Analysis-of-Faults-AREVA.ppt
- 2. Fault Analysis Alan Wixon Senior Applications Engineer
- 3. > Fault Analysis – January 2004 3 3 Power System Fault Analysis (1) TO :- Calculate Power System Currents and Voltages during Fault Conditions Check that Breaking Capacity of Switchgear is Not Exceeded Determine the Quantities which can be used by Relays to Distinguish Between Healthy (i.e. Loaded) and Fault Conditions Appreciate the Effect of the Method of Earthing on the Detection of Earth Faults Select the Best Relay Characteristics for Fault Detection Ensure that Load and Short Circuit Ratings of Plant are Not Exceeded Select Relay Settings for Fault Detection and Discrimination Understand Principles of Relay Operation Conduct Post Fault Analysis All Protection Engineers should have an understanding
- 4. > Fault Analysis – January 2004 4 4 Power System Fault Analysis (2) Consider Stability Conditions Required fault clearance times Need for 1 phase or 3 phase auto-reclose Power System Fault Analysis also used to :-
- 5. > Fault Analysis – January 2004 5 5 Computer Fault Calculation Programmes Widely available, particularly in large power utilities Powerful for large power systems Sometimes overcomplex for simple circuits Not always user friendly Sometimes operated by other departments and not directly available to protection engineers Programme calculation methods:- understanding is important Need for ‘by hand’ spot checks of calculations
- 6. > Fault Analysis – January 2004 6 6 Pocket Calculator Methods Adequate for the majority of simple applications Useful when no access is available to computers and programmes e.g. on site Useful for ‘spot checks’ on computer results
- 7. > Fault Analysis – January 2004 7 7 Vectors Vector notation can be used to represent phase relationship between electrical quantities. V Z I V = Vsinwt = V 0 I = I - = Isin(wt-)
- 8. > Fault Analysis – January 2004 8 8 j Operator Rotates vectors by 90° anticlockwise : Used to express vectors in terms of “real” and “imaginary” parts. 1 90 90 90 90 j = 1 90 j2 = 1 180 = -1 j3 = 1 270 = -j
- 9. > Fault Analysis – January 2004 9 9 a = 1 120 ° Rotates vectors by 120° anticlockwise Used extensively in “Symmetrical Component Analysis” 120 120 1 120 2 3 j 2 1 - 120 1 a 2 3 j 2 1 240 1 a2
- 10. > Fault Analysis – January 2004 10 10 a = 1 120 ° Balanced 3Ø voltages :- VA VC = aVA a2 + a + 1 = 0 VB = a2VA
- 11. > Fault Analysis – January 2004 11 11 Balanced Faults
- 12. > Fault Analysis – January 2004 12 12 Balanced (3Ø) Faults (1) RARE :- Majority of Faults are Unbalanced CAUSES :- 1. System Energisation with Maintenance Earthing Clamps still connected. 2. 1Ø Faults developing into 3Ø Faults 3Ø FAULTS MAY BE REPRESENTED BY 1Ø CIRCUIT Valid because system is maintained in a BALANCED state during the fault Voltages equal and 120° apart Currents equal and 120° apart Power System Plant Symmetrical Phase Impedances Equal Mutual Impedances Equal Shunt Admittances Equal
- 13. > Fault Analysis – January 2004 13 13 Balanced (3Ø) Faults (2) LINE ‘X’ LOADS LINE ‘Y’ 3Ø FAULT ZLOAD ZLY IbF IcF IaF ZLX ZT ZG Ec Ea Eb GENERATOR TRANSFORMER
- 14. > Fault Analysis – January 2004 14 14 Balanced (3Ø) Faults (3) Positive Sequence (Single Phase) Circuit :- Ea Ec Ea F1 N1 Eb IbF Ia1 = IaF IcF IaF ZT1 ZLX1 ZLX2 ZG1 ZLOAD
- 15. > Fault Analysis – January 2004 15 15 Representation of Plant
- 16. > Fault Analysis – January 2004 16 16 Generator Short Circuit Current The AC Symmetrical component of the short circuit current varies with time due to effect of armature reaction. Magnitude (RMS) of current at any time t after instant of short circuit : where : I" = Initial Symmetrical S/C Current or Subtransient Current = E/Xd" 50ms I' = Symmetrical Current a Few Cycles Later 0.5s or Transient Current = E/Xd' I = Symmetrical Steady State Current = E/Xd Ι Ι Ι Ι Ι Ι )e - ' ( )e ' - " ( t/Td' - t/Td" - ac i TIME
- 17. > Fault Analysis – January 2004 17 17 Simple Generator Models Generator model X will vary with time. Xd" - Xd' - Xd X E
- 18. > Fault Analysis – January 2004 18 18 Parallel Generators 11kV 20MVA XG=0.2pu 11kV 11kV j0.05 j0.1 XG=0.2pu 20MVA If both generator EMF’s are equal they can be thought of as resulting from the same ideal source - thus the circuit can be simplified.
