# Direction of travel of an Arc across 2 distribution lines

Can anyone explain to me why an arc between 2 conductors on a bare aluminium overhead line will always travel away from the source towards the load. Many thx,
Daniel
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Hello Daniel
In article

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
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John Rye
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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.
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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
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On 1/24/2011 18:32 PM, Daniel wrote:

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

You are though.
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ask a serious question, in good faith... and get insulted. pathetic....
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I recently replaced an old (original to the house) light switch. It was rated for use at 120V AC/DC. I don't remember the exact ratings, but the DC current rating was less than the AC current rating. It had a definite Snap when switched.
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"Michael Moroney" wrote in message writes:

I recently replaced an old (original to the house) light switch. It was rated for use at 120V AC/DC. I don't remember the exact ratings, but the DC current rating was less than the AC current rating. It had a definite Snap when switched.
The old ones did better than the new ones - that is the cost of silence.However, the old switches were definitely downrated for 120VDC
Don Kelly cross out to reply
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8X--------------

well at least it looks like i'm crapping in reasonable company...that's something i guess :(

ok..understood..
ok
ok
ok
agreed
ahhh ...ok...so they will in fact lag behind...but my reasoning was wrong. i forgot about the emmisivity with heat factor.

yup. live and learn,,,thx
now i hav to think whether or not i should do a revised presentation to my section....
bummer! :(
regards
Daniel
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Hello All

We are wandering somewhat off track now, and if we continue we probably ought to change the subject line.
The essential thing to remember about interrupting dc is that there is no natural current zero, There are then 4 possible approaches :-
(a) Use a tuned circuit to produce a current zero. This is theoretically possible, but I can not quickly think of a practical example.          (b) Stretch the arc so that the arc voltage is higher than the supply voltage.          (c) Cool the arc so that the arc voltage is higher than the supply voltage
(d) Split the arc into several arcs in series so that the total arc voltage is greater than the supply voltage because of the extra anode and cathode voltage drops.      In practice you can use a mixture of these techniques, but as been said the higher the voltage the more difficult it becomes to make a compact circuit-breaker.
John
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John Rye
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"John Rye" wrote in message
We are wandering somewhat off track now, and if we continue we probably ought to change the subject line.
The essential thing to remember about interrupting dc is that there is no natural current zero, There are then 4 possible approaches :-
(a) Use a tuned circuit to produce a current zero. This is theoretically possible, but I can not quickly think of a practical example.
(b) Stretch the arc so that the arc voltage is higher than the supply voltage.
(c) Cool the arc so that the arc voltage is higher than the supply voltage
(d) Split the arc into several arcs in series so that the total arc voltage is greater than the supply voltage because of the extra anode and cathode voltage drops. In practice you can use a mixture of these techniques, but as been said the higher the voltage the more difficult it becomes to make a compact circuit-breaker.
John ..................................................................................... (b),( c) don't work with HV AC simply because the space needed is so large. After all, an uncontrolled arc at 240KV can flare for a length of the order of 40-50 ft, judging by movies of faults on lines. Hence even in the early days work breakers would have short gaps in oil, trying to get enough pressure to sweep the arc away at current zero. Later this was improved upon (in England) by use of "explosion pots" and the bulk oil breaker probably peaked with a Westinghouse breaker (De-Ion?) where the pressure forced transverse flow of oil across the arc path. The history of circuit breakers is quite fascinating. Air blast breakers pretty well put paid to bulk oil breakers in the 50's. and these were modular bringing in your point (d) Typically a 69KV breaker had 2 gaps of the order of 1 inch in series and used a blast of 600psi air (which at heavy currents wasn't going to break the active arc but could blow out the arc products and replace them by clean air at the current zero. Modern breakers still depend on this whether air, minimum oil or SF6 etc.
As for (a) this, with an AC breaker, has been done with some success on HVDC systems. I am not up to date on progress but it appears that it is easier to simply turn off the rectifier inverters on both side of the fault- hence shutting down the line at the supply current zero. check:
http://www.claverton-energy.com/ttechnical-feasibility-of-complex-multi-terminal-hvdc-and-ideological-barriers-to-inter-country-exchanges.html
There have been other ones than what is cited. 100ms is pretty slow by today's standard. This one is a thyrister based unit at lower voltages.
http://sciencestage.com/d/3602337/two-stage-thyristor-circuit-breaker-short-circuit-characteristics.html
Don Kelly cross out to reply -------------------------
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And in addition....Flemings LH Rule..sums it all up nicely. Thanks for jogging my memory
Daniel
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wrote:

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.
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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.
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?"Rich." wrote in message wrote:

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
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Op 25-1-2011 5:36, Don Kelly schreef:

And now explain lightning.
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pim.

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And now explain lightning.
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pim.
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