Question

Question 2: A 120 miles long medium length transmission line operating at 60 Hz delivers 46MW of power to a load with 0.707 power factor lagging. If the receiving end voltage is 69 KVLL, and line parameters are: r= 0.15 S2/Km, x=0.70 S2/Km, and y= 5.2 uS/Km, use nominal a model for the line and calculate the following values: a) Series impedance and shunt admittance of the line, b) Sending end voltage and current (Vs, Is) c) The voltage regulation of the transmission line, and the efficiency of the line.

Answer 1

line dength = 120 miles con (120* 1.809) = 193.08 km NOTE:- We are provieled with lines resistance, recictance, admittance for slines. Tarper phase) = 0.15 n-/km 0.05 11km (per phase) Y 3 Xline per phase) = 0.70.2/km 3 = 0.233 0/km.Cperphase) Yshunt = 5.2.X106 Siemen. km. This are admittance formed with respect to ground hence are considered per phase only. Sending end and recievengend - 2.6 x 106 70/km. Note Yshunt is distributed along evenly 5 5.2x106 siemen km

a z the senes impedance As the transmission and shunt aclmitance line is 193.08 km are ces follows:- (193-08)km(0.05) n/km = Kline Total 3 pex phase) 90654. 80.233.14 km)( 193.08 km = .4449876 Xline total perphase) Yahunt = ( 5.2X106 10/km) (193-08)Km. 4 million total

(6) Psd = 46 MIN - (on PB6 = 46 MM V3 VL IL Cesd = 46 MW 3 Vphe x Iph cest = 46 MW. We can make use of either of the formula to calculate phase Currect Ps6 = 46 MW 3.Vph. Iph cesø = 46x100. - Iph = 46x100 (37 (39.8383x103) (0.707) Iph = 544.39 Amperes.

-> 9.654 +j 44.9876 I544.39 IX) = 544.39 Cos - 0.707 lagi $ = -45° Toy = 544.39 -45° ICH ILIC2 'smo Il phase Tjamo * lagi Jc (D)*(8X) Cos¢£0.707 (398883 Loº) x103 x [ 1 x10 %] Lgoº. = 39-83831 Trine - using kcl at ® we can find Itxuline leo. . Ilme $39.8383 + 544639 2-450 = 39-8383_909 + 544.39L-45° , 384.94 - 384.94 = 384.94 - ; 345.1017 Iline = 384.94 - 345-1017 June e . . = 516.98 516.48 1 - 141.87 - 141.87 Amperes . V = Ve + I, 7 Vs = : 39.8383 X!0% LE + (516-98 L-41-87°) ( 7 line) Nere: Aline = 9.654 +j 44-9876. = 46.011 127.85° = 39.8383 X10° 20° + 99986-766 [ 36.01º. 39838.3 + 19241.457 + 13984886 . = 59079.3+13984.86 Vs = 60711.94 | 13-31 V lusl= 60.711 kv I, (6)(7) - (60711-94 119.317 (103) 190 60.7119 103-31

Applying Kel at node @ Is = Ies + IT 60.71191103.31 + 51 6.98 -41.87 -j 345.0545 = -13.9770 + j 59-081 + 384.974 - 370,997 - 285.9735 = 468.422 -37.625° Is = 60.711 13.31° KV Is = 468-422 L-37.625 Amps. Vsenda Shunt shunt + Zline ! - Regulation and efficiency :- Voltage regulation - To find Voltage regulation we must know the No load secieving end voltage Vnclace ? (secieving) VNe load Zshunt X Vsend. I Nolaad = - LOAD is Nor connected Kshant_%ahunt. - -Saccon VAL = -11000 xits 1. - 31000 -J1000 + 9.654 +j 44-9876 9.654 - 955.0124 Die =) 1.047 10:574° x Vs VAL = (1.047 | -0.579°) ( 60.711 113.3°) KW NL = 63.5644 112-731 Ky imu Crecientes Vregulation" 163.56441 - 139-83831 139483831 0.5955 Or) 59.55%:

efficiency - Pout Pout +losses Losses = 3 I line 1² Ra = 3 15160981(9-654). = 7.74 Mwalt. (for 3phases) 46 MW 4647-44) MW = 0.8559 9% = 85.59%.