- 19. > Fault Analysis – January 2004 19 19 P.U. Diagram IF j0.05 j0.1 j0.2 1.0 1.0 j0.2 IF j0.05 j0.1 j0.2 j0.2 1.0
- 20. > Fault Analysis – January 2004 20 20 Positive Sequence Impedances of Transformers 2 Winding Transformers ZP = Primary Leakage Reactance ZS = Secondary Leakage Reactance ZM = Magnetising impedance = Large compared with ZP and ZS ZM Infinity Represented by an Open Circuit ZT1 = ZP + ZS = Positive Sequence Impedance ZP and ZS both expressed on same voltage base. S1 P1 P S P1 S1 ZP ZS ZM N1 N1 ZT1 = ZP + ZS
- 21. > Fault Analysis – January 2004 21 21 Motors Xd" M 1.0 Fault current contribution decays with time Decay rate of the current depends on the system. From tests, typical decay rate is 100 - 150mS. Typically modelled as a voltage behind an impedance
- 22. > Fault Analysis – January 2004 22 22 Induction Motors – IEEE Recommendations Small Motors Motor load <35kW neglect Motor load >35kW SCM = 4 x sum of FLCM Large Motors SCM motor full load amps Xd" Approximation : SCM = locked rotor amps SCM = 5 x FLCM assumes motor impedance 20%
- 23. > Fault Analysis – January 2004 23 23 Synchronous Motors – IEEE Recommendations Large Synchronous Motors SCM 6.7 x FLCM for Assumes X"d = 15% 1200 rpm 5 x FLCM for Assumes X"d = 20% 514 - 900 rpm 3.6 x FLCM for Assumes X"d = 28% 450 rpm or less
- 24. > Fault Analysis – January 2004 24 24 Analysis of Balanced Faults
- 25. > Fault Analysis – January 2004 25 25 Different Voltages – How Do We Analyse? 11kV 20MVA ZG=0.3pu 11/132kV 50MVA ZT=10% ZL=40 O/H Line 132/33kV 50MVA ZT=10% ZL=8 Feeder
- 26. > Fault Analysis – January 2004 26 26 Referring Impedances Consider the equivalent CCT referred to :- Primary Secondary R1 X1 N : 1 Ideal Transformer R2 X2 R1 + N2R2 X1 + N2X2 R1/N2 + R2 X1/N2 + X2
- 27. > Fault Analysis – January 2004 27 27 Per Unit System Used to simplify calculations on systems with more than 2 voltages. Definition : P.U. Value = Actual Value of a Quantity Base Value in the Same Units
- 28. > Fault Analysis – January 2004 28 28 Base Quantities and Per Unit Values Particularly useful when analysing large systems with several voltage levels All system parameters referred to common base quantities Base quantities fixed in one part of system Base quantities at other parts at different voltage levels depend on ratio of intervening transformers 11 kV 20 MVA O/H LINE 11/132 kV 50 MVA ZT = 10% ZT = 10% 132/33 kV 50 MVA FEEDER ZL = 8 ZL = 40 ZG = 0.3 p.u.
- 29. > Fault Analysis – January 2004 29 29 Base Quantities and Per Unit Values (1) Base quantites normally used :- BASE MVA = MVAb = 3 MVA Constant at all voltage levels Value ~ MVA rating of largest item of plant or 100MVA BASE VOLTAGE = KVb = / voltage in kV Fixed in one part of system This value is referred through transformers to obtain base voltages on other parts of system. Base voltages on each side of transformer are in same ratio as voltage ratio.
