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ECE 576 – Power System Dynamics and Stability Lecture 28: Power System Stabilizer Prof. Tom Overbye University of Illinois at Urbana-Champaign overbye@illinois.edu 1 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 Functions of a PSS • • • • PSS adds a signal to the excitation system proportional to speed deviation. This adds positive damping – Other inputs, like power, voltage or acceleration, can be used – Signal is generated locally from the shaft. Both local mode and inter-area mode can be damped. When oscillation is observed on a system or a planning study reveals poorly damped oscillations, use of participation factors helps in identifying the machine(s) where PSS has to be located. Tuning of PSS regularly is important. Modal Analysis technique forms the basis for multimachine systems. 3 Example PSS • An example single input stabilizer is shown below (IEEEST) – The input is usually the generator shaft speed deviation, but it could also be the bus frequency deviation, generator electric power or voltage magnitude VST is an input into the exciter 4 Example PSS • Below is an example of a dual input PSS (PSS2A) – Combining shaft speed deviation with generator electric power is common – Both inputs have washout filters to remove low frequency components of the input signals IEEE Std 421.5 describes the common stabilizers 5 PSS Tuning: Basic Approach • • • The PSS parameters need to be selected to achieve the desired damping through a process known as tuning The next several slides present a basic method using a single machine, infinite bus (SMIB) representation Start with the linearized differential, algebraic model with controls u added to the states Δx = AΔx BΔy EΔu 0 = CΔx + DΔy • If D is invertible then Δx = A BD-1C Δx EΔu A sys Δx EΔu 6 SMIB System (Flux Decay Model) • Process is introduced using an SMIB with a flux decay machine model and a fast exciter Vt e j Re ~ ( I d jIq )e j ( / 2) jX e V 0 pus 1 pu [TM ( Eq' I q ( X d X d' ) I d I q D pu )] 2H 1 ' Eq ' ( Eq' ( X d X d' ) I d E fd ) Tdo 7 Stator Equations • Assume Rs=0, then the stator algebraic equations are: X q I q Vt sin( ) 0 (1) Eq' Vt cos( ) X d' I d 0 ( 2) (Vd jVq )e j ( / 2 ) Vt e j (3) Vd jVq Vt e j e j ( / 2 ) (4) Expand right hand side of (4) Vd jVq Vt sin( ) jVt cos( ) (5) Vd Vt sin( ) and Vq Vt cos( ) X q I q Vd 0 ( 6) Eq' Vq X d' I d 0 (7 ) (1) and (2) become 8 Network Equations • The network equation is (assuming zero angle at the infinite bus and no local load) ( I d jI q )e j ( / 2 ) ( I d jI q ) (Vd jVq )e j ( / 2 ) V 0 Re jX e (Vd jVq ) V e j ( / 2 ) Re jX e Re I d X e I q Vd V sin Simplifying, X e I d Re I q Vq V cos Vt e j ~ Re ( I d jIq )e j ( / 2) Xe V 0 9 Complete SMIB Model 1 E ' ( Eq' ( X d X d' ) I d E fd ) Tdo ' q pus Machine equations 1 pu [TM ( Eq' I q ( X q X d' ) I d I q D pu )] 2H X q I q Vd 0 Stator equations Eq' Vq X d' I d 0 Re I d X e I q Vd V sin Network equations X e I d Re I q Vq V cos Vt e ~ j Re ( I d jIq )e j ( / 2) V 0 jX e 10 Linearization of SMIB Model Step 1: Linearize the stator equations X q I d 0 Vd 0 V ' I E ' q X d 0 q q Step 2: Linearize the network equations Vd Re X e I d V cos V I q X e Re q V sin Step 3: Equate the righthand sides of the above equations Re ( X e X q ) I d 0 V cos I E ' ( X X ' ) Re d e q q V sin Notice this is equivalent to a generator at the infinite bus with modified resistance and reactance values 11 Linearization (contd) Solve for I d , I q ReV cos V sin ( X q X e ) E ' I d 1 ( X e X q ) q I (1) ' Re ReV sin V cos ( X d X e ) q The determinant is Re2 ( X e X q )( X e X d' ) • 1. 2. Final Steps involve Linearizing Machine Equations . x f ( x, I d , I q , E fd , TM ) (2) Substitute (1) in the linearized equations of (2). 12 Linearization of Machine Equations 1 Eq' Tdo' 0 pu I qo 2 H 0 E ' q 0 s (1) Ds 0 2 H pu 0 T'1 ( X d X d' ) do 0 1 o ' 2 H I q ( X d X q ) 0 0 1 2H ( X d' X q ) I qo 21H T'1 I d do 0 I q 'o Eq 0 0 E fd 0 T M 1 2H Symbolically we have x Ax Bu C I d q (2 ) I d q Dx ( 3) • Substitute (3) in (2) to get linearized model. 13 Linearized SMIB Model K4 1 1 ' E Eq ' ' E fd ' K 3Tdo Tdo Tdo ' q s pu pu • • Ds K2 K1 1 ' Eq pu TM 2H 2H 2H 2H Excitation system is yet to be included. K1 – K4 constants are defined on next slide 14 K1 – K4 Constants 1 1 K3 ( X d X d' )( X q X e ) V ( X d X d' ) K4 ( X q X e ) sin Re cos 1 o K 2 I q I qo ( X d' X q )( X q X e ) Re ( X d' X q ) I do Re Eq'o 1 o K1 [ I q V ( X d' X q ){( X q X e ) sin Re cos } V {( X d' X q ) I do Eq'o }{( X d' X e ) cos Re sin }] • K1 – K4 only involve machine and not the exciter. 15 Including Terminal Voltage • The change in the terminal voltage magnitude also needs to be include since it is an input into the exciter Vref Vref Vt Vt Vt Vd2 Vq2 Differentiating Vt 2 Vd2 Vq2 2Vt Vt 2Vdo Vd 2Vqo Vq Vt o d Vqo V Vd Vq Vt Vt 16 Computing • • While linearizing the stator algebraic equations, we had Vd 0 V ' X q d X q Eq' 0 E ' 0 q Substitute this in expression for Vt to get Vt K 5 K6 Eq' 1 Vdo K 5 X q [ ReV sin V cos ( X d' X e )] Vt [ X ( ReV cos V ( X q X e ) sin )] Vt Vqo ' d o o o V V 1 V q q K6 d X q Re X d' ( X q X e ) Vt Vt Vt 17 Heffron–Phillips Model • Add a fast exciter with a single differential equation TA E fd E fd K A (Vref Vt ) • Linearize TA E fd E fd K A (Vref Vt ) TA E fd E fd K A Vref K A ( K 5 K 6 Eq' ) • This is then combined with the previous three differential equations to give x A sys x BΔu Δu [TM Vref ] T SMIB Model 18 Block Diagram • K1 – K6 are affected by system loading and Xe 19 Numerical Example • Consider an SMIB system with Zeq = j0.5, Vinf = 1.05, in which in the power flow the generator has S = 54.34 – j2.85 MVA with a Vt of 115 – Machine is modeled with a flux decay model with (pu, 100 MVA) H=3.2, T'do=9.6, Xd=2.5, Xq=2.1, X'd=0.39, Rs=0, D=0 Saved as case B2_PSS_Flux (but only available in v18) 20 Initial Conditions • The initial conditions are j I G e ( I d jI q )e j ( / 2 ) 115 1.050 0.544318 j0.5 (0) angle of E where