Phase-to-earth short-circuit calculation

[radug]

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.

 

[jghrist]

Yes, the Ø-grd fault current can be greater than the 3Ø fault. This is because on a delta-wye connection, the zero-sequence current is limited only be the transformer impedance, not by the system impedance.

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.

 

[radug]

Jghrist,

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?

 

[jghrist]

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?

Basically, yes. If this were a two-winding transformer, the following would hold (everything in per unit). With your three-winding transformer, things are a bit more complicated, but similar.

I_3Ø=V_s/(Z_s+Z_t)
I_1Ø=3·V_s/(2·Z_s+3·Z_t)

If you assume Z_s=0, then I_3Ø and I_1Ø would be equal and a bit larger than actual.

 

[mykh]

You may want also consider double line to ground. In this example, it will be even higher than line to ground. Check
[URL unfurl="true"]http://www.arcadvisor.com/example_2.html[/url] and
[URL unfurl="true"]http://www.arcadvisor.com/faq/mva_to_ka.html[/url] on how to resolve unsymmetrical faults using MVA method for short circuit analysis.

 

[7anoter4]

hi radug,
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

 

[7anoter4]

sorry,4/0 I meant,of course!

 

[radug]

Hi 7anoter4,

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.

 

[7anoter4]

hi radug
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