Question

A 3-ph, 50 Hz overhead transmission line 100 km long delivers 20 MW at 0.9 p.f lagging and at 110 kV. The resistance and reactance per phase per km are (0.2) Ω and (j0.4) Ω respectively, while capacitive admittance per phase per km is (j2.5 * 10-6) siemen. Using nominal T circuit, Find:

1- Constants A, B, C, and D. 2- Sending end voltage.       

3- Sending end current 4- Sending end power factor   

5- Voltage regulation of this T.L.    6- Transmission line efficiency.

Hint: For T equivalent circuit: A = D = (1+ZY/2)   , B= Z(1+ZY/4)   , C= Y

Answer 1

Gliven- felcoHz, lookm, P= 20MW, V = 110ku, pf - oglag. - Ro Total Resistance R = 0.2 x 100km = 20 XL=04x100 soon. Ys 2'5x16bx lion = 2:586045 z = 20+ HOT !.00, A = lt zy ImmaM YIC =lt ko+340)11 2:5X164) Bivs 2 A - 0.99 20:14 =) liv bliv Bu zl1+zy ) +620401) ( 146204401) 42.5x604)) Je = 44.60 463.50] SH* lovi) [C=Y= 2.611048) Receiving End Voltage VR F10x109,565508 v Load current IR = 20x0 = 116.6A V3 x 110 x lofx09 het lokale 1.581 i cosQR -0.9. Load current - IR = IR (Losar-i Sim OR) = 116.610.9-jo:435) TIR = 105-55071 Vi="ve+Iez - 63508+ (05-1507), (101120) v=65572 +31593

Charging current' -' Ic - JYT, -42.5x604 (655+2 +31593) Il = 0.4+1641 Sending End current Iss Ip+Tc = (10.5-5 504) + (-0.4+] 164) Hisapo litain V Is F110.08 L18:15 . Sending send power factor = cos (18.15) vpif = 0.95033 de 2 Sending End voltage, Ts = Ve + Isz =165572+51593) +104:6-134-3) (101520) = 64304 +9 3342 Ilvsl =>67367V Line Value Ja67387x13=116717V=116:41 kv voltage regulation = VS - VR X100 = 116.414 - 110 Hii VR w XOS = 6.1069% 110 Total Line Losses for 34: -315R + 3IR2 N 3X0103*x 10 + 3 x616-6)* xlo I o.fFOMW Transmission Efficiency F 20 8100 20+0°770 Transmission Efficieny 96.299 /