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NEW FAULT LOCATION ALGORITHMS FOR TRANSMISSION LINES IN INTERCONNECTED POWER SYSTEMS
K. Ramar∗ and A.A.A. Eisa∗
References
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[13] H. Saadat, Power system analysis (WCB/McGraw-Hill, Sin-gapore, 1999).Appendix AA non-iterative procedure to determine the fault locationin the case of double-line-to-ground fault shown in Fig. 7is derived below. Referring to Fig. 7:247VF B = VpB − x[ z21 z22 z23 ]IpAIpBIpC= VpB − xZ2I1(A-1)VF C = VpC − x[ z31 z32 z33 ]IpAIpBIpC= VpC − xZ3I1(A-2)Also: VF B − VF C = 0 (A-3)From (A-1)–(A-3), we get:VpB − VpC = x(Z2 − Z3)I1 (A-3)Hence: x =(VpB − VpC)(Z2 − Z3)I1(A-5)Appendix BThe line connecting buses P and Q with the L-G fault inphase A of circuit 1 is shown in Fig. 11.Figure 11. Double-circuit line with the L-G fault in cir-cuit 1.Referring to the ﬁgure:VpA = x[ z1A1A z1A1B z1A1C ]I1+ x[ z1A2A z1A2B z1A2C ]I2 + RF IF A (B-1)Figure 12. Positive sequence equivalent circuit.where:z1A1A, z1A1B, z1A1C, z1A2A, z1A2B, and z1A2C are the ele-ments of the ﬁrst row of the 6 × 6 line impedance matrixZ, per unit length of the double-circuit line.Z =z1A1A z1A1B z1A1C z1A2A z1A2B z1A2Cz1B1A z1B1B z1B1C z1B2A z1B2B z1B2Cz1C1A z1C1B z1C1C z1C2A z1C2B z1C2Cz2A1A z2A1B z2A1C z2A2A z2A2B z2A2Cz2B1A z2B1B z2B1C z2B2A z2B2B z2B2Cz2C1A z2C1B z2C1C z2C2A z2C2B z2C2C(B-2)I1 =I1AI1BI1Cis the vector of phase components of currentin line 1 at bus P.I2 =I2AI2BI2Cis the vector of phase components of currentin line 2 at bus P.Because the fault currents in the other two phases arezeros,IF B = 0, IF C = 0. (B-3)In terms of symmetrical components:I0FA = I1FA = I2FA =13IFA (B-4)where the superscripts 0, 1, and 2 refer to zero-, positive-,and negative-sequence components.The positive sequence equivalent circuit of the systemat during-fault condition will be as shown in Fig. 12.It may be noted that the positive sequence selfimpedances Z11 and Z12 and the mutual impedance Z112 ofthe line are equivalent to the per-phase impedances of theline Z1, Z2, and Z12 respectively.From loop 1:xZ11 I11A + (l − x)Z11 (I11A − I1F A) + lZ112I12A= lZ12 I12A + xZ112I11A + (l − x)Z112(I11A − I1FA) (B-5)248From (B-5):I1FA =l(l − x)I11A −(Z12 − Z112)(Z11 − Z112)I12A=l(l − x)Ie1(B-6)where:Ie1 = I11A −Z12 − Z112(Z11 − Z112)I12A (B-7)From (B-4) and (B-6):IFA = 3I1FA = 3l(l − x)Ie1 (B-8)Substituting (B-8) in (B-1):VpA = x[z1A1A z1A1B z1A1C]I1+ x[z1A2A z1A2B z1A2C]I2 + 3RFl(l − x)Ie1= xVe1 + 3RFl(l − x)Ie1 (B-9)where:Ve1 = [ z1A1A z1A1B z1A1C ]I1+ [ z1A2A z1A2B z1A2C ]I2 (B-10)Pre-multiplying both sides of (B-9) by I∗e1:I∗e1VpA = xI∗e1Ve1 + 3RFl(l − x)|Ie1|2(B-11)Equating imaginary parts on both sides of (B-11):x =imag(I∗e1VpA)imag(I∗e1Ve1)(B-12)Table 3Gen. No. Voltage Z1Z0Mag. (kV) Phase (deg.) Mag. (ohm) Phase (deg.) Mag. (ohm) Phase (deg.)1 31.80417 45.2445 1.0580 80 1.1109 8010 28.63868 13.6943 0.7935 80 1.0051 8011 30.53802 47.1793 1.3225 80 0.4232 80Figure 13. Tower conﬁguration of single-circuit transmission lines (PSCAD/EMTDC).Appendix CThe parameters of the 11-bus system used to test thealgorithms are given here.System frequency = 50 HzC.1 GeneratorsBase MVA = 100 MVABase voltage = 23 kVVoltage, positive sequence impedance, and zero sequenceimpedance of each generator are given in Table 3.C.2 TransformersTransformer MVA = 100 MVAWinding voltages = 23/230 kVThe positive sequence leakage reactance of each trans-former is given in Table 4.Table 4Transformer X1(pu)1–2 0.0310–4 0.0411–7 0.05The transformer between buses 11 and 7 is groundedthrough a reactor of reactance 0.08 per unit.C.3 Transmission LinesAll transmission lines, except the faulted line 6–8, aresingle-circuit, untransposed lines. They have the towerconﬁguration shown in Fig. 13, the lengths are given in249Table 5, and the conductor and ground wire data are givenin Table 6.Table 5Transmission Line Length (km)2–3 1002–5 502–6 1503–4 1303–6 1304–6 2004–9 2305–7 1407–8 1208–9 160Table 6Conductors Ground WiresRadius (m) 0.0203454 0.0055245DC resistance (ohm/km) 0.03206 2.8645Sag (m) 10 ¸ 10 ¸Figure 14. Tower conﬁguration of double-circuit transmission line (PSCAD/EMTDC).The faulted line 6-8 is untransposed and has a lengthof 200 km. In the single-circuit case it has the towerconﬁguration shown in Fig. 13, and the conductor andground wire data are given in Table 6. In the double-circuitcase it has the tower conﬁguration shown in Fig. 14, andthe conductor and ground wire data are given in Table 6.C.4 LoadsThe load data are given in Table 7.Table 7Bus No. P (MW) Q (MVAR)3 150 1205 120 606 140 908 110 909 80 50250
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Abstract
DOI:
10.2316/Journal.203.2008.3.203-3531
From Journal
(203) International Journal of Power and Energy Systems - 2008
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