- 30. > Fault Analysis – January 2004 30 30 Base Quantities and Per Unit Values (2) Other base quantites :- kA in kV . 3 MVA Current Base Ohms in MVA ) (kV Z Impedance Base b b b b 2 b b Ι
- 31. > Fault Analysis – January 2004 31 31 Base Quantities and Per Unit Values (3) Per Unit Values = Actual Value Base Value Current Unit Per ) (kV MVA . Z Z Z Z Impedance Unit Per KV KV kV Voltage Unit Per MVA MVA MVA MVA Unit Per b a p.u. 2 b b a b a p.u. b a p.u. b a p.u. Ι Ι Ι
- 32. > Fault Analysis – January 2004 32 32 Transformer Percentage Impedance If ZT = 5% with Secondary S/C 5% V (RATED) produces I (RATED) in Secondary. V (RATED) produces 100 x I (RATED) 5 = 20 x I (RATED) If Source Impedance ZS = 0 Fault current = 20 x I (RATED) Fault Power = 20 x kVA (RATED) ZT is based on I (RATED) & V (RATED) i.e. Based on MVA (RATED) & kV (RATED) is same value viewed from either side of transformer.
- 33. > Fault Analysis – January 2004 33 33 Example (1) Per unit impedance of transformer is same on each side of the transformer. Consider transformer of ratio kV1 / kV2 Actual impedance of transformer viewed from side 1 = Za1 Actual impedance of transformer viewed from side 2 = Za2 MVA 1 2 kVb / kV1 kVb / kV2
- 34. > Fault Analysis – January 2004 34 34 Example (2) Base voltage on each side of a transformer must be in the same ratio as voltage ratio of transformer. Incorrect selection of kVb 11.8kV 132kV 11kV Correct selection 132x11.8 132kV 11kV of kVb 141 = 11.05kV Alternative correct 11.8kV 141kV 141x11 = 11.75kV selection of kVb 132 11.8kV 11.8/141kV 132/11kV OHL Distribution System
- 35. > Fault Analysis – January 2004 36 36 Example kVb MVAb 132 50 349 219 A 33 50 21.8 874 A Z V 11 50 2625 A 2.42 11 kV 20 MVA 132/33 kV 50 MVA 10% 40 11/132 kV 50 MVA 10% 8 3 FAULT 0.3p.u. Zb = 2 MVAb kVb Ib = MVAb 3kVb p.u. 0.3 x 50 20 = 0.75p.u. 0.1p.u. 40 349 = 0.115p.u. 0.1p.u. 8 21.8 = 0.367p.u. 1p.u. 1.432p.u. IF = 1 = 0.698p.u. 1.432 I11 kV = 0.698 x Ib = 0.698 x 2625 = 1833A I132 kV = 0.698 x 219 = 153A I33 kV = 0.698 x 874 = 610A
- 36. > Fault Analysis – January 2004 37 37 Fault Types Line - Ground (65 - 70%) Line - Line - Ground (10 - 20%) Line - Line (10 - 15%) Line - Line - Line (5%) Statistics published in 1967 CEGB Report, but are similar today all over the world.
- 37. > Fault Analysis – January 2004 38 38 Unbalanced Faults
- 38. > Fault Analysis – January 2004 39 39 Unbalanced Faults (1) In three phase fault calculations, a single phase representation is adopted. 3 phase faults are rare. Majority of faults are unbalanced faults. UNBALANCED FAULTS may be classified into SHUNT FAULTS and SERIES FAULTS. SHUNT FAULTS: Line to Ground Line to Line Line to Line to Ground SERIES FAULTS: Single Phase Open Circuit Double Phase Open Circuit
- 39. > Fault Analysis – January 2004 40 40 Unbalanced Faults (2) LINE TO GROUND LINE TO LINE LINE TO LINE TO GROUND Causes : 1) Insulation Breakdown 2) Lightning Discharges and other Overvoltages 3) Mechanical Damage
- 40. > Fault Analysis – January 2004 41 41 Unbalanced Faults (3) OPEN CIRCUIT OR SERIES FAULTS Causes : 1) Broken Conductor 2) Operation of Fuses 3) Maloperation of Single Phase Circuit Breakers DURING UNBALANCED FAULTS, SYMMETRY OF SYSTEM IS LOST SINGLE PHASE REPRESENTATION IS NO LONGER VALID
