Dynamics of the synchronous machine

ELEC0047 - Power system dynamics, control and stability Dynamics of the synchronous machine Thierry Van Cutsem t.vancutsem@ulg.ac.be www.montefiore.ulg.ac.be/~vct October 2018 1 / 38

Time constants and characteristic inductances Time constants and characteristic inductances Objective define accurately a number of time constants and inductances characterizing the machine electromagnetic transients the latter appeared in the expression of the short-circuit current of the synchronous machine : see course ELEC0029 use these expressions to derive from measurements the inductances and resistances of the Park model Assumption As we focus on electromagnetic transients, the rotor speed θ is assumed constant, since it varies much more slowly. 2 / 38

Time constants and characteristic inductances Laplace transform of Park equations V d(s) + θ r ψ q (s) V f (s) 0 V q(s) θ r ψ d (s) 0 0 = R a + sl dd sl df sl dd1 sl df R f + sl ff sl fd1 sl dd1 sl fd1 R d1 + sl d1d 1 +L d = }{{} R d + sl d i d (0) i f (0) i d1 (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 d(s) I f (s) I d1 (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

Time constants and characteristic inductances Time constants and inductances Eliminating I f, I d1, I q1 and I q2 yields: where : V d (s) + θ r ψ q (s) = Z d (s)i d (s) + sg(s)v f (s) V q (s) θ r ψ d (s) = Z q (s)i q (s) Z d (s) = R a + sl dd [ ] [ ] 1 [ ] R sl df sl f + sl ff sl fd1 sldf dd1 sl fd1 R d1 + sl d1d 1 sl dd1 = R a + sl d (s) l d (s) : d-axis operational inductance 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 inductance ] 1 [ slqq1 sl qq2 ] 4 / 38

Time constants and characteristic inductances Considering the nature of RL circuits, l d (s) and l q (s) can be factorized into: (1 + st d )(1 + st d l d (s) = L dd (1 + st d0 )(1 + st d0 ) with 0 < T d < T d0 < T d < T d0 (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 d(s) = L dd d-axis synchronous inductance s 0 lim l d(s) = L T d d = L T d dd s T d0 T d0 d-axis subtransient inductance lim l q(s) = L qq q-axis synchronous inductance s 0 lim l q(s) = L s q = L qq T q T q T q0 T q0 q-axis subtransient inductance 5 / 38

Time constants and characteristic inductances : Direct derivation of L d elimin. of f and d 1 R d + sl d R a + sl d (s) s s sl d sl d elimin. of f and d 1 L d = L dd [ ] [ ] 1 [ ] L L df L ff L fd1 Ldf dd1 L fd1 L d1d 1 L dd1 and similarly for the q axis. = L dd L2 df L d 1d 1 + L ff L 2 dd 1 2L df L fd1 L dd1 L ff L d1d 1 L 2 fd 1 6 / 38

Time constants and characteristic inductances Transient inductances If damper winding effects are neglected, the operational inductances simplify into : l d (s) = L dd 1 + st d 1 + st d0 l q (s) = L qq 1 + st q 1 + st q0 and the limit values become : lim l d(s) = L T d d = L dd s T d0 q T q0 lim l T q(s) = L q = L qq s d-axis transient inductance q-axis transient inductance Using the same derivation as for L d, one easily gets: L d = L dd L2 df L ff L q = L qq L2 qq 1 L q1q 1 7 / 38

Time constants and characteristic inductances Typical values machine with machine with round rotor salient poles round rotor salient poles (pu) (pu) (s) (s) L d 1.5-2.5 0.9-1.5 T d0 8.0-12.0 3.0-8.0 L q 1.5-2.5 0.5-1.1 T d 0.95-1.30 1.0-2.5 L d 0.2-0.4 0.3-0.5 T d0 0.025-0.065 0.025-0.065 L q 0.2-0.4 T d 0.02-0.05 0.02-0.05 L d 0.15-0.30 0.25-0.35 T q0 2.0 L q 0.15-0.30 0.25-0.35 T q 0.8 T q0 0.20-0.50 0.04-0.15 T q 0.02-0.05 0.02-0.05 T α 0.02-0.60 0.02-0.20 inductances in per unit on the machine nominal voltage and apparent power 8 / 38

