Bài giảng Electromechanical energy conversion - Chapter IV: Synchronous Machines - Nguyễn Công Phương

NguyễnCôngPhương ELECTROMECHANICAL ENERGY  CONVERSION Synchronous Machines Contents I. Magnetic Circuits and Magnetic Materials II. Electromechanical Energy Conversion Principles III. Introduction to Rotating Machines IV. Synchronous Machines V. Polyphase Induction Machines VI. DC Machines VII.Variable – Reluctance Machines and Stepping Motors VIII.Single and Two – Phase Motors IX. Speed and Torque Control sites.google.com/site/ncpdhbkhn 2 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 3 Introduction to Polyphase Synchronous Machines (1) Synchronous machine: • An AC machine. • Speed under steady – state conditions is Field winding proportional to the frequency of the (AC) c current in its armature winding (on the a b stator, & usually a three – phase N winding). Rotor • The rotor (field winding): S – There is a magnetic field created by the a DC current on the motor. b Stator – Rotates at the same speed as the rotating c magnetic field produced by the armature Armature winding winding • The DC power required for excitation is supplied by the excitation system sites.google.com/site/ncpdhbkhn 4 Introduction to Polyphase Synchronous Machines (2) • In older machines, the excitation current is supplied through slipping rings from a DC machine. • In modern machines, the excitation is supplied from AC exciters & solid – state rectifiers. • In other systems (brushless excitation system), the alternator of the AC exciter is on the rotor, & the current is supplied directly to the field winding without any slipping ring. sites.google.com/site/ncpdhbkhn 5 Introduction to Polyphase Synchronous Machines (3) A single synchronous generator supplying power to an impedance load: • Acts as a voltage source. • Its frequency is determined by the speed of its mechanical drive (the prime mover). • The amplitude of the generated voltage is proportional to the frequency & the field current. • The current & power factor are determined by the generator field excitation & the impedance of the generator & the load. sites.google.com/site/ncpdhbkhn 6 Introduction to Polyphase Synchronous Machines (4) Synchronous generators: • Can be readily operated in parallel. • The electricity supply system has hundreds of them operating in parallel, interconnected by thousands of kilometres of transmission lines. • Must be coordinated both technically & administratively. • When a synchronous generator is connected to a large interconnected system: – The voltage & frequency at its armature terminals are substantially fixed by the system. – Armature currents will produce a component of the air – gap magnetic field which rotates at synchronous speed as determined by the system electrical frequency fe. – The fields of the stator & rotor must rotate at the same speed, & therefore the rotor must turn at precisely synchronous speed. – It is useful to represent the remainder of the system as a constant – frequency, constant – voltage source, referred to as an “infinte bus”. sites.google.com/site/ncpdhbkhn 7 Introduction to Polyphase Magnetic axis Synchronous Machines (5) of rotor 2  poles a TFsin R fRF c b 22 f  RF • Ф : resulting air – gap flux per pole. R Magnetic axis • Ff: mmf of the DC field winding. b  f c of phase a • δRF: electrical phase angle