1. ECE 576 – Power System Dynamics and Stability Prof. Tom Overbye University of Illinois at Urbana-Champaign overbye@illinois.edu 1 Lecture 28: Power System Stabilizers, Energy Functions

2. Announcements Be reading Chapter 9 Homework 8 is posted; it should be done before the final but need not be turned in Final exam is on Monday May 9 from 8 to 11am in regular room. Closed book, closed notes, with two 8.5 by 11 inch note sheets and regular calculators allowed 2

3. Damping of Electromechanical Modes Damping can be considered using either state-space analysis or frequency analysis With state-space analysis the equations can be written as The change in E fd comes from the exciter 3

4. State-Space Analysis With no exciter there are three loops – Top complex pair of eigenvalues loop – Bottom loop through  E q ’ contributes positive damping Adding the exciter gives 4

5. B2_PSS_Flux Example The SMIB can be used to plot how the eigenvalues change as the parameters (like K A ) are varied 5 Values for K 1 to K 6 can be verified for previous case using the PowerWorld SMIB matrix

6. Verified K Values The below equations verify the results provided on the previous slide SMIB matrix with the earlier values 6

7. Root Locus for Varying K A This can be considered a feedback system, with the general root locus as shown 7

8. Frequency Domain Analysis Alternatively we could use frequency domain analysis Ignoring T A gives If K 5 is positive, then the response is similar to the case without an exciter If K 5 is negative, then with a sufficiently large K A the electromechanical modes can become unstable 8

9. Frequency Domain Analysis Thus a fast acting exciter can be bad for damping, but has over benefits, such as minimizing voltage deviations Including T A Including the torque angle gives 9

10. Torque-Angle Loop The torque-angle loop is as given in Figure 8.11; with damping neglected it has two imaginary eigenvalues 10

11. Torque-Angle Loop with Other Dynamics The other dynamics can be included as in Figure 8.12 11 H(s) contributes both synchronizing torque and damping torque

12. Torque-Angle Loop with Other Dynamics Previously we had If we neglect K 4 (which has little effect at 1 to 3 Hz) and divide by K 3 we get 12

13. Synchronizing and Damping Torque Let s = j , and then rewrite H(j  ) as 13 The real part is the contribution to the synchronizing torque and the imaginary part the contribution to the damping torque

14. Synchronizing Torque The synchronizing torque is determined by looking at the response at low frequencies (s=j  0) With high K A, this is approximately 14 The synchronizing torque is dominated by K 1 ; it is enhanced if K 5 is negative

15. Damping Torque The damping torque is enhanced when the imaginary part of H(j  ) is positive, and made more unstable when it is negative Since y and K A are positive, this depends on the sign of K 5 If K5 is negative, then high K A can cause problems 15

16. Numerical Example (Effect of K 5 ) Test system 1Test System 2 Eigen Values Test System 1 Test System 2 unstable 16

17. Addition of PSS Loop The impact of a PSS can now be considered in which the shaft speed signal is fed through the PSS transfer function G(s) to the excitation system To analyze effect of PSS signal assume ,  V REF =0 17

18. PSS Contribution to Torque Approximation is for usual range of constants) 18

19. PSS Contribution to Torque For high gain K A If PSS were to provide pure damping (in phase with  pu ), then ideally, Practically G(s) is a combination of lead and lag blocks. This is a pure lead function that is not physically realizable 19

20. PSS Block Diagram Provide damping over the range of frequencies (0.1-2 Hz) “Signal wash out” (T w ) is a high pass filter – TW is in the range of 5 to 20 seconds – Allows oscillation frequencies to pass unchanged. – Without it, steady state changes in speed would modify the terminal voltage. 20

21. Criteria for Setting PSS Parameters K PSS determines damping introduced by PSS. T 1, T 2, T 3, T 4 determine phase compensation for the phase lag present with no PSS. A typical technique is to compensate for the phase lag in the absence of PSS such that the net phase lag is: – Between 0 to 45 from 0.3 to 1 Hz. – Less than 90 up to 3 Hz. Typical values of the parameters are: – K PSS is in the range of 0.1 to 50 – T 1 is the lead time constant, 0.2 to 1.5 sec – T 2 is the lag time constant, 0.02 to 0.15 sec – T 3 is the lead time constant, 0.2 to 1.5 sec – T 4 is the lag time constant, 0.02 to 0.15 sec 21

