Myths

Yes, as we've often discussed, I understand that.

As a matter of interest, are there any situations in which you feel it acceptable to omit overload protection? If not, is it your feeling that 433.3.1(II) should not exist?

Kind Regards, John
If I don't have total control over something, then no. (EDIT: Caveat; I don't hold a copy of latest regs so unable to see exactly what is current)

However I am comfortable with downstream protection under some circumstances, such as RFC spurs when correctly installed.
 
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If I don't have total control over something, then no.
In the real world, I doubt that it would ever be totally impossible to think up some obscure and incredibly improbable way in which a particular load could result in an 'overload current';, so I'll take that as a 'no'!
(EDIT: Caveat; I don't hold a copy of latest regs so unable to see exactly what is current)
The reg which permits omission of overload protection in the sort of situation we've been discussing is ...
433.3 Omission of devices for protection against overload
....433.3.1 General

A device for protection against overload need not be provided:
......... (ii) for a conductor which, because of the characteristics of the load or the supply, is not likely to carry overload
current, provided that the conductor is protected against fault current in accordance with the requirements
of Section 434
However I am comfortable with downstream protection under some circumstances, such as RFC spurs when correctly installed.
That's a totally different matter. As far as I can see, the only potential 'downside' of downstream overload protection is that it obviously does not afford protection in the case of faults in the cable itself, upstream of the protective device.

Kind Regards, John
 
However, you must surely be aware of the fact that, whenever that concept is mentioned, people like SUNRAY and bernard scrape thee barrel and postulate scenarios (I would think incredibly improbable, since certainly not 'likely') in which an overload current (not a short-circuit or fault to earth) can theoretically arise in the elements of things like immersions and ovens - primarily situations in which three is a 'looped' healing conductor, two parts of which come into contact (thereby 'short-circuiting' a portion of the resistance/impedance).
Question - in such a scenario, the altering of the "fixed load" resistance be down to a "short circuit" of part of that element? Food for thought.
I do agree that such an instant is often a possiblillity of occouring thereby our fixed load is no longer a fixed load but the chances are extremely rare but not totally absolutely impossible.
In much the same way we calculate a short circuit or an earth fault as a dead short between the conductors involved and it is quite clear that such a "bolted fault" is deemed to always occour when we apply our calculations although we are all mindfull that this is not always the case.
Limited touch area of contact and corrosion between touching parts very often could make those shorting parts quite resistive in reality.
We always ignore that possibility although we know it exists.
 
Question - in such a scenario, the altering of the "fixed load" resistance be down to a "short circuit" of part of that element?
Yep, that's the scenario which 'they always' talk about
I do agree that such an instant is often a possiblillity of occouring thereby our fixed load is no longer a fixed load but the chances are extremely rare but not totally absolutely impossible.
As you say, "not absolutely impossible" (but "incredibly rare"), but that doesn't stop them raising the 'possibility' whenere thge omision of overload protection is mentioned. Even in regulatory terms, one is 'allowed' to omit overload protection if the nature of the load is that it is "not likely" that there will ever be n 'overload current' - and I'm sure that "incredibly unlikley" qualifies as "not likely".

I think bernard usually argues purely on a theoretical basis, but a 'problem' (sorry!)( with SUNRAY is that he has apparently experienced such "incredibly unlikely" occurrences, which influences his view. However, all that proves is that "incredibly unlikely" does not mean 'impossible' - after all, people do get 'struck by lightning'!
In much the same way we calculate a short circuit or an earth fault as a dead short between the conductors involved and it is quite clear that such a "bolted fault" is deemed to always occour when we apply our calculations although we are all mindfull that this is not always the case.
Limited touch area of contact and corrosion between touching parts very often could make those shorting parts quite resistive in reality.
We always ignore that possibility although we know it exists.
Very much so. I think that is something which is widely overlooked and the consequence of it is that the protection against real world (L-E) 'faults' and 'short-circuits' (L-N) is probably often nowhere near as 'good' (qutre probably 'not adequate' in regulatory terms) as we would like to think.

