[Sunil]

We are using grounding transformer of following specs
rated voltage 33 kV, connection Zn(zigzag) , fault current/ph -266.7/30s, Zo=52.7ohms/ph Po=1244 W , Io=0.07 A
We are using grounding transformer of following specs
rated voltage 33 kV, connection Zn(zigzag) , fault current/ph -266.7/30s, Zo=52.7ohms/ph Po=1244 W , Io=0.07 A

How can I calculate the positive sequence impedance and the x/r ratios? Is there any data missing to calculate the same?

Topic

[Hamid]

The zigzag connection is a three-phase connection and is constructed as shown in Figure below. There are three pairs of windings, each having a 1:1 turns ratio. The left-hand set of windings shown in the figure is a conventional Y connection, a?-b?-c?, with the neutral N brought out. The open ends of the Y are electrically connected to the right-hand set of windings as follows: a?connects to the right-hand winding paired with to the b?-N winding, b?connects to the right-hand winding paired to c?-N winding, and c?connects to the right-hand winding paired to the a?-N winding. The opposite ends of the right-hand windings are brought out as the phase terminals a, b, and c. If three currents, equal in magnitude and phase, are applied to the three terminals, the ampere-turns of the a?-N winding cancel the ampere-turns of the c?- c winding, the ampere-turns of the b?-N winding cancel the ampere turns of the a?-a winding, and the ampere-turns of the c?-N winding cancel the ampere turns of the b?-b winding. Therefore, the transformer allows the three in-phase currents to easily flow to neutral. If three currents, equal in magnitude but 120

[sunil]

Thank you for the explanation. Here again I have a problem in calculating the Xo since Io specified has been very low from the data. Is it normal? And what to be done with the negative value in calculation?

[Hamid]

Ok, it seems the mentioned zero sequence current is grounding transformer no load current and must not mistake with fault zero sequence current. Indeed you can write: 1244 = 3R₀.(266.7/2)² R₀=0.023 Ohm

[NADER]

266.7 Amp. doesn’t need to be divided by 2 . Why do you think because of Zigzag configuration , we are bound to consider half of fault current !?

[Hamid]

Sorry,thanks a lot for your good comment, indeed it should be 3 not 2, because we can write: I_n= 3 x I₀

[Author]

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