Help with magnetic force from fault current on overhead wire

[Utility Engineer]

I'm trying to resolve an issue I'm experiencing with some overhead power distribution lines, was wondering if anyone can help me out.

Here is the scenario: three phase, 4 wire, wye connected 13.4kV power distribution system. 556.5kcmil bare ACRS wire is mounted on a crossarm 24" apart over a 200' span between poles. When a fault (short circuit) is experienced downstream of this location due to a separate unrelated event (equipment failure etc.), the wires at this location are coming together and creating a second fault (short circuit). I can see visual evidence of this happening in the form of burn marks and pitting on the wires, as well as I have eye witness accounts of this happening. What I suspect is happening is that the fault current from the original fault (~20,000 amps) is passing through this location and is creating enough magnetic force between the wires to draw them together and make contact, thereby creating a second fault.

What I want to know is how do I create a formula that tells me how much current is required to make the wires touch, or conversely how far apart I have to separate the wires in order to have them not touch for a given fault current.

I assume the variable inputs are:
Wire spacing
Wire tension
Wire sag
Span length
wire diameter
wire weight per ft

Can somebody help me get pull together the formulas I need to figure this out.

Attached is a diagram to help hopefully.

 

[NascentOxygen]

When a fault (short circuit) is experienced downstream of this location due to a separate unrelated event (equipment failure etc.), the wires at this location are coming together and creating a second fault (short circuit).

So this downstream fault is not sufficient to throw breakers anywhere? And it persists for a sufficiently long time that two overhead wires can be drawn together.

 

[Baluncore]

This question is more complex than you might at first expect.

There will only be an attraction between two conductors if the currents flow in the same direction.
If two conductors form a return circuit, the opposite direction currents will repel.

Consider a short between two phases that results in a loop current that returns to the generator. The phases will have high opposite direction currents, so the wires will repel. The third phase will not have any unusual current, but one of the two shorted phases might be pushed into it by the other. If all wires have the same catenary, then if only one is forced it will rise inside and above the other, so contact is still possible, but less likely.

A lightning strike to one phase will induce half the strike current in each of the other phases. Those two phases might then be attracted together.

If you look along a catenary in a strong cross-wind you will see the wire is hanging more sideways than down, the wind drag force is greater than the gravitational force. Could it be the wires swinging with different periods in gusty winds that results in contact ?The sag of 61” and the separation of 24” identify the angle of swing needed for one or two conductors at the middle of the span to contact.
For one conductor only, the swing angle will need to be = Asin( 24” / 61” ) = 23.17°.
For two conductors attracting each other, the angle will be = Asin( 12” / 61” ) = 11.35°

The mass of the conductor is 0.717 lbs per foot. That is 1.067 kg per metre. The vertical gravitational force will be 1.067 * 9.8 = 10.457 Newtons per metre length. If the line swings sideways to hang at an angle, theta, from the vertical, then the horizontal restoring force, Fg, will be;
F_g = 10.457 * Sin( theta ).

For one conductor, F_g = 10.457 * ( 24” / 61” ) = 4.114 N.
For two conductors, F_g = 10.457 * ( 12” / 61” ) = 2.057 N.
Note that the arcsine and sine cancelled.

That restoring force will be countered or balanced by Ampere's force law between conductors.
Where the separation, d, is measured in metres, the magnetic force per metre length is;
F_m = 2*10^-7 * I₁ * I₂ / d Newton
https://en.wikipedia.org/wiki/Ampère's_force_law#Equation

Then there is a complexity. As the wire begins to be attracted, the conductor swings closer, so d is reduced and the magnetic force increases. As the conductor swings, the sine function increases the gravitational restoring force. Does F_g or F_m rise fastest? Under what circumstances can the wires get close enough to overcome F_g and contact.

In your picture you show two wires on one side of the poles, only one on the other. The middle wire should switch sides at each pole. That way the clearance at mid-span will be increased.

To work out what is happening we need a better description of the fault and the direction of fault currents in the three phases.

 

[magoo]

There's an IEEE paper on this subject from 2003, Overhead Distribution Conductor Motion Due to Short-Circuit Forces by D. J. Ward.

What I found was that the forces from a phase to phase fault causes the two faulted conductors to swing apart and the subsequent swing causes them to contact each other. The currents in the two faulted phases are the same back to the source, and the force/unit length is the same. So if you have a longer span ahead of the initial fault location, this is where the contact is most likely. I had seen cases where we had the fault occur, a recloser operated and then the upstream circuit breaker operated. If you see the burn marks, you become a believer. We verified this with digital relays at the substation and were able to see that the subsequent fault currents were higher than the initial ones.

So the relative considerations - high fault currents, long spans, short phase spacings, large amounts of sag, large wire size, etc.- all factor into the analysis.

It is not a quick formula type of solution since it involves calculating the swinging motion of the two faulted phase conductors for the duration of the fault.

 

[Baluncore]

The code here apparently comes from the D.J. Ward paper.
https://github.com/powerdistributio...blob/master/calculators/conductor_slapping.md

 

[magoo]

Yes, that is the code I used to perform the analysis and write the paper.

 

[Baluncore]

@ Magoo, now I understand. Many thanks for your post and the reference.

In my initial analysis I had not considered the return swing and slap after the wires were pushed far apart.

Unfortunately your paper is well secured behind a paywall, so it cannot be read by people in my position.
Only the github code and the abstract are available to me.