As you say, all of our calculations of the adequacy of fault/SC protection are based on 'bolted shorts' (of "negligible impedance") - indeed, our calculations assume a zero impedance. In the real world, I imagine that faults/shorts of, say, 'an Ohm or three' are far from uncommon - and nothing realistic can provide protection against them - it would require a device which was a lot more 'trigger-happy' than even a B-Curve MCB.

Kind Regards, John
 
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In the real world, I imagine that faults/shorts of, say, 'an Ohm or three' are far from uncommon - and nothing realistic can provide protection against them - it would require a device which was a lot more 'trigger-happy' than even a B-Curve MCB.
Please explain.
 
Please explain.
We calculate the 'maximum Zs' that will 'just' result in magnetic tripping of the MCB (in the 'worse case' for a B-Curve MCB requiring a current of 5 x In) with a supply voltage of 218.5V (230V x Cmin). i.e. , for a B-curve MCB ......
max Zs = (230 x Cmin) / (5 x In)
For any given (measured or calculated) Zs, the current during fault conditions (with a supply voltage of 218.5V) is therefore:
fault current = (230 x Cmin) / Zs
... for example, for a B32 MCB and a Zs of about 1.36Ω ...
fault current = (230 x Cmin) / 1.36 = 160.7 A
... i.e. just enough to trip a B32 (in 'worst case')

However, that assumes the the actual loop impedance of the fault circuit is equal to the (measured or calculated) Zs, and therefore assumes that the impedance of the actual fault is zero. If,with the same circuit, the fault actually had an impedance of, say, 0.5Ω, then we would have:
fault current = (230 x Cmin) / (1.36 + 0.5) = 117.5 A
... clearly not enough to trip a B32 in the worst case (magnetic trip threshold = 5 x In)

Hence, to get magnetic tripping in that circuit with a fault of just 0.5 impedance would require an MCB with a 'worst case' trip threshold of less than 117.5 A, maybe one with a spec requiring tripping at "2-3 x In".

Things obviously get 'worse' if the impedance of the fault is even greater (than 0.5Ω ). For example, if, with this circuit ,the fault had an impedance of 1Ω, then we would have:
fault current = (230 x Cmin) / (1.36 + 1.0) = 92,6 A

,... which would mean that we would need an MCB with a ('worst case') magnetic trip threshold below 92.6 A - maybe one with a spec requiring tripping at "1.0 - 2.5 x In" ...and so on, depending upon how high a fault impedance one wanted to guarantee would result in a magnetic trip.

Does that help?

Kind Regards, John
 
Are you saying that MCBs (45A or 50A likely being the highest in most properties) will not trip with 160.7A, 117.5A or 92.6A running through them?
 
Are you saying that MCBs (45A or 50A likely being the highest in most properties) will not trip with 160.7A, 117.5A or 92.6A running through them?
They would, of course, 'trip' eventually (with any current greater than 1.13 x In), but 'thermally', not magnetically, hence not quickly enough to satisfy required disconnection times fort fault protection.

As you know ...
A B45 needs at least 135A (and possibly as much as 225A) to trip magnetically.
A B50 needs at least 150A (and possibly as much as 250A) to trip magnetically.

Kind Regards, John
 
Yes, I know but perhaps that is why we do not consider such faults as you mention when talking about fault protection.
 
'trip' eventually (with any current greater than 1.13 x In)
I`d say trip possible if 1.13 x In exceeded and must trip if 1.45 x In reached but that is "within the conventional time (which might be 2 hours. It could be 4 hours in some instances).

Well a fuse blowing anyway!
 
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Yes, I know but perhaps that is why we do not consider such faults as you mention when talking about fault protection.
Sure - but, as I was 'cautioning' it inevitably means that some (probably many) circuits which satisfy our standard ("maximum Zs") calculations will not provide the disconnection times required by BS7671 for fault protection in the face of a 'real-world' (rather than a hypothetical 'zero impedance') fault.

Look at it like this. If I told you (or an EICR inspector discovered) that the Zs of a circuit was "just a little too high" ('just a little above' the 'maximum Zs' ") to satisfy the fault protection requirements, then you (or the inspector) would very probably conclude that such was 'not acceptable' (maybe even an EICR C2), wouldn't you?

