In reference to the power system reactance diagram, all values are given in pu. Determine the followings:
- Construct the bus admittance matrix for the power system shown in Fig.
- 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.
Bus code p-q
|
Impedance zpq
|
Shunt admittance ypq/2
|
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
|