IGEE402_Galiana_Winter_2006_Midterm_Solutions

IGEE402_Galiana_Winter_2006_Midterm_Solutions - Power...

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Power System Analysis ECSE 464B Solution to Mid-Term Test ( ) max 0 0 1) ( ) 100(3 sin cos ) 100 3 2 sin 100 3 2 cos 4 4 100 3 2 441.421 1 100 2 2 ( ) 3 sin 100 3 363.662 4 D T T aver d t wt wt wt wt D MW D D t dt wt MW T T w LF π π π π = + + = + + = + = + = = = + = + = = aver max 82.37 % D = 2) P = 0.8 , U = 0.2 LOLP = 1-P(all three Generators available) = 1 3 0.8 0.488 = 3) Units out Load loss Probability 0 0 3 0 3 0 0.512 C P U = 1 50 2 1 3 1 0.384 C P U = 2 150 1 2 3 2 0.096 C P U = 3 250 0 3 3 3 0.008 C P U = ELOL = = 0(0512)+50(0.384)+150(0.096)+250(0.008) = 35.6 MW k k k d P 4) ( ) 0 1 1 abc 1 I a I = + = Zero sequence currents 0 1 abc V = = 1 V a Zero sequence voltages A load is balanced if and only if its 012 Z matrix is diagonal (the three sequence networks are decouple) By injecting a zero sequence currents 0 1 abc I I = we get a zero sequence voltages 0 1 abc V V = . So it is possible that this load is balanced. However, this is a necessary
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