ECE 422/522 Power System Operations & Planning/Power Systems Analysis II : 7 - Transient Stability

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1 ECE 4/5 Power System Operations & Planning/Power Systems Analysis II : 7 - Transient Stability Spring 014 Instructor: Kai Sun 1

2 Transient Stability The ability of the power system to maintain synchronism when subjected to a severe transient disturbance such as a fault on transmission facilities, loss generation, or loss of a large load. The system response to such disturbances involves large excursions of generator rotor angles, power flows, bus voltages, and other system variables. Stability is influenced by the nonlinear characteristics of the system If the resulting angular separation between the machines in the system remains within certain bounds, the system maintains synchronism. If loss of synchronism because of transient instability occurs, it will usually be evident within -3 seconds of the initial disturbances

3 Single-Machine Infinite Bus (SMIB) System P e = =P max sinδ P max = E E B /X T Swing Equation: Hd dt 0 P P P P m e m max sin P a is called Accelerating Power P m = P e The rotor will accelerate if P m increases, or P e decreases due to, e.g., a contingency on the transmission path 3

4 Power Angle Relationship P e = =P max sinδ Question: What if both circuits are out of service? 4

5 Response to a step change in P m Hd P a PmPe PmP dt 0 max sin Consider a sudden increase in P m : P m0 P m1. New equilibrium point b satisfying P e (δ 1 )=P m1 a b: Due to the rotor s inertia, δ cannot jump from δ 0 to δ 1, so P a =P m1 -P e (δ 0 )>0 and ω r increases from ω 0. When b is reached, P a =0 but ω r >ω 0, so δ continues to increase. b c: δ>δ 1 and P a <0, so ω r decreases until c. At c, ω r =ω 0 and δ reaches the peak value δ max. P a ω r δ a >0 =ω 0, b 0 =ω r, max, c <0 =ω 0, =δ max, c: At c, the rotor starts to decelerate (since P a <0) with ω r <ω 0 and δ decreases. With all resistances (damping) neglected, δ and ω r oscillate around the new equilibrium point b with a constant amplitude. 5

6 Equal-Area Criterion (EAC) If δ max exists where dδ/dt=0: r ( PmPd e) H H r ( Pm Pd e) ( m e) 0 max r J P Pd 0 max At δ max, ω r =0 and the integral=0 (Note: all losses are neglected) Moment of inertia in p.u. (Kundur s Page 131) 6 1 max ( P Pd ) ( P P) d m e e m 0 1 = area A area A 0 1

7 Equal-Area Criterion (EAC): The stability is maintained only if a decelerating area A the accelerating area A 1 can be located above P m (before d, i.e. the Unstable Equilibrium Point or UEP). SEP UEP If A < A 1, δ will continue increasing at UEP (since ω r >ω 0 ), so it will lose stability. For the case with a step change in P m, P m or the new P m does matter for transient stability. 7

8 Transient stability limit for a step change of P m Following a step change P m0 P m, solve the transient stability limit of P m : Assume A 1 = A : P P d P dp m 1 max ( ) sin sin ( ) 1 0 max max m max ( ) P P (cos cos ) max 0 m max 0 max P P (cos cos ) P (cos cos ) P m 0 max 1 0 max max 1 m max At the UEP, Pm Pmax sinmax ( )sin cos cos max 0 max max 0 SEP UEP Solve δ max to calculate the transient stability limit for a step change of P m : P P sin m max 1 max max Question: Can we increase P m beyond that transient limit (with P m <P max )? 8

9 Solve δ max by the Newton-Raphson method ( )sin cos cos max 0 max max 0 The nonlinear function form: f( ) cos max 0 c Select an initial estimate: / ( k ) max Calculate iterative solutions by the N-R algorithm: ( k 1) ( k) ( k) max max max where c f( ) c f( ) ( k) ( k) ( k ) max max max ( k) ( k) df ( max 0 )cosmax d max ( k ) max Give a solution when a specific accuracy ε is reached, i.e. ( k 1) ( k) max max 9

10 Response to a three-phase fault P e = =P max sinδ P e,during fault << P e, post-fault P e,post-fault <P e,pre-fault for a permanent fault (cleared by tripping the fault circuit) or P e, post-fault =P e,pre-fault for a temporary fault 10

11 Stable Unstable 11

12 Critical Clearing Angle (CCA) Saadat s Sec.11.6 (Example 11.5) Consider a simple case A three-phase fault at the sending end P e, during fault =0 if all resistances are neglected Critical Clearing Angle δ c ( sin ) c max Pd m Pmax Pm d 0 c A 1 A Integrating both sides: P P (cos cos ) P ( ) m c 0 max c max m max c Pm cos c ( max c) cos P max max Pm ( ) cos P max 0 c 0 = π-δ 0 1

13 Critical Clearing Time (CCT) Solve the CCT from the CCA: Since P e, during fault =0 for this case, during the fault: Hd dt 0 P m d t 0 0 Pm dt Pmt dt H 0 H 0 4H Pt m 0 t c 4 H ( c 0) P 0 m 13

