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Symmetrical Components Fault Calculations

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michaeljmack

This document provides an outline and overview of a hands-on relay school covering transmission protection theory including symmetrical components and fault calculations. It discusses power system problems like faults and disturbances, different fault types, and introduces symmetrical components and how they are used to analyze balanced and unbalanced systems. It also covers per unit systems, calculating sequence networks for faults, and evaluating transmission line, generator, and transformer impedances.

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1 Hands-On Relay School Jon F. Daume Bonneville Power Administration March 14-15, 2011 Theory Track Transmission Protection Theory Symmetrical Components & Fault Calculations
2 Class Outline Power system troubles Symmetrical components Per unit system Electrical equipment impedances Sequence networks Fault calculations
3 Power System Problems Faults Equipment trouble System disturbances
4 Fault Causes Lightning Wind and ice Vandalism Contamination External forces Cars, tractors, balloons, airplanes, trees, critters, flying saucers, etc. Equipment failures System disturbances Overloads, system swings
5
6 Fault Types One line to ground (most common) Three phase (rare but most severe) Phase to phase Phase to phase to ground
7 Symmetrical Components
8 Balanced & Unbalanced Systems Balanced System: 3 Phase load 3 Phase fault Unbalanced System: Phase to phase fault One line to ground fault Phase to phase to ground fault Open pole or conductor Unbalanced load
9 Balanced & Unbalanced Systems A C B Balanced System A C B Unbalanced System
10 Sequence Currents for Unbalanced Network Ia2 Ic2Ib2 Negative Sequence Ic0 Ib0 Ia0 Zero Sequence Ia1 Ic1 Ib1 Positive Sequence
11 Sequence Quantities Condition + - 0 3 Phase load - - 3 Phase fault - - Phase to phase fault - One line to ground fault Two phase to ground fault Open pole or conductor Unbalanced load
12 Phase Values From Sequence Values Currents: IA = Ia0 + Ia1 + Ia2 IB = Ib0 + Ib1 + Ib2 IC = Ic0 + Ic1 + Ic2 Voltages: VA = Va0 + Va1 + Va2 VB = Vb0 + Vb1 + Vb2 VC = Vc0 + Vc1 + Vc2
13 a Operator a = -0.5 + j √3= 1 ∠ 120° 2 a2 = -0.5 – j √3= 1 ∠ 240° 2 1 + a + a2 = 0 1 a a2
14 Phase Values From Sequence Values Currents: IA = Ia0 + Ia1 + Ia2 IB = Ia0 + a2Ia1 + aIa2 IC = Ia0 + aIa1 + a2Ia2 Voltages: VA = Va0 + Va1 + Va2 VB = Va0 + a2Va1 + aVa2 VC = Va0 + aVa1 + a2Va2
15 Sequence Values From Phase Values Currents: Ia0 = (IA + IB + IC)/3 Ia1 = (IA + aIB + a2IC)/3 Ia2 = (IA + a2IB + aIC)/3 Voltages: Va0 = (VA + VB + VC)/3 Va1 = (VA + aVB + a2VC)/3 Va2 = (VA + a2VB + aVC)/3
16 Zero Sequence Filter 3Ia0 = Ig = Ir = IA + IB + IC and: 1 + a + a2 = 0 IA = Ia0 + Ia1 + Ia2 +IB = Ia0 + a2Ia1 + aIa2 +IC = Ia0 + aIa1 + a2Ia2 = Ig = 3Iao + 0 + 0
17 Ia Ic Ib 3I0 = Ia + Ib + Ic Zero Sequence Current Filter
18 Zero Sequence Voltage Filter 3V0 3 VO Polarizing Potential Ea Eb Ec
19 Negative Sequence Filter Some protective relays are designed to sense negative sequence currents and/or voltages Much more complicated than detecting zero sequence values Most modern numerical relays have negative sequence elements for fault detection and/or directional control
20 Example IA = 3 + j4 IB = -7 - j2 IC = -2 + j7 +j -j IA = 3+j4 IB = -7-j2 IC = -2+j7
21 Zero Sequence Ia0 = (IA + IB + IC)/3 = [(3+j4)+(-7-j2)+(-2+j7)]/3 = -2 + j3 = 3.61 ∠ 124° Ia0 = Ib0 = Ic0 Ic0 Ib0 Ia0 Zero Sequence
22 Positive Sequence Ia1 = (IA + aIB + a2IC)/3 = [(3+j4)+(-0.5+j√3/2)(-7-j2) +(-0.5-j√3/2)(-2+j7)]/3 = [(3+j4)+(5.23-j5.06)+(7.06-j1.77)]/3 = 5.10 - j 0.94 = 5.19 ∠ -10.5° Ib1 is rotated -120º Ic1 is rotated +120º
23 Positive Sequence Ia1 Ic1 Ib1
24 Negative Sequence Ia2 = (IA + a2IB + aIC)/3 = [(3+j4)+(-0.5-j√3/2)(-7-j2) +(-0.5+j√3/2)(-2+j7)]/3 = [(3+j4)+(1.77+j7.06)+(-5.06-j5.23)]/3 = -0.1 + j 1.94 = 1.95 ∠ 92.9° Ib2 is rotated +120º Ic2 is rotated -120º
25 Negative Sequence Ia2 Ic2 Ib2
26 Reconstruct Phase Currents Ia Ic Ib Ic1 Ib1 Ia1 Ib0 Ia0 Ic0 Ia2 Ib2 Ic2
27 Positive, Negative, and Zero Sequence Impedance Network Calculations for a Fault Study
28 +, -, 0 Sequence Networks Simple 2 Source Power System Example Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
29 Impedance Networks & Fault Type Fault Type + - 0 3 Phase fault - - Phase to phase fault - One line to ground fault Two phase to ground fault
30 Per Unit
31 Per Unit Per unit values are commonly used for fault calculations and fault study programs Per unit values convert real quantities to values based upon number 1 Per unit values include voltages, currents and impedances Calculations are easier Ignore voltage changes due to transformers Ohms law still works
32 Per Unit Convert equipment impedances into per unit values Transformer and generator impedances are given in per cent (%) Line impedances are calculated in ohms These impedances are converted to per unit ohms impedance
33 Base kVA or MVA Arbitrarily selected All values converted to common KVA or MVA Base 100 MVA base is most often used Generator or transformer MVA rating may be used for the base
34 Base kV Use nominal equipment or line voltages 765 kV 525 kV 345 kV 230 kV 169 kV 138 kV 115 kV 69 kV 34.5 kV 13.8 kV 12.5 kV etc.