between magnetic axes of ФR & Ff. • In a generator: the prime – mover torque a acts in the direction of rotation of the rotor, pushing the rotor mmf wave ahead of the T resultant air – gap flux, the g electromechanical torque then opposes rotation. Generator o o  • In a synchronous motor: the 180 90 RF electromechanical torque is in the direction 0 90o 180o of rotation, in opposition to the retarding Motor torque of the mechanical load on the shaft. m sites.google.com/site/ncpdhbkhn 8 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances & Equivalent Circuits 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 9 Synchronous – Machines Inductances & Equivalent Circuits (1) v Magnetic axis a of rotor ia a a aaiiii a  ab b  ac c  af f c b f m  t  0 bbaabbbbccbffiiii     iiii    Magnetic axis c ca a cb b cc c cf f b  f c of phase a  f faiiii a  fb b  fc c  ff f a ia va sites.google.com/site/ncpdhbkhn 10 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances & Equivalent Circuits a) Rotor Self – Inductance b) Stator – to – Rotor Mutual Inductances c) Stator Inductances & Synchronous Inductance d) Equivalent Circuit 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 11 Rotor Self – Inductance aaaaabbaccaff iiii bbaabbbbccbffiiii    c caiiii a  cb b  cc c  cf f  f faiiii a  fb b  fc c  ff f  ffffffflL LL0  (rotor self – inductance) sites.google.com/site/ncpdhbkhn 12 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances & Equivalent Circuits a) Rotor Self – Inductance b) Stator – to – Rotor Mutual Inductances c) Stator Inductances & Synchronous Inductance d) Equivalent Circuit 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 13 Stator – to – Rotor Mutual Inductance a  aaiiii a ab b ac c af f b baiiii a  bb b  bc c  bf f c caiiii a  cb b  cc c  cf f  f faiiii a  fb b  fc c  ff f af faL afcos me mst 0 af  fa Lt afcos( e  e0 ) poles (stator – to – rotor mutual inductance)  t me2 m e e0 sites.google.com/site/ncpdhbkhn 14 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances & Equivalent Circuits a) Rotor Self – Inductance b) Stator – to – Rotor Mutual Inductances c) Stator Inductances & Synchronous Inductance d) Equivalent Circuit 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 15 Stator Inductances & Synchronous Inductance (1) a  aaiiii a ab b ac c af f b baiiii a  bb b  bc c  bf f c caiiii a  cb b  cc c  cf f  f faiiii a  fb b  fc c  ff f aa bb ccL aa LL aa0  al ab ba ac caL aa0 /2 (stator inductance & stator mutual inductance) sites.google.com/site/ncpdhbkhn 16 Stator Inductances & Synchronous Inductance (2) a  aaiiii a ab b ac c af f af faLt afcos( e  e0 ) aa bb ccL aa LL aa0  al ab ba ac caL aa0 /2 LL ()L Li aa00 i  aa i  L cos() t  i aaaalab0022 cafeef L ()()cos()L Li aa0 i  i L t  i aa00 al a2 b c af e e f sites.google.com/site/ncpdhbkhn 17 Stator Inductances & Synchronous Inductance (3) L ()()cos()L Li aa0 i  i L t  i aaaala002 bcafeef iiiabc 0 L ()L Li aa0 i  L cos() t  i aaaalaaafeef002 3Laa0 LaliL a afcos( e t  e0 ) i f 2 LsaiL afcos( e t  e0 ) i f 3L L  aa0  L (synchronous inductance) sal2 sites.google.com/site/ncpdhbkhn 18 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances & Equivalent Circuits a) Rotor Self – Inductance b) Stator – to – Rotor Mutual Inductances c) Stator Inductances & Synchronous Inductance d) Equivalent Circuit 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 