- 41. > Fault Analysis – January 2004 42 42 Unbalanced Faults (4) Analysed using :- Symmetrical Components Equivalent Sequence Networks of Power System Connection of Sequence Networks appropriate to Type of Fault
- 42. > Fault Analysis – January 2004 43 43 Symmetrical Components
- 43. > Fault Analysis – January 2004 44 44 Symmetrical Components Fortescue discovered a property of unbalanced phasors ‘n’ phasors may be resolved into :- (n-1) sets of balanced n-phase systems of phasors, each set having a different phase sequence plus 1 set of zero phase sequence or unidirectional phasors VA = VA1 + VA2 + VA3 + VA4 - - - - - VA(n-1) + VAn VB = VB1 + VB2 + VB3 + VB4 - - - - - VB(n-1) + VBn VC = VC1 + VC2 + VC3 + VC4 - - - - - VC(n-1) + VCn VD = VD1 + VD2 + VD3 + VD4 - - - - - VD(n-1) + VDn - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Vn = Vn1 + Vn2 + Vn3 + Vn4 - - - - - Vn(n-1) + Vnn (n-1) x Balanced 1 x Zero Sequence
- 44. > Fault Analysis – January 2004 45 45 Unbalanced 3-Phase System VA = VA1 + VA2 + VA0 VB = VB1 + VB2 + VB0 VC = VC1 + VC2 + VC0 Positive Sequence Negative Sequence VA1 VC1 120 VB1 VA2 VB2 VC2 240
- 45. > Fault Analysis – January 2004 46 46 Unbalanced 3-Phase System Zero Sequence VA0 VB0 VC0
- 46. > Fault Analysis – January 2004 47 47 Symmetrical Components VA VA1 + VA2 + VA0 VB VB1 + VB2 + VB0 VC VC1 + VC2 + VC0 VA VB VC + + VB1 VC1 VA1 VB2 VC2 VC0 VB0 VA0 VA2 VB1 = a2VA1 VB2 = a VA2 VB0 = VA0 VC1 = a VA1 VC2 = a2VA2 VC0 = VA0 = = = Phase Positive + Negative + Zero
- 47. > Fault Analysis – January 2004 48 48 Converting from Sequence Components to Phase Values VA0 VC1 VC VA2 VA1 VA VC0 VC2 VB2 VB0 VB1 VB VA = VA1 + VA2 + VA0 VB = VB1 + VB2 + VB0 = a2VA1 + a VA2 + VA0 VC = VC1 + VC2 + VC0 = a VA1 + a2VA2 + VA0
- 48. > Fault Analysis – January 2004 49 49 VA1 = 1/3 {VA + a VB + a2VC} VA2 = 1/3 {VA + a2VB + a VC} VA0 = 1/3 {VA + VB + VC} Converting from Phase Values to Sequence Components VC 3VA0 VB VA0 VA
- 49. > Fault Analysis – January 2004 50 50 Summary VA = VA1 + VA2 + VA0 VB = 2VA1 + VA2 + VA0 VC = VA1 + 2VA2 + VA0 IA = IA1 + IA2 + IA0 IB = 2IA1 + A2 + IA0 IC = IA1 + 2IA2 + IA0 VA1 = 1/3 {VA + VB + 2VC} VA2 = 1/3 {VA + 2VB + VC } VA0 = 1/3 {VA + VB + VC } IA1 = 1/3 {IA + IB + 2IC } IA2 = 1/3 {IA + 2IB + IC } IA0 + 1/3 {IA + IB + IC }
- 50. > Fault Analysis – January 2004 51 51 Residual Current IA IRESIDUAL = IA + IB + IC = 3I0 IB IC E/F Used to detect earth faults IRESIDUAL is Balanced Load IRESIDUAL is /E Faults zero for :- 3 Faults present for :- /Ø/E Faults Ø/ Faults Open circuits (with current in remaining phases)
- 51. > Fault Analysis – January 2004 52 52 Residual Voltage Residual voltage is measured from “Open Delta” or “Broken Delta” VT secondary windings. VRESIDUAL is zero for:- Healthy unfaulted systems 3 Faults / Faults VRESIDUAL is present for:- /E Faults //E Faults Open Circuits (on supply side of VT) VRESIDUAL = VA + VB + VC = 3V0 Used to detect earth faults
- 52. > Fault Analysis – January 2004 53 53 Example Evaluate the positive, negative and zero sequence components for the unbalanced phase vectors : VA = 1 0 VB = 1.5 -90 VC = 0.5 120 VC VA VB
