In reference to the power system reactance diagram, all values are given in pu. Determine the followings:

1. Construct the bus admittance matrix for the power system shown in Fig. 1
2. Use Gauss-Seidel method to solve the load flow problem

Table 1: Impedances and line charging (in p.u.) for the system shown in Fig.1

 Bus code p-q Impedance Shunt admittance 1-12 0.02+j0.16 j0.03 1-4 0.015+j0.12 j0.024 1-3 0.015+j0.13 j0.024 1-2 0.03+j0.25 j0.031 12-11 0.025+j0.20 j0.021 12-5 0.016+j0.09 j0.015 12-4 0.016+j0.10 j0.015 2-8 0.06+j0.60 j0.04 3-8 0.03+j0.28 j0.04 3-4 0.018+j0.16 j0.03 4-5 0.014+j0.1 j0.03 4-9 0.008+j0.065 j0.012 9-8 0.045+j0.40 j0.038 9-7 0.02+j0.18 j0.025 9-6 0.04+j0.38 j0.036 9-10 0.035+j0.32 j0.034 8-7 0.045+j0.40 j0.042 7-6 0.045+j0.40 j0.042 6-10 0.015+j0.13 j0.020

Table 2: Bus data for the system shown in Fig. 1

 Bus code Vinitial Pg Qg Pl Ql 1(slack) 1.0+j0 0 0 0 0 2 1.0+j0 0 0 0.66 0.26 3 1.0+j0 0 0 0.50 0.15 4 1.0+j0 0 0 0.54 0.16 5 1.0+j0 0 0 0.42 0.18 6 1.0+j0 0 0 0.64 0.28 7 1.0+j0 0. 0 0.60 0.26 8 1.0+j0 0.68 0.18 0.06 9 1.02+j0 0.84 0.24 0.09 10 1.01+j0 0.97 0.20 0.05 11 1.01+j0 0.85 0 0 12 1.03+j0 0.72 0.12 0.03

In reference to the power system reactance diagram, all values are given in pu. Determine the followings:

1. Construct the bus admittance matrix for the power system shown in Fig. 1
2. Use Gauss-Seidel method to solve the load flow problem

Fig. 1

Table 1: Impedances and line charging (in p.u.) for the system shown in Fig.1

 Bus code p-q Impedance Shunt admittance 1-12 0.02+j0.16 j0.03 1-4 0.015+j0.12 j0.024 1-3 0.015+j0.13 j0.024 1-2 0.03+j0.25 j0.031 12-11 0.025+j0.20 j0.021 12-5 0.016+j0.09 j0.015 12-4 0.016+j0.10 j0.015 2-8 0.06+j0.60 j0.04 3-8 0.03+j0.28 j0.04 click here for more information on this paper 3-4 0.018+j0.16 j0.03 4-5 0.014+j0.1 j0.03 4-9 0.008+j0.065 j0.012 9-8 0.045+j0.40 j0.038 9-7 0.02+j0.18 j0.025 9-6 0.04+j0.38 j0.036 9-10 0.035+j0.32 j0.034 8-7 0.045+j0.40 j0.042 7-6 0.045+j0.40 j0.042 6-10 0.015+j0.13 j0.020

Table 2: Bus data for the system shown in Fig. 1

 Bus code Vinitial Pg Qg Pl Ql 1(slack) 1.0+j0 0 0 0 0 2 1.0+j0 0 0 0.66 0.26 3 1.0+j0 0 0 0.50 0.15 4 1.0+j0 0 0 0.54 0.16 5 1.0+j0 0 0 0.42 0.18 6 1.0+j0 0 0 0.64 0.28 7 1.0+j0 0. 0 0.60 0.26 8 1.0+j0 0.68 0.18 0.06 9 1.02+j0 0.84 0.24 0.09 10 1.01+j0 0.97 0.20 0.05 11 1.01+j0 0.85 0 0 12 1.03+j0 0.72 0.12 0.03

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