The bus admittance matrix and power flow 2 Solution

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EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 1 Class 7b - solution The Bus admittance matrix and Power Flow Xl=0.43; L=30; vsc = 10; AnT=60; VnT1=150; VnT2=20 VRE=147; VG=153; Reference values: Vrif1=150 Vrif2=20 Arif=100 Zrif1=Vrif1^2/Arif = 225 Zrif2=Vrif2^2/Arif = 4 Impedances: Line Z12=Xl*L1*1i /Zrif1= 0.0533i Transformer The electric transformer can be modeled in one of the two following ways: a. b. Taking into account that the parameter vsc represents the impedance of the transformer in p.u. with respect to the transformer nominal data, the reactance of the transformer, in absolute values, is ZT=XT=vsc/100*ZrefT [Ω] where ZrefT is the reference impedance of the transformer and is computed with respect to the nominal data of the transformer. Therefore Bus a Bus b jX T zT kT jX T zT kT Bus a Bus b EXERCISES FOR ELECTRIC POWER SYSTEMS COURSE ENERGY DEPARTMENT, Electrical section 2 ZT=XT=vsc/100*(VnT k^2/AnT) where VnTk is the nominal voltage of the transformer winding connected at Bus b if model a. was chosen or, the nominal voltage of the transformer winding connected at Bus a if model b. was chosen (in other words, VnT k is the nominal voltage of the winding connected at the bus at which the transformer impedance is connected). In what regards the tap ratio of the transformer kT, this is computed as: kT = VnT,bus_a / VnT,bus_b , if model a. was adopted kT = VnT,bus_b / VnT,bus_a , if model b. was adopted When expressing the XT and the kT in p.u., the following simplifying hypothesis is made: the reference voltages of the network are equal to the nominal voltages of the transformer, i.e. Vref, bus_a = VnT, bus_a and Vref, bus_b = VnT, bus_b. Therefore, the network reference quantities are Zref,bus_a = Vref,bus_a^2/Aref…