Zero Sequence fault currents 4

[Electoman]

Could anybody please explain the following: I am interested in the current distribution of a single-phase to ground fault on a YY transformer with both neutrals earthed. The PRAG explains that the currents for a A-E fault will be Ia=(2C1 + C0)I0, Ib= Ic= -(C1-C0)I0, where C0 and C1 are zero and positive sequence distribution factors.
The fault is on the secondary side of the TX and I want to know the impact on the primary side.
What I could gather, is that Ia'=2*Io and Ib'= Ic' = -Io.
E.g. if Ia' = 0.5A, Ib'= Ic'= 0.25A.
If this is the right assumption, how is this derived from first principals. The PRAG (Protective Relay Application Guide) is not very clear on this. Any explanation will be much appreciated.

 

[jbartos]

Comment on PRAG: Visit

http://www.areva-td.com/static/largefiles/04-fault_calculations.pdf

for:
1. Equation 4.1
2. Equation 4.35 under paragraph:
4.5 CURRENT AND VOLTAGE DISTRIBUTION
IN A SYSTEM DUE TO A FAULT
Practical fault calculations involve the examination of
the effect of a fault in branches of network other than
the faulted branch, so that protection can be applied
correctly to isolate the section of the system directly
involved in the fault. It is therefore not enough to
calculate the fault current in the fault itself; the fault
current distribution must also be established.

This calculation is per system and involves all three phases that are loaded when a single phase to ground materializes.

Some postings above were addressing:
4.4.5 Single-phase Open Circuit Fault
The single-phase open circuit fault is shown
diagrammatically in Figure 4.12(a).
Equation 4.27