ECE4762011_Lect12

# ECE4762011_Lect12 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

This preview shows pages 1–9. Sign up to view the full content.

Lecture 12 Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = =
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 + - = = + =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 = = = = = = = = = = ÷ = = = = = + = - ÷ ÷

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 32

ECE4762011_Lect12 - ECE 476 POWER SYSTEM ANALYSIS Lecture...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online