Time constants and characteristic inductances Comments in the direct axis: pronounced time decoupling : T d0 T d0 T d T d subtransient time constants T d and T d0: short, originate from damper winding transient time constants T d and T d0: long, originate from field winding in the quadrature axis: less pronounced time decoupling because the windings are of quite different nature! salient-pole machines: single winding in q axis the parameters L q, T q and do not exist. T q0 9 / 38

Rotor motion Dynamics of the synchronous machine Rotor motion θ m angular position of rotor, i.e. angle between one axis attached to the rotor and one attached to the stator. Linked to electrical angle θ r through: θ r = p θ m p number of pairs of poles ω m mechanical angular speed: ω m = d dt θ m ω electrical angular speed: ω = d dt θ r = pω m Basic equation of rotating masses (friction torque neglected): I d dt ω m = T m T e T m T e I moment of inertia of all rotating masses mechanical torque provided by prime mover (turbine, diesel motor, etc.) electromagnetic torque developed by synchronous machine 10 / 38

Motion equation expressed in terms of ω: Rotor motion I d p dt ω = T m T e Dividing by the base torque T B = S B /ω mb : Defining the speed in per unit: I ω mb ps B d dt ω = T mpu T epu ω pu = ω ω N = 1 ω N d dt θ r and taking ω mb = ω B /p = ω N /p, the motion equation becomes: Defining the inertia constant: I ω 2 mb S B the motion equation is rewritten as: d dt ω pu = T mpu T epu H = 1 2 I ω2 mb S B 2H d dt ω pu = T mpu T epu 11 / 38

Rotor motion Inertia constant H called specific energy ratio kinetic energy of rotating masses at nominal speed apparent nominal power of machine has dimension 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 hydro plant 1.5 3 s 12 / 38

Rotor motion Relationship between H and launching time t l t l : time to reach the nominal angular speed 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 accelerated 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 define t l with reference to T B, not T N. In this case, t l = 2H. 13 / 38

Compensated motion equation Rotor motion In some simplified models, the damper windings are neglected. To compensate for the neglected damping torque, a correction term can be added: 2H d dt ω pu + D(ω pu ω sys ) = T mpu T epu D 0 where ω sys is the system angular frequency (which will be defined in Power system dynamic modelling under the phasor approximation ). Expression of electromagnetic torque Using the base defined in slide # 16 : T e = p(ψ d i q ψ q i d ) T epu = T e = ω mb ω p(ψ d i q ψ q i d ) = B (ψ d i q ψ q i d ) T B S B 3VB 3IB = ψ d 3VB i q 3IB ψ q 3VB i d 3IB = ψ dpu i qpu ψ qpu i dpu ω B In per unit, the factor p disappears. ω B 14 / 38

Per unit system for the synchronous machine model Per unit system for the synchronous machine model Recall on per unit systems Consider two magnetically coupled 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 and S 1B = S 2B is called reciprocal 15 / 38

Per unit system for the synchronous machine model in the single phase in each in each rotor circuit equivalent to of the d, q winding, stator windings windings 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 impedance 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

Per unit system for the synchronous machine model The equal-mutual-flux-linkage per unit system For two magnetically coupled 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-inductance of coil 1 L 22 self-inductance of coil 1 L l1 leakage inductance of coil 1 L l2 leakage inductance 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 followed by the magnetic field lines which cross both windings; the field is created by i 1 and i 2. 17 / 38

Assume we choose V 1B and V 2B such that: V 1B V 2B = n 1 n 2 In order to have the same base power in both circuits: Per unit system for the synchronous machine model 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 created by I 2B in coil 1 is equal to the flux created by I 1B in the same coil 1, after removing the part that corresponds to leakages. 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 said to yield equal mutual flux linkages (EMFL) 18 / 38

