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Synchronous machines
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- 2. 2 Classroom Rules: 1. Italk, you listen 2. I don’t allow: A. Heads on desk B. Feet on desk C. Going to sleep in class D. MP3 players 3. Mobile phones are to be off, or on “silent” unless OK’d with me, and in you bag (not on desk) 4. Should they be OK’d, and you receive a call, you will answer it outside the classroom 5. No SMS-ing in class 6. Keith Butler’s number is: 0417 637 909
- 3. 3 Topics (from LearningOutcomes) 10. Three-phase synchronous machines • operating principles • construction feature • application 11. Three-phase synchronous machines • effects of load changes • effects of excitation change • load/current characteristics 12. Single phase synchronous machines • alternators • motors • applications
- 4. 4 Assessment (a)All prac’s mustbe completed • A final written exam will be given on the learning outcomes covered. Should a student fail, they will be allowed one further attempt during block. Should they fail this they will be allowed one further attempt within six weeks of completing the block. Should they fail this they will be in a “Show Cause” situation.
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- 6. 6500 KW Genset/ Generator Set, powered with a Cummins VTA28G1 Engine
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- 9. 9 2 x 14MWSynchronous Motors …apparently, they use permanent magnets!
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- 11. 11 3 phase Dunlitemachine
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- 13. 13 Synchronous machines arenot just the big units, but they can be small also.
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- 15. 15 V AC Supply R V I I Currentis in phase with voltage. Time->
- 16. 16 LAC supply V I V I Currentlags the Voltage by 90o
- 17. 17 CAC Supply But ifan ammeter were placed in series it would most definitely read a current. Current appears to pass through the capacitor. In reality, it is charging in one direction, and then discharging and recharging in the other direction.
- 18. 18 V I CAC Supply V I Current leadsthe Voltage by 90o
- 19. 19 This opposition tocurrent flow is called: Inductive reactance, in inductors. (XL) Capacitive reactance in capacitors. (XC) Both Inductors and Capacitors oppose, or “resist” current flow when connected to AC supplies. While it opposes current flow, it is NOT called resistance. Current flow through resistance produces HEAT. Current flow in inductors and capacitors doesn’t!
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- 21. 21 Generator / Transformer/ Motor S F S F S F S F S F S F B A C AC B
- 22. 22 AC B Motor Why isn’t aneutral run to a balanced three phase Star connected load? A B C N ?????
- 23. 23 AC B Motor Because the Starpoint is at 0V A B C 0V0VN And the neutral is at zero volts also. So if they were joined no current would flow. 0A So why join it?
- 24. 24 AC B Motor A B C The neutral isnot connected to a balanced three phase star connected load. Only connected to unbalanced loads!!!
- 25. 25 Transformer S F B S F C Generator/ Transformer / Motor A B C F S F S FS S F A
- 26. 26 Transformer SF B SF C Generator / Transformer/ Motor A B C S F S F SF SF A Swapped
- 27. 27 3-phase Transformer Secondary SF B C S F A S F A B C Note that all windings are connected in series, with the two ends joined together.
- 28. 28 If we didthat with three batteries, there would be major problems!
- 29. 29 The voltmeter shouldread the sum of the three voltages? Right? VA VB VC A C B F S F S FS Transformer V The voltmeter reads the phasor sum of the voltages.
- 30. 30 The voltmeter reads,in effect, the distance between the beginning of VA and the end of VC. ie. 0V VA VB VC A C B F S F S FS Transformer V We can connect the two ends together because the phasor sum adds up to zero!
- 31. 31 Transformer VA VB VC A C B F S F S FS Thevoltmeter reads, in effect, the distance between the beginning of VA and the end of VC. ie. 0V We can connect the two ends together because the phasor sum adds up to zero! No Arc!
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- 33. 33 Generator Load STAR VL =√3 VPH IL = IPH DELTA IL = √3 IPH VL = VPH P = 3 x VPH x IPH x Cosφ = 3 x VL/√3 x IL x Cosφ = 3/√3 x VL x IL x Cosφ = √3 VL x IL x Cosφ P = 3 x VPH x IPH x Cosφ = 3 x VLx IL/√3 x Cosφ = 3/√3 x VL x IL x Cosφ = √3 VL x IL x Cosφ
- 34. 34 P = √3VL x IL x Cosφ NOT: P = 415 x I x pf.
