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Unit 2 samplesolutions

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Symmetrical Components, Per Unit Representation

Symmetrical Components, Per Unit Representation

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  • 1. 1. Three resistances of 10, 15 and 20 ohms are connected in star across a three-phase supply of 200V per phase as shown in the figure. The supply neutral is earthed while the load neutral is isolated. Find the currents in each load branch and the voltage of load neutral above earth. Use the method of symmetrical components.VAN = 200VBN = 200a2VCN = 200aLet us represent it this way, it will be easier.Apply KVL to the loop AacCVAN – 10 Ia +20Ic – VCN = 0And also apply KVL to the loop AabBVAN – 10Ia +15Ib – VBN = 0Now write the currents in symmetrical components and also we know that the zerosequence components will be zero as the load neutral current is not grounded.And if youwe apply KCL at load neutral we get Ia0 = 0.Hence we get,Ia1(20a - 10) + Ia2(20a2 - 10) = 200a – 200AndIa1(15a2 – 10 ) + Ia2(15a – 10 ) = 200a2 – 200Solving the above equation in terms of ‘a’ and in the end substitute ‘a’ value.Ia1 = 13.8461 - 0.0000iIa2 = 2.3077 + 1.3324iHence Ia = Ia1+Ia2 = 16.1539 + 1.3323iIb = Ib1 + Ib2 = - 9.2302 - 10.6584iIc = Ic1 +Ic2 = - 6.9230 + 9.3262iTo get Vn(voltage of the neutral with respect to ground) – start from ground and come toneutral of load
  • 2. Vn = 200 - 10*IaVn = 38.4614 -13.3234iMagnitude = 40.7037 2. The voltages at the terminals of a balanced load consisting of three 20 ohm Y-connected resistors are 200(0◦), 10(255.5◦) and 200(151◦) V. Find the line currents from the symmetrical components of the line voltages if the neutral of the load is isolated. What relation exists between the symmetrical components of the line and……….. Solution : Convert star to delta and you get the value of all resistances which are balanced as 60 ohms Iab = 200/60 = 10/3 (00) Ibc = 100(255.50)/60 = 5/3(255.50) Ica = 200(1510)/60 = 10/3(1510) Using symmetrical components resolution formula involving A-1 From this we get Iab1 = 2.7(15.10) Iab2 = 1.03(- 44.530) Ia1 = Iab1 – Ica1 = 4.67(-150) Ia2 = Iab2 – Ica2 = 1.784(-14.50) Ia = Ia1 + Ia2(zero seq. Comp. Currents will not appear in the lines of delta connected or star connected load with neutral isolated) Ia = 6.453(-14.870) Ib1 = a^2Ia1 = 4.67(-1350) Ib2 = aIb2 = 1.784(105.50) Ib = Ib1+Ib2 = 4.105(-156.90) Ic1 = a*Ia1 = 4.67(1050) Ic2 = a^2 Ia2 = 1.784(-134.50) Ic = Ic1+ Ic2 = 4.066(127.40) There the power dissipated in 3 phase 20 ohms resistors is
  • 3. P = Ia^2(20)+ Ib^2(20) + Ic^2(20) = (6.453^2 + 4.1^2+ 4.066^2)(20) P = 1500W

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