Question

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 and total reactance is 66.5 ohm and the admittance due to capacitance is 0.442*10-3 mho. Determine: (i) (ii) (iii) A,B, C and D constants of long T.L Sending end voltage, current and power factor Transmission efficiency, Voltage regulation

Answer 1

Sol& Given that x. Receiving end voltage & r = 124 kv * Receiving end power Py = 60 MVA + 0.8 (34) = 48 MWI 1: power factor cos@=0.8 l= 300km lagging] * Length of the transmission line * Total Resistance R = 25.3 & total Reactance XL = 66.5 t * Shunt adml Htance *c = 0.442.4 to ov (*) Transmission line is a long line, so we need exact Transmission line equation *. Exact ABCD parameters for long transmission line [vs ] [ I ] [ coshal sinhol ze sinhot Ivo 7 coshol (Ir] :: A=D = (os bb), B = 7c sinhol, c sinho)

Here da propagation constant 2. Vzy 7e = characteristic impedance of the line 7 = Series impedance / km 3=8r+ful eelkm = shunt admittance Ikm y = gtful wolkm : 008432/km were here = 0.2217.2/km was not 0:4438 0931.473410 v/km 300 in 3= 8+ hot = 0.0843 +90.2217 selkm y = qtfwc = 0+81.473010 vlkm L.9-07

* 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 0:1744 * 7 - 0.0843+80.2017 ✓ $0.473 #com = 394.6719-872.5034 r - 401.2762 L-10.4094 & calculate el 81 0.0320+$0.1744 ere = 2.0320 30.1744 = 1.0325 [0.1744 vad - 1.0325 / 9.9923°

1.0325199923° = 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 * sinhol = (1.0325 (१.११23) - (०.१685 -१.११23) - 0.03151+30-1736 = 0.1764 179.7/09

4. 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. 2762 L-10.4094 - 4.3959 410" (90./203 :. The Agep PцYamelir s are A = Da cosh8) 0.9853 L0-3228° 8 = Ze sinhal = 70.8011_69.3015 ca Ž sinh31 = 4.39597 võ4 190./203° * Sending end Voltage Vsph = A Vr ph + 8 Ir ph Veph = vil bex = 71.5914 KV = 71.5914 Le kv (refference)

*. 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 Irph = 279.3630 L-36.8698° Amp .. Sending, end voltage Vs ph = A Vrph + 8 Irph = (0.9853 (0.3228 ) (71.591470 Lo) +60.8011 269.3015°) ( 279.3630 L-36.8698°) = 87232.1581+; 11004.8767 Volt Vaph = $7.9 23 L7.1702 ky VS LL = 13 * V sph = 152.288 47.1902 KV

* Sending end current Isph = (Voph +D Ir ph = (4.3959 *rā9 L90./203°) (71.5914 *1ồ LO ) +(0.9853 L0.3228°) ( 279.3630 L-36.8698°) = 221.0662-8132.4394 Amp Isph = 257.7023 L-30.9255° Amp Isph = Ist = 257.7023 L-30.9255° Amp * sending power factor= cosas power factor angle 0s = angle between sendingend Voltage and current = 7.1902- (-30.9255) - 38.1157 :. Cos os = COS 38.1157 cosas = 0.7867 lagging,

(LLC) input power Ps= 3/vsph) |Isph) cosas Ps = 3* 87.923 #1ồ * 257.7023 * 0.7867 Ps = 53.475MWI output & Transmission efficiency = Input 48 Mill 53.475 MW = 0.897614 = 89.7614 % * Voltage Regulation - 1-Vr|| | | | 1152.288 0.98531 -//241 - 0.9853 11241 = 0.2464 = 24.64 %

NOTE We can solve this problem by using approximate ABCD parameters of tong transmission line L 131121 A=D= 1+ 21 B= 2 (1+ 2 c=Y (1+ Zr) 1611217 H. by using Taylor's series coshol = 1+ (315" (31.34 21 74! + Cosho - + - +... neglect higher order terms stabil = 1 + enten det sinhol = - - - 51 T - neglect higher order Terms = 81 ( 17 )