fault calculation

FAULT CALCULATIONSompol CPower System Fault AnalysisAll protection Engineers should have and understanding to :- Calculate power system currents and voltages during fault condition Check the breaking capacity of switchgear is not exceededDetermine the quantities which can be used by relays to distinguish between healthy (i.e. loaded ) and fault conditionsAppreciate the effect of the method of earthing on the detection of earth faultsSelected the best relay characteristics for fault detectionEnsure that load and short circuit ratings of plant are not exceeded Select relay settings for fault detection and discriminationUnderstand principles of relay operationConduct post fault analysisPower System Fault AnalysisPower System Fault Analysis also used to :- Consider stability conditionsRequired fault clearance timesNeed for 1 phase or 3 phase autorecloseComputer Fault Calculation ProgrammesWidely available, particularly in large power utilitiesPowerful for large power systemsSometimes overcomplex for simple circuitsNot always use friendlySometimes operated by other department and not directly available to protection engineersProgramme calculation methods :-understanding is important Need for ‘by hand’spot checks of calculationPer-Unit System Per-unit impedance of transformer is the same regardless ofwhether it is determined from ohmic values referred to the high or low side of transformer .Only one MVA Base is used for all equipments in the network . The impedance of each equipment can be merged directly . Manufacturer usually specify the impedance in per-unit on thebase of the nameplate rating .~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40ΩY 132/33kV 50MVAZ T = 10%O/H Line Z L = 8ΩFeederDifferent voltageHow do we Analyses ?Referring Impedance R 1X 1R 2X 2N :1IdealTransformerR 1+N 2R 2X 1+N 2X 2R 1/N 2+R 2X 1/N 2+X 2Primary SecondaryConsider the equivalent CCT refer to :-Base Quantities and Per Unit Values Particularly useful analyzing large systems with several voltagelevels .All system parameters referred to common base quantities .Base quantities fixed in on part of system .Base quantities at other parts at different voltage levels dependon ratio of intervening transformers .~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40ΩY 132/33kV 50MVAZ T = 10%O/H Line Z L = 8ΩFeederBase Quantities and Per Unit Values Base quantities normal used :-Base MVA =MVA b=3ØMVAConstant all voltage levels Value –MVA rating of largest item ofplant or 100 MVABase voltage =kV b=Ø/Øvoltage in kVFixed in one part of systemThis value is referred through transformers to obtain basevoltage on other parts of systemBase voltage on each side of transformer are in same ratio asvoltage ratio.Base Quantities and Per Unit Values Other base quantities :-Base Impedance =Z b= (kV b)2 in OhmsMVAbBase Current=I b= MVA b in kA√3 x kV bPer Unit SystemUse to simplify calculation on system with more than two voltage Definition :P.U.Value =Actual ValueBase ValueBase Quantities and Per Unit Values Per unit values=Actual ValueBase ValuePer unit MVA =MVA p.u.