However, in terms of fault protection (i.e. achieving, or not, the required disconnection times) the situation would be just the same if the Zs were close to (but not above) the 'maximum Zs), but the impedance of the fault was 'just a little above zero'. In electrical terms, that would really be just as 'not acceptable', wouldn't it?

As I've tried to explain, there's not a lot we can do about this with available devices, since there are not MCBs/RCBOs (at least, not 'readily available) that will magnetically trip at less than 3 x In.

I suppose all I'm doing (because ebee raised the issue) is reminding people that we are fooling ourselves if we believe that satisfying the 'maximum Zs' requirements will ensure that the required disconnection times will necessarily be achieved with real-world faults (which will never be of 'exactly zero' impedance, and could bee of 'significant' impedance').

Kind Regards, John
 
I`d say trip possible if 1.13 x In exceeded ...
As I understand it, it must trip (eventually, even if it takes hours) for any current above 1.13 x In.
and must trip if 1.45 x In reached but that is "within the conventional time (which might be 2 hours. It could be 4 hours in some instances).
As above, it should trip eventually at any current above 1.13 x In, and, as you say, within a period (usually taken as 1 hour, but it could be longer) at 1.45 x In.

However, the required disconnection time (in TN installations) to provide an acceptable degree of fault protection is 0.2 seconds, so I don't think we, or the regs, are likely to be very satisfied by disconnection which takes an hour or two :)

Kind Regards, John
 
i know this is very different to what is being discussed but a while back I was asked to look at a problem with a 28day RSL transmission on band II, the transmitter site being in woodland on top of the North Downs and the studio was some 1/2mile down the hill. I had supplied the cable for the stereo audio (4 pair CAT 2) and as it sounded fine with the transmitter in the studio on a dummy load and crap when in situ the fualt was promptly placed on the audio cable not being screened.

I turned up and met customer in the woods, first test I made was listening to the audio directly on the input terminals with a battery powered amplifier which sounded fine, I'd already heard how bad the radio signal sounded.

Transmitter had 3 leds (power, stereo both lit & alarm not lit) and 2 volume controls, so all looking good.

Next test was measure the mains voltage which was way down, near 150V, transmitter spec ~120W 230V and contained a linear power supply.

Customer had purchased cable to do the same run from the studio. AKA 0.5mm² 3 core flex.

I checked the maximum current by shorting the supply with current range of multimeter, which was only a couple or 3amps.

The quickest fix was a generator followed by 2.5, 4 and 6mm² cables
 
However, the required disconnection time (in TN installations) to provide an acceptable degree of fault protection is 0.2 seconds, so I don't think we, or the regs, are likely to be very satisfied by disconnection which takes an hour or two :)
So what can be done about the situation for the infinite number of possible fault currents that people seem to think might exist when one stray wire touches something it should not?

I feel we must be missing something.

I have said before I once blue the DNO 60A fuse when a stray T&E CPC happened to spring into the CU main switch line incomer screw. It did not stay there, of course, but was blown back immediately by the flash.
Obviously the current must have been great enough for the instant disconnection with no damage to anything apart from a small globule melted on the end of the CPC and a mark on the screw.

How then do these faults of quite considerable resistance manage to maintain contact for long enough if the disconnection is not 'instantaneous'?

Edit: The CPC was only 1mm².
 
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Look at it like this. If I told you (or an EICR inspector discovered) that the Zs of a circuit was "just a little too high" ('just a little above' the 'maximum Zs' ") to satisfy the fault protection requirements, then you (or the inspector) would very probably conclude that such was 'not acceptable' (maybe even an EICR C2), wouldn't you?
We need to consider are the readings we take with our meters may not be that accuracy. Example of 1.111 of anything on a meter may prompt some to assume it might be accurate to +/- 0.001 , this is far from reality.
Accuracy is often far worse than resolution might induce some to think it might be.
Treat meter accuracy on most equipment as a bit better than a length of damp string. It`s only a guide. Two readings seeming far apart might actually be very closely spaced apart. A ball park figure.
If a reading is well within what we expect it to be or well without what we are expecting might be an indication of whether it looks OK or not.
 

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