Alternative definition of base currents Per unit system for the synchronous machine model From respectively (1) and (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 1B I 1B = (L 11 L l1 ) L 1B = L 11pu L l1pu (3) L 21pu = L 21I 1B = (L 22 L l2 ) = L 22pu L l2pu L 2B I 2B L 2B In this pu system, self-inductance = mutual inductance + leakage reactance. Does not hold true for inductances in Henry! The inductance 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

Per unit system for the synchronous machine model Application to synchronous machine we have to choose a base voltage (or current) in each rotor winding. Let s first consider the field winding f (1 f, 2 d) we would like to use the EMFL per unit system we do not know the number of turns of the equivalent circuits f, d, etc. instead, we can use one of the alternative definitions of base currents: I fb 3IB = L dd L l L df I fb = 3I B L dd L l L df (4) L dd, L l can be measured L df can be obtained by measuring the no-load voltage E q produced by a known field current i f : E q = ω NL df 3 i f L df = the base voltage is obtained from V fb = S B I fb 3Eq ω N i f (5) 20 / 38

Per unit system for the synchronous machine model What about the other rotor windings? one cannot access the d 1, q 1 and q 2 windings to measure L dd1, L qq1 et L qq2 using formulae similar to (5) it can be assumed that there exist base currents I d1b, I q1b et I q2b leading 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 windings (only in pu) not a big issue in so far as we do not have to connect anything to those windings (unlike the excitation system to the field winding)... 21 / 38

Numerical example Dynamics of the synchronous machine Numerical example A machine has the following characteristics: nominal frequency: 50 Hz nominal apparent power: 1330 MVA stator nominal voltage: 24 kv X d = 0.9 Ω X l = 0.1083 Ω field current giving the nominal stator voltage at no-load (and nominal frequency): 2954 A 1. Base power, voltage, impedance, inductance and current at the stator S B = 1 330 MVA U B = 24 000 V V B = 24 000 3 = 13 856 V Z B = 3 V B 2 = 0.4311 Ω L B = Z B = Z B = S B ω B ω N I B = S B = 3 V B Z B 2 π 50 = 1.378 10 3 H 1 330 106 3 13 856 = 31 998 A 22 / 38

Numerical example 2. Base power, current, voltage, impedance and induct. in field winding S fb = S B = 1 330 MVA 3 Eq 3 (24/ 3) 10 3 L df = = = 2.586 10 2 H ω N i f 2 π 50 2 954 Using Eq. (4) : L dd = X d ω N = 0.9 2 π 50 = 2.865 10 3 H L l = X l ω N = 0.1083 2 π 50 = 3.447 10 4 H I fb = 3 I B L dd L l L df V fb = S fb 1 330 106 = = 246 251 V I fb 5 401 = 5 401 A A huge value! This is to be expected since we use the machine nominal power S B in the field winding, which is not designed to carry such a high power! Z fb = V fb I fb = 246 251 5 401 = 45.594 Ω L fb = Z fb ω B = 45.594 2 π 50 = 0.14513 H 23 / 38

Numerical example 3. Convert L dd and L df in per unit L dd = X d = 0.9 = 2.078 pu 0.4331 L l = 0.1083 = 0.25 pu 0.4331 In per unit, in the EMFL per unit system, Eq. (3) can be used. Hence : L df = L dd L l = 2.078 0.25 = 1.828 pu (6) Remarks Eq. (6) does not hold true in Henry : L dd L l = 2.865 10 3 3.447 10 4 = 2.5203 10 3 H L df = 2.586 10 2 H in the EMFL per unit system, L dd and L df have comparable values 24 / 38

Dynamic equivalent circuits of the synchronous machine Dynamic equivalent circuits of the synchronous machine In the EMFL per unit system, the Park inductance matrices take on the simplified form: L l + M d M d M d L d = M d L lf + M d M d M d M d L ld1 + M d 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 inductance L l in d and q windings 25 / 38

