SIMPLE ONE LINE DIAGRAM FAULT IMPEDANCE INPUT DATA

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1 Exam June Heidi Krohns Student number: nnnnnn SIMPLE ONE LINE DIAGRAM FAULT IMPEDANCE INPUT DATA δ :=. Voltage Phase Angle er := es := 7 e j δ Source S Voltage es = i er := 7 Source R Voltage ZS := j.76 Source S Positive Sequence Impedance ZS := j Source S Zero Sequence Impedance ZR :=.58 + j.93 Source R Positive Sequence Impedance ZR := 3 ZR Source S Zero Sequence Impedance ZL :=.35 + j Positive Sequence Line Impedance Fault Indicator Charles Kim /5

2 ZL := 3 ZL Zero Sequence Line Impedance Fault Location m :=.5 Fault Impedances (for AG fault case) INF := Fault Resistance ZFA := + j ZFB := INF + j ZFC := INF + j ZFG :=.85 + j ZFG :=. ZFG :=.4 + j ZFG :=.57 + j ZFG := + j CONSTANTS rad := Operator := π 8 rad a := e j a = i BAL := a one := a Three phase voltages at S and R zero := i ES := es BAL ES = i i ER := er BAL ER = i i CIRCUIT EQUATION In 3-phase matrix form, the equation loos lie this: Fault Indicator Charles Kim /5

3 How do we form the soure impedance ZS and ZR? Let us consider the lin between 3-phase circuit and symmetrical components Coversion of positive sequence and zero sequence impedances to Self and Mutual impedances zs( z, z) := Conversion Matrix Format z + 3 z zm( z, z) := z z 3 Zzz (, ) := zs( z, z) zm( z, z) zm( z, z) zm( z, z) zs( z, z) zm( z, z) zm( z, z) zm( z, z) zs( z, z) Now Conversion ZS := Z( ZS, ZS) ZL := Z( ZL, ZL) ZR := Z( ZR, ZR) ZS = i i i i i i i i i Source and Line Impedances to the Fault ZSS := ZS + m ZL ZSS = ZRR := ZR + ( m) ZL ZRR = i i i i i i i i i i i i i i i i i i Fault Indicator Charles Kim 3/5

4 Build System Part of the Impedance Matrix ZTOP := augment( augment( ZSS, zero), one) ZTOP = i i i i i i i i i ZMID := augment( augment( zero, ZRR), one) ZMID = i i i i i i i i i ZSYS := stac( ZTOP, ZMID) ZSYS = i i i i i i i i i i i i i i i i i i Pre-fault conditions: ZPRE := ZS + ZL + ZR ZPRE = i i i i i i i i i ISPRE := ZPRE ( ES ER) ISPRE =.7 +.4i.4.73i i Fault Indicator Charles Kim 4/5

5 IRPRE := ZPRE ( ER ES) IRPRE =.7.4i i.56.49i Pre_fault voltage at S end VSP := ES ZS ISPRE VRP := ZS IRPRE ER VSP = VRP = i i i i i i ES = i i i Build the voltage Vector null := ( ) ( ) E := stac stac( ES, ER), null T TS := augment( augment( one, zero), zero) TR := augment( augment( zero, one), zero) Building Fault Part of the Impedance Matrix: E = i i i i i TS = TR = Fault Indicator Charles Kim 5/5

6 ZFAG := ZFA + ZFG ZFBG := ZFB + ZFG ZFCG := ZFC + ZFG ZFAG =.57 ZFBG = ZF := ZFAG ZFG ZFG ZFG ZFBG ZFG ZFG ZFG ZFCG ZF = FABCG := augment( augment( ZF, ZF), one) FABCG = FINAL Z MATRIX Fault Indicator Charles Kim 6/5

7 ZABCG := stac( ZSYS, FABCG) ZABCG = i i i i i i i i i i i i i i i i i i YABCG := ZABCG Fault Currents: IABCG := YABCG E E = i i i i i IABCG = i i i i.5.69i.66.9i i i i Fault Indicator Charles Kim 7/5

