LG and LL fault 1

[22574]

Hi,

Anyone can provide me some formula or
computation why a L-G fault is higher that L-L
fault in distribution system mostly with XLPE feeders?

Existing Transformers are solidly ground.

Thanks

 

[7anoter4]

Ik3=VoltL-L/sqrt(3)/Z1 Z1=approx.X1
Ik3=VoltL-L/sqrt(3)/X1
Ik1=SQRT(3)*VoltL-L/sqrt((X1+X2+Xo)^2+(3*Rg)^2)
If Rg [grounding resistance]=0 and X1=Xo then Ik1=Ik3.
If the Xo is less than X1 then Ik1>Ik3 - If there is a capacitance connected between live phase to ground –or grounded neutral-for instance[20-30 microF].
Xresult=-L*w/(w^2*L*C-1) L=circuit to ground inductance C= circuit to ground capacitance.
Up to resonance w^2*L*C<1 then Xresult inductive. From resonance Xresult capacitive.
The XLPE cable capacitance is less than PVC as the XLPE permittivity is only 2.3 and of PVC 4-8. For a usual medium voltage cable the parallel resonance occurs only for 400-500 km length.
But, also the transformer winding capacitance to ground may reduce the phase to ground reactance. Usually Xo=>X1 then Ik1=<Ik3.

 

[dpc]

Delta-wye core form transformers can have zero sequence impedance that is lower than the positive sequence impedance. This will give higher current for L-G fault at the transformer terminals.

 

[7anoter4]

I agree with dpc. In core-type transformers Xo = .85 to .9 times X1.In shell type Xo=X1, indeed.

 

[jghrist]

Zero-sequence impedance does not include primary system impedance for a DYn transformer. Pos- and Neg-sequence impedance does. Assuming primary system impedance Zs1=Zs2=Zs and transformer impedance Zt1=Zt2=Zt0=Zt, then
I3Ø = 3·V/(3·Zs + 3·Zt)
I1Ø = 3·V/(2·Zs + 3·Zt)
I1Ø = I3Ø*(3·Zs + 3·Zt)/(2·Zs + 3·Zt)

 

[7anoter4]

Correct. But Zs<<Zt usually so may be neglected. And in the short circuit case is also an arc resistance, almost always, which reduces the Ik1.