2 phase to ground fault current splits I_0

[acog]

Hi,

I have received a fault study document. Using this document, I am trying to work out the maximum fault current which actually travels through the earthing system and returns to the upstream star point.

Typically I use the worst case single line to ground fault current for this purpose.

However, for this project I notice that the author of the network study has included a 3*I[sub]0[/sub] column for the two phase to ground fault scenario. The magnitude of this is higher than the 3*I[sub]0[/sub] single line to ground column.

I was previously under the impression that under a two phase to ground fault, majority of the fault current will return to the upstream star point through the two faulted phase conductors (and therefore not contribute to ground potential rise and can be ignored for the purposes of earthing design); however, the 3*I[sub]0[/sub] heading indicates that this is actually zero sequence current, which indicates the worst case for soil current could actually be the 2 phase to ground scenario.

Has the network study author made a mistake or is my interpretation incorrect?

If anyone can please provide some clarity on the use of two phase to ground faults in earthing studies, this would be greatly appreciated.

 

[7anoter4]

According to IEC 60909-0 “Short-circuit currents in three-phase a.c. systems - Part 0: Calculation of currents” Figure 10 – “Diagram to determine the short-circuit type (figure 3) for the highest short-circuit current referred to the symmetrical three-phase short-circuit current at the short-circuit location when the impedance angles of the sequence impedances Z(1),Z(2), Z(o) are identical”, if Z(1)=Z(2) then, if Z(o)<Z(1) then phase-to-phase-to ground is the maximum short-circuit.
Indeed, according to formula (51) ch. 4.2.3 Line-to-line short circuit with earth connection
I”kE2E=sqrt(3)*c*Un/|Z(1)+2*Z(0)|
Then if Z(o)/Z(1)<1 I”kE2E>I”k1>I”k3

 

[ijl]

A short circuit calculation typically gives only the phase currents in the case of a LLE-fault. The current to earth must be calculated separately from the current balance.

Consider an example: V = 1000V, Z1 = Z2 = Z0 = 0.1 Ohm (first picture). The A- and B-phase currents and the current to earth are all equal to 5773A in the case of the LLE-fault (as are also the currents in the cases of the LLL, LL, and LE-faults).

If the negative sequence impedance Z2 is increased to 1 Ohm, for example, then the phase currents of the LLE-fault decrease to 5017A, but the current to earth increases to 8250A (!) (second picture)

 

[ijl]

The first picture, with Z = 0.1 Ohm. The second one is included in the main message.