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# 1. Given the power system per-unit parameters and oneline diagram shown below (Note: modi-

fied version of Power World Example6_14; both generators are Y-connected and solidly grounded,

i.e., Zng=0),

G1: 400 MVA, 15 kV, Xd'=0.24, X2=Xd"=0.12, X0=0.08, PG1~220 MW, |V1|= 1.02

G3: 800 MVA, 15 kV, Xd'=0.24, X2=Xd"=0.12, X0=0.08, PG3=520 MW, |V3|= 1.02

T1: 400 MVA, 15 kV A / 345 kV Y, X=0.08, X/R=13.33

T2: 800 MVA, 15 kV A / 345 kV Y, X=0.08, X/R=13.33

Line 2-4: R24-0.0090, X24-0.1, B24=1.72, Rate A = 1000 MVA, 200 mi

Line 2-5: R25-0.0045, X25-0.05, B25-0.88, Rate A = 1000 MVA, 100 mi

Line 4-5: R45-0.00225, X45-0.025, B45-0.44, Rate A = 1000 MVA, 50 mi

assume a 200-Mvar capacitor has been added to bus 2, where the original load is 720 MW,

252 Mvar, and the load at bus 3 is 20 MW, 10 Mvar. The transmission network is rated

345 kV and the system base is 100 MVA. The zero sequence line reactances are three times

the positive sequence reactances. The negative sequence line reactances are the same as the

positive sequence values. Solve the original case power flow using Power World. Create a table

of bus #, V (pu) and angle (deg). Create another table of branch flows: From, To, MW,

Mvar, MVA, % loading (based on Rate A rating).

Fig: 1