35 Base Ohms, Amps Base ohms: kV2 1000 = kV2 base kVA base MVA Base amps: base kVA = 1000 base MVA √3 kV √3 kV
36 Base Ohms, Amps (100 MVA Base) kV Base Ohms Base Amps 525 2756.3 110.0 345 1190.3 167.3 230 529.0 251.0 115 132.3 502.0 69 47.6 836.7 34.5 11.9 1673.5 13.8 1.9 4183.7 12.5 1.6 4618.8
37 Conversions Percent to Per Unit: base MVA x % Z of equipment 3φ MVA rating 100 = Z pu Ω @ base MVA If 100 MVA base is used: % Z of equipment = Z pu Ω 3φ MVA rating
38 Ohms to Per Unit pu Ohms = ohms / base ohms base MVA x ohms = pu Ω @ base MVA kV2 LL
39 Per Unit to Real Stuff Amps = pu amps x base amps kV = pu kV x base kV Ohms = pu ohms x base ohms
40 Converting Between Bases Znew = Zold x base MVAnew x kV2 old base MVAold kV2 new
41 Evaluation of System Components Determine positive, negative, and zero sequence impedances of various devices (Z1, Z2, Z0) Only machines will act as a voltage source in the positive sequence network Connect the various impedances into networks according to topography of the system Connect impedance networks for various fault types or other system conditions
42 Synchronous Machines ~ Machine values: Machine reactances given in % of the machine KVA or MVA rating Ground impedances given in ohms
43 Synchronous Machines Machine values: Subtransient reactance (X"d) Transient reactance (X'd) Synchronous reactance (Xd) Negative sequence reactance (X2) Zero sequence reactance (X0)
44 Synchronous Machines Machine neutral ground impedance: Usually expressed in ohms Use 3R or 3X for fault calculations Calculations generally ignores resistance values for generators Calculations generally uses X”d for all impedance values
45 Generator Example Machine nameplate values: 250 MVA, 13.8 kV X"d = 25% @ 250 MVA X'd = 30% @ 250 MVA Xd = 185% @ 250 MVA X2 = 25% @ 250 MVA X0 = 10% @ 250 MVA
46 Generator Example Convert machine reactances to per unit @ common MVA base, (100): X"d = 25% / 250 = 0.1 pu X'd = 30% / 250 = 0.12 pu Xd = 185% / 250 = 0.74 pu X2 = 25% / 250 = 0.1 pu X0 = 10% / 250 = 0.04 pu base MVA x % Z of equipment = Z pu Ω @ base MVA 3φ MVA rating 100
47 Generator Example ~ R1 jX1” = 0.1 R0 jX0 = 0.04 R2 jX2 = 0.1
48 Transformers Zx X Ze Zh H 1:N Vh Ih Zhx H X Equivalent Transformer - Impedance in % Zhx Ω = Vh /Ih = Zh + Zx /N2 Zhx % = Vh /Ih x MVA/kV2 x 100
49 Transformers Impedances in % of the transformer MVA rating Convert from circuit voltage to tap voltage: %Xtap = %Xcircuit kV2 circuit kV2 tap
50 Transformers Convert to common base MVA: %X @ base MVA = base MVA x %X of Transformer MVA of Measurement %X of Transformer = pu X @ 100 MVA MVA of Measurement X1 = X2 = X0 unless a special value is given for X0
51 Transformer Example 250 MVA Transformer 13.8 kV Δ- 230 kV Yg 10% Impedance @ 250 MVA X = 10% = 0.04 pu @ 100 MVA 250 X1 = X2 = X0 = X Assume R1, R2, R0 = 0
52 Transformer Example R1 jX1 = 0.04 R0 jX0 = 0.04 R2 jX2 = 0.04 Zero sequence connection depends upon winding configuration.
53 Transformer Connections Winding Connection Sequence Network Connections Z1, Z2 Z0 Z1, Z2 Z0
54 Transformer Connections Winding Connection Sequence Network Connections Z1, Z2 Z0 Z1, Z2 Z0 Z1, Z2 Z0 Z1, Z2 Z0
55 Delta Wye Transformer A B C b a Ia Ic Ib IA IB IC nIA nIC 3I0 = IA+IB+IC nIB
56 Delta Wye Transformer Ia = nIA - nIC = n(Ia0+Ia1+Ia2- Ia0-aIa1-a2Ia2 ) = n(Ia1 - aIa1 + Ia2 - a2Ia2 ) Ib = nIB - nIA = n(Ia0+a2Ia1+aIa2 -Ia0-Ia1-Ia2 ) = n(a2Ia1 - Ia1 + aIa2 - Ia2 ) Ic = nIC - nIB = n(Ia0+aIa1+a2Ia2 -Ia0-a2Ia1-aIa2 ) = n(aIa1 - a2Ia1 + a2Ia2 - aIa2 ) No zero sequence current outside delta
57 Transformer Connections A YG / YG connection provides a series connection for zero sequence current A Δ / YG connection provides a zero sequence (I0) current source for the YG winding Auto transformer provides same connection as YG / YG connection Use 3R or 3X if a Y is connected to ground with a resistor or reactor
58 Three Winding Transformer Impedances ZHL, ZHM, & ZML given in % at corresponding winding rating Convert impedances to common base MVA Calculate corresponding “T” network impedances: ZH = (ZHL+ ZHM - ZML)/2 ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2
59 “T” Network Calculate corresponding “T” network impedances: ZH = (ZHL+ ZHM - ZML)/2 ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2 ZHL= ZH + ZL ZHM = ZH + ZM ZML= ZM+ ZL ZH ZM ZL
60 Transformer Example 230 kV YG/115 kV YG/13.2 kV Δ Nameplate Impedances ZHL= 5.0% @ 50 MVA ZHM = 5.75% @ 250 MVA ZML = 3.15% @ 50 MVA
61 Transformer Example Convert impedances to per unit @ common MVA Base (100) ZHL= 5.0% @ 50 MVA = 5.0 / 50 = 0.10 pu ZHM = 5.75% @ 250 MVA = 5.75 / 250 = 0.023 pu ZML = 3.15% @ 50 MVA = 3.15 / 50 = 0.063 pu
62 Transformer Example Convert impedances to “T” network equivalent ZH = (ZHL+ ZHM - ZML)/2 = (0.1 + 0.023 - 0.063)/2 = 0.03 pu ZM = (- ZHL+ ZHM + ZML)/2 = (-0.1 + 0.023 + 0.063)/2 = - 0.007 pu ZL = (ZHL- ZHM + ZML)/2 = (0.1 - 0.023 + 0.063)/2 = 0.07 pu
63 Transformer Example 0.03 -0.007 0.07 H M 0.03 -0.007 0.07 H M LL H, 230 kV L, 13.8 kV M, 115 kV +, - Sequence 0 Sequence
Problem Calculate pu impedances for generators and transformers Use 100 MVA base Ignore all resistances
65 Problem Fault 13.8 kV 13.8 kV230 kV230 kV 115 kV
66 Problem - Generator Data Machine nameplate values: 300 MVA Nameplate rating X"d = 25% @ 300 MVA X'd = 30% @ 300 MVA Xd = 200% @ 300 MVA X2 = 25% @ 300 MVA X0 = 10% @ 300 MVA Left generator: 13.8 kV Right generator: 115 kV
67 Problem - Transformer Data Two winding transformer nameplate values 300 MVA Transformer 13.8 kV Δ- 230 kV Yg 10% Impedance @ 300 MVA Three winding transformer nameplate values 230 kV Yg/115 kV Yg/13.8 kV Δ ZHL= 5.0% @ 50 MVA (230 kV – 13.8 kV) ZHM = 6.0% @ 300 MVA (230 kV –115 kV) ZML = 3.2% @ 50 MVA (115 kV – 13.8 kV)
68 Transmission Lines R jX
69 Positive & Negative Sequence Line Impedance Z1 = Z2 = Ra + j 0.2794 f log GMDsep 60 GMRcond or Z1 = Ra + j (Xa + Xd) Ω/mile Ra and Xa from conductor tables Xd = 0.2794 f log GMD 60
70 Positive & Negative Sequence Line Impedance f = system frequency GMDsep = Geometric mean distance between conductors = 3√(dabdbcdac) where dab, dac, dbc = spacing between conductors in feet GMRcond = Geometric mean radius of conductor in feet Ra = conductor resistance, Ω/mile
71 Zero Sequence Line Impedance Z0 = Ra + Re + j 0.01397 f log De _______ 3√(GMRcond GMDsep 2) or Z0 = Ra + Re + j (Xa + Xe - 2Xd) Ω/mile
72 Zero Sequence Line Impedance Re = 0.2862 for a 60 Hz. system. Re does not vary with ρ. De = 2160 √(ρ /f) = 2788 @ 60 Hz. ρ = Ground resistivity, generally assumed to be 100 meter ohms. Xe = 2.89 for 100 meter ohms average ground resistivity.
