Question

Problem 5.11 A three-phase, 60 Hz transmission line with a length of 180 miles consists of three Cardinal conductors per bundle with equilateral spacing of 35 cm between conductors. The bundles are arranged vertically with spacing between centers, as sketched in Figure 5.94 (x = 5 m, y = 6.5 m). A 500 kV network supplies the line. The supply short circuit current is 22 kA. (a) Determine the total resistance, inductive reactance, and capacitive admittance of the transmission line. (b) Draw the equivalent circuit. (c) Obtain the Thévenin equivalent circuit as viewed from the load terminals. (d) Plot the short-circuit current versus distance, assuming that a short circuit may occur at any point on the transmission line. Explain the tendency of the curve. (e) If the short circuit occurs at the end of the line, determine the magnitude of the current.

Answer 1

Answer:

Answer, - 500k Phase Given that Three Cardinal conductors per bundle per phase are arranged cu shown below d=35cm = 0.35 m x=5m ; Y= 6.5m Line length d= 180 miles-289.682 km Supply voltage line toline UsLe = 500kV ; Uson - Soo ku Voltage For Cardinal conductors, radius, 8=0.0158m Supply short circuit current, Is (sc) = 22 KA (a) Resistance, R= 1 ; P = Resistivity of conductor l=289.68 km A = 18²= ITX (0.0158)² m2 [R = 7.155 Inductance per phase per km, Lpg = Molly in / GMD | Him Ph 21 self (mo) Geometric mean Distance Gimo - (0, x02x03) - (82x82 x 13) %> D,= (x² + y² = √5 + 65 = 8.2m = 9.56 AM D2 = √n?ty= = √5 +6.52 = 8.2m

D3=24=2x6.5=13m Self GMD for Inductance = (0.7788 8xdxd)3-(0788 X0.0158 X0-35x035)'S -0.1146m Self GMD for Capacitance =(x xd xd)3 = 0.1246m so, 4pn= ult x 16 7 x 1 Ln / 9:56 lm 21 10.1146) = 8.8478 107 HIM = 8.847X104 H/km Total Inductance, L-Leux 289.68 = 8.847x 289.68x104 = 0.25625H Inductance Reactance, x = 217fL XL = 96.60 Capacitance per km per phase, Clen = 24 Eotr Ilm Ln/GMDL Self GMD) Clem 21x8-854x tol? &= 8.8 16.1246 Egal = 1.28 18 x 16 "flm = 1.2818 x 108 Fl km

Total capacitance, c= clen x 289.68 = 1.28488108 x 289.68 = 3.713x106 C= 1.8565 MF Yc=ja1fc = 21x60x1.858 15 =jo.698 x 16² vons (siemens) Ye= jo.698ms (6) Given transmission line length 180 miles so medium Line represented as Nominal - 11 model JXL IR R sin 7.15 2 J XL m 196. 64 mo 7.150 $96.60 - - by VR con vi Vs 10.698 ms my Equivalent circuit (C) Zth=8.38 ti 1196 4 n=315 kv (d) short circait fault Current verses length up to fault Short circuit fault Carrent Distance

zth end of line 8.38+jlige Vih y Iresc) Short circuit current 315KV Uth IV (SC) 315 kv 8838 +; 119_21 S9 12+h) - 2.64KA