously translates to a PEFC at the origin of 2300A, and that is roughly the current that will flow through a B6 if a zero impedance fault arises more-or-less at its output terminal (i.e. at the very start of the final circuit) - and that ~2300A will flow until something clears the fault.

If (as is likely with a B6 circuit) the CSA of the final circuit cable is 1mm², then, in the limiting case (disconnection just fast enough to satisfy the requirement), the adiabatic equation as presented in BS7671 can be reduced/rearranged to:

t = (k/I)² ... and if we assume the 2300A fault current and k=115, that becomes;

t = (115/2300)² = 0.0025 seconds = 2.5 ms

What I question is whether (given all the mechanical factors I have mentioned), an MCB can operate in 2.5 ms (even if the fault arises at the 'optimal' point of the cycle).
My assumption is that with a transformer of up to 16kA they have tested or calculated that even if you shorted at the output at the peak the fault current doesn't instantly rise to the maximum, and if it's not at the peak it will never get there.
I'm not clear as to your point here. Are you suggesting that, say, the PEFC (for calculation purposes) with Uo=230V and Ze=0.1 should be considered as something less than 2300A? If not, then the above arithmetic (and consequential maximum acceptable disconnection time) surely applies, doesn't it?

Kind Regards, John

Are you suggesting that, say, the PEFC (for calculation purposes) with Uo=230V and Ze=0.1 should be considered as something less than 2300A?
yes I'm saying that if measured Ze=0.1 at 50Hz then the impedance on an instantaneous short circuit occurring at the peak of the cycle (effectively a lot higher than 50Hz) would be higher.
Not to mention that the voltage may be dragged down depending on the size of the transformer.

yes I'm saying that if measured Ze=0.1 at 50Hz then the impedance on an instantaneous short circuit occurring at the peak of the cycle (effectively a lot higher than 50Hz) would be higher.
OK. In that case, for a start, what did you mean when you wrote (which is what really started this whole discussion) "For adiabatic the pefc is used rather than the zs because pefc is the worst case."? As I said at the time, I presumed that what you meant was that for an adiabatic calculation one uses the PEFC at the origin of the circuit, not the PEFC based on a measurement of Zs at the furthest point of the circuit - and, as I said, that seems totally logical, if one wants to take into account the possible 'worst case' situation of a fault very close to the origin of the circuit.

So, I thought I understood (and essentially agreed), but I was assuming that "the PEFC at the origin of the circuit" would be very close to Uo/Ze. However, you now seem to be saying that, for that adiabatic calculation, the "the PEFC at the origin of the circuit" one should use is appreciably less than Uo/Ze. What "PEFC at the origin of the circuit" are you therefore saying should be used?

Also .... I need to think about it a bit more, but I'm not sure that the 'frequency' (really 'rate of rise') of the transient at the onset of the fault would necessarily be very different at different points in the cycle, would it?

Kind Regards, John

Yes your first and most of your second paragraphs are exactly what I meant. But the caveat is adiabatic equation only works sensibly for disconnection times between 0.01 and 5s.
So now that we've got to talking about sub 0.01s disconnection times, you don't calculate I2t using I and t, you just read it from the charts provided for the protective devices.

Regarding the rate of rise, as you know Ze includes reactance as well as resistance. The inductance component of that would increase as the frequency increased (and vice versa for capacitance). Since the reactive component of the earth loop is mostly inductance, that would follow that the impedance would be higher at a higher frequency.
Following from that, if the fault occurred at the zero point of the cycle, there would only be a 50Hz (sine) component, meaning the impedance of the loop would be similar to Ze. However the MCB would disconnect before the peak.
However if the fault occurred closer to the peak, it would effectively be a square wave, which has harmonics up to infinity. As a result of that the square rise would be smoothed out to closer to a sine wave as most of the higher harmonics would be lost, therefore giving the MCB time to operate before the current climbs close to the peak value, if it even gets there on the first cycle.

Now some of that may be rubbish, but it seems sensible and logical, however it's nice that the standard makers have produced the information for us so we don't have to calculate why!

Yes your first and most of your second paragraphs are exactly what I meant. But the caveat is adiabatic equation only works sensibly for disconnection times between 0.01 and 5s.
Yes, and it's really only the bottom end we're concerned about. A truly adiabatic process would have to be of infinitely short duration, and it becomes progressively less adiabatic as the duration increases. AS you say (and, indeed, as BS7671 says) the convention is to regard a process lasting more than 5s as being non-adiabatic, but even at 5s it is becoming a doubtful assumption (hence inaccurate answers from calculations). Anyway, as said, it is sub-0.01s that interests us ....
So now that we've got to talking about sub 0.01s disconnection times, you don't calculate I2t using I and t, you just read it from the charts provided for the protective devices.
We're back to what I said about people regarding I²t as an 'entity', whereas it is, and always will be "current squared times the duration of that current". If "the charts provided for the protective devices" (and see ** below) gave a different answer from what one would get by calculating I²t from I and t (t for the I in question being obtained from a graph which covered the relevant range of values), then either something would be very wrong or else something has gone wrong with my brain!

