Synchronous Machines Models and Equations Overview

ECE 576 Power System Dynamics and Stability Lecture 10: Synchronous Machines Models Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu 1

Announcements Homework 2 is due now Homework 3 is on the website and is due on Feb 27 Read Chapters 6 and then 4 2

Single Machine, Infinite Bus System (SMIB) Usually infinite bus angle, vs, is zero Book introduces new variables by combining machine values with line values = = = + + de d ed X X X de d + ep etc R R R se s e 3

Introduce New Constants ( ) = T Transient Speed t s s 2 H Mechanical time constant = T s s 1 = A small parameter s We are ignoring the exciter and governor for now; they will be covered in much more detail later 4

Stator Flux Differential Equations d ( ) de = + + + 1 sin se d R I V t qe s vs dt T s d ( ) qe = + + 1 cos se q R I V t de s vs dt T s d oe = se o R I dt 5

Special Case of Zero Resistance d ( ) de = + + 1 sin V Without resistance this is just an oscillator t qe s vs dt T s d ( ) qe = + + 1 cos V t de s vs dt T s An exact integral manifold (for any sized ): ( ) ( ) sin qe s vs V = = cos V d dt de s vs = ( : ) Note T s t 6

Direct Axis Equations q dE dt ( ) do q d = T E X X d d d X X ( ) ( ) d q + + I X X I E E 1 d d s d fd ( ) 2 d X X s d ( ) do q d 1 d = + T E X X I 1 d s d dt 7

Quadrature Axis Equations d dE dt ( ) qo d q = + T E X X q q q X X ( ) ( ) q d + + I X X I E 2 q q s q ( ) 2 q X X s d ( ) 2 q qo d q = T E X X I 2 q s q dt 8

Swing Equations 2 d dt d dt H ( ) = = (recall and T = ) T T s s t t s s s ( ) = t T T de q I qe d I T s M FW These are equivalent to the more traditional swing expressions d dt H d T I dt = s 2 ( ) = qe d I T M de q FW s 9

Stator Flux Expressions ( ( ) ) ( ( ) ) d d d X X X X X X X X s de d X I q = + + E 1 de d d s d s ( ( ) ) ( ( ) ) q q q X X X X s qe q X I d = + E 2 qe q X X X X q s q s = oe o X I oe 10

Network Expressions 2 2 = + V V V t d q d ( ) ed = + + + 1 sin V R I V d e d t eq s vs T dt s d ( ) eq = + + 1 cos V R I V q e q t ed s vs T dt s = = ep d X I X I ed eq ep q 11

Machine Variable Summary 3 fast dynamic states We'll get to the exciter and governor shortly; for now Efd is fixed , , de qe oe 6 not so fast dynamic states q d , , , , , E E 1 2 d q t 8 algebraic states I I V V V , , , , , , , I d q o d q t ed eq 12

Elimination of Stator Transients If we assume the stator flux equations are much faster than the remaining equations, then letting go to zero creates an integral manifold with ( ) = + + 0 sin se d R I V qe s vs ( ) = + 0 cos se q R I V de s vs = 0 se o R I 13

Impact on Studies Image Source: P. Kundur, Power System Stability and Control, EPRI, McGraw-Hill, 1994 14

Stator Flux Expressions ( ( ) ) ( ( ) ) d d d X X X X X X X X s de d X I q = + + E 1 de d d s d s ( ( ) ) ( ( ) ) q q q X X X X s qe q X I d = + E 2 qe q X X X X q s q s = oe o X I oe 15

Network Constraints ( ( ) ) ( ( ) ) q q d X X X X X X s qe q X I d = + 0 se d R I E 2 q X X q s q s ( ) + sin V s vs ( ( ) ) ( ( ) ) d d d X X X X X X X X s de d X I q = + 0 se q R I E 1 d d s d s ( ) + cos V s vs 16

"Interesting" Dynamic Circuit ( ( ) ) ( ( ) ) q q q X X X X ( ) s d q d + E X X I 2 q q X X X X q s q s ( ) ( ( ) ) ( ( ) ) 2 j d d d = e X X X X X X X X s q + + j E 1 d d s d s )( ) ( ) ( 2 j d + + R jX I jI e s d q ( )( ) ( ) 2 j j + + + + R jX I jI e V e vs e ep d q s 17

"Interesting" Dynamic Circuit ( ) = + sin V R I X I V d e d ep q s vs ( ) = + + cos V R I X I V q e q ep d s vs These last two equations can be written as one complex equation. ( e q d R e jV V + = + ) ( )( ) ( ) ( ) 2 2 j j + jX I jI e ep d q j + vs V e s 18

Subtransient Algebraic Circuit E ( ( ) ) ( ( ) ) q q q X X X X ( ) s E = d q d + E X X I 2 q q X X X X q s q s ( ( ) ) ( ( ) ) d d d X X X X X X X X ( ) 2 j s q + + j E e 1 d d s d s 19

Subtransient Algebraic Circuit Subtransient saliency use to be ignored (i.e., assuming X"q=X"d). However that is increasingly no longer the case ) ( () () = + + E E X X I 2 d q q d q often neglected q ( ) 2 j () () + + j E e 1 q d d 20