73 Transmission Lines Ra j(Xa+Xd) Ra+Re j(Xa+Xe-2Xd) Ra j(Xa+Xd) Z1 Z2 Z0
74 Transmission Line Example 230 kV Line 50 Miles long 1272 kcmil ACSR Pheasant Conductor Ra = 0.0903 Ω /mile @ 80° C Xa = 0.37201 Ω /mile GMR = 0.0466 feet Structure: horizontal “H” frame
75 Transmission Line Example Structure “H” frame: GMD = 3√(dabdbcdac) = 3√(23x23x46) = 28.978 feet Xd = 0.2794 f log GMD 60 = 0.2794 log 28.978 = 0.4085 Ω /mile A CB 23 Feet 23 Feet J6 Configuration
76 Transmission Line Example Z1 = Z2 = Ra + j (Xa + Xd) = 0.0903 + j (0.372 + 0.4085) = 0.0903 + j 0.781 Ω /mile Z1 Line = 50(0.0903 + j 0.781) = 4.52 + j 39.03 Ω = 39.29 Ω ∠ 83.4 ° Per unit @ 230 kV, 100 MVA Base base MVA x ohms = pu Ω @ base MVA kV2 LL Z1 Line = (4.52 + j 39.03)100/2302 = 0.0085 + j 0.0743 pu
77 Transmission Line Example Z0 = Ra + Re + j (Xa + Xe – 2Xd) = 0.0903 + 0.286+ j (0.372 + 2.89 - 2 x0.4085) = 0.377 + j 2.445 Ω /mile Z0 Line = 50(0.377 + j 2.445) = 18.83 + j 122.25 Ω = 123.69 Ω ∠ 81.2 ° Per unit @ 230 kV, 100 MVA Base Z0 Line = (18.83 + j 122.25)100/2302 = 0.0356 + j 0.2311 pu
78 Transmission Line Example Z1 Z2 Z0 0.0085 j0.0743 0.0356 j0.2311 0.0085 j0.0743
79 Long Parallel Lines Mutual impedance between lines
80 Mutual Impedance Result of coupling between parallel lines Only affects Zero sequence network Will affect ground fault magnitudes Will affect ground current flow in lines Line #1 Line #2 3I0, Line #1 3I0, Line #2
81 Mutual Impedance ZM = Re + j 0.838 log De Ω/mile GMDcircuits or ZM = Re + j (Xe − 3Xd circuits) Ω/mile Re = 0.2862 @ 60 Hz De = 2160 √(ρ /f) = 2788 @ 60 Hz Xe = 2.89 for 100 meter ohms average ground resistivity
82 Mutual Impedance GMDcircuits is the ninth root of all possible distances between the six conductors, approximately equal to center to center spacing GMDcircuits = 9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) Xd circuits = 0.2794 log GMDcircuits
83 Mutual Impedance Example A CB 23 Feet 23 Feet A CB 23 Feet 23 Feet Circuit #1 Circuit #2 46 Feet 46 Feet 92 Feet 69 Feet 69 Feet 92 Feet 115 Feet 138 Feet 115 Feet 92 Feet
84 Mutual Impedance Example GMDcircuits = 9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) = 9√(92x115x138x69x92x115x46x69x92) = 87.84 feet Xd circuits = 0.2794 log GMDcircuits = 0.2794 log 87.84 = 0.5431 Ω/mile ZM = Re + j (Xe − 3Xd circuits) = 0.2862 + j (2.89 - 3x0.5431) = 0.2862 + j 1.261 Ω/mile (Z0 = 0.377 + j 2.445 Ω /mile)
85 Mutual Impedance Model Bus 1 Bus 2 Z0 Line 1 Z0 Line 2 ZM Bus 1 Bus 2Z02 - ZM Z01- ZM ZM
86 Mutual Impedance Model Model works with at least 1 common bus ZM Affects zero sequence network only ZM For different line voltages: pu Ohms = ohms x base MVA kV1 x kV2 Mutual impedance calculations and modeling become much more complicated with larger systems
87 Mutual Impedance Fault Example Taft Taft 645 Amps 1315 Amps645 Amps 1980 Amps 920 Amps 260 Amps920 Amps 1370 Amps 1LG Faults With Mutual Impedances 1LG Faults Without Mutual Impedances Garrison Garrison Taft Garrison Taft Garrison
88 Problem Calculate Z1 and Z0 pu impedances for a transmission line Calculate R1, Z1, R0 and Z0 Calculate Z1 and Z0 and the angles for Z1 and Z0 Calculate Z0 mutual impedance between transmission lines Use 100 MVA base and 230 kV base
89 Problem Fault 13.8 kV 13.8 kV230 kV230 kV 115 kV
90 Transmission Line Data 2 Parallel 230 kV Lines 60 Miles long 1272 kcmil ACSR Pheasant conductor Ra = 0.0903 Ω /mile @ 80° C Xa = 0.37201 Ω /mile GMR = 0.0466 feet H frame structure - flat, 23 feet between conductors Spacing between circuits = 92 feet centerline to centerline
91 Fault Calculations and Impedance Network Connections
92 Why We Need Fault Studies Relay coordination and settings Determine equipment ratings Determine effective grounding of system Substation ground mat design Substation telephone protection requirements Locating faults
93 Fault Studies Fault Types: 3 Phase One line to ground Phase to phase Phase to phase to ground Fault Locations: Bus fault Line end Line out fault (bus fault with line open) Intermediate faults on transmission line
94 Fault Study Assumptions Ignore loads Use generator X”d Generator X2 equal X”d Ignore generator resistance Ignore transformer resistance 0 Ω Fault resistance assumed Negative sequence impedance = positive sequence impedance
95 Positive Sequence Network Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 Fault
96 Negative Sequence Network Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + I2 Fault
97 Zero Sequence Network Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + I0 Fault
98 Network Reduction Simple 2 Source Power System Example Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
99 Three Phase Fault Only positive sequence impedance network used No negative or zero sequence currents or voltages Simple 2 Source Power System Example Fault
100 Three Phase Fault 1PU Z1 0.084 I1=11.9 I2=0 I0=0 V0 - + V2 - + V1 - + Z0 0.081 Z2 0.084
101 Three Phase Fault Simple 2 Source Power System Example Fault Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 Sequence Network Connection for 3 Phase Fault I1 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007
102 Three Phase Fault Positive Sequence Network Reduced Simple 2 Source Power System Example Fault V1=1-I1Z1 + Vl = 1 Vr = 1 I1 0.177 0.160
103 Three Phase Fault Vectors Va Vc Vb Ia Ic Ib
104 Three Phase Fault MVAFault = MVABase ZFault pu or I pu Fault current = 1 pu ESource ZFault pu
105 Three Phase Fault I1 = E / Z1 = 1 / Z1 I2 = I0 = 0 IA = I1 + I2 + I0 = I1 IB = a2I1 IC = aI1 V1 = 1 – I1Z1 = 0 V2 = 0, V0 = 0 VA = VB = VC = 0
106 Phase to Phase Fault Positive and negative sequence impedance networks connected in parallel No zero sequence currents or voltages Simple 2 Source Power System Example Fault
107 Phase to Phase Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
108 Phase to Phase Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + I2 = -I1 Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 I1 + Vl = 1 Vr = 1 Sequence Network Connection for Phase to Phase Fault Fault
