Are there formulae to calculate fuse/mcb disconnection times at a given fault current instead of having to look up graphs? The same question as being asked here: https://electronics.stackexchange.c...ermine-a-formula-for-fuse-disconnection-times
Disconnection time formula?
- Thread starter skenk
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I don't know for certain, but I doubt it, because I think the graphs we see are almost certainly empirically-based, rather than based on theory. One could (approximately) mathematically model those empirical curves and thereby produce 'formulae', but those formulae might well be complicated and would quite probably be completely different for each rating of MCB/fuse, so it probably would not be worthwhile to create them.
However, maybe someone can prove me wrong!
Kind Regards, John
No.
The person at that other forum has missed the point entirely. It's not necessary to calculate the actual disconnection time for a certain current.
All that is required is to ensure the fault current is greater than the current required to cause disconnection within the time required.
As there are only a certain number of fuse ratings, and only a few disconnection times, the small chart to the right of the graph is really all that is required.
If that is really what they meant, it is entirely the wrong process - fuse ratings and type are not selected based on the disconnection time.https://electronics.stackexchange.com/questions/129632/can-i-determine-a-formula-for-fuse-disconnection-times said:
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Even the log-log graphs we are used to seeing are far from linear (other than for some types of fuses - but even they are certainly not truly linear). Any continuous curve without discontinuities can theoretically be modelled (to create a formula), so that could be done for a particular type/rating of fuse or the thermal part of the curve for an MCB of a particular type/rating, but, as I said, the resulting formulae could be very complex and would probably be different for every device/rating.
Quite so - and much simpler.
More to the point, as flameport has said, I really can't see why anyone would want such formulae. In fact, I'm not really sure why an electrician would even ever need to refer to the graphs, since all that matters for design is knowledge of the maximum current needed for operation of the device within particular (essentially arbitrary) disconnection times - and those currents are defined in the relevant product Standards (certainly for MCBs).
Kind Regards, John
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Not every installation has been designed though, some have just been chucked together! And when assessing existing installations and temporary installations it would be useful to know whether the fault current would be sufficient to operate the cpd before the cable overheated, for example (at the end of a 80m extension cable from a submain perhaps!). I'm just trying to make up some formulae in Excel and this is leaving a hole.
I can derive how long a cable will take to overheat from the fault current but not how long it will take to operate a given cpd. I can tell the client that a fault won't disconnect within the required time, but I can't tell them how long it will take. I guess due to the nature of the devices it's not simple to calculate and would be likely to be inaccurate in real life scenarios anyway.
Not all circuits are designed in the classic way, for example a shed could be supplied using that spare 6mm cable in the lockup, Will that be ok for a 32A sub-main? and what fuse could I go to if the client wanted to pay more for 10 or 16mm cable? (max demand for the whole installation and other factors accounted for). If the cable took that trickier route that was 10m shorter, or longer because the sketch someone did isn't accurate how would that affect things? With the classic dumb cable calculators the only unknown is cable size, whereas often a more likely scenario is that you know from experience (or client's budget) it's going to end up being one of 3 sizes, there's 2 possible routes for the cable, and the max demand on the submain is theoretical anyway because they're only having a socket and light at the moment and you just want to know the maximum breaker/capacity you can give them with their 2 or 3 options. There are only certain sizes of cable and breaker so that shorter cable route can make all the difference in providing sufficient power without breaking the bank and your back. It's easier to see the limits of each cable when each variable can be adjusted. I appreciate this is not directly related to the fault current-disconnection time relationship but just a general comment on design.
I can derive how long a cable will take to overheat from the fault current but not how long it will take to operate a given cpd. I can tell the client that a fault won't disconnect within the required time, but I can't tell them how long it will take. I guess due to the nature of the devices it's not simple to calculate and would be likely to be inaccurate in real life scenarios anyway.
