Short Circuit Calculation - Impedance Method - Three Winding Power Transformer

[pdias]

Hello All!

I need help to calculate the impedance of the Power Transformer in order to determine the Isc3k (Three-phase fault) by the impedance method.

I have the bellow circuit:

The Power Transformer have 3 windings: Primary (1N), Secondary (2N) and a Tertiary (3N). An Auxiliary Transformer is connected to the Power Transformer Tertiary.

The Auxilary Transformar will provide the AC current to the Auxiliary Supplies of the Substation.

To determine the Impedance of the Power Transformer, ZPT, since we have a Terciary (3N) I should consider only the impedance from the Primary to the Tertiary? (in that case the manufacturer must provide the short circuit impedance %, base on 450MVA, 1N-3N)

Thank you in advance

 

[Kiribanda]

Why you need to calculate?
It is very strange to know why you cannot find the name plate of such a large critical transformer operating at 400kV?

 

[pdias]

The power transformer is not yet manufactured and the nameplate drawing is not completed until the FAT.

Nevertheless, I want to know how I should do it.

 

[waross]

Power transformer = 450 MVA
Auxilliary transformer = 0.5 MVA
Ratio = 900:1
With that ratio, considering the Power transformer as an infinite source adds a negligible error.
Available Short Circuit Current of a 500 KVA transformer with an infinite source equals Rated Current divided by percent impedance.
500000 KVA / root 3 / 400 Volts = 722 Amps
722 Amps / 6% = 12,028 Amps, ASCC.
The actual first cycle transient current and the resulting magnetic forces will depend on the X:R ratio of the transformer and the point on wave of the short circuit.
The actual first half cycle transient current may approach a peak of 2.82 times the ASCC

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[7anoter4]

I agree with waross : you need only the last transformer impedance and the low voltage feeder to CB, if you need the three-phase short circuit current at CB [400V] busbar- then the base voltage is 0.4 kV.
The System impedance will be 0.4^2/Scc=4.57E-6 ohms
According to IEC 60909-2 the transformer impedance between high and medium windings may be 13-16% then Zac=0.4^2/110*16%=0.00022 ohm
-110 MVA= the third winding rated power [my approx.]-
Usually for a small transformer [of 500 kVA for ex.] all three windings
are of the same rated power-then 500 KVA.
The impedance Z2ac =0.4^2/0.5*6%=0.0192 ohm
Isc=400/sqrt(3)/0.0192=12028.13 A

 

[pdias]

Thank you waross and 7anoter4!