Author Topic: Tripping circuit breakers  (Read 7486 times)
Medved
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Re: Tripping circuit breakers « Reply #15 on: January 25, 2017, 12:45:40 PM » Author: Medved
I maybe said it not really correctly:
Normally (= by default) the breakers are sized according to wiring and are supposed to protect that wiring (and eventually switches and sockets/connectors):
4mm^2 -> B25A
2.5mm^2 -> B16A
...

Only in special cases, when you wan some additional protection (e.g. a motor against overheating by overload,...), you may use for a branch circuit a slower breaker, but its characteristic envelope should always be within the "B" characteristic corresponding to the used wiring.
But that is very rare, as the generic breakers are usually not accurate enough to really provide reliable protection, so a dedicated protection system is needed anyway (e.g. a thermal cutout, motor temperature simulator based protections,...).
So in most installations are usually the B-types of the highest rating the used wiring allows.
Anyway, the wiring should be rated for the higher inrush current (4mm^2 for a motor requiring C16A breaker), the installed breaker rating is based on those wires (B25A for the same example)

And because mainly modern units with either soft start and/or inverter drive do not exhibit any significant surge current, the higher rated B type allows higher installed power for the appliance when eventually upgraded.
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Re: Tripping circuit breakers « Reply #16 on: January 25, 2017, 02:25:55 PM » Author: Ash
This makes no sense

Initially lets consider the steady working current. The unit draws 10A

 - With 10A breaker it would work fine, and 1.5mm^2 cable would handle the 10A fine. This is fine for the unit and for the cable, but not fine for the breaker - i can imagine a 10A breaker loaded to the limit getting eventually damaged (or parameters drift etc) from the bimetal inside it staying hot all the time

 - With 16A breaker we need 2.5mm^2 cable. The 16A breaker and 2.5mm^2 would handle the 10A fine and both have good safety margins

The momentary starting current (80A for <1 sec) does not put any significant amount of heat energy into the cable. Even for an 1.5mm^2 cable



So :

 - For this unit 2.5mm^2 is okay

 - For momentary 80A the min breakers would be D10, C16, B32 (10*10 = 100, 5*16 = 80, 3*32 = 96), while B25 would be marginal (3*25 = 75, that might trip at normal startup if the breaker you got is on the "low" side)

 - D10 is irrelevant because of the 10A, no further questions

 - C16 protects 2.5mm^2

 - B32 protects 6mm^2, not 4mm^2 !!. B25 protects 4mm^2

"protects" means protects under any conditions (will trip faster than the cable overheats, for any value of current, from 0 up to the magnetic tripping range of the breaker and higher)



All the 6mm^2 or 4mm^2 overkill just because a rule that says specifically that C cannot be used in a home ??!!
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Re: Tripping circuit breakers « Reply #17 on: January 26, 2017, 06:45:17 AM » Author: Medved
The 10A steady and 80A for 1s means the 1.5mm^2 is way insufficient (talking about cable in a conduit or so, not a flex in free air).
The thing is, not only the bi-metal in the breaker, but as well the wires in the cable are heated by the steady 10A to its limit too, causing its degradation (same as the breaker).

The thing is, the cable operating temperature shall be all the time low enough, so even an eventual short circuit happening during the peak operating temperature and the related heat surge can not cause the wire to exceed safe temperature. So the operating peak temperature should still be sufficiently below the cable max rating, such margin is needed for the surge from the short circuit event.

So the fact the 80A for 1s does not overheat the cable is still not enough for the system to be considered safe. The cable should be so thick, it won't overheat even when a short circuit (so a short circuit current equal to the breaker capacity for the time it takes for the breaker to really interrupt the current) occurs just at the moment of the peak temperature from the inrush current surge.

The "B" curves are engineered exactly according to these rules, to just match the thermal characteristics of the typical cables.
So the maximum current profile you may get from a 1.5mm^2 wire cable, while still maintaining the safeguard margin needed for the fault events, just corresponds to the B10A breaker characteristic.

Cables in free air have better cooling, so generally allow higher loading, but still the allowed current profiles and relative fault margins are the same, so a 1.5mm^2 free air cable matches a B16A breaker.

