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Synchronous Machines POWER SYSTEM MODELLING AND CONTROL (EEEN40550) Prof. Federico Milano Email: [email protected] Tel.: 01 716 1844 Room 157a – Engineering & Materials Science Centre School of Electrical & Electronic Engineering University College Dublin Dublin, Ireland Dublin, 2021 Synchronous Machines - 1 Synchronous Machine Overview • Generator Overview • Model of the Synchronous Machine • Models for Stability Analysis and Control: ◦ Subtransient Model ◦ Transient Model ◦ Electro-Mechanical Model • Steady-State Model Dublin, 2021 Synchronous Machines - 2 Generator Components Steam, Pressure Fuel Synchronous Machine Torque Speed Ω Entropy Boiler Valve V, I Turbine Power Control field Primary Frequency Control Governor voltage Excitation Primary Voltage Control AVR V ref − Ω Power set-point Grid GEN. pressure meas. Firing Control Elec Power + − V + Control Center Ωref Signals from control centers, market and AGC Ω OEL, UEL, PSS, etc, V Auxiliary Controllers Dublin, 2021 Synchronous Machines - 3 3-Phase Synchronous Machines • Complete dynamic equations of 3-phase synchronous machines • Characterisation of machine inductances • Why transforming machine variables? • Park’s transformation • Transformed (d-q -0) circuit equations • Steady-state open circuit operation Dublin, 2021 Synchronous Machines - 4 Synchronous Machine Types (I) • Round Rotor ◦ Solid state rotor with (generally) only one pair of poles (Ωn = 3,000 rpm at 50Hz). This is used in steam or gas turbine groups. • Salient Pole Rotor ◦ Laminated rotor with several pairs of poles (Ωn 3,000 rpm at 50 Hz). This is used in big hydro plants coupled with Kaplan’s turbines. • Permanent Magnet/Brushless Rotor ◦ Mainly used for motors or for “small” generators (∼ 1 MVA). Since there is no need for brushes, the machine is compact and robust. However, no reactive power regulation is possible. Dublin, 2021 Synchronous Machines - 5 Synchronous Machine Scheme a ωr Damper windings c b Effects of induced currents f in the rotor core q2 b dr q1 q2 d d θr ar f DC Field a Dublin, 2021 q1 c qr Synchronous Machines - 6 Synchronous Machine Windings • a − a , b − b , c − c → stator windings (ac) • f − f → field winding (dc) • q1 − q1 → effect of induced currents in the rotor core • d − d , q2 − q2 → damper windings (fictitious!) Dublin, 2021 Synchronous Machines - 7 Circuit Equations for 3-Phase Machine • 3 stator coils (a, b, c) • 2 rotor coils in direct axis: ◦ Field winding ◦ Direct-axis damper • 1 rotor coil in quadrature (damper) • The machine is represented by 6 coils • There is no mutual coupling between the circuits in the direct axis and those in the quadrature axis • Rotor-stator and stator-stator inductances are a function of θr Dublin, 2021 Synchronous Machines - 8 Stator Flux Linkage Equations ψa = + Laa ia − Lab ib − Lac ic − Laf if − Lad id1 − Laq iq1 ψb = − Lab ia + Lbb ib − Lbc ic − Lbf if − Lbd id1 − Lbq iq1 ψc = − Lac ia − Lcb ib + Lcc ic − Lcf if − Lcd id1 − Lcq iq1 Dublin, 2021 Synchronous Machines - 9 ψa −lab laa ψb = −lab ψc −lac −ψa lbb −lcb −lab laa −ψb = − −lba −ψc −lca ψs Dublin, 2021 lbb −lcb In Matrix Form ia laf −lac ib − lbf −lbc lcc ic lcf lad lbd lcd ia −lac laf ib + lbf −lbc lcc ic lcf lad lbd lcd laq if id1 lbq lcq iq1 laq if id1 lbq lcq iq1 = − Ls is + Lsr ir Synchronous Machines - 10 Rotor Flux Linkage Equations ψf = − Laf ia − Lbf ib − Lcf ic + Lf f if + Lf d id1 ψd1 = − Lad ia − Lbd ib − Lcd ic + Lf d if + Ldd id1 ψq1 = − Laq ia − Lbq ib − Lcq ic + Lqq iq1 Dublin, 2021 Synchronous Machines - 11 ψf In Matrix Form ia lf f lf b lf c ib + ldf ldb ldc lqb lqc ic 0 lf a ψd1 = − lda ψq1 lqa ψr Dublin, 2021 lf d ldd 0 0 if id1 0 iq1 lqq = − Lrs is + Lr ir Synchronous Machines - 12 Stator Voltage Equations • • • Dublin, 2021 dψa − Ra i a ea = − dt dψb − Rb i b eb = − dt dψc − Rc i c ec = − dt Synchronous Machines - 13 In Matrix Form −ψa ea Ra 0 eb = d −ψb − 0 Rb dt ec −ψc 0 0 ia 0 ib 0 ic Rc d ψ + Rs i s es = dt s Dublin, 2021 Synchronous Machines - 14 Rotor Voltage Equations • dψf + Rf i f ef = + dt • ed1 dψd1 =+ + Rd1 id1 dt • eq1 = + Dublin, 2021 dψq1 + Rq1 iq1 dt Synchronous Machines - 15 In Matrix Form ψf ef Rf 0 ed1 = d ψd1 + 0 Rd1 dt eq1 ψq1 0 0 0 if id1 0 iq1 Rq1 d ψ + Rr i r er = dt r Dublin, 2021 Synchronous Machines - 16 General Equations in Matrix Form (I) • Magnetical Equations: • Electrical Equations • Where: ψ = L i d ψ + R i e = dt T e = ea , eb , ec , ef , ed1 , eq1 T i = ia , ib , ic , if , id1 , iq1 T ψ = −ψa , −ψb , −ψc , ψf , ψd1 , ψq1 Dublin, 2021 Synchronous Machines - 17 General Equations in Matrix Form (II) lac laf lad laq −laa lab l −lbb lbc lbf lbd lbq ab l lcb −lcc lcf lcd lcq ca L = −lf a −lf b −lf c lf f lf d 0 −lda −ldb −ldc ldf ldd 0 0 lqq −lqa −lqb −lqc 0 −[Ls ] L = −[Lrs ] Dublin, 2021 [Lsr ] [Lr ] Synchronous Machines - 18 General Equations in Matrix Form (III) 0 0 0 0 0 −Ra 0 0 0 0 0 −Rb 0 0 0 0 −Rc 0 R = 0 0 0 0 0 Rf 0 0 0 0 0 Rd1 0 0 0 0 0 Rq1 [Rs ] R = [0] Dublin, 2021 [0] [Rr ] Synchronous Machines - 19 Partitioned Matrix Equations • Flux Linkage: • Voltage Equations: Dublin, 2021 ψs = − Ls is + Lsr ir ψr = − Lrs is + Lr ir d ψs + Rs i s es = dt d ψr + Rr i r er = dt Synchronous Machines - 20 Self & Mutual Inductances • We make the following assumptions: 1. There is no saturation →The system is linear 2. Stator surface is smooth →We ignore the tooth ripple 3. We have a sinusoidally distributed MMF in the air gap. The winding and pole shape are such that there are no space harmonics. In this case only fundamental frequency EMF will be induced in the windings at steady-state. Dublin, 2021 Synchronous Machines - 21 Stator Self Inductance • For the round rotor, the self inductance is constant i.e. not a function of θr . Hence: laa = lbb = lcc = constant • For the salient pole rotor, there is a constant (average) and a variable part of the self inductance