= MVA aMVAbPer unit voltage =kV p.u.=kV akVbPer unit impedance=Z p.u.= Z a=Z a MVA bZ b (kVb)2Per unit current=I p.u.= I aIbTransformer percentage impedance If Z T = 5%with secondary short circuit5% V (rated)produces I (rated)in secondaryV (rated)produce 100 x I (rated)5If Source impedance Z S = 0Fault current =20 x I (rated)Fault power =20 x kVA (rated)Z T is base on I (rated)& V (rated)i.e. Base on MVA (rated)& kV (rated)Is same value viewed from either side of transformer Conversion of Per Unit from one Set Quantities to AnotherZ b1Z b2MVA b1 MVA b2kV b1kV b2Actual Z = Z aZ P.U.1=Z aZ b1Z P.U.2=Z a = Z P.U.1 x Z b1Z b2Z b2= Z P.U.1x (kV b1)2x MVA b2MVA b1 (kV b2)2= Z P.U.1x MVA b2 x (kV b1)2MVA b1 (kV b2)2Z P.U.1Z P.U.2Example 0Incorrect selection of kV b11.8kV 132kV 11kVCorrect selection of kV b132x11.8 132kV 11kV141=10.05kVAlternative correct selection11.8kV 141kV 141x11 = 11.75kV 13211.8/141kV OHL Distribution System 11.8kV 132/11kV ~Base voltage on each side of transformer must be in the same ratio as voltage ratio of transformer Example 1kV b 1113233MVA b 505050Z b = 2.42 Ω348.5 Ω21.8 Ω(kV b )2MVA bI b =2624 A 219 A 875 A MVA b√3kV bZ P.U.O.3 x 5040820348.521.8= 0.75 P.U.= 0.115 P.U.= 0.367 P.U.~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40ΩY 132/33kV 50MVAZ T = 10%O/H Line Z L = 8ΩFeeder3-ØFaultI 11kV =0.698 x 2624=1832 A I 132kV =0.698 x 219=153 A I 33kV =0.698 x 875=611 A I F = 1/1.432 = 0.698 P.U.~ 1.432 p.u.1 p.u.Example 2Generator Transformer Line Motor50 MVA 11kV 50MVA X1=X2= 1.3225 ohm 20 MVA 115 kV Xd”= 20% 11/115kV Ynd1X0= 6.6125 ohm Xd”= 20%X2 = 20% Imp voltage =12.5%X2 = 20%X0 = 10% at 50 MVA Base X0 = 10%Neutral resister = 12.1 ohm~Y 11kV Bus Transformer Xd”M O/H LineRn 115kV Bus 115kV BusXd”Convert Impedance to Per-Unit GeneratorMVAb= 100 MVA, kVb= 11 kVZb= kVb2/MVAb = 112/100 = 1.21 ohmXd”= j0.2 x 100/50 = j0.4 p.u.X2 = j0.2 x 100/50 = j0.4 p.u.X0 = j0.1 x 100/50 = j0.2 p.u.Rn= 12.1/1.21 = 10 p.u.TransformerMVAb= 100 MVA, kVb= 11&115 kVImp voltage = 0.125 x 100/50 = 0.25 p.u.X1 = X2 = X0 = j0.25 p.u.Convert Impedance to Per-Unit LineMVAb= 100 MVA, kVb= 115 kVZb= kVb2/MVAb = 1152/100 = 132.25 ohmX1 = X2 = j1.3225/132.5 = j0.01 p.u.X0 = j6.6125/132.5 = j0.05 p.u.MotorMVAb= 100 MVA, kVb= 115 kVZb= kVb2/MVAb = 1152/100 = 132.25 ohmXd”= j0.2 x 100/20 = j1.0 p.u.X2 = j0.2 x 100/20 = j1.0 p.u.X0 = j0.1 x 100/20 = j0.5 p.u.Symmetrical Components VectorsVector notation can be used to represent phase relationship between 2 electrical quantities~IV Zv = V sin ωt = V/0°i = I sin (ωt-θ)=I/θ°θLagging = -Leading = +“j ”operator : j = 1/90° Rotates vector by 90°Anti-clockwise :-90°90°90°90°Used to express vectors in terms of ‘real’and ‘imaginary’parts 1/0°j 2= 1/180°= -1j = 1/90°j 3= 1/270°= -j ‘a ’operator : a = 1/120° Rotates vectors by 120°Anti-clockwise Used extensively in ‘Symmetrical Component Analysis’120°120°120°a = 1/120°= -½+j √3/2a 2= 1/240°= -½-j √3/21 , a 3Balance 3Øvoltages :-120°120°120°V C = a V A V AV B = a 2V Aa 2 + a + 1 = 0‘a ’operator : a = 1/120°Symmetrical Components Property of unbalanced phasors ‘n’phasors may be resolved into :-(n-1) sets of balanced n-phase systems of phasors, each set having a different phase sequence, plus1set of zero phase sequence or unidirectional phasorsSymmetrical Components V A = V A1+ V A2+ V A3 + V A4 + ………. +V A(n-1)+ V An V B = V B1+ V B2+ V B3 + V B4 + ………. +V B(n-1)+ V Bn V C = V C1+ V C2+ V C3 + V C4 + ………. +V C(n-1)+ V Cn V D = V D1+ V D2+ V D3 + V D4 + ………. +V D(n-1)+ V Dn ………………………………………………………….V N = V N1+ V N2+ V N3 + V N4 + ………. +V N(n-1)+ V Nn(n-1) x Balanced 1 x Zero Sequence Consider 5-phase unbalanced System V a1V b1First Set of Balanced Phasors V c1Vd1V e1V a2V d2Second Set of Balanced PhasorsV b2Ve2V c2V a3V c3Third Set of Balanced Phasors V e3Vb3V d3V a4V e4Fourth Set of Balanced PhasorsV d4Vc4V b4V a0V b0V c0V d0V e0Fifth Set of Balanced Phasors( 360°/ 5 )Unbalanced 3 Phase System V A = V A1+ V A2+ V A0V B = V B1+ V B2+ V B0V C = V C1+ V C2+ V C0V A1V B1V C1Positive Sequence V A2V C2Negative Sequence V B2V A0V B0V C0Zero Sequence ( 360°/ 3 )Symmetrical ComponentsV A1V B1V C1+V A0V B0V C0+V A2V C2V B2V AV B V CV A = V A1+ V A2+ V A0V B = V B1+ V B2+ V B0V C = V C1+ V C2+ V C0Converting from sequence Component to Phase Values V A = V A1+ V A2+ V A0V B = V B1+ V B2+ V B0=a 2V A1+ aV A2+ V A0V C = V C1+ V C2+ V C0=aV A1+ a 2V A2+ V A0V A1V B1V C1V A2V B2V B0V C2V A0V C0V A V BV CV A1V B1V C1V A2V C2VB2V A0V B0V C0Converting from Phase Values to Sequence ComponentVA1 = 1/3 { V A + aV B + a 2V C }V A2= 1/3 { V A + a 2V B + aVC }V A0= 1/3 { V A + V B + V C }V BV A VCV A0V C0= 3V 0V B0SummaryV A = VA1+ VA2+ VA0V B = a2VA1+ aVA2+ VA0V C = aVA1+ a2VA2+ VA0I A = IA1+ IA2+ IA0I B = a2IA1+ aIA2+ IA0I C = aIA1+ a2IA2+ IA0VA1= 1/3 { VA+ aVB+ a2VC}VA2= 1/3 { VA+ a2VB+ aVC}VA0= 1/3 { VA+ VB+ VC}IA1= 1/3 { IA+ aIB+ a2IC}IA2= 1/3 { IA+ a2IB+ aIC}IA0= 1/3 { IA+ IB+ IC}Residual CurrentI RESIDUAL is Zero for :-Balanced Load3ØFaultsØØFaults Residual Current = IA+ IB+IC=3II RESIDUAL is present for :-Ø/E FaultsØØ/E FaultsOpen circuits (with currentremaining phases )Use to detect earth faultsE/F I AI BI CResidual Voltage RESIDUAL Voltage is measuredfrom Open Delta or Broken DeltaVT Secondary windings.V RESIDUAL is Zero for :-Healthy Unfaulted System3ØFaultsØØFaultsV Residual = V A + V B + V C=3V 0 V RESIDUAL is present for :-Ø/E FaultØØ/E FaultOpen circuits (on supply sideof VT)Use to detect earth faults R g = nGround Plane A B C AVb Vc Va n BC VanVbn VcnNormal Balanced ConditionPrimary voltageSecondary voltage Va+Vb+Vc = 0g = n Ground Plane A B C AVb Vc Va n BC Van VbnVcn Fault Condition3V0Va+Vb+Vc = 3V03V0Secondary voltage Primary