19 Equivalent Circuits (1) d ei() afdt af f afLt afcos( e  e0 )  LI eLIt sin(  ), E  eaff af e af f e e0 af 2 d vRia aaadt asaafeLi Lcos( t  e0 ) di vRiL a  LIsin( t  ) aaasdt eaffee0 di vRiL a  e aaasdt af sites.google.com/site/ncpdhbkhn 20 Equivalent Circuits (2) di vRiLa  e Iˆ aaasdt af a  X s + Ra ˆ – vVaa Vˆ ˆ Motor a Eaf  Ri RIˆ aa a a ˆˆ ˆˆ VRIaaasaaf jXI  E di LjLIjXIa  ˆˆ s esa sa ˆˆˆˆ dt VRIjXIEaaasaa   f Iˆ eaffLI a eEjˆ eje0 af af  2 X s + Ra – Vˆ ˆ Generator a Eaf  sites.google.com/site/ncpdhbkhn 21 Equivalent Circuits (3) ˆˆ ˆˆ ˆˆˆˆ VRIjXIEaaasaaf  VRIjXIEaaasaaf   ˆ ˆ Ia Ia   X s X s + Ra + Ra – ˆ – ˆ Va Va ˆ Motorˆ Generator Eaf  Eaf  ˆˆ ˆˆˆˆ3 ˆˆ VRIaaasaafaaaealaaaafjXI  E V RI j LLIE 0  2  XLal  al : the armature leakage reactance X  X al + Ra 3 – XL  : the effective magnetizing reactance Eˆ ˆ  aa0 R Va 2 Eˆ Eˆ : the air-gap voltage af   R sites.google.com/site/ncpdhbkhn 22 Equivalent Circuits (4) Ex. A 50-Hz, three – phase synchronous motor has a terminal voltage of 380V (line – line) & a terminal current of 100A at a power factor of 0.96 lagging. The field – current is 38A. The machine synchronous reactance is 1.68Ω. Calculate: a) The generated voltage Eaf; b) The mutual inductance Laf; c) The electrical power? VRIˆˆjXI ˆˆ  E aaasaaf ˆˆ ˆ Eaf VasajXI Ra  0 ˆ Va 380 / 3 220V, line-to-neutral 1oˆ o pf0.96 lagging   cos (0.96)  16.3 Ia 100  16.3 A ˆ o o Ejaf 220 1.68(100  16.3 ) 236.4 43.0 V Eaf 236.4 V eaLIff 2Eaf 2 236.4 Eaf  Laf   28.0mH 2 e I f 50 2  38 PVIpfin 3 a a 3 236.4  100  0.96  68.08kW sites.google.com/site/ncpdhbkhn 23 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics a) Open – Circuit Saturation Characteristic and No – Load Rotational Losses b) Short – Circuit Characteristic and Load Loss c) Steady – State Power – Angle Characteristics d) Steady – State Operating Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 24 Open – Circuit Saturation Characteristic and No – Load Rotational Losses (1) • The open – circuit saturation Air – gap line characteristic (a.k.a. the open – circuit occ saturation curve, OCC) of a synchronous Rated machine is a curve of the open – circuit voltage armature terminal voltage as a function of the field excitation when the machine is running at synchronous speed. • It represents the relation between he space – fundamental component of the air – gap flux and the mmf acting on the Field current magnetic circuit when the field winding creates the only mmf source. Open – circuit armature voltage 0 b a • It is initially linear (the air – gap line) as the field current is increased from zero. • The air – gap line represents the machine open – circuit voltage characteristic corresponding to unsaturated operation. • OCC is a measurement of the relationship between the field current If & the generated voltage Eaf, providing a direct measurement of the field – to – armature mutual inductance Laf. sites.google.com/site/ncpdhbkhn 25 Open – Circuit Saturation Characteristic and No – Load Rotational Losses (2) • The open – circuit saturation characteristic (a.k.a. the open – circuit saturation curve, OCC) of a synchronous machine is a curve of the open – circuit armature terminal voltage as a function of the field excitation when the machine is running at synchronous speed. • The no – load rotational losses: the Open – circuit loss core mechanical power