2 Example Problem Using the voltages of the converged Newton Raphson load flow

2 example problem using the voltages of the converged

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𝐿𝐿𝑗𝑗𝑗𝑗 + 𝑗𝑗𝑄𝑄 𝐿𝐿𝑗𝑗𝑗𝑗 )
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2 Example Problem: Using the voltages of the converged Newton Raphson load flow solution, the single line diagram below shows these voltages. Let’s us compute for the line flows, line losses and total system loss. G G 2 1 3 0.0125 + j0.025 pu V 3 =1.01927 0.09111° pu SWING BUS V 1 =1.03 0° pu 4 G V 4 =1.01145 0.01926° pu 2.8 + j2.1 pu 0.0125 + j0.0375 pu 0.06 + j0.12 pu 0.03 + j0.06 pu 0.01 + j0.03 pu 0.01 + j0.02 pu V 2 =0.98609 -1.03741° pu Solving for the line currents, then subsequently the line flows and line losses: 𝐼𝐼 12 = 𝑉𝑉 1 − 𝑉𝑉 2 𝑍𝑍 12 = 1. 7012∠ − 41.3823° 𝑝𝑝𝑝𝑝 𝑆𝑆 12 = 𝑉𝑉 1 𝐼𝐼 12 = 1.3147 + 𝑗𝑗 1.1584 𝑝𝑝𝑝𝑝 𝑆𝑆 21 = 𝑉𝑉 2 𝐼𝐼 21 = 1.2786 − 𝑗𝑗 1.0860 𝑝𝑝𝑝𝑝 𝑆𝑆 𝐿𝐿12 = 𝑆𝑆 12 + 𝑆𝑆 21 = 0.0361 + 𝑗𝑗 0.0724 𝑝𝑝𝑝𝑝 𝑆𝑆 𝐿𝐿12 = ( 𝑉𝑉 1 − 𝑉𝑉 2 ) 𝐼𝐼 12 = ( 𝑉𝑉 2 − 𝑉𝑉 1 ) ( 𝑉𝑉 2 − 𝑉𝑉 1 ) 𝑍𝑍 12 = | 𝑉𝑉 2 − 𝑉𝑉 1 | 2 𝑍𝑍 12 = 0.03618 + 𝑗𝑗 0.07235 𝐼𝐼 13 = 𝑉𝑉 1 − 𝑉𝑉 3 𝑍𝑍 13 = 2. 7380∠ − 80.1783° 𝑝𝑝𝑝𝑝 𝑆𝑆 13 = 𝑉𝑉 1 𝐼𝐼 13 = 0.04811 + 𝑗𝑗 0.27788 𝑝𝑝𝑝𝑝 𝑆𝑆 31 = 𝑉𝑉 3 𝐼𝐼 31 = 0.04717 − 𝑗𝑗 0.27507 𝑝𝑝𝑝𝑝 𝑆𝑆 𝐿𝐿13 = 𝑆𝑆 13 + 𝑆𝑆 31 = 0.000937 + 𝑗𝑗 0.00281 𝑝𝑝𝑝𝑝 𝐼𝐼 32−1 = 𝑉𝑉 3 − 𝑉𝑉 2 𝑍𝑍 23−1 = 2. 8798∠ − 33.1681° 𝑝𝑝𝑝𝑝 𝑆𝑆 32−1 = 𝑉𝑉 3 𝐼𝐼 32−1 = 0.24546 + 𝑗𝑗 0.16098 𝑝𝑝𝑝𝑝 𝑆𝑆 23−1 = 𝑉𝑉 2 𝐼𝐼 23−1 = 0.24048 − 𝑗𝑗 0.15103 𝑝𝑝𝑝𝑝 𝑆𝑆 𝐿𝐿32−1 = 𝑆𝑆 32−1 + 𝑆𝑆 23−1 = 0.004980 + 𝑗𝑗 0.009950 𝑝𝑝𝑝𝑝
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3 𝐼𝐼 32−2 = 𝑉𝑉 3 − 𝑉𝑉 2 𝑍𝑍 23−2 = 5. 7596∠ − 33.1681° 𝑝𝑝𝑝𝑝 𝑆𝑆 32−2 = 𝑉𝑉 3 𝐼𝐼 32−2 = 0.49091 + 𝑗𝑗 0.32197 𝑝𝑝𝑝𝑝 𝑆𝑆 23−2 = 𝑉𝑉 2 𝐼𝐼 23−2 = 0.48096 − 𝑗𝑗 0.30206 𝑝𝑝𝑝𝑝 𝑆𝑆 𝐿𝐿32−2 = 𝑆𝑆 32−2 + 𝑆𝑆 23−2 = 0.009950 + 𝑗𝑗 0.019910 𝑝𝑝𝑝𝑝 𝐼𝐼 34 = 𝑉𝑉 3 − 𝑉𝑉 4 𝑍𝑍 34 = 0. 35433∠ − 54.1318° 𝑝𝑝𝑝𝑝 𝑆𝑆 34 = 𝑉𝑉 3 𝐼𝐼 34 = 0.21114 + 𝑗𝑗 0.29301 𝑝𝑝𝑝𝑝 𝑆𝑆 43 =
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  • Summer '19
  • Power, 0°, Power flow study, Sgen, JOHN CARLO G. PERION, JOHN KENNETH T. SINSON

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