Power System Dynamics and Stability: Modal Analysis and PSSs

This lecture discusses modal analysis and Power System Stabilizers (PSSs) in power system dynamics and stability, covering topics such as eigenvalue calculations, generator models, and exciter effects. Key papers and examples are provided to enhance understanding and application in power systems engineering.

• Power System Dynamics
• Modal Analysis
• Power System Stabilizers
• Eigenvalues
• Exciter Effects

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1. ECE 576 Power System Dynamics and Stability Lecture 26: Modal Analysis, Power System Stabilizers (PSSs) Prof. Tom Overbye University of Illinois at Urbana-Champaign overbye@illinois.edu Special Guest Lecture by TA Soobae Kim 1

2. Announcements Read Chapters 8 and 9 Homework 8 should be completed before final but need not be turned in Final is Wednesday May 14 at 7 to 10pm Key papers for book's approach on stabilizers are F.P. DeMello and C. Concordia, "Concepts of Synchronous Machine Stability as Affected by Excitation Control, IEEE Trans. Power Apparatus and Systems, vol. PAS-88, April 1969, pp. 316-329 W.G. Heffron and R.A. Philips, "Effects of Modern Amplidyne Voltage Regulator in Underexcited Operation of Large Turbine Generators," AIEE, PAS-71, August 1952, pp. 692-697 2

3. Example: Bus 4 with GENROU Model The eigenvalues can be calculated for any set of generator models This example replaces the bus 4 generator classical machine with a GENROU model There are now six eigenvalues, with the dominate response coming from the electro-mechanical mode with a frequency of 1.83 Hz, and damping of 6.92% 3

4. Example: Bus 4 with GENROU Model and Exciter Adding an relatively slow EXST1 exciter adds additional states (with KA=200, TA=0.2) As the initial reactive power output of the generator is decreased, the system becomes unstable Case is saved as B4_GENROU_Sat_SMIB 4

5. Example: Bus 4 with GENROU Model and Exciter With Q4 = 25 Mvar the eigenvalues are And with Q4=0 Mvar the eigenvalues are 5

6. Example: Bus 4 with GENROU Model and Exciter Graph shows response following a short fault when Q4 is 0 Mvar 88 86 84 82 80 78 76 74 72 70 68 66 64 62 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Rotor Angle_Gen Bus 4 #1 g f e d c b This motivates trying to get additional insight into how to increase system damping, which is the goal of modal analysis 6

7. Modal Analysis - Comments Modal analysis or analysis of small signal stability through eigenvalue analysis is at the core of SSA software In Modal Analysis one looks at: Eigenvalues Eigenvectors (left or right) Participation factors Mode shape Goal is to determine how the various parameters affect the response of the system Power System Stabilizer (PSS) design in a multi- machine context is done using modal analysis method. 7

8. Eigenvalues, Right Eigenvectors For an n by n matrix A the eigenvalues of A are the roots of the characteristic equation: det[ ] = A I A I = 0 Assume 1 n as distinct (no repeated eigenvalues). For each eigenvalue i there exists an eigenvector such that: = Av v i i i vi is called a right eigenvector If i is complex, then vi has complex entries 8

9. Left Eigenvectors For each eigenvalue i there exists a left eigenvector wi such that: = i t i t i w A w Equivalently, the left eigenvector is the right eigenvector of AT; that is, = A w w t i i i 9

10. Eigenvector Properties The right and left eigenvectors are orthogonal i.e. , i i i j 0 0 = w v w v t t ( ) i j We can normalize the eigenvectors so that: , i i i j 1 = w v w v = t t ( ) 0 i j 10

11. Eigenvector Example 1 3 4 2 1 4 = = = A A I , 0 0 3 2 + + 2 + 3 (3) 4(10) 3 49 = = = = 5, 2 2 3 10 0 1,2 2 2 1= Right Eigenvectors 5 + + = = 4 2 5 5 v v v v v = = v v choose 1 1 11 11 v 21 v 11 v = = Av v v 5 21 11 1 1 1 3 1 1 21 11 21 21 = v , Similarly 1 4 = = v 2 2 2 3 11

12. Eigenvector Example Left eigenvectors 5 w A 1 3 4 2 1= = = t t w 5 [ ] 5[ ] w w w w 1 1 11 21 11 21 + + = = 3 2 5 w w w w = w v w w 11 21 11 = = 4, 3 Let w then w 21 11 4 5 11 21 3 4 21 1 = = w w 2 1 2 2 1 w v 1 1 4 3 4 1 = = = = v v w w 1 2 1 2 3 1 = = = = t t t t w v w v 7 , 7 , w v = 0 , 0 Verify 1 1 2 2 2 1 1 2 t i 1. We would like to make This can be done in many ways. i 12

13. Eigenvector Example 3 4 = 1 1 7 = W Let 1 T W V I Then 3 1 4 1 4 1 0 0 1 1 7 = Verify 1 1 3 It can be verified that WT=V-1 . The left and right eigenvectors are used in computing the participation factor matrix. 13

