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Electrical Machines

Electrical Machines. LSEGG216A 9080V. Transformer Losses & Efficiency. Week 3. Objective s. Describe the power losses which occur in a transformer Describe the tests which allow the power losses of a transformer to be calculated

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Electrical Machines

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  1. Electrical Machines LSEGG216A 9080V

  2. TransformerLosses & Efficiency Week 3

  3. Objectives • Describe the power losses which occur in a transformer • Describe the tests which allow the power losses of a transformer to be calculated • Calculate transformer losses and efficiency using test results • Define the all day efficiency of a transformer • Calculate the all day efficiency of a transformer • Describe the relationship between transformer cooling and rating

  4. Objectives • Calculate the all day efficiency of a transformer • Describe the relationship between transformer cooling and rating • Describe the methods of cooling • List the properties of transformer oil • Describe the tests conducted on transformer oil

  5. Transformer Ratings Transformers are rated to supply a given output in Volt Amps or VA at a specified frequency and terminal voltage.

  6. Transformer Ratings They are NOT rated in Watts The load power factor is unknown

  7. Transformer Ratings They are NOT rated in Watts The load power factor is unknown

  8. Problem S = 2 kVA V2 = 230 V V1 = 6,351 V Power output at unity PF ? Power = S x PF P = 2 kVA x 1 P = 2 kW

  9. Problem S = 2 kVA V2 = 230 V V1 = 6,351 V Full load secondary current at 0.8 PF ? I = 10.87 A

  10. Student Exercise 1

  11. S = 20 kVA V1 = 1270 V V2 = 200 V (a) Power output at unity power factor P = 20 kW

  12. S = 20 kVA V1 = 1270 V V2 = 200 V (b) Power output at 0.8 power factor P = 16 kW

  13. S = 20 kVA V1 = 1270 V V2 = 200 V (c) Full-load secondary current at unity power factor I = 100 A

  14. S = 20 kVA V1 = 1270 V V2 = 200 V (d) Secondary current when transformer supplies 10 kW at 0.8 power factor I = 62.5 A

  15. Efficiency Ratio between Input power and Output Power

  16. Efficiency Efficiency is normally expressed as a percentage

  17. Transformer Efficiency PowerIn PowerOut η = 95% η = 90% η = 100% Overcome Copper Losses OvercomeIron Losses Some Poweris used to:

  18. Student Exercise 2

  19. S = 20 kVA V1 = 230 V η = 90% V2 = 32 V PF = 0.85 (a) Power output of transformer P = 85 W

  20. S = 20 kVA V1 = 230 V η = 90% V2 = 32 V PF = 0.85 (b) Power input P = 94.4 W

  21. S = 20 kVA V1 = 230 V η = 90% V2 = 32 V PF = 0.85 (c) Losses P = 9.4W

  22. Transformer Losses Copper Losses (Cu) • Varies with load current • Produces HEAT • Created by resistance of windings • Short circuit test supplies copper losses

  23. Short Circuit Test Copper Losses (Cu) Limited Supply Voltage ≈ 5-10 % Secondary Short Circuited Wattmeter indicates Copper Losses (Cu)

  24. Short Circuit Test Copper Losses (Cu) • Finds Cooper losses at full load • Copper losses vary with the square of the load Full load Cu loss = 100 W Transformer loaded at 50% PCu = 25 W

  25. Copper Losses (Cu) Cu Losses (W) % Load

  26. Transformer Losses Iron Losses (Fe) • Fixed • Always present • Related to transformers construction Eddy Currents Hysteresis Reduced by laminations Produces HEAT Reduced by using special steels in laminations

  27. Open Circuit Test Finds Iron Losses (Fe) Full Supply Voltage Secondary Open Circuit Wattmeter indicates Iron Losses (Fe)

  28. Transformer Efficiency Student Exercise 3

  29. Cu FL= 840 W Fe = 220 W Sout = 30 kVA Calculate η%at Full Load η% = 96.6%

  30. Cu FL= 840 W Fe = 220 W Sout = 30 kVA Calculate η%at 75%Load η% = 97%

  31. Cu FL= 840 W Fe = 220 W Sout = 30 kVA Calculate η%at 50%Load η% = 97.21%

  32. Cu FL= 840 W Fe = 220 W Sout = 30 kVA Calculate η%at 25%Load η% = 96.49%

  33. 100% η = 96.6% 75% η = 97% 50% η = 97.21% 25% η = 96.49%

  34. η% Cu Losses Fe Losses Losses (W) η% % Load Fe = Cu =Max η

  35. Maximum Efficiency Fe = 220 Cu = 840 Fe = Cu =Max η Load %= 51.18% η%= 97.21%

  36. All Day Efficiency • Most Transformers are connected permanently • The time that the transformer has to be calculated when determining efficiency • Able to determine the best transformer for the application by its efficiency

  37. All Day Efficiency Sout = 300 kVA Fe = 1.25 kVA Cu = 3.75 kVA Transformer A

  38. All Day Efficiency Sout = 300 kVA Fe = 2.5 kVA Cu = 2.5 kVA Transformer B

  39. Transformer Cooling • Transformer ratings can be increased if their windings are cooled by some external means • The most common cooling mediums are in direct with transformer windings; Air Oil and/or • The most common methods of circulation are Forced Natural and/or

  40. Transformer Classification • Transformers are allocated symbols which indicate the type of cooling used • Can consist of up to 4 letters indicating the cooling system

  41. Transformer Classification Type AN Air Natural Dry Transformer withNatural Air Flow

  42. Transformer Classification Type AF Air Forced Dry Transformer withForced Air Flow

  43. Transformer Classification Type ONAN Oil Natural Air Natural Oil Tank Cooling Natural Oil Flow - Natural Air Flow

  44. Type ONAF Transformer Classification Oil Natural Air Forced Oil Tank Cooling Natural Oil Flow - Forced Air Flow

  45. Transformer Classification Type OFAF Oil Forced Air Forced Oil Tank Cooling Forced Oil Flow – Forced Air Flow

  46. Transformer Oil Acts as Coolant & Insulator • Low Viscosity • High Flash point • Chemically inert • Good insulator

  47. Transformer Oil Tests • Dielectric Strength • Acidity • Power factor • Interfacial tension • Dissolved Gas

  48. THE END

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