2) A three-phase 765 kV, 60Hz, 300 km, line has the following positive sequence impedance and admittance: 2=0.0165+j0.33

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2 A Three Phase 765 Kv 60hz 300 Km Line Has The Following Positive Sequence Impedance And Admittance 2 0 0165 J0 33 1 (54.63 KiB) Viewed 64 times

2) A three-phase 765 kV, 60Hz, 300 km, line has the following positive sequence impedance and admittance: 2=0.0165+j0.3306=0.3310<87.14° 0/km, V-j4.674x105 S/km. Identical shunt reactors (inductors) are connected from each phase conductor to neutral at both ends of the 300-km line during light load conditions, providing 50% compensation. The reactors are removed during heavy load conditions. Full load is 1.8 kA at unity p.f. and at 730 kV. Assuming that the sending-end voltage (Vs) is constant, determine the following: 1 Cos 0-1 a) Percent voltage regulation of the (using nominal xt circuit for ABCD parameters Juncompensated line (20p) b) Neglecting line losses, find the theoretical steady-state stability limit for the line (Assume a 5000-km wavelength and Vs-Vr=765 kV) (10p.) c) The equivalent shunt admittance and series impedance of the compensated line (10p) d) Percent voltage regulation of the compensated line (10p)