- 35. 35 Three Single PhasePower Equations: True Power = Watts = V x I x Cosφ Apparent Power = VA = V x I Reactive Power = VAR’s = V x I x Sinφ Power Factor = Cosφ where Cosφ = Cosine of the angle between Voltage and Current
- 36. 36 VA Watts Var’s φ Phase angle between current andvolts This can be put as a triangle: VA2 = Watts2 + Var’s2
- 37. 37 V=240V Alternators, where thewindings are limited by the current through them, are rated in VA. To rate them in watts, (ie. watts delivered to the load) would give no idea of the current through them. Load 3 = 14.14A at 45º Load 1 = 10A Load 2 = 20A at 60º P = V x I x Cos 45º = 240 x 14.14 x 0.707 = 2.4kW P = V x I x Cos 0º = 240 x 10 x 1 = 2.4kW P = V x I x Cos 60º = 240 x 20 x 0.5 = 2.4kW
- 38. 38 Q What dictatesthe phase angle of the current supplied by a single alternator supplying a single load? V=240V Load 3 = 14.14A at 45º Load 1 = 10A Load 2 = 20A at 60º P = V x I x Cos 45º = 240 x 14.14 x 0.707 = 2.4kW P = V x I x Cos 0º = 240 x 10 x 1 = 2.4kW P = V x I x Cos 60º = 240 x 20 x 0.5 = 2.4kW A The load
- 39. 39 V Al currents heretake the same power Constant power line
- 40. 40 V Al currents heretake the same power Higher power
- 41. 41 V Al currents heretake the same power Constant power line
- 42. 42 V Al currents heretake the same power Lower power
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- 45. 45 Alternator Mechanical Energy ElectricalEnergy Losses AlternatorPrime Mover - Diesel Engine - Steam Turbine - Small petrol engine Alternator: Pout Eff% = x 100 Pin Alternator: Pin = Pout + Losses
- 46. 46 Losses Synchronous Motor Mechanical Energy MSBElectricalEnergy Motor: Pout Eff% = x 100 Pin Motor Pin = Pout + Losses Motor Load
- 47. 47 Synchronous Machine Stator - Identicallywound to an induction motor. - Connected to supply. Rotor - Constant DC field - Connected to supply via sliprings. Electrical Power DC Supply
- 48. 48 • The statorproduces a rotating magnetic field exactly the same as an induction motor. • The rotor is a magnet and locks in to the RMF • Rotor travels at SYNCHRONOUS SPEED. SYNCHRONOUS MOTOR
- 49. 49 Synchronous Machine If asynchronous motor is OVER driven by the load (eg electric train going down a hill), then it will generate power, still at synchronous speed. If an alternator coupled to the grid is UNDER driven by the prime mover (eg steam stops), then it will motor, and drive the turbine at synchronous speed. Electrical Power DC Supply
- 50. 50 Synchronous Machine In otherwords, the two machines are identical in construction. Electrical Power DC Supply
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- 52. 52 3000RPM 1500RPM 1000RPM750RPM 185kw 310A, .88pf 315A, .86pf 343A, .80pf 348A, .78pf 220kw 362A, .89pf 375A, .86pf 408A, .78pf 412A, .78pf 150kw 242A, .90pf 265A, .87pf 279A, .80pf 278A, .77pf Lower the RPM, • Larger value IS • More lagging IS Characteristic of WEG® Induction Motors. 110kw 182A, .90pf 200A, .84pf 205A, .80pf 203A, .81pf 22kw 39A, .87pf 41A, .83pf 42A, .80pf 47A, .74pf 4kw 7.8A, .87pf 8.2A, .82pf 9A, .74pf 11A, .63pf What is the tendency as RPM gets lower?
