Step-up transformer = Star-grounded LV / Star-grounded HV. How to calculate EPR?

[tofulover]

Hi,

I am looking at a pre-existing electrical system which is an off-grid microgrid.

This consists of 3 off low voltage 415V 500kVA generators feeding into 1 off 2000kVA 415V/11000V step up transformer to then connect to 11kV infrastructure for a small village.

The 3 off generators at LV 415V are connected to an LV electrical switchboard which has the TN-C-S earthing topology. In Australia we called this the MEN system.
We have been informed that the SLG at the LV side of the TX is 20kA (being LV) and at the HV side of the TX is 0.5kA (being 11kV).
The step-up transformer is Wye-neutral-grounded at the LV side and also Wye-neutral-grounded at the HV side. Due to the proximity of the existing earthing, the client has connected the two earthing together.

The task that I am looking at is to establish what is the EPR of the system.
Do I consider only the 0.5kA SLG fault on the HV side as the EPR?
I am not too sure what is the impact to EPR by the cross-bonding between the TX HV earth and LV earth.

Any thoughts?

 

[Kiribanda]

You have mentioned that that two earthing systems are bonded together. No worries. That means any earth fault current on the 415V side will rise the EPR at the 11kV side too. Therefore, you have to consider 20kA earth fault current which is maximum out of the two values and calculate the EPR. If it is higher than the allowable level then you have to reduce the earth grid resistance.

 

[tofulover]

You have mentioned that that two earthing systems are bonded together. No worries. That means any earth fault current on the 415V side will rise the EPR at the 11kV side too. Therefore, you have to consider 20kA earth fault current which is maximum out of the two values and calculate the EPR. If it is higher than the allowable level then you have to reduce the earth grid resistance.

may be a silly question from me - wouldnt the 20kA SLG at LV side - most of the fault current will return via the earthing cable and then via the MEN back to the neutral of the generators being the source?

 

[Kiribanda]

What you say is correct. If the 415V system is totally with cables, yes the earth fault currents will go through the cable amour etc and the fault current discharged to the earth is very small meaning you are taking the split factor as almost zero. Unless I see the detailed dwg of the earthing system or model it on CDEGS, I cannot confirm the correct split factor.

 

[tofulover]

What you say is correct. If the 415V system is totally with cables, yes the earth fault currents will go through the cable amour etc and the fault current discharged to the earth is very small meaning you are taking the split factor as almost zero. Unless I see the detailed dwg of the earthing system or model it on CDEGS, I cannot confirm the correct split factor.

Thanks. If there is a SLG on the HV side - in this case 0.5kA, does this 0.5kA then go back to the neutral of the star side of the HV winding only?

Does the 0.5kA HV SLG fault affect the LV side at all?

 

[cherryg]

In TN-C-S system, the earth fault path is through the least impedance path i.e the PEN conductor back to the source. So, for the solidly earthed Generators (each contributing 6.67kA; calculated % reactance = 10.43%), the earth fault current will not enter earth and contribution to GPR is negligible (unless you want to design for damage to the PEN conductor). Assuming that one generator (out of three) has a damaged PEN conductor, the individual contribution will now see the earth resistance. If the earth resistance is say 0.5ohm, the earth fault current is 240V/0.5A = 480A which is negligible compared to 6.67kA. Thus GPR for the 415V side is negligible and true for TN-C-S systems. Now coming to the 11kV system, if the power is being transmitted by overhead line without a grounding wire, then the L-G current of 0.5kA will be flowing back to the source for any remote end L-G fault. GPR for this small current will be low.
Summary: If the fault current does not flow through the earth (e.g., in TN-C-S systems with intact PEN/PE conductors), GPR is not a concern because the return path is via the conductor, not the earth

 

[tofulover]

In TN-C-S system, the earth fault path is through the least impedance path i.e the PEN conductor back to the source. So, for the solidly earthed Generators (each contributing 6.67kA; calculated % reactance = 10.43%), the earth fault current will not enter earth and contribution to GPR is negligible (unless you want to design for damage to the PEN conductor). Assuming that one generator (out of three) has a damaged PEN conductor, the individual contribution will now see the earth resistance. If the earth resistance is say 0.5ohm, the earth fault current is 240V/0.5A = 480A which is negligible compared to 6.67kA. Thus GPR for the 415V side is negligible and true for TN-C-S systems. Now coming to the 11kV system, if the power is being transmitted by overhead line without a grounding wire, then the L-G current of 0.5kA will be flowing back to the source for any remote end L-G fault. GPR for this small current will be low.
Summary: If the fault current does not flow through the earth (e.g., in TN-C-S systems with intact PEN/PE conductors), GPR is not a concern because the return path is via the conductor, not the earth

Thanks for your reply. I agree with what you said.
Much appreciated.

 

[waross]

Summary: If the fault current does not flow through the earth (e.g., in TN-C-S systems with intact PEN/PE conductors), GPR is not a concern because the return path is via the conductor, not the earth

In TN-C-S system, the earth fault path is through the least impedance path i.e the PEN conductor back to the source.

Not safe assumptions.
Is the EPR not considering the instance when all or most of the fault current does return through the earth?
And remember that as well as increasing impedance with distance, the line conductor and the earth return conductor form a voltage divider and the voltage driving current through the earth may be significantly greater than 1/2 of line to ground voltage.
Additionally, the cross sectional area of earth supporting a fault current increases with a corresponding decrease in resistance/impedance per unit length at greater distances.

 

[tofulover]

Not safe assumptions.
Is the EPR not considering the instance when all or most of the fault current does return through the earth?
And remember that as well as increasing impedance with distance, the line conductor and the earth return conductor form a voltage divider and the voltage driving current through the earth may be significantly greater than 1/2 of line to ground voltage.
Additionally, the cross sectional area of earth supporting a fault current increases with a corresponding decrease in resistance/impedance per unit length at greater distances.

Hi Waross, what would be the steps then for me to determine the EPR in my case?
Some high-level steps would be most appreciated.