Question

a 50 Hz,132kv transmission line in 100 km long and having following distributed r=0.17ohm/km L=0.0011h/km C=0.0082uf/km.it is deliver in 70mv power at 0.8 lagging . neglecting the leakage conductance calculate a)voltage at sending end B) current at sending end c)sending end power d)regulation of line e) efficiency​

Answer 1

Explanation:

For a 100-km length of the line, R = 0.15 × 100 = 15 Ω ; XL = 314 × 1.2 × 10−3 × 100 = 37.7 Ω XC = 106/314 × 0.01 × 100 = 3187 Ω Using the nominal T-method, the equivalent circuit is shown in Fig. 41.33 (a) VR = 132/√3 = 76.23 kV = 76,230 V. Hence, VR = 76,230 + j 0 Load current, IR = 72 × 102/√3 × 132 × 103 × 0.8 = 394 A ∴ IR = 394 (0.8 − j 0.6) = 315 − j 236 A ; ZBC = (7.5 + j 18.85) W Drop/phase over BC = IRZBC = (315 − j 236) (7.5 + j 18.85) = 6802 + j 4180 V1 = VR + IRZBC = (76,230 + j0) + ( 6802 + j 4180) = 88,030 + j 4180 IS = IC + IR = (−1.31 + j 26) + (315 − j 236) = (313.7 − j 210) = 377.3 ∠− 33.9° Drop/phase over AB = ISZA B = (313.7 − j 210) (7.5 + j 18.85) = 6320 + j 4345 ∴ VS = V1 + ISZA B = (83,030 + j 4180) + (6320 + j 4345) = 89,350 + j 8525 = 89,750 ∠ 5.4° Line value of sending-end voltage = √3 × 89,750 × 10−3 = 155.7 kV Phase difference between VS and IS = 33.9° + 5.4° = 39.3° with current lagging as shown in Fig. 41.33 (b) cos φS = cos 39.3° = 0.774 (lag)Read more on Sarthaks.com - https://www.sarthaks.com/487360/phase-overhead-transmission-line-long-with-between-lines-receiving-following-constants