Dynamics of the synchronous machine

Transcription

1 ELEC Power system ynamics, control an stability Dynamics of the synchronous machine Thierry Van Cutsem t.vancutsem@ulg.ac.be These slies follow those presente in course ELEC0029 October / 38

2 Time constants an characteristic inuctances Time constants an characteristic inuctances Objective efine accurately a number of time constants an inuctances characterizing the machine electromagnetic transients the latter appeare in the expression of the short-circuit current of the synchronous machine : see course ELEC0029 use these expressions to erive from measurements the inuctances an resistances of the Park moel Assumption As we focus on electromagnetic transients, the rotor spee θ is assume constant, since it varies much more slowly. 2 / 38

3 Time constants an characteristic inuctances Laplace transform of Park equations V (s) + θ r ψ q (s) V f (s) 0 V q(s) θ r ψ (s) 0 0 = R a + sl sl f sl 1 sl f R f + sl ff sl f1 sl 1 sl f1 R 1 + sl 1 1 +L = }{{} R + sl i (0) i f (0) i 1 (0) R a + sl qq sl qq1 sl qq2 sl qq1 R q1 + sl q1q 1 sl q1q 2 sl qq2 sl q1q 2 R q2 + sl q2q 2 I (s) I f (s) I 1 (s) }{{} R q + sl q i q(0) i q1 (0) i q2 (0) +L q I q (s) I q1 (s) I q2 (s) 3 / 38

4 Time constants an characteristic inuctances Time constants an inuctances Eliminating I f, I 1, I q1 an I q2 yiels: where : V (s) + θ r ψ q (s) = Z (s)i (s) + sg(s)v f (s) V q (s) θ r ψ (s) = Z q (s)i q (s) Z (s) = R a + sl [ ] [ ] 1 [ ] R sl f sl f + sl ff sl f1 slf 1 sl f1 R 1 + sl 1 1 sl 1 = R a + sl (s) l (s) : -axis operational inuctance Z q (s) = R a + sl qq [ ] [ R sl qq1 sl q1 + sl q1q 1 sl q1q 2 qq2 sl q1q 2 R q2 + sl q2q 2 = R a + sl q (s) l q (s) : q-axis operational inuctance ] 1 [ slqq1 sl qq2 ] 4 / 38

5 Time constants an characteristic inuctances Consiering the nature of RL circuits, l (s) an l q (s) can be factorize into: (1 + st )(1 + st l (s) = L (1 + st 0 )(1 + st 0 ) with 0 < T < T 0 < T < T 0 (1 + st q)(1 + st q ) l q (s) = L qq (1 + st q0 )(1 + st q0 ) with 0 < T q < T q0 < T q < T q0 Limit values: lim l (s) = L -axis synchronous inuctance s 0 lim l (s) = L T = L T s T 0 T 0 -axis subtransient inuctance lim l q(s) = L qq q-axis synchronous inuctance s 0 lim l q(s) = L s q = L qq T q T q T q0 T q0 q-axis subtransient inuctance 5 / 38

6 Time constants an characteristic inuctances : Direct erivation of L elimin. of f an 1 R + sl R a + sl (s) s s sl sl elimin. of f an 1 L = L [ ] [ ] 1 [ ] L L f L ff L f1 Lf 1 L f1 L 1 1 L 1 an similarly for the q axis. = L L2 f L L ff L 2 1 2L f L f1 L 1 L ff L 1 1 L 2 f 1 6 / 38

7 Time constants an characteristic inuctances Transient inuctances If amper wining effects are neglecte, the operational inuctances simplify into : l (s) = L 1 + st 1 + st 0 l q (s) = L qq 1 + st q 1 + st q0 an the limit values become : lim l (s) = L T = L s T 0 q T q0 lim l T q(s) = L q = L qq s -axis transient inuctance q-axis transient inuctance Using the same erivation as for L, one easily gets: L = L L2 f L ff L q = L qq L2 qq 1 L q1q 1 7 / 38

8 Time constants an characteristic inuctances Typical values machine with machine with roun rotor salient poles roun rotor salient poles (pu) (pu) (s) (s) L T L q T L T L q T L T q0 2.0 L q T q 0.8 T q T q T α inuctances in per unit on the machine nominal voltage an apparent power 8 / 38