- 53. 53 So why usea Synchronous Motor? Uses: – Low Speed Drives. Low speed induction motors draw very large currents at poor power factors. This cannot be altered or corrected. In synchronous motors, the p.f. can be altered to cause the motor to draw minimum current. (The alternative is to use a high speed induction motor through a gearbox.) – Power Factor Correction – Constant Speed drives
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- 57. 57 Salient Pole RotorCylindrical Rotor 2 basic types: Cylindrical rotor Salient Pole -Low speed -Diesel Prime Mover -Hydro systems -High speed -Steam Turbine
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- 64. 64 www.tecowestinghouse.com Small salient polesynchronous machine rotor
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- 75. 75 Losses Synchronous Motor Mechanical Energy MSBElectricalEnergy Motor: Pout Eff% = x 100 Pin Motor Pin = Pout + Losses Motor Load
- 76. 76 Same as aninduction motor.
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- 79. 79 A1 A2 B1B2 C1 C2 2-Pole Machine ie. 3000RPM Inreality, the coils span more slots in a 2-pole motor. N S Notice that for a two pole stator we have a 2-pole rotor
- 80. 80 N N S S A A AA B B B B C C C C 4-pole machine Afour pole stator must have a four pole rotor
- 81. 81 Flux + - Time-> 1 Resultant flux = 1.5x flux of one phase N SN S S N
- 82. 82 Flux + - Time-> 2 Resultant flux = 1.5x flux of one phase N S N S N S
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- 84. 84 Flux + - Time-> 3 4 561 2 So the flux rotates one full rev in one cycle, for our two pole machine.
- 85. 85 Flux + - Time-> 3 4 561 2 Because the flux is a constant value, it gives: 1. Very quiet operation 2. Constant torque as the rotor rotates.
- 86. 86 Flux + - Time-> 3 4 561 2 This rotating magnetic field rotates at: 3000RPM for a 2-pole motor 1500RPM for a 4-pole motor
- 87. 87 Flux + - Time-> 3 4 561 2 To reverse the direction of rotation: reverse any two phases to the motor.
- 88. 88 where N =RPM f = frequency P = Number of poles (per phase). N = 120f/P So the speed of the rotating magnetic field is affected by: Frequency, and Number of poles.
- 89. 89 As the rotatingmagnetic field rotates, the rotor is locked in synchronism with it and is dragged along for the ride.
- 90. 90 N S As the rotatingmagnetic field rotates, the rotor is locked in synchronism with it and is dragged along for the ride.
- 91. 91 What will happenas a load is put on the shaft? N S
- 92. 92 N S What will happenas a load is put on the shaft?
- 93. 93 The load triesto slow it down. But it must do synchronous speed! So it stretches the lines of flux. N S
- 94. 94 N S C/L of RMF C/Lof Rotor Field Torque Angle α
- 95. 95 If the linesstretch to breaking point (ie too much load), then the rotor stalls This is referred to as “Pull Out Torque”. N S
- 96. 96 What would theTorque Curve look like? RPM Ns0 Torque Curve for an induction motor Torque
- 97. 97 What would theTorque Curve look like? Torque RPM Ns0 Torque “Curve” for a Synchronous Motor Pull out Torque Zero Torque below synchronous speed
- 98. 98 1. Amortisseur winding Startinga Synchronous Motor?
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- 103. 103 Starting a SynchronousMotor? 1. Amortisseur winding This gets the motor up to speed as an induction motor. When it is close to synchronous speed it will lock in. 2. Shorting the rotor DC winding and starting it as a wound rotor motor. When it is close to synchronous speed, the short is removed and DC is applied to the rotor. It will (hopefully) lock in.
- 104. 104 3. Using apony motor to get the synchronous motor up to speed, then applying AC to the stator and DC to the rotor. (Not applicable if there is a high starting torque load connected) Note that these starting methods will only work if the load on the motor at start can be reduced or eliminated. Starting a Synchronous Motor?
- 105. 105 • Amortisseur windingsalso reduce hunting. • Hunting is rhythmic fluctuations of the RPM around an average value. • If not subdued, hunting can cause the rotor to swing out of synchronism.
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- 107. 107 N S N S N S N S N S And all thiswhile it is whizzing around at synchronous speed!
