Exam June Heidi Krohns Student number: nnnnnn heidi.rohns@tut.fi SIMPLE ONE LINE DIAGRAM FAULT IMPEDANCE INPUT DATA δ :=. Voltage Phase Angle er := es := 7 e j δ Source S Voltage es = 69.987 +.344i er := 7 Source R Voltage ZS := 4.4 + j.76 Source S Positive Sequence Impedance ZS := 5.357 + j 54.378 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 3.864 Positive Sequence Line Impedance Fault Indicator Charles Kim /5
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 =.5 +.866i BAL := a one := a Three phase voltages at S and R zero := 69.987 +.344i ES := es BAL ES = 33.83 6.83i 36.57 + 59.939i ER := er BAL ER = 7 35 6.6i 35 + 6.6i CIRCUIT EQUATION In 3-phase matrix form, the equation loos lie this: Fault Indicator Charles Kim /5
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 =.88 + 5.643i 7.84 + 4.367i 7.84 + 4.367i 7.84 + 4.367i.88 + 5.643i 7.84 + 4.367i 7.84 + 4.367i 7.84 + 4.367i.88 + 5.643i Source and Line Impedances to the Fault ZSS := ZS + m ZL ZSS = ZRR := ZR + ( m) ZL ZRR =.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i.76 + 6.44i.69 +.576i.69 +.576i 7.49 + 5.655i.5 + 8.863i 7.49 + 5.655i.69 +.576i.76 + 6.44i.69 +.576i 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i.69 +.576i.69 +.576i.76 + 6.44i Fault Indicator Charles Kim 3/5
Build System Part of the Impedance Matrix ZTOP := augment( augment( ZSS, zero), one) ZTOP =.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i ZMID := augment( augment( zero, ZRR), one) ZMID =.76 + 6.44i.69 +.576i.69 +.576i.69 +.576i.76 + 6.44i.69 +.576i.69 +.576i.69 +.576i.76 + 6.44i ZSYS := stac( ZTOP, ZMID) ZSYS =.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i.76 + 6.44i.69 +.576i.69 +.576i.69 +.576i.76 + 6.44i.69 +.576i.69 +.576i.69 +.576i.76 + 6.44i Pre-fault conditions: ZPRE := ZS + ZL + ZR ZPRE = 3.777 + 35.33i 8. + 8.3i 8. + 8.3i 8. + 8.3i 3.777 + 35.33i 8. + 8.3i 8. + 8.3i 8. + 8.3i 3.777 + 35.33i ISPRE := ZPRE ( ES ER) ISPRE =.7 +.4i.4.73i.56 +.49i Fault Indicator Charles Kim 4/5
IRPRE := ZPRE ( ER ES) IRPRE =.7.4i.4 +.73i.56.49i Pre_fault voltage at S end VSP := ES ZS ISPRE VRP := ZS IRPRE ER VSP = VRP = 69.97 +.447i 34.597 6.89i 35.37 + 6.37i 7.7.896i 34.3 + 6.85i 35.785 6.89i ES = 69.987 +.344i 33.83 6.83i 36.57 + 59.939i 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 = 69.987 +.344i 33.83 6.83i 36.57 + 59.939i 7 35 6.6i 35 + 6.6i TS = TR = Fault Indicator Charles Kim 5/5
ZFAG := ZFA + ZFG ZFBG := ZFB + ZFG ZFCG := ZFC + ZFG ZFAG =.57 ZFBG = ZF := ZFAG ZFG ZFG ZFG ZFBG ZFG ZFG ZFG ZFCG ZF =.57.57.57.57.57.57.57 FABCG := augment( augment( ZF, ZF), one) FABCG =.57.57.57.57.57.57.57.57.57.57.57.57.57.57 FINAL Z MATRIX Fault Indicator Charles Kim 6/5
ZABCG := stac( ZSYS, FABCG) ZABCG =.5 + 8.863i 7.49 + 5.655i 7.49 + 5.655i.57.57.57 7.49 + 5.655i.5 + 8.863i 7.49 + 5.655i.57.57 7.49 + 5.655i 7.49 + 5.655i.5 + 8.863i.57.57.76 + 6.44i.69 +.576i.69 +.576i.57.57.57.69 +.576i.76 + 6.44i.69 +.576i.57.57.69 +.576i.69 +.576i.76 + 6.44i.57.57 YABCG := ZABCG Fault Currents: IABCG := YABCG E E = 69.987 +.344i 33.83 6.83i 36.57 + 59.939i 7 35 6.6i 35 + 6.6i IABCG =.394.848i.5 +.69i.66 +.9i 4.487 8.8i.5.69i.66.9i 9.33 5.489i 6.93 67.38i 6.44 + 53.86i Fault Indicator Charles Kim 7/5
