ECE 422/522 Power System Operations & Planning/ Power Systems Analysis II 2 Synchronous achine odeling Spring 214 Instructor: Kai Sun 1
Outline Synchronous achine odeling Per Unit Representation Simplified odels for Stability Studies 2
Synchronous Generators Salient-pole rotor Cylindrical/round 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 Round rotors 7% of large synchronous generators (15~15VA) 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 36 or 18rpm (2 or 4-pole) No special damper windings but eddy in the solid steal rotor gives damping effects 4 16 poles salient-pole rotor (12 W) Round rotor generator under construction (Source: http://emadrlc.blogspot.com)
Generator odel d F r n Flux linkage with coil a (leading the axis of a by ωt) ψa Nφcosωt Induced voltage: dψ a ea ωnφsinωt Emax sinωt dt (reaches the maximum at the current position) f P n 2 6 (n: synchronous speed in rpm; P: the number of poles) m N S γ γ F s e a Axis of coil a (reference) Assume: i a is lagging e a by γ (i a reaches the maximum when mn aligns with aa ) 2 4 ia Imax sin( ωt γ) ib Imax sin( ωt γ π) ic Imax sin( ωt γ π) 3 3 agnetic motive forces (F s) of three phases: F Ki F sin( ωt γ) a a m 2 Fb Kib Fmsin( ωt γ π) 3 4 Fc Kic Fmsin( ωt γ π) 3 F s 3 2 F m F s is orthogonal to mn and revolving synchronously with F F r due to the rotor 5
Under Steady-State Conditions F r + F s gives F F sr in air gap F s induces EF E ar F sr results air gap flux φ sr to induce EF N γ d F r F sr n Axis of coil a E sr E a +E ar For a round rotor, define the reactance of the armature reaction X ar -E ar /(ji a ) Terminal voltage V, resistance R a and leakage and reactance X l satisfy m S γ e a F s (reference) E V + [ R + j( X + X )] I V + ( R + jx ) I a a ar a a s a F r X s X l +X ar is known as the synchronous reactance F sr γ E a E a Load E sr E ar F s 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 1-1977: the quadrature (q) axis is defined to lead the d-axis by 9 7
Winding Circuits Note: we define opposite directions for the current and flux Equations on the EF (electromotive force) and flux of each winding e a dψ a /dt R a i a ψ Q F ψ F ψ a l aa i a l aa i b l aa i c + l aa i F + l aa i D + l aa i Q ψ D e F dψ F /dt + R F i F ψ F l Fa i a l Fb i b l Fc i c + l FF i F + l FD i D + l FQ i Q ψ a ψ b ψ c ψ F ψ D ψ Q l aa l aa l aa l aa l aa l aa l bb l bb l bb l bb l bb l bb l cc l cc l cc l cf l cd l cq l FF l FF l FF l FF l FF l FF l DD l DD l DD l DD l DD l DD l QQ l QQ l QQ l QQ l QQ l QQ i a i b i c i F i D iq Stator self-inductances (l aa, l bb, l cc ) Stator mutual inductances (l ab, l bc, l ac ) Stator-to-rotor mutual inductances (l af, l bd, l aq ) Rotor self-inductances (l FF, l DD, l QQ ) Rotor mutual inductances (l FD, l DQ, l FQ ) ψ aaa ψ FFF L SS L SS L RR L RR i aaa i FFF ost of the efforts in synchronous machine modeling is to find constants and make the EF and flux equations be simpler 8
