From now, we ignore the superbar - with variables in per unit. ψ ψ. l ad ad ad ψ. ψ ψ ψ

Transcription

1 From now, we ignore the superbar - with variables in per unit. ψ 0 L0 i0 ψ L + L L L i d l ad ad ad d ψ F Lad LF MR if = ψ D Lad MR LD id ψ q Ll + Laq L aq i q ψ Q Laq LQ iq 41

2 Equivalent Circuits for d- and q-axes e d = -ω r ψ q -R a i d + pψ d = -ω r ψ q -R a i d +L ad d(-i d +i F +i D )/dt -L l di d /dt e F = R F i F + pψ F = R F i F -L ad di d /dt +L F di F /dt +M R di D /dt e D =0= R D i D +pψ D =R D i D -L ad di d /dt +M R di F /dt +L D di D /dt L l +L ad e q = ω r ψ d -R a i q + pψ q =ω r ψ d -R a i q +L aq d(-i q +i Q )/dt -L l di q /dt e Q =0=R Q i Q +pψ Q =R Q i Q -L aq di q /dt + L Q di Q /dt L l +L aq 42

3 Equivalent Circuits for d- and q-axes e d = L ad d(-i d +i F +i D )/dt -L l di d /dt -ω r ψ q -R a i d e F = -L ad di d /dt +L F di F /dt +M R di D /dt +R F i F i D +i F = L ad d(-i d +i F +i D )/dt +(M R - L ad ) d(i D +i F ) /dt +(L F - M R ) di F /dt + R F i F e D =0= -L ad di d /dt +M R di F /dt +L D di D /dt+ R D i D = L ad d(-i d +i F +i D )/dt +(M R - L ad ) d(i D +i F ) +(L D - M R ) di D /dt+ R D i D e q = L aq d(-i q +i Q )/dt -L l di q /dt +ω r ψ d -R a i q e Q =0= -L aq di q /dt + L Q di Q /dt +R Q i Q = L aq d(-i q +i Q )/dt +(L Q -L aq ) di Q /dt +R Q i Q 43

4 Equivalent Circuits with Multiple Damper Windings (e.g. Round-rotor Machines) d axis: L fd L F - M R R fd R F e fd e F L 1d (or L kd1 ) L D - M R R 1d (or R kd1 ) R D M R -L ad 0 ψ fd ψ F (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 1q (or L kq1 ) L Q L aq R 1q (or R kq1 ) R Q L 2q (or L kq2 ) L G - L aq R 2q (or R kq2 ) R G L 1d =L D - M R R 1d =R D L 1q =L Q L aq R 1q =R Q L fd =L F - M R R fd =R F e fd =e F L 2q =L G L aq R 2q =R G 44

5 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 45

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7 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(2θ) mh l ab = cos(2θ+π/3) mh l af =40.0 cosθ mh L F = mh R a = Ω R F = Ω 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 cos2θ = cos2θ mh l ab = -M s - L m cos(2θ+π/3) = cos(2θ+π/3) mh l af = M F cosθ =40.0 cosθ 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 47

8 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 = /( ) = Ω ω base =2π 60 =377 elec. rad/s L s base =Z s base /ω 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 = Ω L F base = Z F base / ω 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 =0.4129/2.753 =0.15 pu M F =M F /M F base =400/241 =1.66 pu L ad =4.5696/2.753 =1.66 pu L F =576.92/ =1.825 pu L aq =4.432/2.753 =1.61pu L fd = L F -M R =L F -L ad = =0.165 pu L d = L l + L ad = =1.81 pu R a =0.0031/ =0.003 pu L q = L l + L aq = =1.76 pu R F =0.0715/ = pu 48

9 Steady-state Analysis All flux linkages, voltages and currents are constant: pψ 1d =0 R 1d i 1d =0 i 1d =0 pψ 1q =0 R 1q i 1q =0 i 1q =0 (Damper winding currents are all zero due to no change in the magnetic field) ψ d = -( L l +L ad )i d +L ad i fd +L ad i 1d = -L d i d +L ad i fd ψ q = -( L l +L aq )i q +L aq i 1q = -L q i q pψ d =0 e d = -ω r ψ q -R a i d =ω r L q i q -R a i d pψ q =0 e q =ω r ψ d -R a i q =-ω r L d i d +ω r L ad i fd -R a i q pψ fd =0 e fd = R fd i fd ω r =1 and L=X in p.u. e d =X q i q -R a i d e q =-X d i d +X ad i fd -R a i q i fd =(e q + R a i q +X d i d )/X ad Can we have a single equivalent circuit representing both d and q axes circuits under balanced steady-state conditions? 49