voltageExample 3Evaluate the positive,negative and Zero Sequencecomponents for the unbalanced phase vectorsV BV AV CV A = 1 /0°V B = 1.5/-90°V C = 0.5/120°Solution V A1= 1/3 {V A + aV B + a 2VC }=1/3 {1 + (1 /120°)(1.5/-90°)+(1 /240°)(0.5/120°}= 0.965 /15°V A2= 1/3 {VA + a 2VB + aVC }=1/3 {1 + (1 /240°)(1.5/-90°)+(1 /120°)(0.5/120°}=0.211 /150°V A0= 1/3 {VA + VB + VC}=1/3 {1 + (1.5/-90°) + (0.5/120°}=0.434 /-55°Positive Sequence Voltage V A1 = 0.965/15°15°V B1= a 2V A1V C1 = aV A1Negative and Zero Sequence VoltageV A2 = 0.211/150°V B2 = aV A2V C2 = a 2V A2150°-55°V A0= 0.434 /-55°V B0= 0.434 /-55°V C0= 0.434 /-55°Symmetrical ComponentsV BV AVC VA2V B2V C2V B1V C1V A1V A0V B0V C0Example 4Evaluate the phase quantities Ia, Ib and Ic from the sequencecomponentI A1=0.6 /0°I A2=-0.4/0°I A0=-0.2/0°SolutionI A=I A1+ I A2+ I A0= 0I B=a2I A1+ aI A2+ I A0=0.6/240°–0.4/120°-0.2/0°=0.91/-109°I C=aI A1+ a2I A2+ I A0=0.6/120°–0.4/240°-0.2/0°=0.91/109°ICIB109-109Balance FaultFault TypesLine –ground (65-70%)Line –Line–Ground (10-20%)Line –Line(10-15%)Line –Line –Line(5%)Statistics published in 1967 CEGB report, but are similar today all over the worldBalance Faults (3Ø)Rare :-Majority of Fault are unbalancedCauses :-System Energization with Maintenance Earthing Clamps still connected1-Øfault developing to 3-Øfault3-Øfault may be represented by 1-Øcircuit valid because system is maintained in a balanced state during the faultVoltages equal and 120°apartCurrents equal and 120°apartPower System Plant SymmetricalPhase impedance EqualMutual impedance EqualShunt admittance EqualBalance Faults (3Ø)~Y 3-ØFaultGenerator TransformerLine ’X’Line ’Y’~Ea Z G ZT Z LX Z LY~Eb Z G ZT Z LX Z LY~Ec ZG Z T Z LX Z LYZ LOADBalance Faults (3Ø) E a E bE c I aFI bF I cF~Ea ZG Z T Z LX Z LYZ LOADN 1Positive Sequence (Single Phase) circuit :-Unbalance FaultUnbalanced FaultIn three phase fault calculation, a single phase representation is adopted.Three phase fault are rareMajority of faults are unbalanced faultUNBALANCED FAULTS may be classified into SHUNT FAULTS and SERIES FAULTSShunt Fault :-Line to GroundLine to LineLine to line to GroundSeries FaultSingle Phase Open CircuitDouble phase open CircuitUnbalanced FaultLine to GroundLine to LineLine to line to GroundCauses ;Insulation BreakdownLightning Discharge and Other OvervoltageMechanical DamageUnbalanced FaultOpen Circuit or Series FaultCausesBroken ConductorOperation of FusesMal-operation of Single Phase Circuit BreakerDuring unbalanced Faults, Symmetry of system is lost Single Phase Representation is no Longer ValidUnbalanced FaultAnalyzed using :-Symmetrical componentEquivalent Sequence Networks of Power SystemConnection of Sequence Networks approximate to Type of FaultSequence NetworksSequence NetworksIt can be shown that providing the system impedances are balanced from the point of generation right up to the fault, each sequence current causes voltage drop of its own