E Vt e j ( Rs jX q ) I G e j . E 115 ( j2.1)(0.544318) 1.4788 65.5 I d jI q I G e j e j ( / 2 ) 0.544342.48, I d 0.4014, and I q 0.3676. Vd jVq Ve j e j ( / 2 ) 139.48 Hence, Vd 0.7718, Vq 0.6358 21 Add an EXST1 Exciter Model • • Set the parameters to KA = 400, TA=0.2, all others zero with no limits and no compensation Hence this simplified exciter is represented by a single differential equation V is the input TA E fd E fd K A (VREF Vt Vs ) s from the stabilizer, with an initial value of zero 22 Initial Conditions (contd) • From the stator algebraic equation, Eq' Vq X d' I d 0.63581 (0.39 )(0.4014 ) 0.7924 E fd Eq' ( X d X d' ) I d 0.7924 (2.5 0.39)0.4014 1.6394 VREF E fd 1.6394 Vt 1 1.0041 KA 400 s 377, TM Eq' I q ( X q X d' ) I d I q (0.7924)(0.3676) (2.1 0.39)(0.4014)(0.3676) 0.5436 checks with (VtV sin ) / X e 23 Computation of K1 – K6 Constants • The formulas are used. Re2 ( X e X q )( X e X d' ) 2.314 1 1 K3 ( X d X d' )( X q X e ) 3.3707 K 3 0.296667 V ( X d X d' ) K4 [( X q X e ) sin Re cos ] 2.26555 1 K 2 [ I qo I qo ( X d' X q )( X q X e ) Re ( X d' X q ) I do Re Eq' o ] 1.0739 Similarly K1 , K 5 , and K 6 are calculated as K1 0.9224 , K 5 0.005 , K 6 0.3572 24 Effect of Field Flux on System Stability • With constant field voltage, i.e. Efd=0, from the block diagram Te K 2 K3 K 4 1 K3Tdo' s – K2, K3, and K4 are usually positive. On diagram v is used to indicate what we've been calling pu 25 Effect of Field Flux on System Stability • For low frequencies and in steady-state, as s=j goes to zero, then Te K 2 K 3 K 4 Te K 2 K 3 K 4 ' 1 K 3Tdo s • Hence field flux variation due to feedback (armature reaction) introduces negative synchronizing torque 26 Effect of Field Flux on System Stability • At higher frequencies the denominator is dominated by the K3T'dos term so K2 K3 K4 K2 K4 Te j ' ' jTdo K 3 Tdo • • This is a phase with (j) Hence Efd results in a positive damping component. 27 Effect of Field Flux on System Stability • In general, positive damping torque and negative synchronizing torque due to Efd at typical oscillation frequencies (1-3 Hz). Te Ts TD K3 (E fd K 4 ) K 2 E K 2 ' ( 1 K T s ) 3 do ' q 28 Effect of Field Flux • • • • There are situations when K4 can become negative. V ( X d X d' ) K4 ( X q X e ) sin Re cos may become negative Case1. A hydraulic generator without a damper winding, light load ( is small) and connected to a line with high Re/Xe ratio and a large system. Case 2. Machine is connected to a large local load, supplied partly by generator and partly by remote large system. If K4 < 0 then the damping due to Efd is negative 29 Effect of Excitation System • • • • • Ignore dynamics of AVR & KA >> 0 Gain through T'do is -(K4+KAK5)K3 If K5 < 0 then gain becomes positive. VREF = 0 Large enough K5 may make system unstable Te ( K 4 K A K5 ) K3 K 2 (1 sK 3'Tdo' K A K3 K 6 ) 30 Numerical Example (Effect of K5) • Test system 1 Test System 2 K A 50, K 5 0 K A 50, K 5 0 K1 3.7585 K 2 3.6816 K1 0.9813 K 3 0.2162 K 4 2.6582 K 3 0.3864 K 