ψ a ψ b ψ c ψ F ψ D ψ Q l aa l aa l aa l aa l aa l aa l bb l bb l bb l bb l bb l bb l cc l cc l cc l bb l bb l bb l FF l FF l FF l FF l FF l FF l DD l DD l DD l DD l DD l DD l QQ l QQ l QQ l QQ l QQ l QQ i a i b i c i F i D iq Wb Turns 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. N x ~ turns of winding x l xy N x Φ my / i y N x N y P xy l yx Φ my ~ mutual flux linking windings x and y due to current in winding y P xy ~ permeance of the mutual flux path A salient pole machine has significantly different permeances in d and q axes, such that the P xy involving a stator winding (e.g. P ab and P af ) is a function of the rotor position α and reaches the maximum twice during o ~36 o P xy P +P 2 cos2α It is advisable to consider d- and q-axis components of P xy individually 9
Stator self-inductances (l aa, l bb, l cc ) l aa is equal to the ratio of flux linking phase a winding to current i a, with zero currents in all other circuits, and an be approximated as l aa L aa + L aa2 cos2θ Detailed calculation: F a has a sinusoidal distribution in space with its peak centered on phase a axis. Resolve F a into two Fs centered on d and q axes F ad has peak N a i a cosθ F aq has peak -N a i a sinθ Air-gap fluxes Φ gad (N a i a cosθ)p d Φ gaq (-N a i a sinθ)p q Φ gaa Φ gad cosθ - Φ gaq sinθ N a i a (P d cos 2 θ + P q sin 2 θ) l gaa N a i a ( P d+pq 2 + P d P q 2 N a Φ gaa /i a N 2 a( P d+pq 2 L g + L m cos2θ cos2θ) + P d P q 2 cos2θ) 1 Add the leakage inductance: l aa l al + l gaa L al + L g + L m cos2θ L s + L m cos2θ L s >L m l aa L s + L m cos2θ l bb L s + L m cos2(θ-2π/3) l cc L s + L m cos2(θ+2π/3)
Stator utual Inductances (l ab, l bc, l ac ) l ab < since windings a and b have12 o (>9 o ) displacement b q a Has the maximum absolute value when θ -3 or 15. S N θ -3 o d Detailed calculation: Φ gba Φ gad cos(θ-2π/3) - Φ gaq sin(θ-2π/3) N a i a [P d cosθcos(θ-2π/3)+p q sinθsin(θ-2π/3)] b c a N a i a [ - P d+pq 4 + P d P q cos(2θ-2π/3)] 2 l gba N a Φ gba /i a -L g /2 + L m cos(2θ-2π/3) d N θ15 o S q Add leakage flux: c l ab l ba s L s /2 L al - L g /2 + L m cos(2θ-2π/3) - s + L m cos(2θ-2π/3) - s - L m cos2(θ+π/6) l ab - s - L m cos2(θ+π/6) l bc - s - L m cos2(θ-π/2) l ca - s - L m cos2(θ+5π/6) 11
Stator to Rotor utual Inductances (l af, l bf, l cf, l ad, l bd, l cd, l aq, l bq, l cq ) The rotor sees a constant permeance if neglecting variations in the air gap due to stator slots b q (Q) θ d (F, D) a When the flux linking a stator winding and a rotor winding reaches the maximum when they aligns with each other and is when they are displaced by 9 o c d-axis l af l Fa F cosθ l bf l Fb F cos(θ-2π/3) l cf l Fc F cos(θ+2π/3) l ad l Da D cosθ l bd l Db D cos(θ-2π/3) l cd l Dc D cos(θ+2π/3) q-axis l aq l Qa - Q sinθ l bq l Qb - Q sin(θ-2π/3) l cq l Qc - Q sin(θ+2π/3) 12
For Salient-pole Rotors Which of the curves will be different for round rotors? 13 No 2 nd harmonic
Rotor Inductances (l FF, l DD, l QQ, l FD, l FQ, l DQ ) They are all constant Rotor self inductances l FF L F l DD L D l QQ L Q Rotor mutual inductances l FD l DF R l FQ l QF l DQ l QD 14
Summary ψ aaa ψ FFF L SS L SS L RR L RR i aaa i FFF L RS L T SR L RR L F R R L D L Q What if we define q-axis lagging d-axis by 9 o? Only L RR is constant L SS and L SR are θ or time dependent How to simplify L SS and L SR? Diagonalize L SS Remove time dependency 15
