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NFPA 70E 130.6(M) states: After a circuit is de-energized by the automatic operation of a circuit protective device, the circuit shall not be manually re-energized until it has been determined that the equipment and circuit can be safely energized.

29 CFR 1910.334(b)(2) states: After a circuit is deenergized by a circuit protective device, the circuit may not be manually reenergized until it has been determined that the equipment and circuit can be safely energized.

My first thought would be, "If the breaker is rated for the available short circuit current, I should be able to close it into a short-circuit without a problem." However, I've heard of cases where it seems that an appropriately-rated molded-case breaker was closed into a short-circuit, which resulted in an arc starting inside the breaker, leading to arc flash injuries.

What is the mechanism of such a malfunction? Age and abuse are possible contributing factors. Is there more to it? Is there some aspect of the construction of a molded-case circuit breaker that can result in an arc flash when closed into a short circuit, even if it was a brand new breaker of appropriate rating?

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  • \$\begingroup\$ You are asking about doing things that are 100% being told not to do in the two specs that you have snipped. No circuit breaker should ever activated when the load is known to be shorted out - That is just common sense. \$\endgroup\$ – Michael Karas Apr 18 '18 at 21:16
  • \$\begingroup\$ Any 'quality' breaker will trip on overload, however repeated trips will burn out the contacts of any breaker. You are doing the equivalent of 'destructive testing'. That is your answer. \$\endgroup\$ – Sparky256 Apr 18 '18 at 21:24
  • \$\begingroup\$ @MichaelKaras The certification process for IEC 60898-rated breakers (and likely others) requires it to withstand that happening multiple times. \$\endgroup\$ – Someone Somewhere Dec 19 '18 at 5:24
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Depending on the supply, the prospective short circuit current could well be in the kilo-amps range, even if the breaker is only rated for tens of amps.

In such extreme cases, the breaker is simply required to break the current. It may not survive the damage. While it is breaking such a large current, there may be a substantial flash.

I wouldn't want to be the person holding the lever of a circuit breaker while it's in the process of breaking a 10kA fault.

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I left out of my question what may be a relevant bit of info: a breaker I heard of was closed into a dead short repeatedly. The third or fourth actuation resulted in an arc in the panel.

The path of an arc through air is through ionized gas. The breaker extinguishes the arc, but the ionized gas is still there, and still ionized. After repeated attempts to actuate the breaker, the pressure produced by the successive arcs could expel the residual ionized gas out of the breaker. That gas can then bridge other conductors, such as the upstream bus bars in the panel, resulting in a new arc.

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  • \$\begingroup\$ This is one possible mechanism. The second is that the circuit breaker may take some time (i.e. 2 seconds) to trip after closing-onto-fault - depends on the protection settings in use. A 2-second duration fault will release plenty of energy. I have seen incident reports where shorting wire was left on a circuit breaker following testing - CB was then closed onto the shorting wire which vapourised and initiated an arc fault. \$\endgroup\$ – Li-aung Yip Dec 19 '18 at 4:07
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Here is my "theory". When attempting to engage (energize) a circuit breaker into known shorted equipment, the instant current change is huge, so the inductance of AC transmission lines would result in infinite voltage potentials across the breaker contacts, which would produce arc flash. Mechanical contact bouncing will boost the effect, since the breaker has a finite time to disengage. So the main problem is in parasitic inductance of supply wires on both ends of the breaker. No breaker construction with molded case can fix this.

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If the breaker is rated for the available short circuit current, I should be able to close it into a short-circuit without a problem." However, I've heard of cases where it seems that an appropriately-rated molded-case breaker was closed into a short-circuit,

Why do you think that the short-circuit current was within the rating of the circuit breaker? Short circuits, by their nature, draw more than the breaker is rated for - lower currents don't cause a trip.

With very large currents available, the contacts of the breaker (which cannot open as designed due to the reset forces) may vaporize. The resulting metal plasma will continue to ionize, with resulting flash damage.

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  • \$\begingroup\$ The OP is referring to the breaker's interrupting rating, not its trip setting -- i.e. how much fault current the breaker's contacts can safely clear when it trips \$\endgroup\$ – ThreePhaseEel Dec 19 '18 at 3:23
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The circuit breaker is rated indeed for the rated fault. However, an fault will generate a lot of heat with the current dissipation inside the device and over-voltage that results from cutting the current. This put a lot of stress on the device.

This mean that there is an chance that the device could be destroyed on a same fault next time depending on the nature of the fault

Also there is issues linked to the load itself like residual voltage. A capacitor bank could have a lot of residual voltage which take a lot of time to decay. If by accident, you would manually re-energized the load while the voltage on the bus is at opposite phase with the voltage inside the capacitor bank. You would end up producing massive over-voltage that could injure you.

So the norms says that you must be sure that both the protective and the loads are in a state which will not produce even more problems when energized.

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I'm not knowledgeable enough to answer questions in this field, but I see so many confusing answers, and critical comments to the question itself, that I had to jump in. I'm knowledgeable enough to understand a well-written answer (which should be comprehensible to the OP or anyone else who can't figure out the answer on their own; I see so many answers that could only be understood by someone who already knows the answer). My terminology is my own. If I'm dead wrong, I apologise for wasting your time, and for this rant.

When a circuit is powered on, it evolves to its operating state, as designed, under the control of its components, including protective components. Completing a circuit by closing a breaker does not allow the circuit to evolve in the way it was designed to do; the protective components might not have time to handle the surge of current, the breaker itself might not have time to trip, and other components can be damaged.

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