ECE 422/522 2 – Synchronous Machine Modeling Spring 2014
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ECE 422/522 Power System Operations & Planning/ Power Systems Analysis II 2 – Synchronous Machine Modeling Spring 2014 Instructor: Kai Sun 1 Outline • Synchronous Machine Modeling • Per Unit Representation • Simplified Models for Stability Studies 2 Synchronous Generators Cylindrical/round rotor Salient-pole rotor Field winding Armature winding Stator Field current 3 Types of Rotors • Salient pole rotors – Concentrated windings on poles and nonuniform air gap – Short axial length and large diameter – Hydraulic turbines operated at low speeds (large number of poles) – Have damper/amortisseur windings to help damp out speed oscillations 16 poles salient-pole rotor (12 MW) • Round rotors – 70% of large synchronous generators (150~1500MVA) – Distributed winding and uniform air gap – Large axial length and small diameter to limit the centrifugal forces – Steam and gas turbines, operated at high speeds, typically 3600 or 1800rpm (2 or 4-pole) – No special damper windings but eddy in the solid steal rotor gives damping effects 4 Round rotor generator under construction (Source: http://emadrlc.blogspot.com) d Generator Model Fr n Flux linkage with coil a (leading the axis of a by ωt) • ψ a = Nφ cos ωt • N γ Induced voltage: dψ ea = − a = ω Nφ sin ωt = Emax sin ωt dt γ ea S (reaches the maximum at the current position) f = • P n 2 60 Axis of coil a (reference) Fs m (n: synchronous speed in rpm; P: the number of poles) Assume: ia is lagging ea by γ (ia reaches the maximum when mn aligns with aa’) 2 4 ) ic I max sin(ωt − γ − π ) = ia I max sin(ωt − = γ ) ib I max sin(ωt − γ − π= 3 3 • Magnetic motive forces (MMF’s) of three phases: = Fa Ki = Fm sin(ωt − γ ) a 2 Fb= Kib= Fm sin(ωt − γ − π ) 3 4 Fc= Kic= Fm sin(ωt − γ − π ) 3 3 Fs = Fm 2 5 Fs is orthogonal to mn and revolving synchronously with MMF Fr due to the rotor d Under Steady-State Conditions • Fr + Fs gives MMF Fsr in air gap • Fs induces EMF Ear • Fsr results air gap flux φsr to induce EMF Esr=Ea+Ear • For a round rotor, define the reactance of the armature reaction Xar= -Ear/(jIa) • Terminal voltage V, resistance Ra and leakage and reactance Xl satisfy Ea =V + [ Ra + j ( X ๏ฌ + X ar )]I a =V + ( Ra + jX s ) I a Fr n Fsr N γ γ ea S Axis of coil a (reference) Fs m Fr Fsr Xs=Xl+Xar is known as the synchronous reactance γ Ea Ea Esr Ear Load Fs 6 Stator and Rotor Windings Armature windings: • a-a’, b-b’ and c-c’ windings Rotor windings: • Field windings – Field winding F-F’ produces a flux on the d-axis. Damper windings – Two damper windings D-D’ and Q-Q’ respectively on d- and q-axes – For a round-rotor machine, consider a second