Autotransformer calculation short circuit impedance

[buskos]

Hello all:

We have a Autotranformer with the data from FAT:
Sn=300/300/1MVA, 400/115/10.5kV
Connection: Yyod5
Short circuit imepedance:
HV-LV u%=12.31 at 300MVA
LV-MV, u%=0.28 at 1MVA
HV-MV, u%=0.33 at 1MVA
LV-MV, u%=84.86at 300MVA
HV-MV, u%=100.08at 300MVA
For short circuit calculation at 10.5kV I have use the uk% between HV-LV uk%=0.33 and LV-MV uk%=0.33 at base 1MVA.Please I want to have comment from your.
Thank you

 

[7anoter4]

I think you do need three of them. I'd prefer all of them in absolute units, referred to 300 MVA and low voltage [115 kV].
The winding reactance will be:
XHV=0.5*(XHV_MV+XHV_LV-XMV_LV)
XMV=0.5*(XHV_MV+XMV_LV-XHV_LV)
XLV=0.5*(XHV_LV+XMV_LV-XHV_MV)
If the short-circuit is in low-voltage terminals the equivalent reactance will be:
Xeq=XHV*XLV/(XHV+XLV)+XMV
The direct, inverse and homopolar reactance: X1=X2=XO =Xeq.

 

[prc]

buskoslease check the combinations.There seems some abnormality in the reported impedance values.MV& LV interchanged.

 

[7anoter4]

You are right prc. First of all I was wrong and I interchange the low-voltage 10.5 Kv with
115 Kv. So Xeq=XHV*XMV/(XHV+XMV)+XLV has to be.
Second the LV-MV and HV-MV reactance are very elevated. It seems to me too, very odd.

 

[7anoter4]

Still I think it is a little more complicate. There are 3 possibilities to connect the medium voltage winding [when the main supply is from high-voltage terminals and low-voltage short-circuitated]:
1)Usually connected on the rated load. V'mv=Vhv; Vlv=0. [XHV||XMV and XLV in series].
If we shall neglect the magnetic current and the load impedance: Xeq=XHV*XMV/(XHV+XMV)+XLV
2) Connected in short-circuit V'mv=0;V'lv=0. XMV||XLV and XHV is in series.
then Xeq=XLV*XMV/(XLV+XMV)+XHV
3) Open circuit then XMV=infinite : Xeq=XLV+XHV