14 For a more general case: P e (during fault)>0 P 3max (pre-fault) A 3 A (post-fault) P 3max P max A 1 (fault-on) P max δ s (δ u ) 0 max sn sin ( ) c P P i d P d P m c 0 max 3max m max c c cos c P ( ) P cos P cos P P m max c 3max max max 0 3max max A 1 = V ke (δ c ), the kinetic energy at δ c A 1 + A 3 = V(δ c )=V ke (δ c )+V pe (δ c ), total energy at δ c A + A 3 = V pe (δ u )=V cr, i.e. the largest potential energy If and only if V(δ c ) V cr (i.e. A 1 A ), the generator is stable 14

15 Factors influencing transient stability How heavily the generator is loaded. The generator output during the fault. This depends on the fault location and type The fault-clearing time The post-fault transmission system reactance The generator reactance. A lower reactance increases peak power and reduces initial rotor angle. The generator inertia. The higher the inertia, the slower the rate of change in angle. This reduces the kinetic energy gained during fault; i.e. the accelerating area A 1 is reduced. The generator internal voltage magnitude (E ). This depends on the field excitation The infinite bus voltage magnitude E B 15

16 EAC for a Two-Machine System Two interconnected machines respectively with H 1 and H The system can be reduced to an equivalent SMIB system d dt H H ( P m1pe1) Pa1 1 1 d dt H H 0 0 ( P mpe) Pa d d d P P ( ) dt dt dt H H a1 a1 1 HH d HP HP HP HP HP HP 1 1 a1 1 a m1 1 m e1 1 e = 0 H1H dt H1H H1H H1H H d dt P P m,1 e,1 P e1 1,max 1,0 H 0 1 ( P P ) d 0 m,1 e,1 P m1 δ 1 16 δ 1,0 δ 1,max

17 Single Machine Equivalent Method (EEAC or SIME) [1]-[3] Based on generator trajectories obtained from time-domain simulations, split all generators into two groups (fixed or dynamic) and build a -machine equivalent and consequently a single machine equivalent, such that EAC can be applied. 1. Y. Xue, et al, "A Simple Direct Method for Fast Transient Stability Assessment of Large Power Systems". IEEE Trans. on PWRS, PWRS3: , Y. Xue, et al, "Extended Equal-Area Criterion Revisited". IEEE Trans. on PWRS, PWRS7: , M. Pavella, et al, Transient Stability of Power Systems: a Unified Approach to Assessment and Control, Kluwer,

18 Some examples from [3] 18

19 Methods for Transient Stability Analysis Analyzing a system s transient stability following a given contingency Time-domain simulation: At present, the most practical available method of transient stability analysis is time-domain simulation in which the nonlinear differential equations are solved by using step-by-step numerical integration techniques. Direct methods: Those methods determine stability without explicitly solving the system differential equations. The methods are based on Lyapunov s second method, define a Transient Energy Function (TEF) as a possible Lyapunov function, and compare the TEF to a critical energy V cr to judge stability EAC is a direct method for a SMIB or two-machine system 19

20 Numerical Integration Methods The differential equations to be solved in power system stability analysis are nonlinear ODEs (ordinary differential equations) with known initial values x=x 0 and t=t 0 dx dt = f(,) x t where x is the state vector of n dependent variables and t is the independent variable (time). Our objective is to solve x as a function of t Explicit Methods In these methods, the value of x at any value of t is computed from the knowledge of the values of x from only the previous time steps, e.g. Euler method and R-K methods Implicit Methods These methods use interpolation functions (involving future time steps) for the expression under the integral, e.g. the Trapezoidal Rule 0

21 Euler Method The Euler method is equivalent to using the first two terms of the Taylor series expansion for x around the point (x 0, t 0 ), referred to as a first-order method (error is on the order of t ) Approximate the curve at x=x 0 and t=t 0 by its tangent x(t 1 ) x 1 dx dt = f(,) xt dx dt x 0 = f ( x, t ) 0 0 x 0 t dx x t dt x 0 x 1 dx = x0 + x = x0 + t dt x 0 t 0 t 1 At step i+1 dx x x t = + i+ 1 i dt x The standard Euler method results in inaccuracies because it uses the derivative only at the beginning of the interval as though it applied throughout the interval i 1

22 Modified Euler (ME) Method dx dt = f( xt,) Modified Euler method consists of two steps: (a) Predictor step: p x = x + t 1 0 dx dt x 0 Slope at the beginning of t x (t 1 ) x c 1 x p 1 x 0 t The derivative at the end of the t is estimated using x 1 p dx dt x p 1 = p f( x, t ) Estimated slope at the end of t 1 1 t 0 t 1 (b) Corrector step: dx dx dx dx + + dt p x dt dt p 0 x x dt c c i x 1 i+ 1 x = x t x = x i 1 i + t + It is a second-order method (error is on the order of t 3 ) Step size t must be small enough to obtain a reasonably accurate solution, but at the same time, large enough to avoid the numerical instability with the computer, e.g. increasing round-off errors.

