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Grounding Riddle No.36 - Grounding Transformers
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?
Author : Sunil - From: India
 
#1
Tue, March 27th, 2012 - 16:56
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°out of phase with each other, are applied to the three terminals, the ampere-turns in the windings cannot cancel and the transformer restricts the current flow to the negligible level of magnetizing current. Therefore, the zigzag winding provides an easy path for in-phase currents but does not allow the flow of currents that are 120° out of phase with each other and we can assume Z1=Z2 = very big.
Also for obtaining of R0 and X0 you can use the following equation:

P0 =3 R0.I02

Z02=X02+R02 
Author : Hamid - From: Iran
 
#2
Wed, March 28th, 2012 - 11:21
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? 
Author : sunil - From: India
 
#3
Wed, March 28th, 2012 - 16:42
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
 
Author : Hamid - From: Iran
 
#4
Thu, April 5th, 2012 - 11:19
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 !? 
Author : NADER - From: IRAN - MALAYER
 
#5
Thu, April 5th, 2012 - 14:52
Sorry,thanks a lot for your good comment, indeed it should be 3 not 2, because we can write:
In= 3 x I0 
Author : Hamid - From: Iran
 
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