10 e a = E m cos(ω s t+α) e b = E m cos(ω s t +α-2π/3) e c = E m cos(ω s t +α+2π/3) θ=ω r t+θ 0 e d = E m cos(α -θ 0 )= E t cos(α -θ 0 ) e q = E m sin(α -θ 0 ) = E t sin(α -θ 0 ) where E t is the per unit RMS value of the armature terminal voltage, which equals the peak value E m in per unit. E t =e d +je q I t=i d +ji q e d =X q i q -R a i d e q =-X d i d +X ad i fd -R a i q E t = X q i q -R a i d +-jx d i d +jx ad i fd -jr a i q = -R a I t +X q i q -jx d i d +jx ad i fd E t = E R a I t jx S I t X S =X q E =j[x ad i fd -(X d -X q )i d ] E q E E I t E t Load E t =E q R a I t jx q I t 50

11 Steady-state equivalent circuit Under no-load or open-circuit conditions, i d =i q =0. E t = E q =jx ad i fd δ i =0 (load angle) E q =j[x ad i fd -(X d -X q )i d ] For round rotor machines or machines with neglected saliency X d =X q =X s (synchronous reactance) E t =E q (R a +jx s )I t E q =X ad i fd 51

12 Computing steady-state values 52

13 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 L adu and L aqu denote the unsaturated values The leakage fluxes do not contribute to the iron saturation. Thus, saturation is determined only by the air-gap flux linkage ψ at = ψ 2 2 aa + ψ aa The saturation relationship ψ at ~ i fd or ψ at ~ MMF under loaded conditions is the same as under no-load conditions. This allows the open-circuit characteristic (OCC) to be considered No magnetic coupling between d and q axes, such that their effects of saturation can be modeled individually 53

14 Air-gap flux and voltage ψ = ψ + jψ at ad aq = ψ + Li + jψ + jli d l d q l q = eq + Raiq jed jraid + LlI t = je jr I + L I t a t l t = j[ E + ( R + jx ) I ] = je t a l t at For steady state conditions: e d = -ψ q -R a i d ψ q = - e d -R a i d e q = ψ d - R a i q ψ d = e q + R a i q ψ = E = E + ( R + jx ) I at at t a l t 54

15 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 q axis saturation data is usually not available Assume K sq =K sd Thus, we focus on estimating K sd K sd = L ad /L adu =ψ at /ψ at0 = ψ at /(ψ at + ψ I ) = I 0 / I (See Kundur s Example 3.3 on Estimating K sd for different loading conditions) 55

16 Estimating the Saturation Characteristic Modeled by 3 approximate functions Segment I (ψ at <ψ T1 ): ψ I =ψ at0 - ψ at =0 Segment II (ψ T1 <ψ at <ψ T2 ): ψ I = A sss e B sss(ψ aa ψ TT ) Segment III (ψ at >ψ T2 ): ψ I nonlinear linear ψ I =ψ G2 +L ratio (ψ at -ψ T2 )- ψ at Note that segments I and II are not connected since when ψ at =ψ T1, ψ I =A sat 0 (usually small) Assume that segments II and III are connected at ψ at =ψ T2 to solve ψ G2 A sss e B sss(ψ TT ψ TT ) =ψ G2 +L ratio (ψ T2 -ψ T2 )- ψ T2 ψ G2 = ψ T2 +A sss e B sss(ψ TT ψ TT ) linear We need to know A sat, B sat, ψ T1, ψ T2 and L ratio 56

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18 P186 of Anderson s book S(1.0) S(1.2) ψ I = A sss e B sss(ψ aa ψ TT ) ~ S G = Ae G B G ( V 0.8) t 58

19 Example 3.2 in Kundur s Book R 1d >>R fd L fd /R fd >>L 1d /R 1d M R L 1q /R 1q >>L 2q /R 2q 59

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21 39.1 o 1.565pu 61

22 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 62

23 Transient and sub-transient parameters R 1d >>R fd L fd /R fd >>L 1d /R 1d d axis circuit L 1q /R 1q >>L 2q /R 2q q axis circuit Considered rotor windings Time constant (open circuit) Time constant (short circuit) T d0 = 8.07(s) T d = Only field Winding L aa +L ff R ff L aa //L l + L ff R ff Add the damper winding Only 1 st damper winding T d = L aa //L ff //L l + L 1d T q = L aa //L l + L 1q T q = R 1d R 1q Add the 2 nd damper winding L T d0 = aa //L ff + L 1d T q0 = L aa +L 1q T q0 = L aa //L 1q + L 2q R 1d R 1q R 2q 0.03(s) 1.00(s) 0.07(s) L aa //L 1q //L l + L 2q R 2q Inductance (Reactance) L d = L l +L ad //L fd L d = L l +L ad //L fd //L 1d L q = L L q = l +L aq //L 1q L l +L aq //L 1q //L 2q 0.30(pu) 0.23(pu) 0.65(pu) 0.25(pu) Based on the parameters of Example

