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Electrical Machines

Electrical Machines. LSEGG216A 9080V. Synchronous Motors. Week 14. Introduction. State the principles of operation of a synchronous motor. Identify the main parts of a synchronous alternator/motor. List the methods used to provide the excitation of a synchronous alternator/motor.

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Electrical Machines

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  1. Electrical Machines LSEGG216A 9080V

  2. Synchronous Motors Week 14

  3. Introduction • State the principles of operation of a synchronous motor. • Identify the main parts of a synchronous alternator/motor. • List the methods used to provide the excitation of a synchronous alternator/motor. • List the starting methods of synchronous

  4. Motor Types 3 Phase 1 Phase • Reluctance • Hysteresis • Permanent Magnet • Inductor

  5. Characteristics • High operating efficiency • Smooth constant starting & accelerating torque • Versatile power factor control • Constant speed • Considerably more expensive than induction motors • Zero starting torque

  6. Stator Same as an induction motor’s stator Some books may call this the Armature

  7. Rotor • Wound simular to a wound rotor induction motor • When operating DC voltage is placed across this coil to produce an electromagnet Some books may call this the Field Windings

  8. Motor Speed • Magnetic force is obtained from an external source (In an induction machine rotor’s magnetism is induced from the stator) • Rotor poles lock onto the RMF • Rotor operates at synchronous speed Called “Excitation” = Nrotor

  9. Torque Angle No Load

  10. Torque Angle Torque Angle • Dependant on: • Load torque • Excitation • The magnetic link between the stator & the rotor can be thought of as a connecting spring. • The excitation can be used to strengthen the spring Full Load

  11. Torque Angle If the load Torque fluctuates Or Changes rapidly • The magnetic “Spring” will bounce • Causing large fluctuations in supply current • Amortisseur windings are added to the rotor • Also known as hunting

  12. Amortisseur Windings Similar to the squirrel cage found in induction machines • When relative movement between the stator and the rotor poles occurs • Voltage is induced into these windings. • Subsequent induced magnetic field tends to slow movement and act like a “shock absorber” • Can be used to aid starting in a simular way to that of the squirrel cage conductors Also Known as “Damper “ windings

  13. Starting Zero starting Torque • Number of methods: • Pony Motor • Low Frequency

  14. Starting Pony Motor An auxiliary smaller motor is used to spin the main motor up to or near Synchronous speed

  15. Starting Low Frequency • The rotor’s excitation windings are short circuited • Act like a wound rotor induction motor • Supply is applied at reduced voltage & frequency • Rotor builds up speed • Excitation is then applied to windings and rotor locks in Tumit 3 and the Shoalhaven hydro schemes use this system

  16. Motor Uses • Can be used as a standard motor similar to an induction motor • Main use as a power factor correction device As an induction machine is cheaper it is seldom used just as a motor Sometimes called a rotary capacitor

  17. Power Factor Correction And the Power Factor Improves The stator current will drop As we increase the excitation

  18. Power Factor Correction And the Power Factor detieriates The stator current will increase If we continue to increase the excitation

  19. Power Factor Correction Curve A = Stator Current Curve B = Power Factor These curves are known as “Vee Curves”

  20. Power Factor Correction These curves are only applicable for a set load torque A different load will produce a different set of curves

  21. Power Factor Correction If the bride is over excited she will lead you to the marriage bed Unity Lagging Leading Under Excited Over Excited

  22. Power Factor Correction A motor has full load of 100A and an excitation current of 8A what will be: • The stator current? • PF of the motor?

  23. Power Factor Correction PF = 0.9 Lagging Stator Current 38% X 100A = 38A

  24. Example • A load of 450 kVA operates at a power factor of 0.65 lagging. • An additional synchronous motor is added having an input power of 90 kW and a maximum power factor of 0.85 leading. • Determine reactive power and the overall power factor 450 x cos 49.5= 292.5kW 49.5 450 x sin 49.5= 450kVA 342kVar

  25. Example • A load of 450 kVA operates at a power factor of 0.65 lagging. • An additional synchronous motor is added having an input power of 90 kW and a maximum power factor of 0.85 leading. • Determine reactive power and the overall power factor tan-1 x 286/382.5 = 36.8 292.5 + 90 = 382.5kW PF = 0.8 Lag 90 x tan 31.8 = 292.5kW 31.8 55.8kVar 342 – 55.8 = 286kVar 49.5 90kW 450kVA 342kVar

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