109 Phase to Phase Fault Vectors Va Vc Vb Ic Ib
110 Phase to Phase Fault I1 = - I2 = E = ___1___ I0 = 0 (Z1 + Z2) (Z1 + Z2) IA = I0 + I1 + I2 = 0 IB = I0 + a2I1 + aI2 = a2I1 - aI1 IB = (a2 - a) E = _-j √3 E_ = -j 0.866 E (Z1 + Z2) (Z1 + Z2) Z1 IC = - IB (assume Z1 = Z2)
111 Phase to Phase Fault V1 = E - I1Z1 = 1 - I1Z1 V2 = - I2Z2 = V1 V0 = 0 VA = V1 + V2 + V0 = 2 V1 VB = V0 + a2V1 + aV2 = a2V1 + aV1 = -V1 VC = -V1 Phase to phase fault = 86.6% 3 phase fault
112 Single Line to Ground Fault Positive, negative and zero sequence impedance networks connected in series Simple 2 Source Power System Example Fault
113 Single Line to Ground Fault 1PU Z1 .084 I0=4.02 V0 - + V2 - + V1 - + Z2 .084 Z0 .081 I2=4.02I1=4.02
114 Single Line to Ground Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 I2 I0 Sequence Network Connection for One Line to Ground Fault I1 = I2 = I0 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007 0.04 0.1160.04 0.116 0.04 0.03 0.07 -0.007 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007
115 Single Line to Ground Fault Vectors Va Vc Vb Ia
116 Single Line to Ground Fault I1 = I2 = I0 = ____E_____ = ____1_____ (Z1 + Z2 + Z0) (Z1 + Z2 + Z0) IA = I1 + I2 + I0 = 3 I0 IB = I0 + a2I1 + aI2 = I0 + a2I0 + aI0 = 0 IC = 0 I Ground = I Residual = 3I0
117 Single Line to Ground Fault V1 = E - I1Z1 = 1 - I1Z1 V2 = - I2Z2 V0 = - I0Z0 VA = V1 + V2 + V0 = 0 VB = V0 + a2V1 + aV2 = (Z1 - Z0 ) + a2 (Z0+Z1+Z1) VC = V0 + aV1 + a2V2 = (Z1 - Z0 ) + a (assumes Z1 = Z2) (Z0+Z1+Z1)
118 Two Phase to Ground Fault Positive, negative and zero sequence impedance networks connected in parallel Simple 2 Source Power System Example Fault
119 Two Phase to Ground Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
120 Two Phase to Ground Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 I2 I0 Sequence Network Connection for Phase to Phase to Ground Fault
121 Two Phase to Ground Fault Vectors Va Vc Vb Ic Ib
122 Other Conditions Fault calculations and symmetrical components can also be used to evaluate: Open pole or broken conductor Unbalanced loads Load included in fault analysis Transmission line fault location For these other network conditions, refer to references.
123 References Circuit Analysis of AC Power Systems, Vol. 1 & 2, Edith Clarke Electrical Transmission and Distribution Reference Book, Westinghouse Electric Co., East Pittsburgh, Pa. Symmetrical Components, Wagner and Evans, McGraw-Hill Publishing Co. Symmetrical Components for Power Systems Engineering, J. Lewis Blackburn, Marcel Dekker, Inc.
124 The end Jon F. Daume Bonneville Power Administration Retired!
1 Hands-On Relay School Jon F. Daume Bonneville Power Administration March 14-15, 2011 Theory Track Transmission Protection Theory Transmission System Protection
2 Discussion Topics • Protection overview • Transmission line protection – Phase and ground fault protection – Line differentials – Pilot schemes – Relay communications – Automatic reclosing • Breaker failure relays • Special protection or remedial action schemes
3 Power Transfer Vs VrX Power Transfer 0 0.5 1 0 30 60 90 120 150 180 Angle Delta TransmittedPower P = Vs Vr sin δ / X
4 Increase Power Transfer • Increase transmission system operating voltage • Increase angle δ • Decrease X – Add additional transmission lines – Add series capacitors to existing lines
5
6 Power Transfer During Faults Power Transfer 0 0.2 0.4 0.6 0.8 1 1.2 0 30 60 90 120 150 180 Angle Delta TransmittedPower Normal 1LG LL LLG 3 Phase
7 Vs Vr Power Transfer 0 0.2 0.4 0.6 0.8 1 1.2 0 30 60 90 120 150 180 Angle Delta Power B P1 3 21 P2 6 4 5 A
8 System Stability • Relay operating speed • Circuit breaker opening speed • Pilot tripping • High speed, automatic reclosing • Single pole switching • Special protection or remedial action schemes
9 IEEE Device Numbers Numbers 1 - 97 used 21 Distance relay 25 Synchronizing or synchronism check device 27 Undervoltage relay 32 Directional power relay 43 Manual transfer or selector device 46 Reverse or phase balance current relay 50 Instantaneous overcurrent or rate of rise relay (fixed time overcurrent) (IEEE C37.2)
10 51 AC time overcurrent relay 52 AC circuit breaker 59 Overvoltage relay 62 Time delay stopping or opening relay 63 Pressure switch 67 AC directional overcurrent relay 79 AC reclosing relay 81 Frequency relay 86 Lock out relay 87 Differential relay (IEEE C37.2) IEEE Device Numbers
11 Relay Reliability • Overlapping protection – Relay systems are designed with a high level of dependability – This includes redundant relays – Overlapping protection zones • We will trip no line before its time – Relay system security is also very important – Every effort is made to avoid false trips
12 Relay Reliability • Relay dependability (trip when required) – Redundant relays – Remote backup – Dual trip coils in circuit breaker – Dual batteries – Digital relay self testing – Thorough installation testing – Routine testing and maintenance – Review of relay operations
13 Relay Reliability • Relay security (no false trip) – Careful evaluation before purchase – Right relay for right application – Voting • 2 of 3 relays must agree before a trip – Thorough installation testing – Routine testing and maintenance – Review of relay operations
14 Transmission Line Protection
15 Western Transmission System Northwest includes Oregon, Washington, Idaho, Montana, northern Nevada, Utah, British Columbia and Alberta. WECC is Western Electricity Coordinating Council which includes states and provinces west of Rocky Mountains. Voltage, kV Northwest WECC 115 - 161 27400 miles 48030 miles 230 20850 miles 41950 miles 287 - 345 4360 miles 9800 miles 500 9750 miles 16290 miles 260 - 500 DC 300 miles 1370 miles
16 Transmission Line Impedance • Z ohms/mile = Ra + j (Xa + Xd) • Ra, Xa function of conductor type, length • Xd function of conductor spacing, length Ra j(Xa+Xd)
17 Line Angles vs. Voltage Z = √[Ra 2 + j(Xa+Xd)2] ∠θ ° = tan-1 (X/R) Voltage Level Line Angle (∠θ °) 7.2 - 23 kV 20 - 45 deg. 23 - 69 kV 45 - 75 deg. 69 - 230 kV 60 - 80 deg. 230 - 765 kV 75 - 89 deg.