24 Identifying Terminal Quantities Apply Laplace transform to the incremental forms of flux linkage equations (see Kundur s Chapter 4.1 for details): ψ d (s) = G(s) e fd (s) L d (s) i d (s) ψ q (s) = L q (s) i q (s) L d (1 + st4)(1 + st6) (1 + st )(1 + st ) 1 3 G 0 (1 + st kd ) (1 + st )(1 + st ) 1 3 Since R 1d >>R fd T 2, T 3 <<T 1, T 5,T 6 <<T 4 >> T 1 +T 2 T 1 +T 3, T 4 +T 5 T 4 +T 6 >> >> (1 + st )(1 + st ) (1 + st )(1 + st ) d d 4 6 (1 + st )(1 + st ) (1 + st )(1 + st ) d0 d (Classical Expressions)

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26 Summary of Transient and Sub-transient Parameters (Classical Expressions) d axis circuit q axis circuit OC Time Constant SC Time Constant T d0 = L aa +L ff R ff T d = L aa //L l + L ff R ff T d0 = T q0 = T q0 = L aa //L ff + L L aa +L 1q 1d R R 1q L aa //L 1q + L 2q 1d R 2q T d = T q = T q = L aa //L ff //L l + L L aa //L l + L 1q 1d L R R aa //L 1q //L l + L 2 1q 1d R 2q Inductance (Reactance) L d = L d = L l +L ad //L fd L l +L ad //L fd //L 1d L q = L q = L l +L aq //L 1q L l +L aq //L 1q //L 2q Note: time constants are all in p.u. To be converted to seconds, they have to be multiplied by t base =1/ω base (i.e. 1/377 for 60Hz). 66

27 Synchronous, Transient and Sub-transient Inductances L d ( s) = L d (1 + st d )(1 + (1 + st )(1 + st d ) st ) d 0 d 0 Under steady-state condition: s=0 (t ) L d (0)=L d (d-axis synchronous inductance) During a rapid transient: s TT L = L ( ) = L = L + L L L d d ad fd 1d d d d l T d 0T d 0 Lad Lfd + Lad L1 d + Lfd L1 d (d-axis sub-transient inductance) Without the damper winding :s>>1/t d and 1/T d0 but << 1/T d and 1/T d0 T L = L ( ) = L = L + L L d ad fd d d d l T d 0 Lad + Lfd (d-axis transient inductance) 67

28 X d X q X q X d X q X d T d0 > T d > >T d0 > T d T q0 > T q > >T q0 > T q 68

29 Parameter Estimation by Frequency Response Tests ψ d (s) = G(s) e fd (s) L d (s) i d (s) T d0 > T d > >T d0 > T d > T kd 1/T d0 <1/ T d << 1/T d0 <1/T d <1/T kd Bode Plot -20dB/decade 69

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31 Swing Equations J ω m = Ta = Tm T J combined moment of inertia of generator e and turbine, kg m 2 ω m angular velocity of the rotor, mech. rad/s t time, s T a accelerating torque in N.m T m mechanical torque in N.m electromagnetic torque in N.m d dt T e Define per unit inertia constant H = 2 1 Jω 2 VA 0m base (s) (s) J = 2H ω 0 2 VA base m Some references define T M or M=2H, called the mechanical starting time, i.e. the time required for rated torque to accelerate the rotor from standstill to rated speed 71

32 dωm J = T = T T dt 2H dωm VA = T T 2 ω dt 0m a m e base m e d ω m Tm Te 2H = dt ω VA / 0m base ω0m d ω m Tm Te 2H = dt ω 0m T base dωr 2 H = Tm Te (in per unit) dt where ω ωr / p m f ωr ω r (in per unit) = = = ω ω / p ω 0m 0 f 0 Angular position of the rotor in electrical radian with respect to a synchronously rotating reference δ = ω ω t+ δ in rad r t 0 0 dδ = ωr ω0 = ωr in rad/s dt d d δ dω d( ωr) = = in rad/s 2 2 r 2 dt dt dt δ d( ω ) d( ωr) = = in rad/s dt r ω0 dt ω0 dt ω ω 1 ω r r = = ω0 0 dδ dt If adding a damping term proportional to speed deviation: 2 2Hdδ = T 2 m T ω dt 0 2 ω 72 e Hd 2 δ D = T - = - 2 m Te K K d D ωr Tm Te 0 dt ω0 δ dt