22. PSS Design Using Example 8.72 Without the PSS, the A matrix is The electromechanical mode 1,2 is poorly damped. Instead of a two-stage lag lead compensator, assume a single-stage lag-lead PSS. Assume that the damping D in the torque-angle loop is zero. The input to the stabilizer is  pu. An extra state equation will be added. The washout stage is omitted. 22 Example is Saved as Case B2_PSS_Flux_Design This is the prior example; we’ll add a stabilizer

23. Book Example 8.7 Response Without PSS Plot shows the rotor angle for a 0.01 second self- clearing fault without a PSS 23

24. Design Procedure The desired stabilizer gain is obtained by finding the gain at which the system becomes unstable by root locus study. The washout time constant T W is set at 10 sec. K PSS is set at (1/3)K * PSS, where K * PSS is the gain at which the system becomes unstable. There are many ways of adjusting the values of T 1, T 2, T 3, T 4 Frequency response methods are recommended. 24

25. Design Procedure Using the Frequency-Domain Method Step 1: Neglecting the damping due to all other sources, find the undamped natural frequency in rad/sec of the Step 2: Find the phase lag of 25 For example K 1 =0.9224, H=3.2, giving  n = 7.37 or 1.17 Hz

26. PSS Design (contd) Step 3: Adjust the phase lead of G(s) such that Ignoring the washout filter whose net phase contribution is approximately zero. 26 In the example let T 2 be set to 0.1, then T 1 of 0.5 gives 3.073  38.4 

27. PSS Design (contd) Knowing  n and  GEP(j  n ) we can select T 1. T 2 can be chosen as some value between 0.02 to 0.15 sec. Step 4: Compute K * PSS, the gain at which the system becomes unstable, using the root locus. Then have Step 5: Select washout T W ; the washout filter should not have any effect on phase shift or gain at the oscillating frequency; 27 In the example the system becomes unstable for K PSS =3.1

28. Example (contd) The added differential equation is then 28

29. Example (contd) Note the improvement of the electromechanical mode 1,2 29

30. PSS Design Example Response Without PSS Plot compares the generator 1 rotor angle for no PSS, and two PSS designs for a 0.01 second self-clearing fault 30 The dual lead-lag adds a second 0.5, 0.1 lead-lag, giving a phase angle shift of 76.8 

31. Stability Phenomena and Tools Large Disturbance Stability (Non-linear Model) Small Disturbance Stability (Linear Model) Structural Stability (Non-linear Model) Loss of stability due to parameter variations. Tools Simulation Repetitive time-domain simulations are required to find critical parameter values, such as clearing time of circuit breakers. Direct methods using Lyapunov-based theory (Also called Transient Energy Function (TEF) methods) Can be useful for screening Sensitivity based methods. 31

32. Transient Energy Function (TEF) Techniques No repeated simulations are involved. Limited somewhat by modeling complexity. Energy of the system used as Lyapunov function. Computing energy at the “controlling” unstable equilibrium point (CUEP) (critical energy). CUEP defines the mode of instability for a particular fault. Computing critical energy is not easy. 32

33. Judging Stability / Instability Stability is judged by Relative Rotor Angles. Monitor Rotor Angles (b) Stable(a) Stable (c) Unstable (d) Unstable 33

34. Mathematical Formulation A power system undergoing a disturbance (fault, etc), followed by clearing of the fault, has the following model – (1) Prior to fault (Pre-fault) – (2) During fault (Fault-on or faulted) – (3) After the fault (Post-fault) XX Faulted Post-Fault (line-cleared) 34 T cl is the clearing time

35. Critical Clearing Time Assume the post-fault system has a stable equilibrium point x s All possible values of x(t cl ) for different clearing times provide the initial conditions for the post-fault system – Question is then will the trajectory of the post fault system, starting at x(t cl ), converge to x s as t   Largest value of t cl for which this is true is called the critical clearing time, t cr The value of t cr is different for different faults 35

36. Region of Attraction (ROA).. The region need not be closed; it can be open: All faulted trajectories cleared before they reach the boundary of the ROA will tend to x s as t  (stable) 36

37. Methods to Compute RoA Had been a topic of intense research in power system literature since early 1960’s. The stable equilibrium point (SEP) of the post-fault system, x s, is generally close to the pre-fault EP, x 0 Surrounding x s there are a number of unstable equilibrium points (UEPs). The boundary of ROA is characterized via these UEPs 37

38. Characterization of RoA Define a scalar energy function V(x) = sum of the kinetic and potential energy of the post-fault system. Compute V(x u,i ) at each UEP, i=1,2,… Defined V cr as – RoA is defined by V(x) < V cr – But this can be an extremely conservative result. Alternative method: Depending on the fault, identify the critical UEP, x u,cr, towards which the faulted trajectory is headed; then V(x) < V(x u,cr ) is a good estimate of the ROA. 38

