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Symmetrical Components I 2004.ppt
- 1. Symmetrical Components I AnIntroduction to Power System Fault Analysis Using Symmetrical Components Dave Angell Idaho Power 21st Annual Hands-On Relay School
- 2. What Type ofFault? -25 0 25 -25 0 25 -25 0 25 -2500 0 2500 -2500 0 2500 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 V A V B V C I A I B I C Cycles VA VB VC IA IB IC
- 3. What Type ofFault? -10000 0 10000 -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 4. What Type ofFault? -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C Cycles IA IB IC
- 5. What Type ofFault? -5000 0 5000 -5000 0 5000 -5000 0 5000 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 6. What Type ofFault? -100 0 100 -100 0 100 -100 0 100 -200 -0 200 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 7. What Type ofFault? -200 0 200 -200 0 200 -200 0 200 -500 0 500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 8. What Type ofFault? -250 0 250 -250 0 250 -250 0 250 -100 0 100 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 9. Basic Course Topics Terminology Phasors Equations Fault Analysis Examples Calculations
- 10.
- 11. Symmetrical Component Phasors Theunbalanced three phase system can be transformed into three balanced phasors. – Positive Sequence – Negative Sequence – Zero Sequence
- 12. Positive Phase Sequence(ABC) -1.0 -0.5 0.0 0.5 1.0 0.000 0.017 0.033 0.050 Time Magnitude Va Vb Vc
- 13. Positive Phase Sequence Each have the same magnitude. Each positive sequence voltage or current quantity is displaced 120° from one another. Va1 Vb1 Vc1
- 14. Positive Phase Sequence The positive sequence quantities have a- b-c, counter clock- wise, phase rotation. Va1 Vb1 Vc1
- 15. Reverse Phase Sequence(ACB) -1.0 -0.5 0.0 0.5 1.0 0.000 0.017 0.033 0.050 Time Magnitude Va Vb Vc
- 16. Negative Phase Sequence Each have the same magnitude. Each negative sequence voltage or current quantity is displaced 120° from one another. Va2 Vc2 Vb2
- 17. Negative Phase Sequence The negative sequence quantities have a- c-b, counter clock- wise, phase rotation. Va2 Vc2 Vb2
- 18. Zero Phase Sequence -1.0 -0.5 0.0 0.5 1.0 0.0000.017 0.033 0.050 Time Magnitude Va Vb Vc
- 19. Zero Phase Sequence Each zero sequence quantity has the same magnitude. All three phasors with no angular displacement between them, all in phase. Va0 Vb0 Vc0
- 20. Symmetrical Components Equations Eachphase quantity is equal to the sum of its symmetrical phasors. Va = Va0 + Va1 +Va2 Vb = Vb0 + Vb1 +Vb2 Vc = Vc0 + Vc1 +Vc2 The common form of the equations are written in a-phase terms.