dependent on θr . Hence: laa =laao + laa2 cos(2θr ) lbb lcc Dublin, 2021 2π =laao + laa2 cos(2(θr − )) 3 2π =laao + laa2 cos(2(θr + )) 3 Synchronous Machines - 22 Stator Self Inductance of Salient Pole Rotors • Variation of laa (θr ) in arbitrary units: laa2 laao 0 θr π 2 π laa (θr ) = laao + laa2 cos(2θr ) Dublin, 2021 Synchronous Machines - 23 Stator Mutual Inductances (I) • For the round rotor case, the stator-stator mutual inductances are constant, i.e. not a function of θr . Hence: lab = lbc = lca = labo = constant • Observe that for symmetry: lab = lba , Dublin, 2021 lbc = lcb , lca = lac Synchronous Machines - 24 Stator Mutual Inductances (II) • For the salient pole rotor, the stator-stator mutual inductances is a constant plus a sinusoidal function: Hence: lab = lbc = lca = π )) 6 π labo + lab2 cos(2(θr − )) 2 5π labo + lab2 cos(2(θr + )) 6 labo + lab2 cos(2(θr + • Note that the argument of the cosine depends on 2θr , not θr . Dublin, 2021 Synchronous Machines - 25 Stator Mutual Inductances (II) • Variation of lab (θr ): lab labo 0 θr π 2 π lab (θr ) = labo + lab2 cos(2(θr + π6 )) Dublin, 2021 Synchronous Machines - 26 Mutual Inductance Between Rotor and Stator (I) • Stator to rotor mutuals vary with rotor position • d-axis rotor circuits (field and armortisseurs) vary as cos(θr ) • q -axis rotor circuits vary as sin(θr ) • Hence: laf Dublin, 2021 = lbf = lcf = laf 0 cos(θr ) 2π laf 0 cos(θr − ) 3 2π laf 0 cos(θr + ) 3 Synchronous Machines - 27 Mutual Inductance Between Rotor and Stator (II) • Variation of laf (θr ): laf laf 0 π 0 θr 2π laf (θr ) = laf 0 cos θr Dublin, 2021 Synchronous Machines - 28 Mutual Inductance Between Rotor and Stator (III) • For the damper winding we have: lad = lbd = lcd = lad0 cos(θr ) 2π lad0 cos(θr − ) 3 2π lad0 cos(θr + ) 3 • In the quadrature axis, we have: Dublin, 2021 laq = lbq = lcq = π ) = 2 2π ) −laq0 sin(θr − 3 2π −laq0 sin(θr + ) 3 laq0 cos(θr + −laq0 sin(θr ) Synchronous Machines - 29 Rotor Self and Mutual Inductances • For both the round rotor and the salient pole case, rotor self-inductances are all constant: lf f = lf f 0 = constant ldd = ldd0 = constant lqq = lqq0 constant = • For both the round and the salient pole rotor mutual inductances are: lf q = ldq = 0 i.e., there is no mutual coupling between d- and q -axis circuits • Moreover, we have: lf d Dublin, 2021 = lf d0 = constant Synchronous Machines - 30 Inductance Matrices (I) • Round Rotor: Ls laao = −labo −labo −labo laao −labo −labo −labo laao • Salient pole rotor: laao + laa2 cos 2(θr ) π Ls = −labo − lab2 cos 2(θr + 6 ) −labo − lab2 cos 2(θr + 5π ) 6 Dublin, 2021 −labo − lab2 cos 2(θr + π6 ) laao + laa2 cos 2(θr − 2π ) 3 −labo − lab2 cos 2(θr − π2 ) −labo − lab2 cos 2(θr + 5π ) 6 −labo − lab2 cos 2(θr − π2 ) laao + laa2 cos 2(θr + 2π ) 3 Synchronous Machines - 31 Inductance Matrices (II) • Rotor-rotor inductances are all constant: lf f 0 LR = lf d0 0 lf d0 ldd0 0 0 0 lqq0 • Stator and rotor inductances are the same in the round rotor and salient pole cases. lad0 cos(θr ) −laq0 sin(θr ) laf 0 cos(θr ) 2π 2π 2π Lsr = laf 0 cos(θr − 3 ) lad0 cos(θr − 3 ) −laq0 sin(θr − 3 ) 2π 2π ) l cos(θ + ) −l sin(θ + laf 0 cos(θr + 2π ad0 r aq0 r 3 3 3 ) • Moreover, due to symmetry: [Lsr ] = [Lrs ]T Dublin, 2021 Synchronous Machines - 32 Variable Transformation (I) • Looking at the flux linkage equations, we see a possible transformation: ψr = − Lrs is + Lr ir ⇓ T ia lad0 cos(θr ) −laq0 sin(θr ) ψf laf 0 cos(θr ) ψd1 = laf 0 cos(θr − 2π ) lad0 cos(θr − 2π ) −laq0 sin(θr − 2π ) ib 3 3 3 2π 2π ψq1 ic laf 0 cos(θr + 2π 3 ) lad0 cos(θr + 3 ) −laq0 sin(θr + 3 ) if 0 lf f 0 lf d0 id1 + lf d0 ldd0 0 iq1 0 0 lqq0 Dublin, 2021 Synchronous Machines - 33 Variable Transformation (II) • The rotor flux linkages are then: 2π 2π ψf = − laf 0 [ia cos(θr ) + ib cos(θr − ) + ic cos(θr + )] + lf f 0 if + lf d0 id1 3 3 2π 2π ψd1 = − lad0 [ia cos(θr ) + ib cos(θr − ) + ic cos(θr + )] + lf d0 if + ldd0 id1 3 3 2π 2π ψq1 = − laq0 [ia sin(θr ) + ib sin(θr − ) + ic sin(θr + )] + lqq0 iq1 3 3 Dublin, 2021 Synchronous Machines - 34 Variable Transformation (III) • If we define a new variable proportional to: 2π 2π ia cos(θr ) + ib cos(θr − ) + ic cos(θr + ) 3 3 • then the dependence on θr of the flux linkage could be eliminated from the direct-axis flux linkage equations. • Let: id = K(ia cos(θr ) + ib cos(θr − 2π 2π ) + ic cos(θr + )) 3 3 • id is the direct axis quantity. Dublin, 2021 Synchronous Machines - 35 Variable Transformation (IV) • Similarly, if we define a new current proportional to: 2π 2π −(ia sin(θr ) + ib sin(θr − ) + ic sin(θr + )) 3 3 the dependence on θr is eliminated. • Let iq = −K(ia sin(θr ) + ib sin(θr − 2π 3 ) + ic sin(θr + 2π 3 )) • iq is the quadrature axis quantity. Dublin, 2021 Synchronous Machines - 36 Zero Sequence Variables • We have introduced id and iq in place of ia , ib and ic . • To retain the complete information of a, b, c quantities, we introduce a third variable. • A possible choice is as follows: io = 1 (ia + ib + ic ) 3 • io is called the zero-sequence current and has no mutual coupling to any circuit on the d and q axes. • The zero sequence is null for balanced conditions. Dublin, 2021 Synchronous Machines - 37 Variable Transformation (V) • For balanced 3-phase steady-state operation: ia =IM cos(ωt) 2π ib =IM cos(ωt − ) 3 2π ic =IM cos(ωt + ) 3 • Substituting into the expression of id : 3 id = K IM cos(θr − ωt) 2 • Hence the maximum value for id is: so we select: K = 23 id = K 32 IM • The constant K is arbitrary, but we choose it to make id and ia numerically equivalent. Dublin, 2021 Synchronous Machines - 38 Physical Interpretation • id is a fictitious current that can be interpreted as the instantaneous current in a winding that is rotating with the rotor and is symmetrical to the direct axis. • id produces the same MMF on the direct axis as does the 3-phase currents in the real stator