C16 for such 1.5mm^2 free air may not exceed the maximum temperature at the motor start, but when a short occurs at the moment of peak temperature, the 6..10kA short circuit current running for the 5..10ms causes the cable to overheat. And that is considered here as not acceptable.
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Re: Tripping circuit breakers « Reply #18 on: January 26, 2017, 04:19:10 PM » Author: Ash
The 1.5mm^2 is brought as extreme example. The cable i consider correct for the application is 2.5mm^2. However, lets do all the math for 1.5mm^2



Values for Iz for 1.5mm^2 in our electrical code is 16A for single Phase (2 heating conductors) cable in conduit :

http://www.nekudat-coach.co.il/image/users/194922/ftp/my_files/template/Docs/doc19_חוק ותקנות בנושא חשמל - המשך_.pdf?id=9841738 - Page 11 Table 70.1

Googling for an European document i found this. Looks more restrictive at 13A, but notice that here it is for 3 heating conductors while we only have 2 in a single Phase circuit :

http://www.schneider-electric.com.au/documents/electrical-distribution/en/local/electrical-installation-guide/EIG-G-sizing-protection-conductors.pdf - Page 15 Fig G20

So 10A steady does not pose any problem for 1.5mm^2



Lets see what happens when overcurrents happen :

1. If the short circuit is in the kA range, i expect the immediate reaction of B10 vs C10 to be largely determined by the speed of breaker mechanics and arc extinguishing after the moment of spring release, not on how fast the solenoid in there pulls in.... It would pull in very fast in both of them. In short, the difference will probably be bigger between B10 vs B10 from different manufacturers (with different designs of all breaker components) than between B10 vs C10 of the same manufacturer (so only different in the solenoid pulling force)

2. If the short circuit is weaker (in the 100's A range maybe ?), there might be some difference in how fast the solenoid pulls.. But at what current exactly would damage to the cable become possible ?

3. We are interested then in what happens in the thermal region, for which the worst case is the max current to which the breaker still might not yet trip magnetically. That is 10In for C i.e. 100A

If an overcurrent like that happens after long time normal load, the cable is hotter to start with, but so is also the bimetal in the breaker - it will trip faster

Anyway, to the math :



Calculate for unit length = 1m

C10 breaker at 100A, still tripping only thermally. According to plot (my post on previous page) tmax = 2 sec



R = rx / A = 16.78e-9[ohm*m] * 1[m] / 1.5e-6[m^2] = 0.011 ohm

E = I^2Rt = ( 100[A] )^2 * 0.011[ohm] * 2[sec] = 220 J

V = Ax = 1.5e-6[m^2] * 1[m] = 1.5e-6 m^3

dt = E / VC = 220[J] / ( 1.5e-6[m^3] * 3.45e6[J/m^3K] ) = 43 K



Now lets guess what is the temperature of the cable after long operation at 10A, and what will happen after 2 sec at 100A :

Take the table from the Israeli code. The table is for 70 degC rated cable, at 35C ambient, with heat dissipation capability of a cable in conduit. Lets assume that if it says 16A, this means that after long time at 16A the cable will reach 70C exactly

dt (16A) = 70 - 35 = 35 deg

dt (10A) ~= dt (16A) * ( 10 / 16 )^2 ~= 14 deg

t (10A) = 35 + 14 = 49 deg

t (10A then SC) = 49 + 43 = 92 deg

70deg rated cable will handle 93deg peak momentary fault



Now all the same with 2.5mm^2 cable, C16 breaker :

R = rx / A = 16.78e-9[ohm*m] * 1[m] / 2.5e-6[m^2] = 0.0067 ohm

E = I^2Rt = ( 160[A] )^2 * 0.0067[ohm] * 2[sec] = 343 J

V = Ax = 2.5e-6[m^2] * 1[m] = 2.5e-6 m^3

dt = E / VC = 343[J] / ( 2.5e-6[m^3] * 3.45e6[J/m^3K] ) = 40 K



Lets calculate "2" from above (short circuit on 1.5mm^2, magnetic trip within half cycle). Question : What current would still be acceptable ?

