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A 3-phase, 50 Hz, long 300 Km transmission line delivers 60 MVA at 124 kV and 0-8 p.f. lagging. The total resistance 25.3 ohm
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Sol& Given that x. Receiving end voltage & r = 124 kv * Receiving end power Py = 60 MVA + 0.8 (34) = 48 MWI 1: power factor cHere da propagation constant 2. Vzy 7e = characteristic impedance of the line 7 = Series impedance / km 3=8r+ful eelkm = shun* propagation constant d-day = 10.0843+30.2217) (91-473-*17* = 0.0001 +$0.0006 = 0.0006 L79.5905 *. 8l=8*300 = 0.0320 +j 0:171.0325199923° = 0.१685 -१.११२३ 8) -1 x. Cosb e+e (1.0325 L१.११ 23) +(०.१685 - १.११23 ) = 0.9853 + 500055 = 0.१853 103228 * si4. c sinh81 - (no). 276 2 I-lo-40% 4) ( 0-17 4 7. 7/) - 25.024.5 +966-2311 = 70.8011 L69.3015 0.1764 279.7109 sinhal Zc 401.*. Pr = 3 1 Vorph) | Irph I cosør Pr . (IrphI= 2 Vyph) coser = 48+10% 38 124*1870.8 VB T = 279.3630 Amp i Irph = | Irphl Edo* Sending end current Isph = (Voph +D Ir ph = (4.3959 *rā9 L90./203°) (71.5914 *1ồ LO ) +(0.9853 L0.3228°) ( 279.3630 L-36.86(LLC) input power Ps= 3/vsph) |Isph) cosas Ps = 3* 87.923 #1ồ * 257.7023 * 0.7867 Ps = 53.475MWI output & Transmission efficiNOTE We can solve this problem by using approximate ABCD parameters of tong transmission line L 131121 A=D= 1+ 21 B= 2 (1+ 2

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