MV grounding with 'broken/open' delta circuit

[djrobinson]

I have a HRGS system using a wye primary (star point connected to ground) and delta secondary. The three phase delta is 'open/broken' at one point and a resistor included to limit the ground fault current. The question I have is, what happens to the ground impedance if the resistor is put in series with a contactor and the contactor is opened thus there will be no current flow in the secondary circuit? How does this affect the impedance to ground and hence the potential ground fault current

 

[djrobinson]

I think I have convinced my self that the megnetising current is significantly larger so the ground current should be much smaller

 

[DanDel]

I don't see what magnetizing current has to do with GF current.
Depending on where this resistor is connected between the delta system and ground, it may limit the GF current. If this is the only connection to ground in the delta system, then opening this connection with a contactor would theoretically introduce infinite impedance to ground. The only reference to ground would be capacitive coupling.

 

[djrobinson]

Thanks DanDel for the response. I have probably not explained myself too well. The only ground connection is from the star point of the wye transformer. What I probably should have said is the magnetising 'impedance' is so large that the potential ground current is smaller. I have tried to add a diagram of the circuit (badly drawn) if it helps.

You might need to turn the page the other way up!?!

 

[waross]

Your ground current will be the no-load current of the two unfaulted transformers. Because of the phase angles, the vector sum of the two transformers will equal the numerical current of one transformer.

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

 

[djrobinson]

Thanks for the posts gentlemne. I am pretty clear about what is happening now. Much appreciated