Zero sequence symmetrical fault currents (L-G) don't lend themselves to the simple method of calculating fault current, but here is a nice document for simple methods for generalizations for L-G faults. http://www.cooperindustries.com/cont...l_Formulas.pdf
Cooper Bussmann does not indicate how to calculate 3 phase transformer phase-to-neutral short-circuit current, indeed.
IEEE 80/2013 pp.76 [for instance] symmetrical one phase short-circuit
Io=E/(Z1+Z2+Zo)≈E/(X1+X2+Xo)
Ineutral=3*Io [in all three phase the same symmetrical component zero sequence current ]
The reactances X1,X2 and Xo are the total reactance of the supply system and the transformer. Xo is different and depends on transformer connection and the magnetic core construction [3 limbs or 5 limbs].
If we consider the short-circuit power of the supply system as infinite [Xsystem=0] then if the transformer connection is Delta/Yn then Xo≈X1 X2=X1[far from generators].
Isc=3*Io=3*277/[3*X1]=277/X1
X1≈Z1=VL_L^2/S*Z%/100=0.041088 ohm
where VL_L=0.48 kV S=0.3 MVA
Isc1=277/0.041088=6741.6 A
Ifla=S[kVA]*1000/1.73/VL_L=360.8 A
That means Isco[symmetric component per phase] is indeed Ifla/Z/3=2248.25 A
But you have 3 Io[one per phase] then IN=3*2248.25=6744.75
Also IEC 60909-0[formula 29]:
Ik=c*Un/sqrt(3)/Z
c=1 Un=380 V
Z=0.041088 ohm
Ik=1*480/1.732/.041088=6744.9 A
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