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PSC higher than MCB breaking capacity

Hi all,

Looking for a practical answer here;

I've read a couple of threads on this but to no clear solution:

  • MCB rated at 3kA breaking capacity
  • 7kA recorded
  • Upstream is a BS88 315A fuse

434.5.2 effectively says something around you should consider the supply side protective devices without giving any further advice (that I understand at least)

After reading discussions on here it seems that the downstream MCB will only be damaged if it tries to operate whilst the fault is present.

Am I correct that to prove this I need to find the worst case operating times at 3kA for the fuses and that should be bigger than the best case operating time for the MCB for this to be safe?

Regards,

Parents
  • The upstream BS88 will be enough for the fault current protection for the installation. It would be better to get a 7kA rated breaker, but, in the worst case, the breaker will be damaged, but, would it really be the worst case that does happen?

    There is added resistance from the source to the end of the final circuit. To get that fault current (and how has it been tested, MFT's are not reliable at higher currents) the fault would have to be at the source. On your final circuit, the impedance will reduce the fault current somewhat, and may even comply at the far end. 

    If it bothers you that much, and is likely to cause more than an inconvenience, an easy solution is to add a fuse (BS88 - 16kA+) in an enclosure,and run the final circuit through that, but I think that is going too far, your upstream BS88 is enough for most people.

  • Hi Alan, thank you for your response.

    We probably have at least 50 of these installations on site, so I need to either say we replace for breakers with appropriate ratings or prove why "The upstream BS88 will be enough for the fault current protection for the installation." 

    How can I prove this?

  • Hi Alan, thank you for your response.

    We probably have at least 50 of these installations on site, so I need to either say we replace for breakers with appropriate ratings or prove why "The upstream BS88 will be enough for the fault current protection for the installation." 

    How can I prove this?

Reply
  • Hi Alan, thank you for your response.

    We probably have at least 50 of these installations on site, so I need to either say we replace for breakers with appropriate ratings or prove why "The upstream BS88 will be enough for the fault current protection for the installation." 

    How can I prove this?

Children
  • I am surprised that what sounds like an industrial installation has MCBs with such a low breaking capacity, but perhaps they are old.

    You need to ensure that the PFC is high enough at the extremity of a circuit to trip in the event of a fault there, but low enough to ensure that the protective devices are not damaged by a fault at the supply end.

    It may be that the let through energy of the fuse will not be high enough to damage the MCBs, but I shall leave it to those cleverer than I to discuss that.

    In the mean time, it sounds as if the installation does not comply with 432. Note that if the MCBs were changed for 10 kA ones, and if the CPCs have been selected by calculation, you would need to ensure that they still comply because I²t is likely to be higher.

  • let though energy - the 'I squared t' or joules per ohm, is the makers info that you need to know to see what might happen in what order.

    In the event of a silver stake fault near the origin (rare) either you must accept that the fuse will race the breaker and  may operate  betore the breaker. If this happens, is it acceptable in your case- I presume more than one thing loses power if the fuse fails ?

    If you can get the I2t data on the breakers and fuse, I, and I'm sure many others, would be happy to walk through the calculation with you. typical fuse data  shows the pre-arc and let-through in that case as a 'bull-rush' graph, but it may be tables (page 6 of the PDF) . Any current current time product event as a lower rated fuse or breaker blows  that ends up below the "head" of the bull-rush will leave the fuse intact, any event involving more current or more time than the top of the head will always  be cut-off early by the fuse opening. Any event in the middle is undetermined.

    Even if it comes out on the wrong side, and the MCB could be  damaged at the energy  needed to ensure the fuse blows, this may be accepted, as such a fault will be a rare event, and the breaker is presumably in an enclosure that would catch any bits that fell off,  but you may need a spare breaker !!

    A 3kA breaker has been tested is you can be sure it can interrupt a 3000A fault current and live to fight another day, With more current it may work fine, but it may break and stay broken, or it may get hot enough at the contacts to weld up in the on position, (and in the extreme case of welding and no suitable upstream energy limiting device it may split open and emit plumes of fire and molten metal... ). In either of the latter cases it needs changing, but so long as something else cuts the power off in that case,  the outcome is not necessarily dangerous - depending what else happens as a consequence .

    The fuse does not limit the current like a resistor would but it does cut it off mid-cycle, reducing the energy dissipated and so reducing the heat damage to the breaker, and anything else  getting hot in the fault path.

    Mike

    edit, to give numbers to conjure with that are the right sort of size, an I2t of a million joules per ohm is 1000A for a second, or about 3000A for 1/3 of a second, or 10kA for 1/100 of a second or any similar combination.

    edit to edit, A way to think of  that million joules per ohm means that every ohm in the fault loop gets heated by 1 megajoule - about a single stick of dynamite worth of bang, or the energy in a large mars bar. Do  not confuse the two, the speed of energy release is a big factor here, you do not eat a mars bar adiabatically - that is to say there is time for the energy to spread about the body and be dissipated as it is slowly released.. A kilo joule or two in contrast is the energy in completely burning one smokers match, approx. (reference)

    Now that figure would be  plain bonkers on an LV system as if you really had a whole ohm in series the fault current would have to  be less than 250 amps, but you can still consider that some grubby contacts with a more credible resistance of say 0.01 ohm could get a heat damage equivalent to 1% of a stick of dynamite, more like a small box of matches going off,  and as such you might still not like it.