8 S - End Fault Currents: IS := TS IABCG IS = i i i R - End Fault Currents: IR := TR IABCG IR = i.5.69i.66.9i S - End Voltages i VS := ES ZS IS VSP = VS = i i R End Voltages * Additional Component for FI VR := ( ZR IR ER) VR = i VRP = i Line Prefault Load Currents from S Bus i i i i i i i Ia := ISPRE Ia =.75 Ib := ISPRE Ib =.75 Ic := ISPRE Ic =.75 arg( Ia) arg( Ib) arg( Ic) = = =.7 = ISPRE = IRPRE =.7 +.4i.4.73i i.7.4i i.56.49i Fault Indicator Charles Kim 8/5

9 Line Prefault Voltages at S Bus Va := VSP Va = Vb := VSP Vb = arg( Va) arg( Vb) =.366 = VSP = i i i Vc := VSP Vc = arg( Vc) =.366 Line Fault Currents from S Bus Iasf := IS Iasf =.35 Ibsf := IS Ibsf =. arg( Iasf) arg( Ibsf ) = 5.97 = IS = i i i Icsf := IS Icsf =.336 arg( Icsf) = 9.75 Line Fault Currents from R Bus Iarf := IR Iarf = 9.88 Ibrf := IR Ibrf =. arg( Iarf ) arg( Ibrf) = = IR = i.5.69i.66.9i Icrf := IR Icrf =.336 Line Fault Voltages at S Bus arg( Icrf ) = 6.95 VSP = i i i Fault Indicator Charles Kim 9/5

10 Vasf := VS Vasf =.56 arg( Vasf) = 39.9 Vbsf := VS Vbsf = Vcsf := VS Vcsf = 8.66 Line Fault Voltage at R Bus arg( Vbsf ) arg( Vcsf) = 3.6 = VS = i i i Varf := VR Vbrf := VR Vcrf := VR Varf = arg( Varf ) Vbrf = arg( Vbrf ) Vcrf = arg( Vcrf ) = = 53.5 = Residual Current and Voltage Vsr, Vrr, Isr, Irr Isrf := IS =.3.387i Irrf := IR = j j j = j = Vsrf := VS = i Vrrf := VR = j j j = j = arg( Isrf ) =.899 IRPREr := j = IRPRE = j arg( Irrf ) =.58 ISPREr := j = i ISPRE = j i IS = arg( Vsrf ) = i i i IR = VS = arg( Vrrf) =.5 VR i.5.69i.66.9i = i i i i i i Fault Indicator Charles Kim /5

11 VRPr VRP i 4 := = VSPr := j j = j = VSP = i 4 j Vsrf Zs := = i Zr := Isrf Vrrf Irrf = i Wattmeteric Method???? SS := Vsrf Isrf SR := Vrrf Irrf Re( SS) = Re( SR) = So How do we generate digital signals of Voltage and Current of the Simulation 4 Cycles with 768 samples per second (8 samples per cycle in 6HZ system)? For S side :=.. 5 delt :=.3 Van := VSP sin π 6 delt + arg VSP T := delt delt = = 8 T := 5 delt + delt T3 := 4 delt + delt Vbn := VSP sin π 6 delt + arg VSP ( 6 delt + arg( VS )) Vcn := VSP sin π 6 delt + arg VSP Vaf := VS sin π Vbf := VS sin π 6 delt + arg VS Vcf := VS sin π 6 delt + arg VS T = Fault Indicator Charles Kim /5

12 Ian := ISPRE sin π 6 delt + arg ISPRE Ibn := ISPRE sin π 6 delt + arg ISPRE Icn := ISPRE sin π 6 delt + arg ISPRE Iaf := IS sin π 6 delt + arg IS Ibf := IS sin π 6 delt + arg IS Icf := IS sin π 6 delt + arg IS Van Vbn Vaf 5 Vbf Ian Ibn Iaf Ibf Fault Indicator Charles Kim /5

