# Symmetrical Components I 2004.ppt

Mar. 28, 2023

### Symmetrical Components I 2004.ppt

1. Symmetrical Components I An Introduction to Power System Fault Analysis Using Symmetrical Components Dave Angell Idaho Power 21st Annual Hands-On Relay School
2. What Type of Fault? -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 of Fault? -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 of Fault? -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 of Fault? -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 of 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
7. What Type of Fault? -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 of Fault? -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. Unbalanced Fault Ia Ib Ic Ia Ib Ic Ia Ib Ic
11. Symmetrical Component Phasors  The unbalanced 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 0.000 0.017 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  Each phase 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 the a Operator  120° counter clock-wise rotation.  A vector multiplied by 1 /120° results in the same magnitude rotated 120°. Va aVa
23. Rotation of the a2 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 we produced?  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 Unbalanced Voltage Va Vc Vb  Va = 13.4 /0°  Vb = 59.6 /- 104°  Vc = 59.6 /104°
39. Analysis Results in These 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 Summing the Positive, Negative and … Va2 Vb2 Vc2 Va1 Vb1 Vc1
41. Zero Sequences Va2 Vb2 Vc2 Va0 Vb0 Vc0 Va1 Vb1 Vc1
42. The Synthesis Equation Results in the Original Unbalanced Voltage Va2 Vb2 Vc2 Va0 Vb0 Vc0 Va1 Vb1 Vc1 Va Vc Vb
43. Symmetrical Components Present During Shunt Faults  Three phase fault – Positive  Phase to phase fault – Positive – Negative  Phase to ground fault – Positive – Negative – Zero
44. Symmetrical Component Review of Faults 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 Component View 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 Ground Fault, 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 Component View 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 to Ground Fault  Voltage – Negative and zero sequence 180 out of phase with positive sequence  Current – All sequence are in phase
50. A to B Fault, 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 Symmetrical Component 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 Symmetrical Component 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 Line Fault  Voltage – Negative in phase with positive sequence  Current – Negative sequence 180 out of phase with positive sequence
54. B to C to 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 Component View 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 Line to 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 Type of 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 Component View 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 This One? -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 with Reverse 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 Last One! -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 Ic0 45 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 Sequence Quantities in Relays  Zero Sequence filters – Current – Voltage  Relay operating quantity  Relay polarizing quantity
65. Zero Sequence Current Ia Ib Ic Direction of the protected line Ia+Ib+Ic 3I0
66. Zero Sequence Voltage (Broken Delta) Va Vb Vc 3V0
67. Zero Sequence Voltage Va Va Vc Vb Va Vb 3Vo Va Vb
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  Given three current sources  How can zero sequence be produced to test a relay?  How can negative sequence produced?
72. How can zero sequence 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 negative sequence 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 Components for 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 components provide: – balanced analysis of an unbalanced system. – a measure of system unbalance – methods to detect faults – an ability to distinguish fault direction

### Editor's Notes

1. A Phase to Ground Fault
2. A to B Phase Fault
3. B to C to Ground Fault
4. C Phase Open Fault
5. A to B Phase Fault
6. B to C to Ground Fault
7. C Phase Open Fault