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Power factor improvement method using synchronous condenser
- 1. POWER FACTOR IMPROVEMENT IN GRIDUSING SMALL HYDRO GENERATING STATION PRESENTED BY, AKHIL.A.G TJAKEEE003 ANUSHA.K.BINO TJAKEEE010 VIJISHA.E.V TJAKEEE051 VISHAG RAVEENDAN TJAKEEE052 Guide: Anil sir 27-03-2014 1
- 2. INTRODUCTION Power Powerfactor Synchronization Synchronous machine Synchronous condenser Current situation 27-03-2014 2
- 3. PF IMPROVEMENT EQUIPMENT Staticcapacitor Synchronous condenser Phase advancers 27-03-2014 3
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- 6. 11 K V 11/110KV TRANSFOMER 110KV SUBSTATION 110/33KV TRANSFOMER 33KV SUBSTATION 33/11KV TRANSFORMER 11KV/415V TRANSFORMER LOAD LOAD LOAD LOADLOAD 11KV LINE PRIMARY TRANSMISSION SECONDARY TRANSMISSION GENERATING STATION PRIMARY DISTRIBUTI ON POWER SYSTEM NETWORK LOAD 27-03-2014 6
- 7. 11 K V 11/110KV TRANSFOMER 110KV SUBSTATION 110/33KV TRANSFOMER 33KV SUBSTATION 33/11KV TRANSFORMER LOAD 11KV/415V TRANSFORMER LOAD LOAD LOAD LOADLOAD SYNCHRONO US SCOPE 3.3/11KV TRANSFORM ER 3.3 KV 11KV LINE PRIMARY TRANSMISSION SECONDARY TRANSMISSION SMALL GENERATING STATION GENERATING STATION PRIMARY DISTRIBUTION PRIMARY DISTRIBUTI ON 27-03-2014 7
- 8. 11 K V 11/110KV TRANSFOMER 110KV SUBSTATION 110/33KV TRANSFOMER 33kV SUBSTATION 33/11KV TRANSFORMER LOAD 11KV/415V TRANSFORMER LOAD LOAD LOAD LOADLOAD SYNCHRONO US SCOPE 3.3/11KV TRANSFORM ER SYNCH RONOU S CONDE NSER 11KV LINE PRIMARY TRANSMISSION SECONDARY TRANSMISSION GENERATING STATION PRIMARY DISTRIBUTION PRIMARY DISTRIBUTI ON 27-03-2014 8
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- 11. DESIGN Generating capacity ofPeechi generating station is 1.25MW. The. We know that , If act as generator Active power = 3 𝑉𝐼 𝐶𝑜𝑠∅ = 1.250MW. PF=0.9 lagging; ∅ = 25.840 Apparent power = 3 𝑉𝐼 = 1.388𝑀𝑉𝐴. Reactive power = 3 𝑉𝐼 𝑆𝑖𝑛 ∅ = 0.604𝑀𝑉𝐴𝑅 Full load current I = 1.388MVA = 243 A 3 ∗ 3.3 KV Full load active current = 218.70A. Full load reactive current = 105.91A. 27-03-2014 11
- 12. If act assynchronous condenser An over excited synchronous motor operated at no load called synchronous condenser. This MVA we are going to run as synchronous condenser with 0.1pf leading. Apparent power = 1.388 MVA. cos−1 (.1) = 84.260 27-03-2014 12
- 13. This 1.388 MVAcan be supplied to 33KV Substation in kannara where Load =3Mw, pf = .6 that means it draws 5 MVA from feeding Ollur 33kV Substation Pf =0.6 ∅ = 53.130 27-03-2014 13
- 14. Power factor =cos 〖(39.72)= 0.769.〗 So Power factor improved from 0.6 to 0.769. With 3 MW = 0.169. 5MVA – 4.08MVA = .92 MVA. I.e. The working of synchronous motor produces .92 MVA as profit. 27-03-2014 14
- 15. Consider now amachine that: 1 Is operated at successively smaller and smaller torque angle 2. Greater and greater field excitation 27-03-2014 15