9 Time constants an characteristic inuctances Comments in the irect axis: pronounce time ecoupling : T 0 T 0 T T subtransient time constants T an T 0: short, originate from amper wining transient time constants T an T 0: long, originate from fiel wining in the quarature axis: less pronounce time ecoupling because the winings are of quite ifferent nature! salient-pole machines: single wining in q axis the parameters L q, T q an o not exist. T q0 9 / 38

10 Rotor motion Dynamics of the synchronous machine Rotor motion θ m angular position of rotor, i.e. angle between one axis attache to the rotor an one attache to the stator. Linke to electrical angle θ r through: θ r = p θ m p number of pairs of poles ω m mechanical angular spee: ω m = t θ m ω electrical angular spee: ω = t θ r = pω m Basic equation of rotating masses (friction torque neglecte): I t ω m = T m T e T m T e I moment of inertia of all rotating masses mechanical torque provie by prime mover (turbine, iesel motor, etc.) electromagnetic torque evelope by synchronous machine 10 / 38

11 Motion equation expresse in terms of ω: Rotor motion I p t ω = T m T e Diviing by the base torque T B = S B /ω mb : Defining the spee in per unit: the motion eq. becomes: Defining the inertia constant: the motion eq. becomes: I ω mb ps B I ω 2 mb S B t ω = T mpu T epu ω pu = ω ω N = 1 ω N t θ r t ω pu = T mpu T epu H = 1 2 I ω2 mb S B 2H t ω pu = T mpu T epu 11 / 38

12 Rotor motion Inertia constant H calle specific energy ratio kinetic energy of rotating masses at nominal spee apparent nominal power of machine has imension of a time with values in rather narrow interval, whatever the machine power. H thermal plant p = 1 : 2 4 s p = 2 : 3 7 s hyro plant s 12 / 38

13 Rotor motion Relationship between H an launching time t l t l : time to reach the nominal angular spee ω mb when applying to the rotor, initially at rest, the nominal mechanical torque: T N = P N ω mb = S B cos φ N ω mb P N : turbine nominal power (in MW) cos φ N : nominal power factor Nominal mechanical torque in per unit: Uniformly accelerate motion: T Npu = T N T B = cos φ N ω mpu = ω mpu (0) + cos φ N 2H At t = t l, ω mpu = 1 t l = 2H cos φ N t = cos φ N 2H t Remark. Some efine t l with reference to T B, not T N. In this case, t l = 2H. 13 / 38

14 Compensate motion equation Rotor motion In some simplifie moels, the amper winings are neglecte. To compensate for the neglecte amping torque, a correction term can be ae: 2H t ω pu + D(ω pu ω sys ) = T mpu T epu D 0 where ω sys is the system angular frequency (which will be efine in Power system ynamic moelling uner the phasor approximation ). Expression of electromagnetic torque T e = p(ψ i q ψ q i ) Using the base efine in the slie # 16 : T epu = T e = ω mb ω p(ψ i q ψ q i ) = B (ψ i q ψ q i ) T B S B 3VB 3IB = ψ 3VB i q 3IB ψ q 3VB i 3IB = ψ pu i qpu ψ qpu i pu ω B In per unit, the factor p isappears. ω B 14 / 38

15 Per unit system for the synchronous machine moel Per unit system for the synchronous machine moel Recall on per unit systems Consier two magnetically couple coils with: ψ 1 = L 11 i 1 + L 12 i 2 ψ 2 = L 21 i 1 + L 22 i 2 For simplicity, we take the same time base in both circuits: t 1B = t 2B In per unit: ψ 1pu = ψ 1 ψ 1B = L 11 L 1B i 1 I 1B + L 12 L 1B i 2 I 1B = L 11pu i 1pu + L 12I 2B L 1B I 1B i 2pu ψ 2pu = ψ 2 ψ 2B = L 21I 1B L 2B I 2B i 1pu + L 22pu i 2pu In Henry, one has L 12 = L 21. We request to have the same in per unit: L 12pu = L 21pu I 2B = I 1B S 1B t 1B = S 2B t 2B S 1B = S 2B L 1B I 1B L 2B I 2B A per unit system with t 1B = t 2B an S 1B = S 2B is calle reciprocal 15 / 38

16 Per unit system for the synchronous machine moel in the single phase in each in each rotor circuit equivalent to of the, q wining, stator winings winings for instance f time t B = 1 ω N = 1 2πf N power S B = nominal apparent 3-phase voltage V B : nominal (rms) phase-neutral 3VB V fb : to be chosen current impeance I B = S B 3V B 3IB S B V fb Z B = 3V 2 B S B 3V 2 B VfB 2 S B S B flux V B t B 3VB t B V fb t B 16 / 38