windings. • iq has the same interpretation but in the quadrature axis. • io is the homopolar current. Dublin, 2021 Synchronous Machines - 39 Steady State • For balanced operation: id =IM cos(θr − ωt) iq = −IM sin(θr − ωt) where ω is the frequency of the current and θr is the angle of the rotor: θr = ωr t + θr0 • For steady state synchronous speed operation id = ⇒ ω = ωr , hence: IM cos θr0 iq = − IM sin θr0 Dublin, 2021 Synchronous Machines - 40 Park’s Transformation • The transformation we have described is known as Park’s Transformation • For example: 3 ψf = − Laf id + Lf f if + Lf d id1 2 3 ψd1 = − Lad id + Lf d if + Ldd id1 2 3 ψq1 = − Laq iq + Lqq iq1 2 Dublin, 2021 Synchronous Machines - 41 Park’s Transformation • Projecting the phase quantities (abc) onto the (dq0) axis, we obtain: 2 fd = (fa cos θr + fb cos(θr − 120◦ ) + fc cos(θr + 120◦ )) 3 2 fq = − (fa sin θr + fb sin(θr − 120◦ ) + fc sin(θr + 120◦ )) 3 1 fo = (fa + fb + fc ) 3 Dublin, 2021 Synchronous Machines - 42 Inverse Park’s Transformation • The inverse Park’s Transformation (dq0) to (abc) is: fa = fd cos θr − fq sin θr + fo fb = fd cos(θr − 120◦ ) − fq sin(θr − 120◦ ) + fo fc = fd cos(θr + 120◦ ) − fq sin(θr − 120◦ ) + fo Dublin, 2021 Synchronous Machines - 43 Park’s Transformation Matrix • In shorthand notation: f d fdq = fq ; fo • hence: fdq • where: cos θr 2 P = − sin θr 3 1 2 Dublin, 2021 fa fs = fb ; fc = P fs cos(θr − 120◦ ) ◦ − sin(θr − 120 ) 1 2 cos(θr + 120◦ ) − sin(θr + 120 ◦ 1 2 Synchronous Machines - 44 Inverse Park’s Transformation Matrix • In shorthand notation: • where: −1 fs = P fdq cos θr −1 ◦ = P cos(θr − 120 ) cos(θr + 120◦ ) Dublin, 2021 − sin θr 1 ◦ − sin(θr − 120 ) − sin(θr + 120◦ ) 1 1 Synchronous Machines - 45 Alternative Park’s Transformation Matrices • Sometimes it is required that: T P̂ = [P̂ ]−1 • To this aim we can define: cos θr 2 − sin θr P̂ = 3 √1 2 ◦ cos(θr − 120 ) ◦ − sin(θr − 120 ) √1 2 ◦ cos(θr + 120 ) − sin(θr + 120 ◦ √1 2 • Using this formulation, the formal expression of instantaneous power is conserved: P (t) =[Vs ]T [Is ] = [[P̂ ]−1 [Vs ]]T [[P̂ ]−1 [Is ]] =[Vs ]T {[P̂ ]−1 }T [P̂ ]−1 [Is ] = [Vpq ][Ipq ] Dublin, 2021 # Synchronous Machines - 46 Derivatives of the Transformed Variables • We define the time derivative of the transformed variables: d [fs (t)] dt d = [P ] {[P ]−1 [fdq (t)]} dt d d = {[P ] [P ]−1 }[fdq (t)] + [P ][P ]−1 [fdq (t)] dt dt [P ] • where [P ][P ]−1 = [I3 ] Dublin, 2021 Synchronous Machines - 47 Derivatives of the Transformed Variables d • and where: [P ] dt [P ]−1 = cos θr 2 − sin θr 3 cos(θr − − sin(θr − cos(θr + 120◦ ) 0 2 3 ωr 2 3 0 − 32 0 0 120◦ ) − cos θr − sin θr ◦ − sin(θr + 120◦ ωr − sin(θr − 120 ) 1 ◦) − sin(θ + 120 r 2 1 2 1 2 = 120◦ ) 0 0 0 = 0 ωr 1 0 −1 0 0 0 0 0 − cos(θr − − cos(θr + 120◦ ) = 120◦ ) ωr Pω where dθr = ωr dt Dublin, 2021 Synchronous Machines - 48 0 0 0 Derivatives of the Transformed Variables • Therefore: d d [P ] [fs (t)] = [fdq (t)] + ωr [Pω ][fdq (t)] dt dt • In scalar form: d f˙d (t) = fd + ωr fq dt d f˙q (t) = fq − ωr fd dt d f˙o (t) = fo dt Dublin, 2021 Synchronous Machines - 49 Rotor Equations ψf ψd1 = ψq1 ef − 32 Laf − 3 Lad 2 0 0 Laf Lf d 0 Lf d Ldd − 32 Laq 0 0 ψf Rf ed1 = d ψd1 + 0 dt eq1 ψq1 0 Dublin, 2021 id 0 Rd 0 0 0 iq 0 if Lqq i d1 iq1 if 0 id1 iq1 Rq Synchronous Machines - 50 Stator Equations • We now transform the stator flux linkages and currents to the (dqo) frame. i a i b ψa i c ψb = −[LS ][LSR ] if ψc id1 iq1 Dublin, 2021 Synchronous Machines - 51 Expanding the Flux Linkage Equations ψa = −[Laao + Laa2 cos 2θr]ia π +[Labo + Laa2 cos 2(θr + )]ib 6 5π +[Labo + Laa2 cos 2(θr + )]ic 6 +Laf cos θr if +Lad cos θr id1 +Laq sin θr iq1 Dublin, 2021 Synchronous Machines - 52 Applying the Inverse Transformation ψd cos θr − ψq sin θr + ψo = −[Labo + Laa2 cos 2θr ](id cos θr − iq sin θr + io ) π 2π 2π +[Labo + Laa2 cos 2(θr + )](id cos(θr − ) − iq sin(θr − ) + io ) 6 3 3 5π 2π 2π +[Labo + Laa2 cos 2(θr + )](id cos(θr + ) − iq sin(θr + ) + io ) 6 3 3 +Laf cos θr if + Lad cos θr id1 − Laq sin θr iq1 Dublin, 2021 Synchronous Machines - 53 Some Identities • We now use the identities: 2π 2π cos(θr − ) − cos(θr + ) = − cos θr 3 3 2π 2π sin(θr − ) − sin(θr + ) = − sin θr 3 3 √ 2π 2π ) − cos(θr + ) = 3 sin θr cos(θr − 3 3 √ 2π 2π sin(θr − ) − sin(θr + ) = − 3 cos θr 3 3 Dublin, 2021 Synchronous Machines - 54 Stator Equations (contd.) ψd cos θr − ψq sin θr + ψo = −Laao id cos θr + Laao iq sin θr − Labo id cos θr + Labo iq sin θr π 2π 5π 2π −Laa2 id [cos 2θr cos θr − cos 2(θr + ) cos(θr − ) − cos 2(θr + ) cos(θr + )] 6 3 6 3 π 2π 5π 2π −Laa2 iq [cos 2θr sin θr − cos 2(θr + ) sin(θr − ) − cos 2(θr + ) sin(θr + )] 6 3 6 3 +(−Laao + 2Labo )io +Laf cos θr if + Lad cos θr id1 + Laq sin θr iq1 Dublin, 2021 Synchronous Machines - 55 Stator Equations (contd.) • The first term in square brackets can be simplified as: π 2π 5π 2π ) cos(θr − ) − cos 2(θr + ) cos(θr + ) 6 3 6 3 3 3 cos 2θr cos θr + sin 2θr sin θr 2 2 3 = cos θr 2 cos 2θr cos θr − cos 2(θr + = • The second term in square brackets can be simplified as: π 2π 5π 2π cos 2θr sin θr − cos 2(θr + ) sin(θr − ) − cos 2(θr + ) sin(θr + ) 6 3 6 3 3 3 cos 2θr sin θr + sin 2θr cos θr = 2 2 3 = − sin θr 2 Dublin, 2021 Synchronous Machines - 56 Stator Equations (contd.) • Substituting, we find: ψd cos θr − ψq sin θr + ψo = 3 [Laf if + Lad id1 − (Laao + Labo + Laa2 )id ] cos θr 2 3 +[−Laq iq1 + (Laao + Labo − Laa2 )iq ] sin θr 2 −(Laao − 2Labo )io • Equating similar terms in each side, we obtain: 3 ψd = −(Laao + Labo + Laa2 )id + Laf if + Lad id1 2 3 ψq = −(Laao + Labo − Laa2 )iq + Laq iq1 2 ψo = −(Laao − 2Labo )io Dublin, 2021 Synchronous Machines - 57 Definition of Inductances • We now define the inductances: 3 Ld =Laao + Labo + Laa2 2 3 Lq =Laao + Labo − Laa2 2 Lo =Laao − 2Labo • The stator flux linkage equations can be rewritten as: ψd = −Ld id + Laf if + Lad id1 ψq = −Lq iq + Laq iq1 ψo = −Lo io Dublin, 2021 