damper winding G-G’ on the q-axis (two windings on each axis) Total number of windings: • Salient pole: 3+3 (discussed here) • Round-rotor: 3+4 ANSI/IEEE standard 100-1977: the quadrature (q) axis is defined to lead the d-axis by 900 7 Note: we define opposite directions for the current and flux Winding Circuits F • Equations on the EMF (electromotive force) and flux of each winding ψ๐ ψ๐น ea=dψa/dt – Ra×ia ψ๐ = −๐๐๐ ๐๐ − ๐๐๐ ๐๐ − ๐๐๐ ๐๐ + ๐๐๐ ๐๐น + ๐๐๐ ๐๐ท + ๐๐๐ ๐๐ ψ๐ท eF=dψF/dt + RF×iF ψ๐น = −๐๐น๐ ๐๐ − ๐๐น๐ ๐๐ − ๐๐น๐ ๐๐ + ๐๐น๐น ๐๐น + ๐๐น๐ท ๐๐ท + ๐๐น๐ ๐๐ ๐๐๐ ψ๐ ๐๐๐ ψ๐ ψ๐ ๐๐๐ − = − ψ๐น ๐๐น๐น ψ๐ท ๐๐ท๐ท ψ๐ ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ψ๐๐๐ ๐๐๐ = ψ๐น๐น๐น ๐๐๐ | | | | | | | ๐๐๐ ๐๐๐ ๐๐๐น − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ท − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ −๐ข๐๐๐ ๐๐๐ ๐ข๐น๐น๐น ๐๐๐ −๐ ๐ ๐๐๐ −๐๐ ๐๐๐ −๐๐ − ๐๐น๐น ๐๐น ๐๐ท๐ท ๐๐ท ๐๐๐ ๐๐ • Stator self-inductances (laa, lbb, lcc) • Stator mutual inductances (lab, lbc, lac) • Stator-to-rotor mutual inductances (laF, lbD, laQ) • Rotor self-inductances (lFF, lDD, lQQ) • Rotor mutual inductances (lFD, lDQ, lFQ) Most of the efforts in synchronous machine modeling is to find constants and make the EMF and flux equations be simpler 8 ๐๐๐ ψ๐ ๐๐๐ ψ๐ ψ๐ ๐๐๐ − = − ψ๐น ๐๐น๐น ψ๐ท ๐๐ท๐ท ψ๐ ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ | | | | | | | ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ − ๐๐น๐น ๐๐ท๐ท ๐๐๐ ๐๐๐ −๐ ๐ ๐๐๐ −๐๐ ๐๐๐ −๐๐ − Wb⋅Turns − ๐๐น๐น ๐๐น ๐๐ท๐ท ๐๐ท ๐๐๐ ๐๐ • The matrix is symmetric because the mutual inductance by definition is the flux linkage with one winding per unit current in the other winding, i.e. lxy โ Nx ×Φmy / iy = Nx×Ny ×Pxy = lyx Nx ~ turns of winding x Φmy ~ mutual flux linking windings x and y due to current in winding y Pxy ~ permeance of the mutual flux path • A salient pole machine has significantly different permeances in d and q axes, such that the Pxy involving a stator winding (e.g. Pab and PaF) is a function of the rotor position α and reaches the maximum twice during 0o~360o Pxy ≈ P0+P2cos2α It is advisable to consider d- and q-axis components of Pxy individually 9 Stator self-inductances (laa, lbb, lcc) • laa is equal to the ratio of flux linking phase a winding to current ia, with zero currents in all other circuits, and an be approximated as laa= Laa0 + Laa2cos2θ • Detailed calculation: – MMFa has a sinusoidal distribution in space with its peak centered on phase a axis. Resolve MMFa into two MMFs centered on d and q axes MMFad has peak = Naiacosθ MMFaq has peak = -Naiasinθ – Air-gap fluxes Φgad = (Naiacosθ)Pd Φgaq = (-Naiasinθ)Pq − Add the leakage inductance: Φgaa= Φgadcosθ - Φgaqsinθ= Naia(Pdcos2θ + Pqsin2θ) lgaa laa = lal + lgaa = Lal + Lg0 + Lm cos2θ Ls >Lm ≥0 P +Pq Pd−Pq = N a ia ( d + cos2θ) 2 2 = Ls + Lm cos2θ P +Pq Pd−Pq = NaΦgaa/ia = N2a( d + cos2θ) 2 2 laa=Ls + Lm cos2θ = Lg0 + Lm cos2θ lcc=Ls + Lm cos2(θ+2π/3) 10 lbb=Ls + Lm cos2(θ-2π/3) Stator Mutual Inductances (lab, lbc, lac) q have120o • lab <0 since windings a and b (>90o) displacement • Has the maximum absolute value when θ = -300 or 1500. a b θ= -30o S N d • Detailed calculation: c Φgba = Φgadcos(θ-2π/3) - Φgaqsin(θ-2π/3) a b =Naia[Pdcosθcos(θ-2π/3)+Pqsinθsin(θ-2π/3)] = N a ia [ - Pd+Pq Pd−Pq + cos(2θ-2π/3)] 4 2 d N S θ=150o lgba = NaΦgba/ia = -Lg0/2 + Lmcos(2θ-2π/3) q c – Add leakage flux: lab = lba Ms ≈ Ls/2 = Lal - Lg0/2 + Lmcos(2θ-2π/3) lab = -Ms - Lmcos2(θ+π/6) = -Ms + Lmcos(2θ-2π/3) lbc = -Ms - Lmcos2(θ-π/2) = -Ms - Lmcos2(θ+π/6) lca = -Ms - Lmcos2(θ+5π/6) 11 Stator to Rotor Mutual Inductances (laF, lbF, lcF, laD, lbD, lcD, laQ, lbQ, lcQ) d (F, D) b q (Q) a • The rotor sees a constant permeance if neglecting variations in the air gap due to stator slots • When the flux linking a stator winding and a rotor winding reaches the maximum when they aligns with each other and is 0 when they are displaced by 90o • d-axis laF = lFa = MF cosθ laD = lDa = MD cosθ lbF = lFb = MF cos(θ-2π/3) lbD = lDb = MD cos(θ-2π/3) lcF = lFc = MF cos(θ+2π/3) lcD = lDc = MD cos(θ+2π/3) • q-axis laQ = lQa = - MQ sinθ lbQ = lQb = - MQ sin(θ-2π/3) lcQ = lQc = - MQ sin(θ+2π/3) 12 θ c For Salient-pole Rotors Which of the curves will be different for round rotors? 13 No 2nd harmonic Rotor Inductances (lFF, lDD, lQQ, lFD, lFQ, lDQ) • They are all constant – Rotor self inductances lFF โ LF lDD โ LD lQQ โ LQ – Rotor mutual inductances lFD = lDF โ MR lFQ = lQF = 0 lDQ = lQD = 0 14 Summary ψ๐๐๐ ๐ณ๐บ๐บ = ψ๐น๐น๐น ๐ณ๐น๐น ๐ณ๐บ๐บ −๐ข๐๐๐ ๐ณ๐น๐น ๐ข๐น๐น๐น LRS = LTSR ๐ณ๐น๐น ๐ฟF ๐R 0 = ๐R ๐ฟD 0 0 0 ๐ฟQ • Only LRR is constant • LSS and LSR are θ or time dependent • How to simplify LSS and LSR? – Diagonalize LSS – Remove time dependency What if we define q-axis lagging d-axis by 90o? 15 Observations ψ๐๐๐ ๐ฟ๐๐ ψ๐น๐น๐น = ๐ฟ๐ ๐ ๐ฟ๐๐ −๐๐๐๐ ๐ฟ๐ ๐ ๐๐น๐น๐น • All harmonic terms in LSS (1st harmonic) and LSR (2nd harmonic) are due to the rotor rotating relative to a, b and c to cause variations in permeance • Constant LRR doesn’t have harmonic terms because it is in a reference frame rotating with the rotor ψFDQ = - LSR iabc + LRR iFDQ LSR iabc = -ψFDQ + LRR iFDQ − LSR iabc may be represented by functions independent of θ – Represent stator currents and flux linkages also in a reference frame rotating with the rotor. 