required to drive 0 Open – circuit voltage the synchronous machine during the open – circuit test. • These losses consist of: – Friction loss: constant – Windage loss: constant – Core loss: a function of the flux sites.google.com/site/ncpdhbkhn 26 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics a) Open – Circuit Saturation Characteristic and No – Load Rotational Losses b) Short – Circuit Characteristic and Load Loss c) Steady – State Power – Angle Characteristics d) Steady – State Operating Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 27 Short – Circuit Characteristic and Load Loss (1) • The short – circuit Air – gap line occ characteristic: ) a ) – A plot of armature current occ versus field current. scc – Obtained by applying a three – scc phase short circuit to the armature terminals of a b (ordinates for ordinates for synchronous machine, with the ( machine driven at synchronous Open – circuit voltage speed. 0 f 0’ Short – circuit armature current Field excitation VRIjXIEˆˆˆˆ   aaasaafERjXIˆˆ() ˆ af a s a Va  0 sites.google.com/site/ncpdhbkhn 28 Short – Circuit Characteristic and Load Loss (2) Air – gap line occ ERjXIˆˆ() ) a R aala ) occ ˆˆ scc ERaf() ajXI s a scc Ra  0 b (ordinates for ordinates for ( ˆ Open – circuit voltage EVaf a, air gap 0 f 0’ Short – circuit armature current IIˆ  Field excitation a a, short circuit ˆ Ia Vaag, X s, unsaturated X  X al + R Iasc, a – ˆ ER ˆ Eaf  sites.google.com/site/ncpdhbkhn 29 Short – Circuit Characteristic and Load Loss (3) p Air – gap line occ occ ) ) a ) Va,rated ) occ scc occ scc scc scc Ia,rated c b Ia (ordinates for (ordinates for ordinates for (ordinates for ( Open – circuit voltage Open – circuit voltage 0 0’ 0 f 0’ Short – circuit armature current f ' f '' Short – circuit armature current Field excitation Field excitation Varated, X s  Ia Of Amperes Field No Load Short Circuit Ratio Of Amperes Field Short Circuit sites.google.com/site/ncpdhbkhn 30 Short – Circuit Characteristic and Ex. Load Loss (4) Given data taken from the open- and short-circuit characteristics of a three-phase, Y- connected, 220-V synchronous machine: The open – circuit characteristic: line-to-line voltage = 220V; field current = 2.84A The short – circuit characteristic: Armature current, A 118 152 Field current, A 2.20 2.84 The air – gap line: field current = 2.20A; line-to-line voltage = 202V. Find the unsaturated value of the synchronous reactance, and the short-circuit ratio? 202 V 116.7V aag, 3 Vaag, 116.7 X su,  0.987Ω/phase Iasc, 118 2.84 SCR 1.29 2.20 sites.google.com/site/ncpdhbkhn 31 Short – Circuit Characteristic and Load Loss (5) • The machine flux level is low  the core loss is negligible. • The mechanical power (A) required to drive the synchronous machine consists of: – Friction & windage losses (B), and – Losses caused by the armature current (C). • C = A – B: the short – circuit load loss. • The short – circuit load loss consists of: – I2R loss in the armature winding, and – Local losses by the armature leakage flux, and – The very small core loss caused by the resultant flux. sites.google.com/site/ncpdhbkhn 32 Short – Circuit Characteristic and Load Loss (6) RT234.5  T  Rtt 234.5  short circuit load loss R  aeff, ()short circuit armature current 2 sites.google.com/site/ncpdhbkhn 33 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics a) Open – Circuit