The bus admittance matrix and power flow 1 Solution

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EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 1 Class 7 - solution The Bus admittance matrix and Power Flow 1. The branches of the network are modeled using the /g2024-model for an electric line. Therefore, the electrical circuit of the network is: Considering the following data: Xl=0.3 Ω/ km; bl=160*10^-6 S/km; L1=150 km; L2=37.5 km; L3=75 km; s The electric parameters of the circuit are Z12=Xl*L1*1i=45i Ω Z13=Xl*L2*1i=11.5i Ω Z23=Xl*L3*1i=22.5i Ω σ Y120=bl*L1/2*1i = bl1/ 2 *1i =0.0120i S s Y130=bl*L2/2*1i= bl2/ 2 *1i =0.0030i S s Y230=bl*L3/2*1i= bl3/ 2 *1i =0.0060i S s 1 2 3 jX 12 jb L1 2 jb L1 2 y120 y120 z12 jX 13 z13 jb L2 2y130 jb L2 2y130 jb L3 2 y230 jb L3 2 y230 jX 23 z23 EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 2 To compute in p.u. we need to define a couple of reference parameters. We choose Aref=100 MVA; Vref = 150 kV s and, we compute: Zref=Vref^2/Aref = 225 Ω σ In p.u., the electrical network parameters are z12=Z12/Zref =0.2000i s z13=Z13/Zref = 0.0500i s z23=Z23/Zref = 0.1000i s and y12 = 1/z12=-5i s y13=1/z13=-20i s y23=1/z23=-10i s y120=Y120*Zref=2.700i s y130= Y130*Zref=0.6750i s y230=Y230*Zref= 1.3500i s The electric circuit of the network can be further reduced by noticing that the bus to ground connections consist of parallel branches: 1 2 3 z12 z13 y y y10 120 130 = + z23 y y y30 120 230 = + y y y20 120 230 = + 0 EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 3 where y10=y120+y130= 3.3750i s y20=y120+y230= 4.0500i s y30=y130+y230=2.0250i s 1.a. The graph associated with the above circuit is: and, assigning arbitrary directions to the branches (edges) of the graph, the orientated graph is obtained: Now, it is possible to build the…