16 ψFDQ ψF = - LSR iabc + LRR iFDQ = -laFia - lbFib – lcFic +lFFiF + lFDiD + lFQiQ = -MFcosθ ia – MFcos(θ-2π/3) ib – MF cos(θ+2π/3)ic +LFiF + MRiD +0 = -MF [ia cosθ + ib cos(θ-2π/3) + ic cos(θ+2π/3)] + LFiF + MRiD =K1×id ψD = -laDia - lbDib – lcDic +lDFiF + lDDiD + lDQiQ = -MD cosθ ia - MD cos(θ-2π/3) ib - MD cos(θ+2π/3)ic +MRiF + LDiD + 0 = -MD [ia cosθ + ib cos(θ-2π/3) + ic cos(θ+2π/3)] + MRiF + LDiD =K2×id ψQ = -laQia - lbQib – lcQic +lQFiF + lQDiD + lQQiQ = MQ sinθ ia + MQ sin(θ-2π/3) ib + MQ sin(θ+2π/3) ic +0 + 0 + LQiQ = MQ [ia sinθ + ib sin(θ-2π/3) + ic sin(θ+2π/3) ] +LQiQ =K3×iq 17 Park’s (dq0) Transformation For balanced steady-state conditions: ia= Im sinωst ib= Im sin(ωst - 2π/3) • Define id= kd [ia cosθ +ib cos(θ-2π/3) +ic cos(θ+2π/3)] ic= Im sin(ωst + 2π/3) iq= -kq Im cos(ωst-θ)×3/2 iq= - kg [ia sinθ +ib sin(θ-2π/3) +ic sin(θ+2π/3) ] θ=ωrt+θ0, ωr≈ωs id= -kd Im sinθ0×3/2 iq= -kq Im cosθ0×3/2 • id= kd Im sin(ωst-θ)×3/2 Constant Define i0= k0(ia + ib + ic) What if we define q-axis lagging d-axis by 90o? 18 Park’s Transformation Matrix - P ψ๐๐๐ ๐๐๐ = ψ๐ ๐ ๐ ๐๐๐ ๐๐๐ −๐ข๐๐๐ ๐๐๐ ๐ข๐ ๐ ๐ ψdq0= P ψabc ψ๐๐๐ ๐๐๐๐ โ ψ๐ ๐ ๐ ๐๐๐๐ ๐๐๐๐ −๐ข๐๐๐ ๐๐๐ ๐ข๐ ๐ ๐ We hope L’RS= (L’SR)T like LRS= (LSR)T L’RS= LRS P-1 = LTSR P-1 =(P-T LSR)T (L’SR)T = (P LSR)T P-T=P or PTP=I kd=kq= 2 3 and k0= (P is a unitary matrix) 1 3 19 idq0= P iabc Flux Equations after Park’s Transformation Ld ψd 0 ψq ψ0 0 − = − ψF ๐MF ψD ๐MD ψQ 0 L0 ψ0 ψd ψF ψD = ψq ψQ 0 Lq 0 − 0 0 ๐MQ Ld ๐MF ๐MD 0 0 L0 − 0 0 0 | | | | | | | ๐MF LF MR ๐MF 0 0 − LF MR 0 ๐MD MR LD ๐MD 0 0 − MR LD 0 0 −id ๐MQ −iq 0 −i0 − − iF 0 iD 0 iQ LQ Lq ๐MQ 20 ๐MQ LQ k= 3 2 Ld= Ls + Ms + 3Lm/2 Lq= Ls + Ms - 3Lm/2 L0= Ls - 2Ms −i0 −id iF iD −iq iQ Voltage Equations e=dψ/dt ± R×i Stator: ψ and i are in opposite directions Rotor: ψ and i are the same directions ๏ฃฎ ea ๏ฃน ๏ฃฎen ๏ฃน ๏ฃฏ e ๏ฃบ ๏ฃฏ ๏ฃบ ๏ฃฏ b ๏ฃบ ๏ฃฏen ๏ฃบ ๏ฃฏ ec ๏ฃบ ๏ฃฏen ๏ฃบ ๏ฃฏ ๏ฃบ−๏ฃฏ ๏ฃบ = e ๏ฃฏ F ๏ฃบ ๏ฃฏ0๏ฃบ ๏ฃฏeD = 0 ๏ฃบ ๏ฃฏ 0 ๏ฃบ ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃบ = e 0 ๏ฃฏ๏ฃฐ Q ๏ฃบ๏ฃป ๏ฃฐ 0 ๏ฃป ๏ฃฎψ๏ฆ a ๏ฃน ๏ฃฎ Ra ๏ฃฏψ๏ฆ ๏ฃบ ๏ฃฏ Rb ๏ฃฏ b๏ฃบ ๏ฃฏ ๏ฃฏψ๏ฆ c ๏ฃบ ๏ฃฏ Rc + ๏ฃฏ ๏ฃบ ๏ฃฏ RF ๏ฃฏψ๏ฆ F ๏ฃบ ๏ฃฏ ๏ฃฏψ๏ฆ D ๏ฃบ ๏ฃฏ 0 ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃฏ๏ฃฐψ๏ฆ Q ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ ๐๐๐๐ ๐ ๐๐๐ = ๐๐ ๐ ๐ ๐ Neutral line: ๐ ๐ ๐ ๐ ๐ 0 RD ๏ฃน ๏ฃฎ −ia ๏ฃน ๏ฃบ ๏ฃฏ −i ๏ฃบ ๏ฃบ๏ฃฏ b๏ฃบ ๏ฃบ ๏ฃฏ −ic ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃบ ๏ฃฏ iF ๏ฃบ ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ RQ ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ iQ ๏ฃบ๏ฃป ψฬ ๐๐๐ −๐ข๐๐๐ ๐๐ง + + ๐ข๐ ๐ ๐ ψฬ ๐ ๐ ๐ ๐ ๏ฃฎ en ๏ฃน ๏ฃฎ1๏ฃน ๏ฃฎ Rn ๏ฃฏ e ๏ฃบ = −( R i + L din ) ๏ฃฏ1๏ฃบ = ๏ฃฏR en = − n n n n ๏ฃฏ ๏ฃบ ๏ฃฏ n dt ๏ฃฏ ๏ฃบ ๏ฃฏ๏ฃฐ en ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ1๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ Rn d = − R n i abc − Ln i abc dt โ (A neutral line is added compared