Question

and a m line, with characteristic impedance of 258.2 , B 80<80°, С 0.003<90". The receiving end load is at 0.99 hen at full load receiving current, the voltage drop limit,0.95 ] 1. A 500 kV, 200 k propagation constant of j0.0146 m1, has the following A, B, an lagging p.f. is not satisfied. 1) [6] determine the practical line loadability in MW; 2) [3] determine the SIL; 3) [6] for the lossless line equivalent, at n lossless line equivalent, at no-load, estimate the receiving end voltage

Answer 1

Given that A 500 kv, 200 km line with characteristic impedance of 2582 Ohm and a propagation constant of j 0.0146 m^-1 A = 0.95 angle 0.2 degree B = 80 angle 80 degree C = 0.003 angle 90 degree v_R = 500 kv voltage drop limit = 0.95 characteristic impendence z_c = 258.2 ohm the receiving and load = 0.99 lagging p.f 1) the practical line load ability in mw: - Sending end voltage v_s = v_e/0.95 v_s = 500 times 10^3/ 0.95 = 526.315 k v practical loading - v_s v_R/ z_c = 526.31 times 10^3 times 500 times 10^3/258.2 Therefore practical loading = 1019.2 m w \r\n 2) surge impendence loading (SL): - surge impendence loading = v^2_R/z_c sl = (500 times 10^2)^2/ 258.2 SiL = 968.24 m w 3) for the losses line equipralent at no load Receiving end voltage = Spiting end voltage Because voltage drop = 0 at no load & loss then transmission