17 Per unit system for the synchronous machine moel The equal-mutual-flux-linkage per unit system For two magnetically couple coils, it is shown that (see theory of transformer): L 11 L l1 = n2 1 R L 12 = n 1n 2 R L 22 L l2 = n2 2 R L 11 self-inuctance of coil 1 L 22 self-inuctance of coil 1 L l1 leakage inuctance of coil 1 L l2 leakage inuctance of coil 2 n 1 number of turns of coil 1 n 2 number of turns of coil 2 R reluctance of the magnetic circuit followe by the magnetic fiel lines which cross both winings; the fiel is create by i 1 an i / 38

18 Assume we choose V 1B an V 2B such that: Per unit system for the synchronous machine moel V 1B V 2B = n 1 n 2 In orer to have the same base power in both circuits: We have: V 1B I 1B = V 2B I 2B I 1B I 2B = n 2 n 1 (L 11 L l1 )I 1B = n2 1 R I 1B = n2 1 n 2 I 2B = n 1n 2 R n 1 R I 2B = L 12 I 2B (1) The flux create by I 2B in coil 1 is equal to the flux create by I 1B in the part of coil 1 crosse by the magnetic fiel lines common to both coils. Similarly in coil 2: (L 22 L l2 )I 2B = n2 2 R I 2B = n2 2 n 1 I 1B = n 1n 2 R n 2 R I 1B = L 12 I 1B (2) This per unit system is sai to yiel equal mutual flux linkages (EMFL) 18 / 38

19 Alternative efinition of base currents Per unit system for the synchronous machine moel From respectively (1) an (2) : I 1B I 2B = L 12 L 11 L l1 I 1B I 2B = L 22 L l2 L 12 A property of this pu system L 12pu = L 12I 2B = (L 11 L l1 ) = L 11pu L l1pu L 1B I 1B L 1B L 21pu = L 21I 1B = (L 22 L l2 ) = L 22pu L l2pu L 2B I 2B L 2B In this pu system, self-inuctance = mutual inuctance + leakage reactance. Does not hol true for inuctances in Henry! The inuctance matrix of the two coils takes on the form: [ ] [ L11 L L = 12 Ll1 + M M = L 12 L 22 M L l2 + M ] 19 / 38

20 Per unit system for the synchronous machine moel Application to synchronous machine we have to choose a base voltage (or current) in each rotor wining. Let s first consier the fiel wining f (1 f, 2 ) we woul like to use the EMFL per unit system we o not know the number of turns of the equivalent circuits f,, etc. instea, we can use one of the alternative efinitions of base currents: I fb 3IB = L L l L f I fb = 3I B L L l L f L, L l can be measure L f can be obtaine by measuring the no-loa voltage E q prouce by a known fiel current i f : E q = ω NL f 3 i f L f = the base voltage is obtaine from V fb = S B I fb 3Eq ω N i f (3) 20 / 38

21 Per unit system for the synchronous machine moel What about the other rotor winings? one cannot access the 1, q 1 an q 2 winings to measure L 1, L qq1 et L qq2 using formulae similar to (3) one may assume there exist base currents I 1B, I q1b et I q2b leaing to the EMFL per unit system, but their values are not known hence, we cannot compute voltages in Volt or currents in Ampere in those winings (only in pu) not a big issue in so far as we o not have to connect anything to those winings (unlike the excitation system to the fiel wining) / 38

22 Dynamic equivalent circuits of the synchronous machine Dynamic equivalent circuits of the synchronous machine In the EMFL per unit system, the Park inuctance matrices take on the simplifie form: L l + M M M L = M L lf + M M M M L l1 + M L q = L l + M q M q M q M q L lq1 + M q M q M q M q L lq2 + M q For symmetry reasons, same leakage inuctance L l in an q winings 22 / 38

23 Dynamic equivalent circuits of the synchronous machine 23 / 38

24 Exercises Dynamics of the synchronous machine Exercises Exercise No. 1 A machine has the following characteristics: nominal frequency: 50 Hz nominal apparent power: 1330 MVA stator nominal voltage: 24 kv X = 0.9 Ω (value of per phase equivalent) X l = Ω (value of per phase equivalent) fiel current giving the nominal stator voltage at no-loa: 2954 A choose the base power, voltage an current in the stator winings choose the base power, voltage an current in the fiel wining, using the EMFL per unit system compute X, X l an L f in per unit. 24 / 38