Synchronous Machines - 58 Round Rotor Machine • For round rotor machines: Laa2 = 0 • Hence, the inductances become: Ld = Lq = Laao + Labo and Lo = Laa2 − 2Labo • Note that: Dublin, 2021 L d > Lq > Lo Synchronous Machines - 59 Stator Voltage Equations • The transformation is done as follows. • We begin with the voltage equations: ψa ea Ra e b = d ψb − 0 dt ec ψc 0 • We now expand the equation ea = d dt ψa 0 Rb 0 ia 0 ib 0 ic Rc − Ra i a and use the inverse transformation of variables. Dublin, 2021 Synchronous Machines - 60 Stator Voltage Equations (contd.) • This gives: ed cos θr − eq sin θr + eo d (ψd cos θr − ψq sin θr + ψo ) − Ra (id cos θr − iq sin θr + io ) dt d d dψo = − ωr ψd sin θr + ψd cos θr − ωr ψq cos θr − ψq sin θr + dt dt dt − Ra (id cos θr − iq sin θr + io ) = Dublin, 2021 Synchronous Machines - 61 Stator Voltage Equations (contd.) • Equating similar terms on each side gives: d e d = ψd − ω r ψq − Ra i d dt d e q = ψq + ω r ψd − Ra i q dt d e o = ψo − Ra i o dt Dublin, 2021 Synchronous Machines - 62 Summary of Transformed Equations - Rotor id ψf − 32 Laf ψd1 = − 3 Lad 2 ψq1 0 0 0 Lf f 0 0 Lf d − 32 Laq 0 0 ψf ef Rf ed1 = d ψd1 + 0 dt eq1 ψq1 0 Dublin, 2021 0 Rd 0 iq Lf d 0 io Ldd 0 if 0 Lqq id1 iq1 if 0 0 id1 iq1 Rq Synchronous Machines - 63 Summary of Transformed Equations - Stator id i q 3 3 ψd 0 0 0 −Ld 2 Laf 2 Lad io 3 ψq = 0 0 0 0 L −L q 2 aq if ψo 0 0 −Lo 0 0 0 id1 iq1 ψd −ψq id 0 ed Ra 0 d eq = dt ψq + ωr ψd − 0 Ra 0 iq eo ψo 0 io 0 0 Ra Dublin, 2021 Synchronous Machines - 64 Mechanical Equations • Mechanical equations are: d θ r = ωr dt d J ωr = Tm − Te dt • In the equation above, ωr is the mechanical rotor speed. If the number of pairs of poles p = 1, mechanical and electrical rotor speeds are equal. Otherwise: ωr,elec = p · ωr Dublin, 2021 Synchronous Machines - 65 Electrical Torque (I) • In general, the electrical torque of a machine is given by: 1 T d Te = −p [I] [L(θr )][I] 2 dθr • Imposing the structure of the synchronous machine, we obtain: d [Is ] −[Ls ] [Lsr ] 1 T T Te = −p [Is ] [−Ir ] 2 dθr −[Lsr ]T [Lr ] [Ir ] • Observe that [Lr ] does not depend on θr . Hence: p Te = + {+[Is ]T [Ls,θr ][Is ] + [Is ]T [Lsr,θr ][Ir ] + [Ir ]T [Lsr,θr ]T [Is ]} 2 Dublin, 2021 Synchronous Machines - 66 Electrical Torque (II) • Observe that, since Te ∈ R: [Ir ]T [Lsr,θr ]T [Is ] = ([Ir ]T [Lsr,θr ]T [Is ])T • Moreover: ([Ir ]T [Lsr,θr ]T [Is ])T = [Is ]T [Lsr,θr ][Ir ] • Finally, we obtain: p Te = + [Is ]T [Ls,θr ][Is ] + p[Is ]T [Lsr,θr ][Ir ] 2 Dublin, 2021 Synchronous Machines - 67 Electrical Power and Torque (I) • The instantaneous 3-phase power output of the stator is: P (t) = ea ia + eb ib + ec ic • Substituting in terms of dqo components we have: P (t) = 3 (ed id + eq iq + 2eo io ) 2 • Under balanced conditions: P (t) = Dublin, 2021 3 (ed id + eq iq ) 2 Synchronous Machines - 68 Electrical Power and Torque (II) • The electromagnetic torque Te can be determined using: e = dψ −ψ ω −R i d q r a d dt d e q = d ψ q + ψd ω r − R a i q dt • and substituting ed and eq in P (t): P (t) = d 3 [(i d 2 dt ψd d + iq dt ψq ) ← Rate of change of armature magnetic energy +(ψd iq − ψq id )ωr ← Power transferred across the air gap −(i2d + i2q )Ra ] Dublin, 2021 ← Armature losses Synchronous Machines - 69 Electrical Power and Torque (III) • The air-gap torque Te is obtained by dividing the power transferred across the air-gap by the rotor speed: Te = = 3 ωr (ψd iq − ψq id ) 2 ωr,mec 3 (ψd iq − ψq id )p 2 • where p is the number of pairs of field poles. Dublin, 2021 Synchronous Machines - 70 Reducing Rotor Quantities to the Stator • Let’s define 3 Ns =vj ( ) 2 Nj Nj 1 s ) ij =ij ( Ns 3 vjs 2 ψjs =ψj ( 3 Ns ) 2 Nj • where: ◦ j is the index of the j -th rotor winding; ◦ Ns is the number of turns of stator windings; and ◦ Nj is the number of turns of the j -th rotor winding. Dublin, 2021 Synchronous Machines - 71 Summary of Reduced Equations - Rotor ψfs − 32 L̂af ψ s = − 3 L̂ad d1 2 s ψq1 0 esf 0 3 L̂ 2 ff 3 L̂ 2 fd 3 L̂ 2 fd 3 L̂ 2 dd 0 0 0 0 0 0 − 32 L̂aq ψfs 3 R̂ 2 f e s = d ψ s + 0 d1 dt d1 s esq1 ψq1 0 • where L̂f f = Dublin, 2021 Ns2 L , L̂dd Nf2 f f = Ns2 2 Ldd Nd1 0 3 R̂ 2 d 0 0 id iq 0 io 0 s i f 3 L̂ qq 2 isd1 isq1 isf i s 0 d1 s 3 R̂ i q q1 2 etc. Synchronous Machines - 72 Summary of Reduced Equations - Stator id i q 3 3 ψd 0 0 0 −Ld 2 L̂af 2 L̂ad io 3 ψq = 0 L̂ 0 0 0 −L q 2 aq s if ψo 0 0 −Lo 0 0 0 s id1 isq1 ψd −ψq id 0 ed Ra 0 d eq = dt ψq + ωr ψd − 0 Ra 0 iq eo ψo 0 io 0 0 Ra Dublin, 2021 Synchronous Machines - 73 Per Unit System for the Stator Quantities • Es,base = peak value of rated line-to-neutral voltage [V] • Is,base = peak value of rated line current [A] • fbase = fn = rated frequency [Hz] ⇒ Derived quantities: ωn = ωbase = 2πfbase Zs,base = Zn = Dublin, 2021 Es,base Is,base [ω] ωm,base = ωmn = p1 ωbase Ls,base = Ln = Zn ωn [H] Synchronous Machines - 74 Other Derived Per Unit Quantities Es,base • Flux base: Ψs,base = Ψn = Ls,base Is,base = [Wb·t] ωbase 3 Es,base Is,base √ = Es,base Is,base • 3-phase power base: Sbase = Sn = 3 √ 2 2 2 • Torque base: Tbase Dublin, 2021 Sn 3 = Tn = = pΨs,base Is,base ωmn 2 [VA] [Nm] Synchronous Machines - 75 Summary of Per Unit Equations - Rotor −xaf ψ s = −xad d1 s ψq1 0 ψfs esf es = 1 d d1 ωn dt s eq1 • where xsaf = 32 L̂saf /Ln , etc. Dublin, 2021 0 0 xsf f xsf d 0 0 xsf d xsdd −xaq 0 0 0 ψfs rfs ψ s + 0 d1 s ψq1 0 rfs = 3 R̂f , 2 Zn 0 s rd1 0 id iq 0 io 0 s i f xsqq s id1 isq1 0 isf i s 0 d1 s isq1 rq1 etc. Synchronous Machines - 76 Summary of Per Unit Equations - Stator id ψd −xd ψq = 0 ψo 0 0 0 xsaf xsad −xq 0 0 0 0 −xo 0 0 ψd ed −ψq ra e q = 1 d ψq + ω ψd − 0 ωn dt eo ψo 0 0 where ω Dublin, 2021 = iq 0 i o 3 s x 2 aq s if 0 s id1 isq1 id 0 0 ra 0 iq io 0 ro ωr ωn Synchronous Machines - 77 Mechanical Equations in Per Unit (I) • Torque equation in per unit: τ e = ψd i q − ψq i d • Mechanical equation: ⇒ ⇒ Dublin, 2021 Tm = Tm Tn = τm = d ωr,mec dt