13 Let us mae Normal (4 cycle)+ Fault (4 cycle) +Normal (4 cycle) Seg := augment( T, Ian, Ibn, Icn, Van, Vbn, Vcn) Seg := augment( T, Iaf, Ibf, Icf, Vaf, Vbf, Vcf) Seg3 := augment( T3, Ian, Ibn, Icn, Van, Vbn, Vcn) Final := stac( Seg, Seg, Seg3) T := Final IaS := Final IrS := IaS + IbS + IcS VrS := VaS + VbS + VcS IbS := Final IcS := Final 3 VaS := Final 4 VbS := Final 5 VcS := Final 6 3 IaS IbS IcS IrS T Fault Indicator Charles Kim 3/5

14 VaS VbS VcS VrS T VrS IrS.6.8. T VrS = IrS = dt :=.38 ω := π 6 ω = mm := 536 window := 8 wind := window Now for all the calculations Fault Indicator Charles Kim 4/5

15 dd := mm.. window :=.. mm window mm window :=.. 8 UrS := submatrix( VrS, 8, 8 + wind,, ) ArS := submatrix( IrS, 8, 8 + wind,, ) UaS := submatrix( VaS, 8, 8 + wind,, ) AaS := submatrix( IaS, 8, 8 + wind,, ) UbS := submatrix( VbS, 8, 8 + wind,, ) AbS := submatrix( IbS, 8, 8 + wind,, ) UcS := submatrix( VcS, 8, 8 + wind,, ) AcS := submatrix( IcS, 8, 8 + wind,, ) PrS := FrS := FFT( UrS ) FFT( ArS ) CompS := PrS ( ) FrS, ( ) CompS = i CompS = i, WattS := Re( CompS ) Real Current Component Method Fault Indicator Charles Kim 5/5

16 ( ) ( ) ( ) ( ) ( ) ( ) PaS := FFT UaS PbS := FFT UbS PcS := FFT UcS FaS := FFT AaS FbS := FFT AbS FcS := FFT AcS CaS := ( PaS ) FaS, ( ), Re( CaS ) RaS := RbS := Re CbS CbS := PbS ( ) FbS, ( ) CcS := PcS, ( ) RcS := Re( CcS ) ( ) FcS, ( ), 5 RaS RbS RcS Phase a: Ima := ( FaS ), ( FbS ), ( FcS ), ZL = i ZL = i αa:= e i π 3 Fault Indicator Charles Kim 6/5

17 Am := 3 αa ( αa) ( αa) αa Phase a: Iaseq := Am Ima Ia := Iaseq ( ) Ia := Iaseq ( ) Yang method: K := ZL ZL ZL Compensated current: Phase a: Ia := FaS + K Ia αa := arg Ia ( ) arg( Ia ) Resistance and Reactance: X := R := Im( ZL) Re( ZL) X = R =.35 Phase Impedance: Phase a: Fault Indicator Charles Kim 7/5

18 Za := PaS ( Ia ), ( ), ( ), Xa := Im Za Ra := Re Za Fault Distance: ma := ( Xa ) + Ra ( ) tan ( αa ), ( ), X + R tan αa m =.5 Fault Indicator Charles Kim 8/5

19 ma ma =.95 6 ma = ma =.44 7 ma =.43 ma =.44 Results and Discussion Part: When changing the m value (place of the fault) at start of the file ma is changing in same relation. However, ma values seem to come more exact when m is bigger. m=.5 m=.45 m=.5 m=.75 m=.9 ma=.7 ma=.4 ma=.459 ma=.695 ma=.83 Fault resistance ZFG was changed and fault place m was constant (m=.5). If the change of resistance is minor, algorithm wors. If value of ZFG increase much, algorithm stops woring. m=.5 ZFG=.4+j ZFG=.85+j ZFG=.57+j ZFG= +j ma=.48 ma=.458 ma=.43 ma=7.378*^-3 Fault Indicator Charles Kim 9/5