Assume : 70degC rated cable can handle a single event of 160degC (with immediate breaker trip and subsequent cooling). Before the short circuit the cable was heated to 49deg by long term 10A

dt = 160 - 49 = 111 degC

E = dtVC = 111[K] * 1.5e-6[m^3] * 3.45e6[J/m^3K] = 574 J

I = sqrt ( E / Rt ) = sqrt ( 574[J] / 0.011[ohm] / 0.01[sec] ) = 2284 A

Well, the "couple 100's A" range is covered fully and beyond. At 2.2kA i think there isnt any difference between the reaction speed of B10 vs C10....

If anything, i'd ask other question - According to this calculation, how come B10 (or for that matter, any ordinary MCB with 1/2 cycle guaranteed reaction speed) is considered as being ok to protect 1.5mm^2 at 6kA or 10kA ? (assume short piece of 1.5mm^2, so that 6/10kA is possible at the end of it)
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Re: Tripping circuit breakers « Reply #19 on: January 27, 2017, 06:07:48 AM » Author: Medved

Lets see what happens when overcurrents happen :

1. If the short circuit is in the kA range, i expect the immediate reaction of B10 vs C10 to be largely determined by the speed of breaker mechanics and arc extinguishing after the moment of spring release, not on how fast the solenoid in there pulls in.... It would pull in very fast in both of them. In short, the difference will probably be bigger between B10 vs B10 from different manufacturers (with different designs of all breaker components) than between B10 vs C10 of the same manufacturer (so only different in the solenoid pulling force)

2. If the short circuit is weaker (in the 100's A range maybe ?), there might be some difference in how fast the solenoid pulls.. But at what current exactly would damage to the cable become possible ?

3. We are interested then in what happens in the thermal region, for which the worst case is the max current to which the breaker still might not yet trip magnetically. That is 10In for C i.e. 100A

If an overcurrent like that happens after long time normal load, the cable is hotter to start with, but so is also the bimetal in the breaker - it will trip faster

Indeed, the short circuit gives always the same amount of extra heat, regardless of the breaker rating, so always heats up the wire by the same temperature difference. In that the preheating of the bimetal plays no role at all.
But what is important is the peak wire temperature after this event and this depend on how hot the wire was just before the fault happened.

The worst case (for a given breaker) is then just on the border, what the breaker allows when just not yet tripping.
And this heat comes from two sources and mainly their combination: Steady constant current (that is dictated by the bimetal characteristic) and the eventual faster overcurrent surge, which still does not trip the electromagnet trigger.
The standard here really looks, which peak temperature the breaker actually allows just without tripping (at the worst case tolerance border; it is based solely on what worst case the breaker allows, not on any real load profile) and use that as a base for the extra overheat coming from the eventual short circuit.




If anything, i'd ask other question - According to this calculation, how come B10 (or for that matter, any ordinary MCB with 1/2 cycle guaranteed reaction speed) is considered as being ok to protect 1.5mm^2 at 6kA or 10kA ? (assume short piece of 1.5mm^2, so that 6/10kA is possible at the end of it)

The 10kA one has to have guaranteed shorter breaking time. The standard (based on wire cross section, resistivity, unity heat capacity and thermal budget) defines the acceptable integral(i^2dt), so the total allowed amount of heat energy going into the conductor during the short circuit is the same for 3kA current with 3kA rated breaker, 6kA for 6kA rated, 10kA for 10kA one, etc.
So in fact the 10kA types have to be faster than the 1/2 period, just because of this limit...

 So for 1.5mm^2 wire and 6kA short circuit current with 1/2 cycle switch off time (so 10ms) the wire will warm up by about 90degC (copper resistance, its temperature dependence, heat capacity, copper density; you may find these in Wiki or so). So when the wire could stand max 160degC peak temperature, the breaker characteristic should not allow the 1.5mm^2 wire to reach more than 70degC peak temperature under any current profile the breaker allows without tripping.
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Re: Tripping circuit breakers « Reply #20 on: January 27, 2017, 07:21:00 PM » Author: Ash
We can assume from the Iz table, that at exactly Iz current for long time, the cable (at the described installation conditions and ambient temp) will reach 70C exactly. Iz for 1.5mm2 is 16A, which will trip a 10A breaker after not very long

The max current the C10 would allow "forever" in the worst case is 14.5A. Its below 16A therefore the final temperature after long work at 14.5A will be below 70C

If we are talking about something like slowly ramping up current - i am not sure how to calculate the temp rise for that (how to choose the "steps" to really make it represent a worst case ?)