25 Exercises Exercise No. 2 A 1330 MVA, 50 Hz machine has the following characteristics (values in pu on the machine base): X l = 0.20 pu X = 2.10 pu = 0.30 pu X X X = 0.25 pu X q R a = pu X q = 2.10 pu q = 0.73 pu = pu T o = 0.03 s T qo = 0.20 s T o = 9.10 s T qo = 2.30 s Determine the inuctances an resistances of the Park moel, using the EMFL per unit system. Check your answers by computing X an X q from the Park inuctance matrices. Hints: time constants must be converte in per unit! ientify first the parameters of the equivalent circuits. 25 / 38

26 Moelling of material saturation Moelling of material saturation Saturation of magnetic material moifies: the machine inuctances the initial operating point (in particular the rotor position) the fiel current require to obtain a given stator voltage. Notation parameters with the upperscript u refer to unsaturate values parameters without this upperscript refer to saturate values. 26 / 38

27 Open-circuit magnetic characteristic Moelling of material saturation Machine operating at no loa, rotating at nominal angular spee ω N. Terminal voltage E q measure for various values of the fiel current i f. saturation factor : k = OA OB = O A O A < 1 a stanar moel : k = m(e q ) n m, n > 0 characteristic in axis (fiel ue to i f only) In per unit: E qpu = ω NL f i f 3 VB = ω N Z B (L L l )i fpu = M pu i fpu Dropping the pu notation an introucing k: = ω NL f I fb i fpu = ω NL f L L l 3IB 3 VB 3 VB L f i fpu E q = M i f = k M u i f 27 / 38

28 Leakage an air gap flux Dynamics of the synchronous machine Moelling of material saturation The flux linkage in the wining is ecompose into: ψ = L l i + ψ a L l i : leakage flux, not subject to saturation (path mainly in the air) ψ a : irect-axis component of the air gap flux, subject to saturation. Expression of ψ a : ψ a = ψ L l i = M (i + i f + i 1 ) Expression of ψ aq : ψ aq = ψ q L l i q = M q (i q + i q1 + i q2 ) Consiering that the an q axes are orthogonal, the air gap flux is given by: ψ ag = ψa 2 + ψ2 aq (4) 28 / 38

29 Moelling of material saturation Saturation characteristic in loae conitions Saturation is ifferent in the an q axes, especially for a salient pole machine (air gap larger in q axis!). Hence, ifferent saturation factors (say, k an k q ) shoul be consiere in practice, however, it is quite common to have only the irect-axis saturation characteristic in this case, the characteristic is use along any axis (not just ) as follows in no-loa conitions, we have ψ a = M i f an ψ aq = 0 ψ ag = M i f M = km u = M u 1 + m(e q ) n = M u 1 + m(m i f ) n = M u 1 + m(ψ ag ) n it is assume that this relation still hols true with the combine air gap flux ψ ag given by (4). 29 / 38

30 Complete moel (in per unit) Moelling of material saturation ψ = L l i + ψ a ψ f = L lf i f + ψ a ψ 1 = L l1 i 1 + ψ a ψ a = M (i + i f + i 1 ) ψ q = L l i q + ψ aq ψ q1 = L lq1 i q1 + ψ aq ψ q2 = L lq2 i q2 + ψ aq ψ aq = M q (i q + i q1 + i q2 ) M = 1 + m M u ( ψ 2 a + ψ2 aq ) n M q = 1 + m M u q ( ψ 2 a + ψ2 aq ) n v = R a i ωψ q ψ t t ψ f = v f R f i f t ψ 1 = R 1 i 1 2H t ω = T m (ψ i q ψ q i ) v q = R a i q + ωψ ψ q t t ψ q1 = R q1 i q1 t ψ q2 = R q2 i q2 1 ω N t θ r = ω 30 / 38

31 Moel simplifications Moel simplifications Constant rotor spee approximation θr ω N (ω = 1 pu) Examples showing that θ r oes not epart much from the nominal value ω N : 1 oscillation of θ r with a magnitue of 90 o an perio of 1 secon superpose to the uniform motion at synchronous spee: θ r = θ o r + 2πf N t + π 2 sin 2πt θ r = 100π + π 2 cos 2πt cos 2πt at its maximum, it eviates from nominal by 10/314 = 3 % only. 2 in a large interconnecte system, after primary frequency control, frequency settles at f f N. f f N = 0.1 Hz is alreay a large eviation. In this case, machine spees eviate from nominal by 0.1/50 = 0.2 % only. 3 a small isolate system may experience larger frequency eviations. But even for f f N = 1 Hz, the machine spees eviate from nominal by 1/50 = 2 % only. 31 / 38

