Question

Phase of a three-phase, 60 Hz, 300 km long power transmission line with a nominal operating line voltage of 500 kV. its inductance per 291 mH/phase, and its capacitance is 3.45uF/phase. Voltage between 500 kV phases from the end of the energy transmission line Under it, a load of 1000 MVA is fed with a power factor of 0.8 back. Power transmission line lossless assumption. This according to the operating conditions, a.) The line's phase constant (61), characteristic impedance (ZC) and natural visible(Sc). 6.) Find the A, B, C and D constants of the line using hyperbolic operation c.) Calculate the value of the maximum active power that the energy transmission line can transmit. d.) Hat to keep the load fed from the end of the line U1 = U2 = 500 kV constant Find the impedance and bias.

Answer 1

Soy 30 to 60 Hz), l= 300km) ( 291 H /prase), ( Burut I prase 198= 500 (ku) SPL= 1000 MUA), Pt= 0.8 a) ßt a wrea 2014.60 52a1x103 x3 4x106 Bez 3 174.35 x 154 rad Bez 0.38 rad Zc= 2914103 impedance 290.43(n) 3.45x106 Vg2 loada naturatural 601? 860-80 (mm) Zc 290.43 b ) Cos(BP) COSL0-38) 0.93 11 D= A= Ba j ze sin (B1)= j 290.43 x sin(0.38) j 107.73 ca j sin(0.33) ý sin(kel ; 1.28/03 Ze 290.43 c) Prop P= 13x vox IL x P. f Jsx 500 x 108 x Icxoo 800x106

11 1.155 (KA) Q2 Cont(0.8) - 36.870 1 - 1.155 -36.82° (ka) ve Jopp: JOFP-500133 = 500/38 = 288,68 (kW) usip A Joop +BTC Vs 0.934 288.68 x103 + 9 107.734 1.15541634-3687 (0.93 ¥28368 + J 109.93 x1,1554–36.47 ] Guy usipa 268.47 + 124.434 53.13 Jsip 357.27 <16-180 (kv) lus Nripl Pmax= 38 IBI 34 357. 24 x 288 68 - 10.4.73 Pman = 2.872 kawl end the usl=(N) Load bank at D) By connecting capacitors be cachevied can

0 CU Try A A B 이 cl C [1] - 미 AAH 연 el 이 CADY D A = ATBY = 1 니스 A 3 니 - 093 j6.493계 510의3 j1539 (M | phase