b) The 6-bus power system network in Figure Q4b shows the positive-sequence and
ID: 2293453 • Letter: B
Question
b) The 6-bus power system network in Figure Q4b shows the positive-sequence and zero-sequence reactance's respectively. Resistances, shunt reactance's, and loads are neglected. Build all power system components and considering the generators and transformers connection in Figure Q4b admittance matric, [Y°1 by calculating admittances for BUS 6 X-0.45 pu x-0.00 X-0.60 pu X0.40 X0.80 x-0.06 p0.15 pu X-0.06 X-0 08 pu X0.40 pu X0.30 pu 0.60 p -080 pu BUS BUS 1 BUS 2 BUs 3 Figure Q4b (10 marks) c) Positive-sequence, Z' and zero-sequence, Z° impedances are given in Figure Q4c The negative-sequence matric is assumed equal to the positive-sequence matric Determine the fault current and 3-phase bus voltage at bus 2 in per unit value consideringa single-line to ground fault occur at bus 4 j0.156 j0.1385 j0.0989 j0.0462 po,0330 j00989 01385 j0.1938 j0.1385 0.0642 0.0462 0.1385 0.0462 j0.0646 j0.1062 j0.1615 j0.1154 jo.1062 0.0330 j0.0462 j0.0758 j0.1154 j0.1253 j0.0758 0.0989 j0.1385 ?0.1778 j0.1062 /00758 /03020 0.0600 /00735 /0.0449 ?0.0081 /0.0032 /00434 2004490.3146 0.0439 0.0176 jo.1449 0 j0.0081 0.0439 0.0899 0.03590.0458 0 0.0032 0.0176 0039 0.0384 0.0183 /0.0434 j0.1449 0.0458 ?0.0183 j0.3794? Figure Q4c (6 marks)Explanation / Answer
b) Yo
0.0000 -16.6667i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.0000 -15.2810i 0.0000 + 1.6700i 0.0000 + 0.0063i 0.0000 - 0.0273i 0.0000 + 1.1108i
0.0000 + 0.0000i 0.0000 + 1.6700i 0.0000 - 4.1667i 0.0000 + 1.2420i 0.0000 + 0.0139i 0.0000 + 1.2497i
0.0000 + 0.0000i 0.0000 + 0.0063i 0.0000 + 1.2420i 0.0000 -18.8663i 0.0000 +16.5906i 0.0000 + 1.0022i
0.0000 + 0.0000i 0.0000 - 0.0273i 0.0000 + 0.0139i 0.0000 +16.5906i 0.0000 -41.5560i 0.0000 - 0.0005i
0.0000 + 0.0000i 0.0000 + 1.1108i 0.0000 + 1.2497i 0.0000 + 1.0022i 0.0000 - 0.0005i 0.0000 - 3.3610i
C=
fault current in pu = -7.28ipu
voltage in pu =0.47pu
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