S - End Fault Currents: IS := TS IABCG IS =.394.848i.5 +.69i.66 +.9i R - End Fault Currents: IR := TR IABCG IR = 4.487 8.8i.5.69i.66.9i S - End Voltages 5.69.8i VS := ES ZS IS VSP = VS = 59.48 66.593i 59.93 + 54.598i R End Voltages * Additional Component for FI VR := ( ZR IR ER) VR = 47.96 + 64.i VRP = 48. 57.5i Line Prefault Load Currents from S Bus 39.64 + 7.74i 69.97 +.447i 34.597 6.89i 35.37 + 6.37i 7.7.896i 34.3 + 6.85i 35.785 6.89i Ia := ISPRE Ia =.75 Ib := ISPRE Ib =.75 Ic := ISPRE Ic =.75 arg( Ia) arg( Ib) arg( Ic) = 8.883.3 8 = 8.344 3.4 =.7 = 38.883 ISPRE = IRPRE =.7 +.4i.4.73i.56 +.49i.7.4i.4 +.73i.56.49i Fault Indicator Charles Kim 8/5
Line Prefault Voltages at S Bus Va := VSP Va = 69.97 Vb := VSP Vb = 69.97 arg( Va) arg( Vb) =.366 = 9.634 VSP = 69.97 +.447i 34.597 6.89i 35.37 + 6.37i Vc := VSP Vc = 69.97 arg( Vc) =.366 Line Fault Currents from S Bus Iasf := IS Iasf =.35 Ibsf := IS Ibsf =. arg( Iasf) arg( Ibsf ) = 5.97 = 6.344 IS =.394.848i.5 +.69i.66 +.9i 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) = 6.768 = 53.656 IR = 4.487 8.8i.5.69i.66.9i Icrf := IR Icrf =.336 Line Fault Voltages at S Bus arg( Icrf ) = 6.95 VSP = 69.97 +.447i 34.597 6.89i 35.37 + 6.37i Fault Indicator Charles Kim 9/5
Vasf := VS Vasf =.56 arg( Vasf) = 39.9 Vbsf := VS Vbsf = 89.68 Vcsf := VS Vcsf = 8.66 Line Fault Voltage at R Bus arg( Vbsf ) arg( Vcsf) = 3.6 = 37.66 VS = 5.69.8i 59.48 66.593i 59.93 + 54.598i Varf := VR Vbrf := VR Vcrf := VR Varf = 4.364 arg( Varf ) Vbrf = 79.978 arg( Vbrf ) Vcrf = 74.833 arg( Vcrf ) = 68.943 = 53.5 = 49.88 Residual Current and Voltage Vsr, Vrr, Isr, Irr Isrf := IS =.3.387i Irrf := IR = j j j = j = Vsrf := VS = 3.38 4.86i Vrrf := VR = j j j = j = arg( Isrf ) =.899 IRPREr := j = IRPRE = j arg( Irrf ) =.58 ISPREr := j = 4.778 8.479i ISPRE = j 56.569 + 4.56i IS = arg( Vsrf ) =.96.394.848i.5 +.69i.66 +.9i IR = VS = arg( Vrrf) =.5 VR 4.487 8.8i.5.69i.66.9i = 5.69.8i 59.48 66.593i 59.93 + 54.598i 39.64 + 7.74i 47.96 + 64.i 48. 57.5i Fault Indicator Charles Kim /5
VRPr VRP 7.5 5.84i 4 := = VSPr := j j = j = VSP =.4 4 +.3i 4 j Vsrf Zs := = 5.357 54.378i Zr := Isrf Vrrf Irrf =.554 + 5.796i Wattmeteric Method???? SS := Vsrf Isrf SR := Vrrf Irrf Re( SS) = 79.63 Re( SR) = 47.95 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 = 7.68 3 768 6 = 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 =.3-4.64-4 3.96-4 5.8-4 Fault Indicator Charles Kim /5
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 6.5-4 7.8-4 9.4-4.4-3.7-3.3-3.43-3.56-3.693-3.83-3... Van Vbn Vaf 5 Vbf 5 4 6 3 Ian Ibn Iaf Ibf 3 4 6 Fault Indicator Charles Kim /5
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 3.5..5. T Fault Indicator Charles Kim 3/5
VaS VbS VcS VrS.5..5. T VrS IrS.6.8. T VrS = 4.999 6 IrS =.488 6 dt :=.38 ω := π 6 ω = 376.99 mm := 536 window := 8 wind := window Now for all the calculations Fault Indicator Charles Kim 4/5
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 = 9.9 4.683i CompS = 9.9 4.683i, WattS := Re( CompS ) Real Current Component Method Fault Indicator Charles Kim 5/5
( ) ( ) ( ) ( ) ( ) ( ) 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 5 5 5 5 Phase a: Ima := ( FaS ), ( FbS ), ( FcS ), ZL =.35 + 3.864i ZL = 3.5 +.59i αa:= e i π 3 Fault Indicator Charles Kim 6/5
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 = 3.864 R =.35 Phase Impedance: Phase a: Fault Indicator Charles Kim 7/5
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
ma.8.6.4. 5 5 ma =.95 6 ma = 69.834 4 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