Te ωmn d +J ω Tn Tn dt 2 d ωmn τe + J ω Tn ωmn dt Te + J Synchronous Machines - 78 Mechanical Equations in Per Unit (II) 2 ωmn Sn is called • The quantity J M = start − up time (observe that Tn · ωmn = Sn ) • It is often defined H = inertia constant as: 2H = M • hence: τm = τe + 2H d ω dt • If considering damping: τm Dublin, 2021 d = τe + D(ω − ωs ) + 2H ω dt Synchronous Machines - 79 Leakage Reactance • Let’s define: xd = x + xmd xq = x + xmq where x is the leakage reactance. • hence ψd = −x id + ψmd ψq = −x iq + ψmq where: ψmd = −xmd id + xaf if + xad id1 ψmq = −xmq iq + xaq iq1 • hence: τ = ψd iq − ψq id = ψmd iq − ψmq id Note that flux leakage does not contribute to the air-gap torque! Dublin, 2021 Synchronous Machines - 80 Steady-State Conditions • In steady-state, we put all d dt ψj terms to zero • Therefore all amortisseur currents are zero: id1 = iq1 = 0 • Then: Dublin, 2021 ed = −ωr ψq − ra id eq = ω r ψd − ra i q esf d = rfs d isf ψd = −xd id + xsad isf ψq = −xq iq ψfs = xsf f isf − xsad id ψd1 = xsf d isf − xsad id ψq1 = −xsaq iq Synchronous Machines - 81 Steady-state Field Current • According to the previous equations, the steady-state field current is: isf ψd + x d i d = xsad • Then, substituting ψd in terms of ed and iq in terms of ψd and eq : isf eq + ra iq + ωxd id = ωxsad • Finally, observing that in steady state ω = ωs = 1 pu, one has: isf = Dublin, 2021 e q + ra i q + x d i d xsad Synchronous Machines - 82 Phasor Representation (Steady-State) • Given: ea =eT cos(ωs t + α) 2π + α) 3 2π ec =eT cos(ωs t + + α) 3 eb =eT cos(ωs t − • Applying the d-q axis transformation, we obtain: ed =eT cos(ωs t + α − θr ) eq =eT sin(ωs t + α − θr ) where θr Dublin, 2021 = ωr t + θo , where θr is the angle by which the d axis leads the a phase. Synchronous Machines - 83 Phasor Representation (II) • We can define the terminal bus voltage phasor as: q -axis ēT = ed + jeq eq ēT where: ed = eT sin δ eq = eT cos δ δ α−θ ed d-axis where δ is the angle by which the q axis leads the phasor ēT Dublin, 2021 Synchronous Machines - 84 Phasor Representation (III) • The terminal current īT is defined as: īT = id + j iq i = i sin(δ + φ) d T iq = iT cos(δ + φ) where: q -axis ēT īT iq φ id Dublin, 2021 d-axis Synchronous Machines - 85 Example - Park’s Vectors (I) • Let consider a 3-phase symmetrical, time-invariant linear load: a b c L R n Dublin, 2021 Synchronous Machines - 86 Example - Park’s Vectors (II) • Let assume to apply three-phase voltages at the terminals a, b, c. • The circuit equations, in time domain, are: Va (t) = Vb (t) = Vc (t) = In (t) = dIa (t) + RIa (t) dt dIb (t) L + RIb (t) dt dIc (t) L + RIc (t) dt Ia (t) + Ib (t) + Ic (t) L • The equations above are always valid. Dublin, 2021 Synchronous Machines - 87 Example - Park’s Vectors (III) • We can rewrite circuit equations using Park’s vectors. • Let define: V̄p (t) I¯p (t) • Then: V̄p (t) = ( = Vd (t) + jVq (t) = Id (t) + jIq (t) d + jω(t))LI¯p + RI¯p dt • Where ω(t) is the speed of the Park’s transformation. • Note that ω(t) can be any function of time (i.e., it is not necessarily constant). Dublin, 2021 Synchronous Machines - 88 Example - Park’s Vectors (IV) • The vector representation can be rewritten as: Vd (t) = Vq (t) = dId (t) − Lω(t)Iq (t) + RId (t) dt dIq (t) + Lω(t)Id (t) + RIq (t) L dt L • The equations above are always valid, but incomplete. Dublin, 2021 Synchronous Machines - 89 Example - Park’s Vectors (V) • To complete the set of equations, we write the zero component equation: Vo (t) = L dIo (t) + RIo (t) dt • and the current balance: In (t) = 3Io (t) if we use [P ] • and In (t) = Dublin, 2021 √ 3Io (t) if we use [P̂ ] Synchronous Machines - 90 Example - Park’s Vectors (VI) • Hence, in Park’s coordinates we have: Vd (t) = Vq (t) = Vo (t) = In (t) = dId (t) L − Lω(t)Iq (t) + RId (t) dt dIq (t) L + Lω(t)Id (t) + RIq (t) dt dIo (t) L + RIo (t) dt 3Io (t) • The equations above are always valid. Dublin, 2021 Synchronous Machines - 91 Example - Park’s Vectors (VII) • If we assume balanced and symmetrical conditions (In = Io = 0), then the Park’s vector equation fully describes the three-phase system for any transient condition. • If we assume also steady-state conditions and that Va , Vb and Vc are symmetric and √ sinusoidal, then, the Park’s vector coincides, except possibly for a factor 2, with the well-known phasor representation if the reference speed ω is constant and equal to the pulsation of the voltages Va , Vb and Vc . • Hence, in steady-state, absolute values and for ω constant: V̄p = jωLI¯p + RI¯p Dublin, 2021 Synchronous Machines - 92 Example - Park’s Vectors (VIII) • Let rewrite the Park’s vector equations in per unit. • We use the bases Vn , In and ωn , then: Vn Zn = In and Zn Ln = ωn • If ω = ωn , we have: 1 dīp v̄p (t) = + (r + jx)īp ωn dt • Using d- and q -axis quantities: Dublin, 2021 vd (t) = vq (t) = 1 did (t) x − xiq (t) + rid (t) ωn dt 1 diq (t) + xid (t) + riq (t) x ωn dt Synchronous Machines - 93 No-load or open circuit conditions • Let’s go back to the synchronous machine model. • In open circuit conditions id = iq = 0 • Substituting in steady-state equations: ψd = xsad isf ψq = 0 ed = 0 eq = xsad isf • Therefore the terminal voltage is: ēT = ed + jeq = jxsad isf Dublin, 2021 Synchronous Machines - 94 Steady-State Equivalent Circuit • If saliency is neglected: xd = xq = xs where xs is defined as synchronous reactance. • We have: ēq = ēT + (ra + j xs )īT where eq = xsad isα rα xs ēT ∠0 eq ∠δ Dublin, 2021 Synchronous Machines - 95 Operational Impedances • Most rotor circuits are short-circuited • So, voltages are zero (eq1 = ed1 = 0) and currents of short circuited circuits can be eliminated from the system • This leads to a formulation of stator equations in an operational form, i.e., transfer functions • Hence: Dublin, 2021 ψd (s) = −xd (s)id (s) + Gf (s)vfs (s) ψq (s) = −xq (s)iq (s) Synchronous Machines - 96 Operational