I really need to have a look at one of these "charts provided for the protective devices" before I can make any particularly intelligent comments, but do I take it that they give different values of It for different currents? If so, I don't really see how they differ conceptually from 'adequate' t/I curves (i.e. ones which cover the range of values of interest). In other words, if the charts could give me a value of I²t for a particular 'I', I could obviously determine what 't' was. ... and it still seems to leave the question of 'what current one uses for the chart look-up'.

[** since the curves reproduced in BS7671 do not usually provide the relevant information needed to undertake the adiabatic calculation which BS7671 specifies, I've never understood why they don't include "the charts provided for the protective devices" {or usable curves, if they exist} in BS7671 ]

Regarding the rate of rise, as you know Ze includes reactance as well as resistance. The inductance component of that would increase as the frequency increased (and vice versa for capacitance). Since the reactive component of the earth loop is mostly inductance, that would follow that the impedance would be higher at a higher frequency.
Needless to say, there's no argument about any of that (and I assumed that we both know that much!), but ....
Following from that, if the fault occurred at the zero point of the cycle, there would only be a 50Hz (sine) component, meaning the impedance of the loop would be similar to Ze. However the MCB would disconnect before the peak. However if the fault occurred closer to the peak, it would effectively be a square wave, which has harmonics up to infinity. As a result of that the square rise would be smoothed out to closer to a sine wave as most of the higher harmonics would be lost, therefore giving the MCB time to operate before the current climbs close to the peak value, if it even gets there on the first cycle.
Those are the bits that I'm 'not so sure about'. Although I've yet to do all my thinking, the sudden appearance of a 'short circuit' will surely result in a very rapid rate of rise of current (a 'transient', with high frequency components), regardless of the point in the cycle, won't it? [the zero-crossing point is somewhat of a special case, since there would initially be no fault current, so I'm 'ignoring' that for the moment!].

Kind Regards, John

We're back to what I said about people regarding I²t as an 'entity', whereas it is, and always will be "current squared times the duration of that current".
Well, technically it is the area under the i2 curve seen during the fault, which would not be the rms fault current i2 times t, on a short duration fault.
the sudden appearance of a 'short circuit' will surely result in a very rapid rate of rise of current (a 'transient', with high frequency components), regardless of the point in the cycle, won't it?
Rapid yes and more rapid then the normal 50hz cycle, but not instant.

Well, technically it is the area under the i2 curve seen during the fault, which would not be the rms fault current i2 times t, on a short duration fault.
Indeed, and that illustrates what I'm been saying about the potential confusion about the 'entity' which (confusingly, and usually strictly incorrectly) is usually denoted by "I²t".

As you say, what we really should be using is the area under the I²/t curve for the duration of the fault current - something which seems to be called 'let through', and it is in terms of that which a 'proper adiabatic equation' would be written. In other words, the equation should be written in terms of ... ... rather than I²t. The reason for our interest in this quantity is clear enough, since if one multiplies it by the total resistance of a conductor, it then gives the total amount of energy dissipated in that conductor during passage of the current and, more important to our context, if one multiplies it by, say, the resistance per metre of the conductor, one will get the total amount of energy peer metre dissipated in the conductor - which, under adiabatic conditions, will, in turn, be directly proportional to temperature rise in that conductor.

I²t is nearly always an approximation to the 'let through' (AUC) we really should be using. It is only exactly the same as 'let through' if 't' is an exact number of half cycles. If t is large in relation to a half cycle (i.e. 'many half cycles, probably plus a bit of a half-cycle'), then it will be approximately correct. You have suggested (as is often suggested) that that equation gives reasonably correct results down to t=0.01s. However, that is only one half-cycle, which I think is a bit iffy (unless it were exactly 0.01s). In other words, I think that It would give a pretty poor approximation to 'let through' (AUC) with, say, t=0.015s, or even 0.025s.

Having said all that, we are presumably, as always, expected to consider 'worst case' scenarios, and I think that makes things a lot simpler. With a very short disconnection time (only a small proportion of a half-cycle) that will presumably occur when the period of fault current is centred at the peak point of the voltage waveform. We are then back to the situation in which the true 'let through' (AUC) actually is (approximately) a sort-of I²t - the 'I' now actually being √2 times what would be the RMS current (calculated from V and Z, perhaps Uo and PEFC) if the fault current were sustained for many (or an exact number of) half-cycles. In other words, if we knew 't' (from adequate curves), we could undertake a worst-case adiabatic calculation. It's interesting that, for such a short disconnection time, the situation would actually be 'worse', by a factor of √2 than would be the case if one undertook ('naive'?) 'RMS calculations'.

Rapid yes and more rapid then the normal 50hz cycle, but not instant.
That's what I wrote, isn't it? I wrote "... very fast rate of rise ... with high frequency components", not "infinitely fast rate of rise ... with infinitely high frequency components".

Kind Regards, John

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