Saturation Characteristic and No – Load Rotational Losses b) Short – Circuit Characteristic and Load Loss c) Steady – State Power – Angle Characteristics d) Steady – State Operating Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 34 Steady – State Power – Angle Characteristics (1) Iˆ R X   ˆˆˆ ˆ ˆ ER12()jXI  E E1 E2   ˆˆ EE12 ˆ Iˆ E1 R  jX jXIˆ j Ee12 E Eˆ   2 o jZ 90 Ze  ˆ I RIˆ EE 12eejj() Z Z ZZ EE I cos 12 cos(  ) cos(  ) ZZZ Z sites.google.com/site/ncpdhbkhn 35 Steady – State Power – Angle Characteristics (2) Iˆ R X   EE12 I cos  cos(Z ) cos( Z ) ˆ ˆ ZZ E1 E2   PEI22 cos ˆ R E1 cos( ) cos  ZZZ jXIˆ 2 EE12 E 2 R Eˆ P cos(  )   2 o 2 Z 2 90 Z Z  Iˆ o1 R RIˆ ZZ90 tan  X 2 EE12 E 2 R P2 sin( Z )  Z Z 2 2 EE12 E 1 R P1 sin(Z ) Z Z 2 sites.google.com/site/ncpdhbkhn 36 Steady – State Power – Angle Characteristics (3) Iˆ R X   EE E2 R P 12sin( ) 1 1 Z 2 Eˆ Eˆ Z Z 1 2   2 EE12 E 2 R P2 sin(Z ) Eˆ Z Z 2 1 jXIˆ  ZX ˆ RZ  E2   90o Z  0  Iˆ ˆ EE RI PP 12sin 12 X (Power –angle characteristic) If  90o EE PP 12 1,max 2,max X sites.google.com/site/ncpdhbkhn 37 Steady – State Power – Angle Characteristics (4) Iˆ R X X s  X eq   + ˆ + – – Va Eˆ Eˆ 1 2 Eˆ Vˆ   af  eq Generator Thevenin equivalent for the external system EE12 EVaf eq PP12 sin P  sin X XXseq EVaf eq Pmax  XXseq sites.google.com/site/ncpdhbkhn 38 Steady – State Power – Angle Ex. 1 Characteristics (5) Given a 2000-hp, 2300-V, unity-power-factor, three-phase, Y-connected, 30-pole, 60- Hz synchronous motor has a synchronous reactance of 1.95 Ω/phase. It is supplied with power directly from a 60-Hz, 2300-V infinite bus. Its field excitation is constant. All losses are neglected. Find the maximum power & torque which this motor can deliver? P 2000 0.746  1492 kVA ˆ three phase Ia 1492  X sm Pa 497 kVA + R 3 a – ˆ Va 2300 ˆ V 1328 V Eafm  a 3 Pa 497 PVIaaacos VI aaIa  374A Va 1328 2222 EVXIafm a() sm a 1328 (1.95 374) 1515V sites.google.com/site/ncpdhbkhn 39 Steady – State Power – Angle Ex. 1 Characteristics (6) Given a 2000-hp, 2300-V, unity-power-factor, three-phase, Y-connected, 30-pole, 60- Hz synchronous motor has a synchronous reactance of 1.95 Ω/phase. It is supplied with power directly from a 60-Hz, 2300-V infinite bus. Its field excitation is constant. All losses are neglected. Find the maximum power & torque which this motor can deliver? VEaafm1328V; 1515V VEaafm 1328 1515 Pa,max  1032 kW X sm 1.95 PPthree phase,max3 a ,max 3 1032 3096kW 2 2    2608rad/s sepoles 30 3 Pthree phase,max 3096 10 Tmax 123.2 kNm s 8 sites.google.com/site/ncpdhbkhn 40 Steady – State Power – Angle Ex. 2 Characteristics (7) Given a 2000-hp, 2300-V, unity-power-factor, three-phase, Y-connected, 30-pole, 60- Hz synchronous motor has a synchronous reactance of 1.95 Ω/phase. It is supplied with power from a three-phase, Y-connected, 2300-V, 1500-kVA, 2-pole, 3600 r/min turbin generator whose synchronous reactance is 2.65 Ω/phase. Find the maximum power & torque which could be supplied? VEaafma1328V; 1515V; I 374A Iˆ a X  X 22 sg sm EjXIVˆˆˆEXIV()  + + afg sg a a afg sg a a ˆ – – Va 22 Eafg (2.65 374) 1328 1657V ˆ Eˆ Eafg  afm EEafg afm 1657 1515 Pmax  546kW XXsg sm 2.65 1.95 Pthree phase,max 3 546  1638kW P 1638 103 T three phase,max 65.2 kNm max 8 s sites.google.com/site/ncpdhbkhn 41 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics a) Open – Circuit Saturation Characteristic