39. Lyapunov’s Method Defining the function V(x) is a key challenge Consider the system defined by Lyapunov's method: If there exists a scalar function V(x) such that 39

40. Ball in Well Analogy The classic Lyapunov example is the stability of a ball in a well (valley) in which the Lyapunov function is the ball's total energy (kinetic and potential) For power systems, defining a true Lyapunov function often requires using restrictive models 40 SEP UEP

41. Power System Example Consider the classical generator model using an internal node representation (load buses have been equivalenced) 41 C ij are the susceptance terms, D ij the conductance terms

42. Constructing the Transient Energy Function (TEF) 1. Relative rotor angle formulation. 2. COI reference frame. 3. It is preferable since we measure angles with respect to the “mean motion” of the system. TEF for conservative system With center of speed as whereWe then transform the variables to the COI variables as It is easy to verify 42

43. TEF If one of the machines is an infinite bus, say, m whose inertia constant M m is very large, then all variables can be referenced with respect to m, whose speed stays constant In the literature m is simply taken as zero. Equation is modified accordingly, and there will be only (m-1) equations after omitting the equation for machine m. The swing equations with D i = 0 become (omitting the algebra): 43

44. TEF We consider the general case in which all M i 's are finite. We have two sets of differential equations: Let the post fault system has a SEP at This SEP is found by solving 44

45. TEF Steps for computing the critical clearing time are: – Construct a Lyapunov (energy) function for the post-fault system. – Find the critical value of the Lyapunov function (critical energy) for a given fault – Integrate the faulted equations until the energy is equal to the critical energy; this instant of time is called the critical clearing time Idea is once the fault is cleared the energy can only decrease, hence the critical clearing time is determined directly Methods differ as to how to implement steps 2 and 3. 45

46. TEF Integrating the equations between the post-fault SEP and the current state gives 46 C ij are the susceptance terms, D ij the conductance terms

47. TEF contains path dependent terms. Cannot claim that is p.d. If conductance terms are ignored then it can be shown to be a Lyapunov function Methods to compute the UEPS are – Potential Energy Boundary Surface (PEBS) method. – Boundary Controlling Unstable (BCU) equilibrium point method. – Other methods (Hybrid, Second-kick etc) (a) and (b) are the most important ones. 47

48. Equal Area Criterion and TEF For an SMIB system with classical generators this reduces to the equal area criteria – TEF is for the post-fault system – Change notation from T m to P m 48

49. TEF for SMIB System since i.e The right hand side of (1) can be written as,where Multiplying (1) by, re-write Hence, the energy function is 49

50. TEF for SMIB System (contd) The equilibrium point is given by This is the stable e.p. Verify by linearizing. Eigenvalues on j  axis. (Marginally Stable) With slight damping eigenvalues are in L.H.P. TEF is still constructed for undamped system. 50

51. TEF for SMIB System The energy function is There are two UEP:  u1 =  -  s and  u2 = -  -  s A change in coordinates sets V PE =0 for  =  s With this, the energy function is The kinetic energy term is 51

52. Energy Function V( ,  ) is equal to a constant E, which is the sum of the kinetic and potential energies. It remains constant once the fault is cleared since the system is conservative (with no damping) V( ,  ) evaluated at t=t cl from the fault trajectory represents the total energy E present in the system at t=t cl This energy must be absorbed by the system once the fault is cleared if the system is to be stable. The kinetic energy is always positive, and is the difference between E and V PE (  s ) 52

53. Potential Energy “well” Potential energy “well” or P.E. curve How is E computed? 53

54. Computation of E E is the value of total energy when fault is cleared. At  =  s is the post-fault SEP, both V KE and V PE are is zero, since  =0 and  =  s Suppose, at the end of the faulted period t=t cl the rotor angle is  cl and the velocity is  cl Then This is the value of E. 54

55. UEPs There are two other equilibrium points of In general these are the solutions that satisfy the network power balance equations (i.e. "low voltage" power flow solutions) 55

56. Energy Functions for a Large System Need an energy function that at least approximates the actual system dynamics – This can be quite challenging! In general there are many UEPs; need to determine the UEPs for closely associated with the faulted system trajectory (known as the controlling UEP) Energy of the controlling UEP can then be used to determine the critical clearly time (i.e., when the fault- on energy is equal to that of the controlling UEP) For on-line transient stability, technique can be used for fast screening 56