sequence only.Regard each current flowing within own network thro’Impedance of its own sequence only, with no interconnection between the sequence networks right up to the point of fault.Sequence Networks+ ve,-ve and Zero sequence networks are drawn for a‘reference’phase. This is usually taken as the ‘A’phase Fault are selected to be ‘balanced’relative to the reference ‘A’Phasee.g.For Ø/E faults consider an A-E faultFoe Ø/Øfaults consider B-C faultSequence networks interconnection is the simplest for the reference phasePositive Sequence Diagram Start with the natural point N 1All generator and load neutrals are connected to N 1 Include all source EMF’sPhase-Neutral voltageImpedance NetworkPositive sequence impedance per phase Diagram finishes at fault point F 1Z1~E 1F 1N 1Example 5V 1 =Positive sequence Ph-N voltage at a fault point I 1= Positive sequence phase current flowing into F 1 V 1=E 1-I 1(Z G1+ Z T1+ Z L1)R~Y Generator TransformerLine F E N ~E 1Z T1F 1Z G1ZL1I 1( N 1 )V 1N 1Negative Sequence Diagram Start with the natural point N 2All generator and load neutrals are connected to N 2 No EMF’s includedNo negative sequence voltage is generated !Impedance NetworkNegative sequence impedance per phaseDiagram finishes at fault point F 2Z 2F 2N 2Example 6V 2 =Negative sequence Ph-N voltage at a fault point I 2= Negative sequence phase current flowing into F 2 V 2=-I 2(Z G2+ Z T2+ Z L2)R~Y Generator TransformerLine F EN Z T2F 2Z G2ZL 2I 2( N 2 )V 2N 2Zero Sequence DiagramFor “in phase”(Zero Phase Sequence ) currents to flow in each phase of the system, there must be a fourth connection (this is typically the neutral or earth connection).Z T0I A0NZ T0I B0Z T0I C0E IA0+IB0+IC0= 3IA0Zero Sequence Diagram Resistance Earthed SystemZ T0I A0NZ T0I B0Z T0I C0E 3IA0RZero sequence voltage between N & E Given byVN= 3IA0. RZero Sequence impedance of neutral to earth pathZ= V/IA0. = 3RVN0Transformer Zero Sequence Impedance ( Z T0)P QZ T0a b a bP QN 0General Zero Sequence Equivalent Circuit for Two Winding Transformer On appropriate side of transformer :Earthed star winding ÎClose link ‘a’Open link ‘b’Delta winding ÎOpen link ‘a’Close link ‘b’Unearthed star winding ÎBoth links ‘open’Z T0‘a’‘b’‘a’‘b’N 0Primary Terminal Secondary TerminalZero Sequence Equivalent CircuitP SZT0 aba bP 0S 0NZero Sequence Equivalent CircuitP SZT0 aba bP 0S 0NZero Sequence Equivalent CircuitP SZT0 aba bP 0S 0NZero Sequence Equivalent CircuitP SZT0 aba bP 0S 0NExample 7 V 0=Zero sequence Ph-N voltage at a fault point I 0= Zero sequence phase current flowing into F 0 V 0=-I 0(Z G0+ Z T0+ Z L0)R~Y Generator TransformerLine F EN Z T0F 0Z G0ZL 0I 0( N 0 )V 03R E 0N 0 V 0 =Zero sequence Ph-N voltage at a fault point I 0= Zero sequence phase current flowing into F 0 V 0=-I 0(Z T0+ Z L0 + 3R T )R ~Generator TransformerLine F E N Z T0F 0Z G0ZL 0I 0N 0V 0R TSystem Single Line Diagram3R 3R T Zero Sequence Network Example 8Network