Hydro Available Fault

[Fuselink]

Dear Friends,
I am trying to figure out how hydro has calculated the L-L-G fault based on the following info. My answers do not match.
Based on IEEE 80.
L-L-G I0=E*X0/( X1*(X0+X2) + X2*X0) ( Page 96)
L-G I0=E/(X0+X1+X2)

X1=X2
Then I get
L-L I0=1763.98
L-G I0=2400.2

Hydro Info:
Maximum Minimum
L-L-G 3467 2832
L-G 2452 2111
Z1 1.66+j6.765 1.6973+j8.36
Z0 3.0442+j18.21 3.117+j20.31

 

[DanDel]

For L-G, it should be Ia0 = Ea/(Z0+Z1+Z2)
For L-L-G, it should be Ia1 = Ea/(Z1+((Z2*Z0)/(Z2+Z0)));
Ia2 = -Ia1(Z0/(Z2+Z0)); Ia0 = -Ia1(Z2/(Z2+Z0))

Without your voltage, it's tough to follow your calculations.

 

[Fuselink]

Sorry about that.
Voltage is 44000V.

 

[Fuselink]

For L-G fault If=3*I0
and I am getting the right answer
Ia0=800A
Ia0=3*I0=2400

but for L-L-G If=3*I0
is giving me 1763A for L-L-G
Ia1=2171A
Ia0=588A

Sorry, It's been long since I have used these equations.