Simplified Machine Models Often more simplified models were used to represent synchronous machines These simplifications are becoming much less common Next several slides go through how these models can be simplified, then we'll cover the standard industrial models 21

Two-Axis Model If we assume the damper winding dynamics are sufficiently fast, then T"do and T"qo go to zero, so there is an integral manifold for their dynamic states ( ( 2 q d q s q E X X I = ) q q = E X X I 1 d s d ) 22

Two-Axis Model Then d ( ) do q d 1 = + = d 0 T E X X I 1 d s d dt q dE dt ( ) do q d = T E X X d d d X X ( ) ( ) d q + + I X X I E E 1 d d s d fd ( ) 2 d X X s q dE dt ( ) do q d = + T E X X I E d d fd 23

Two-Axis Model And d ( ) 2 q qo d q = = 0 T E X X I 2 q s q dt dE dt d ( ) qo d q = + T E X X q q q X X ( ) ( ) q d + + I X X I E ( ) 2 q q s q 2 q X X s d dE dt ( ) qo d q = + T E X X I q q 24

Two-Axis Model ( ) ( ) ( ) q d = + + + 0 sin R R I X X I E V s e d ep q s vs ( ) ( ) ( ) d q = + + + + 0 cos R R I X X I E V s e q ep d s vs 25

Two-Axis Model q d E ( ) do q d = + T E X X I E d d fd dt No saturation effects are included with this model d ( ) d E qo d q = + T E X X I q q dt d = s dt ( ) 2 H d d q q d = T E I E I X X I I FW T M d q d q dt s 26

Two-Axis Model ( ) ( ) ( ) q d = + + + 0 sin R R I X X I E V s e d ep q s vs ( ) ( ) ( ) d q = + + + + 0 cos R R I X X I E V s e q ep d s vs ( ) = + sin V R I ep q X I V d e d s vs ( ) = + + cos V R I ep d X I V q e q s vs 2 2 = + V V V t d q 27

Flux Decay Model If we assume T'qo is sufficiently fast then = + dE dt dE dt d dt ( ) q = d T E X X I 0 qo d q q q ( ) do q d = + T E X X I E d d fd = s 2H d ( ) ( d q q q d = T E I E I X X I I T M d d q FW dt s ( ) ) q q q q d = T X X q d I I E I X X I I T M q d q FW ( ) q q d = T E I X X I I T M q d q FW 28

Flux Decay Model This model is no longer common 29

Classical Model Has been widely used, but most difficult to justify From flux decay model d do = = = X X T q E 0 q = 0 E Or go back to the two-axis model and assume q d do qo = = = = X X T T d = ( const const) E E q 2 2 0 0 q d = + E E E 0 q E 0 1 = tan 2 0 d E 30

Classical Model Or, argue that an integral manifold exists for q , q = such that ) q d I X , , , const. E E E R V E d fd f ( R d q + = const E X ( ) ( ) 2 2 0 0 0 q 0 d q d q = + + E E X X I E () 0 1 = tan 2 31

Classical Model ( ) ( ) j - 2 I + jI e d q d = s dt 0 E V 2 H d ( ) 0 s = sin T T M vs FW d + dt X X 0 ep This is a pendulum model 32

Summary of Five Book Models a) Full model with stator transients 1 qo T 0 b) Sub-transient model c) Two-axis model s = ( ) 0 do T = ( ) 0 qo = d) One-axis model T X ) e) Classical model (const. E behind d 33

Damping Torques Friction and windage Usually small Stator currents (load) Usually represented in the load models Damper windings Directly included in the detailed machine models Can be added to classical model as D( - s) 34

Industrial Models There are just a handful of synchronous machine models used in North America GENSAL Salient pole model GENROU Round rotor model that has X"d = X"q GENTPF Round or salient pole model that allows X"d X"q GENTPJ Just a slight variation on GENTPF We'll briefly cover each one 35

Network Reference Frame In transient stability the initial generator values are set from a power flow solution, which has the terminal voltage and power injection Current injection is just conjugate of Power/Voltage These values are on the network reference frame, with the angle given by the slack bus angle V V jV V V = + = + , In book i j jV , j r j i Di Qi Voltages at bus j converted to d-q reference by V V V V V V V V sin cos cos sin sin cos cos sin , , r j d j , , d j r j = = , , i j q j , , q j i j Similar for current; see book 7.24, 7.25 36

Network Reference Frame Issue of calculating , which is key, will be considered for each model Starting point is the per unit stator voltages (3.215 and 3.216 from the book) V R I = = d q s d V R I q d s q ( ) ( ) ( ) + = + j Equivalently, V +jV +jI R I d q d q s q d Sometimes the scaling of the flux by the speed is neglected, but this can have a major impact on the solution 37

