Question

Question 2 (a) Define the term "surge impedance loading". [2 marks] (b) Prove that when the loss less line delivers load equal to the surge impedance loading voltages at both ends of the line are equal in magnitude. [4 marks) (c) A 400 kV, 50 Hz, 300 km transmission line delivers 700 MW at 0.85 power factor lagging. Parameters of the line are given below;

Answer 1

a) Surge impedence Loading It is used in the prediction of maximum Loading capacity of transmission lines. the characteristics impedence of an ideal or lossless Trains onission line is known as surge impedence of line. Meathematically it can be represented as follows: RAWL ZE= 1 / 1 Gt juc voltage SILFUR end voltage is SIL is defined as the power delivered by a line to a purely resistive load Equal in value to the surge impedence of that line where VR = receiving In this receining Equal to sending and voltage is called flat voltage, receiving End Current is equal to sending End current Also, power factor is unity b) In the surge impedence leading analysis let us consider ideal or lossless transmission line supplying the power to purely resistive load of impedence Equal to surge impedence Vako Za Zg 32 p = G co QUISTO V8 Lossless ToL

has R=6FO a=0 Hence, the lossless line delivers load equal to Lossless line R = wie under this, we know that NS=Cospl Vat jzs singl .IR Ve = Cosrl. Vet jzgsingh VR { 3R = Ve .:VS = VR Cos Bl. Vere + j Eus Sin RI NR (24-7] SVR (Cos feltjsin B1) Conclusion. IVs I = VRI Ze Vs = VR e Re : 14s8 - I VRTCBA -! Ends are the surge im pedence loading, voltages at both of the line Equal in magnitude C). givena: 40oku, 50 H2, Bookm delivers 700mW at 85 Pfl 7=0.025 tj.3 2/km y=j4X106 Blkm lagging

Receiving end phase wolltegel VR)= = 400 kv 13 231 KV (05 (85)=31.80 6 How, receiving and current (IR)= 700x10* (3 x 4008103 xo85 IR = (1.188 k 14)<-31.8 Ve= (vzy 1) = (06.025 +3.3) (4X160) )x 300 F(17:23,10% )* 300 <87.615 - 328 (87061 Zc = { / We know ·025 +33 = 287.665-1.2" 4x100 j A= 0 = Cosh (ves= (1+ ya! =.946(:272 sinh (wd.) = ryz (1+ y z ľ 3245 <87:67 2.324 (84.1 B = Ec sinhlre = (287.66(-1-2)(-324687-70) = 92-65<86-5 Now Vs = A VR + BIR =4.94676242")x 231x10 + (92065(86-5) 1-1881173 Vs -(296.4<17.8')kv (31.8' Phase

As load is disconnected IR=0 : Vsleine) Zc :: A (13 4296.4 <17-8")kv Uscrines (513.3%<19.8")kv for Is (sending luncnt) we have to bring C and D c = sinhwl -324 (87.7" 287.66(-12 =(1.128163) (29 D= .946 (-272² is =C VR - (12x103)<89" x 231x1024.9466•272 XI-12810 Isline (1.01972-18.2) KA Now, power factor is vos of angle between Vs and Is 017-8-(-18-2) 36" Coso = cos 36" [/f = .8 lag + D IR Is phase . 8 laggin Power factor ii). "Vs = AVR 542.68 A 513.38 Now VR= 996 542-68> 213.38 VR) Vs so, high voltage is appeared at receiving end that may cause rupture of Insulation It is also called as Ferranti Effect