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Calc L-N Fault

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    Calc L-N Fault

    I have a 300kVa with a 480Y/277 secondary, 5.35% z.

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

    Thanks

    #2
    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
    Ron

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      #3
      Thanks Ron

      Comment


        #4
        Ron -
        I didn't see 3phase, single line to ground (neutral) fault.
        Without data you’re just another person with an opinion – Edwards Deming

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          #5
          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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