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ECE 422/522 Power System Operations & Planning/Power Systems Analysis II : 7 - Transient Stability Spring 2014 Instructor: Kai Sun 1 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 2-3 seconds of the initial disturbances 2 Single-Machine Infinite Bus (SMIB) System Pe= =Pmaxsinδ Pmax= E’EB/XT • Swing Equation: 2 H d 2 Pm Pe Pm Pmax sin 2 0 dt Pa is called Accelerating Power Pm= Pe • The rotor will accelerate if Pm increases, or Pe decreases due to, e.g., a contingency on the transmission path 3 Power Angle Relationship Pe= =Pmaxsinδ Question: What if both circuits are out of service? 4 Response to a step change in Pm 2 H d 2 Pa Pm Pe Pm Pmax sin 0 dt 2 Consider a sudden increase in Pm: Pm0→ Pm1. New equilibrium point b satisfying Pe(δ1)=Pm1 • a→ b: Due to the rotor’s inertia, δ cannot jump from δ0 to δ1, so Pa=Pm1-Pe(δ0)>0 and ωr increases from ω0. When b is reached, Pa=0 but ωr >ω0, so δ continues to increase. • b→ c: δ>δ1 and Pa<0, so ωr decreases until c. At c, ωr=ω0 and δ reaches the peak value δmax. Pa ωr a >0 =ω0, ↑ b 0 =ωr, max,↑ c <0 =ω0 , ↓ δ ↑ ↑ =δmax, ↓ • ← c: At c, the rotor starts to decelerate (since Pa<0) with ωr<ω0 and δ decreases. • With all resistances (damping) neglected, δ and ωr oscillate around the new equilibrium point b with a constant amplitude. 5 Equal-Area Criterion (EAC) r2 0 H ( Pm Pe )d 2 H 0 r2 ( Pm Pe )d 2 20 If δmax exists where dδ/dt=0: 1 r J 2 0 2 max 0 ( Pm Pe )d At δmax, ∆ωr=0 and the integral=0 max 1 Moment of inertia in p.u. (Kundur’s Page 131) ( Pm Pe )d 0 max 1 ( Pe Pm )d = |area A1 | |area A2 | 0 (Note: all losses are neglected) 6 • Equal-Area Criterion (EAC): The stability is maintained only if a decelerating area |A2| ≥ the accelerating area |A1| can be located above Pm (before d, i.e. the Unstable Equilibrium Point or UEP). UEP SEP • If |A2|<|A1|, δ will continue increasing at UEP (since ωr>ω0), so it will lose stability. • For the case with a step change in Pm, ∆Pm or the new Pm does matter for transient stability. 7 Transient stability limit for a step change of Pm Following a step change Pm0→Pm, solve the transient stability limit of Pm: • Assume | A1 |=| A2 |: 1 Pm (1 0 ) Pmax sin d 0 max 1 Pmax sin d Pm (max 1 ) Pm0 Pmax (cos 1 cos 0 ) Pmax (cos max cos 1 ) Pmmax (max 0 ) Pm Pmax (cos 0 cos max ) • At the UEP, Pm Pmax sin max SEP UEP (max 0 )sin max cos max cos 0 • Solve δmax to calculate the transient stability limit for a step change of Pm: Pm Pmax sin max 1 max Question: Can we increase Pm beyond that transient limit (with Pm<Pmax)? 8 Solve δmax by the Newton-Raphson method (max 0 )sin max cos max cos 0 • The nonlinear function form: f (max ) cos 0 c (k ) / 2 max • Select an initial estimate: • Calculate iterative solutions by the N-R algorithm: ( k 1) max (k ) max (k ) max where (k ) max (k ) c f (max ) df d max (k ) c f (max ) (k ) (k ) (max 0 ) cos max (k ) max • Give a solution when a specific accuracy ε is reached, i.e. ( k 1) (k ) max max 9 Response to a three-phase fault Pe= =Pmaxsinδ • Pe,during fault << Pe, post-fault • Pe,post-fault <Pe,pre-fault for a permanent fault (cleared by tripping the fault circuit) or Pe, post-fault =Pe,pre-fault for a temporary fault 10 Unstable Stable 11 Critical Clearing Angle (CCA) • Saadat’s Sec.11.6 (Example 11.5) • Consider a simple case – A three-phase fault at the sending end – Pe, during fault=0 if all resistances are neglected – Critical Clearing Angle δc c 0 Pm d max c ( Pmax sin Pm )d |A1| |A2| Integrating both sides: Pm c 0 Pmax (cos c cos max ) Pm (max c ) cos c Pm (max c ) cos max Pmax Pm ( 0 c ) cos 0 Pmax = π-δ0 12 Critical Clearing Time (CCT) • Solve the CCT from the CCA: Since Pe, during fault=0 for this case, during the fault: 2 H d 2 Pm 2 0 dt t d 0 Pm dt 0 Pmt 0 dt 2H 2H tc 0 Pmt 2 0 4H 4 H (c 0 ) 0 Pm 13 • For a more general case: Pe (during fault)>0 (pre-fault) P3max (post-fault) ≤P3max A3 A2 (fault-on) ≤ P2max A1 P2max δs c (δu) Pm c 0 P2 max sin d 0 cos c max c P3max sin d Pm (max c ) Pm (max c ) P3max cos max P2 max cos 0 P3max P2 max • |A1| = Vke(δc), the kinetic energy at δc • |A1|+|A3| = V(δc)=Vke(δc)+Vpe(δc), total energy at δc • |A2|+|A3| = Vpe(δu)=Vcr, i.e. the largest potential energy • If and only if V(δc)≤Vcr (i.e. |A1|≤|A2|), the generator is stable 14 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 A1 is reduced. • The generator internal voltage magnitude (E’). This depends on the field excitation • The infinite bus voltage magnitude EB 15 EAC for a Two-Machine System • Two interconnected machines respectively with H1 and H2 – The system can be reduced to an equivalent SMIB system d 21 0 0 ( ) P P Pa1 m1 e1 2 H1 2 H1 dt 2 d 212 d 21 d 22 0 Pa1 Pa1 2 2 ( ) 2 2 H1 H 2 dt dt dt d 2 2 0 0 ( ) P P Pa 2 m2 e2 2H 2 2H 2 dt 2 H 2 Pa1 H1Pa 2 H 2 Pm1 H1Pm 2 H 2 Pe1 H1Pe 2 2 H1H 2 d 212 = 2 0 H1 H 2 dt H1 H 2 H1 H 2 H1 H 2 Pe12 2 H12 d 212 Pm ,12 Pe ,12 2 0 dt 12,max 12,0 0 ( Pm,12 Pe ,12 )d 0 H12 Pm12 16 δ12,0 δ12,max δ12 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 2-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: 400–412, 1988. 