4 1.4746 K 5 0.0544 K 6 0.3616 K 5 0.1103 K 6 0.4477 Tdo' 5 sec Tdo' 5 sec H 6 sec TA 0.2 sec K 2 1.093 H 6 sec TA 0.2 sec Eigen Values Test System 1 Test System 2 0.353 j10.946 0.015 j 5.38 2.61 j 3.22 2.77 j 2.88 unstable 31 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 , VREF=0 32 PSS Contribution to Torque pu TPSS G (s ) PSS K2 K3 1 sK 3Tdo' KA 1 sTA vs Vref 0 K6 TPSS G ( s) K 2 K A K 3 v K A K 3 K 6 (1 sK 3Tdo' )(1 sTA ) G ( s) K 2 K A 1 G ( s )GEP( s ) TA ' 2 ' K 3 K A K 6 s K 3 Tdo s TdoTA (for usual range G ( s) K 2 K A of constants) 1 ' 33 K K 1 s T A 6 do / K A K 6 1 sT A K3 PSS Contribution to Torque • For high gain KA TPSS K 2 G ( s) v K 6 1 s Tdo' / K A K 6 1 sTA • If PSS were to provide pure damping (in phase with pu), then ideally, G ( s) K PSS 1 s Tdo' / K A K 6 1 sTA • K PSS Gain of PSS i.e. pure lead function Practically G(s) is a combination of lead and lag blocks. 34 PSS Block Diagram ymax pu sTw 1 sTw 1 sT1 1 sT2 1 sT3 1 sT4 K PSS v s ymin G ( s ) K PSS • • • • (1 sT1 ) (1 sT3 ) sTw K PSS G1 ( s ) (1 sT2 ) (1 sT4 ) (1 sTw ) Provide damping over the range of frequencies (0.1-2 Hz) “Signal wash out” (Tw) 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 terminal voltage. 35 Criteria for Setting PSS Parameters • • • • KPSS determines damping introduced by PSS. T1, T2, T3, T4 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: – – – – – KPSS is in the range of 0.1 to 50 T1 is the lead time constant, 0.2 to 1.5 sec T2 is the lag time constant, 0.02 to 0.15 sec T3 is the lead time constant, 0.2 to 1.5 sec T4 is the lag time constant, 0.02 to 0.15 sec 36 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 TW is set at 10 sec. KPSS is set at (1/3)K*PSS, where (1/3)K*PSS is the gain at which the system becomes unstable. There are many ways of adjusting the values of T1, T2, T3, T4 Frequency response methods are recommended. 37 Design Procedure Using the Frequency-Domain Method • TPSS K 2 K A K3 GEP( s )G ( s ) , GEP( s ) v K A K 3 K 6 (1 sTdo' K 3 )(1 sTA ) Step 1: Neglecting the damping due to all other sources, find the undamped natural frequency n in rad/sec of the torque - angle loop from: 2H 2 s K1 0, i.e. s1, 2 j n K1 s Te s 2Hs 1 s where n D TM 0 Te Step 2: Find the phase lag of 2H s 1 s 2 Ds K1 K1 s 2H GEP(s) at s j n 38 PSS Design (contd) • • Step 3: Adjust the phase lead of G(s) such that G ( s ) s j n GEP( s ) s j n 0 • • 1 sT1 Let G ( s ) K PSS 1 sT2 k Ignoring the washout filter whose net phase contribution is approximately zero. k 1 or 2 with T1 T2 . Thus, k 1 : 1 j nT1 1 j nT2 GEP( j n ) 39 PSS Design (contd) • • Knowing n and GEP(j n) we can select T1. T2 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 * K PSS 13 K PSS . Alternative Procedure The char. equation is X n n 1 2 2H cos 1 s S 2 Ds K1 0 (1) The damping ratio is X Damping ratio 12 D / MK1 where M 2H / s 40 PSS Design (contd) • The characteristic roots are: s1,2 MD D M 2 4MK1 2 2 2 K1 D 4K1 D D j if 2M M 2M M M n jn 1 2 Note that n K1 M . Therefore