- 108. 108 And all thiswhile it is whizzing around at synchronous speed! N S
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- 110. 110 Vsupply Vinduced Induced in the statorfrom the rotor Phasor Diagram of Synchronous Motor
- 111. 111 Vsupply Vinduced Torque angle α Isupply VR Phasor Diagramof Synchronous Motor
- 112. 112 Phasor diagram forincreased load: (Excitation current held constant) Vsupply Vinduced α Isupply VR Increased load = Increased Torque Angle Vinduced Increasing the load increases the power taken from supply
- 113. 113 Vsupply Vinduced Vinduced Phasor diagram forincreased excitation: (Constant Load) α Constant load = Constant Power line
- 114. 114 Phasor diagram forincreased excitation: (Constant Load) Vsupply Vinduced α Constant load = Constant Power line
- 115. 115 Vsupply Phasor diagram forincreased excitation: (Constant Load) Vinduced VR Isupply So to force the supply current leading, we INCREASE excitation α Constant load = Constant Power line
- 116. 116 Vsupply Vinduced Isupply VR Phasor diagram fordecreased excitation: (Constant Load) α Constant load = Constant Power line
- 117. 117 Vsupply α Isupply VR Phasor diagram fordecreased excitation: (Constant Load) Vinduced So to force the supply current lagging, we DECREASE excitation Constant load = Constant Power line
- 118. 118 Vsupply Vinduced Isupply VR α Vsupply α I VR Vinduced Vinduced Isupply VR α Constant load =Constant Power line Vsupply
- 119. 119 Vsupply With a constantload, changing excitation changes the phase angle and value of supply current. Isupply By increasing the DC excitation current to the rotor, the synchronous motor can act as a capacitor It can be used for power factor correction. Constant load = Constant Power line
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- 121. 121 Deductions From VeeCurves • At any particular load there is a certain value of rotor current which gives a minimum value of stator current and unity pf. • If the rotor current is altered either way, the stator current will increase, and pf will decrease away from 1. • For any given load there is a certain value of rotor current below which the rotor will fall out of synchronism. • For any given load there are two values of rotor current that will give identical values of stator current. The lower value gives a lagging pf, and the higher value gives a leading pf.
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- 123. 123 0.8 pf lagUnity 0.8 pf leadPer unit Power output
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- 126. 126 Points: • At aset load there is a value of excitation that will give minimum line current. • Reducing OR increasing excitation from this value will only increase line current. • At any other value of line current, there are two values of excitation current that can produce this. • If a synchronous motor is heavily loaded, supply current may not be able to be driven highly leading. • If a synchronous motor is lightly loaded, supply current can be driven highly leading.
- 127. 127 Single Phase SynchronousMotors • Used when constant speed is critical, with low torque requirements. They have low efficiency, hence made in small sizes. • Application: • clocks • record players • timers • recorders • communications • servo installations • Two main types: • Reluctance motor • Hysteresis motor
- 128. 128 Reluctance Motor Stator • Statorsame as a single phase, split phase motor. • Centrifugal Switch operates at 75% synchronous speed to open circuit the start winding. Rotor • Assembled from laminated sheets with defined teeth cut away. This forms salient poles. • Windings are of the squirrel-cage type. • Number of rotor poles equals the number of stator poles. Two pole, 3000 RPM rotor
- 129. 129 Reluctance Motor • Operation –Starts as an induction motor, with slip. – A single phase stator has a “Start” and “Run” winding. At 75% centrifugal the centrifugal switch operates. – As the load is light there is small slip – The salient poles become permanently magnetised by the stator field – The salient poles will then lock to the stator field. – Once locked into synchronism the motor will continue to operate at synchronous speed. – Not as much power output as a similar physical size 1-phase motor.
- 130. 130 Hysteresis Motor Rotor • Constructedfrom hardened steel rings, instead of thin, magnetically soft, silicon steel laminations. • “Hysteresis” opposes any change once the flux is created, so the rotor will lock into the RMF like a permanent magnet. Stator • Often a shaded pole stator principle is used. • If the shaded pole principle is used then the motor is self starting. • Magnetic poles are established in the rotor. • These poles lock to the stator poles. • The rotor runs at synchronous speed determined by the poles and frequency.
