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 44 100 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 DM W DD t d t w t M W TT w LF 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 30 0.512 CPU = 1 50 21 31 0.384 = 2 150 12 32 0.096 = 3 250 03 33 0.008 = ELOL = = 0(0512)+50(0.384)+150(0.096)+250(0.008) = 35.6 MW kk k dP 4) 0 11 abc 1 I aI = Zero sequence currents 0 1 abc V 1 Va 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 VV = . So it is possible that this load is balanced. However, this is a necessary condition but it is not sufficient, to prove that the load is balanced we have to check for a positive and negative sequence as well.
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5) Yes, the load is balanced since
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This note was uploaded on 09/16/2011 for the course IGEE 402 taught by Professor Galiana during the Fall '10 term at McGill.

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IGEE402_Galiana_Winter_2006_Midterm_Solutions - Power...

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