Symmetrical Components. References

Transcription

1 Symmetrical Components Review of basics Sequence Equivalents Fault Analysis Symmetrical Components Fall 28 References Your power systems analysis class text book NPAG: Chapter 4 (analysis) Chapter 5 (equipment models) J.L. Blackburn, Protective Relaying: Principles and Applications, Any Edition: Chapter 4 P.M. Anderson, Analysis of Faulted Power Systems, IEEE Press 995 J.L. Blackburn, Symmetrical Components for Power Systems Engineering, Marcel-Dekker, 993. Nasser Tleis, Power Systems Modelling and Fault Analysis: Theory and Practice. Newnes Power Engineering Series. 28 Symmetrical Components Fall 28

2 History Fortescue, 98» Unbalanced n-phase system can be broken down into sets of balanced n-phase systems» Add using superposition Symmetrical Components Fall 28 What is it? Method for analysis of multiphase systems under unbalanced conditions» Steady-state conditions» Phasor Analysis» Visualization Allows a fast, efficient way to analyze unbalanced condition in real time Relays set to look at specific components Symmetrical Components Fall 28 2

3 Basic Equations De-couple voltage or current into 3 balanced 3 phase sets» Positive phase sequence (ABCABC ): V» Negative phase sequence (ACBACB ): V 2» Zero phase sequence (A=B=C), V» All RMS phasors» Per phase analysis Symmetrical Components Fall 28 Per Phase Symmetrical Component Equations a 2 V A = V A + V A + V A2 V B = V B + V B + V B2 = V A + a 2 V A + av A2 V C = V C + V C + V C2 = V A + av A + a 2 V A2 Symmetrical Components Fall 28 3

4 Analysis Equations V A =V =(V A + V B + V C )/3 V A =V =(V A + av B + a 2 V C )/3 V A2 =V 2 =(V A +a 2 V B + av C )/3 Same expressions for current Can express in Matrix form too Symmetrical Components Fall 28 Analysis Equations Va Va Va2 3 a a 2 a 2 a 4 Va Vb Vc Va Vb Vc a 3 a 32 a 32 a 3 Va Va Va2 a 2 a a a 2 Va Va Va2 Symmetrical Components Fall 28 4

5 Circuit Analysis Represent power system in per phase sequence networks Each network contains voltage and impedance elements for the sequence Reduce network to Thevenin Equivalent in each phase sequence All sources generally positive sequence Symmetrical Components Fall 28 Basic Sequence Networks Zero Sequence Positive Sequence Negative Sequence Ea Va + jx Va + + jx Va2 + jx2 + Ia + Ia + Ia2 Symmetrical Components Fall 28 5

6 Basic Sequence Networks Impedance will differ between sequences Zero sequence will also include ground impedance Connect them as appropriate for different fault types Symmetrical Components Fall 28 Fault Analysis Fault Types:» Single line to ground» Line to line» Double line to ground» Three phase (positive sequence unless unbalanced fault impedances)» Phase open» Phase open and line to ground» Simultaneous faults Symmetrical Components Fall 28 6

7 Fault Detection For Protection Purposes Basic fault analysis techniques calculate currents/voltages at fault location» ABC or symmetrical components Need fault signature as seen at the relay location» Rough location of fault for correct response Symmetrical Components Fall 28 Single Line to Ground Connections Constraints at fault location (phase A):» V A =» I B =I C Therefore:» V A = V + V + V 2 =» I =(I A + + )/3 and» I = I 2 =(I A + + )/3 so» I = I = I 2 =(I A )/3 Symmetrical Components Fall 28 7

8 SLG Connections To Satisfy these constraints we connect the three networks in series Connect reference point of one network to fault location of next one F F F2 N N N2 Symmetrical Components Fall 28 SLG Faults Solve for I, can calculate I A Next solve for V, V 2, and V and calculate phase voltage Note that in general jx o will be replaced with Z =jx + 3Z gr + 3Z f» The factor of 3 results since sequence is the same to all 3 phases Symmetrical Components Fall 28 8

9 Phase to Phase Faults Two phases (often B and C for analysis) shorted together» Not shorted to ground» V B =V C» I B = -I C From symmetrical component equations:» I =» I = -I 2 and V =V 2 Symmetrical Components Fall 28 Phase to Phase Faults Fault impedance (if any) appears between the networks No Zero Sequence Network V B =I B *Z f + V C F N F2 N2 Symmetrical Components Fall 28 9