Observations ψ aaa ψ FFF L SS L SS L RR L RR i aaa i FFF All harmonic terms in L SS (1 st harmonic) and L SR (2 nd harmonic) are due to the rotor rotating relative to a, b and c to cause variations in permeance Constant L RR doesn t have harmonic terms because it is in a reference frame rotating with the rotor ψ FDQ - L SR i abc + L RR i FDQ L SR i abc -ψ FDQ + L RR i FDQ L SR i abc may be represented by functions independent of θ Represent stator currents and flux linkages also in a reference frame rotating with the rotor. 16
ψ FDQ - L SR i abc + L RR i FDQ ψ F -l af i a - l bf i b l cf i c +l FF i F + l FD i D + l FQ i Q - F cosθ i a F cos(θ-2π/3) i b F cos(θ+2π/3)i c +L F i F + R i D + - F [i a cosθ + i b cos(θ-2π/3) + i c cos(θ+2π/3)] + L F i F + R i D K 1 i d ψ D -l ad i a - l bd i b l cd i c +l DF i F + l DD i D + l DQ i Q - D cosθ i a - D cos(θ-2π/3) i b - D cos(θ+2π/3)i c + R i F + L D i D + - D [i a cosθ + i b cos(θ-2π/3) + i c cos(θ+2π/3)] + R i F + L D i D K 2 i d ψ Q -l aq i a - l bq i b l cq i c +l QF i F + l QD i D + l QQ i Q Q sinθ i a + Q sin(θ-2π/3) i b + Q sin(θ+2π/3) i c + + + L Q i Q Q [i a sinθ + i b sin(θ-2π/3) + i c sin(θ+2π/3) ] +L Q i Q K 3 i q 17
Park s (dq) Transformation Define For balanced steady-state conditions: i a I m sinω s t i b I m sin(ω s t - 2π/3) i c I m sin(ω s t + 2π/3) i d k d [i a cosθ +i b cos(θ-2π/3) +i c cos(θ+2π/3)] i d k d I m sin(ω s t-θ) 3/2 i q - k g [i a sinθ +i b sin(θ-2π/3) +i c sin(θ+2π/3) ] i q -k q I m cos(ω s t-θ) 3/2 θω r t+θ, ω r ω s Define i k (i a + i b + i c ) i d -k d I m sinθ 3/2 i q -k q I m cosθ 3/2 Constant What if we define q-axis lagging d-axis by 9 o? 18
Park s Transformation atrix - P ψ aaa ψ FFF L SS L SS L RR L RR i aaa i FFF ψ dq P ψ abc i dq P i abc ψ ddd ψ FFF LL SS LL SS LL RR L RR i ddd i FFF We hope L RS (L SR ) T like L RS (L SR ) T L RS L RS P -1 L T SR P -1 (P -T L SR ) T (L SR ) T (P L SR ) T P -T P or P T PI (P is a unitary matrix) k d k q 2 3 and k 1 3 19
Flux Equations after Park s Transformation ψ d ψ q ψ ψ F ψ D ψ Q L d k F k D L q k Q L k F L F R k D R L D k Q L Q i d i q i i F i D i Q k 3 2 L d L s + s + 3L m /2 L q L s + s - 3L m /2 L L s - 2 s ψ ψ d ψ F ψ D ψ q ψ Q L L d k F k D k F L F R k D R L D L q k Q k Q L Q i i d i F i D i q i Q 2
Voltage Equations edψ/dt ± R i Stator: ψ and i are in opposite directions Rotor: ψ and i are the same directions ea en ψ a Ra ia e b e ψ b Rb i n b e c e ψ n c R c i c + ef ψ F RF if ed ψ D R D i D eq ψ Q RQ iq e aaa e FFF R aaa R FFF i aaa i FFF + ψ aaa ψ FFF + e n (A neutral line is added compared to slide #8) Neutral line: e n 1 Rn Rn Rn ia Ln Ln Ln ia di n d e e n ( Rnin Ln )1 Rn Rn R n i b Ln Ln L n i n + b dt dt e n 1 Rn Rn R n i c Ln Ln L n i c d Ri n abc Ln iabc dt 21
e aaa e FFF R aaa R FFF i aaa i FFF + ψ aaa ψ FFF + e n Assume R a R b R c ψ dq P ψ abc i dq P i abc 22
P ψ aaa ψ ddd d Pψ aaa dd P ψ aaa + Pψ aaap P 1 ψ ddd + Pψ aaa θω r t+θ 23
Transformer voltages due to flux change in time ( under steady-state conditions) Speed voltages due to flux change in space S ω r ψ q ω r ψ d ω r ( L q i q + k Q i Q ) ω r ( L d i d + k F i F + k D i D ) ωrl q ( i q ) ω r k Q i Q ω r L d ( i d ) + ω r k F i F + ω r k D i D ) 24
P e n P e n P ( -R n i abc - L n di abc /dt ) -P R n P -1 P i abc - P L n P -1 P di abc /dt -P R n P -1 i dq - P L n P -1 (di dq /dt dp/dt i abc ) PR P n PL P 3R n 3L n 1 1 n Note: P L n P -1 dp/dt i abc Pen 3Ri n d 3Ln i d ndq R L n n R R R n n n Rn Rn R n Rn Rn R n L L L n n n Ln Ln L n L L L n n n 25
