Direction of travel of an Arc across 2 distribution lines

First it is great to see a serious question on this newsgroup for a change !

The answer is straightforward once you realise that the arc is a conductor, and is free to move.

The current flowing out and back in the two "fixed" conductors produces a magnetic field between the two conductors perpendicular to the plane of the conductors.

The current in the arc is flowing in this magnetic field and will therefore be subject to force at right angles to the current and the field. That is along the line of the conductors.

Fleming's left hand rule (See Wiki) provides a simple way of showing that the direction of the force on the arc is away from the source of the current.

If you wish you can get to the same answer by more fundamental analysis.

The size of the force is proportional to the current and the flux density.

Since the flux density is also proportional to the current the force is actually proportional to the current squared.

This means that the effect is much more dramatic as current levels increase.

I spent part of my career investigating failures of high voltage switchgear, and one always had to remember that the major damage was where the arc ended up not where the fault started.

John

I agree that a real question is a welcome change. Now for the short version of what John just wrote. :-) Electricity seeks ground. The circuit path to ground is through the load. So in order to reach ground the flow is from the source, across the gap, through the load, and finally back through the grounded conductor.

Thanks for that, John. I came to that conclusion myself....but it's always nice to have someone confirm it. I made some nice slides showing how the arc would tend to move...

Without confusing the target audience with maths I first showed how current prefers to travel in a straight line. Then showed how the arc across the 2 conductors would attempt to straighten out...bowing slightly towards the load. But then due to the arc prefering the shortest distance between the 2 conductors, the ends of the arc would kind of edge along to keep the arc straight. And that way the arc would continuely be moving towards the load.

Seemed to make sense to me anyway.

Kind regards

Daniel

And in addition....Flemings LH Rule..sums it all up nicely. Thanks for jogging my memory

Daniel

Another effect you can see in many high-power arcs from switches and such is that the current-carrying air is very hot so is buoyant and tends to rise upward. This is sometimes enough 'stretching' of the arc to extinguish it.

Or just study the electric arc form of a "Jacob's Ladder" for a moment :-)

daestrom

Nice to hear from you again daestrom. a name from the days when this was a really good and interesting newsgroup before it was hijacked by some idiots and junk advertisers...

Daniel

thedirectionof the force on thearcis away from the source of the current.

So, you are saying that the arc can only initiate and occur *into* a "load"?

As far as attractors go, when there is a potential at the end of a conductor, an "exposed node" as it were, the only attractor for that node would be a conductor which is common to the "return" of the source for that "exposed node" potential. From the POV of those two "co-attractors" it matters not which node has the load, if any in line with it. The arc forms due to current flow direction rules and standard atmospheric air dielectric breakdown rules. So, from negative to positive.

Just as in the flipping of a coin, however, the arc can form from either node and travel toward either node as it cannot be determined when the air will ionize and arc over from one to the other, nor can it be declared as to which part of the cycle is peaking when the arc initiates, and that is what determines direction. From the POV of the two exposed nodes where the arc will occur, it matters not which is the directly connected node, and which has a load in line with it. To those nodes, the air is the very high value resistor that finally breaks down to a near short.

No. An electrical power source seeks ONLY its return conductor/ path. That in no way means Earth "ground" or any other "ground" has to be that path.

Current flow is a very specific thing, and its direction has nothing to do with where a "load" is located.

That would depend on the type of jacob's ladder. A free air version rises because the air that the arc heats rises off near it and pulls it along as it rises. The arc itself is NOT air, it is an electron stream the air gets out of the way for more than anything else.

Those that are in glass columns have very little gas in them, if not even a vacuum.

nder VARPC>---

He had his asshole moments then too. He is certainly not like some of the dopes that were here.

You are though.

I agree that a real question is a welcome change. Now for the short version of what John just wrote. :-) Electricity seeks ground. The circuit path to ground is through the load. So in order to reach ground the flow is from the source, across the gap, through the load, and finally back through the grounded conductor.

----------------------------- Actually, your interpretation of what John correctly said is quite wrong and is misleading. :-( a) two conductors - no "ground" is needed. (you can call one conductor "ground" but that is meaningless as it is simply the return path to the source as part of a closed circuit- which is required for any current to flow). b) The arc is parallel to the load so current through the arc doesn't go through the "load". Draw a diagram.

John has it right (I generally forget Flemings rules due to which is right hand and which is left hand and go to a "goose rule" atop the "rubber band theorem" - which work for simple situations like this avoiding the math involved in a vector cross-product). What John left out is that the current in the arc interacts with the magnetic field between the conductors to produce a force on the arc. In a Jacob's ladder, both thermal forces and magnetic forces make the arc move upward - but the magnetic forces are generally dominant.

Don Kelly cross out to reply

under VARPC>---

ask a serious question, in good faith... and get insulted. pathetic....

And now explain lightning.

Which kind? What part of dielectric breakdown don't you understand?

Draw the closed circuit.

It won't help, if you can't already see the circuit in your mind.

+-------+-------+ | | | | | .-. | | | | | | | | /^\ --- '-' ( | ) --- | \|/ | --- | | v ^ | | --- === === | GND GND === GND

And now explain lightning.