14. Modal Matrices The deviation away from an equilibrium point can be defined as = x A x From this equation it is difficult to determine how parameters in A affect a particular x because of the variable coupling To decouple the problem first define the matrices of the right and left eigenvectors (the modal matrices) 1 2 1 [ , ..... ] & [ when Diag = = AV V = = V v v v W w w w , ,..... ] 2 n n ( ) i 14

15. Modal Matrices It follows that = 1 V AV To decouple the variables define z so = = x Vz = = x Vz A x AVz Then = = = 1 z V AVz WAVz z Since is diagonal, the equations are now uncoupled with i i i z z = So = x Vz ( ) t ( ) t 15

16. Modal Matrices Thus the response can be written in terms of the individual eigenvalues and right eigenvectors as Note, we are requiring that the eigenvalues be distinct! n = z 0 e t x v ( ) t ( ) i i i i 1 = Furthermore with x= VZ = = 1 T z V x W x So z(t) can be written as using the left eigenvectors as ( ) x t 1 = = t t z W x w w w ( ) t ( ) [ t .... ] 1 2 n ( ) x t n 16

17. Modal Matrices We can then write the response x(t) in terms of the modes of the system = = t ( ) ( ) ( ) ( ) z t z 0 w x t w x 0 i i t i C i i n = t so ( ) x v t C e i i i i 1 = x t = + + t t t Expanding ( ) ... v C e v C e v C e n 1 2 i i1 1 i2 2 in n So Ci represents magnitude of excitation of the ith mode resulting from the initial conditions. 17

18. Numerical example x x x x = 0 8 1 1 1 1 = = x , ( ) 0 2 4 2 2 = Eigenvalues are , 4 2 1 2 1 1 2 = = v v Eigenvectors are , 1 2 4 1 1 2 = V Modal matrix 4 . . . 0 2425 0 9701 0 8944 0 4472 = V Normalize so . 18

19. Numerical example (contd) Left eigenvector matrix is: = = T z = W AVz . . . 1 3745 1 4908 0 6872 0 3727 T 1 W V . z z z z 4 0 2 1 1 = 0 2 2 19

20. Numerical example (contd) = = 1 z x , ( ) ( ) ( ) 2 z 0 ( ) . 4 123 0 z 4z 0 V 0 1 1 z 0 1 = = , z 2z 2 2 . 4 123 0 = = = = 4t 2t T C W x ( ) ( ) ; ( ) ( ) , ( ) 0 z t z 0 e z t z 0 e 1 1 2 2 x = Vz Because of the initial condition, the 2nd mode does not get excited ( ) ( ) ( ) ( ) x t x t z t z t 1 1 2 1 1 = 4 2 2 . . . 0 2425 0 9701 0 4472 0 8944 2 t = + = ( ) ( ) ( ) C z t C z t i i i Cv z 0 e i 1 1 2 2 . i 1 = 20

21. Mode Shape, Sensitivity and Participation Factors So we have ) ( ), t t = x( Vz = t z W x ( ) t ( ) t x(t) are the original state variables, z(t) are the transformed variables so that each variable is associated with only one mode. From the first equation the Right Eigenvector gives the mode shape i.e. relative activity of state variables when a particular mode is excited. For example the degree of activity of state variable xk in vi mode is given by the element Vki of the the Right Eigenvector matrix V 21

22. Mode Shape, Sensitivity and Participation Factors The magnitude of elements of vi give the extent of activities of n state variables in the ith mode and angles of elements (if complex) give phase displacements of the state variables with regard to the mode. The left eigenvector wi identifies which combination of original state variables display only the ith mode. 22

23. Eigenvalue Parameter Sensitivity To derive the sensitivity of the eigenvalues to the parameters recall Avi = ivi; take the partial derivative with respect to Akj by using the chain rule + = + A v v Av A v i i i i i i A A A kj kj kj kj t i w Multiply by A A v v A + = + t i t i t i t i w v w A w v w i i i i i i A A A kj kj kj kj v + = t i t i t i w v w A I w v [ ] i i i i i A A A kj kj kj 23

24. Eigenvalue Parameter Sensitivity This is simplified by noting that by the definition of wi being a left eigenvector Therefore = = t i w A I ( ) 0 i A t i w v i i A A kj kj A Since all elements of are zero, except the kth row, jth column is 1 Thus i ki ji kj A A kj = W V 24

25. Sensitivity Example In the previous example we had 1 3 4 2 1 1 4 3 4 1 1 7 = = 5, 2, = = A V W , , 1,2 3 1 Then the sensitivity of 1 and 2 to changes in A are 3 4 3 4 4 3 A A 1 7 1 7 = = , 1 2 4 3 For example with 1 3 4 3 = = 5.61, 1.61, A , 1,2 25

26. Participation Factors The participation factors, Pki, are used to determine how much the kth state variable participates in the ith mode = P V W ki ki ki The sum of the participation factors for any mode or any variable sum to 1 The participation factors are quite useful in relating the eigenvalues to portions of a model For the previous example P would be 3 4 4 3 1 7 = P 26

27. PowerWorld SMIB Participation Factors The magnitudes of the participation factors are shown on the PowerWorld SMIB dialog The below values are shown for the four bus example with Q4 = 0 27