and No – Load Rotational Losses b) Short – Circuit Characteristic and Load Loss c) Steady – State Power – Angle Characteristics d) Steady – State Operating Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 42 Steady – State Operating Iˆ Characteristics (1) a  X s ˆˆˆ + R EjXIVaf s a a a – Vˆ P  jQ ˆ a VIˆˆ  P jQ Iˆ Eaf  aa a ˆ Va PjQ EjXˆˆ  V Q af sˆ a Va Field heating limit ˆˆ ˆ2 Machine rating EVaf a jXP s()()  jQ V a P ˆˆ 2 0 EVaf a jXP s XQ s V a VEaaf 22 22 X s ()()(EVaf a XP s XQ s V a ) Armature heating limit 22 2 VIaa 2 Va VEaaf PQ  2  Va XXss  X s sites.google.com/site/ncpdhbkhn 43 Steady – State Operating Iˆ Characteristics (2) a  X s + Ra – Vˆ ˆ a Eaf  Q Field heating limit Machine rating 0 P VEaaf X s Armature heating limit VIaa V 2  a X s sites.google.com/site/ncpdhbkhn 44 Steady – State Operating Ex. Characteristics (3) Data are losses of a 45-kVA synchronous motor, the terminal voltage is 220V, the power factor is 0.80 lagging, If = 5.50A, the armature & field windings are at a temperature of 75oC. Find its losses? 45000 I 118.1A a 220 3 R 234.5 75 f R 35.5 29.8 234.5 25 f R 234.5 75 a R 0.0339Ω/phase 0.033 234.5 25 a 22 IRff5.50 35.5 1.07 kW 22 3IRaa 3 118.1  0.0399  1.67 kW sites.google.com/site/ncpdhbkhn 45 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 46 Effect of Salient Poles (1) Fundamental Axis of field pole field flux Actual field flux Direct axis ˆ Quadrature axis f Armature surface ˆ Ia Eˆ af Pole ˆ ar Fundamental Actual armature flux armature flux sites.google.com/site/ncpdhbkhn 47 Effect of Salient Poles (2) Fundamental Fundamental field flux field flux Fundamental Actual armature flux Actual field flux field Actual flux armature flux Armature surface Armature surface Pole  Third harmonic armature flux Fundamental Actual armature flux armature flux sites.google.com/site/ncpdhbkhn 48 Effect of Salient Poles (3) Fundamental field flux Fundamental armature flux Actual Axis of field pole field Actual flux armature flux Direct axis ˆ Armature surface  f Quadrature axis  Eˆ ˆ af ar ˆ Ia Third harmonic armature flux sites.google.com/site/ncpdhbkhn 49 Effect of Salient Poles (4) ˆ ar ˆ ad ˆ Iq ˆ Eaf ˆ aq Quadrature axis ˆ ˆ  R Id ˆ Ia ˆ  f Direct axis sites.google.com/site/ncpdhbkhn 50 Effect of Salient Poles (5) Iˆ ˆ q Eaf Quadrature axis  ˆ ˆ Va jXIqq X : Direct – axis synchronous reactance  d X : Quadrature – axis synchronous reactance Iˆ RIˆ jXIˆ q d ˆ aa dd Ia ˆˆ ˆ ˆ ˆ Direct axis EafVRI a a a jXI d d jXI q q ˆ Eaf DE jX Iˆ F qd Iˆ G q ˆ A  Va ˆ E EFjXI qq  D Iˆ Iˆ a DF DE EF jX() Iˆˆ  I jX I ˆ d C ˆ qd q qa B RIaa ˆˆ ˆ AFVaaaqa RI jXI ˆˆ ˆ ˆ The sum(VRIjXIaaaqa )locates the angular position of the generated voltage Eaf sites.google.com/site/ncpdhbkhn 51 Effect of Salient Poles (6) Ex. 1 Given a salient – pole synchronous generator, its reactances Xd & Xq are 1.00 & 0.60 per unit, respectively, Rarmature ≈ 0, power factor is 0.85 lagging. Compute the generated voltage? ˆˆ ˆ ˆ The sum(VRIjXIaaaqa )locates the angular position of the generated voltage Eaf 1o  cos (0.85) 31.8 Eˆ F af  j31.8o ˆ G Ieˆ  1.00 Iq a Vˆ A  a E ˆˆ  AFVaqa jXI D Iˆ o Iˆ a 1.00je 0.60  1.00  j31.8 d C ˆ B RIaa o  1.41e j21.2  21.2o oo o IIdasin(21.2 31.8 ) 1.00sin53 0.80 oo o IIqacos(21.2 31.8 ) 1.00cos53 0.60 sites.google.com/site/ncpdhbkhn 52 Effect of Salient Poles (7) Ex. 1 Given a salient – pole