e ddd e FFF R R R i ddd i FFF + ψ ddd ψ FFF + S + n ddd e e d ef e D eq e Q R a + 3R n R a R F R D R a R Q i i d i F i D i q i Q ω r L q ω r k Q ω r L d ω r k F ω r k D + L + 3L n i i d i F i D i q i Q 26 L d k F k D k F L F R k D R L D L q k Q k Q L Q d i i d i F i D i q i Q /dt +
Voltage Equations after Park s Transformation e e d ef e D eq e Q L + 3L n R a + 3R n R a ω r L q ω r k Q R F R D ω r L d ω r k F ω r k D R a R Q L d k F k D k F L F R k D R L D L q k Q k Q L Q d i i d i F i D i q i Q /dt i i d i F i D i q i Q + 27
Winding Circuits after Park s Transformation e e d ef e D eq e Q R a + 3R n R a R F R D R a R Q i i d i F i D i q i Q + L + 3L n L d k F k D k F L F R k D R L D L q k Q k Q L Q d i i d i F i D i q i Q /dt + ω r ψ q ω r ψ d d axis flux causes a speed voltage ω r ψ d in the q axis winding q axis flux causes a speed voltage ω r ψ q in the d axis winding 28
29 If k d k q 2/3 and k 1/3, a unit-to-unit relationship holds between abc and dq variables. i a I m sinω s t i b I m sin(ω s t - 2π/3) i c I m sin(ω s t + 2π/3) i d k d I m sin(ω s t-θ) 3/2 i q -k q I m cos(ω s t-θ) 3/2 i k (i a + i b + i c ) P By defining proper base inductances, the matrix may become symmetric in per unit Alternative Park s Transformation Q D F q d Q Q D R D R F F Q q D F d Q D F q d i i i i i i L L L L L L 2 3 2 3 2 3 ψ ψ ψ ψ ψ ψ
Per Unit Representation Quantity in p.u. Actual quantity / Base quantity x x x bbbb p.u. 3
Base Quantities for Synchronous achines S base ~i base e base ψ base ~ L base i base Z base ~ e base /i base L base ~ Z base /ω base T base ~ S base / ω base For steady-state conditions, only two base quantities for each voltage level should be provided, e.g. e base and i base, or S base and e base Considering dynamics, 3 base quantities are needed, e.g. f base, e base, i base S base, Z base, L base, ψ base, T base f base, e base, S base i base, Z base, L base, ψ base, T base Base d q F D Q 1 f base f base f base f base f base f base 2 3 31
Stator Base Quantities Using the machine ratings as the base values e s base (V) peak value of rated line-to-neutral voltage i s base (A) peak value of rated line current f base (Hz) rated frequency Accordingly: S 3φ base (VA) 3E RS base I RS base 3(e s base / 2) (i s base / 2) 3 2 e s base i s base Z s base (Ω ) e s base /i s base L s base (H) Z s base /ω base ω base (elec. rad/s) 2πf base ω mbase (mech. rad/s) ω base (2/p f ) t base (s) 1/ ω base 1/(2πf base ) ψ s base (Wb turns) L s base i s base e s base /ω base T base (N m) S 3φ base / ω mbase 3 2 (p f 2 s base i s base Base d q F D Q 1 f base f base f base f base f base f base 2 e s base e s base e s base S 3φ base S 3φ base S 3φ base 3 i s base i s base i s base 32
33 Base d q F D Q 1 f base f base f base f base f base f base 2 e s base e s base e s base 3 i s base i s base i s base i F base i D base i Q base S 3φ base S 3φ base S 3φ base Q D F q d Q Q D R D R F F Q q D F d Q D F q d i i i i i i L L L L L L 2 3 2 3 2 3 ψ ψ ψ ψ ψ ψ i Fbase, i Dbase and i Qbase should enable a symmetric per-unit inductance matrix How to select rotor base quantities?