18 Typical Line Protection
19 Distance Relays (21, 21G)
20 Distance Relays • Common protective relay for non radial transmission lines • Fast and consistent trip times – Instantaneous trip for faults within zone 1 – Operating speed little affected by changes in source impedance • Detect multiphase faults • Ground distance relays detect ground faults • Directional capability
21 CT & PT Connections 21 67N I Phase 3I0 = Ia + Ib + Ic 3V0 V Phase I Polarizing
22 Instrument Transformers • Zsecondary = Zprimary x CTR / VTR • The PT location determines the point from which impedance is measured • The CT location determines the fault direction – Very important consideration for • Transformer terminated lines • Series capacitors • Use highest CT ratio that will work to minimize CT saturation problems
23 Saturated CT Current -100 -50 0 50 100 150 -0.017 0.000 0.017 0.033 0.050 0.06
24 Original Distance Relay • True impedance characteristic – Circular characteristic concentric to RX axis • Required separate directional element • Balance beam construction – Similar to teeter totter – Voltage coil offered restraint – Current coil offered operation • Westinghouse HZ – Later variation allowed for an offset circle
25 Impedance Characteristic R X Directional
26 mho Characteristic • Most common distance element in use • Circular characteristic – Passes through RX origin – No extra directional element required • Maximum torque angle, MTA, usually set at line angle, ∠θ ° – MTA is diameter of circle • Different techniques used to provide full fault detection depending on relay type – Relay may also provide some or full protection for ground faults
27 3 Zone mho Characteristic X R Zone 1 Zone 2 Zone 3 3 Zone Distance Elements Mho Characteristic
28 Typical Reaches 21 Zone 1 85-90% 21 Zone 2 125-180%, Time Delay Trip 21 Zone 3 150-200%, Time Delay Trip Typical Relay Protection Zones 67 Ground Instantaneous Overcurrent 67 Ground Time Overcurrent 67 Ground Time Permissive Transfer Trip Overcurrent
29 Coordination Considerations, Zone 1 • Zone 1 – 80 to 90% of Line impedance – Account for possible errors • Line impedance calculations • CT and PT Errors • Relay inaccuracy – Instantaneous trip
30 Coordination Considerations • Zone 2 – 125% or more of line impedance • Consider strong line out of service • Consider lengths of lines at next substation – Time Delay Trip • > 0.25 seconds (15 cycles) • Greater than BFR clearing time at remote bus • Must be slower if relay overreaches remote zone 2’s. – Also consider load encroachment – Zone 2 may be used with permissive overreach transfer trip w/o time delay
31 Coordination Considerations • Zone 3 – Greater than zone 2 • Consider strong line out of service • Consider lengths of lines at next substation – Time Delay Trip • > 1 second • Greater than BFR clearing time at remote bus • Must be longer if relay overreaches remote zone 3’s. – Must consider load encroachment
32 Coordination Considerations • Zone 3 Special Applications – Starter element for zones 1 and 2 – Provides current reversal logic for permissive transfer trip (reversed) – May be reversed to provide breaker failure protection – Characteristic may include origin for current only tripping – May not be used
33 Problems for Distance Relays • Fault in front of relay • Apparent Impedance • Load encroachment • Fault resistance • Series compensated lines • Power swings
34 3 Phase Fault in Front of Relay • No voltage to make impedance measurement-use a potential memory circuit in distance relay • Use a non-directional, instantaneous overcurrent relay (50-Dead line fault relay) • Utilize switch into fault logic – Allow zone 2 instantaneous trip
35 Apparent Impedance • 3 Terminal lines with apparent impedance • Fault resistance also looks like an apparent impedance • Most critical with very short or unbalanced legs • Results in – Short zone 1 reaches – Long zone 2 reaches and time delays • Pilot protection may be required
36 Apparent Impedance Bus A Bus BZa = 1 ohm Ia = 1 Zb = 1 Ib = 1 Z apparent @ Bus A = Za + ZcIc/Ia = 3 Ohms Apparent Impedance Ic = Ia + Ib = 2 Zc = 1 Bus C
37 Coordination Considerations • Zone 1 – Set to 85 % of actual impedance to nearest terminal • Zone 2 – Set to 125 + % of apparent impedance to most distant terminal – Zone 2 time delay must coordinate with all downstream relays • Zone 3 – Back up for zone 2
38 Load Encroachment • Z Load = kV2 / MVA – Long lines present biggest challenge – Heavy load may enter relay characteristic • Serious problem in August, 2003 East Coast Disturbance • NERC Loading Criteria – 150 % of emergency line load rating – Use reduced voltage (85 %) – 30° Line Angle • Z @ 30° = Z @ MTA cos (∠MTA° -∠30° ) for mho characteristic
39 Load Encroachment • NERC Loading Criteria – Applies to zone 2 and zone 3 phase distance • Other overreaching phase distance elements – All transmission lines > 200 kV – Many transmission lines > 100 kV • Solutions – Don’t use conventional zone 3 element – Use lens characteristic – Use blinders or quadrilateral characteristic – Tilt mho characteristic toward X axis – Utilize special relay load encroachment characteristic
40 Load Encroachment X R Zone 1 Zone 2 Zone 3 Load Consideration with Distance Relays Load Area
41 Lens Characteristic • Ideal for longer transmission lines • More immunity to load encroachment • Less fault resistance coverage • Generated by merging the common area between two mho elements
42 Lens Characteristic
43 Tomato Characteristic • May be used as an external out of step blocking characteristic • Reaches set greater than the tripping elements • Generated by combining the total area of two mho elements
44 Quadrilateral Characteristic • High level of freedom in settings • Blinders on left and right can be moved in or out – More immunity to load encroachment (in) – More fault resistance coverage (out) • Generated by the common area between – Left and right blinders – Below reactance element – Above directional element
45 Quadrilateral Characteristic R X Quadrilateral Characteristic
46 Special Load Encroachment X R Zone 1 Zone 2 Zone 4
47 Fault Resistance • Most severe on short lines • Difficult for ground distance elements to detect • Solutions: – Tilt characteristic toward R axis – Use wide quadrilateral characteristic – Use overcurrent relays for ground faults
48 Fault Resistance X R Zone 1 Zone 2 Zone 3 Fault Resistance Effect on a Mho Characteristic Rf
49 Series Compensated Lines • Series caps added to increase load transfers – Electrically shorten line • Negative inductance • Difficult problem for distance relays • Application depends upon location of capacitors
50 Series Caps 21 21 Zl Zc Zl > Zc
51 Series Caps Bypass MOD Bypass Breaker Discharge Reactor Damping Circuit Metal-Oxide Varistor (MOV) Capacitor (Fuseless) Triggered Gap Isolating MOD Isolating MOD Platform Main Power Components for EWRP Series Capacitors
52 Coordination Considerations • Zone 1 – 80 to 90% of compensated line impedance – Must not overreach remote bus with caps in service • Zone 2 – 125% + of uncompensated apparent line impedance – Must provide direct tripping for any line fault with caps bypassed – May require longer time delays
53 Power Swing • Power swings can cause false trip of 3 phase distance elements • Option to – Block on swing (Out of step block) – Trip on swing (Out of step trip) • Out of step tripping may require special breaker • Allows for controlled separation • Some WECC criteria to follow if OOSB implemented
54 Out Of Step Blocking X R Zone 1 Zone 2 Typical Out Of Step Block Characteristic OOSB Outer Zone OOSB Inner Zone t = 30 ms?
