Question

3.8 A 275 kV three-phase transmission line of length 96 km is rated at 800 A. The values of resistance, inductance and capacitance per phase per kilometre are 0.078 12, 1.056 mH and 0.029 F, respectively. The receiving-end voltage is 275 kV when full load is transmitted at 0.9 power factor lagging. Calculate the sending-end voltage and current, and the transmission efficiency, and com- pare with the answer obtained by the short-line representation. Use the nomi- nal n and T methods of solution. The frequency is 60 Hz. (Answer: Vs = 179 kV per phase)

Answer 1

Given data of Tr. line = 96 km. Total Resistance R = 96% 0.078 = 7.4882 ( for 96 km length) Total inductance, L= 96 x 1.056 = 101.376 MH 11 Total Capacitance, C = 96 x 0.029 = 9.784 LIF ve = 975 KV - Cos 0 = 0.9 lag, IR = 800 A = 0.8 KA <Line 2 I'mpedance , Z = R+jwl = 7.488+j 21 X 60X 101,376x10 = 38,94 (78911122 Adflavce 3 = +2 c = 0 + 30 x 60 x 884 x10 = 1,044 x 10 lo Using Nominal T- method B7 fogs 0.23 38.55 [19,05 Oulxco 490 0.98(0,23 Vs phase = 10.98 0.23 )( 275 )+(38.55 (79,0570.8 L-Guit 69 ) Vs, phase = 175,02 20:276 Kvě I vs, come le = 304.52 KVE 19.X V3! (0.98 (0.23) C0.8 - 63.0.9), 15 = (1.049 vi* _40" )( 245 ) + (3-soba-naffo 0.7276 (-13.708 KA CSScanned with CamScanner

11s1 = 727.6 A. 2011 cacat sanator efficiency in PR x 100 = 3 V -√3 VR IP COS OR BVs Is as a = 3 X 975 X800 *0.9 x 100 2. Par V3 X 304,52 x 727.6 x 0.927 | | 0 = 48,39 71 NHÀ Nominal TT method CA B i (29+'?) Lyczy +1) (2+1)] = 1 000 38.94 1 7809 3 8.94 [78.9 = r 0.98 (0.93 1.038 Xlő3_9021., 0.98 %0.23] JA Vs.phase = (0.98 (0.23)(275.) + (38.94 L78-9) (0.8L-Cos'.9 ) = 176.08 18.33 KV [ Vs, line l = 304,98 kv Tg = (1.038x10*490.11 ) ( 245 ) + (0.98 (0.28) (0.8 [456.) = 0.727 L-13.248 KA. ITs] = 7278A Efficiency M = P x 100 = x 3 x 275 x 800X0.9 Vioo X100 3 X 304.98X727.870.926 n = 98.33% CScanned with CamScanner

for short transmission line :- I D 38.94 L78.91 faol - Co PI 54.94_16.99 's phase = 375 + (38.94 78.91)(0.8 L-G5 "0.9) V3 = 179,23 % 7,98 kv. I vs ),... = 310,43 kv In = In = 800 A det E x 275 x 800 x0.9 100 x 100 - 13X310.43 X 800 x0.63 M = 95.96 % Scanned with CamScanner