to slide #8) Rn Rn Rn 21 Rn ๏ฃน ๏ฃฎia ๏ฃน ๏ฃฎ Ln Rn ๏ฃบ๏ฃบ ๏ฃฏ๏ฃฏib ๏ฃบ๏ฃบ − ๏ฃฏ๏ฃฏ Ln Rn ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ ic ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ Ln Ln Ln Ln Ln ๏ฃน d Ln ๏ฃบ๏ฃบ dt Ln ๏ฃบ๏ฃป ๏ฃฎia ๏ฃน ๏ฃฏi ๏ฃบ ๏ฃฏ b๏ฃบ ๏ฃฏ๏ฃฐ ic ๏ฃบ๏ฃป ๐๐๐๐ ๐ ๐๐๐ = ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ Assume Ra=Rb=Rc ψฬ ๐๐๐ −๐ข๐๐๐ ๐๐ง + + ๐ข๐ ๐ ๐ ψฬ ๐ ๐ ๐ ๐ ψdq0= P ψabc idq0= P iabc 22 ๐⋅ψฬ ๐๐๐ ψฬ ๐๐๐ = ๐ ๐ψ๐๐๐ ๐๐ = ๐ฬψ๐๐๐ + ๐ψฬ ๐๐๐ =๐ฬ๐ −๐ ψ๐๐๐ + ๐ψฬ ๐๐๐ θ=ωrt+θ0 23 Transformer voltages due to flux change in time (=0 under steady-state conditions) Speed voltages due to flux change in space −ω๐ ψq −ω๐ (−Lq ๐๐ + ๐MQ ๐๐ ) −ω๐ Lq (−๐๐ ) − ω๐ ๐MQ ๐๐ ๐ โ ω๐ ψd = ω๐ (−Ld ๐๐ + ๐MF ๐๐น + ๐MD ๐๐ท ) = ω๐ Ld (−๐๐ ) + ω๐ ๐MF ๐๐น + ω๐ ๐MD ๐๐ท ) 0 0 0 24 P⋅en P en= P ( -Rn iabc - Ln diabc /dt ) = -P Rn P-1 P iabc - P Ln P-1 P diabc/dt = -P Rn P-1 idq0 PR n P −1 - P Ln P-1 (didq0/dt – dP/dt ×iabc) ๏ฃฎ0 0 0 ๏ฃน ๏ฃฏ= ๏ฃบ −1 PL P 0 0 0 n ๏ฃฏ ๏ฃบ ๏ฃฏ๏ฃฐ0 0 3Rn ๏ฃบ๏ฃป ๏ฃฎ0 0 0 ๏ฃน ๏ฃฏ0 0 0 ๏ฃบ ๏ฃฏ ๏ฃบ ๏ฃฏ๏ฃฐ0 0 3Ln ๏ฃบ๏ฃป ๏ฃฎ Rn R n = ๏ฃฏ๏ฃฏ Rn ๏ฃฏ๏ฃฐ Rn ๏ฃฎ Ln L n = ๏ฃฏ๏ฃฏ Ln ๏ฃฏ๏ฃฐ Ln Note: P Ln P-1× dP/dt ×iabc =0 ๏ฃฎ ๏ฃน ๏ฃฎ 0 ๏ฃน ๏ฃฏ 0 ๏ฃบ ๏ฃฏ ๏ฃบ − ๏ฃฏ๏ฃฏ 0 ๏ฃบ๏ฃบ − ๏ฃฏ 0 ๏ฃบ โ ndq0 Pen = ๏ฃฏ๏ฃฐ3Rni0 ๏ฃบ๏ฃป ๏ฃฏ d ๏ฃบ ๏ฃฏ3Ln i0 ๏ฃบ d ๏ฃป ๏ฃฐ 25 Rn Rn Rn Rn ๏ฃน Rn ๏ฃบ๏ฃบ Rn ๏ฃบ๏ฃป Ln Ln Ln Ln ๏ฃน Ln ๏ฃบ๏ฃบ Ln ๏ฃบ๏ฃป ๐๐๐๐ ψฬ ๐๐๐ ๐ง๐๐๐ ๐ ๐ −๐ข๐๐๐ ๐ = + + + ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ข๐ ๐ ๐ ψฬ ๐ ๐ ๐ ๐ ๐ −๐0 −๐๐ Ra ๐๐น RF ๐๐ท + RD −๐๐ Ra R Q ๐๐ 0 0 0 0 0 0 −ω๐ Lq −ω๐ ๐MQ 0 0 0 0 0 0 0 0 ω๐ ๐MF ω๐ ๐MD 0 0 0 0 0 0 ๐0 R a + 3R n ๐๐ ๐๐น ๐๐ท = ๐๐ ๐๐ 0 0 0 0 0 0 0 0 0 0 ω๐ Ld 0 L0 + 3Ln −๐0 −๐๐ ๐๐น ๐๐ท −๐๐ ๐๐ 26 Ld ๐MF ๐MD ๐MF LF MR ๐MD MR LD Lq ๐MQ ๐MQ LQ −๐0 −๐๐ ๐๐น ×d ๐ /dt + ๐ท −๐๐ ๐๐ Voltage Equations after Park’s Transformation R a + 3R n 0 0 0 0 0 ๐0 0 0 −ω๐ Lq −ω๐ ๐MQ 0 Ra ๐๐ ๐๐น 0 0 RF 0 0 0 = ๐๐ท 0 0 0 0 0 RD ๐๐ 0 ω๐ Ld ω๐ ๐MF ω๐ ๐MD Ra 0 ๐๐ 0 0 0 0 0 RQ L0 + 3Ln −๐0 Ld ๐MF ๐MD −๐๐ MR ๐MF LF ๐๐น ×d ๐ /dt LD ๐MD MR ๐ท −๐๐ Lq ๐MQ ๐๐ ๐MQ LQ 27 −๐0 −๐๐ ๐๐น ๐๐ท + −๐๐ ๐๐ Winding Circuits after Park’s Transformation ๐0 R a + 3R n ๐๐ ๐๐น ๐๐ท = ๐๐ ๐๐ Ra RF RD Ra RQ L0 + 3Ln −๐0 −๐๐ ๐๐น ๐๐ท + −๐๐ ๐๐ Ld ๐MF ๐MD ๐MF LF MR ๐MD MR LD Lq ๐MQ ๐MQ LQ −๐0 0 −๐๐ −ω๐ ψq ๐๐น 0 ×d ๐ /dt + ๐ท 0 −๐๐ ω๐ ψd ๐๐ 0 • d−axis flux causes a speed voltage ωr ψd in the q−axis winding • q−axis flux causes a speed voltage − ωr ψq in the d−axis winding 28 Alternative Park’s Transformation ia= Im sinωst id= kd Im sin(ωst-θ)×3/2 ib= Im sin(ωst - 2π/3) P iq= -kq Im cos(ωst-θ)×3/2 ic= Im sin(ωst + 2π/3) i0= k0(ia + ib + ic) • If kd=kq=2/3 and k0=1/3, a unit-to-unit relationship holds between abc and dq0 variables. ๏ฃฎ Ld ๏ฃฎψ d ๏ฃน ๏ฃฏ 0 ๏ฃฏψ ๏ฃบ ๏ฃฏ ๏ฃฏ q๏ฃบ ๏ฃฏ 0 ๏ฃฏψ 0 ๏ฃบ ๏ฃฏ 3 ๏ฃฏ ๏ฃบ = ๏ฃฏ2 MF ๏ฃฏψ F ๏ฃบ ๏ฃฏ 3 ๏ฃฏψ D ๏ฃบ ๏ฃฏ M D ๏ฃฏ ๏ฃบ ๏ฃฏ2 ๏ฃฏ๏ฃฐψ Q ๏ฃบ๏ฃป ๏ฃฏ ๏ฃฏ 0 ๏ฃฐ MF MD 0 0 0 L0 0 0 0 0 0 0 LF MR 0 0 MR LD 3 MQ 2 0 0 0 0 Lq 0 ๏ฃน M Q ๏ฃบ๏ฃบ ๏ฃฎ− id ๏ฃน ๏ฃฏ− i ๏ฃบ 0 ๏ฃบ๏ฃฏ q ๏ฃบ ๏ฃบ๏ฃฏ− i ๏ฃบ 0 ๏ฃบ๏ฃฏ 0 ๏ฃบ ๏ฃบ ๏ฃฏ iF ๏ฃบ 0 ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃบ ๏ฃฏ๏ฃฐ iQ ๏ฃบ๏ฃป LQ ๏ฃบ ๏ฃป 29 • By defining proper base inductances, the matrix may become symmetric in per unit Per Unit Representation Quantity in p.u. = Actual quantity / Base quantity ๏ฟฝ= ๐ ๐ ๐๐๐๐๐ 30 p.u. Base Quantities for Synchronous Machines Sbase~ibase ebase Zbase ~ ebase/ibase Lbase ~ Zbase/ωbase ψbase ~ Lbase×ibase Tbase~ Sbase / ωbase • For steady-state conditions, only two base quantities for each voltage level should be provided, e.g. ebase and ibase, or Sbase and ebase • Considering dynamics, 3 base quantities are needed, e.g. – fbase , ebase, ibase → Sbase, Zbase, Lbase, ψbase, Tbase – fbase , ebase, Sbase → ibase, Zbase, Lbase, ψbase, Tbase Base 1 d q 0 F D Q fbase fbase fbase fbase fbase fbase 2 3 31 Stator Base Quantities • Using the machine ratings as the base values – es base (V) peak value of rated line-to-neutral voltage – is base (A) peak value of rated line current – fbase (Hz) rated frequency • Accordingly: – – – – – – Zs base (โฆ ) Ls base (H) ωbase (elec. rad/s) ωmbase (mech. rad/s) tbase (s) ψs base (Wb⋅turns) =es base/is base =Zs base/ωbase =2πfbase =ωbase×(2/pf) =1/ ωbase =1/(2πfbase) =Ls base×is base= es base/ωbase – Tbase (N⋅m) Base 3 2 = 3ERMS base× IRMS base = 3(es base/ 2)×(is base/ 2)= es base×is base – S3φ base (VA) 0 3 ๐๐ 2 F D Q = S3φ base / ωmbase = ( 2 )ψs base×is base d q 1 fbase fbase fbase fbase fbase fbase 2 es base es base es base S3φ base S3φ base S3φ base 3 is base is base is base 32 How to select rotor base quantities? Base d q 0 F D Q 1 fbase fbase fbase fbase fbase fbase 2 es base es base es base S3φ base S3φ base S3φ base 3 is base is base is base iF base iD base iQ base ๏ฃฎ Ld ๏ฃฎψ d ๏ฃน ๏ฃฏ 0 ๏ฃฏψ ๏ฃบ ๏ฃฏ ๏ฃฏ q๏ฃบ ๏ฃฏ 0 ๏ฃฏψ 0 ๏ฃบ ๏ฃฏ 3 ๏ฃฏ ๏ฃบ = ๏ฃฏ2 MF ๏ฃฏψ F ๏ฃบ ๏ฃฏ 3 ๏ฃฏψ D ๏ฃบ ๏ฃฏ M D ๏ฃฏ ๏ฃบ ๏ฃฏ2 ๏ฃฏ๏ฃฐψ Q ๏ฃบ๏ฃป ๏ฃฏ ๏ฃฏ 0 ๏ฃฐ MF MD 0 0 0 L0 0 0 0 0 0 0 LF MR 0 0 MR LD 3 MQ 2 0 0 0 0 Lq 0 ๏ฃน M Q ๏ฃบ๏ฃบ ๏ฃฎ− id ๏ฃน ๏ฃฏ ๏ฃบ 0 ๏ฃบ ๏ฃฏ − iq ๏ฃบ ๏ฃบ๏ฃฏ− i ๏ฃบ 0 ๏ฃบ๏ฃฏ 0 ๏ฃบ ๏ฃบ ๏ฃฏ iF ๏ฃบ 0 ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃบ ๏ฃฏ iQ ๏ฃบ LQ ๏ฃบ ๏ฃฐ ๏ฃป ๏ฃป 33 iFbase, iDbase and iQbase should enable a symmetric per-unit inductance matrix ๏ฃฎ Ld ๏ฃฎψ d ๏ฃน ๏ฃฏ 0 ๏ฃฏψ ๏ฃบ ๏ฃฏ ๏ฃฏ q๏ฃบ ๏ฃฏ 0 ๏ฃฏψ 0 ๏ฃบ ๏ฃฏ 3 ๏ฃฏ ๏ฃบ = ๏ฃฏ2 MF ๏ฃฏψ F ๏ฃบ ๏ฃฏ 3 ๏ฃฏψ D ๏ฃบ ๏ฃฏ M D ๏ฃฏ ๏ฃบ ๏ฃฏ2 ๏ฃฐ๏ฃฏψ Q ๏ฃป๏ฃบ ๏ฃฏ 0 ๏ฃฏ ๏ฃฐ ψ๐ =− ψ๐ ๐๐๐๐ = ψ๐น MF MD 0 0 0 L0 0 0 0 0 0 0 LF MR 0 Lq 0 0 MR LD 3 MQ 