- 53. > Fault Analysis – January 2004 54 54 Solution VA1 = 1/3 (VA + aVB + a2VC) = 1/3 1 + (1 120) (1.5 -90) + (1 240) (0.5 120) = 0.965 15 VA2 = 1/3 (VA + a2VB + aVC) = 1/3 1 + (1 240) (1.5 -90) + (1 120) (0.5 120) = 0.211 150 VA0 = 1/3 (VA + VB + VC) = 1/3 (1 + 1.5 -90 + 0.5 120) = 0.434 -55
- 54. > Fault Analysis – January 2004 55 55 Positive Sequence Voltages VA1 = 0.96515º VC1 = aVA1 VB1 = a2VA1 15º
- 55. > Fault Analysis – January 2004 56 56 Zero Sequence Voltages Negative Sequence Voltages VA2 = 0.211150° VB2 = aVA2 150º VC2 = a2VA2 -55º VA0 = 0.434-55º VB0 = - VC0 = -
- 56. > Fault Analysis – January 2004 57 57 Symmetrical Components VC1 VA1 VB1 VC2 VA2 VB2 VC0 VA0 VB0 VA2 VC2 VB2 VC VA V0 VB
- 57. > Fault Analysis – January 2004 58 58 Example Evaluate the phase quantities Ia, Ib and Ic from the sequence components IA1 = 0.6 0 IA2 = -0.4 0 IA0 = -0.2 0 Solution IA = IA1 + IA2 + IA0 = 0 IB = 2IA1 + IA2 + IA0 = 0.6240 - 0.4120 - 0.20 = 0.91-109 IC = IA1 + 2IA2 + IA0 = 0.6120 - 0.4240 - 0.20 = 0.91-109
- 58. > Fault Analysis – January 2004 62 62 Representation of Plant Cont…
- 59. > Fault Analysis – January 2004 63 63 Transformer Zero Sequence Impedance P Q P Q a a ZT0 b b N0
- 60. > Fault Analysis – January 2004 64 64 General Zero Sequence Equivalent Circuit for Two Winding Transformer On appropriate side of transformer : Earthed Star Winding - Close link ‘a’ Open link ‘b’ Delta Winding - Open link ‘a’ Close link ‘b’ Unearthed Star Winding - Both links open Secondary Terminal 'a' 'a' Primary Terminal 'b' 'b' N0 Z T0
- 61. > Fault Analysis – January 2004 65 65 Zero Sequence Equivalent Circuits (1) S0 ZT0 N0 P0 P S a a b b
- 62. > Fault Analysis – January 2004 66 66 Zero Sequence Equivalent Circuits (2) S0 ZT0 N0 P0 P S a a b b
- 63. > Fault Analysis – January 2004 67 67 Zero Sequence Equivalent Circuits (3) S0 ZT0 N0 P0 P S a a b b
- 64. > Fault Analysis – January 2004 68 68 Zero Sequence Equivalent Circuits (4) S0 ZT0 N0 P0 P S a a b b
- 65. > Fault Analysis – January 2004 69 69 3 Winding Transformers ZP, ZS, ZT = Leakage reactances of Primary, Secondary and Tertiary Windings ZM = Magnetising Impedance = Large Ignored T S P S N1 ZM ZT ZS ZP P T S N1 ZT ZS ZP P T ZP-S = ZP + ZS = Impedance between Primary (P) and Secondary (S) where ZP & ZS are both expressed on same voltage base Similarly ZP-T = ZP + ZT and ZS-T = ZS + ZT
- 66. > Fault Analysis – January 2004 70 70 Auto Transformers ZHL1 = ZH1 + ZL1 (both referred to same voltage base) ZHT1 = ZH1 + ZT1 (both referred to same voltage base) ZLT1 = ZL1 + ZT1 (both referred to same voltage base) H L T L N1 ZM1 ZT1 ZL 1 ZH1 H T L N1 ZT1 ZL1 ZH1 H T Equivalent circuit is similar to that of a 3 winding transformer. ZM = Magnetising Impedance = Large Ignored
- 67. > Fault Analysis – January 2004 71 71 Sequence Networks
- 68. > Fault Analysis – January 2004 72 72 Sequence Networks (1) It can be shown that providing the system impedances are balanced from the points of generation right up to the fault, each sequence current causes voltage drop of its own sequence only. Regard each current flowing within own network thro’ impedances of its own sequence only, with no interconnection between the sequence networks right up to the point of fault.
