Load Flow Assignment In the power network shown below, bus 1...

Load Flow Assignment In the power network shown below, bus 1 is the slack (reference) bus, where the voltage is defined at (128.0$\angle$0°)$^\circ$ kV, buses 2, 4, 5, 6, 7 and 8 are PQ buses and bus 3 is a PV bus, as shown. bus 1 $V_1$=128 kV T$_1$ 132kV/33kV 80 MVA X=0.16 pu Load 5 MVA 0.94 lagging bus 2 T$_2$ 132kV/33kV 80 MVA X=0.16 pu 60 km OHL R=0.1 ?/km, X=0.3 ?/km C=0 25 km OHL R=0.1 ?/km, X=0.4 ?/km C=0 $V_3$=32.4 kV T$_3$ 33kV/11kV 10 MVA X=0.1 pu bus 6 8 km cable R=0.27 ?/km X=0.32 ?/km C=0.3 µF/km bus 3 Load 9 MW Load 8 MW 0.88 lagging 35 km OHL R=0.1 ?/km, X=0.4?/km C=0 bus 4 T$_4$ 33kV/11kV 12 MVA X=0.12 pu Load 7 MVA 0.9 lagging bus 7 Load 8 MVA 0.96 lagging Load 26 MVA 0.86 lagging bus 5 T$_5$ 132kV/11kV 15 MVA X=0.15 pu Load 12 MVA 0.92 lagging bus 8 Use the Gauss-Seidel method to calculate the network bus voltages, and all network currents, line losses and line power & reactive power flows.

Transcript

00:01 We are given a power system power system we have the network power system network so here we are given here voltage v1 which is equal to 1 and phase here is 0 degree per unit and the load bus is h2 so the h2 is 280 mw plus we can write j multiplied by 60 mw power or we can say line impedance we are given mva so all these things we have z which is equal to 0 .02 plus we can write j multiplied by here j multiplied by 0 .04 per unit now in this equation in the first sub part of this equation we want to find so here in this first sub part let's find out dv2 of 2 so let's find out this value so first thing we understand all these things we have so here we can write v2 of 0 which is equal to 1 0 degree z12 which is equal to 0 .02 plus we can write j 0 .04 and the y12 which is equal to 1 divided by 0 .02 plus we can write j 0 .04 which is simplified and we can write answer is j 10 minus j 20 now here for the h2 which is equal to we can write minus 280 plus we can write j 60 divided by we can write here 100 which is simplified and we can write answer is minus 2 .8 minus j 0 .60 for initial condition the v2 of 0 which is equal to we can write 1 plus we can write j 0 .0 and here we can write v2 of 0 which is equal to we can write here minus 2 .8 plus we can write j 0 .60 divided by we can write 1 plus j 0 .0 plus we can write 10 minus we can write j 20 multiplied by we can write 1 divided by we can write 10 minus j 20 now v2 of 1 which is equal to we just simplified and we can write that is 0 .92 minus we can write j 0 .1 let's find out the v2 of 2 so here we can write v2 of 2 which is equal to we can write minus 2 .8 plus we can write j 0 .60 divided by we can write 0 .92 plus we can write j 0 .1 plus we can write 10 minus j 20 multiplied by 1 divided by we can write 10 minus j 20 now v2 of 2 which is equal to we just simplified and we can write answer is minus 0 .99 minus we can write j 0 .09 this is our right answer now let's solve our second part of this question second part we want to find s121 so this value we want to find m bar so that line loss also we…

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