Connection Interconnection of SequenceNetworksConsider sequence networks as blocks with a fault terminals F & N for external connectionPositive Sequence NetworkF1N1 I1V1Negative Sequence NetworkF2N2 I2Zero Sequence NetworkF0N0 IV2VInterconnection of Sequence NetworksFor any given fault there are 6 quantities to be considered at the fault pointV A V B V C I A I B I CRelationships between these for any type of fault can be converted into an equivalent relationship between sequence components V 1V 2V 0I 1I 2I 0This is possible if :-1. Any 3Øquantities are known (provided they are not all voltages or all currents) or2. 2 are known and 2 others are known to have a specific relationshipFrom the relationship between sequence V’s and I’s, the manner in which the isolation sequence networks are connected can be determinedThe connection of the sequence networks provides a single phase representation (in sequence terms) of the faultTo derive the system constraints at the fault terminals :-Terminals are connected to represent the faultV A V B V CFI A I B I CLine to Ground Fault on Phase ‘A’At fault point :-V A =0VB =?V C =?I A =?I B =0I C =0V A V B V C FI A I B I C Phase to Earth Fault on Phase ‘A’At fault pointV A =0; I B = 0, I C = 0V A =V 1+ V 2+ V 0= 0V 1=-(V 2+ V 0)------------------------------(1)I 0=1/3 ( I A + I B + I C )=1/3 I AI 1=1/3 ( I A + aI B + a 2I C )=1/3 I AI 2=1/3 ( I A + a 2I B + aI C )=1/3 I AI 1= I 2= I 0-------------------------(2)To comply with (1) & (2) the sequence networks must be connectedin series :-Phase to Earth Fault on Phase ‘A’To comply with (1) & (2) the sequence networks must be connectedin series :-Positive Sequence Network F 1N 1I 1Negative Sequence Network F 2N 2I 2Zero Sequence Network F 0N 0I 0V 1V 2V 0Example : Phase to Earth FaultI 1 = I 2=I 0= 132000√3 x 81.1= 940 AI F = I A = I 1+ I 2+ I 0 = 3I 0I F = 2820 A~SourceLine F 132kV 2000 MVA Z S1= 8.7ΩZ S0= 8.7ΩZL1= 10ΩZ L0= 35ΩA-G FaultI F~10F 18.7I1N 110F 28.7I 2N 235F 08.7I 0N 0Total Impedance = 81.1 ΩE 1Earth Fault with Fault ResistancePositive Sequence NetworkF1N1I1NegativeSequenceNetworkF2N2I2ZeroSequenceNetworkFNIV1V2V3Z FPhase to Phase Fault : B-C FaultPositive Sequence NetworkF1N1I1V1NegativeSequenceNetworkF2N2I2V2ZeroSequenceNetworkFNI0V0Example : Phase to Phase Fault I 1 = 132000√3 x 37.4= 2037 A I 2= -2037 AI B = a 2I 1+ aI 2= a 2I 1–aI 1= (a 2–a)I 1=(-j)√3 x 2037I F = 3529 A~10F 18.7I1N 110F 28.7I 2N 235F 08.7I 0N 0Total Impedance = 37.4 Ω132000/√3~SourceLine F132kV 2000 MVA Z S1= 8.7ΩZ S0= 8.7ΩZ L1= 10ΩZ L0= 35ΩB-C FaultI FPhase to Phase Fault withResistancePositive Sequence Network F 1N 1I 1V 1Negative Sequence NetworkF 2N 2I 2V 2Zero Sequence NetworkF 0N 0I 0V 0Z FPhase to Phase to Earth Fault : B-C-E FaultPositive Sequence NetworkF1N1I1V1NegativeSequenceNetworkF2N2I2V2ZeroSequenceNetworkFNI0V0Phase to Phase to Earth Fault : B-C-E Fault with ResistorPositive Sequence