What happens to the load current of healthy phase on L-L fault?

[Roleci]

As far as I know no ground current flows during line to line faults. But when there is a load current on healthy phase,it must flow as ground current according to the formula.Let’s Say there is b-c fault. I0= 1/3(Ia+Ib-Ic). In this case 3I0 equals to Ia which is the load current before fault. Are there zero sequence currents on L-L faults or not? Can someone explain this please?

Thank you in advance.

 

[cuky2000]

In contrast with LLG fault, there is no zero sequence current on a L-L fault.

Since I_a=0 & I_b=-I_c, the circuit can be represented with the (+) & (-) sequence in parallel through the fault impedance Z_f.

Therefore, the zero-sequence is inactive for L-L fault since there is no a path to ground.

 

[7anoter4]

If the system presents a working neutral the healthy phase may continue to work-mainly if the other phases are disconnected.
Before the disconnection a voltage sag may be present on this phase also.
However, as cuky2000 said no grounding current will circulate due to L-L short-circuit.

 

[Roleci]

In calculations,it is assumed there is no load current before the fault.Because load current is too small compared to fault current. So that's why IA=0. But I'm saying what if Ia is not equal to zero. As I said before it must flow as ground current in a grounded system.

 

[waross]

It will depend on a number of things:
The nature of the load.
The impedance of the fault.
The impedance of the conductors from the source to the fault.
The impedance of the source.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[cuky2000]

The Imbalance load current during ground fault will be split in the neutral, another ground path, and a fraction of the load current is injected into the earth. Usually, this portion of the current traveling in the earth is usually small creating not issues with the step and touch potentials while tripping the protective devices.

 

[davidbeach]

Where was Ia going before the fault? If it was returning to the source on B and/or C it still does that. There are three parts to the circuit; the source, the load, and the fault. The only difference between pre-fault and post-fault is the fault. You can also use superposition to analyze them separately and add the results. If neither load nor the fault have a ground component then the sum won't either.

 

[protoslash]

I think we are over-complicating things.

The healthy phase is not involved in the fault, so its current is just load current. As long as the health phase voltage is good during the fault, you will see full pre fault load current after the fault occurs. (assuming load is all single phase for simplicity)

Whatever symmetrical component formula you are using for L-L fault, it is ignoring load. As soon as you add in load, there is no magic formula anymore, you have to convert the load into impedance and solve the system case by case.

I just modeled a simple 4 wire multi grounded distribution system. A L-L fault anywhere does not affect load current on the healthy phase. And yes there is zero sequence current on the line since load is converted to impedance and they provide zero sequence path to ground.