Question 1 Given the positive sequence network for a 3-bus power system for a symmetrical three phase to ground with π§π = π0.1 π. π’ is shown below: The three port generator-transmission is given to be: π£1 0.1274 π£ [ 2 ] = π [0.1061 π£3 0.0981 0.1061 0.1345 0.1151 0.0981 πΌ1 1 0.1151] [πΌ2 ] + [1] π0π 0.1215 πΌ3 1 For a three phase bolted fault, only positive sequence network and only positive sequence bus impedance is required. The off-diagonal terms are not used to calculate the fault currents flowing into the fault (they are used to calculate branch currents.) The main diagonal terms are the Thevenin Impedances. Since the assumption is 1.0 pu voltage then each fault current is the reciprocal of the Thevenin Impedance.Thus for the faulted system, wherein both the current injection and generator sources are simultaneously present, the bus voltages can be obtained by adding the voltages. πΌπ (π) = −π0 πππ + ππ ππ (π) = −ππ ∗ π0 πππ + ππ ππ (π) =∗ π0 − πππ ∗ π0 πππ + ππ πππ - diagonal element of impedance matrix for p bus ππ π0 Fault impedance - pre fault voltage usually 1 < 0 πΌπ (π) - fault current at bus p ππ (π) - fault voltage at bus p ππ (π) - bus i voltage due to fault at bus p Fault Current At Bus 1 πΌπ (π) = −π0 πππ + ππ π0 = 1 , πππ = π0.1274 , ππ = π0.1 −π°π (π) = −π = −ππ. πππ π. π ππ. ππππ + ππ. π Fault Current At Bus 2 πΌπ (π) = −π0 πππ + ππ π0 = 1 , πππ = π0.1345 , ππ = π0.1 −π°π (π) = −π = −ππ. πππ π. π ππ. ππππ + ππ. π Fault Current At Bus 3 πΌπ (π) = −π0 πππ + ππ π0 = 1 , πππ = π0.1215 , ππ = π0.1 −π°π (π) = −π = −ππ. πππ π. π ππ. ππππ + ππ. π Due to fault at Bus 1 Bus voltages are ππ (π) = −ππ ∗ π0 πππ + ππ P=1, π0 = 1 , π11 = π0.1274 , ππ = π0.1 −ππ.π π½π (π) = ππ.ππππ+ππ.π = π. ππππ π. π π ∗π0 ππ (π) = π0 − π ππ+π ππ π i=2, p=1, π0 = 1 , π11 = π0.1274 , ππ = π0.1 , π21 = π0.1061 π2 (π) = 1 − π π21 11 +ππ π½π (π) = π − ππ. ππππ = π. ππππ π. π ππ. ππππ + ππ. π i=3, p=1, π0 = 1 , π11 = π0.1274 , ππ = π0.1 , π31 = π0.0981 π3 (π) = 1 − π π31 11 +ππ π½π (π) = π − Bus Currents are ππ. ππππ = π. ππππ π. π ππ. ππππ + ππ. π Fault current at bus 1 is given by −π°π (π) = −π = −ππ. πππ π. π ππ. ππππ + ππ. π From bus 2 to bus 1 πΌπ−π = ππ − ππ πππ πΌ2−1 = π2 − π1 π21 πΌ2−1 = π2 − π1 π21 π1 = π. ππππ π. π , π2 = π. ππππ π. π , π3 = π. ππππ π. π , πππ = ππ. ππ πΌ2−1 = −0.4398 + 0.5334 = −ππ. ππ π. π π0.08 From bus 3 to bus 2 πΌ3−2 = π32 = π3 − π2 π32 0.06 ∗ 0.06 = π0.03 2 ∗ 0.06 πΌ3−2 = π3 − π2 π32 π1 = π. ππππ π. π , π2 = π. ππππ π. π , π3 = π. ππππ π. π , πππ = ππ. ππ πΌ3−2 = −0.5334 + 0.5686 = −ππ. ππππ π. π π0.03 From Bus 3 to bus 2 the currents are equally between route 3-21 and 3-22 π°π−ππ = −ππ. πππππ π. π π°π−ππ = −ππ. πππππ π. π From bus 3 to bus 1 πΌ3−1 = π3 − π1 π31 π31 = π0.13 πΌ3−1 = π3 − π1 π31 π1 = π. ππππ π. π , π2 = π. ππππ π. π , π3 = π. ππππ π. π , πππ = ππ. ππ πΌ3−1 = −0.4398 + 0.5686 = −ππ. ππππ π. π π¨πππ π0.13 Current from G1 to bus 1 πΌπ1−1 = 1 − π1 π01 π01 = π0.25 πΌπ1−1 = 1 − π1 π01 π1 = π. ππππ π. π , π2 = π. ππππ π. π , π3 = π. ππππ π. π , πππ = ππ. ππ πΌπ1−1 = 1 − 0.4398 = −ππ. ππππ π¨πππ π0.25 Current from G2 to bus 3 πΌπ2−3 = 1 − π3 π02 π02 = π0.2 πΌπ2−3 = 1 − π3 π02 π1 = π. ππππ π. π , π2 = π. ππππ π. π , π3 = π. ππππ π. π , πππ = ππ. π π°ππ−π = π − π. ππππ = −ππ. πππ π¨πππ ππ. π