Dynamic equivalent circuits of the synchronous machine 26 / 38

Modelling of material saturation Modelling of material saturation Saturation of magnetic material modifies: the machine inductances the initial operating point (in particular the rotor position) the field current required to obtain a given stator voltage. Notation parameters with the upperscript u refer to unsaturated values parameters without this upperscript refer to saturated values. 27 / 38

Open-circuit magnetic characteristic Modelling of material saturation Machine operating at no load, rotating at nominal angular speed ω N. Terminal voltage E q measured for various values of the field current i f. saturation factor : k = OA OB = O A O A < 1 a standard model : k = 1 1 + m(e q ) n m, n > 0 characteristic in d axis (field due to i f only) In per unit: E qpu = ω NL df i f 3 VB = ω N Z B (L dd L l )i fpu = M dpu i fpu Dropping the pu notation and introducing k: = ω NL df I fb i fpu = ω NL df L dd L l 3IB 3 VB 3 VB L df i fpu E q = M d i f = k M u d i f 28 / 38

Leakage and air gap flux Dynamics of the synchronous machine Modelling of material saturation The flux linkage in the d winding is decomposed into: ψ d = L l i d + ψ ad L l i d : leakage flux, not subject to saturation (path mainly in the air) ψ ad : direct-axis component of the air gap flux, subject to saturation. Expression of ψ ad : ψ ad = ψ d L l i d = M d (i d + i f + i d1 ) Expression of ψ aq : ψ aq = ψ q L l i q = M q (i q + i q1 + i q2 ) Considering that the d and q axes are orthogonal, the air gap flux is given by: ψ ag = ψad 2 + ψ2 aq (7) 29 / 38

Modelling of material saturation Saturation characteristic in loaded conditions Saturation is different in the d and q axes, especially for a salient pole machine (air gap larger in q axis!). Hence, different saturation factors (say, k d and k q ) should be considered in practice, however, it is quite common to have only the direct-axis saturation characteristic in this case, the characteristic is used along any axis (not just d) as follows in no-load conditions, we have ψ ad = M d i f and ψ aq = 0 ψ ag = M d i f M d = km u d = M u d 1 + m(e q ) n = M u d 1 + m(m d i f ) n = M u d 1 + m(ψ ag ) n it is assumed that this relation still holds true with the combined air gap flux ψ ag given by (7). 30 / 38

Summary: complete model Summary: complete model (variables in per unit) ψ d = L l i d + ψ ad ψ f = L lf i f + ψ ad ψ d1 = L ld1 i d1 + ψ ad ψ ad = M d (i d + i f + i d1 ) 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 d = 1 + m M u d ( ψ 2 ad + ψ2 aq ) n M q = 1 + m M u q ( ψ 2 ad + ψ2 aq ) n v d = R a i d ωψ q dψ d dt d dt ψ f = v f R f i f d dt ψ d1 = R d1 i d1 2H d dt ω = T m (ψ d i q ψ q i d ) v q = R a i q + ωψ d dψ q dt d dt ψ q1 = R q1 i q1 d dt ψ q2 = R q2 i q2 1 ω N d dt θ r = ω 31 / 38

Model simplifications Model simplifications Constant rotor speed approximation θr ω N (ω = 1 pu) Examples showing that θ r does not depart much from the nominal value ω N : 1 oscillation of θ r with a magnitude of 90 o and period of 1 second superposed to the uniform motion at synchronous speed: θ r = θ o r + 2πf N t + π 2 sin 2πt θ r = 100π + π 2 cos 2πt 314 + 10 cos 2πt at its maximum, it deviates from nominal by 10/314 = 3 % only. 2 in a large interconnected system, after primary frequency control, frequency settles at f f N. f f N = 0.1 Hz is already a large deviation. In this case, machine speeds deviate from nominal by 0.1/50 = 0.2 % only. 3 a small isolated system may experience larger frequency deviations. But even for f f N = 1 Hz, the machine speeds deviate from nominal by 1/50 = 2 % only. 32 / 38