- 69. > Fault Analysis – January 2004 73 73 +ve, -ve and zero sequence networks are drawn for a ‘reference’ phase. This is usually taken as the ‘A’ phase. Faults are selected to be ‘balanced’ relative to the reference ‘A’ phase. e.g. For Ø/E faults consider an A-E fault For Ø/Ø faults consider a B-C fault Sequence network interconnection is the simplest for the reference phase. Sequence Networks (2)
- 70. > Fault Analysis – January 2004 74 74 Positive Sequence Diagram 1. Start with neutral point N1 - All generator and load neutrals are connected to N1 2. Include all source EMF’s - Phase-neutral voltage 3. Impedance network - Positive sequence impedance per phase 4. Diagram finishes at fault point F1 N1 F1 E1 Z1
- 71. > Fault Analysis – January 2004 75 75 Example V1 = Positive sequence PH-N voltage at fault point I1 = Positive sequence phase current flowing into F1 V1 = E1 - I1 (ZG1 + ZT1 + ZL1) Generator Transformer Line F N R E N1 E1 ZG1 ZT1 ZL1 I1 F1 V1 (N1)
- 72. > Fault Analysis – January 2004 76 76 Negative Sequence Diagram 1. Start with neutral point N2 - All generator and load neutrals are connected to N2 2. No EMF’s included - No negative sequence voltage is generated! 3. Impedance network - Negative sequence impedance per phase 4. Diagram finishes at fault point F2 N2 Z2 F2
- 73. > Fault Analysis – January 2004 77 77 Example V2 = Negative sequence PH-N voltage at fault point I2 = Negative sequence phase current flowing into F2 V2 = -I2 (ZG2 + ZT2 + ZL2) Generator Transformer System Single Line Diagram Negative Sequence Diagram Line F N R E N2 ZG2 ZT2 ZL2 I2 F2 V2 (N2)
- 74. > Fault Analysis – January 2004 78 78 Zero Sequence Diagram (1) For “In Phase” (Zero Phase Sequence) currents to flow in each phase of the system, there must be a fourth connection (this is typically the neutral or earth connection). IA0 + IB0 + IC0 = 3IA0 IA0 N IB0 IC0
- 75. > Fault Analysis – January 2004 79 79 Zero Sequence Diagram (2) Zero sequence voltage between N & E given by V0 = 3IA0.R Zero sequence impedance of neutral to earth path Z0 = V0 = 3R IA0 3IA0 N E R Resistance Earthed System :-
- 76. > Fault Analysis – January 2004 80 80 Zero Sequence Diagram (3) (N0) E0 Generator Transformer System Single Line Diagram Zero Sequence Network F N R E N0 ZG0 ZT0 ZL0 I0 F0 V0 Line RT 3R 3RT V0 = Zero sequence PH-E voltage at fault point I0 = Zero sequence current flowing into F0 V0 = -I0 (ZT0 + ZL0)
- 77. > Fault Analysis – January 2004 81 81 Network Connections
- 78. > Fault Analysis – January 2004 82 82 Interconnection of Sequence Networks (1) Consider sequence networks as blocks with fault terminals F & N for external connections. F1 POSITIVE SEQUENCE NETWORK N1 F2 NEGATIVE SEQUENCE NETWORK N2 F0 ZERO SEQUENCE NETWORK N0 I2 V2 I0 V0
- 79. > Fault Analysis – January 2004 83 83 Interconnection of Sequence Networks (2) For any given fault there are 6 quantities to be considered at the fault point i.e. VA VB VC IA IB IC Relationships between these for any type of fault can be converted into an equivalent relationship between sequence components V1, V2, V0 and I1, I2 , I0 This is possible if :- 1) Any 3 phase quantities are known (provided they are not all voltages or all currents) or 2) 2 are known and 2 others are known to have a specific relationship. From the relationship between sequence V’s and I’s, the manner in which the isolation sequence networks are connected can be determined. The connection of the sequence networks provides a single phase representation (in sequence terms) of the fault.
- 80. > Fault Analysis – January 2004 84 84 IA VA IB IC VB VC F To derive the system constraints at the fault terminals :- Terminals are connected to represent the fault.