My calculation was with 49C initial. If we do it with 70C initial (that is what our standard requires) then add +21 to all temps, it will be 113 deg instead of 92 deg

Maybe for longer time (as it is heating to this temp within 2 sec and not within 1/2 cycle), but the cooling takes time too and if cooling is slow then the 2 sec difference maybe is not too big compared to the overall time the wire is at >>70C..

Either way, it is still below 160C so ok for the standard requirements

Then where does the objection to C in the "thermal for C" region come from ?



Calculated in the post above, got : 1.5mm2 10msec 2.2kA 111deg. Where the 10msec 6kA 90deg figure came from ?

Temperature dependence of Copper resistivity : In case of a short cable (where Isc is determined mostly by other stuff of the circuit, or by bad contact in the point of short circuit), rising resistance of the wire as it heats won't make the short circuit current go down.. It will only make the I2Rt energy go up. So temperature dependence does not make the case any better anyway



I still dont see, where C breaker is a problem
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Re: Tripping circuit breakers « Reply #21 on: January 28, 2017, 03:07:10 AM » Author: Medved
My calculation was with 49C initial. If we do it with 70C initial (that is what our standard requires) then add +21 to all temps, it will be 113 deg instead of 92 deg

Maybe for longer time (as it is heating to this temp within 2 sec and not within 1/2 cycle), but the cooling takes time too and if cooling is slow then the 2 sec difference maybe is not too big compared to the overall time the wire is at >>70C..

Either way, it is still below 160C so ok for the standard requirements

Then where does the objection to C in the "thermal for C" region come from ?

These temperatures are, what the breaker allows just before tripping. But then when the short occurs at this point, it will take the breaker some time to break it and during this time the wire will be exposed to the short circuit current. And this time is given mainly by the mechanics opening the contacts, so does not depend at all on how the breaker was preheated or so, it is always the same. That means this short circuit current will heat up the wire, rising its temperature.
As the current and time are not dependent on any load profile before, with given resistance it always creates the same amount of heat energy, which has nowhere else to go than the thermal capacity of the wire itself (the heat carried away by other cooling is negligible, as the event takes just very short time)

The exact numbers were somehow made up, I just wanted to explain the concept.



Calculated in the post above, got : 1.5mm2 10msec 2.2kA 111deg. Where the 10msec 6kA 90deg figure came from ?

The 90degC is what will be added to the lets say 70degC by the short circuit event (I hope I did the calculus right).

It is just a calculation of given amount of heat transferred into thermal capacance of a given amount of copper:
The power:
P = I^2 * R = I^2 * Resistivity * Length / CrossSection
The heat:
W = time * P = Time * Length * I^2 * Resistivity * Length / CrossSection
The wire mass:
Mass = Density * Length * CrossSection

And the temperature rise:
deltaT = W / (Mass * SpecificHeatCapacity)
= (Time * Length * I^2 * Resistivity * Length / CrossSection)/(Density * Length * CrossSection * SpecificHeatCapacity)

After compacting:
deltaT = (Time * I^2 * Resistivity)/(Density * CrossSection^2 * SpecificHeatCapacity)

You see, the wire length is out of the equation. It makes sense, as longer the wire, higher the resistance so the generated heat, but more mass where it is absorbed, so the length should indeed cancel itself out.

And if I did my calculus correctly, for the 1.5mm^2 copper (assume 0.02uOhm*m because of the higher temperatures), 6kA short circuit current (breaker rating) and 10ms switch off time (what I guessed from the standard curves) it yields about 90degC (that is, what I would expect, given the plastic tolerances use to be around the 150..200degC range for a short time exposure).



Temperature dependence of Copper resistivity : In case of a short cable (where Isc is determined mostly by other stuff of the circuit, or by bad contact in the point of short circuit), rising resistance of the wire as it heats won't make the short circuit current go down.. It will only make the I2Rt energy go up. So temperature dependence does not make the case any better anyway

The heat created within the wire is given just by the current and resistance of that wire alone. Of course, the current is, what depends on all resistances in the circuit, but I assumed the current as given.