32 Moel simplifications The phasor (or quasi-sinusoial) approximation Unerlies a large class of power system ynamic simulators consiere in etail in the following lectures for the synchronous machine, it consists of neglecting the transformer voltages ψ an ψ in the stator Park equations t t this leas to neglecting some fast varying components of the network response, an allows the voltage an currents to be treate as a sinusoial with time-varying amplitues an phase angles (hence the name) at the same time, three-phase balance is also assume. Thus, the stator Park equations become (in per unit, with ω = 1): v = R a i ψ q v q = R a i q + ψ an ψ an ψ q are now algebraic, instea of ifferential, variables. Hence, they may unergo a iscontinuity after of a network isturbance. 32 / 38

33 Exercise No. 3 Exercise No. 3 Show that the moel of slie No. 42, uner the phasor approximation of slie No. 44, can be written in the form: with the vector of ifferential states: the vector of algebraic states: an the vector of inputs : x t = f(x, y, u) 0 = g(x, y, u) x = [ψ f ψ 1 ψ q1 ψ q2 θ ω] T, y = [i i q ψ a ψ aq ] T, u = [v v q v f T m ] T 33 / 38

34 The classical moel of the synchronous machine The classical moel of the synchronous machine Very simplifie moel use : in some analytical evelopments in qualitative reasoning ealing with transient (angle) stability for fast assessment of transient (angle) stability. Classical refers to a moel use when there was little computational power. Approximation # 0. We consier the phasor approximation. Approximation # 1. The amper winings 1 et q 2 are ignore. The amping of rotor oscillations is going to be unerestimate. Approximation # 2. The stator resistance R a is neglecte. This is very acceptable. The stator Park equations become : v = ψ q = L qq i q L qq1 i q1 v q = ψ = L i + L f i f 34 / 38

35 The classical moel of the synchronous machine Expressing i f (resp. i q1 ) as function of ψ f an i (resp. ψ q1 an i q ) : ψ f = L ff i f + L f i i f = ψ f L f i L ff ψ q1 = L q1q1 i q1 + L qq1 i q1 i q1 = ψ q1 L qq1 i q L q1q1 an introucing into the stator Park equations : v = (L qq L2 qq1 ) i q L qq1 ψ q1 = X q i q + e (5) L q1q1 L q1q1 }{{}}{{} L q e v q = (L L2 f ) i + L f ψ f = X i + e q (6) L }{{ ff L }} ff {{} L e q e an e q : are calle the e.m.f. behin transient reactances are proportional to magnetic fluxes; hence, they cannot vary much after a isturbance, unlike the rotor currents i f an i q1. 35 / 38

36 The classical moel of the synchronous machine Approximation # 3. The e.m.f. e an e q are assume constant. This is vali over no more than - say - one secon after a isturbance; over this interval, a single rotor oscillation can take place; hence, amping cannot show its effect (i.e. Approximation # 1 is not a concern). Equations (5, 6) are similar to the Park equations in steay state, except for the presence of an e.m.f. in the axis, an the replacement of the synchronous by the transient reactances. Approximation # 4. The transient reactance is the same in both axes : X = X q. Questionable, but experiences shows that X q has less impact... If X = X q, Eqs. (5, 6) can be combine in a single phasor equation, with the corresponing equivalent circuit: V + jx Ī = Ē = E δ 36 / 38

37 Rotor motion. This is the only ynamics left! The classical moel of the synchronous machine e an e q are constant. Hence, Ē is fixe with respect to an q axes, an δ iffers from θ r by a constant. Therefore, 1 ω N t θ r = ω can be rewritten as : 1 ω N t δ = ω The rotor motion equation: 2H t ω = T m T e is transforme to involve powers instea of torques. Multiplying by ω: 2H ω t ω = ωt m ωt e ωt m = mechanical power P m of the turbine ωt e = p r s = p T (t) + p Js + W ms P (active power prouce) t since we assume three-phase balance AC operation, an R a is neglecte 37 / 38

38 The classical moel of the synchronous machine Approximation # 5. We assume ω 1 an replace 2Hω by 2H very acceptable, alreay justifie. Thus we have: 2H t ω = P m P where P can be erive from the equivalent circuit: P = E V X sin(δ θ) 38 / 38