- 81. > Fault Analysis – January 2004 85 85 Line to Ground Fault on Phase ‘A’ At fault point :- VA = 0 VB = ? VC = ? IA = ? IB = 0 IC = 0 IA VA IB IC VB VC
- 82. > Fault Analysis – January 2004 86 86 Phase to Earth Fault on Phase ‘A’ At fault point VA = 0 ; IB = 0 ; IC = 0 but VA = V1 + V2 + V0 V1 + V2 + V0 = 0 ------------------------- (1) I0 = 1/3 (IA + IB + IC ) = 1/3 IA I1 = 1/3 (IA + aIB + a2IC) = 1/3 IA I2 = 1/3 (IA + a2IB + aIC) = 1/3 IA I1 = I2 = I0 = 1/3 IA ------------------------- (2) To comply with (1) & (2) the sequence networks must be connected in series :- I1 F1 N1 V1 +ve Seq N/W I2 F2 N2 V2 -ve Seq N/W I0 F0 N0 V0 Zero Seq N/W
- 83. > Fault Analysis – January 2004 87 87 Example : Phase to Earth Fault SOURCE LINE F 132 kV 2000 MVA ZS1 = 8.7 ZS0 = 8.7 A - G FAULT ZL1 = 10 ZL0 = 35 IF 8.7 10 I1 F1 N1 8.7 10 I2 F2 N2 8.7 35 I0 F0 N0 Total impedance = 81.1 I1 = I2 = I0 = 132000 = 940 Amps 3 x 81.1 IF = IA = I1 + I2 + I0 = 3I0 = 2820 Amps
- 84. > Fault Analysis – January 2004 88 88 Earth Fault with Fault Resistance F1 POSITIVE SEQUENCE NETWORK N1 F2 NEGATIVE SEQUENCE NETWORK N2 F0 ZERO SEQUENCE NETWORK N0 I2 V2 I0 V0 I1 V1 3ZF
- 85. > Fault Analysis – January 2004 89 89 Phase to Phase Fault:- B-C Phase I1 F1 N1 V1 +ve Seq N/W I2 F2 N2 V2 -ve Seq N/W I0 F0 N0 V0 Zero Seq N/W
- 86. > Fault Analysis – January 2004 90 90 Example : Phase to Phase Fault Total impedance = 37.4 IB = a2I1 + aI2 I1 = 132000 = 2037 Amps = a2I1 - aI1 3 x 37.4 = (a2 - a) I1 I2 = -2037 Amps = (-j) . 3 x 2037 = 3529 Amps. SOURCE LINE F 132 kV 2000 MVA ZS1 = ZS2 = 8.7 B - C FAULT ZL1 = ZL2 = 10 8.7 10 8.7 10 I1 I2 F1 N1 F2 N2 132000 3
- 87. > Fault Analysis – January 2004 91 91 Phase to Phase Fault with Resistance I1 F1 N1 V1 +ve Seq N/W I2 F2 N2 V2 -ve Seq N/W I0 F0 N0 V0 Zero Seq N/W ZF
- 88. > Fault Analysis – January 2004 92 92 Phase to Phase to Earth Fault:- B-C-E I1 F1 N1 V1 +ve Seq N/W I2 F2 N2 V2 -ve Seq N/W I0 F0 N0 V0 Zero Seq N/W
- 89. > Fault Analysis – January 2004 93 93 Phase to Phase to Earth Fault:- B-C-E with Resistance I1 F1 N1 V1 +ve Seq N/W I2 F2 N2 V2 -ve Seq N/W I0 F0 N0 V0 Zero Seq N/W 3ZF
- 90. > Fault Analysis – January 2004 94 94 Maximum Fault Level Can be higher than 3 fault level on solidly- earthed systems Check that switchgear breaking capacity > maximum fault level for all fault types. Single Phase Fault Level :
- 91. > Fault Analysis – January 2004 95 95 3Ø Versus 1Ø Fault Level (1) Xg XT E Xg XT E Z1 IF 3Ø 1 T g F Z E X X E Ι
- 92. > Fault Analysis – January 2004 96 96 3Ø Versus 1Ø Fault Level (2) Z0 IF 1Ø Xg XT E Z2 = Z1 Z1 Xg2 XT 2 Xg0 XT 0 0 1 F Z 2Z 3E Ι
- 93. > Fault Analysis – January 2004 97 97 3Ø Versus 1Ø Fault Level (3) LEVEL FAULT LEVEL FAULT 1 0 0 1 LEVEL FAULT 1 1 1 1 LEVEL FAULT 3 1 Z Z IF Z 2Z 3E 1 Z 2Z 3E 3Z 3E Z E 3
- 94. > Fault Analysis – January 2004 98 98 Open Circuit & Double Faults
- 95. > Fault Analysis – January 2004 99 99 Series Faults (or Open Circuit Faults) P2 P Q OPEN CIRCUIT FAULT ACROSS PQ Q2 N2 P0 Q0 N0 P1 Q1 N1 NEGATIVE SEQUENCE NETWORK POSITIVE SEQUENCE NETWORK ZERO SEQUENCE NETWORK
- 96. > Fault Analysis – January 2004 100 100 Interconnection of Sequence Networks P1 POSITIVE SEQUENCE NETWORK Q1 P2 NEGATIVE SEQUENCE NETWORK Q2 P0 ZERO SEQUENCE NETWORK Q0 I1 V1 I2 V2 I0 V0 N3 N2 N1 Consider sequence networks as blocks with fault terminals P & Q for interconnections. Unlike shunt faults, terminal N is not used for interconnections.
