• Apparent source power at B side: SsB = 1280 MVA• Base current of differential current protection: IBase = 42 A• Apparent power of transformer: Sn = 10 MVA• Short circuit impedance of transformer: ek = 10%• Nominal voltage on transformer high voltage winding: Un = 138 kVFault current on the high voltage (HV) side of the tap transformer is calculated fora three-phase fault on the low voltage (LV) side. 138 kV is chosen as calculationvoltage.ZsA ZsBZtrfEZlA ZlBIEC14000046-1-en.vsdIEC14000046 V1 EN-USFigure 47: Thevenin equivalent of the tap transformerConverting the sources into impedances gives:ZS A = =1381700 11 22. ΩEQUATION14000034 V1 EN-US (Equation 20)ZSB = =1381200 15 92. ΩEQUATION14000035 V1 EN-US (Equation 21)Calculating the short circuit impedance of the transformer gives:Z e UStrfk nn= × = × =100 0 1 13810 190 42 2. . ΩEQUATION14000036 V1 EN-US (Equation 22)Based on the Thevenin equivalent, it is possible to calculate the fault current on theHV side of the transformer:If Zres1381383= ×EQUATION14000037 V1 EN-US (Equation 23)Section 6 1MRK 505 393-UEN BDifferential protection102 Line differential protection RED650 2.2 IECApplication manual