55 Ground Distance Protection and Kn (21G)
56 Fault Types • 3 Phase fault – Positive sequence impedance network only • Phase to phase fault – Positive and negative sequence impedance networks in parallel • One line to ground fault – Positive, negative, and zero sequence impedance networks in series • Phase to phase to ground fault – Positive, negative, and zero sequence impedance networks in parallel
57 Sequence Networks
58 What Does A Distance Relay Measure? • Phase current and phase to ground voltage Zrelay = VLG/IL (Ok for 3 phase faults only) • Phase to phase current and phase to phase voltage Zrelay = VLL/ILL (Ok for 3 phase, PP, PPG faults) • Phase current + compensated ground current and phase to ground voltage Zrelay = VLG/(IL + 3KnI0) (Ok for 3 phase, 1LG, PPG faults)
59 Kn - Why? • Using phase/phase or phase/ground quantities does not give proper reach measurement for 1LG fault • Using zero sequence quantities gives the zero sequence source impedance, not the line impedance • Current compensation (Kn) does work for ground faults • Voltage compensation could also be used but is less common
60 Current Compensation, Kn Kn = (Z0L - Z1L)/3Z1L Z0L = Zero sequence transmission line impedance Z1L = Positive sequence transmission line impedance IRelay = IA + 3I0(Z0L- Z1L)/3Z1L = IA + 3KnI0 ZRelay = VA Relay/IRelay = VA/(IA + 3KnI0) = Z1L Reach of ground distance relay with current compensation is based on positive sequence line impedance, Z1L
61 Current Compensation, Kn • Current compensation (Kn) does work for ground faults. • Kn = (Z0L – Z1L)/3Z1 – Kn may be a scalar quantity or a vector quantity with both magnitude and angle • Mutual impedance coupling from parallel lines can cause a ground distance relay to overreach or underreach, depending upon ground fault location • Mutual impedance coupling can provide incorrect fault location values for ground faults
62 Ground Fault Protection (67N)
63 Ground Faults • Directional ground overcurrent relays (67N) • Ground overcurrent relays – Time overcurrent ground (51) – Instantaneous overcurrent (50) • Measure zero sequence currents • Use zero sequence or negative sequence for directionality
64 Typical Ground Overcurrent Settings • 51 Time overcurrent Select TOC curve, usually very inverse Pickup, usually minimum Time delay >0.25 sec. for remote bus fault • 50 Instantaneous overcurrent >125% Remote bus fault • Must consider affects of mutual coupling from parallel transmission lines.
65 Polarizing for Directional Ground Overcurrent Relays • I Residual and I polarizing – I Polarizing: An autotransformer neutral CT may not provide reliable current polarizing • I Residual and V polarizing – I Residual 3I0 = Ia + Ib + Ic – V Polarizing 3V0 = Va + Vb + Vc • Negative sequence – Requires 3 phase voltages and currents – More immune to mutual coupling problems
66 Current Polarizing I Polarizing Auto Transformer Polarizing Current Source CT H1 X1 H3 X3 H2 X2 Y1 Y2 Y3 H0X0
67 Voltage Polarizing 3 VO Polarizing Potential Ea Eb Ec
68 Mutual Coupling • Transformer affect between parallel lines – Inversely proportional to distance between lines • Only affects zero sequence current • Will affect magnitude of ground currents • Will affect reach of ground distance relays
69 Mutual Coupling Line #1 Line #2 3I0, Line #1 3I0, Line #2
70 Mutual Coupling vs. Ground Relays Taft Taft 645 Amps 1315 Amps645 Amps 1980 Amps 920 Amps 260 Amps920 Amps 1370 Amps 1LG Faults With Mutual Impedances 1LG Faults Without Mutual Impedances Garrison Garrison Taft Garrison Taft Garrison
71 Other Line Protection Relays
72 Line Differential 87 87
73 Line Differential Relays • Compare current magnitudes, phase, etc. at each line terminal • Communicate information between relays • Internal/external fault? Trip/no trip? • Communications dependant! • Changes in communications paths or channel delays can cause potential problems
74 Phase Comparison • Compares phase relationship at terminals • 100% Channel dependant – Looped channels can cause false trips • Nondirectional overcurrent on channel failure • Immune to swings, load, series caps • Single pole capability
75 Pilot Wire • Common on power house lines • Uses metallic twisted pair – Problems if commercial line used – Requires isolation transformers and protection on pilot wire • Nondirectional overcurrent on pilot failure • Newer versions use fiber or radio • Generally limited to short lines if metallic twisted pair is used
76 Pilot Wire
77 Current Differential • Similar to phase comparison • Channel failure? – Distance relay backup or – Non directional overcurrent backup or – No backup – must add separate back up relay • Many channel options – Changes in channel delays may cause problems – Care required in setting up digital channels
78 Current Differential • Single pole capability • 3 Terminal line capability • May include an external, direct transfer trip feature • Immune to swings, load, series caps
79 Transfer Trip
80 Direct Transfer Trip • Line protection • Equipment protection – Transformer terminated lines – Line reactors – Breaker failure • 2 or more signals available – Analog or digital tone equipment
81 Tone 1 Xmit Tone 2 Xmit PCB Trip Coil PCB Trip Coil Tone 1 Rcvd Tone 2 Rcvd Direct Transfer Trip Protective Relay Protective Relay Direct Transfer Trip
82 Direct Transfer Trip Initiation • Zone 1 distance • Zone 2 distance time delay trip • Zone 3 distance time delay trip • Instantaneous ground trip • Time overcurrent ground trip • BFR-Ring bus, breaker & half scheme • Transformer relays on transformer terminated lines • Line reactor relays
83 Tone 2 Xmit Tone 2 Rcvd Permissive Relay PCB Trip Coil PCB Trip Coil Tone 2 Xmit Tone 2 Rcvd Permissive Transfer Trip Permissive Relay Permissive Transfer Trip
84 Permissive Keying • Zone 2 instantaneous • Permissive overcurrent ground (very sensitive setting) • PCB 52/b switch • Current reversal can cause problems
85 PRT Current Reversal A C D B Ib Id Ia Ic Fault near breaker B. Relays at B pick up Relays at B key permissive signal to A, trip breaker B instantaneously Relays at A pick up and key permissive signal to B. Relays at C pick up and key permissive signal to C. Relays at D block I Fault, Line AB I Fault, Line CD
86 PRT Current Reversal A C D B Id Ia Ic Breaker B opens instantaneously. Relays at B drop out. Fault current on line CD changes direction. Relays at A remain picked up and trip by permissive signal from B. Relays at C drop out and stop keying permissive signal to C. Relays at D pick up and key permissive signal to D. I Fault, Line AB I Fault, Line CD