2. Y. Xue, et al, "Extended Equal-Area Criterion Revisited". IEEE Trans. on PWRS, PWRS7: 10101022,1992. 3. M. Pavella, et al, “Transient Stability of Power Systems: a Unified Approach to Assessment and Control”, Kluwer, 2000. 17 Some examples from [3] 18 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 Vcr to judge stability – EAC is a direct method for a SMIB or two-machine system 19 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=x0 and t=t0 dx = f ( x, t ) dt 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 20 dx = f ( x, t ) dt Euler Method • The Euler method is equivalent to using the first two terms of the Taylor series expansion for x around the point (x0, t0), referred to as a first-order method (error is on the order of ∆t2) – Approximate the curve at x=x0 and t=t0 by its tangent x(t1) x1 x0 dx dt ∆t = f ( x0 , t0 ) x0 t0 ∆x ≈ dx dt ∆t x1= x0 + ∆x= x0 + x0 dx dt ∆t x0 – At step i+1 xi +1 = xi + dx ∆t dt xi • 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 21 t1 dx = f ( x, t ) dt Modified Euler (ME) Method • Modified Euler method consists of two steps: (a) Predictor step: dx x1p = x0 + dt x (t1) x1c x1p ∆t x0 ∆t x0 Slope at the beginning of ∆t t0 t1 The derivative at the end of the ∆t is estimated using x1p dx dt xp = f ( x1p , t1 ) 1 Estimated slope at the end of ∆t dx dx dx dx + + dt xi dt x p dt dt x p x0 c i +1 c 1 ∆t x i +1 = xi + x1 = x0 + ∆t 2 2 • It is a second-order method (error is on the order of ∆t3) • 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. (b) Corrector step: 22 dx = f ( x, t ) dt Runge-Kutta (R-K) Methods • General formula of the 2nd order R-K method: (error is on the order of ∆t3) = k1 f ( x0 , t0 ) ∆t k2 x(t1) x0+αk1 f ( x0 + α k1 , t0 + β∆t ) ∆t x0 ∆t x1 =x0 + a1k1 + a2k2 At Step i+1: t0 = k 1 f ( xi , ti ) ∆t The ME method is a special case with a1=a2=1/2, α=β=1 k= f ( xi + α k1 , ti + β∆t ) ∆t 2 xi +1 =xi + a1k1 + a2k2 • General formula of the 4th order R-K method: (error is on the order of ∆t5) 1 xi +1 =xi + ( k1 + 2k2 + 2k3 + k4 ) 6 = k1 f ( xi , ti ) ∆t k2 = f ( xi + k1 ∆t , ti + ) ∆t 2 2 k2 ∆t , ti + ) ∆t 2 2 = k4 f ( xi + k3 , ti + ∆t ) ∆t k3 = f ( xi + 23 t1 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 |λmax/λmin| 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 24 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 t1 • 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(xn,tn) and f(xn+1,tn+1) are neglected. Δt f ( x0 ,t0 ) + f ( x1 ,t1 ) 2 Δt xn+1 = xn + f ( xn ,tn ) + f ( xn+1 ,tn+1 ) 2 x1 = x0 + x(t1)=x(t0)+|A|+|B| f B f(x1,t1) f(x0,t0) Compared to ME method: A x n += xn + 1 ∆t t0 t1 25 ∆t f ( xn ,tn ) + f ( xnp+1 ,tn+1 ) 2 Kundur’s Example 13.1 26 27 28 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 YN x = f(x, V) I(x, V) = YN V 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. 6.2.3) 29 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, 2nd order R-K, 4th order R-K and Trapezoidal Rule methods 30 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 Pmi 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 • Solve the initial power flow and determine the initial bus voltage phasors Vi. • Terminal currents Ii of m generators prior to disturbance are calculated by their terminal voltages Vi and power outputs Si, 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): X’ E’1 E’2 E’m d1 X’d2 X’dm reduce Ybus nxn Ybus m×m 32 • Node voltage equation with ground as reference Ibus is the vector of the injected bus currents Vbus is the vector of bus voltages measured from the reference node Ybus is the bus admittance matrix : Yii (diagonal element) is the sum of admittances connected to bus i Yij (off-diagonal element) equals the negative of the admittance between buses i and j Compared to the Ybus for power flow analysis, additional m internal generator nodes are added and Yii (i≤n) is modified to include the load admittance at node i 33 • To simplify the analysis, all nodes other than the generator internal nodes are eliminated as follows I1 I2 where Im needs to be updated whenever the network is changed. where θij is the angle of Yij 2 ω0 34 Homework #6 • Saadat’s 11.14-11.17, due on 4/24 (Thur) 35