Model simplifications The phasor (or quasi-sinusoidal) approximation Underlies a large class of power system dynamic simulators considered in detail in the following lectures for the synchronous machine, it consists of neglecting the transformer voltages dψ d and dψ d in the stator Park equations dt dt this leads to neglecting some fast varying components of the network response, and allows the voltage and currents to be treated as a sinusoidal with time-varying amplitudes and phase angles (hence the name) at the same time, three-phase balance is also assumed. Thus, the stator Park equations become (in per unit, with ω = 1): v d = R a i d ψ q v q = R a i q + ψ d and ψ d and ψ q are now algebraic, instead of differential, variables. Hence, they may undergo a discontinuity after of a network disturbance. 33 / 38

The classical model of the synchronous machine The classical model of the synchronous machine Very simplified model used : in some analytical developments in qualitative reasoning dealing with transient (angle) stability for fast assessment of transient (angle) stability. Classical refers to a model used when there was little computational power. Approximation # 0. We consider the phasor approximation. Approximation # 1. The damper windings d 1 et q 2 are ignored. The damping of rotor oscillations is going to be underestimated. Approximation # 2. The stator resistance R a is neglected. This is very acceptable. The stator Park equations become : v d = ψ q = L qq i q L qq1 i q1 v q = ψ d = L dd i d + L df i f 34 / 38

The classical model of the synchronous machine Expressing i f (resp. i q1 ) as function of ψ f and i d (resp. ψ q1 and i q ) : ψ f = L ff i f + L df i d i f = ψ f L df i d L ff ψ q1 = L q1q1 i q1 + L qq1 i q1 i q1 = ψ q1 L qq1 i q L q1q1 and introducing into the stator Park equations : v d = (L qq L2 qq1 ) i q L qq1 ψ q1 = X q i q + e d (8) L q1q1 L q1q1 }{{}}{{} L q e d v q = (L dd L2 df ) i d + L df ψ f = X d i d + e q (9) L }{{ ff L }} ff {{} L e d q e d and e q : are called the e.m.f. behind transient reactances are proportional to magnetic fluxes; hence, they cannot vary much after a disturbance, unlike the rotor currents i f and i q1. 35 / 38

The classical model of the synchronous machine Approximation # 3. The e.m.f. e d and e q are assumed constant. This is valid over no more than - say - one second after a disturbance; over this interval, a single rotor oscillation can take place; hence, damping cannot show its effect (i.e. Approximation # 1 is not a concern). Equations (8, 9) are similar to the Park equations in steady state, except for the presence of an e.m.f. in the d axis, and the replacement of the synchronous by the transient reactances. Approximation # 4. The transient reactance is the same in both axes : X d = X q. Questionable, but experiences shows that X q has less impact... If X d = X q, Eqs. (8, 9) can be combined in a single phasor equation, with the corresponding equivalent circuit: V + jx d Ī = Ē = E δ 36 / 38

Rotor motion. This is the only dynamics left! The classical model of the synchronous machine e d and e q are constant. Hence, Ē is fixed with respect to d and q axes, and δ differs from θ r by a constant. Therefore, 1 ω N d dt θ r = ω can be rewritten as : 1 d ω N dt δ = ω The rotor motion equation: 2H d dt ω = T m T e is transformed to involve powers instead of torques. Multiplying by ω: 2H ω d dt ω = ωt m ωt e ωt m = mechanical power P m of the turbine ωt e = p r s = p T (t) + p Js + dw ms P (active power produced) dt since we assume three-phase balanced AC operation, and R a is neglected 37 / 38

The classical model of the synchronous machine Approximation # 5. We assume ω 1 and replace 2Hω by 2H very acceptable, already justified. Thus we have: 2H d dt ω = P m P where P can be derived from the equivalent circuit: P = E V X sin(δ θ) 38 / 38