Ld ψ d ψ q ψ 3 ψ 2 F 3 ψ D 2 ψ Q F D 3 2 L q Q L L F F R L D R D L Q Q i i i if id iq d q ψ d Ld F D id ψ q Lq Q iq ψ L i ψ F F LF R if ψ D D R LD i D ψ Q Q LQ iq ψ d ψ s bbbb L d i d L s bbbb i s bbbb + L d i d L s bbbb i s bbbb + F i F bbbb L s bbbb i s bbbb F i F L s bbbb i s bbbb + i F i F bbbb + ψ d L d i d + F i F + D i D D i D L s bbbb i s bbbb D i D bbbb L s bbbb i s bbbb i D i D bbbb F F F i F bbbb L s bbbb i s bbbb 3 2 F i S bbbb L F bbbb i F bbbb L F base i 2 F base 3 2 L S base i 2 S base ω base L F base i 2 F base 3 2 ω base L S base i 2 S base ψ F 3 F i d L F i F R i D + + ψ F bbbb 2 L F bbbb i F bbbb L F bbbb i F bbbb L F bbbb i F bbbb 3 2 F i S bbbb L F bbbb i F bbbb i d i s bbbb + L F i F L F bbbb i F bbbb + R i D bbbb L F bbbb i F bbbb i D i D bbbb e F base i F base 3 2 e S base i S base S 3φ base e D base i D base e Q base i Q base ψ F F i d +L F i F + R i D 34
Rotor Base Quantities ψ d Ld F D id ψ q Lq Q iq ψ L i ψ F F LF R if ψ D D R LD i D ψ Q Q LQ iq Stator self-inductance L d or L q can be split into two parts: Leakage inductance due to flux that does not link any rotor circuit utual inductance due to flux that links the rotor circuits Stator leakage inductances in d and q axes are nearly equal. Then L d L l + L aa L q L l + L aa Assume that all the per unit mutual inductances between the stator and rotor circuits in each axis are equal F D L aa Q L aa Some references suggest rotor mutual inductance R L aa to further simplify equivalent circuits 35
L aa L s bbbb L aa F D F i F bbbb D i D bbbb L s bbbb i s bbbb L s bbbb i s bbbb i F bbbb L aa F i S bbbb, A e F base S 3φ base / i F base, V Z F base e F base / i F base S 3φ base / i 2 F base, Ω L F base Z F base / ω base, H ψ F base L F base i F base, Wb turns F base L S base i S base / i F base L aa Q L aa Q i Q bbbb L s bbbb L s bbbb i s bbbb i Q bbbb L aa Q i S bbbb, A e Q base S 3φ base / i Q base, V Z Q base e Q base / i Q base S 3φ base / i 2 Q base, Ω L Q base Z Q base / ω base, H ψ Q base L Q base i Q base, Wb turns Q base L S base i S base / i Q base i D bbbb L aa D i S bbbb, A e D base S 3φ base / i D base, V Z D base e D base / i D base S 3φ base / i 2 D base, Ω L D base Z D base / ω base, H ψ D base L D base i D base, Wb turns D base L S base i S base / i D base 36 L ad -L aq based per unit system
Per Unit Voltage Equations e ddd Ri ddd + ψ ddd + S + n ddd e FFF R R i FFF + ψ FFF e d eq e R a R a R a i d iq i + pψ d pψ q pψ + ω r ψ q ω r ψ d 3R n i 3L n pi e F R F R D R Q i F i D iq + pψ F pψ D pψ Q pd/dt differential operator Divide both sides of each equation by one of the following: e S base ω base ψ S base ω base L S base i S base Z S base i S base e F base ω base ψ F base ω base L F base i F base Z F base i F base e D base ω base ψ D base ω base L D base i D base Z D base i D base e Q base ω base ψ Q base ω base L Q base i Q base Z Q base i Q base 37
For example: e d e s bbbb R a i d Z s bbbb i s bbbb + pψ d ω bbbb ψ s bbbb ω r ψ q ω bbbb ψ s bbbb Note: p ω bbbb d ω bbbb dd t bbbbd dd d dt p Per unit differential operator e d R a i d +pψ d ω r ψ q e F e F bbbb R F i F Z F bbbb i F bbbb + pψ F ω bbbb ψ F bbbb e F R F i F +pψ F e d e q e R a R a R a i d i q + i pψ d pψ q + pψ ω r ψ q ω r ψ d 3R n i 3L n pi e F R F R D R Q i F i D + i Q pψ F pψ D pψ Q 38
Per Unit Power and Torque Instantaneous power at the machine terminal: P t e a i a +e b i b +e c i c [e a e b e c ] [i a i b i c ] T [e d e q e ] P -T P -1 [i d i q i ] T [e d e q e ] (P T P) -1 [i d i q i ] T P t 3 2 (e d i d + e q i q +2e i ) P T P 1 3 2 2 1, (P T P) -1 3 2 1 1 2 3 2 (e d i d + e q i q ) (under balanced conditions) Divided by S 3φ base 3 2 e s base i s base P t e d i d + e q i q P t 3 2 [ (i d pψ d + i q pψ q ) +(ψ d i q - ψ q i d ) ω r - (i 2 d+ i 2 q)r a ] Power transferred across the air-gap The air-gap torque (i.e. electrical torque): T e 3 2 (ψ d i q - ψ q i d ) ω r /ω mech 3 2 (ψ d i q - ψ q i d ) p f /2 Divided by T base 3 2 (p f 2 )ψ s base i s base T e ψ d i q ψ q i d 39
Per Unit Reactance X2π f L X Z bbbb 2πf 2πf bbbb L L bbbb If ff base X L The per unit reactance of a winding is numerically equal to the per unit inductance. 4