IEC60909-0 Impedance correction factors for Transformer
IEC60909-0 Impedance correction factors for Transformer
(OP)
For both transformers and synchronous generators,IEC60909-0 specifies an impedance correction factor proportional to 1 / ( 1 + x * sin pre-fault powerfactor angle ).
Where x = per-unit reactance of transformer and x"d for generator.
For generators I already calculate the subtransient voltage E" = (((V+x"d*I*sin(load pf angle ))^2+(x"d*I*cos(load pf angle))^2)^0.5
Why introduce the IEC60909-0 transformer impedance correction factor ?
Why is the IEC60909-0 transformer impedance factor required ?
Where x = per-unit reactance of transformer and x"d for generator.
For generators I already calculate the subtransient voltage E" = (((V+x"d*I*sin(load pf angle ))^2+(x"d*I*cos(load pf angle))^2)^0.5
Why introduce the IEC60909-0 transformer impedance correction factor ?
Why is the IEC60909-0 transformer impedance factor required ?






RE: IEC60909-0 Impedance correction factors for Transformer
if you consider only the part 0 of the IEC 60909, you can't find an explanation of the correction factor K.
Some indications seem to appear from the note at the end of point 3.3.3 but they are misleading.
You can find the true explanation in the IEC 909-1 / 1991, even if it is referred to the old 909 / 1988.
At point 2.2 of 909-1 you can find that the K factor is introduced to correct the approximated calculation method of 909 and to make it similar to the superposition method.
Actually the superposition method makes a vectorial sum of the pre-fault currents and the fault currents. While the 909 takes into account only the faulted conditions.
Regards
Alex