Force on conductors during a short circuit fault

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Summary:

How do I relate fault current to the force on two conductors?
If I have two parallel bars serving as a DC bus with a capacitor across the output and there is a short fault event, I know the peak current is limited by V/X. The instantaneous current would be a function of the time RLC time constant. How do I relate this to the force on the conductors due to this transient?
 

Answers and Replies

  • #2
Baluncore
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Welcome to PF.

Where is the short fault ? At the capacitor, somewhere along the transmission line, or at the generator.
The current will be limited by fault protection.
The peak force will occur at peak current.
 
  • #3
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If the fault was at the capacitor, I would expect that to produce the highest current since the resistance and inductance is minimized. If it occurred at the generator I would expect di/dt to be the minimum since the inductance would be highest. The peak would also be the lowest since the resistance would be the highest. Is this line of thinking correct?

I'm interested in knowing how to calculate the peak force but I'm not sure how to do this.

Regardless of if the peak is limited by resistance or protection I'm still interested in knowing how it varies with time time in general. I know this isn't realistic for an actual design, but I'm just trying to understand it right now. If there is no protection, my thought is that the current should look sinusoidal. It would increase as a sinusoid with a resonant frequency of f=1/(2pi*sqrt(LC)). The peak would be limited by I=V/R.

The current would initially be DC. As the short happens the current increases so there is a time varying current. That would also induce a time varying magnetic field so I figured I could use maxwells equations but I'm not sure how or if that is even correct. Am I on the right path here or is there something else that I'm not considering? Again, I'm not really considered with the generator in this case. I'm only trying to get the theory where the capacitor is across the bus and the bus is shorted at any point along the bus.
 
  • #4
anorlunda
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Short circuit calculations can be simple or very complicated. There are software packages for the complicated ones.

I searched "short circuit calculation" and I found a wealth of sources, and tutorials for both simple and complicated cases. One of them said this:

A capacitor in an AC system charges and discharges in a controlled manner every half cycle, based on the sinusoidal driving voltage and system impedances. When a fault occurs, the system voltage is suddenly changed and the capacitor discharges at a rapid rate, with a high discharge current. The current is greatest if the fault occurs when the capacitor is charged to the maximum at a voltage peak. Only the impedance between the capacitor and the fault limits the discharge current.

The current will “ring down” based on circuit resistance and reactance. The resistance provides damping and the interaction between the system reactance and capacitor determines the frequency of the oscillating current.
 
  • #5
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anorlunda,

Thanks for the link. That's good to see my It would work pretty much as described. I've found stuff on how to calculate the short circuit current, but not so much the force generated during the short circuit event. Software is one way to calculate the force.

The Lorentz force seems like it may be helpful here since that's what I would use to show wires attracting or repelling. Since the magnetic field is time varying I'm not sure if I also need one or more of maxwells equations. It is a DC bus so there would be some e-field between conductors. That would also fit in with maxwells equations.

Basically, everything seems coupled and I'm trying to figure out how to tackle this.
 
  • #6
Tom.G
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Just as a worst case first approximation:

Since any inductance will slow the discharge, and hence the peak current, try calculating the Lorentz force using the peak current accounting for only the resistance of the busbars and the ESR (Equivalent Series Resistance) of the capacitor.

If the assembly withstands that, you have a starting point to decide on a safety factor you are comfortable with. Since this seems to be a pulse operation, include the possibility of mechanical fatigue in the mounting.

Cheers,
Tom
 
  • #7
Baluncore
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If the short circuit is between the conductors then the currents will flow in opposite directions so the conductors will be pushed apart. The DC situation is the simple computation case. Obviously, the direction of the force is the same with AC, although it is then pulsed.

There is another thread on this subject;
https://www.physicsforums.com/threa...ce-from-fault-current-on-overhead-wire.895215

Code for the computation on transmission lines is here.
https://github.com/powerdistributio...blob/master/calculators/conductor_slapping.md
 
Last edited:
  • #8
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Just as a worst case first approximation:

Since any inductance will slow the discharge, and hence the peak current, try calculating the Lorentz force using the peak current accounting for only the resistance of the busbars and the ESR (Equivalent Series Resistance) of the capacitor.
A larger inductance will slow the discharge but what if the protection never enabled for some reason and the current was able to reach the peak just in a longer time compared to a much smaller inductance. Since the peak is still relatively the same would the peak force be the same? If it's dependent on di/dt then it should be higher right? That's part of why I'm trying to determine how to build in the time varying portion of this.

To calculate the Lorentz force, F=q(E + v x B), I need the B field. I'm not sure which one of maxwells equations to use to get that. Is E the E-field between the conductors during normal operation found by E=V/d? Charge q would be the electron charge so is v then the peak current? I kind of doubt but that's where I'm lost and why I have these questions. Thanks for your help.
 
  • #9
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If the short circuit is between the conductors then the currents will flow in opposite directions so the conductors will be pushed apart. The DC situation is the simple computation case. Obviously, the direction of the force is the same with AC, although it is then pulsed.
Thanks for that reference. I'll read it and see if I can relate it here.

If the fault is not cleared, shouldn't the DC fault be the same as an AC fault in this case? You would basically have an LC tank so there would be some resonant frequency with some damping due to the bus and ESR of the caps. I figured the current would then continue to ring down at the resonant frequency until the fault was cleared, something burnt up, or all energy was dissipated.

I'm still not sure what equations to use here to even calculate the peak force in this case. As I mentioned above, to do F=q(E + v x B) don't I need to find the Bfield from one of maxwell's equations? And what E and v do I use? This is a new area to me but I just sort of remember these equations from physics classes a long time ago.
 
  • #10
Tom.G
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A larger inductance will slow the discharge but what if the protection never enabled for some reason and the current was able to reach the peak just in a longer time compared to a much smaller inductance.
With inductance included, the peak will be smaller than for the purely resistive case because the capacitor will be partly discharged while rising to the peak.

Try these videos:

(above found with:
https://www.google.comsearch?&q=magnetic+force+between+two+parallel+conductors)
 

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