87 Directional Comparison Blocking • Overreaching relays • Delay for channel time • Channel failure can allow overtrip • Often used with “On/Off” carrier
88 Block Xmit Block Rcvd PCB Trip Coil PCB Trip Coil Block Xmit Block Rcvd Directional Comparison Blocking Scheme Time DelayTime Delay Forward Relay Reverse Relay Reverse Relay Forward Relay TDTD Directional Comparison
89 Directional Comparison Relays • Forward relays must overreach remote bus • Forward relays must not overreach remote reverse relays • Time delay (TD) set for channel delay • Scheme will trip for fault if channel lost – Scheme may overtrip for external fault on channel loss
90 Tone Equipment • Interface between relays and communications channel • Analog tone equipment • Digital tone equipment • Security features – Guard before trip – Alternate shifting of tones – Parity checks on digital
91 Tone Equipment • Newer equipment has 4 or more channels – 2 for direct transfer trip – 1 for permissive transfer trip – 1 for drive to lock out (block reclose)
92 Relay to Relay Communications • Available on many new digital relays • Eliminates need for separate tone gear • 8 or more unique bits of data sent from one relay to other • Programmable functions – Each transmitted bit programmed for specific relay function – Each received bit programmed for specific purpose
93 Telecommunications Channels • Microwave radio – Analog (no longer available) – Digital • Other radio systems • Dedicated fiber between relays – Short runs • Multiplexed fiber – Long runs • SONET Rings
94 Telecommunications Channels • Power line carrier current – On/Off Carrier often used with directional comparison • Hard wire – Concern with ground mat interconnections – Limited to short runs • Leased line – Rent from phone company – Considered less reliable
95 Automatic Reclosing (79) • First reclose ~ 80% success rate • Second reclose ~ 5% success rate • Must delay long enough for arc to deionize t = 10.5 + kV/34.5 cycles 14 cycles for 115 kV; 25 cycles for 500 kV • Must delay long enough for remote terminal to clear • 1LG Faults have a higher success rate than 3 phase faults
96 Automatic Reclosing (79) • Most often single shot • Delay of 30 to 60 cycles following line trip is common • Checking: – Hot bus & dead line – Hot line & dead bus – Sync check • Utilities have many different criteria for transmission line reclosing
97 More on Reclosing • Only reclose for one line to ground faults • Block reclose for time delay trip (pilot schemes) • Never reclose on power house lines • Block reclosing for transformer fault on transformer terminated lines • Block reclosing for bus faults • Block reclosing for BFR • Do not use them
98 Breaker Failure Relay (50BF)
99 Breaker Failure • Stuck breaker is a severe impact to system stability on transmission systems • Breaker failure relays are recommended by NERC for transmission systems operated above 100 kV • BFRs are not required to be redundant by NERC
100 Breaker Failure Relays 1. Fault on line 2. Normal protective relays detect fault and send trip to breaker. 3. Breaker does not trip. 4. BFR Fault detectors picked up. 5. BFR Time delay times out (8 cycles) 6. Clear house (open everything to isolate failed breaker)
101 Breaker Failure Relay Typical Breaker Failure Scheme with Retrip BFR Fault Detector PCB Trip Coil #1 TD Protective Relay 86 Trip Block Close TD PCB Trip Coil #2 BFR Retrip BFR Time Delay, 8~
102 Typical BFR Clearing Times Proper Clearing: 0 Fault occurs +1~ Relays PU, Key TT +2~ PCB trips +1~ Remote terminal clears 3-4 Cycles local clearing time 4-5 Cycles remote clearing time Failed Breaker: 0 Fault occurs +1~ BFR FD PU +8~ BFR Time Delay +1~ BFR Trips 86 LOR +2~ BU PCBs trip +1~ Remote terminal clears 12-13 Cycles local back up clearing time 13-14 Cycles remote backup clearing
103 Remedial Action Schemes (RAS) aka: Special Protection Schemes
104 Remedial Action Schemes • Balance generation and loads • Maintain system stability • Prevent major problems (blackouts) • Prevent equipment damage • Allow system to be operated at higher levels • Provide controlled islanding • Protect equipment and lines from thermal overloads • Many WECC & NERC Requirements
105 Remedial Action Schemes • WECC Compliant RAS – Fully redundant – Annual functional test – Changes, modifications and additions must be approved by WECC • Non WECC RAS – Does not need full redundancy – Local impacts only – Primarily to solve thermal overload problems
106 Underfrequency Load Shedding • Reduce load to match available generation • Undervoltage (27) supervised (V > 0.8 pu) • 14 Cycle total clearing time required • Must conform to WECC guidelines • 4 Steps starting at 59.4 Hz. • Restoration must be controlled • Must coordinate with generator 81 relays • Responsibility of control areas
107 Undervoltage Load Shedding • Detect 3 Phase undervoltage • Prevent voltage collapse • Sufficient time delay before tripping to ride through minor disturbances • Must Conform to WECC Guidelines • Primarily installed West of Cascades
108 Generator Dropping • Trip generators for loss of load • Trip generators for loss of transmission lines or paths – Prevent overloading
109 Reactive Switching • On loss of transmission lines – Trip shunt reactors to increase voltage – Close shunt capacitors to compensate for loss of reactive supplied by transmission lines – Close series capacitors to increase load transfers – Utilize generator var output if possible – Static Var Compensators (SVC) provide high speed adjustments
110 Direct Load Tripping • Provide high speed trip to shed load – May use transfer trip – May use sensitive, fast underfrequency (81) relay • Trip large industrial loads
111 Other RAS Schemes • Controlled islanding – Force separation at know locations • Load brake resistor insertion – Provide a resistive load to slow down acceleration of generators • Out of step tripping – Force separation on swing • Phase shifting transformers – Control load flows
112 Typical RAS Controller
113 Typical RAS Controller Outputs • Generator tripping • Load tripping • Controlled islanding and separation (Four Corners) • Insert series caps on AC Intertie • Shunt capacitor insertion • Shunt reactor tripping • Chief Jo Load Brake Resister insertion • Interutility signaling • AGC Off
114 Chief Jo Brake 1400 Megawatts @ 230 kV
115 RAS Enabling Criteria • Power transfer levels • Direction of power flow • System configuration • Some utilities are considering automatic enabling/disabling based on SCADA data • Phasor measurement capability in relays can be used to enable RAS actions
