Ground grid design slg or dlg faults

[lkincer]

Should I use the single line to ground or double line to ground fault value in substation ground grid design? The DLG fault is somewhat higher but IEEE 80 is confusing to me.

 

[Kiribanda]

lkincer,
From your system studies if you find Z1> Z0, then the double line to ground fault will give the maximum ground fault current. A tansformer with solidly grounded WYE on the HV side (DELTA on low side) is
a typical example which gives a higher line to line to ground fault current on the high side than single line to ground fault.

 

[lkincer]

The substation is 69-13 kV. Transformer is 30 MVA, HV delta and LV wye solidly grounded.

 

[7anoter4]

If the transformer is located on the substation ground grid it has to be an OHL or MV cable supplying 69kV. It could be Zo<Z1[=Z2].Usually Zo=Z1/3.Then-as Kiribanda already said-the Line-to-line-to ground will be the maximum short-circuit current.
You have to get the actual data from the Utility.
As a drill let’s take an example:
From IEC 60076-5 Power Transformers- Ability to withstand short-circuit Table 2 – Short-circuit apparent power of the system:
For up to 72.5 kV in Europe S”k=3000 MVA in USA 5000 MVA.
If Z1=69^2/5000=0.9522 ohm Z2=Z1 and Zo=Z1/3
Neglecting resistances:
I''kE2=3*69/SQRT(3)*X2/(X1*(Xo+X2)+X2*Xo)
I”kE= 3*Io=3*69/SQRT(3)*0.9522/(0.9522*(0.9522*1.5)+0.9522^2/3)=68.46 kA.
If we’ll consider the resistances:
Usually R1=R2=R=X1/10. Ro=0.25 ohm [an average substation grounding grid resistance.]
The line-to-line-to-ground current will be [simplifying by (R+j*X)]:
3*69/sqrt(3)/sqrt [(2*R+3*Ro)^2+(2*X+Xo)^]=49.54 kA.
If the transformer is not located on the substation grounding grid but at a remote substation connected with an OHL or MV cable of 13[or 15] kV then the line-to-ground will be the maximum short-circuit current.
From the same IEC 60076-5
Table 1 – Recognized minimum values of short-circuit impedance for transformers with two separate windings:
25-40 MVA 10% [minimum].
X1=X2=Xo=10/100*13^2/30=0.5633 ohm
According to ANSI C37-010 X/R=10-25 R1=R2=Ro=0.5633/20=0.028 ohm.
Grounding resistance Rg=0.25 ohm
3*Io=3*13/sqrt(3)/sqrt((3*0.5633)^2+(3*0.028+3*0.25)^2)=11.94 kA.

 

[jghrist]

Use whichever fault current gives the highest zero-sequence current.