Phase-to-earth short-circuit calculation
Phase-to-earth short-circuit calculation
(OP)
Hello,
I have the following system:
System voltage: 30 kV
3-phase symmetrical short-circuit: 10553 A
phase-earth short-circuit: 5508
Transformer: 30kV/0,69kV/0,4kV
1600kVA/1250kVA/350kVA
I need to calculate the maximum phase-to-earth short-circuit in 0,69kV and 0,4kV. Can it be greater than the 3-phase maximum short circuit? if so, in which scenarios?
I need the data to determine the cross section of the ground grid electrode. Many times I have seen studies using the 3-phase short-circuit value but I am not sure if that is correct.
Thanks, Radug.
I have the following system:
System voltage: 30 kV
3-phase symmetrical short-circuit: 10553 A
phase-earth short-circuit: 5508
Transformer: 30kV/0,69kV/0,4kV
1600kVA/1250kVA/350kVA
I need to calculate the maximum phase-to-earth short-circuit in 0,69kV and 0,4kV. Can it be greater than the 3-phase maximum short circuit? if so, in which scenarios?
I need the data to determine the cross section of the ground grid electrode. Many times I have seen studies using the 3-phase short-circuit value but I am not sure if that is correct.
Thanks, Radug.






RE: Phase-to-earth short-circuit calculation
You can assume an infinite source (zero impedance) and calculate the 3Ø fault, and use this for the ground grid sizing. The Ø-grd fault will be less than this.
RE: Phase-to-earth short-circuit calculation
You say that in delta-wye (as is my case, I forgot to say) the zero-sequence current can be greater than 3phase, but then you say that I can assume that the phase-ground fault will be smaller than the 3-phase fault with infinite source.
So, you say that with delta-wye:
3ph short circuit taking the source into account will be smaller than phase-ground, because in that case, the source impedance is ignored in the calculations and so, I can have a phase-ground short-circuit of up to the 3-phase short-circuit ignoring the source?
RE: Phase-to-earth short-circuit calculation
I3Ø=Vs/(Zs+Zt)
I1Ø=3·Vs/(2·Zs+3·Zt)
If you assume Zs=0, then I3Ø and I1Ø would be equal and a bit larger than actual.
RE: Phase-to-earth short-circuit calculation
http://www.arcadvisor.com/example_2.html and
http://www.arcadvisor.com/faq/mva_to_ka.html on how to resolve unsymmetrical faults using MVA method for short circuit analysis.
RE: Phase-to-earth short-circuit calculation
If the transformer is located inside the fences –on the grounding grid- the current which produces potential rise is the phase-earth current on 30 kV [5508 A] . The short circuit current on the 0.69 kV and 0.4 KV side passes only through the grounding grid itself and does not enter the ground beneath. Never the less I propose a minimum 95 sqr.mm copper [lead covered is better] or 1/0 awg from mechanical considerations.You may check also if the copper withstand 3 phase symetrical short-circuit for 1 second[thermal stability].
Regards
RE: Phase-to-earth short-circuit calculation
RE: Phase-to-earth short-circuit calculation
In fact, I was considering 95sqr.mm copper, but I need to check phase-to-ground faults in 0.69 and 0.4 kV. I know that for grid design I have to consider only 30 kV, but for conductor section I have to consider the highest short-circuit value at any voltage level.
RE: Phase-to-earth short-circuit calculation
I suppose the 3 winding transformer parameters are :
S12=1.6 MVA uk12= 7% U1=30 KV Y connection
S13=1.25 MVA uk13= 13% U2=0.69 KV delta connection
S23=0.35 MVA uk23=5% U3=0.4 KV Yn connection
Calculated impedance[ all for U1=30 KV] will be:
Z12= uk12/100*U1^2/S12= 39.4 ohms
Z13= uk13/100*U1^2/S13= 93.6 ohms
Z23= uk23/100*U1^2/S23 = 128.6 ohms
Z1=(Z12+Z13-Z23)/2= 2.2
Z2=(Z12+Z23-Z13)/2= 37.2
Z3=(Z13+Z23-Z12)/2= 91.4
Direct [positive] sequence reactance[approximately=impedance] for short circuit on 0.4 KV transformer terminals is:
X1=Z1+Z3=2.2+ 91.4 =93.6 ohms
The negative-sequence reactance=positive sequence reactance.X2= X1 =93.6 ohms
Let's take Xo=0 [usually Xo=X1 ]
Iko=U1^2*SQRT(3)/U3/(2*X1)= 30^2*1.73/0.4/2/93.6=20.8 KA
If the grounding cable is lead covered the maximum temperature permissible will be 300 degrees C[the start temperature
will be 40 degrees C] The minimum sqr.mm cross section area has to be 93.
For 0.69 KV windings no zero-sequence current[Xo=infinite]
X1=Z1+Z2=2.2+37.2= 39.4 ohms
Ik3=U1^2/U2/sqrt(3)/X1= 30^2/.69/1.73/39.4= 13.9 KA
If you feel necessity of increase cross section you may add a 240 sqr.mm copper grounding cable but only along the
Possible return grounding current[e.g. along cable trays or ducts conveying the supply cables]
As I never succeed to include a sketch in the post see pls.
the attachment.
Regards