three-single phase transformer impedance

[metta0810]

I have three-single phase transformer, each is rated at 250kVA, 34500-2404x7212/12490Y.

The name plate shows if the secondary side is connected in series the voltage is 7212V, 34.7Amps. If the secondary is connected in parallel the voltage is 2404V, 104 amps.

The impedance shows as 5.15% at 34500-7212 VOLTS.

The secondary side of the three-single phase transformer is connected to 2.4kV Bus.

Since, the impedance is obtained based on 34500-7212 Volts, Do I need to convert the impedance based on 34500-2404? If so, how do I convert the impedance?

Thanks.

 

[metta0810]

Attached is the transformer name plate

 

[dpc]

The same impedance can be used, since it is in percent based on the kVA rating. In reality, there may be some slight variation in the leakage reactance at the various secondary connections, but the differences should be minor.



David Castor

http://www.cvoes.com

 

[rbulsara]

My thoughts are same as dpc's. The nameplate says 5.15% itself is "approximate" so it is close enough. Only accurate way to know is to perform the short circuit test, by definition. For the purposes of short circuit calcs, the same value should serve the purpose.

Rafiq Bulsara

http://www.srengineersct.com

 

[metta0810]

Thank you all for your answers.

 

[odlanor]

ZOld=15.5 ; %
NbOld=150; %kVA
VbOld=7.212; %kV
NbNew= 150; %kVA
VbNew= 2.404; % kV

ZNew = ZOld * (VbOld/VbNew)^2 * (NbNew/NbOld)

ZNew => 139.50 %; ?????

 

[odlanor]

ZOld=5.15 ; % ERRATA
NbOld=150; %kVA
VbOld=7.212; %kV
NbNew= 150; %kVA
VbNew= 2.404; % kV

ZNew = ZOld * (VbOld/VbNew)^2 * (NbNew/NbOld)

ZNew => 46.35 %; ????