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

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

(OP)

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

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

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

The two substations are separated by the cables I listed earlier

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

How ever you have to design the secondary busbar in such a way to overcome the actual short-circuit current.

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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.

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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?

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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.

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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.

## RE: Determining The Fault Current Of A Secondary Substation Using Details From A Primary Substation

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.