I still dont see, where C breaker is a problem

The "worst case = steady maximum load" assumes, the thermal time constant of the wire is the same or longer than the one of the thermal trigger of the breaker. But I'm not sure if that is really the case. If the wire is of shorter time constant (that I would guess), higher the current, higher the wire temperature compare to what corresponds to the temperature of the thermal trigger. This mismatch is then restricted by the presence of the fast acting electromagnetic trigger.

If the bimetal could be relied on as being faster than the wire heating, the electromagnetic trigger would not be required at all - if the bimetal would always warm up quicker than the wire, it will always trip the breaker in time before the wire reaches the maximum temperature even on such high overcurrents as a short circuit (What is different for a short circuit compare to an other overload except the current value? I see nothing...).

So the fact the electromagnet is required there just means the bimetal warms up slower than the wire, so some sort of speedup is required.
And that means when you push the level of the fast acting electromagnet up (so allow for higher currents before the thermal part triggers), it results into higher temperature of the wire, giving it less margin for the extreme overcurrent faults (like the short circuit), when the mechanical delay become the dominant contributor.

What I'm quite sure the code here does require the breaker to be of the "B" type with just that explanation of the maximum temperature the breaker allows to the wire and the requirement of sufficient headroom on top of it for the further heating when the short circuit happens just at the moment of the worst case operating temperature.
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Re: Tripping circuit breakers « Reply #22 on: January 28, 2017, 05:25:00 AM » Author: Ash
Lets call some current "points" :

1 - the normal load (10A for 10A breaker)

2 - the "thermal trip worst case" corner (50A 4sec for B, 100A 2sec for C)

3 - short circuit in 100's A / 1's kA range



What profile are we examining ?

(assume 70c initial) --------> 3

1 --------> 2 --------> (just moment before the breaker trips thermally) 3

1 --------> 3

1 --------> some other points (what are they ?) --------> 3



The calculation for heat surge are exactly as i showed above (example of 100A 2sec "thermal corner" of C10)

R = rx / A = 16.78e-9[ohm*m] * 1[m] / 1.5e-6[m^2] = 0.011 ohm

E = I^2Rt = ( 100[A] )^2 * 0.011[ohm] * 2[sec] = 220 J

V = Ax = 1.5e-6[m^2] * 1[m] = 1.5e-6 m^3

dt = E / VC = 220[J] / ( 1.5e-6[m^3] * 3.45e6[J/m^3K] ) = 43 K
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Re: Tripping circuit breakers « Reply #23 on: February 06, 2017, 01:44:31 PM » Author: Medved
Problem is, you never know, what the short circuit current will be.
So if the Code allows 6kA, it has to count on 6kA heating the wires.
And that is 36x more heat than the 1kA (assuming the same switch OFF timing), so needs lower initial temperature, when the upper peak temperature limit is supposed to stay the same.

The thing is, the electromagnet trigger has the same nature of the response time vs current (it needs always the same momentum to trip, while the force accelerating the trigger lever is proportional to I^2, so the momentum to I^2*t the same way as heat buildup in any resistor).
The only timing difference towards the bimetal trigger is, the response of the rest of the mechanism is not negligible, because all timing is way faster.
So for currents just above the electromagnet trip threshold (up to 10x the rated current for B type) the electromagnet alone is still the slowest thing in the full switch off chain (electromagnet->contact moving away->arc extinguish), for currents in the kA range the forces are so great, the "rest of the mechanism" (contacts + arc) becomes the slowest thing in the chain.
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Re: Tripping circuit breakers « Reply #24 on: February 06, 2017, 04:33:28 PM » Author: Ash
What current profile is assumed ?

The absolute worst case i can think of is something like progressing isolation breakdown ending in a full Metal contact short circuit. First the current ramps up (first slow, then fast over the last few seconds) right up to the breaker thermal tripping point - at which point the wire might allready be in the 70..160 deg range, and finally, just before it trips thermally, come the 6kA

But i dont think the code acts like that

Also, can it be that the code line of thought is : "for the absolute worst case, no need to keep the cable intact (160 degC), but only to keep it within the non-ignition temperature of the isolation (~650 degC)" i.e. no immediate dangerous situation (untill the breaker is flipped back on and power applied to the damaged cable....), but the cable has to be replaced ?
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Re: Tripping circuit breakers « Reply #25 on: February 08, 2017, 10:53:12 AM » Author: Medved
Indeed, the code really aim at the absolute hottest wire the breaker allows. Does not matter at all, how realistic that scenario is in the real life, the only important factor is, whether the breaker allows such load (so wire heating) or not. The reason for this approach is, as the new appliances and devices are coming, the load profiles are changing and may easily become way different towards what was anticipated few years ago. And even in such environment the thing should remain unconditionally safe.