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- 132. 132 Why generate AC?…Why not DC? DC cant be “transformed” through a transformer. AC can go through a transformer. Large brushless DC generators are not possible Large brushless AC alternators are! Why do we want to transform it? It is easier to transmit to distant places at higher voltages as the current will be lower. (P=V x I) Induction motors are simpler and cheaper than DC motors
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- 134. 134 N S V Generating an ACVoltage
- 135. 135 N S Generating anAC Voltage Volts + - Time->
- 136. 136 NS Generating an ACVoltage Volts + - Time->
- 137. 137 NS Generating an ACVoltage Volts + - Time->
- 138. 138 NS Generating an ACVoltage Volts + - Time->
- 139. 139 N S Generating anAC Voltage Volts + - Time->
- 140. 140 NS Generating an ACVoltage Volts + - Time->
- 141. 141 N S V Generating a ACVoltage 3-Phase
- 142. 142 N S Volts + - Time-> Generating aAC Voltage 3-Phase Require: Three sets of coils physically displaced from each other by 120º electrical.
- 143. 143 VA VC VB Generating a ACVoltage 3-Phase N S
- 144. 144 Generating a ACVoltage 3-Phase A1 A2 N S
- 145. 145 Generating a ACVoltage 3-Phase N S A1 A2 B1B2 C1 C2
- 146. 146 N N S S A A AA B B B B C C C C 4-pole machine Afour pole stator must have a four pole rotor Generating a AC Voltage 3-Phase
- 147. 147 Alternator • Reasons forhaving the three phase winding on the stator rather than the rotor: – More space on the stator for the three phase winding. – Only one, low voltage winding on the rotor. • Easier to insulate. • Less problems with centrifugal force. – Only two sliprings required rather than four (3-ph + N)
- 148. 148 Alternator Stator - Connected toload. Rotor - Constant DC field - Connected to its own DC supply via sliprings. Electrical Power Mechanical Power MagneticField
- 149. 149 Alternator Q: What keepsan alternator producing 50Hz under all load conditions? A: The governor on the prime mover. It detects any drop in speed, and tries to speed the unit up. Alternator Petrol Engine
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- 153. 153 VOUTILOAD VR VZ VL VGEN VZ =Internal Impedance of the alternator VR = Internal Resistance of the alternator VL = Internal Reactance of the alternator Resistive Load
- 154. 154 Resistive Load VOUTILOAD VR VZ VL VGEN •Notice that terminal volts DROP as load increases • Load current and p.f. are dictated by the LOAD!
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- 156. 156 Inductive Load VOUT ILOAD V R V Z V L VGEN • Nowthere is a greater voltage drop under load
- 157. 157 Capacitive Load VOUT ILOAD V R V Z V L VGEN • Nowthere is a voltage RISE under load Parallel • Because of the voltage rise under load, it is not desirable to run alternators at a leading power factor.
- 158. 158 Leading pf Unity pf Laggingpf Load Current Output Voltage Effect of Power Factor on Output Voltage
- 159. 159 Voltage Regulation (VNL –VFL) %Voltage Regulation = x 100 VFL eg An alternator output falls from 240V to 200V with constant excitation. Calculate the % voltage regulation. (Ans: 20%)
- 160. 160 Summary: When an alternatoris standing by itself with a single load: Output voltage is affected by excitation current Output frequency is affected by input power to the alternator. Alternators - stand alone
- 161. 161 When an alternatoris tied to the grid, you cannot change: Grid voltage Grid frequency So the output voltage of the alternator will not change, and the output frequency of the alternator will not change. Notice that, for a stand alone alternator with stand alone load, these are the two things that changed when: (a) the excitation was altered, and (b) the power input to the alternator was increased (ie. Put the foot down on the prime mover) Alternators tied to the Grid
- 162. 162 Alternators tied tothe Grid VOUT VR VZ VL VGEN If excitation is increased, and VOUT cannot alter, VGEN will increase and push the triangle over. ILOAD 1. Altering Excitation.
