Question

A 220-kV, 150 MVA, 60-Hz, three-phase transmission line is 140 km long with the characteristic parameters of r = 0.09 ohm/km, x = 0.88 ohm/km, y = 4.1 * 10^-6 S/km. 10. Calculate the ABCD constants of this transmission line. 11. What is the sending end voltage if the line is supplying rated voltage and apparent power at 0.85 PF lagging? 12. What is the voltage regulation of the transmission line for the conditions in question 11. 13. What is the efficiency of the transmission line for the conditions in question 11.

Answer 1

Y= ј4.VXls*x 140 七一 3 X 2.20자03 . 393. 648 A .. IR-39 3. 6482-31.788A : @. 9646+5 3. 6162)(13.013 ku)十 (12-6+]123.2)(393.6%2-31-788

Sendin end line liu ニ 오구2-305 KU (13) At no load IA 20 ニ /62·984kv /6 2.984-127.01 Re28 316 -31. 구83 Sendi ed pona cen (14. 389 t 21-y?! 32.82G3 Mu Pe 127 S M

Answer 2

The total impedance is, Z = (0.09 +70.88) x140 -12.6 + j1232 Ω The total shunt admittance of the line is Y (14.1×10-6)×140 =./574 × 10-6 S Since the length of the line is greater than 80 km but less than 200 km Thus, the given transmission line is a medium transmission line. 10 For a medium transmission line, the ABCD parameters are given as. YZ 2 YZ 4 YZ 2

Thus, YZ 2 (,574 x10-6S)(12.6+7123.2 Ω) 2 1+(j3.6162-35.36)x10) 0.9646 + 13.6162×10 Thus, A D 0.9646 3.6162x10 Now B-12.6 +j123.2 Ω and YZ 4 (j574 x10-6)(12.6 + jl232 (2) 4 - i574x106 (0.982 +1.808x10 3 一一1.038x10-6 + j563.67×10-6 S Thus, C=-1.0384 j563.67 μS

The phase angle, φ is, cos-1 (0.85) - COS -31.79°1a Therefore, I, -393.642-31.79° A Write the expression for the sending end voltage as: v, (0.9646+ j3.6162x 10 127.02x10 (12.6+ j123.2) (393.644-31.79°) Vs 122523.49+ j459.33-29764.85+ j38609.32 V 157.22214.39° kV The sending end line voltage is: V3(157.22214.39° V 263.77 14.390 kV

12 For no load conditions, Rnl 157.22 x10 214.39° 0.9646+ 3.6162x10-3 158025.92 + j39913.946 Rnl VRl-162988.69214.17 Write the expression for the voltage regulation. V.R 162.99 kV Rf Rf 162.99-127.02 127.02 V.R 0.2832 Thus, %VI 28.32%

13 The sending end current, Is is, Is = (-1.038x10° +,563.67x 10-6)(127.02x103 +(0.9646 +j3.6162x109) (393.644-31.799) Is 0.132 +j71.59 +323.49-j198.82 1. = 348.63/-21.75° A Calculate the sending end power, = 3(157.22 x10')(348.64) cos( 14.39+21.75) -132797.81x103 W -132.798 MW Calculate the receiving end power, 150 (0.85) -127.5 MW Calculate the efficiency of the line, η. 100 Po 127.5 132.798 0.9601 x100 Thus, %77-96.01%