IEC60909 correction factor

[Murlinor]

Hello Friends.

For two winding transformers IEC60909 describes correction faactor for impedance. The c factor to be used is cmax. When calculating minimum currents, should the correction factors be omitted? Because the formula for correction factor is based on cmax and not cmin?

 

[7anoter4]

The calculation method, used in the IEC Standard 60909 determines the short-circuit currents at the location F using equivalent voltage source: cUn / 3 . This source is defined as the voltage of an ideal source applied at the short-circuit location in the positive sequence system, whereas all other sources are ignored. All network components are replaced by their internal impedances.
Hence, cmax [or cmin] is not an impedance correction.
IEC Standard 60909 classifies short-circuit currents according to their magnitudines (maximum and minimum) and their distances from the generator (far and near).Maximum short-circuit currents determine equipment ratings [and grounding elements], while minimum currents dictate protective device settings. See -for instance:

http://www.sueng.co.kr/pds/technical/IEC_요약본.pdf

 

[Murlinor]

Thanx for answering. I'm sorry for not being clear of what I was asking for. My apologies.

In IEC60909, when you calculate the impedance for transformers Zt a impedance correction factor is to be introduced:

Kt = 0,95*cmax/(1+0,6*xT), formula 12a in IEC60909-0.

This factor is then multiplied by the transformer impedance and hence you have the corrected transformer impedance:

Ztk=Kt*Zt

But as you can see the Kt factor is dependant on cmax. I assume that the correction factor Kt only implies for maximum calculations not minimum. If it was applied to minimum calculations as well IEC should have stated:

Kt = 0,95*c/(1+0,6*xT), where c would be dependant on max or min calculations.

I only want a confirmation, that I understand the standards correctly. That the impedance correction factor for transformers are only used in max calculations.

Best Regards

 

[EddyWirbelstrom]

To add to Murlinor’s question – not answer it – IEC60909-0 formula 12a is expanded to formula 12a as follows :

KT = ( Un / Ub ) * ( cmax / ( 1 _xT ( IT before / IrT ) sin ? T before ) formula 12b

Where
Ub = highest operating voltage before the short-circuit.
IT before = highest current before the short-circuit
? T before = powerfactor angle before the short-circuit

The 0.6 in formula 12a corresponds to a powerfactor before short-circuit of 0.80.

Formulas 12a and 12b for transformers are similar to formula 18 for synchronous generators where the voltage “behind the generator subtransient reactance” ( E”) is greater than the nominal voltage ( c Un / sqrt(3) ) which is used for short-circuit calculation. The lower the pf, the higher the E” and therefore the lower the correction factor KG.

Is this the same for transformers ?

 

[7anoter4]

You are right as 60909-0 introduced a correction for transformer, also. In the previous edition [IEC909] only for generator was a similar correction.
The 60909 Standard calculation way is not an actual but a conventional one. The minimum short-circuit current will state the protection setting. So ,I think, for a minimum short-circuit current cmax is good for this correction too, since doing so the transformer reactance will be higher and then "minimum" short-circuit current will be lower. The Standard does not define another correction.
The correction factors are based on statistical data and operating practice. If you need to be entirely correct you have to calculate the short-circuit current using superposition method [Thevenin's principle].

 

[Murlinor]

Thanks for the reply. Formula 12b can be used as an alternative.

I have to disagree with you 7anoter4. Assume cmax=1,1, then Kt will be <1 when Xt is higher than aprox. 7.5% or

Kt = 0,95*cmax/(1+0,6*xT), Kt-->0 when xT--> Infinite
(I think this is the reason why IEC writes cmax, to make sure that cmin is not used instead)

This means that transformers in high voltage systems and with large short circuit voltage ukr=UXr (Assumed that URr=0). This will eventually give a impedance reduction and increase short circuit currents. Therefore it is in my understanding that this factor shall be omitted when calculating minimum currents.

I hope you agree.

 

[7anoter4]

I agree with you, never-the-less for usual transformer where Xt=11%-14% [400 kV] and 4-8% [medium
voltage] Kt is close to 1.I never used correction for transformer [kt=1] except for Generator-Step-up Transformer Group.

 

[Murlinor]

Thanx, nice to know that I have understood the standard. Thanks for the help, appreciate it.

Best Regards