Equivalent Circuits with Multiple Damper Windings (e.g. Round rotor Machines)

Transcription

1 Equivalent Circuits with Multiple Damper Windings (e.g. Round rotor Machines) d axis: L fd L F - M R fd F L 1d L D - M R 1d D R fd R F e fd e F R 1d R D Subscript Notations: ( ) fd ~ field winding quantities ( ) kd ~ k-th d-axis damper winding quantities ( ) kq ~ k-th q-axis damper winding quantities M R -L ad 0 M R -L ad 0 is named L fkd1 in some literature to model rotor mutual flux leakage, i.e. the flux linking the rotor s field and damper windings but not stator windings) q axis: L 1d =L D - M R R 1d =R D L fd =L F -M R R fd =R F e fd =e F L 1q L Q L aq R 1q R Q 1q Q L 1q =L Q L aq L 2q =L G L aq L 2q L G L aq R 2q R G 2q G R 1q =R Q R 2q =R G 25

2 Example: a model with 3 rotor windings in each of d and q axis equivalent circuits EPRI Report EL 1424 V2, Determination of Synchronous Machine Stability Study Constants, Volume 2, 1980 The proposed equivalent circuits are expected to contain sufficient details to model all machines Parameters are estimated by frequency response tests 26

3 Example 3.1 (Kundur s book) A 555 MVA, 24kV, 0.9p.f., 60Hz, 3 phase, 2 pole synchronous generator has the following inductances an resistances associated with the stator and field windings: l aa = cos(2 ) mh l ab = cos(2 + /3) mh l af =40.0 cos mh L F = mh R a = R F = a. Determine L d and L q in H b. If the stator leakage inductance L l is mH, determine L ad and L aq in H c. Using the machine rated values as the base values for the stator quantities, determine the per unit values of the following in the L ad base reciprocal per unit system (assuming L ad =M F =M R in per unit): L l, L ad, L aq, L d, L q, M F, L F, L fd, R a and R F Solution: a. l aa =L s + L m cos2 = cos2 mh l ab = -M s - L m cos(2 + /3) = cos(2 + /3) mh l af = M F cos =40.0 cos mh L d = L s + M s + 3L m /2 = = mh L q = L s + M s -3L m /2 = = mh b. L ad = L d L l = = mh L aq = L q L l = =4.432 mh 27

4 c. 3-phase VA base = 555 MVA E RMS base =24/ 3 = kv e s base (peak) = 2 E RMS base = = kv I RMS base =3-phaseVAbase / (3 E RMS base )= /( )= A i s base (peak) = 2 I RMS base = = A Z s base =e s base /i s base = /( ) = base =2 60 =377 elec. rad/s L s base =Z s base / base = / =2.753 mh i F base =L ad /M F i s base =4.5696/ = A e F base =3-phaseVAbase/i Fbase = /2158 = kv Z F base = e F base / i F base = /2158 = L F base = Z F base / base = /377 = mh M F base = L S base i S base / i F base = /2158 =241 mh Then per unit values are: L l = L l /L s base =0.4129/2.753 =0.15 pu R F =R F /Z F base =0.0715/ = pu L ad =4.5696/2.753 =1.66 pu M F =M F /M F base =400/241 =1.66 pu L aq =4.432/2.753 =1.61pu L F =L F /L F base =576.92/ =1.825 pu L d = L l + L ad = =1.81 pu L fd = L F -M R =L F - L ad = =0.165 pu L q = L l + L aq = =1.76 pu R a =0.0031/ =0.003 pu 28

5 Steady state Analysis All derivatives (px) are zero: p r =0 r =1 and L=X in p.u. p fd =0 e fd = R fd i fd p 1d =0 i 1d =0 d = -L d i d +L ad i fd p 1q =0 i 1q =0 q = -L q i q p d =0 e d = q -R a i d =L q i q -R a i d p q =0 e q = d -R a i q = -L d i d +L ad i fd -R a i q Voltage and flux equations: e fd = R fd i fd e d =X q i q R a i d e q = -X d i d +X ad i fd R a i q fd =(L ad +L fd )i fd L ad i d 1d =L ad (i fd i d ) 1q = 2q = L aq i q Single equivalent circuit for both d and q axes: =i d +ji q =e d +je q = X q i q R a i d jx d i d +jx ad i fd jr a i q If saliency is neglected: = (R a + jx q where =j[x ad i fd -(X d -X q )i d ] E q =X ad i fd X d =X q =X s (synchronous reactance) 29