- 97. > Fault Analysis – January 2004 101 101 Derive System Constraints at the Fault Terminals Ia Ib Ic P Q Va Vb Vc Va' Vb' Vc' va vb vc The terminal conditions imposed by different open circuit faults will be applied across points P & Q on the 3 line conductors. Fault terminal currents Ia, Ib, Ic flow from P to Q. Fault terminal potentials Va, Vb, Vc will be across P and Q.
- 98. > Fault Analysis – January 2004 102 102 Open Circuit Fault On Phase A (1) At fault point :- va = ? vb = 0 vc = 0 Ia = 0 Ib = ? Ic = ? Ia Ib Ic P Q Va Vb Vc Va' Vb' Vc' va vb vc
- 99. > Fault Analysis – January 2004 103 103 At fault point vb = 0 ; vc = 0 ; Ia = 0 v0 = 1/3 (va + vb + vc ) = 1/3 va v1 = 1/3 (va + vb + 2vc ) = 1/3 va v2 = 1/3 (va + 2vb + vc ) = 1/3 va v1 = v2 = v0 = 1/3 va --------------------- (1) Ia = I1 + I2 + I0 = 0 --------------------------- (2) From equations (1) & (2) the sequence networks are connected in parallel. Open Circuit Fault On Phase A (2) I1 P1 Q1 V1 +ve Seq N/W I2 P2 Q2 V2 -ve Seq N/W I0 P0 Q0 V0 Zero Seq N/W
- 100. > Fault Analysis – January 2004 104 104 Two Earth Faults on Phase ‘A’ at Different Locations (1) At fault point F Va = 0 ; Ib = 0 ; Ic = 0 It can be shown that Ia1 = Ia2 = Ia0 Va1 + Va2 + Va0 = 0 (2) At fault point F' Va‘ = 0 ; Ib' = 0 ; Ic' = 0 It can be shown that Ia'1 = Ia'2 = Ia'0 Va'1 + Va'2 + Va'0 = 0 F F' a-e a'-e N
- 101. > Fault Analysis – January 2004 105 105 F1 Ia1 Va1 N1 F'1 Ia'1 Va'1 N'1 F2 Ia2 Va2 N2 F’2 Ia’2 Va’2 N’2 F0 Ia0 Va0 N0 F’0 Ia’0 Va’0 N’0
- 102. > Fault Analysis – January 2004 106 106 F 1 Ia1 Va1 N1 F'1 Ia'1 Va'1 N'1 F 2 Ia2 Va2 N2 F’2 Ia’2 Va’2 N’2 F 0 Ia0 Va0 N0 F’0 Ia’0 Va’0 N’0 INCORRECT CONNECTIONS As :- Va0 ≠ Va0' Va2 ≠ Va2' Va1 ≠ Va1'
- 103. > Fault Analysis – January 2004 107 107 F1 Ia1 Va1 N1 F'1 Ia'1 Va'1 N'1 F2 Ia2 Va2 N2 F’2 Ia’2 Va’2 N’2 F0 Ia0 Va0 N0 F’0 Ia’0 Va’0 N’0 Ia’2 1/1 Va’2 1/1 Va’0
- 104. > Fault Analysis – January 2004 108 108 Open Circuit & Ground Fault Open Circuit Fault At fault point :- Line to Ground Fault At fault point :- va = ? Va' = 0 vb = 0 Vb' = ? vC = 0 Vc' = ? Ia = 0 Ia + I'a = ? Ib = ? Ib + I'b = 0 Ic = ? Ic + I'c = 0 Ia Ib Ic P Q Va Vb Vc Va' Vb' Vc' va vb vc Ia' Ib' Ic' Ia+Ia' Ib+Ib' Ic+Ic'
- 105. > Fault Analysis – January 2004 109 109 P1 Ia1 va1 Q1 Ia'1 Va1 N1 P2 Ia2 va2 Q2 Ia’2 Va2 N2 P0 Ia0 va0 Q0 Ia’0 Va0 N0 Va’2 Va’0 Va’1 1:1 Ia1 + Ia'1 Va’1 Ia1 Ia1 + Ia'1 Ia2 Ia2 + Ia’2 Va’2 Ia2 + Ia’2 Ia0 Ia0 + Ia’0 Ia0 + Ia’0 Va’0