Impedances (II) • Hence: xd (s) Dublin, 2021 = (1 + sTd )(1 + sTd ) xd )(1 + sT ) (1 + sTdo do xq (s) = (1 + sTq )(1 + sTq ) xq )(1 + sT ) (1 + sTqo qo Gf (s) = Gs (1 + sTf ) )(1 + pT ) (1 + pTdo do Synchronous Machines - 97 Operational Impedances (III) • The order of the transfer functions depends on the number of circuits in the rotor • Definition of time constants: Dublin, 2021 Tdo , Tdo , Tqo , Tqo Open circuit time constants Td , Td , Tq , Tq Shortcircuit time constants Tdo , Tqo , Td , Tq Transient time constants Tdo , Tqo , Td , Tq Sub-transient time constants Synchronous Machines - 98 Definition of Time Constants • For example: Tqo = 1 s (x aq + xmq ) s ωn rq1 Tdo = 1 s (x af + xmd ) s ω n rf = s x x 1 md af s (x + ) ad s s ωn rd1 xmd + xaf Tdo etc. Dublin, 2021 Synchronous Machines - 99 Definition of Machine Parameters • In steady-state: xd (0) = xd xd is the d-axis synchronous reactance • During rapid transients, s → ∞, hence: T d Td xd = xd (∞) = xd Tdo Tdo xd is the sub-transient d-axis reactance Dublin, 2021 Synchronous Machines - 100 Definition of Machine Parameters (II) • If we neglect the damper winding: xd T d ∼ = xd (s) ∼ = xd Tdo • From the definitions of the machine time constants: xd xmd xsaf = x + xmd + xsaf xd is the d-axis transient reactance • Finally: xd = x + Dublin, 2021 xmd xsaf xmd xsaf xsad1 + xmd xsad1 + xsaf xsad1 Synchronous Machines - 101 Definition of Machine Parameters (III) • Similarly we define xq , xq (and xq if we have a second q -axis amortisseur) ◦ Synchronous q -axis reactance xq = xq (0) • Tq xq T qo ◦ Transient q -axis reactance xq ◦ Sub-transient q -axis reactance x q Then: xq = x + = = Tq Tq xq T T qo qo xmq xsaq1 xmq +xsaq1 , etc. • The following inequalities hold: xd xq > xq xd > xq xd Tdo > Td > Tdo > Td > Tf Tqo > Tq > Tqo > Tq Dublin, 2021 Synchronous Machines - 102 Typical Values of Standard Parameters Dublin, 2021 Parameter Hydro Thermal xd 0.6-1.5 1.0-2.3 xq 0.4-1.0 1.0-2.3 xd 0.2-0.5 0.15-0.4 xq - 0.3-1.0 x d 0.15-0.35 0.12-0.25 x q 0.2-0.45 0.12-0.25 Tdo 1.5-9.0 s 3.0-10.0 s Tqo - 0.5-2.0 s Tdo 0.01-0.05 s 0.02-0.05 s Tqo 0.01-0.09 s 0.02-0.05 s x 0.1-0.2 0.1-0.2 ra 0.002-0.02 0.0015-0.005 Synchronous Machines - 103 dq -axis models of Synchronous Machines • We assume the machine is connected to bus h with voltage v̄h = vh ∠θh • Common equations: ◦ Power injections: ph = v d id + v q iq qh = v q id − v d iq ◦ AC-grid interface: vd = vh sin(δ − θh ) vq = vh cos(δ − θh ) ◦ Electromagnetic Torque: τ e = ψd i q − ψq i d Dublin, 2021 Synchronous Machines - 104 dq -axis models of Synchronous Machines • Mechanical Equations: δ̇ = ω̇ = ωn (ω − ωs ) 1 (τm − τe − D(ω − ωs )) 2H where: ◦ D is a damping coefficient. ◦ ωn is the base synchronous frequency in rad/s, e.g. 314.16 rad/s at 50 Hz. ◦ τm is the mechanical torque provided by the turbine. Dublin, 2021 Synchronous Machines - 105 Stator Electrical Equations • Full dynamic equations: ψ˙d ψ˙q = ωn (ra id + ωψq + vd ) = ωn (ra iq − ωψd + vq ) (∗) • If we consider flux dynamics “fast”: ψ˙d = 0 and ψ˙q = 0, hence: 0 = ra id + ωψq + vd 0 = ra iq − ωψd + vq (∗∗) • If we assume that speed deviations are small then ω ≈ 1 pu: Dublin, 2021 0 = r a i d + ψq + v d 0 = r a i q − ψd + v q (∗ ∗ ∗) Synchronous Machines - 106 Sauer-Pai’s Model (I) • Common model typically used for simulating US grids: ėq = (−eq − (xd − xd )(id + γd2 ψ̇d ) + vf )/Tdo ėd = (−ed + (xq − xq )(iq + γq2 ψ̇q ))/Tqo ψ̇d = (−ψd + eq − (xd − x )id )/Tdo ψ̇q = (−ψq − ed − (xq − x )iq )/Tqo • Substituting the expressions of ψ̇d and ψ̇q : Dublin, 2021 ėq = (−eq − (xd − xd )(id − γd2 ψd − (1 − γd1 )id + γd2 eq ) + vf )/Tdo ėd = (−ed + (xq − xq )(iq − γq2 ψq − (1 − γq1 )iq − γd2 ed ))/Tqo ψ̇d = (−ψd + eq − (xd − x )id )/Tdo ψ̇q = (−ψq − ed − (xq − x )iq )/Tqo Synchronous Machines - 107 Sauer-Pai’s Model (II) where: γd1 = γq1 = γd2 = γq2 = xd − x xd − x xq − x xq − x xd − xd 1 − γd1 = (xd − x )2 xd − x xq − xq 1 − γq1 = (xq − x )2 xq − x and with the algebraic constraints: Dublin, 2021 0 = ψd + xd id − γd1 eq − (1 − γd1 )ψd 0 = ψq + xq iq − γq1 ed − (1 − γq1 )ψq Synchronous Machines - 108 Marconato’s Model(I) • An alternative model is the following (used by ENEL, Italian ISO): ėq ėd ėq ėd Dublin, 2021 = = = = (−eq − (xd − xd − γd )id + (1 − Tdo TAA )vf ) Tdo (−ed + (xq − xq − γq )iq ) Tqo (−eq + eq − (xd − xd + γd )id + Tdo TAA vf ) Tdo (−ed + ed + (xq − xq + γq )iq ) Tqo Synchronous Machines - 109 Marconato’s Model(II) where: γd γq = = xd Tdo (x − x d d) Tdo xd xq Tqo (x − x q q) x Tqo q • and the additional algebraic constraints: Dublin, 2021 0 = ψd + xd id − eq 0 = ψq + xq iq + ed Synchronous Machines - 110 Marconato’s Model - d-axis equivalent circuit id (xd − xd ) + γd (xd − xd ) − γd − + TAA 1− Td0 + − 1 sTd0 − eq + + TAA Td0 xd + − 1 sTd0 eq − ψd + d-axis vf Dublin, 2021 Synchronous Machines - 111 Marconato’s Model - q -axis equivalent circuit q -axis iq (xq − xq ) + γq (xq − xq ) − γq + − Dublin, 2021 1 sTq0 + ed + xq + − 1 sTq0 ed + −ψq + Synchronous Machines - 112 Anderson-Fouad’s Model • Most common model that can be found in text books: e˙q = (−eq − (xd − xd )id + vf )/Tdo ėd = (−ed + (xq − xq )iq )/Tqo ėq = (−eq + eq − (xd − xd )id )/Tdo ėd = (−ed + ed + (xq − xq )iq )/Tqo • This model can be considered a simplification of the Sauer Pai’s model with: eq = ψd , ed = −ψd , γd1 ≈ γq1 ≈ 0, γd2 ψ̇d ≈ 0, γq2 ψ̇q ≈ 0, • This model can also be viewed as a simplification of the Marconato’s model with: γd = γq = TAA ≈ 0 Dublin, 2021 Synchronous Machines - 113 Simplified Magnetic Equations (I) • Two d- and one q -axis model • One d- and two q -axis model • One d- and one q -axis model • One d-axis model • Electromechanical model • “Classical” model Dublin, 2021 Synchronous Machines - 114 Two d- and one q -axis model (I) • Let’s assume Tqo ≈ 0 and xq ≈ xq →ed ≈ 0. Hence: ėq = (−eq − (xd − xd )(id − γd2 ψd − (1 − γd1 )id + γd2 eq ) + vf )/Tdo ψ̇d = (−ψd + eq − (xd − x )id )/Tdo ψ̇q = (−ψq − (xq − x )iq )/Tqo • Plus the algebraic equations: Dublin, 2021 0 = ψd + xd id − γd1 eq − (1 − γd1 )ψd 0 = ψq + xq iq − (1 − γq1 )ψq Synchronous Machines - 115 Alternative Two d- and q -axis model • Using Marconato’s model and imposing Tqo ≈ 0, we obtain: ėq = ėq = ėd = TAA )v )/T f do Tdo TAA (−eq + eq − (xd − xd + γd )id + vf )/Tdo Tdo (−eq − (xd − xd − γd )id + (1 − (−ed + (xq − xq )iq )/Tqo • Plus the algebraic equations: Dublin, 2021 0 = vq + ra iq − eq + xd id 0 = vd + ra id − ed − xq iq Synchronous Machines - 116 One d- and Two q -axis model • Let’s assume that: xd ≈ xd ≈ xq , hence we have: ėq = (−eq − (xd − xd )id + vf )/Tdo ėd = (−ed + (xq − xq − γq )iq )/Tqo ėd = (−ed + ed + (xq − xd + γq )iq )/Tqo • Plus the algebraic equations: Dublin, 2021 0 = vq + ra iq − eq + xd id 0 = vd + ra id − ed − xq iq Synchronous Machines - 117 One d- and one q -axis model • If we assume that Tdo ≈ Tqo ≈ 0, then we obtain the so-called two axis model. This is the model used in most stability studies. We have: ėq = (−eq − (xd − xd )id + vf )/Tdo ėd = (−ed + (xq − xq )iq )/Tqo • Plus the algebraic equations: Dublin, 2021 0 = vq + ra iq − eq + xd id 0 = vd + ra id − ed − xq iq Synchronous Machines - 118 One d-axis model • Further simplifying the machine magnetical equations, we can neglect the dynamic on ≈ 0). We obtain: the q-axis (Tqo ėq = (−eq − (xd − xd )id + vf )/Tdo • Plus the algebraic equations: Dublin, 2021 0 = vq + ra iq − eq + xd id 0 = v d + ra i d − x q i q Synchronous Machines - 119 Electromechanical model • A pure electromechanical model neglects all dynamics of magnetical equations. As a consequence, the field voltage is substituted by a constant eq . Another assumption is that ω ≈ 1, hence pe ≈ ωTe ≈ Te . We have: pe = (vq + ra iq )iq + (vd + ra id )id • Moreover if ra ≈ 0, pe = ph (power injected into the grid at bus h). • Finally let assume that xq = xd . These assumptions lead to: eq = constant emf behind the transient reactance ed . We have: Dublin, 2021 0 = vq + ra iq − eq + xd id 0 = vd + ra id − xd iq Synchronous Machines - 120 Electromechanical (classical) model • In summary the most simplified dynamic model is the following: δ̇ = Ωb (ω − 1) ω̇ = (pm − pe − D(ω − 1))/2H 0 = (vq + ra iq )iq + (vd + ra id )id − pe 0 = vq + ra iq − eq + xd id 0 = vd + ra id − xd iq 0 = vh sin(δ − θh ) − vd 0 = vh cos(δ − θh ) − vq ph = v d i d + v q i q qh = v q id − v d iq Dublin, 2021 Synchronous Machines - 121 Sub-transient Electromechanical Model • For very fast transients, it may be convenient to assume constant eq and ed . • Hence: vd = ed − ra id + xq iq vq = eq − ra iq − xd id • This is an alternative electromechanical model where we define a “constant emf” behind the sub-transient reactance • Observe that the so-called classical machine model also assumes that ra ≈ 0 and D ≈ 0. ⇒ Lossless Model Dublin, 2021 Synchronous Machines - 122 Comparison of Machine Models of Different Orders 6.a ⇒ (∗ ∗ ∗) 6.d ⇒ (∗∗) 8.a ⇒ (∗) All models are based on the Sauer-Pai’s model for magnetic equations. Dublin, 2021 Synchronous Machines - 123 Comparison of Models of Different Types All simulations are solved using same integration step. Dublin, 2021 Synchronous Machines - 124 Dynamic Shaft Model (I) δ̇HP ω̇HP = ωn (ωHP −ωs ) = (Tm −DHP (ωHP −ωs )−D12 (ωHP −ωIP ) +KHP (δIP −δHP ))/2HHP We model the shaft as a mass-spring system δ̇IP ω̇IP = ωn (ωIP −ωs ) = (−DIP (ωIP −ωs )−D12 (ωIP −ωHP )−D23 (ωIP −ωLP ) +KHP (δHP −δIP )+KIP (δLP −δIP ))/2HIP δ̇LP ω̇LP τm = ωn (ωLP −ωs ) = (−DLP (ωLP −ωs )−D23 (ωLP −ωIP )−D34 (ωLP −ω) +KIP (δIP −δLP )+KLP (δ−δLP ))/2HLP δ̇ = ωn (ω−1) ω̇ = (−Te −D(ω−ωs )−D34 (ω−ωLP )−D45 (ω−ωEX ) KLP (δLP −δ)+KEX (δEX −δ))/2H τe HP IP LP rotor EX δ̇EX = ωn (ωEX −ωs ) ω̇EX = (−DEX (ωEX −ωs )−D45 (ωEX −ω) +KEX (δ−δEX ))/2HEX Dublin, 2021 Synchronous Machines - 125 Time Domain Simulation using a Dynamic Shaft Dublin, 2021 Synchronous Machines - 126 Subsynchronous Resonance Model τm τe r vd + jvq HP IP LP Rotor EX i̇L,d = i̇ L,q = Line Model → v̇C,d = v̇ = xL xC vh ∠θh + iL,d + jiL,q − h vC,d + jvC,q ωn (iL,q + (vd − riL,d − vC,d − vh sin(δ − θh ))/xL ) ωn (−iL,d + (vq − riL,q − vC,q − vh cos(δ − θh ))/xL ) ωn (xC iL,d + vC,q ) ωn (xC iL,q − vC,d ) ψ̇f = ωn (vf − xf if ) ψ = x i − (x − x )i f f f d L,d Generator Model → ψd = (xd − x )if − xd iL,d ψ = −x i C,q q Dublin, 2021 and flux eqs. (*) q L,q Synchronous Machines - 127 Time domain simulation with subsynchronous resonance The resonance model has a frequency ≈ Dublin, 2021 ωn (1 + xC xL ) Synchronous Machines - 128 Infinite Bus Model • Let’s consider the simplified electromechanical equations of the synchronous machine: δ̇ =ωn (ω − ωs ) ω̇ = 1 (pm − pe (δ)) 2H • A common approximation of network equivalents with “high” energy/power is to consider the network as a machine with H → ∞ and eq = constant. • Observe that if H → ∞ then ω̇ → 0, then ω = constant (typically ω = ωs is assumed). • Then if ω = ωs ⇒ δ̇ = 0 ⇒ δ = constant ⇒ Phase Reference. Dublin, 2021 Synchronous Machines - 129 Electromechanical Oscillations • Let’s consider the One-Machine Infinite-Bus (OMIB) system: e∠δ v∠0 xd xL xTh Infinite Bus • From machine equations we obtain: p δ̇ = ωn (ω − ωs ) 1 ω̇ = 2H (pm − pe (δ)) pe A B pm xeq π 2 Dublin, 2021 π pe = xev sin δ eq = xd + xL + xTh with: δ Synchronous Machines - 130 Equilibrium Points of the OMIB • The OMIB has two equilibrium points: • Let x = [δ, ω]T , then: • Assume: e = v = pm = 1.0 pu xeq = 0.5 pu xA = [0.5236, 1]T xB = [2.6180, 1]T Dublin, 2021 Synchronous Machines - 131 Stability of the equilibrium points of the OMIB • Point xA : Let’s assume a small perturbation ∂δ >0 Then pe (δA + ∂δ) > pm ⇒ ω̇ < 0 ⇒ ω decreases ⇒ ω < ωA = 1 ⇒ δ̇ < 0 ⇒ δ decreases A similar conclusion can be drawn if ∂δ < 0 Point xA is a “sink” ⇒ Stable equilibrium point • Point xB : Let’s assume a small perturbation ∂δ >0 Then