ECE4762011_Lect14 - Lecture 14 Power Flow Professor Tom...

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Unformatted text preview: Lecture 14 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS 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. Exam covers thru the end of lecture 13 (today) An example previous exam (2008) is posted. Note this is exam was given earlier in the semester in 2008 so it did not include power flow, but the 2011 exam will (at least partially) 3 Modeling Voltage Dependent Load So far we've assumed that the load is independent of the bus voltage (i.e., constant power). However, the power flow can be easily extended to include voltage depedence with both the real and reactive l Di Di 1 1 oad. This is done by making P and Q a function of : ( cos sin ) ( ) ( sin cos ) ( ) i n i k ik ik ik ik Gi Di i k n i k ik ik ik ik Gi Di i k V V V G B P P V V V G B Q Q V θ θ θ θ = = +- + =-- + = ∑ ∑ 4 Voltage Dependent Load Example 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 In previous two bus example now assume the load is constant impedance, so P ( ) (10sin ) 2.0 ( ) ( 10cos ) (10) 1.0 Now calculate the power flow Jacobian 10 cos 10sin 4.0 ( ) 10 V V Q V V V V V J θ θ θ θ = + = =- + + = + = x x x 2 2 2 2 2 sin 10cos 20 2.0 V V V θ θ - + + 5 Voltage Dependent Load, cont'd (0) 2 2 2 2 (0) 2 2 2 2 2 2 (0) 1 (1) Again set 0, guess 1 Calculate (10sin ) 2.0 2.0 f( ) 1.0 ( 10cos ) (10) 1.0 10 4 ( ) 12 10 4 2.0 0.1667 Solve 1 12 1.0 0.9167 v V V V V V θ θ- = = + = = - + + = - =- = x x J x x 6 Voltage Dependent Load, cont'd Lin e Z = 0 .1 j O ne Tw o 1 .0 0 0 pu 0 .8 9 4 pu 1 6 0 M W 8 0 M V R 1 6 0 .0 M W 1 2 0 .0 M V R- 1 0 .3 0 4 D e g 1 6 0 .0 M W 1 2 0 .0 M VR- 1 6 0 .0 M W -8 0 .0 M VR With constant impedance load the MW/Mvar load at bus 2 varies with the square of the bus 2 voltage magnitude. This if the voltage level is less than 1.0, the load is lower than 200/100 MW/Mvar 7 Dishonest Newton-Raphson Since most of the time in the Newton-Raphson iteration is spend calculating the inverse of the Jacobian, one way to speed up the iterations is to only calculate/inverse the Jacobian occasionally – known as the “Dishonest” Newton-Raphson – an extreme example is to only calculate the Jacobian for the first iteration ( 1) ( ) ( ) -1 ( ) ( 1) ( ) (0) -1 ( ) ( ) Honest:- ( ) ( ) Dishonest:- ( ) ( ) Both require ( ) for a solution v v v v v v v v ε + + = = < x x J x f x x x J x f x f x 8 Dishonest Newton-Raphson Example 2 1 (0) ( ) ( ) ( ) ( ) 2 (0) ( 1) ( ) ( ) 2 (0) Use the Dishonest Newton-Raphson to solve...
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This note was uploaded on 11/17/2011 for the course ECE 476 taught by Professor Overbye,t during the Fall '08 term at University of Illinois, Urbana Champaign.

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ECE4762011_Lect14 - Lecture 14 Power Flow Professor Tom...

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