6 Computing per unit steady state values r Active and Reactive Powers S EI t 1p.u. * t ( e j e )( i j i ) d q d q ( ei ei) j( ei ei) P e i e i t d d q q d d qq qd d q where e d r q R a i d e i i R i i 2 2 r( d q q d) a( d q) P R i i 2 2 e a( d q) Q e i e i t q d d q R i q r d a q Rotor angle E i t = (R a + jx q Air gap torque (or electric torque) P T i i e e d q q d r 2 2 = Pt Ra( id iq) 30

7 Representation of Magnetic Saturation Assumptions for stability studies The leakage fluxes are not significantly affected by saturation of the iron portion, so L l is constant and only L ad and L aq saturate in equivalent circuits L adu L ad L aqu L aq where L adu and L aqu denote their unsaturated values The leakage fluxes do not contribute to the iron saturation. Thus, saturation is determined only by the air-gap flux linkage j at ad aq Li j jli d l d q l q at = Saturation relationship at vs. i fd (or MMF) under loaded conditions is the same as under no-load conditions, so only the open-circuit characteristic (OCC) is considered. No magnetic coupling between d and q axes, saturations on L ad and L aq can be modeled individually 31

8 Estimating Saturation Factors K sd and K sd L ad =K sd L adu L aq =K sq L aqu Salient pole machines The path for q-axis flux is largely in air, so L aq does not vary significantly with saturation of the iron portion of the path Assume K sq =1.0 for all loading conditions. Round rotor machines There is a magnetic saturation in both axes, but the saturation data in q axis is usually not available Assume K sq =K sd Thus, we focus on estimating K sd K sd = L ad /L adu = at / at0 = at /( at + I ) = I 0 / I (See Kundur s Example 3.3 on Estimating K sd for different loading conditions) 32

9 Modeling of the Saturation Characteristic I is modeled by 3 approximate functions Segment I ( at < T1 ): I = at0 - at =0 Segment II ( T1 < at < T2 ): Segment III ( at > T2 ): I nonlinear linear I = G2 +L ratio ( at - T2 )- at Note: segments I and II are not connected since when at = T1, I =A sat 0 (usually small) Segments II and III are assumed to be connected at at = T2 to solve G2 = G2 +L ratio ( T2 - T2 )- T2 G2 = T2 + linear Five independent parameters: A sat, B sat, T1, T2 and L ratio 33

10 L ad =K sd L adu K sd = at / at0 = I 0 /I Here L = T1, M = T2 and RS=L ratio 34

11 From Anderson s book S(1.0)= S(1.2)= SV Ae G t G B G ( V 0.8) t 35

12 Example 3.2 in Kundur s Book R 1d >>R fd L fd /R fd >>L 1d /R 1d L 1q /R 1q >>L 2q /R 2q M R Figure

13 37

14 39.1 o 1.565pu 38

15 Sub transient and Transient Analysis Following a disturbance, currents are induced in rotor circuits. Some of these induced rotor currents decay more rapidly than others. Sub transient parameters: influencing rapidly decaying (cycles) components Transient parameters: influencing the slowly decaying (seconds) components Synchronous parameters: influencing sustained (steady state) components 39

16 Transient Phenomena Study transient behavior of a simple RL circuit vt () V sin( t ) ut ( ) m di() t Ri() t L vt () dt Unit step function i(t) Apply Laplace Transform RI(s) LsI [ ( s) i(0)] V( s) Apply Inverse Laplace Transform to I(s) Vs () 1 () s 1 Is () where R 1 s R/ s 1 s L R i / ( ) sin( ) t t Im t Ime sin( ) Steady-state (sinusoidal component) dc offset (transient component), tan Saadat s Example 8.1 R=0.125, L=10mH, V m =151V =L/R=0.08s (time for decaying to 37%) 40