- 16. FULL LOAD We knowthat, Output of AC Generator = 1250000 MW 3 𝑉𝐼 𝑐𝑜𝑠∅ = 1250000. Full load current IA =243A Input of AC Generator = 1314434MW 𝟑 ∗ 𝑬 ∗ 𝑰 ∗ 𝒄𝒐𝒔ɸ =1306200 E = 3.44 kV Voltage drop across the generator = .147kV 27-03-2014 16
- 17. 75% LOAD Assume ɸ=20º Output of AC Generator = 9375000 MW √3 VI cos∅=9375000 IA = 174.54A Excitation current = 165.5A Voltage drop across the generator = jIaX = 0.082 kV 27-03-2014 17
- 18. 50% LOAD Assume ɸ=150 Output of AC Generator = 625000 MW 3 𝑉𝐼 𝑐𝑜𝑠∅ = 625000 3 ∗ 3.3 ∗ 𝐼A*.96 = 625000 IA = 113A Excitation current = 147A Voltage drop across the generator = jIaX = 29.24*1.38 = 0.043 kV 27-03-2014 18
- 19. 25% LOAD Assume ɸ=100 Outputof AC Generator = 312500 MW 3 𝑉𝐼 𝑐𝑜𝑠∅ = 312500 3 ∗ 3.3 ∗ 𝐼A*.98 = 312500 IA = 55.51A Excitation current = 128.50A Voltage drop across the generator = jIaX = 9.63*1.38 = 0.013 kV 27-03-2014 19
- 20. NO LOAD Assume ɸ=20 Output of AC Generator = 17136.85 MW 3 ∗ 3.3 ∗ 𝐼A*.99 = 17136.85 IA = 3A Excitation current = 110A Voltage drop across the generator = jIaX = 0.10*1.38 = 0.000138 kV 27-03-2014 20
- 21. SYNCHRONOUS CONDENSER OPERATION Synchronousgenerator has the capability of operating as a synchronous condenser. Increasing the field excitation causes E to increase. We know that Vt = E + jIaX 27-03-2014 21
- 22. SUCCESSIVELY GREATER FIELDEXCITATION Case1 Apparent Power = 317320.81MW Active Power = 31736.65 MVA Reactive Power = 315729.76 MVAR Let we assume that 0.1 power factor ie, ɸ = 84.260 Ia, act =5.50 A Ia, react =54.72 A Resultant current I =55A Voltage drop across the generator, jIaX = 1.38 * 54.72 =75.51 V 27-03-2014 22
- 23. Case 2 Apparent Power= 647047.61MW Active Power = 64714.08 MVA Reactive Power = 643803.30MVAR Assume ɸ = 84.260 Ia, act =11.30 A Ia, react =112.43 A Resultant current I =113A Voltage drop across the generator, jIaX = 1.38 * 113 =155.15 V 27-03-2014 23
- 24. Case 3 Apparent Power= 997666.66 MW Active Power = 99781.03 MVA Reactive Power = 992664.34MVAR Assume ɸ = 84.260 Ia, act =17.45 A Ia, react=173.66 A Resultant current I =174.54A Voltage drop across the generator, jIaX = 1.38 * 173.66 =239.65V 27-03-2014 24
- 25. Case 4 Apparent Power= 1388866.19MW Active Power = 138906.62 MVA Reactive Power = 1381902.4MVAR Assume ɸ = 84.260 Ia , act =24.30A Ia, react =241.76 A Resultant current I =242.98A Voltage drop across the generator, jIaX =155.15 V 27-03-2014 25
- 26. STARTING & EXCITATIONCONTROL 27-03-2014 26
- 27. SYNCHRONOUS MACHINE ®ULATOR 27-03-2014 27
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- 30. CONCLUSION Low powerfactor problems in the power system network can be analysed. Power factor improved from 0.6 to 0.769. With 3MW The working of synchronous motor produces .92 MVA as profit. 27-03-2014 30
- 31. REFERENCE C.JAYARAMAN –Concept of reactive power compensation Juan Dixon and Luis Moran -“ Reactive Power Compensation Technologies”. T.J Millen- “ Reactive Power Control in Electrical Systems.” RICARDO DOMKE -Power factor improvement 27-03-2014 31
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