Lecture 19

# Three gen system fault example for simplicity assume

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Unformatted text preview: us i we get -If Zii = −V f = −Vi(1) 13 ! Determination of Fault Current Hence Vi(1) If = Zii Where Zii @ driving point impedance Zij (i ≠ j ) @ transfer point imepdance Voltages during the fault are also found by superposition Vi = Vi(1) + Vi(2) Vi(1) are prefault values 14 ! Three Gen System Fault Example For simplicity assume the system is unloaded before the fault with E g1 = Eg 2 = Eg 3 = 1.05∠0° Hence all the prefault currents are zero. 15 ! Three Gen Example, cont’d Ybus 0# " −15 10 = j \$ 10 −20 5 % \$ % 5 −9 % \$0 & ' −1 Zbus 0# " −15 10 = j \$ 10 −20 5 % \$ % 5 −9 % \$0 & ' " 0.1088 0.0632 0.0351# = j \$0.0632 0.0947 0.0526 % \$ % \$ 0.0351 0.0526 0.1409 % & ' 16 ! Three Gen Example, cont’d −1.05 For a fault at bus 1 we get I1 = = j9.6 = − I f − j 0.1088 V (2) " 0.1088 0.0632 0.0351# " j 9.6 # = j \$0.0632 0.0947 0.0526 % \$ 0 % \$ %\$ % \$ 0.0351 0.0526 0.1409 % \$ 0 % & '& '...
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