And indeed, the main priority is the safety. Whether the equipment remains usable after the event is not required.
Even if you look what the current breaking capability means: It is the current the breaker is capable to interrupt. It does not mean the breaker will work again.
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Re: Tripping circuit breakers « Reply #26 on: February 08, 2017, 12:14:23 PM » Author: Ash
I want to model the "absolute worst case" as i see it : Progressive isolation breakdown and then good contact short circuit. Lets say for a B16 breaker this time

The wire is overheated as much as possible in the moment exactly before the short circuit. What current profile would you take in the "isolation breakdown" part ? "4 sec x 5 In" is the worst or there is something worse that would fit in the breaker's worst case no-trip characteristic ?



The load profiles (for normal load) should never exceed Irms <= breaker nominal rating. Everything above that would already be a fault. Would the possible fault current profiles really change a lot to the worse in the future ? (Dont we allready have bad connection short circuits, stalled motors and so on ? Arent those the worst things that can happen ?)



How do the codes protect from a condition like keeping damaged devices that "still work" in service ?

Breaker that is badly damaged inside from arcing, but still flips on "normally" and makes contact ? Then this breaker might be by itself a source of overheating and fire (from resistive heating of burned contacts or deformed contacts touching each other in the wrong spot), or it may blow up or burn through the next time it trips for <6kA short circuit. The user may not realise that there is any problem with the breaker

Wiring isolation that melted a little from >160C at a previous fault but does not yet plain break down ? What if the isolation in question is isolation of a flex cable (which is considered double isolation) i.e. the cable is getting closer not only towards short circuit, but towards Pikachu to the user who touches it, possibly without a short circuit happening ?
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Re: Tripping circuit breakers « Reply #27 on: February 08, 2017, 02:19:55 PM » Author: Medved
I want to model the "absolute worst case" as i see it : Progressive isolation breakdown and then good contact short circuit. Lets say for a B16 breaker this time

That is a failure, not an overload.

The wire is overheated as much as possible in the moment exactly before the short circuit. What current profile would you take in the "isolation breakdown" part ? "4 sec x 5 In" is the worst or there is something worse that would fit in the breaker's worst case no-trip characteristic ?



The load profiles (for normal load) should never exceed Irms <= breaker nominal rating. Everything above that would already be a fault. Would the possible fault current profiles really change a lot to the worse in the future ? (Dont we allready have bad connection short circuits, stalled motors and so on ? Arent those the worst things that can happen ?)

Many loads count on some degree of short term overcurrent tolerance of the breaker.
Plus the main overload situations are expected to come from users connecting more loads than the branch is rated for. It is task for the breaker to "kick the user into the @$$" by tripping and shutting down, it is one of its main function in the installation.

How do the codes protect from a condition like keeping damaged devices that "still work" in service ?

The code assumes, the damaged device by itself is not a safety hazard.
But it is the breaker's task is to protect the installation, from being overloaded/damaged by the faulty appliance.



Breaker that is badly damaged inside from arcing, but still flips on "normally" and makes contact ? Then this breaker might be by itself a source of overheating and fire (from resistive heating of burned contacts or deformed contacts touching each other in the wrong spot), or it may blow up or burn through the next time it trips for <6kA short circuit. The user may not realise that there is any problem with the breaker

The breaker should either "not hold ON" (so impossible to arm), work normally or e.g. prematurely trigger. For teh described situation one of the methods is to keep the contacts in proximity to the bimetal, so when the contacts heat up, it causes the bimetal trigger to trip as well, even when the current is way lower than rated.


Wiring isolation that melted a little from >160C at a previous fault but does not yet plain break down ? What if the isolation in question is isolation of a flex cable (which is considered double isolation) i.e. the cable is getting closer not only towards short circuit, but towards Pikachu to the user who touches it, possibly without a short circuit happening ?

