Current's Effect on Electrical Arcs

[flexj624]

TL;DR Summary

How does current physically affect the ability of an electric arc to exist, and how is this mathematically represented.

Hello, I figured this belongs more in a physics thread than an electrical engineering thread, but please advise I am wrong. I am having a hard time finding the correlation between the amount of electric AC current flowing in a conductor and its affect on allowing an electric arc. Obviously the electric arc can only exist if the breakdown voltage of the medium it exists within (in this case just air) is met, but how does current affect this?

For example, say a high voltage power line with a nominal voltage of 7200V has 2 amps flowing through the conductor, if the circuit is opened in open air, a tiny spark may be visible as the conductor is pulled away from its connection point, but a large electric arc may not be pulled. But that same circuit carrying 30 amps would draw a significant arc if the conductor was pulled away from its connection point. Same voltage, same electric field, same surface charge density on the wire, but yet the arc can sometimes be several feet.

Obviously, the current plays into this somehow, but I am failing to see how. All of the equations I find do not have current as a variable, so it isn’t making sense. Does it simply have to do with the fact that as current increases heat also increases which ultimately lowers the breakdown of the air and allows the arc to increase in length? I am missing the “ah ha!” detail and I am hoping someone can point me in the right direction. Related formulas will be very much appreciated!

 

[DaveE]

Current in the wire will create an associated magnetic field which contains energy. EEs would model this as inductance (in the simplest version). Should something like a break in the wire decrease the current magnitude the energy in the B-field will induce a voltage with a polarity to keep the current flowing. This is described by the inductance definition $V=L \frac{di}{dt}$. The induced voltage is related to the rate of change of the current.

So, a break in the wire will cause a voltage spike to help induce an arc. It will also do this to resist the arc extinguishing. Higher current isn't the same as a higher rate of change of the current, but higher current does mean that there is more energy available to continue inducing voltage for longer.

The details of how a fault creates $\frac{di}{dt}$ isn't easily defined. I would guess that in some scenarios a higher starting current would support a higher $\frac{di}{dt}$, but this all depends on unspecified details of the fault.

 

[sophiecentaur]

TL;DR Summary: How does current physically affect the ability of an electric arc to exist, and how is this mathematically represented.

Obviously, the current plays into this somehow,

Current will only flow when the Electric Field (in a particular place) is enough to ionise the air locally and to strike the arc. This can either be due to the shape of a surface or the frequency.

There can be situations where the current flowing in one circuit, combined with the Reactive part of the Impedance, can produce higher local voltages, greater than the nominal voltage in that circuit.

Also (this may be the answer to your question) when a circuit has inductance and the circuit is broken, a high voltage can be induced across the gap. This induced voltage is equal to the Inductance times the rate of change of current. Any change in current will involve some induced voltages in any practical circuit.

The nature of arcs is that they have a negative resistance characteristic so the volts dropped may reduce with increasing current. Not Ohm's Law behaviour!

PS @DaveE snap.

 

[DaveE]

Arcs are initiated by a very high E-field strength that can ionize insulators (air, usually). Enough Ions need to be created to complete the circuit, i.e. jump the gap. Completing the circuit is normally easier than getting the process started because additional ions effectively reduce the remaining gap and increasing the E-filed strength. But there are examples, like corona discharge, where this doesn't happen.

Once an arc is started they will usually switch to more of an avalanche process where the ion generation is due to the kinetic energy of collisions with in the gas. This would normally cause a significant decrease in impedance as more ions are generated, as @sophiecentaur mentioned. So arcs are harder to start than they are to maintain, voltage-wise. The Jacobs Ladder is a nice example of this ionized conductor path.

In the case where the fault doesn't really alter the entire circuit much (eg. not a ground fault), than that availability of higher current from the source will generate more ions, more quickly. So high current arcs can be very hard to extinguish. The primary goal in switching is to kill the arc ASAP because you may not be able to later. In the case of arc lamps, or fluorescent lamps, there is external circuitry to control the current flow.

But this actually has little to do with how the arc is initiated.

 

[DaveE]

https://en.wikipedia.org/wiki/Townsend_discharge

 

[tech99]

An arc, as distinct from a spark, is usually initiated by drawing apart the two electrodes. It can occur with voltages of just a few volts, so there is no need for high voltages or inductive kicks. The positive electrode supplies ions for the flame and becomes eroded. I am not sure about an AC arc as it should extinguish on the zero crossings, but I presume sufficient ions remain to allow it to re-strike.

 

[DaveE]

It can occur with voltages of just a few volts

But still requires a high E-field magnitude.

I am not sure about an AC arc as it should extinguish on the zero crossings, but I presume sufficient ions remain to allow it to re-strike.

Yes. Although they typically won't extinguish and still be called "arcs". Because people won't often notice a 10msec arc, and if it doesn't extinguish at the first zero crossings, it probably won't later with more ions present. But this is exactly why switches have much lower DC ratings than AC. The zero crossings definitely help. Except for some exotic types, switches nearly always have some tiny arc when they open.

 

[flexj624]

Higher current isn't the same as a higher rate of change of the current, but higher current does mean that there is more energy available to continue inducing voltage for longer.

This makes a lot of sense and is exactly what I needed to put me on the right path. Thank you for your response!

 

[flexj624]

Also (this may be the answer to your question) when a circuit has inductance and the circuit is broken, a high voltage can be induced across the gap. This induced voltage is equal to the Inductance times the rate of change of current. Any change in current will involve some induced voltages in any practical circuit.

This is very interesting. I am wondering if all of the real world examples I have seen have been highly inductive, which is likely because of the nature of the circuits I have seen this on. Conversely, it is almost guaranteed for me to draw a significant arc when I de-energize capacitor banks by pulling fuses (and essentially opening the parallel connection). I attributed this to the higher e-field the capacitor was creating, but I wonder if the inductive nature of the circuit has any effect on this because the capacitive reactance is decreased when the capacitor is opened.

 

[Baluncore]

Conversely, it is almost guaranteed for me to draw a significant arc when I de-energize capacitor banks by pulling fuses ...

The arc forms and continues after the circuit is opened, simply due to the inductance of the wires in the circuit. The energy, in the magnetic field that surrounds the wires, must go somewhere when the circuit is opened, while the current falls to zero.