Not all circuits are designed in the classic way, for example a shed could be supplied using that spare 6mm cable in the lockup, Will that be ok for a 32A sub-main? and what fuse could I go to if the client wanted to pay more for 10 or 16mm cable? (max demand for the whole installation and other factors accounted for). If the cable took that trickier route that was 10m shorter, or longer because the sketch someone did isn't accurate how would that affect things? With the classic dumb cable calculators the only unknown is cable size, whereas often a more likely scenario is that you know from experience (or client's budget) it's going to end up being one of 3 sizes, there's 2 possible routes for the cable, and the max demand on the submain is theoretical anyway because they're only having a socket and light at the moment and you just want to know the maximum breaker/capacity you can give them with their 2 or 3 options. There are only certain sizes of cable and breaker so that shorter cable route can make all the difference in providing sufficient power without breaking the bank and your back. It's easier to see the limits of each cable when each variable can be adjusted. I appreciate this is not directly related to the fault current-disconnection time relationship but just a general comment on design.
I still don't really understand. If you are talking about 'fault current' in the correct sense (i.e. a 'short circuit'), then the Zs at the furthest point in the circuit will determine the magnitude of the fault current, and the specification of the OPD will then indicate whether that fault current is enough to produce the required 'immediate detection' - e.g. with a Type B MCB, since it's magnetic tripping has to happen between 3 and 5 times its rating (In), one has to assume that it could take 5 times In (e.g. 160A for a B32 MCB).
As for 'overload' currents, that's just a question whether (as required) all cables (including any flexible extensions!) have a CCC which is no less than the In of the OPD. If that is satisfied, then the cable is regarded as adequately protected from overheating. Length of cables is not a consideration in that respect.
As above, I'm still not clear what you are trying to achieve, but you give a little insight when you go on to say ...
You now seem to be talking about a non-complaint circuit, with an excessive Zs. The required disconnection times are related to safety (primarily protection against electric shock), rather than overheating of the cable. Even if you did determine (from graphs, or a formula if it existed) the time that it would take for a (true) fault current to operate the OPD, given that the fault current was too low for the required 'immediate' disconnection, I don't really know how that information would really help you. What do you think it would tell you, other than that the circuit was non-compliant and therefore not deemed to be 'safe'?
In any event, as has been said, it would be impractical to derive 'formulae'. If there was some perceived point in the exercise (estimating actual, non-compliant, disconnection times) you could do it in software by creating look-up tables containing data read from the published graphs (probably different for each type and rating of OPD), extrapolating between points in that look-up data if you wanted to - but I still don't see what that would achieve.
Kind Regards, John
You mean Zx? Yes, I was wondering if there's a formula for Zx/disconnection time if you prefer that to fault current/disconnection time.
So what's the disconnection time if the fault current is 127A as I recently measured at the end of a long extension, and would that trip the breaker or blow the 13A fuse first?
I'm not talking about overload currents, that's an easy one, check the CCC of the cable in the installed conditions... done.
It would tell me how long it would take to disconnect the fault, which I could then explain to the client instead of saying, "it won't trip fast enough, but I don't know how long it will take to trip".
I could create look-up tables in excel, that would be a pain and so I was wondering if there were formulae. Have you never wondered how long a non-compliant circuit would take to trip? Or how that compares with the maximum time to reach the cable limiting temperature (434.5.2...)?
No, I meant what I typed - Zs. Zx is (in BS7671) the impedance of a floor!
If there were a formula in terms of Zs, it would be trivial to re-write it in terms of fault current, since fault current = 230/Zs. However, as you've been told, there is no such formula.
Given that 127A is so many times the In of the 13A fuse, that would blow virtually instantly, even if the breaker did not operate first. How long the breaker would take would, obviously, depend upon its rating. If it were a B32 then, per graphs, it could take about 20 seconds - but the fuse would probably have blown long before that had a chance to happen..
You could - but I don't know what either you or the client could usefully do with that information.
As you've been told, there aren't formulae.
Yes, on occasions I have wondered that - and, when I have, I've looked at the graphs.
How exactly would you determine that?
Kind Regards, John
Well Zs is the earth loop impedance and for the purposes of my enquiry that's not relevant, I'm looking at L-N faults.
