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 W ound simular to a wound rotor induction motor W hen 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 = N rotor Called “Excitation”

9. Torque Angle No Load

10. Torque Angle Full Load 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

11. 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 Torque Angle

12. Amortisseur Windings W hen relative movement between the stator and the rotor poles occurs V oltage is induced into these windings. S ubsequent induced magnetic field tends to slow movement and act like a “shock absorber” C an be used to aid starting in a simular way to that of the squirrel cage conductors Similar to the squirrel cage found in induction machines 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 As we increase the excitation T h e s t a t o r c u r r e n t w i l l d r o p And the Power Factor Improves

18. If we continue to increase the excitation T h e s t a t o r c u r r e n t w i l l i n c r e a s e And the Power Factor detieriates

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

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 Lagging Unity Leading Under ExcitedOver Excited If the bride is over excited she will lead you to the marriage bed

22. 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 38% PF = X 100A= 38A Stator Current 0.9Lagging

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

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 450kVA 49.5  342kVar 292.5kW 90kW 31.8  90 x tan 31.8  = 55.8kVar 342 – 55.8 =286kVar 292.5 + 90 =382.5kW tan -1 x 286/382.5 =36.8  PF =0.8 Lag