Question

A 230-kV, three-phase transmission line has a per phase series impedance of z = 0.05j0.45 2 per km and a per phase shunt admittance ofy = j3.4 x 10-6 siemens per km. The line is 80 km long. Using the nominal r model, determine (a) The transmission line ABCD constants.

Answer 1

Soluhan. Ia sokm (short Transmission line-7-model) Giren: 7 = (0-05+) 0-45) 12/km Z = 0.05+0.45) x 80 = (4+ 136).2 - y = j34xło -607/km | sở y = 03-415-4) x 80 = (3D-212 X 5-3) " (a) now, ABCD constant for this line in nominal tr Model:- (Tr] =[AB] = [[1+ y2) z Icon M(1+x2) (**2) Ø 1+(42) = 1+(4+336) (10.272.X10-3) - (0.9951 +J0.541 X10-3) <= (4+)36) = (36-22 Samol = (0.995120-0313") Yli + y) = 60.73924 x167 -3 7-3803X10-8) = (7-38810-82-179.98°) Hence 0.9951 20.0313° 3622283-66° 7-38X10-82-179.989 0.9951 20.0313 280X <--- Cost Coa) (b) Now for load: DOOMVA; and at 0.8 lagging pf at 220 kV.' So load Current (IR) = 200X103 Or Recieringend current 524.86_-36-879) Amp ne as Vs LA BIVR l Is C DJL IRI Hene so sending end voltage, Vs = 40.995120-0313") ( xlo?20°) + (36-2228366) (524.862-36-87") Vs = 140104.4125.704 v (phase) Velline) = 140104.41 EV vs Cline) = 42.668 25.704") kev Exlooo osendingen 1 =(7-38x10-62479.98) ( 230*10>) + (0.995|20:0313X524-862-3€23 Is= 522.282-86-839° Ampa

220 ► Now, witage Regulechion: (Ve) = (Us! - {vel 242.66€ - 220 o.ggs1 IVA [(V.R.) = 0.1084 or 10.84 %) The sending end power! Ps = x 242.668 X 5 22-28 X cos ( 36-339+ 5.7049) Ils = 161.736 MW Noca as Recesinin erd power! PR = 13 X 220X103x 524.86 X0.8 TPR = 160 MW) X 103 Hence 4 - Transmission etlichen R Xloo OS - Xloo IGO 161.736 = 0.9892X100 = 98.927.

(C) Now for load: 306 MW et unity pf of 220kv So VR = 220 X 103 Lov IR = 306 x 106 -roana Amb =(803.04 209) Amp 220X33X103 sending roltage rs =(220X109) 6.9951 20.0313)+ (803-04)( 36.22283.66) 13 Vs = 132. Bo6x103 v Cphase) <12.6* "s = 230.027 Z12.6° kul Cline) sending end current: Es (220x103) (438X10-8 420:38°) +(80394) Co.9951 con Fs = 799.095 20.0313° Ampa ► holteege Regulation: VIR = -NR _1 230.027 - 2201 Io 9951 0.0 5072 220 5.072% * Sending and power! Ps = 13X 799.095 X 230.027X103 cos (12-6-0.0313) IPs = 310.7446 MWI o Then efficieney! my.= PR xioo 4 %= 306 X 100 31017446 My = 98.47%. NRI PCX TOO