33 Block diagram representation of swing equations 2 ω Hd 2 δ D = T - = - 2 m Te K K d D ωr Tm Te 0 dt ω0 d( ωr ) 2H = Tm Te -KD ωr dt ω = r 1 ω 0 dδ dt δ dt 73

34 State-Space Representation of a Synchronous Machine So far, we modeled all critical dynamics about a synchronous machine: State variables (px): stator and rotor voltages, currents or flux linkages swing equations (rotor angle and speed) Time constants: Inertia: 2H Sub-transient and transient time constants, e.g. T d0 and T d0 Other parameters Stator and rotor self- or mutual-inductances and resistances Rotor mechanical torque T m and stator electromagnetic torque T e 74

35 Differential Equations on a Salient-pole Machine Consider 5 windings (d, F, D, q and Q) Voltage equations: e=r i+l di/dt+ω r N I pi = - L -1 (R+ω r N) i +L -1 e Swing equations: Since ψ=l i, here is pψ= - (R+ω r N)L -1 ψ+e τ pω r =T m -T e (where τ=2h/ω 0 ) pδ=ω r - ω 0 =ω r - 1 T e = ψ d i q ψ q i d ψ d = L d i d + km F i F + km D i D, ψ q = L q i q + km Q i Q T e = L d i q i d + km F i q i F + km D i q i D + L q i d i q km Q i d i Q 75

36 d State-space Model i i i i i ω r δ 0 0 i 0 0 d i L R+ ω N r i D Le 0 0 = (T d0, T d0, T q0, T q0 ) iq i Q L i km i km i L i km i d q F q D q q d Q d / 0 0 ω r T τ m τ τ τ τ τ δ d F 1 F D / dt q Q Thus, the state-space model of a synchronous machine: dx/dt=f(x, e d, e q and e fd, T m ) where x=[i d i Fd i 1d i q i 1q ω r δ] T or [ψ d ψ fd ψ 1d ψ q ψ 1q ω r δ] T is the state vector. (See the section 4.12 in Anderson s book). e fd and T m are usually known but e d and e q are related to state variables, so additional equations (on the machine outside, e.g. loading conditions) should be introduced 76

37 Simplified Models [ψ d ψ fd ψ 1d ψ q ψ 1q ω r δ] T Neglect pψ d, pψ q and variations of ω r (i.e. ω r =1 pu) in the voltage equations =ψ q = -L q i q [ψ fd ψ 1d ψ 1q ω r δ] T Inertia τ ~ pω r Transient T d0 ~ pψ fd Sub-transient T d0 ~pψ 1d T q0 ~ pψ 1q [ψ fd ω r δ] T Neglect damper windings, i.e. pψ 1d and pψ 1q =ψ d = -L d i q Inertia τ ~ pω r Transient T d0 ~ pψ fd Constant flux linkage assumption [ω r δ] T (classic model) Inertia τ ~ pω r 77

38 Neglect of Stator pψ terms 78

39 Neglecting Damper Windings ψ d = - L d i d + L ad i fd ψ q = - L q i q ψ fd = - L ad i d + (L ad +L fd ) i fd pψ fd = -e fd - R fd i fd (only differential equation left for the equivalent circuits) T d0 pe q = E fd E q (L d L d )i d where E q = L aa L F ψ fd and E fd = L aa R ff e fd T q0 pe d = E d (L q L q )i q (added for round-rotor machines to give the so called Two-Axis Model ) X d =L l +L ad //L fd X q =L l +L aq 79

40 Classic Model Eliminate the differential equations on flux linkages (swing equations are the only differential equations left) Assume X d =X q 2H pω ω r =T m -T e 0 pδ=ω r -1 ~ E t ~ = E ( R a + ~ jx ) I d t E = E + ( R + jx ) I t0 a d t0 E is constant and can be estimated by computing its pre-disturbance value 80

41 Simplified models neglecting pψ and saliency effects Used in short-circuit analysis Used for stability studies Used for steady-state analysis 81

42 Comparison of PSS/E Generator Models 82