- 163. 163 VOUT ILOAD V R V Z V L VGEN If excitation isincreased, and VOUT cannot alter, VGEN will increase and push the triangle over. Alternators tied to the Grid 1. Altering Excitation. Note that input power to the alternator is not changing, so output power does not change either. Constant Power Line (Output power of the alternator has not Changed)
- 164. 164 VOUT ILOAD V R V Z V L If excitation isreduced, and VOUT cannot alter, VGEN will reduce and pull the triangle back. Alternators tied to the Grid 1. Altering Excitation. VGEN This drives the load current lagging
- 165. 165 VOUT ILOAD V R V Z VL VGEN If excitation is reduced, and VOUT cannot alter, VGEN will reduce and pull the triangle back. Alternators tied to the Grid 1. Altering Excitation. This will drive the load current leading
- 166. 166 If input poweris reduced, and frequency and VOUT cannot alter, output power will reduce. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 167. 167 If input poweris reduced, and frequency and VOUT cannot alter, output power will reduce. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 168. 168 If input poweris reduced, and frequency and VOUT cannot alter, output power will reduce. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD Size of triangle reduces
- 169. 169 If input poweris reduced, and frequency and VOUT cannot alter, output power will reduce. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD Size of triangle reduces
- 170. 170 If input poweris increased, and frequency and VOUT cannot alter, output power will increase. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 171. 171 If input poweris increased, and frequency and VOUT cannot alter, output power will increase. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 172. 172 If input poweris increased, and frequency and VOUT cannot alter, output power will increase. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 173. 173 If input poweris increased, and frequency and VOUT cannot alter, output power will increase. Alternators tied to the Grid 2. Altering input power to the alternator. VOUT VR VZ VL VGEN ILOAD
- 174. 174 Alternators - tiedto the Grid Summary: Changing excitation changes the pf of output current. Changing input power changes output power • Increasing excitation drives load current lagging • Reducing excitation drives load current leading • Increasing input power increases output power • Reducing input power reduces output power • Output frequency and voltage do not change.
- 175. 175 Alternators – standalone Summary: • Changing excitation changes output voltage. • Changing input power changes RPM, which changes output frequency. • Here, output frequency and voltage do change
- 176. 176 Paralleling Alternators To parallelalternators (or parallel one onto the grid), the following criteria must be met: • Output voltage must be the same • Output frequency must be the same • Phase rotation must be the same • Supply voltage must be in phase It is understood that they must both produce the same waveform – a sine wave!
- 177. 177 Alternator Rating Alternators arerated according to: Frequency Voltage Current kVA The frequency dictates the RPM (3000, 1500, etc). Voltage and Current give the kVA rating.
- 178. 178 Efficiency Losses: • By farthe main loss in an alternator is HEAT loss. • If an alternator can be kept cool, more power can be obtained from it. ie. Instead of a 300MW machine, it will become a 500MW machine. • More power must be put into it to get this increased output power. • Cooling large alternators is a big deal! They are often cooled using hydrogen.
- 179. 179 Efficiency Losses: • Copper Losses: I2Rlosses in the stator winding I2R losses in the rotor winding • Iron Losses: Hysteresis loss in stator Eddy current Loss in stator • Friction and windage
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- 182. 182 Single Phase Alternators •These are usually low rated units for portable use. • Prime mover is usually a small petrol or diesel engine. • The engine speed is kept constant by a governor. • This speed will usually be either 3000RPM or 1500RPM • Output voltage is kept constant using an automatic voltage regulator. This senses the output voltage and adjusts the rotor excitation current automatically. • They are usually self exciting, so if the load is left on at start, they may not build up output voltage. • Many small alternators are brushless. • Usually, neither side is earthed. This is called a FLOATING system.
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- 184. 184 Rotor Brushless Alternators AC is sampled RegulatorDC Field P.S. Note: Self Excited 3-phase out
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- 186. 186 Small Alternators -Factors whenchoosing: •Voltage: 240V / 415V (1-phase or 3-phase) •kVA rating •RPM (3000RPM or 1500RPM) •Petrol or Diesel •Brushless or brushes •Ability to start loads such as motors •Extras: Soundproofing, starting, power outlets, mounting holes, 12VDC / welding output
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