2 0 0 0 0 ๏ฃน M Q ๏ฃบ๏ฃบ ๏ฃฎ− id ๏ฃน ๏ฃฏ ๏ฃบ 0 ๏ฃบ ๏ฃฏ − iq ๏ฃบ ๏ฃบ๏ฃฏ− i ๏ฃบ 0 ๏ฃบ๏ฃฏ 0 ๏ฃบ ๏ฃบ ๏ฃฏ iF ๏ฃบ 0 ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃบ ๏ฃฏ iQ ๏ฃบ LQ ๏ฃบ ๏ฃฐ ๏ฃป ๏ฃป ๐ฟ๐ × ๐๐ ๐๐น × ๐๐น ๐๐ท × ๐๐ท + + ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ๐ฟ๐ ×๐๐ − ๐ฟ๐ ๐๐๐๐ ×๐๐ ๐๐๐๐ + ๐๐น ×๐๐น ๐๐๐๐ ๐ฟ๐ ๐๐๐๐ ×๐๐ ๐๐๐๐ × ๐๐น ๐๐น ๐๐๐๐ ψ๐ = −๐ฟ๐ × ๐๐ +๐๐น × ๐๐น + ๐๐ท × ๐๐ท ψ๐น ๐๐๐๐ =− =− 3 ๏ฃฎψ d ๏ฃน ๏ฃฎ Ld ๏ฃฏψ ๏ฃบ ๏ฃฏ 0 ๏ฃฏ q๏ฃบ ๏ฃฏ ๏ฃฏψ 0 ๏ฃบ ๏ฃฏ 0 ๏ฃฏ ๏ฃบ=๏ฃฏ ๏ฃฏψ F ๏ฃบ ๏ฃฏ M F′ ๏ฃฏψ D ๏ฃบ ๏ฃฏ M D′ ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃฏ๏ฃฐψ Q ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ 0 + ๐๐ท ×๐๐ท ๐๐๐๐ ๐ฟ๐ ๐๐๐๐ ×๐๐ ๐๐๐๐ × 3 ๐ฟ๐น × ๐๐น ๐๐ × ๐๐ท ๐๐น × ๐๐ + + 2 ๐ฟ๐น ๐๐๐๐ × ๐๐น ๐๐๐๐ ๐ฟ๐น ๐๐๐๐ × ๐๐น ๐๐๐๐ ๐ฟ๐น ๐๐๐๐ × ๐๐น ๐๐๐๐ ๐๐น ×๐๐ ๐๐๐๐ 2 ๐ฟ๐น ๐๐๐๐ ×๐๐น ๐๐๐๐ × ๐๐ ๐๐ ๐๐๐๐ + ๐ฟ๐น ×๐๐น ๐ฟ๐น ๐๐๐๐ ×๐๐น ๐๐๐๐ + ๐๐ ×๐๐ท ๐๐๐๐ ๐ฟ๐น ๐๐๐๐ ×๐๐น ๐๐๐๐ × ๐๐ท ๐๐ท ๐๐ท ๐๐๐๐ ๐๐ท ๐๐๐๐ 0 MF MD 0 0 L0 0 0 0 0 LF 0 0 MR 0 M Q′ 0 0 MR LD 0 0 0 Lq ๐๐น = ๐′๐น ๐๐น ×๐๐น ๐๐๐๐ 3 ๐๐น ×๐๐ ๐๐๐๐ = ๐ฟ๐ ๐๐๐๐ ×๐๐ ๐๐๐๐ 2 ๐ฟ๐น ๐๐๐๐ ×๐๐น ๐๐๐๐ 3 2 LF base i2F base= LS base i2S base 3 ωbaseLF base i2F base=2 ωbase LS base i2S base 3 2 eF base iF base = eS base iS base = S3φ base =eD base iD base =eQ base iQ base ψ๐น = −๐๐๐น × ๐๐ +๐ฟ๐น × ๐๐น +๐๐ × ๐๐ท 34 0 ๏ฃน ๏ฃฎ − id ๏ฃน ๏ฃบ๏ฃฏ ๏ฃบ M Q ๏ฃบ ๏ฃฏ − iq ๏ฃบ 0 ๏ฃบ ๏ฃฏ − i0 ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ 0 ๏ฃบ ๏ฃฏ iF ๏ฃบ 0 ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ LQ ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ iQ ๏ฃบ๏ฃป 0 0 M F M D 0 ๏ฃน ๏ฃฎ − id ๏ฃน ๏ฃฎψ d ๏ฃน ๏ฃฎ Ld ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃฏψ ๏ฃบ ๏ฃฏ 0 L M 0 0 0 q q Q ๏ฃบ ๏ฃฏ − iq ๏ฃบ ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃฏψ 0 ๏ฃบ ๏ฃฏ 0 0 L0 0 0 0 ๏ฃบ ๏ฃฏ − i0 ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃบ=๏ฃฏ ψ M L M 0 0 0 F R ๏ฃบ ๏ฃฏ iF ๏ฃบ ๏ฃฏ F๏ฃบ ๏ฃฏ F ๏ฃฏψ D ๏ฃบ ๏ฃฏ M D 0 0 M R LD 0 ๏ฃบ ๏ฃฏ iD ๏ฃบ ๏ฃบ๏ฃฏ ๏ฃบ ๏ฃฏ ๏ฃบ ๏ฃฏ 0 LQ ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐ iQ ๏ฃบ๏ฃป ๏ฃฏ๏ฃฐψ Q ๏ฃบ๏ฃป ๏ฃฐ๏ฃฏ 0 M Q 0 0 Rotor Base Quantities • Stator self-inductance ๐ฟ๐ or ๐ฟ๐ can be split into two parts: – Leakage inductance due to flux that does not link any rotor circuit – Mutual inductance due to flux that links the rotor circuits • Stator leakage inductances in d and q axes are nearly equal. Then ๐ฟ๐ = ๐ฟ๐ + ๐ฟ๐๐ ๐ฟ๐ = ๐ฟ๐ + ๐ฟ๐๐ • Assume that all the per unit mutual inductances between the stator and rotor circuits in each axis are equal ๐๐น = ๐๐ท = ๐ฟ๐๐ ๐๐ = ๐ฟ๐๐ • Some references suggest rotor mutual inductance ๐๐ = ๐ฟ๐๐ to further simplify equivalent circuits 35 ๐ฟ๐๐ ๐ฟ๐ ๐๐๐๐ ๐ฟ๐๐ = ๐๐น = ๐๐ท ๐๐น × ๐๐น ๐๐๐๐ ๐๐ท × ๐๐ท ๐๐๐๐ = = ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ๐๐น ๐๐๐๐ โ ๐ฟ๐๐ ๐๐น × ๐๐ ๐๐๐๐ , A ๐ด๐ธ × ๐๐บ ๐๐๐๐ , A ZQ base= eQ base / iQ base โ S3φ base / i2Q base, โฆ LQ baseโ ZQ base / ωbase , H H ψF base โ LF base×iF base, Wb⋅turns ψQ base โ LQ base×iQ base, Wb⋅turns MQ base = LS base × iS base / iQ base MF base = LS base × iS base / iF base ๐ฟ๐๐ ๐๐ท ๐ณ๐๐ eQ baseโ S3φ base / iQ base, V ZF baseโ eF base / iF base =S3φ base/ i2F base, โฆ ๐๐ท ๐๐๐๐ โ ๐ฟ๐ ๐๐๐๐ ๐๐ธ ๐๐๐๐ โ eF baseโS3φ base / iF base, V LF baseโ ZF base / ωbase , ๐ฟ๐๐ ๐ฟ๐๐ = ๐๐ ๐๐ × ๐๐ ๐๐๐๐ = ๐ฟ๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ , A eD baseโ S3φ base / iD base, V ZD base= eD base / iD base โ S3φ base / i2D base, โฆ LD baseโ ZD base / ωbase , H ψD base โ LD base×iD base, Wb⋅turns MD base = LS base × iS base / iD base Lad-Laq based per unit system 36 Per Unit Voltage Equations ๐๐๐๐ = −๐๐ข๐๐๐ + ψฬ ๐๐๐ + ๐ + ๐ง๐๐๐ ๐๐ ๐ ๐ = ๐ ๐ ๐ข๐ ๐ ๐ + ψฬ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐๐ = − ๐ ๐ ๐0 ๐ ๐ ๐ ๐น ๐๐น 0 = 0 ๐ ๐ท ๐ ๐ ๐ψd ๐๐ −ω๐ ψq 0 0 ๐๐ + ๐ψq + ω๐ ψd − 0 0 − ๐ψ0 3๐ ๐ ๐0 3๐ฟ๐ ๐๐0 0 ๐0 ๐ψF ๐๐น ๐๐ท + ๐ψD ๐ψQ ๐๐ p=d/dt differential operator • Divide both sides of each equation by one of the following: eS base = ωbase×ψS base = ωbase×LS base×iS base =ZS base×iS base eF base = ωbase×ψF base = ωbase×LF base×iF base =ZF base×iF base eD base = ωbase×ψD base = ωbase×LD base×iD base = ZD base×iD base eQ base = ωbase×ψQ base = ωbase×LQ base×iQ base = ZQ base×iQ base 37 • For example: ๐๐ ๐๐ ๐๐๐๐ =− • Note: ๐ ๐ × ๐๐ ๐ψ๐ ω๐ ψ๐ + − ๐๐ ๐๐๐๐ × ๐๐ ๐๐๐๐ ω๐๐๐๐ × ψ๐ ๐๐๐๐ ω๐๐๐๐ × ψ๐ ๐๐๐๐ ๐ ω๐๐๐๐ = ๐ ω๐๐๐๐ ๐๐ = ๐ก๐๐๐๐ ๐ ๐ = โ๐ ๐๐ ๐๐ก Per unit differential operator ๐๐ = −๐ ๐ × ๐๐ +๐ψ๐ − ω๐ ψ๐ ๐๐น ๐๐น ๐๐๐๐ = ๐ ๐น × ๐๐น ๐ψ๐น + ๐๐น ๐๐๐๐ × ๐๐น ๐๐๐๐ ω๐๐๐๐ × ψ๐น ๐๐๐๐ ๐๐น = ๐ ๐น × ๐๐น +๐ψ๐น ๐ ๐ ๐๐ ๐๐ = − ๐0 ๐ ๐น ๐๐น 0 = 0 ๐ ๐ ๐ ๐ท ๐ ๐ ๐ ๐ ๐ψd ๐๐ −ω๐ ψq 0 0 0 0 − − ๐๐ + ๐ψq + ω๐ ψ d 3๐ ๐ ๐0 3๐ฟ๐ ๐๐0 ๐ψ0 0 ๐0 ๐ψF ๐๐น ๐๐ท + ๐ψD ๐ψQ ๐๐ 38 Per Unit Power and Torque • Instantaneous power at the machine terminal: Pt = eaia+ebib+ecic = [ea eb ec]×[ia ib ic]T = [ed eq e0] ×P-TP-1× [id iq i0]T = [ed eq e0] ×(PTP)-1× [id iq i0]T 2 0 0 1 0 1 3 ๐ ๐ ๐ = 0 2 0 , (PTP)-1 = 0 1 3 3 2 Pt= (ed id + eq iq+2e0 i0) 0 0 1 0 0 2 3 2 = (ed id + eq iq) (under balanced conditions) Divided by S3φ base = P t= 3 e ×i 2 s base s base 3 2 [ (id pψd + iq pψq) +(ψd iq - ψqid) ωr - (i2d+ i2q)Ra] 3 3 ๐ ๐ก = ๐๐ ๐ ๐ + ๐ ๐ ๐ ๐ Power transferred across the air-gap • The air-gap torque (i.e. electrical torque): Te= 2 (ψd iq - ψqid) ωr/ωmech= 2 (ψd iq - ψqid) pf/2 Divided by Tbase = 3 ๐๐ ( )ψs base×is base 2 2 39 ๐๐ = ψ๐ ๐๐ − ψ๐ ๐๐ 0 0 2 Per Unit Reactance X=2π f L • If f=fbase ๐ 2π๐ ๐ฟ = × ๐ ๐๐๐๐ 2π๐๐๐๐๐ ๐ฟ๐๐๐๐ ๐ = ๐ฟ • The per unit reactance of a winding is numerically equal to the per unit inductance. 40