23 Runge-Kutta (R-K) Methods General formula of the nd order R-K method: (error is on the order of t 3 ) k = f( x, t ) t k = f( x + αk, t + β t) t x1= x0+ ak 1 1+ ak At Step i+1: x(t 1 ) x 0 +αk 1 x 0 dx = f(,) xt dt t t 0 t 1 k = f( x, t ) t 1 i 1 i x = i x + + ak + ak i k = f( x + αk, t + β t) t 1 i 1 1 i The ME method is a special case with a 1 =a =1/, α=β=1 General formula of the 4 th order R-K method: (error is on the order of t 5 ) 1 x = i 1 x + + i ( k1 k k3 k4 ) k1 = f( xi, ti) t k1 t k = f( xi +, ti + ) t k t k3 = f( xi +, ti + ) t k = f( x + k, t + t) t 4 i 3 3 i

24 Numerical Stability of Explicit Integration Methods Numerical stability is related to the stiffness of the set of differential equations representing the system The stiffness is measured by the ratio of the largest to smallest time constant, or more precisely by λ max /λ min of the linearized system. Stiffness in a transient stability simulation increases with modeling more details (more small time constants are concerned). Explicit integration methods have weak stability numerically; with stiff systems, the solution blows up unless a small step size is used. Even after the fast modes die out, small time steps continue to be required to maintain numerical stability 4

25 Implicit Methods Implicit methods use interpolation functions for the expression under the integral. Interpolation implies the function must pass through the yet unknown points at t 1 The simplest implicit integration method is the Trapezoidal Rule method. It uses linear interpolation. The stiffness of the system being analyzed affects accuracy but not numerical stability. With larger time steps, high frequency modes and fast transients are filtered out, and the solutions for the slower modes is still accurate. For example, for the Trapezoidal rule, only dynamic modes faster than f(x n,t n ) and f(x n+1,t n+1 ) are neglected. Δt x(t 1 )=x(t 0 )+ A + B x 1=x 0+ ( 0 0) ( 1 1) f x,t +f x,t Δt f B x f(x 0,t 0 ) f(x 1,t 1 ) n+ 1 = x n + f x n,t n + f x n+ 1, tn+ 1 ( ) ( ) A t Compared to ME method: t x = x + f x,t + f x,t n+ + p ( ) ( ) 1 n n n n 1 n+1 t 0 t 1 5

26 Kundur s Example

27 7

28 8

29 Overall System Equations The overall system equations are expressed in the general form comprising a set of first-order differential equations (dynamic devices) and a set of algebraic equations (devices and network) where x V I Y N x = f(x, V) I(x, V) = Y V N DE AE state vector of the system bus voltage vector current injection vector node admittance matrix. It is constant except for changes introduced by network-switching operations; symmetrical except for dissymmetry introduced by phase-shifting transformers (Kundur s Sec. 6..3) 9

30 Solution of the Equations Schemes for the solution of equations DE and AE are characterized by the following factors The manner of interface between the DE and AE. Either a partitioned approach or a simultaneous approach may be used The integration method, i.e. an implicit method or explicit method, used to solve the DE. The method used to solve the AE (power flow analysis), e.g. the Newton-Raphson method. Most commercialized power system simulation programs provide the Modified Euler, nd order R-K, 4 th order R-K and Trapezoidal Rule methods 30

31 A Simplified Model for Multi-Machine Systems Consider these classic simplifying assumptions: Each synchronous machine is represented by a voltage source E with constant magnitude E behind X d (neglecting armature resistances, the effect of saliency and the changes in flux linkages) The mechanical rotor angle of each machine coincides with the angle of E The governor s actions are neglected and the input powers P mi are assumed to remain constant during the entire period of simulation Using the pre-fault bus voltages, all loads are converted to equivalent admittances to ground. Those admittances are assumed to remain constant (constant impedance load models) Damping or asynchronous powers are ignored. Machines belonging to the same station swing together and are said to be coherent. A group of coherent machines is represented by one equivalent machine 31

32 Solve the initial power flow and determine the initial bus voltage phasors V i. Terminal currents I i of m generators prior to disturbance are calculated by their terminal voltages V i and power outputs S i, and then used to calculate E i All loads are converted to equivalent admittances: To include voltages behind X di, add m internal generator buses to the n-bus power system network to form a n+m bus network (ground as the reference for voltages): E 1 X d1 E X d X dm E m Y bus nxn Y reduce bus m m 3

33 Node voltage equation with ground as reference I bus is the vector of the injected bus currents V bus is the vector of bus voltages measured from the reference node Y bus is the bus admittance matrix : Y ii (diagonal element) is the sum of admittances connected to bus i Y ij (off-diagonal element) equals the negative of the admittance between buses i and j Compared to the Y bus for power flow analysis, additional m internal generator nodes are added and Y ii (i n) is modified to include the load admittance at node i 33

34 To simplify the analysis, all nodes other than the generator internal nodes are eliminated as follows I 1 where I I m where θ ij is the angle of Y ij needs to be updated whenever the network is changed. ω 0 34

35 Homework #6 Saadat s , due on 4/4 (Thur) 35

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