IEC shortcircuit two phase faults

[luis7006]

Hello,

I would like to receive your feedback related to shortcircuit analysis using IEC 60909 methodology to calculate two phase and two phase to earth faults because, there are not all of the currents determined by the software.
We already read the standard and it talks about only initial and peak currents for these sort of faults.

This is very different from the three phase and single phase to ground faults where initial, breaking, peak and thermal currents are calculated.

The question is:
Does anybody know the reasson for IEC 60909 not calculate breaking and thermal currents for two phase and two phase to earth faults?

I thank you in advance your comments!

 

[davidbeach]

No idea why that standard specifically doesn't require that, but in general those values will not exceed the larger of the 3-phase or phase-ground fault currents.

 

[luis7006]

Hello,

I thank your answer, however our concern it is not related to the magnitude of the values.
It is oriented mainly to the customer requirement to "see" all of the values for all of the faults.

 

[ijl]

I observed that the breaking current and the thermal current are (set) equal to the initial current in the program Calpos (Neplan) in the cases of asymmetric faults. I do not have the standard at hand, but I found some support for this practice in the literature. Maybe this is the way your customer can "see" all the fault currents.

 

[7anoter4]

The ABB Switchgear Manual sustains [cited]:
"the double earth fault is not included in Fig. 3-7 [Diagram for determining the fault with the highest short-circuit current from IEC 909-the former edition of IEC 60909] ;it result in smaller currents than a two-phase short-circuit"
It seems that is not always true.
From ABB Switchgear Manual Table 3-2 :
Two-phase-to-earth fault:
I"kE2E=1.1*SQRT(3)*ULL/[Z1+ZO+ZO*Z1/Z2]
Phase-to-phase fault -clear of earth=
I"k2=1.1*ULL/[Z1+Z2]
Phase-to-earth fault=
I''k1= 1.1*sqrt(3)*ULL/[Z1+Z2+ZO]
Let say Z=X neglecting R for Z1,Z2 and Zo
Let say: K21=Z2/Z1; KO1=Zo/Z1 then:
K2p=Two-phase-to-earth fault/ Phase-to-earth fault
K2p= I"kE2E/ I'k1=1.1*sqrt(3)*ULL/[Z1+ZO+ZO*Z1/Z2]/ 1.1*sqrt(3)*ULL/[Z1+Z2+ZO] or:
K2p= [1+K21+KO1]/]1 +KO1+KO1/K21]
K2E2P= Two-phase-to-earth fault/ Phase-to-phase fault -clear of earth
K2E2P= I"kE2E/ I'k2=1.1*sqrt(3)*ULL/[Z1+ZO+ZO*Z1/Z2]/ 1.1*ULL/[Z1+Z2] or:
K2E2P=sqrt(3)*(1+K21)/(1+KO1+KO1/K21)
If K21=1 then for KO1=0 to 1
K2p= 0 to 1 K2E2P= 0 to 1.15
If K21=0.5 then for KO1=0 to 1
K2p= 0.5 to 1.25 K2E2P= 0 to 1.73
See Table 3-2 on the attachment.
Regards

 

[luis7006]

Hello,

I thank your responses, they are very usefull!

However, the main question is:
Why does not IEC Std for shorcircuit calculations determine thermal and breaking currents for two phase faults?

What I have seen is that the three phase fault is the highest current magnitude unless, there could be some syncronous motor connected at the same bus, in this last case, the two phase fault gives a higher current magnitude.

I thank in advance your feedback

 

[7anoter4]

In my opinion, the breaking and thermal short-circuit current is taken only for subtransient period of the short-circuit phenomenon as the maximum value. For this period the report Z2/Z1 is very close to 1[even when generators in close location are involved].From Fig. 3-7 [see the attachment] for Z2/Z1=0.9-1 the maximum short-circuit current is the symmetrical 3 phase current. The differences are in transient and mainly in steady state short-circuit. But in these states the current value will be less.
Regards