Quite, but 127A was just an example, and virtually instantly is not a time that may or may not be less than 0.4s for example
That's OK John you don't need to know everything.
Yeah that's why I "was" wondering rather than "am" although you seem more certain than in your first reply.
Well for temperature rise you would use the formula t = (k2*S2)/I2 (nice and easy in excel), then you could compare that with the actual disconnection time, if you could calculate that.
Yes, it would be a bit of a pain - but, as has been said, I'm not sure you would have any alternatives.
Probably even more of a pain but, rather than literally using look-up tables, you could embed them in Excel formulae. I've just very roughly extracted a few values for a B32 from the BS7671 graph. Without bothering about attempts at interpolation (which couldn't/shouldn't really be linear), if you put the following into cell C5 of an Excel worsheet, then it will give you an approximate disconnection time for any fault current you enter into cell A5 ...
=IF(A5<50,">1000 secs",IF(A5<60,"230-1000 secs",IF(A5<70,"100-230 secs",IF(A5<80,"60-100 secs",IF(A5<90,"43-60 secs",IF(A5<100,"33-43 secs",IF(A5<160,"11-33 secs",IF(A5>=160,"<11 secs"))))))))
Kind Regards, John
Not wishing to repeat, but -
If that complies then all is well.
It will nearly always be 'better' than L-E.
https://www.pat-testing-training.net/articles/fuse-operation-characteristics.php
Yes, it is; That's virtually all that matters.
If that complies then all is well.
Then measure the Neutral loop.
It will nearly always be 'better' than L-E.
I don't know if John meant to say "virtually"; instantaneous is 0.1s but the maximum Zs for 1362 fuses is for 0.4s which is all that is required.
https://www.pat-testing-training.net/articles/fuse-operation-characteristics.php
You could use s=[sq.rt.((I sq.) x t)] / k. If it is large enough then all is well - but for that you have to use a graph to find the time.
I've read all of this, and I still have no idea what skenk's difficulty is all about.
I'm not sure why you want to ignore the possibility of L-
With 127A through a 13A BS1362, "virtually instantly" would be an awful lot less than 0.4s - seemingly something around 0.005s according to the curves I have.
When I wrote my firts reply, I was certain that there would/could be no (accurate) theory-based formulae. As I said, it would be possible to mathematically model the empirical curves and, although I was not certain, I found it hard to believe that anyone would have bothered to do that - particular because I don't believe that there would be any generally-perceived 'need' (even if you have one).
That is a re-arrangement of an adiabatic equation, and hence is only valid if adiabatic conditions apply (i.e. if the disconnection time is less than about 5s) - and I thought the whole point of what you seem to be trying to do was to determine disconnection times which could well be quite prolonged? In any event, even for disconnection times ≤5s, it would not tell you what the temp rise was - merely whether or not it was regarded as 'safe'.
Kind Regards, John
Edit: crucial typo corrected
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"Zs is the earth loop impedance and for the purposes of my enquiry that's not relevant"
edited min/max ze/zs error
Zs can be many even hundreds of ohms if the circuit is RCD-protected but large L-N loop impedances (what I was calling Zx) mean low short circuit fault currents which means longer disconnection times and possible melting cables (you get 4-way adapters with 1mm cable for example). In a properly designed circuit volt drop limits mean you never get near that point but I have been seeing very long 13A extension cables and (very long) temporary installations sometimes in 1.5mm cable with C16 overcurrent protection. Even if the cable isn't going to melt I would still like to know approximately how long a fault will take to clear (without have to get the book out).
Yes that's what I'm measuring the L-N loop impedance/PSSC (tester shows both at the same time). Sometimes it's higher than the maximum permitted Zs, automatic disconnection has to come before the cable overheats (343.5.2). You have to check the graphs/tables.
Yes that's another way to write t = (k2*S2)/I2 I have both in the excel sheet I have been playing around with. It doesn't help with calculating disconnection time from fault current/loop impedance.
edited min/max ze/zs error
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