pe (δB + ∂δ) < pm ⇒ ω̇ > 0 ⇒ ω increases ⇒ ω > ωB = 1 ⇒ δ̇ > 0 ⇒ δ increases A similar conclusion can be drawn if ∂δ < 0 Point xB is a “source” ⇒ unstable equilibrium point Dublin, 2021 Synchronous Machines - 132 General Approach to define the stability of E.P.s • Let’s consider an ODE system: ẋ = f (x), x ∈ Rn • Be xo an E.P. of f such that 0 = f (xo ) • Then the solution λ of det(F x|xo − λI n ) = 0 are the eigenvalues of the system (or characteristic roots) • F x|xo ≡ As is the system STATE MATRIX. • I n the identity matrix of order n. Dublin, 2021 Synchronous Machines - 133 First Lyapunov’s Stability Criteria • If all {λj } < 0 for j = 1, . . . , n then the E.P. is stable. • If exists at least one {λj } > 0 for j = 1, . . . , n then the E.P. is unstable. • If exists at least one {λj } = 0 for j = 1, . . . , n then the stability of the E.P. cannot be defined ⇒ the E.P. is a bifurcation point. Dublin, 2021 Synchronous Machines - 134 OMIB Example • Let’s compute the eigenvalues of the OMIB system with e = v = pm = 1 pu and xeq = 0.5 pu, H = 8 MWs/MVA and ωn = 314.16 rad/s • We have: det(As − λI 2 ) = 0 where As = • For xA ⇒ λ1,2 = ± 0 ωn 1 ev − 2H xeq cos δ 0 1 ev −ωn 2H xeq cos δA = ±j5.8317 → stable? • For xB ⇒ λ1,2 = ±5.8317 ⇒ U.E.P. Dublin, 2021 Synchronous Machines - 135 Effect of Damping • Point xA leads to {λ1,2 }=0, hence we do not know if the point is stable or not. • However, let consider the following modification: 1 ω̇ = (pm − pe (δ) − D(ω − ωs )) 2H • Hence: As = For D 0 ωn 1 ev − 2H xeq cos δ D − 2H > 0 ⇒ λ1,2 = −α ± jβ with α > 0 ⇒ hence xA is a weakly stable equilibrium point. Dublin, 2021 Synchronous Machines - 136 Synchronisation of Synchronous Machines • Let’s assume the two-machine system: pA A A • pmA − pA = MA dω dt pB B B pmB − pB = MB dω dt then pA = A sin δAB + B cos δAB + C pB = −A sin δAB + B cos δAB + D where A, B , C and D depend on machine and system parameters: eA , eB , xdA , xdB , xLine , etc. • Finally: Dublin, 2021 dδA dt = ωn (ωA − ωs ), dδB dt = ωn (ωB − ωs ) Synchronous Machines - 137 Synchronisation of Synchronous Machines (II) • If the two machines are synchronous ⇒ ωA = ωB and δAB = δA − δB is constant ⇒ δAB = const • Scheme of the whole system: pmA + − eA pA δAB Grid pB + ωn s − eB − pmB + ωA 1 sMA 1 sMB ωB • In steady-state the input signals to the integrators must be 0. Dublin, 2021 Synchronous Machines - 138 Relative motion between machine rotors • Let’s consider again the two machine system and define: δAB = δA − δB ωAB = ωA − ωB • Then we can combine the differential equations of the two machines: dωAB dt dδAB dt Dublin, 2021 = pmB − pB pmA − pA − MA MB = ωn (ωA − ωB ) Synchronous Machines - 139 Relative Motion between Machine Rotors (II) ⇒ Expanding the dωAB dt equation: dωAB dt MB pmA − MB pA − MA pmB + MA pB MA MB MB pmA − MA pmB MB pA − MA pB − MA MB MA M B MB pmA − MA pmB + ((MA + MB )pmA − (MA + MB )pmA ) MA MB −MB pA − MA pB + ((MA + MB )pA − (MA + MB )pB ) MA MB = = ⇒ dωAB dt = + Dublin, 2021 Synchronous Machines - 140 Relative Motion between Machine Rotors (III) • Let’s define: MAB = pmAB = pAB = MA MB MA + MB pmA − MA (pmA + pmB )/(MA + MB ) MAB pA − MA (pA + pB )/(MA + MB ) MAB • Hence we obtain: dωAB dt dδAB dt Dublin, 2021 = pmAB − pAB MAB = ωn ωAB Synchronous Machines - 141 Relative Motion between Machine Rotors (IV) • The resulting scheme is as follows: pmAB + ωAB 1 sMAB − pAB Grid δAB ωn s pAB • We have obtained a system similar to the OMIB If pmAB = constant, eA and eB are constant, then the stability can be defined based on δAB . (The OMIB can be obtained by imposing MB pmAB → ∞) δ1 Dublin, 2021 δ2 δAB Synchronous Machines - 142 WSCC 9-bus 3-machine system WSCC 3−machine, 9−bus system (Copyright 1977) Example 2.6−2.7, pp. 41−46, Power System Control and Stability, P.M. Anderson and A.A. Fouad Bus 9 Bus 7 Bus 8 Bus 3 Bus 2 Bus 5 Bus 6 Bus 4 Bus 1 Dublin, 2021 Synchronous Machines - 143 Eigenvalues at the E.P. (no damping) 15 10 Imag 5 0 −5 −10 −15 −1 Dublin, 2021 −0.8 −0.6 −0.4 −0.2 0 Real 0.2 0.4 0.6 0.8 1 Synchronous Machines - 144 Eigenvalues at the E.P. (with damping) 15 10 Imag 5 0 −5 −10 −15 −0.025 −0.02 −0.015 −0.01 −0.005 0 Real Dublin, 2021 Synchronous Machines - 145 Transient following a fault 12 10 Rotor angles (rad) 8 6 4 δSyn 1 δSyn 2 2 0 δSyn 3 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time (s) A fault occurs at bus 7 at t Dublin, 2021 = 1 and is cleared at t = 1.083 s Synchronous Machines - 146 Center of Inertia (COI) • It is useful to refer machine angles and speeds to the center of inertia (COI), which is a weighted sum of all machine angles and speeds: δCOI = j∈G ωCOI = Hj δ j j∈G j∈G Hj Hj ω j j∈G Hj where G is the set of generators. • Machine rotor angle equations are modified as follows: δ̇ = ωn (ω − ωCOI ) Dublin, 2021 Synchronous Machines - 147 Transient following a fault (with COI) 1.2 δSyn 1 1 δSyn 2 δ Syn 3 Rotor angles (rad) 0.8 0.6 0.4 0.2 0 −0.2 −0.4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time (s) Dublin, 2021 Synchronous Machines - 148 Small Signal Stability of a General System • In general we have several machines (with their controllers) connected to the grid (power flow equations). • The resulting model is: ẋ = f (x, y) 0 = g(x, y) where ◦ x ∈ Rn , ◦ y ∈ Rm , ◦ f : R(n+m) → Rn , and ◦ g : R(n+m) → Rm . • The equilibrium point is (x0 , y 0 ) Dublin, 2021 s.t. 0 = f (x0 , y 0 ), 0 = g(x0 , y 0 ). Synchronous Machines - 149 State Matrix of a General System • Let’s determine the state matrix As : Fy ∆ẋ ∆x ∆x F = x = Ac Gx Gy 0 ∆y ∆y As = F x − F y G−1 y Gx , • Let with Gy non-singular! D = F y G−1 y Gx which is often called degradation matrix • D provides a “measure” of the effect of the grid on the stability of the dynamic system, i.e., F x (which is generally stable!) → ONLY the eigenvalues of As provide information on the stability of the system. Dublin, 2021 Synchronous Machines - 150