synchronous generator, its reactances Xd & Xq are 1.00 & 0.60 per unit, respectively, Rarmature ≈ 0, power factor is 0.85 lagging. Compute the generated voltage? ˆ  j31.8o o o IeIadq1.00 ; 0.80; I 0.60; 21.2 ; 31.8 Iˆ Eˆ ˆˆ ˆ ˆ ˆ q af Quadrature EafVRI a a a jXI d d jXI q q  Vˆ jXIˆ axis  a qq j21.2o ˆ ˆ jXIˆ Ieˆ  0.60 Id RIaa dd q Iˆ Direct axis a ˆ jj(9021.2)oo  68.6 o Ied 0.80 0.80 e ˆ  jjj68.6ooo 21.2 21.2 Ejejaf 1 1  0.80  0.60  0.60 e  1.73 e sites.google.com/site/ncpdhbkhn 53 Effect of Salient Poles (8) Ex. 2 Given a salient – pole synchronous generator, its reactances Xd = Xq = Xs = 1.00, Rarmature ≈ 0, power factor is 0.85 lagging. Compute the generated voltage? ˆ  j31.8o Iea  1.00 ˆˆ ˆ EafV ajXI s a oo 111je  jj31.8  1.75 e 29.1 sites.google.com/site/ncpdhbkhn 54 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 55 Power – Angle Characteristics of Salient – Pole Machines (1) Eˆ XXXdT d eq af SM X eq XXdq, ˆ XXXqT q eq Veq Iˆ ˆ q Eaf Quadrature VV sin axis deq ˆ ˆ Va jXIqq  Iˆ VV cos a ˆ qeq Iˆ jXIdd d jXIˆ ˆ eq q Direct jXeq I a P VIdd VI qq ˆ Veq axis ˆ jXIeq d VIeq dsin VI eq q cos EVaf eq cos EVaf eqcos XIXIV eq d d d eq cos XI dT d Id X dT Veq sin VXIXIXIeqsin  eq q q q qT q Iq X qT sites.google.com/site/ncpdhbkhn 56 Power – Angle Characteristics of Salient – Pole Machines (2) ˆ Eaf SM X eq XXdq, ˆ Veq PVIeq dsin VI eq q cos 2 EVaf eq cos I  EVaf eq V eq() X dT X qT d P sin  sin 2 X dT XXXdT2 dT qT Veq sin Iq  X qT sites.google.com/site/ncpdhbkhn 57 Power – Angle Characteristics of Salient – Pole Machines (3) EV V2 () X X P af eqsin eq dT qT sin 2 XXXdT2 dT qT 0.6 0.4 0.2 P 0 -0.2 -0.4 -0.6 -3 -2 -1 0 1 2 3  (rad) sites.google.com/site/ncpdhbkhn 58 Power – Angle Characteristics of Salient – Pole Machines (4) EV V2 () X X P af eqsin eq dT qT sin 2 XXXdT2 dT qT EV • af eq sin : for a cylindrical – rotor machine. X dT VX2 () X • eq dT qT sin 2 : 2XXdT qT – Includes the effect of salient poles. – If XdT = XqT (uniform – air – gap machine), then it is zero. • The characteristic for negative values of δ is the same except for a reversal in the sign of P. – For a generator: δ > 0 – For a motor: δ < 0 sites.google.com/site/ncpdhbkhn 59 Power – Angle Characteristics of Salient – Pole Machines (5) EV V2 () X X P af eqsin eq dT qT sin 2 XXXdT2 dT qT Generator (δ > 0): 2 dP EVaf eq V eq() X dT X qT cos 2 cos2  0 max dX dT2 XX dT qT PPmax  max Motor (δ < 0): 2 dP EVaf eq V eq() X dT X qT cos  2 cos 2  0  max dX dT2 XX dT qT PPmax  max sites.google.com/site/ncpdhbkhn 60 Synchronous Machines 1. Introduction to Polyphase Synchronous Machines 2. Synchronous – Machine Inductances and Equivalent Circuits 3. Performance Characteristics 4. Effects of Salient Poles 5. Power – Angle Characteristics of Salient – Pole Machines 6. Permanent – Magnet AC Motors sites.google.com/site/ncpdhbkhn 61 Permanent – Magnet AC Motors • Polyphase synchronous motors with permanent – magnet Rotor rotors. magnetic axis • Similar to the synchronous a machines discussed up to this c   t  point, with the exception that N b m 0 the field windings are replaced  Phase a by permanent magnets  can S be analysed with the techniques b c magnetic axis of this chapter by assuming that a the machine is excited by a Permanent – field current of constant value. magnet rotor • Frequently referred to as “brushless motors”. sites.google.com/site/ncpdhbkhn 62