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Phase to ground fault distribution system

Phase to ground fault distribution system

Phase to ground fault distribution system

(OP)
good morning from Spain.

In my zone the 15-20kv distribution system is grounded by a resistor that limits the earth current to 1000 or 300 amperes depending on the substation (cables or lines).

I'm analyzing an Oscillography of an earth to ground fault. In tis case the transformer has a 13 ohms that limits the 20kv system earth faults to 1000 amperes.

I had the open delta voltage, the high and m.t currents at the transformer and the current that runs through the resistor.

I can't find the relation between the current through the resistor and the open delta or m.t voltages taking in to account the 13 ohms value.

Any clue.

Thanks in advance

RE: Phase to ground fault distribution system

If you are using a grounding transformer such as a zig-zag or a wye delta-bank or if your transformer side in question is a resistance or solidly grounded wye, then you should be using √3 of the line to line voltage.

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

RE: Phase to ground fault distribution system

(OP)
Hello again,

The transformer is a Delta-Wye with the neutral of the M.T side as I said grounded wiht a 13 ohms resistor. For a (22kv/squrt(3))/13=978 Amperes.This is the máximum current trhroug the resistor.


I know int a phase to ground fault there are capcitives thar run through the sane phases and contribute to the end current of the affected line.

In my case only have the oscillography of the transformer. The current through the resistor equals the sum of the three phases of the transformer. The open-delta voltages equals the sum of the voltages. Till that everything ok. Supose the fautl is in the 0 phase. There is still a voltage at the cero phase. Lest call it V0g. How can i Know the V0n voltage since the VOn equals the Vng suposing the fault is near the subestation.

My cuestion is, at the moment of the phase to ground fault. What is formula that can equal the current trhoug the resistor taking into account you only have the open-delta voltage (3V0), the resistor (13ohms) and the V0g (faulted phase), V4g, V8g voltages. Is it posible to get it only with the open delta voltage.

Thanks for your answers.



RE: Phase to ground fault distribution system

antxon,
One clarification please if you do not mind.You have mentioned that your system is having a 13 Ohm resistor which limits the ground fault current to 1000A even though it is
little high value for a 15kV network. Does your 15kV ground fault protective relay system has a "BROKEN DELTA" ground fault detection method? (This is the first time I heard a low
resistance grounded system is having a "BROKEN DELTA" ground fault detection)

RE: Phase to ground fault distribution system

(OP)
In This case the network is 20kv. The broken delta is used to to detect the cut or damage of the neutral or the resistor and only used As an alarm.recently it is also implemented in the transformer Oscillography.

Another matter is that our 20kv transformers are really YNyn0 with the m.t star connected to earth through a 42 or 13 ohms 300 or 1000 amperes limit. although in theory (equivalentcircuit)there is no way for earth faults through the resistor. In Practice there is. I have heard it is because of the transformer construction, I heard the current goes through the core of the transformer. Any written theory to proof this reality?

Good afternoon

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