The mains cable should be properly rated against the breaker. Here the code expects, the eventual overload protection is within the appliance (or are unlikely due to nature of the appliance), so only hard short faults are expected (on 230V the arc developes so fast the low resistance high current state, any isolation fault between conductors yields hard short within miliseconds).
That means you may use 0.5mm^2 appliance cord even after 16A breaker - it only can not be preheated in any way (so the 2.5A limit) in order to safely (for the surrounding; not necessarily the cable itself) endure the eventual short circuit.

And each Pikachu is supposed to check the integrity of the cables, mainly when experiencing any problems with the appliance (this requirement is emphasized in every manual), the overheat damage is well visible on the cable...
No code could prevent problems with heavily neglected maintenance (that is, what not checking the cable is)...
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Re: Tripping circuit breakers « Reply #28 on: February 08, 2017, 03:22:58 PM » Author: Ash
That is a failure, not an overload

Why is there difference ? The breaker must be sized to protect the cable anyway



Many loads count on some degree of short term overcurrent tolerance of the breaker

Then lets choose another initial assumption : The cable is rated for 70degC normal / 160degC emergency temperature. Normal use is not emergency therefore devices cannot plan on exceeding the cable max temp

Now, lets say we have one such appliance with significant starting "heat impulse". The user plugs this thing in and switches it on. But what is plugged into the otehr receptacles on the circuit at the time ? The design of the new appliance cant take that into account, so we still dont know what is the possible initial temp of the wiring



The code assumes, the damaged device by itself is not a safety hazard.
But it is the breaker's task is to protect the installation, from being overloaded/damaged by the faulty appliance.


The breaker should either "not hold ON" (so impossible to arm), work normally or e.g. prematurely trigger. For teh described situation one of the methods is to keep the contacts in proximity to the bimetal, so when the contacts heat up, it causes the bimetal trigger to trip as well, even when the current is way lower than rated.

It might well be a safety hazard. Maybe not through the way of heating contacts but something else : Metal plating coat on everything inside, bad contact when making connection (so bigger arc when the breaker is closed into a short circuit by the user), depleted Plastic shields, ..

In most breakers the bimetal is at about similar distance from the contacts and from one of the terminals, and often connected to the contacts by a stranded Ponytail while it is assembled itself on the terminal (so better heat conduction between it and the terminal). Breakers often dont react to heat at the terminal (from bad connection of the wire), even when the breaker casing allready melts around the terminal. Why would it respond reliably to heat at the contacts ?



The mains cable should be properly rated against the breaker. Here the code expects, the eventual overload protection is within the appliance (or are unlikely due to nature of the appliance), so only hard short faults are expected (on 230V the arc developes so fast the low resistance high current state, any isolation fault between conductors yields hard short within miliseconds).
That means you may use 0.5mm^2 appliance cord even after 16A breaker - it only can not be preheated in any way (so the 2.5A limit) in order to safely (for the surrounding; not necessarily the cable itself) endure the eventual short circuit

Short circuits would probably heat this cable quite a bit without any preheating. Only question is, how much current would a short circuit in such appliance really make : Probably a short of a lamp or a transformer in a radio won't reach the kA range

What about heat from external source preheating the cable : where it connects to the back of a lamp holder and already is at its thermal limit (the isolation in this place is hard and discolored in an old lamp, possibly to the point where only the "2nd" layer of isolation remained in the way between 230V and the user in some worse example). In many lamps the cable right up to this point is going through the arm of the lantern or the like, so the place most prone to damage is not accessible for the user to see
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Re: Tripping circuit breakers « Reply #29 on: February 09, 2017, 05:33:25 AM » Author: Medved
The thing with any overload is, it should either be safe for the cable (and keep the margin for the short), or the breaker should trip and so prevent further heating. Regardless what caused the overload, indeed, for the breaker it does not matter at all. The point what I wanted to

For the cable, the assumption is, the cable is operated within its rated ambient temperature envelope. If not, the fault is not the breaker, but the heat source. Indeed, many such light fixtures are extremely poorly designed in that matter.
But the fact is, the heated part is on the end of the cable, so if the short happens behind that, the current become limited by the whole cable resistance, so cause less heat surge. This then would be the only design aspect making it somehow safe...
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