NetworkF1N1I1V1NegativeSequenceNetworkF2N2I2V2ZeroSequenceNetworkFNI0V03ZF3ØVersus 1ØFault Level I F = E = EX G1+ X T1Z 1~E X GXT~X T1X G1I F 3ØE I FZ 13ØVersus 1ØFault Level I F = 3E2Z 1+ Z 0~E X GX T~X T1F 1X G1I 1N 1X T2F 2X G2I 2N 2X T0F 0X G0I 0N 01ØE I FZ 1Z 2Z 03ØVersus 1ØFault Level3ØFault Level= E = 3E= 3EZ 13Z12Z1+ Z11ØFault Level= 3E2Z1+ ZIf Z0< Z11ØFault Level> 3ØFault LevelCheck that switchgear breaking capacity > maximum fault level for all fault typesExample 9~Y11kV Bus TransformerXd”M O/H LineRn 115kV Bus115kV BusXd”Find 3 phase & single line to ground fault at 115 kV Motor BusGenerator Transformer Line MotorXd”= j0.4 p.u.X1 = X2 = X0 = j0.25 p.u. X1 = X2 = j0.01 p.u. Xd”= j1.0 p.u. X2 = j0.4 p.u.X0= j0.05 p.u. X2 = j1.0 p.u. X0 = j0.2 p.u.X0 = j0.5 p.u. Rn= 10 p.u.Base at 100 MVAG 11 kV bus j0.4Va1Find : Z th 1N 1Positive sequence networkM 115 kV bus 115 kV busj0.25j0.01j1.0N 1N 1Gj0.66J1.0j0.39759j0.397591 /0°Va1Ia 111 kV bus j0.4Va2Find : Z th 2N 2Negative sequence network115 kV bus 115 kV busj0.25j0.01j1.0N 2j0.66J1.0j0.39759Va2Ia 211 kV bus j0.2Va0Find : Z th 0N 0Zero sequence network115 kV bus 115 kV busj0.25j0.05j0.5N 0j0.3j0.5j0.1875Va03Rn = 30Ia 03 Phase faultN 1GZ th 1= j0.39759Va1Ia 1If = Ia 1= Va 1/ Zth 1= 1 / 0.39759p.u.= 2.5152p.u.Ib = MVAb kA√3 x kVb = 100 = 0.502 kA√3 x 115If = 2.5152 x 0.502= 1262.7 ASingle line to ground fault N 1G j0.39759Va1Ia 1N 2j0.39759Va2Ia 2N 0j0.1875Va0Ia 0GZth 1= j0.98268Ia1= Ia 2= Ia 0Ia 1= Va1 / Zth 1 = 1 / 0.98268 p.u.= 1.01763p.u.Ia = Ia 1+ Ia 2+ Ia 0= 3 x 1.01763= 3.0529 p.u.Ibase = 0.502 kAIf = 3.0529 x 0.502= 1532.6 A Example 10Find 3 phase & single line to ground fault at 22 kV BusDriving point Transformer CableZ1 = Z2 = 0.001+ j0.02 p.u.X1 = X2 = X0 = j0.4 p.u. Z1 = Z2 = j1.033 p.u. Z0 = 0.002+ j0.0067 p.u.Z0 = j1.55 p.u. .Base at 100 MVAY115kV Bus Transformer cable22kV Bus 22kV BusMAIN systemPositive sequence network G115 kV bus 0.001+ j0.0222 kV bus 22 kV busj0.4j1.033Negative sequence network115 kV bus 0.001+ j0.0222 kV bus 22 kV busj0.4j1.033Zero sequence network115 kV bus 0.002+ j0.006722 kV bus 22 kV busj0.4j1.553 Phase faultEquivalent Impedance at 22 kV Bus= 0.001+ j0.02 + j0.4 + j1.033 p.u.= 0.001+ j1.453p.u.Fault current = 1/ ( 0.001+j1.43 )= 0.00047-j0.68823 p.u.= 0.68823/-89.96°p.u.Base current = 100 x 103/(1.732x22) A= 2624.3AThree Phase current = 0.68823 x 2624.3A= 1806.12/-89.96°AG0.001+ j1.45322 kV busSingle line to ground fault G 0.001 + j1.453Ia 10.001 + j1.453Ia 2j1.95Ia 0GZth 1= 0.002 + j4.856Ia 1= Ia 2= Ia 01 p.u.Z1Z2Z0Equivalent Impedance at 22 kV Bus= 0.001 + j1.453+0.001 + j1.453+j1.95 p.u.= 0.002 + j4.856 p.u.Ia 1= Va1 / Zth 1 = 1 / ( 0.002 + j4.856 ) p.u.= 0.0000848 -j.20593p.u.= 0.20593/-89.98°p.u.Ia = Ia 1+ Ia 2+ Ia 0= 3 x 0.20593/-89.98°p.u.= 0.61779/-89.98°p.u.Base current = 100 x 103/ ( 1.732 x 22 )= 2624.3ASingle line to ground fault current = 0.61779 x 2624.3A= 1621.27/-89.98°Single line to ground fault。