L-L fault & current in neutral

[lm227]

I understand in symmetrical component formulas there is no zero sequence current for a L-L fault in a 3 phase 4 wire system. Thinking about it practically, during a L-L fault wouldn't there be a significant current imbalance between the two faulted phases and the un-faulted phase? Doesn't unbalanced current in a 3 phase system 4 wire system then return on the neutral in this condition? How can zero sequence current be 0 if there is a large phase imbalance during a L-L fault?

 

[wwhitney]

Thinking about it practically, during a L-L fault wouldn't there be a significant current imbalance between the two faulted phases and the un-faulted phase?

Yes, in the sense that it is not a balanced 3 phase load, it is a single phase load.

Doesn't unbalanced current in a 3 phase system 4 wire system then return on the neutral in this condition?

No, this is true in the sense of unbalance meaning the sum of the 3 line currents is non-zero. That gives you the return current on the neutral. But with an L-L fault, the current on say L1 is negative the current on say L2. So the sum of the currents is zero, and there is no imbalance in this sense.

If you want to decompose the triple of currents (J, -J, 0) from a single phase load or fault in terms of symmetrical components, then the zero sequence current is given by the average of the 3 currents, which is 0.

The positive sequence sequence current is given by taking the 1/3 of the dot product with (1, e^{2pi * i/3}, e^{4pi * i/3}), while the negative sequence current is given by taking 1/3 of the dot product with (1, e^{4pi * i/3}, e^{2pi * i/3}). So the positive sequence current is J/3 * (1 - e^{2pi * i/3}) = J/6 * (3 + sqrt(3)*i). The negative sequence current just works out to be the complex conjugate of that, or J/6 * (3 - sqrt(3)*i).

At least, that's what I get if I've read the relevant parts of https://en.wikipedia.org/wiki/Symmetrical_components correctly.

Cheers, Wayne

 

[David Castor]

Draw it out and try to explain how the L-L fault current would show up in the ground conductor. It doesn't. Imagine an ungrounded delta system with a L-L fault. There will definitely be fault current, but no ground current.

 

[Julius Right]

In my opinion, if there are 3 wire it may be Line-to-line short circuit with Earth connection.

 

[jim dungar]

In my opinion, if there are 3 wire it may be Line-to-line short circuit with Earth connection.

Then it would be called a L-L-G fault which uses a different formula than a simple L-L fault.