- 21. The a Operator Used to shift the a-phase terms to coincide with the b and c-phase Shorthand to indicate 120° rotation. Similar to the j operator of 90°. Va
- 22. Rotation of thea Operator 120° counter clock-wise rotation. A vector multiplied by 1 /120° results in the same magnitude rotated 120°. Va aVa
- 23. Rotation of thea2 Operator 240° counter clock-wise rotation. A vector multiplied by 1 /240° results in the same magnitude rotated 240°. Va a2Va
- 24. B-Phase Zero Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator. Vb0 = Va0 Va0 Vb0 Vc0
- 25. B-Phase Positive Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator Vb1 = a2Va1 Va1 Vb1 Vc1
- 26. B-Phase Negative Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator Vb2 = aVa2 Va2 Vc2 Vb2
- 27. C-Phase Zero Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator. Vc0 = Va0 Va0 Vb0 Vc0
- 28. C-Phase Positive Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator Vc1 = aVa1 Va1 Vb1 Vc1
- 29. C-Phase Negative Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator Vc2 = a2Va2 Va2 Vc2 Vb2
- 30. What have weproduced? Va = Va0 + Va1 + Va2 Vb = Va0 + a2Va1 + aVa2 Vc = Va0 + aVa1 + a2Va2
- 31. Symmetrical Components Equations Analysis –To find out of the amount of the components Synthesis – The combining of the component elements into a single, unified entity
- 32. Symmetrical Components Synthesis Equations Va = Va0 + Va1 + Va2 Vb = Va0 + a2Va1 + aVa2 Vc = Va0 + aVa1 + a2Va2
- 33. Symmetrical Components Analysis Equations Va0 = 1/3 ( Va + Vb + Vc) Va1= 1/3 (Va + aVb + a2Vc) Va2= 1/3 (Va + a2Vb + aVc)
- 34. Symmetrical Components Analysis Equations- 1/3 ?? Where does the 1/3 come from? Va1= 1/3 (Va + aVb + a2Vc) Va = Va0 + Va1 + Va2 When balanced 0 0
- 35. Symmetrical Components Analysis Equations- 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases Va
- 36. Symmetrical Components Analysis Equations- 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases yields Va aVb Vc Va Vb
- 37. Symmetrical Components Analysis Equations- 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases yields 3 Va. Divide by the 3 and now Va = Va1 a2 Vc Va aVb Vc Va Vb
- 38. Example Vectors An UnbalancedVoltage Va Vc Vb Va = 13.4 /0° Vb = 59.6 /- 104° Vc = 59.6 /104°
- 39. Analysis Results inThese Sequence Quantities Va0 Vb0 Vc0 Va2 Vc2 Vb2 Va1 Vb1 Vc1 Va0 = -5.4 s Va1 = 42.9 s Va2 = -24.1
- 40. Synthesize by Summingthe Positive, Negative and … Va2 Vb2 Vc2 Va1 Vb1 Vc1
- 41.
- 42. The Synthesis EquationResults in the Original Unbalanced Voltage Va2 Vb2 Vc2 Va0 Vb0 Vc0 Va1 Vb1 Vc1 Va Vc Vb
- 43. Symmetrical Components Present DuringShunt Faults Three phase fault – Positive Phase to phase fault – Positive – Negative Phase to ground fault – Positive – Negative – Zero
- 44. Symmetrical Component Review ofFaults Types Let’s return to the example fault reports and view the sequence quantities present
- 45. Three Phase Fault,Right? -25 0 25 -25 0 25 -25 0 25 -2500 0 2500 -2500 0 2500 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 V A V B V C I A I B I C Cycles VA VB VC IA IB IC
- 46. A Symmetrical ComponentView of an Three-Phase Fault Component Magnitude Angle Ia0 7.6 175 Ia1 2790 -64 Ia2 110 75.8 Va0 0 0 Va1 18.8 0 Va2 0.7 337 0 45 90 135 180 225 270 315 I1 V1
- 47. A to GroundFault, Okay? -10000 0 10000 -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 48. A Symmetrical ComponentView of an A-Phase to Ground Fault Component Magnitude Angle Ia0 7340 -79 Ia1 6447 -79 Ia2 6539 -79 Va0 46 204 Va1 123 0 Va2 79 178 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- 49. Single Line toGround Fault Voltage – Negative and zero sequence 180 out of phase with positive sequence Current – All sequence are in phase
- 50. A to BFault, Easy? -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C Cycles IA IB IC
- 51. A Phase SymmetricalComponent View of an A to B Phase Fault Component Magnitude Angle Ia0 3 -102 Ia1 5993 -81 Ia2 5961 -16 Va0 1 45 Va1 99 0 Va2 95 -117 0 45 90 135 180 225 270 315 I1 I2 V1 V2
