ECE4762011_Lect12 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

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Lecture 12 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS
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2 Announcements Be reading Chapter 6, also Chapter 2.4 (Network Equations). HW 5 is 2.38, 6.9, 6.18, 6.30, 6.34, 6.38; do by October 6 but does not need to be turned in. First exam is October 11 during class. Closed book, closed notes, one note sheet and calculators allowed
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3 Power Flow Requires Iterative Solution i bus * * * * i 1 1 In the power flow we assume we know S and the . We would like to solve for the V's. The problem is the below equation has no closed form solution: S Rath n n i i i ik k i ik k k k V I V Y V V Y V = = = = = Y er, we must pursue an iterative approach.
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4 Gauss Iteration There are a number of different iterative methods we can use. We'll consider two: Gauss and Newton. With the Gauss method we need to rewrite our equation in an implicit form: x = h(x) To iterate we fir (0) ( +1) ( ) st make an initial guess of x, x , and then iteratively solve x ( ) until we find a "fixed point", x, such that x (x). ˆ ˆ ˆ v v h x h = =
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5 Gauss Iteration Example ( 1) ( ) (0) ( ) ( ) Example: Solve - 1 0 1 Let k = 0 and arbitrarily guess x 1 and solve 0 1 5 2.61185 1 2 6 2.61612 2 2.41421 7 2.61744 3 2.55538 8 2.61785 4 2.59805 9 2.61798 v v v v x x x x k x k x + - = = + =
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6 Stopping Criteria ( ) ( ) ( 1) ( ) A key problem to address is when to stop the iteration. With the Guass iteration we stop when with If x is a scalar this is clear, but if x is a vector we need to generalize t v v v v x x x x ε + < - @ ( ) 2 i 2 1 he absolute value by using a norm max x v j n i i i x x = < = = x x
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7 Gauss Power Flow * * * * i 1 1 * * * * 1 1 * * 1 1, * * 1, We first need to put the equation in the correct form S S S S 1 i i i i n n i i i ik k i ik k k k n n i i i ik k ik k k k n n i ik k ii i ik k k k k i n i i ik k ii k k i V I V Y V V Y V V I V Y V V Y V Y V Y V Y V V V Y V Y V = = = = = = = = = = ÷ = = = = = + = - ÷ ÷
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Gauss Two Bus Power Flow Example A 100 MW, 50 Mvar load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2.
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ECE4762011_Lect12 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

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