Ground Grid Design

[CuriousElectron]

Does available fault contribution from the utility determine the ground grid design of a facility? I'm thinking in terms of GPR, which is a function of fault current. In situations where the available fault duty is unknown, it sounds it may be better to be on the conservative side and design the system with a low ground impedance.
Just sharing my thoughts out loud

Thanks,
EE

 

[7anoter4]

I am not so sure this is always the right thing to do in order to reduce EPR. In IEEE 80/2013 chpt.15.4 “Effect of substation ground resistance” is stated that decreasing the grounding grid resistance it may increase the Ig [the part of the fault current flowing through grounding grid]. Usually when this resistance is negligible with respect the other impedances involved in Ig equivalent impedance it could be o.k. However if the other impedance are negligible with respect to ground resistance then Ig≈V/Rg and EPR=Ig*Rg =V.
First the grounding grid resistance depends mostly on soil resistivity and area and less on the grid itself that means we cannot decrease more the grounding grid by change the grid assemble.
Let's take an example.
Substation area=30x30=900 sqm
Soil resistivity[ρ]=1000 Ωm
System short-circuit power=6000 MVA
System voltage=110 kV
Rs[circular metal plate at zero depth]= Rg=ρ/4√π/A=14.77 Ω
Let’s take a grid of 11 columns and 11 rows at 0.5 m depth. Rg=ρ[1/LT+1/√20A(1+1/1+h√20/A)]
Rg=15.9 Ω
If[the rms value of symmetrical ground fault current]=6000/√3/110=31.49 kA
Zsystem=V^2/MVA=110^2/6000=2.017 Ω
That means if the fault current return entirely to the source then If will be this value.
Since the current is forced to flow through the grounding resistance the current will be
Isys=Ig=V/√3/(Rg+jXsys)=110/√3/(15.9+j2.017)=3.96 kA EPR=Ig*Rg=63 kV
If we take the minimum grounding resistance Rg=14.77 Ig=4.2 kA and EPR=62.9 kV.