#### chris kennedy

##### Senior Member

- Location
- Miami Fla.

- Occupation
- 60 yr old tool twisting electrician

To calculate line to neutral fault current do I use 1/3 FLA and Z?

Thanks

- Thread starter chris kennedy
- Start date

- Location
- Miami Fla.

- Occupation
- 60 yr old tool twisting electrician

To calculate line to neutral fault current do I use 1/3 FLA and Z?

Thanks

- Location
- New York, 40.7514,-73.9925

http://www.cooperindustries.com/con...rary/BUS_Ele_Tech_Lib_Electrical_Formulas.pdf

- Location
- Miami Fla.

- Occupation
- 60 yr old tool twisting electrician

Thanks Ron

- Location
- North of the 65 parallel

- Occupation
- EE (Field - as little design as possible)

Ron -

I didn't see 3phase, single line to ground (neutral) fault.

I didn't see 3phase, single line to ground (neutral) fault.

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