- 52. C Phase SymmetricalComponent View of an A to B Phase Fault Component Magnitude Angle Ic0 3 138 Ic1 5993 279 Ic2 5961 104 Vc0 1 -75 Vc1 99 0 Vc2 95 2.5 0 45 90 135 180 225 270 315 I1 I2 V1 V2
- 53. Line to LineFault Voltage – Negative in phase with positive sequence Current – Negative sequence 180 out of phase with positive sequence
- 54. B to Cto Ground -5000 0 5000 -5000 0 5000 -5000 0 5000 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 55. A Symmetrical ComponentView of a B to C to Ground Fault Component Magnitude Angle Ia0 748 97 Ia1 2925 -75 Ia2 1754 101 Va0 8 351 Va1 101 0 Va2 18 348 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- 56. Line to Lineto Ground Fault Voltage – Negative and zero in phase with positive sequence Current – Negative and zero sequence 180 out of phase with positive sequence
- 57. Again, What Typeof Fault? -100 0 100 -100 0 100 -100 0 100 -200 -0 200 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 58. C Symmetrical ComponentView of a C-Phase Open Fault Component Magnitude Angle Ic0 69 184 Ic1 101 4 Ic2 32 183 Vc0 0 162 Vc1 79 0 Vc2 5 90 0 45 90 135 180 225 270 315 I0 I1 I2 V1 V2
- 59. One Phase Open(Series) Faults Voltage – No zero sequence voltage – Negative 90 out of phase with positive sequence Current – Negative and zero sequence 180 out of phase with positive sequence
- 60. What About ThisOne? -200 0 200 -200 0 200 -200 0 200 -500 0 500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 61. Ground Fault withReverse Load Ic0 164 -22 Ic1 89 -113 Ic2 41 -6 Vc0 4 -123 Vc1 38 0 Vc2 6 -130 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- 62. Finally, The LastOne! -250 0 250 -250 0 250 -250 0 250 -100 0 100 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- 63. Component Magnitude Angle Ic045 40 Ic1 153 -4 Ic2 132 180 Vc0 0.5 331 Vc1 40 0 Vc2 0.5 93 Fault on Distribution System with Delta – Wye Transformer 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- 64. Use of SequenceQuantities in Relays Zero Sequence filters – Current – Voltage Relay operating quantity Relay polarizing quantity
- 65. Zero Sequence Current Ia Ib Ic Directionof the protected line Ia+Ib+Ic 3I0
- 66. Zero Sequence Voltage (BrokenDelta) Va Vb Vc 3V0
- 67.
- 68. Sequence Operating Quantities Zero and negative sequence currents are not present during balanced conditions. Good indicators of unbalanced faults
- 69. Sequence Polarizing Quantities Polarizing quantities are used to determine direction. The quantities used must provide a consistent phase relationship.
- 70. Zero Sequence Voltage Polarizing 3Vo is out of phase with Va -3Vo is used to polarize for ground faults Va Vb 3Vo
- 71. Learning Check Giventhree current sources How can zero sequence be produced to test a relay? How can negative sequence produced?
- 72. How can zerosequence be produced to test a relay? A single source provides positive, negative and zero sequence – Note that each sequence quantity will be 1/3 of the total current Connect the three sources in parallel and set their amplitude and the phase angle equal to one another – The sequence quantities will be equal to each source output
- 73. How can negativesequence produced? A single source provides positive, negative and zero sequence – Each sequence quantity will be 1/3 of the total current Set the three source’s amplitude equal to one another and the phase angles to produce a reverse phase sequence (Ia at /0o , Ib at /120o and Ic at /-120o ) – Only negative sequence will be produced
- 74. Advanced Course Topics Sequence Networks Connection of Networks for Faults Per Unit System Power System Element Models
- 75. References Symmetrical Componentsfor Power Systems Engineering, J Lewis Blackburn Protective Relaying, J Lewis Blackburn Power System Analysis, Stevenson Analysis of Faulted Power System, Paul Anderson
- 76. Conclusion Symmetrical componentsprovide: – balanced analysis of an unbalanced system. – a measure of system unbalance – methods to detect faults – an ability to distinguish fault direction