10 Double Line to Ground Two phases shorted together and to ground Could have several impedances» I A» V B = (Z f + Z gr )*I B + Z f *I f» V c = (Z f + Z gr )*I C + Z f *I f F N F N F2 N2 Symmetrical Components Fall 28 Fault Detection For Protection Purposes Basic fault analysis techniques calculate currents/voltages at fault location» ABC or symmetrical components Need fault signature as seen at the relay location» Rough location of fault for correct response Symmetrical Components Fall 28

11 What Type of Fault? VA VB VA VB VC VC Symmetrical Components Fall 28 What Type of Fault? IR IR - Symmetrical Components Fall 28

12 What Type of Fault? Symmetrical Components Fall 28 What Type of Fault? 5 IR IR -25 Symmetrical Components Fall 28 2

13 What Type of Fault? IR IR - -2 Symmetrical Components Fall 28 What Type of Fault? 2 IR IR -5 Symmetrical Components Fall 28 3

14 What Type of Fault? 25 IR IR - Symmetrical Components Fall 28 Three Phase Fault, Right? VA VB VA VB VC VC Symmetrical Components Fall 28 4

15 A Symmetrical Component View of an Three-Phase Fault 9 A-Ground Fault Component Magnitude Angle Ia Ia Ia Va Va V 225 I 35 Symmetrical Components Fall A to Ground Fault, Okay? IR IR - Symmetrical Components Fall 28 5

16 A Symmetrical Component View of an A-Phase to Ground Fault A-Ground Fault Component Magnitude Angle Ia Ia Ia Va Va 23 8 V2 V V I I I2 35 Symmetrical Components Fall 28 Single Line to Ground Fault Voltage» Negative and zero sequence 8 out of phase with positive sequence Current» All sequence are in phase Symmetrical Components Fall 28 6

17 A to B Fault, Easy? Symmetrical Components Fall 28 A Phase Symmetrical Component View of an A to B Phase Fault A-B Fault Component Magnitude Angle Ia 3-2 Ia Ia Va 45 Va 99 8 V2 V I2 225 I 35 Symmetrical Components Fall

18 C Phase Symmetrical Component View of an A to B Phase Fault 9 35 I2 45 Component Magnitude Angle Ic 3 38 Ic Ic Vc -75 Vc 99 Vc V2 V 225 I 35 Symmetrical Components Fall Line to Line Fault Voltage» Negative in phase with positive sequence Current» Negative sequence 8 out of phase with positive sequence Symmetrical Components Fall 28 8

19 B to C to Ground 5 IR IR -25 Symmetrical Components Fall 28 A Symmetrical Component View of a B to C to Ground Fault 9 Component Magnitude Angle Ia Ia Ia2 754 Va 8 35 Va Va I2 I V2 V 45 V 225 I Symmetrical Components Fall 28 9

20 Line to Line to Ground Fault Voltage» Negative and zero in phase with positive sequence Current» Negative and zero sequence 8 out of phase with positive sequence Symmetrical Components Fall 28 Again, What Type of Fault? IR IR - -2 Symmetrical Components Fall 28 2

21 C Symmetrical Component View of a C-Phase Open Fault Component Magnitude Angle Ic Ic 4 Ic Vc 62 Vc 79 Vc I I2 V2 V I Symmetrical Components Fall 28 One Phase Open (Series) Faults Voltage» No zero sequence voltage» Negative 9 out of phase with positive sequence Current» Negative and zero sequence 8 out of phase with positive sequence Symmetrical Components Fall 28 2

22 What About This One? 2 IR IR -5 Symmetrical Components Fall 28 Ground Fault with Reverse Load Ic Ic 89-3 Ic2 4-6 Vc 4-23 Vc 38 Vc V2 V I2 V I I Symmetrical Components Fall

23 Finally, The Last One! 25 IR IR - Symmetrical Components Fall 28 Fault on Distribution System with Delta Wye Transformer Component Magnitude Angle Ic 45 4 Ic 53-4 Ic Vc.5 33 Vc 4 Vc I2 V2 I V I V Symmetrical Components Fall 28 23