Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation 1

[Edelma_1]

Hello Everyone,

I was given a task by my boss to determine the EARTH FAULT CURRENT of a secondary substation (500kVA,6.6kV/0.433kV) using values from the primary substation, see the values below:

Primary Substation voltage - 6.6kV
Primary Earth Grid Resistance (Ra) - 0.1Ω
Primary Substation Classification - 196V (Cold)
Fault Level of primary substation - 1679A
Disconnection time - 1sec
Underground cables connected from the primary substation and their lengths (in kilometres):
3 x 1c 300mm² XLPE (35mm Screen) - 0.015
3c 0.25in² PILCSWA - 0.281
3c 300 PICAS & ACAS - 0.093
3c 0.25in² PILCSWA - 0.121
3c 300 PICAS & ACAS - 0.986

I would like to know if:

1. It is possible to get the fault current of the secondary substation using the values above.
2. The formula needed to do this.

Your fast responses would really be appreciated

 

[cranky108]

That looks like something I did for a university class.

 

[jghrist]

What do you mean by "primary substation" and "secondary substation"? Is there a transformer between them? Can you post a one-line diagram?

 

[Edelma_1]

there are two substations; the primary substation is the existing network substation from where the 6.6kV line is coming from (33kV/6.6kV) while the secondary substation is going to be the new one to be constructed ; a 500kVA 6.6kV/0.433kV transformer.

The two substations are separated by the cables I listed earlier

 

[7anoter4]

You have to know the actual cable resistance and reactance per km and the actual short-circuit current -or power-on supply source busbar. What do you have now it been only the rated thermal short-circuit for 1sec.
How ever you have to design the secondary busbar in such a way to overcome the actual short-circuit current.

 

[7anoter4]

It is very difficult to understand your problem. In my opinion, if you have to
supply 6.6 kV to a 500 kVA transformer and you have to chose one of these 5 possibilities. However, according to IEC 60502-2 , a 3*16 mm^2 aluminium cable is enough for this current of 43.74 A.

 

[7anoter4]

Another possibility is if the Fault Level of Primary substation 1679A it is actually at the 33 kV System short-circuit current and then all the 6.6 kV cables supply some other installations.
The System short-circuit [maximum allowed]=sqrt(3)*1.679*33 =95.968 MVA.
Then using IEC 60502-2 rated current recommended for aluminum conductors in underground run we get 1847.7 A -total supplied from Primary Station.
Checking for voltage drop we see if the cable cross section was not amplified in order to get the minimum voltage drop.
Then the primary transformer rating is 20 MVA.
Following IEC 60076-5 Ability to withstand short circuit , we get the minimum values of short-circuit impedance =8%
Then we have to calculate the total impedance from the System up to secondary of 6.6/0.433 kV terminal.
The base voltage 0.433 V
Zsystem 0.433^2/MVAsh=0.001953 ohm
Zfirstransformer=0.433^2/20*8/100=0.00075 ohm
According to one of the catalogues for 3.8/6.6 kV 70 sqr.mm aluminum cable Zcab=0.579 ohm/km [the minimum cross-section usually in stock]
Znew cable=0.433^2/6.6^2*0.579=0.002492 ohm
Zsecondtransformer=0.433^2/0.5*5/100=0.018749 ohm
Total Z=0.023945
Then the short-circuit current at second transformer secondary terminal is 0.433/sqrt(3)/0.023945=10.44 kA

 

[Kiribanda]

It is a matter of modeling from 6.6kV down to 433V level including cables.
You have given only the single phase earth fault current at the 6.6kV transformer terminals.
But you need to give the three phase sc current at the same 6.6kV terminals?
Otherwise I could have assumed an infinite bus at the 33kV level and calculate the 3-ph sc current at the 6.6kV terminals if
the MVA rating & Z of 33-6.6kV transformer was given. But those data are also not available in the diagram.
Pl. provide either,
1)MVA rating and the impedance of the 33-6.6kV transformer?
OR
2)3-phase sc current at the 6.6kV terminals of the 33-0.6kV transformer?

 

[Edelma_1]

The Neutral earth resistance at source substation is taken to be 2.31 Ohms, the impedance is at 18% and the fault MVA is 500MVA

 

[7anoter4]

Thank you Kiribanda, for your answer.
I thought the calculated short-circuit current it is for three-phase solid connected. However, if Zo>=Z1 the three-phase short-circuit is the maximum.
In ABB Switchgear Manual 11 Edition chapter 12 .1.3 Impedance voltage, voltage variation and short-circuit withstand it is noted:
With transformers of vector groups Dy and Yd, the single-phase sustained short-circuit current is about the same as the three-phase value.
The System short-circuit apparent power Ssh=sqrt(3)*33*Isc33kV=1.73*33*1.679=95.968 MVA, I think.
The First Transformer MVA=20 and Ukr=8% as noted. The apparent power of this transformer was stated
after summing all the currents supplied from-based on the conductor cross section [aluminum, I suppose] according to IEC 60502-2 standard. The short-circuit voltage [Ukr] it is according to IEC 60076-5 Table 1.
Also, the second transformer power, 500 kVA is transmitted by o.p. and 5% Ukr is also from IEC standard.

 

[7anoter4]

If we consider the last additions, it turns out that the short-circuit current on the 33kV side when the short-circuit is at the 6.6 kV terminals of transformer no.1 is indeed 1.679 kA.
Zsys=Vsys^2/Sshsys=33^2/500=2.178 ohm
Ztrf1=Vsys^2/Strf1*18/100=33^2/20*18/100=9.801 ohm
Ish33kV=1.05*Vsys/√3/(Zsys+Ztrf1) According to IEC
Ish33kV=1.05*33/sqrt(3)/(2.178+9.801)=1.670 kA
Now, the short-circuit impedance at second transformer 0.433 kV terminal will be:
Zsh=Zsys*0.433^2/33^+Ztrf1*0.433^2/33^2+Zcab*0.433^2/6.6^2+Ztrf2
Zsh=0.03099 ohm
Isc=0.433/sqrt(3)/0.03099=8.0668 kA
As you can see this problem is actually a riddle and I think I managed to give an answer.

 

[7anoter4]

As usual, I did not notice that the short-circuit current to the ground was requested in the o.p.
EARTH FAULT CURRENT and not fault current as in title.
Then, I also checked the statement that Zo is equal to Z1for Dy or Yd connected transformers from ABB and I found that it refers to an isolated transformer.
So, I concluded that Zo is much less than Z1 because it only refers to the last transformer.
In these conditions the maximum short-circuit current it is in phase-to-phase to ground fault
Now, single-phase to ground fault current I"k1=sqrt(3)*V/(2*Z1+Zo)=sqrt(3)*0.433/(2*0.03099+0.018749)=9.29 kA
and I"kE=sqrt(3)*V/(Z1+2Zo)=sqrt(3)*0.433/(0.03099+2*0.018749)=10.95 kA
Then, Kiribanda was right.