Two-Axis Model We'll start with the PowerWorld two-axis model (two- axis models are not common commercially, but they match the book on 6.110 to 6.113 Represented by two algebraic equations and four differential equations = + + = + = + The bus number subscript is omitted since it is not used in commercial block diagrams E V R I X I q q s q R I d d E V X I d d 1 s d q q d dE dt 1 dE dt ( ) ( ) d q q = + ( ) , ( ) E X X I E E X X I qo q d d d fd q q T T do 2 d dt H d d q q q d = = , ( ) T E I E I X X I I T s M d d q FW dt s 38

Two-Axis Model Value of is determined from (3.229 from book) ( s q E V R jX I = + + ) Sign convention on current is out of the generator is positive Once is determined then we can directly solve for E'q and E'd 39

Example (Used for All Models) Below example will be used with all models. Assume a 100 MVA base, with gen supplying 1.0+j0.3286 power into infinite bus with unity voltage through network impedance of j0.22 Gives current of 1.0-j0.3286 and generator terminal voltage of 1.072+j0.22 = 1.0946 11.59 Bus 1 Bus 2 X12 = 0.20 Bus 4 Infinite Bus XTR = 0.10 slack Bus 3 100.00 MW 57.24 Mvar 1.0946 pu X13 = 0.10 X23 = 0.20 -100.00 MW -32.86 Mvar 11.59 Deg 1.0463 pu 1.0000 pu 6.59 Deg 0.00 Deg 40

Two-Axis Example For the two-axis model assume H = 3.0 per unit- seconds, Rs=0, Xd = 2.1, Xq = 2.0, X'd= 0.3, X'q = 0.5, T'do = 7.0, T'qo = 0.75 per unit using the 100 MVA base. Solving we get ( )( 1.0946 11.59 2.0 1.052 52.1 = ) = + = 2.814 52.1 18.2 E j V V 0.7889 0.6146 0.6146 1.0723 0.7889 0.7107 0.8326 d = = 0.220 q I I 0.7889 0.6146 0.6146 0.7889 1.000 0.3287 0.9909 0.3553 d = = q 41

Two-Axis Example And ( )( ) q = = + = = 0.8326 0.3 0.9909 1.1299 E 0.7107 (0.5)(0.3553) 1.1299 (2.1 0.3)(0.9909) = + 0.5330 E E d = 2.9135 fd Saved as case B4_TwoAxis 42

Subtransient Models The two-axis model is a transient model Essentially all commercial studies now use subtransient models First models considered are GENSAL and GENROU, which require X"d=X"q This allows the internal, subtransient voltage to be represented as = + + ( d q q d E jE j + = + ( ) E V R jX I s ) 43

Subtransient Models Usually represented by a Norton Injection with ( d q d q s R jX + ) q d + + j + E jE + = = I jI R jX s May also be shown as ( ) ( ) q + d d q + j + j j ( ) + = = = j I jI I jI d q q d + R jX R jX s s In steady-state = 1.0 44

GENSAL The GENSAL model has been widely used to model salient pole synchronous generators In the 2010 WECC cases about 1/3 of machine models were GENSAL; in 2013 essentially none are, being replaced by GENTPF or GENTPJ In salient pole models saturation is only assumed to affect the d-axis 45

GENSAL Block Diagram (PSLF) A quadratic saturation function is used. For initialization it only impacts the Efd value 46

GENSAL Initialization To initialize this model 1. Use S(1.0) and S(1.2) to solve for the saturation coefficients 2. Determine the initial value of with ( s q E V R jX I = + + ) 3. Transform current into dq reference frame, giving id and iq 4. Calculate the internal subtransient voltage as = + + ( ) E V R jX I s 5. Convert to dq reference, giving P"d+jP"q= "d+ "q 6. Determine remaining elements from block diagram by recognizing in steady-state input to integrators must be zero 47

GENSAL Example Assume same system as before, but with the generator parameters as H=3.0, D=0, Ra = 0.01, Xd = 1.1, Xq = 0.82, X'd = 0.5, X"d=X"q=0.28, Xl = 0.13, T'do = 8.2, T"do = 0.073, T"qo =0.07, S(1.0) = 0.05, and S(1.2) = 0.2. Same terminal conditions as before Current of 1.0-j0.3286 and generator terminal voltage of 1.072+j0.22 = 1.0946 11.59 Use same equation to get initial ( 1.072 0.22 (0.01 0.82)(1.0 1.35 1.037 1.70 37.5 j = + = ) + = + + E V R jX I s + q = + 0.3286) j j j 48

GENSAL Example Then I I I I sin cos cos sin d r = q i 0.609 0.793 0.793 0.609 1.0 0.3286 0.869 0.593 = = And V = = + + + + ( ) R jX j j I s + + 1.072 1.174 0.22 (0.01 0.497 0.28)(1.0 0.3286) j j 49

GENSAL Example Giving the initial fluxes (with = 1.0) q 0.609 0.793 0.793 0.609 1.174 0.497 0.321 1.233 = = d To get the remaining variables set the differential equations equal to zero, e.g., ( 0.82 0.28 0.593 1.425, 1.104 q d E = = ) ( )( ) q q = = = 0.321 X X I q q Solving the d-axis requires solving two linear equations for two unknowns 50