Verify that D D 2M n 2M M D K1 2 K1 M K1 D 2 D 2 2 n . M 4K1 M 2M 2 41 PSS Design (contd) • Since the phase lead of G(s) cancels phase lag due to GEP(s) at the oscillatory frequency, the contribution of PSS through GEP(s) is a pure damping torque with a damping coefficient DPSS DPSS K PSS GEP( s ) s jn G1 ( s ) s jn The characteristic equation is DPSS K1 s s 0 M M i.e., s 2 2n M 0 2 DPSS 2n M K PSS GEP ( jn ) G1 ( jn ) We can thus find K PSS , knowing ωn and the desired ξ. A reasonable choice for ξ is between 0.1 and 0.3. 42 PSS Design (contd) • • • • The washout filter should not have any effect on phase shift or gain at the oscillating frequency. A large value of TW is chosen so that sTW is much larger than unity. Choose Gw ( j n ) 1 Hence, its phase contribution is close to zero. The PSS will not have any effect on the steady state behavior of of the system since in steady state pu=0 43 Example • • • • Without the PSS, the A matrix is 0 0.104 0.3511 0.2236 0 0 377 0 0.1678 0.144 0 0 714 . 4 10 0 5 The eigenvalues are l1,2 0.0875 j7.11, l3,4 2.588 j8.495. The electromechanical mode l1,2 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. 44 Example (contd) Vref pu K PSS (1 sT1 ) (1 sT2 ) PSS • Vs KA 1 sTA E fd Vt The added differential equation is then K PSS T1 1 Vs Vs pu K PSS pu T2 T2 T2 K PSS T1 K 2 K1 1 ' Vs pu K PSS Eq T2 T2 T2 2H 2H 45 Example (contd) With a choice of K PSS 0.5, T1 0.5, T2 0.1, the new matrix is 0 0.104 0 0.3511 0.236 0 0 377 0 0 0.1678 0.144 0 0 0 714 . 4 10 0 5 2000 0.42 0.36 5 0 10 The eigenvalues are l1,2 0.8612 j7.7042 l3,4 1.6314 j8.5504, l5 10.3661. Note the improvement of the electromechanical mode l1,2 46 Example (industry) • CONVERT TO SMIB (RETAIN G16) 47 Example (contd) Eigenvalue -0.18929 -0.91111 -0.017097 j3.3105 -10.145 -8.0319 j16.785 -24.531 -29.031 Frequency Hz Damping Ratio 0 0 0.52688 0.0051643 0 2.6714 0.43165 0 0 E.M. Mode has f = 0.5688 Hz The participator for this mode is • State Number 1 2 3 E'q E fd 4 5 E'd E'q 6 E fd 7 tg1 8 9 tg 2 10 tg 3 Participation Factor 0.45177 j0.027549 0.50613 j0.028591 -0.00011 j0.011599 0.00012 j0.00011 -0.004354 j0.01354 -0.00034 j7.033e-6 -0.00177 j0.00075 0.005614 j0.016166 0.040345 j0.023404 0.0025854 j0.0092191 * * Poorly damped E.M . Mode 0.005 Exciter mode well damped Convert these into magnitudes } Turbine Governor States Largest Participation by angle and speed states. 48 Root Locus • If G(s) (Transfer function of PSS) is a real gain, it will add pure damping Root Locus Study (with KPSS = 10). 49 Phase Compensation GEP( s ) s j Ideal phase lead. PSS phase lead. G( s) 10s (1 .03s)(1 .04s) 1 10s (1 .01s)(1 .01s) Washout does not add much phase lead. 50 Root Locus with PSS 51 Eigenvalues with PSS Eigenvalue -0.10078 -0.18895 -0.90803 -1.8476 -0.40649 j3.2883 -10.151 -7.668 j16.996 -24.531 -29.031 -95.914 -104.05 Frequency Hz 0 0 0 0 0.52334 0 2.705 0 0 0 0 Damping Ratio 1 1 1 1 0.12268 1 0.41125 1 1 1 1 Note Improvement in Damping Ratio of E.M Mode from 0.05 to 0.123 52