NEW FAULT LOCATION ALGORITHMS FOR TRANSMISSION LINES IN INTERCONNECTED POWER SYSTEMS

K. Ramar∗ and A.A.A. Eisa∗

References

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  13. [13] H. Saadat, Power system analysis (WCB/McGraw-Hill, Singapore, 1999). Appendix A A non-iterative procedure to determine the fault location in the case of double-line-to-ground fault shown in Fig. 7 is derived below. Referring to Fig. 7: 247 VF B = VpB − x[ z21 z22 z23 ]      IpA IpB IpC      = VpB − xZ2I1 (A-1) VF C = VpC − x[ z31 z32 z33 ]      IpA IpB IpC      = 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 B The line connecting buses P and Q with the L-G fault in phase A of circuit 1 is shown in Fig. 11. Figure 11. Double-circuit line with the L-G fault in circuit 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 elements of the first row of the 6 × 6 line impedance matrix Z, per unit length of the double-circuit line. Z =           z1A1A z1A1B z1A1C z1A2A z1A2B z1A2C z1B1A z1B1B z1B1C z1B2A z1B2B z1B2C z1C1A z1C1B z1C1C z1C2A z1C2B z1C2C z2A1A z2A1B z2A1C z2A2A z2A2B z2A2C z2B1A z2B1B z2B1C z2B2A z2B2B z2B2C z2C1A z2C1B z2C1C z2C2A z2C2B z2C2C           (B-2) I1 =      I1A I1B I1C      is the vector of phase components of current in line 1 at bus P. I2 =      I2A I2B I2C      is the vector of phase components of current in line 2 at bus P. Because the fault currents in the other two phases are zeros, IF B = 0, IF C = 0. (B-3) In terms of symmetrical components: I0 FA = I1 FA = I2 FA = 1 3 IFA (B-4) where the superscripts 0, 1, and 2 refer to zero-, positive-, and negative-sequence components. The positive sequence equivalent circuit of the system at during-fault condition will be as shown in Fig. 12. It may be noted that the positive sequence self impedances Z1 1 and Z1 2 and the mutual impedance Z1 12 of the line are equivalent to the per-phase impedances of the line Z1, Z2, and Z12 respectively. From loop 1: xZ1 1 I1 1A + (l − x)Z1 1 (I1 1A − I1 F A) + lZ1 12I1 2A = lZ1 2 I1 2A + xZ1 12I1 1A + (l − x)Z1 12(I1 1A − I1 FA) (B-5) 248 From (B-5): I1 FA = l (l − x) I 1 1A − (Z1 2 − Z1 12) (Z1 1 − Z1 12) I1 2A = l (l − x) Ie1 (B-6) where: Ie1 = I1 1A − Z1 2 − Z1 12 ( Z1 1 − Z1 12) I1 2A (B-7) From (B-4) and (B-6): IFA = 3I1 FA = 3 l (l − x) Ie1 (B-8) Substituting (B-8) in (B-1): VpA = x[z1A1A z1A1B z1A1C]I1 + x[z1A2A z1A2B z1A2C]I2 + 3RF l (l − x) Ie1 = xVe1 + 3RF l (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 + 3RF l (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 3 Gen. No. Voltage Z1 Z0 Mag. (kV) Phase (deg.) Mag. (ohm) Phase (deg.) Mag. (ohm) Phase (deg.) 1 31.80417 45.2445 1.0580 80 1.1109 80 10 28.63868 13.6943 0.7935 80 1.0051 80 11 30.53802 47.1793 1.3225 80 0.4232 80 Figure 13. Tower configuration of single-circuit transmission lines (PSCAD/EMTDC). Appendix C The parameters of the 11-bus system used to test the algorithms are given here. System frequency = 50 Hz C.1 Generators Base MVA = 100 MVA Base voltage = 23 kV Voltage, positive sequence impedance, and zero sequence impedance of each generator are given in Table 3. C.2 Transformers Transformer MVA = 100 MVA Winding voltages = 23/230 kV The positive sequence leakage reactance of each transformer is given in Table 4. Table 4 Transformer X1 (pu) 1–2 0.03 10–4 0.04 11–7 0.05 The transformer between buses 11 and 7 is grounded through a reactor of reactance 0.08 per unit. C.3 Transmission Lines All transmission lines, except the faulted line 6–8, are single-circuit, untransposed lines. They have the tower configuration shown in Fig. 13, the lengths are given in 249 Table 5, and the conductor and ground wire data are given in Table 6. Table 5 Transmission Line Length (km) 2–3 100 2–5 50 2–6 150 3–4 130 3–6 130 4–6 200 4–9 230 5–7 140 7–8 120 8–9 160 Table 6 Conductors Ground Wires Radius (m) 0.0203454 0.0055245 DC resistance (ohm/km) 0.03206 2.8645 Sag (m) 10 ¸ 10 ¸ Figure 14. Tower configuration of double-circuit transmission line (PSCAD/EMTDC). The faulted line 6-8 is untransposed and has a length of 200 km. In the single-circuit case it has the tower configuration shown in Fig. 13, and the conductor and ground wire data are given in Table 6. In the double-circuit case it has the tower configuration shown in Fig. 14, and the conductor and ground wire data are given in Table 6. C.4 Loads The load data are given in Table 7. Table 7 Bus No. P (MW) Q (MVAR) 3 150 120 5 120 60 6 140 90 8 110 90 9 80 50 250

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