Unit 3 Part 1

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Doubts Clarification

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Unit 3 Part 1

  1. 1. Short and Medium Lines 1. 15000 KVA is received at 33KV at 0.85 pf lagging over an 8km 3 phase OH line. Each line has R = 0.29 Ω/km, and X = 0.65 Ω/km. calculate (b)the voltage at the sending end (b) the power factor at the sending end, (c) the regulation, and (d) the efficiency of the transmission line. [20.31|_2.37°KV, 0.8275 lagging, 6.58%, 96.38%] Radius?and find Now you will get the correct answer (power is inductance? 15000KVA) 2. A 3 phase line, 10km long, delivers 5MW at 11KV, 50Hz, 0.8pf lagging. The power loss in the line is 10% of the power delivered. The line conductors are situated at the corners of an equilateral triangle of 2m side. Calculate the voltage and pf at the sending end. Assume resistivity of copper to be 1.774µΩcm. [7.535|_5.266°KV, 0.7416 lagging] Solution: Vr = 6350V Ir = 328A It si short line The power loss in the line is 3Ir2R Power loss = 10% of power delivered 3Ir2R = 10% 5MW; R = 1.549 ohm = ρl/a l= 10km; a = 1.147cm2; Hence r = 0.6042cm; r’ = 0.7788r Spacing between the conductors, D = 2m = 200cm Inductance L = 2x10-7 ln [200/(0.6042x0.7788)] = 1.21 x 10-6 H/m Therefore total inductance L = 10,000 X 1.21 x 10-6 H Reactance X = 3.8 ohm Vr = 6350|_0°V Rec. pf angle = cos-1(0.8) = 36.87° Z = 1.549 + j3.8 = 4.1|_67.8226° ohm Vs = Vr + IrZ = 7535.13|_5.266°V
  2. 2. The phase difference between Vs and Is will be 5.266° – (-36.87°) = 42.136° Sneding end pf = cos42.136° = 0.7416(lagging) 3. A 3 phase 50Hz, 150km line operates at 110KV between the lines at the sending end. The total inductance and capacitance per phaseare 0.2H and 1.5µF. Neglecting losses calculate the value ??? of receiving –end load having a pf of unity for which the voltage at the receiving end will be the same as that at the sending end. Assume one-half of the total capacitance of the line to e concentrated at each end. [173.275A] Solution : Vr = 110,000/sqrt(3) = 63508.53|_0°V XL =2 πf(0.2) = 62.832 ohms Z = j62.832 ohms Y = j2 πfC = j4.712 x 10-4 S = 4.712 x 10-4|_90° mho or Siemens or S Iab =( Y/2 )Vr = j14.963 A Let the load current be Ir, since the load pf is unity, Ir = Ir |_0° Hence I = Ir + Iab = Ir + j 14.963 Vs = Vr + IZ = Vr|_0° + (Ir + j 14.963)( j62.832) = (Vr – 940.155) + j62.832(Ir)
  3. 3. Vs2 = (Vr – 940.155)2 + [62.832(Ir)]2 63508.532 = (63508.53– 940.155)2 + [62.832(Ir)]2 Ir = 173.275A

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