116 RAS Design Criteria • Generally fully redundant • Generally use alternate route on telecommunications • Extensive use of transfer trip for signaling between substations, power plants, control centers, and RAS controllers
117 UFOs vs. Power Outages
118 the end Jon F. Daume Bonneville Power Administration retired March 15, 2011
Transmission System Faults and Event Analysis Fault Analysis Theory and Modern Fault Analysis Methods Presented by: Matthew Rhodes Electrical Engineer, SRP 1
Transmission System Fault Theory • Symmetrical Fault Analysis • Symmetrical Components • Unsymmetrical Fault Analysis using sequence networks • Lecture material originally developed by Dr. Richard Farmer, ASU Research Professor 2
3 Symmetrical Faults
4 Faults Shunt faults: Three phase a b c Line to line Line to ground 2 Line to ground b a c a b c a b c
5 Faults Series faults One open phase: a b c 2 open phases a b c Increased phase impedance Z a b c
6 Why Study Faults? • Determine currents and voltages in the system under fault conditions • Use information to set protective devices • Determine withstand capability that system equipment must have: – Insulating level – Fault current capability of circuit breakers: • Maximum momentary current • Interrupting current
7 Symmetrical Faults α t=0 2 V i(t) Fault at t = 0AC R L )sin(2)( αω += tVte
8 Symmetrical Faults For a short circuit at generator terminals at t=0 and generator initially open circuited: dt di LRite +=)( dt di LRitVSin +=+ )(2 αω by using Laplace transforms i(t) can be found (L is considered constant)
9 Symmetrical Faults ]/)()([ 2 )( TteSintSin Z V ti −−−−+= θαθαω 2222 )( XRLRZ +=+= ω R X Tan R L Tan 11 −− == ω θ Where: R X R L T ω == Time Constant ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω Where: Iac = ac RMS fault current at t=0 (Examples) Note that for a 3- phase system α will be different for each phase. Therefore, DC offset will be different for each phase
10 t = 0 acI2 iac Idc = 0 ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω o 90== θα V2 e(t) o 90=α
11 ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω 0=α o 90=θ V2 e(t) t = 0 iac02 acI 02 acI idc
12 iac02 acI 02 acI idc 022 acI t 0=α o 90=θ ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω )(ti
13 Symmetrical Faults Iac and Idc are independent after t = 0 22 dc I ac I RMS I += Tt e aco I dc I − = 2 Substituting: Tt e ac IT t e ac I ac I RMS I 2 21)222((max) − += − += ]/)2/([ 2 )( TtetSin Z V ti −+−+= πω
14 Asymmetry Factor IRMS(max) = K(τ) Iac Asymmetry Factor = K(τ) rx eK τπ τ 4 21)( − += Where: τ = number of cycles (Example 7.1) fRXT π2/=
15 Example 7.1 •Fault at a time to produce maximum DC offset •Circuit Breaker opens 3 cycles after fault inception I Fault at t = 0AC R = 0.8 Ώ XL = 8 Ώ V = 20 kVLN - + CB Find: 1. Iac at t = 0 2. IRMS Momentary at = 0.5 cycles 3. IRMS Interrupting Current τ
16 Example 7.1 a. RMSAC kAI 488.2 88.0 20 )0( 22 = + = b. 438.121)5.0( ) 10 5.(4 =+= Π− eK KAImomentary 577.3)488.2)(438.1( == c. 023.121)3( ) 10 3(4 =+= Π− eK KAI ngInterrupti 545.2)488.2)(023.1( ==
17 AC Decrement In the previous analysis we treated the generator as a constant voltage behind a constant impedance for each phase. The constant inductance is valid for steady state conditions but for transient conditions, the generator inductance is not constant. The equivalent machine reactance is made up of 2 parts: a) Armature leakage reactance b) Armature reaction (See Phasor Diagram)
18 AC Decrement Steady state model of generator XL is leakage reactance XAR is a fictitious reactance and XAR>> XL XAR is due to flux linkages of armature current with the field circuit. Flux linkages can not change instantaneously. Therefore, if the generator is initially unloaded when a fault occurs the effective reactance is XL which is referred to as Subtransient Reactance, x”. EI R XL XAR Load
19IL jILXL jILXAR(t) EIField Flux Armature Reaction Resultant Field ET XL XAR - + EI I=IL Load Loaded Generator
20 E”Field Flux Armature Reaction = 0 Resultant Field ET0 t = 0 - XL XAR=0 ET0 - + E” = E’ = E = ET0 I=0 Unloaded Generator
21 XL XAR - + E” = E’ = E = ET0 I=0 t=0 E”Field Flux Armature Reaction = 0 Resultant Field ET0 = 0 Faulted Generator
22 XL XAR=0 - + E” = E’ = E = ET0 I = I” E” = jI”XL t=0+ Field Flux Resultant Field ET = 0 I” Armature Reaction = 0
23 XL XAR’ - + E” = E’ = E = ET0 I = I’ E’ = jI’(XL + XAR’) t ≈ 3Cyc. Field Flux Resultant Field ET = 0 I’ Armature Reaction = 0
24 XL XAR - + E” = E’ = E = ET0 I = I E’ = jI(XL + XAR) t =∞ Field Flux Resultant Field ET = 0 I’ Armature Reaction = 0
25 AC Decrement As fault current begins to flow, armature reaction will increase with time thereby increasing the apparent reactance. Therefore, the ac component of the fault current will decrease with time to a steady state condition as shown in the figure below. "2I '2I I2 "2I
26 AC Decrement For a round rotor machine we only need to consider the direct axis reactance. dX E I " "2 "2 = Subtransient dX E I ' '2 '2 = d X E I 2 2 = Transient Synchronous (steadystate)
27 AC Decrement Can write the ac decrement equation [ ] ([ )])'()'"(2)( '" θαω −++−+−= − − tSinIeIIeIIt ac i dT t dTt For an unloaded generator (special case):TEEEE === '" T”d: Subtransient time constant (function of amortisseur winding X/R) T’d: Transient time constant (function of field winding X/R) Look at equation for t=0 and t=infinity
28 AC Decrement For t = 0 [ ] ([ )])'()'"(2)( '" θαω −++−+−= − − tSinIeIIeIIt ac i dT t dTt For t = ∞ IIiac 2]00[2(max) =++= "2])'()'"[(2(max) IIIIIIiac =+−+−=
29 ac and dc Decrement Transform ac decrement equation to phasor form ] θα −+ − −+ − −= ⎢ ⎢ ⎣ ⎡ /')'(")'"( _ IdT t eIIdT t eIIacI dc decrement equation: A T t eSinIdcI − −= )("2 θα Where TA = Armature circuit time constant (Example 7.2)
30 Example 7.2 I Fault at t = 0AC R = 0 V = 1.05 pu - + CB x”d =.15pu T”d = .035 Sec. x’d = .24pu T’d = 2.0 Sec. xd = 1.1pu TA = 0.2 Sec. No load when 3-phase fault occurs Breaker clears fault in 3 cycles. Find: a) I”, b) IDC(t) c) IRMS at interruption d) Imomentry (max) S 500 MVA, 20kV, 60 Hz Synchronous Generator
31 Example 7.2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 2035. t t AC eetI 2. max "2)( t DC eItI − = KAIBase 434.14 320 500 == kApu dx E I 1010.7 15. 05.1 " " " ==== a DCI 2.2. max 9.9)7(2)( tt DC eetI −− == b
32 Example 7.2 Part c: Find IRMS at interruption (3 cycles) .sec05.0 60 3 ==t ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 205.035. 05. eetIAC ( )[ ] puIAC 92.4909.)975)(.258.3()24(.5.205.1)05(. =++= pueIDC 71.79.9)05(. 2. 05. == − kApuIRMS 132146.971.792.4)05(. 22 ==+= c
33 Example 7.2 Part d: Find IMomentary(max) at t = ½ cycle sec0083. 60 5. ==t ( )[ ] puIAC 43.6909.)996)(.258.3()79(.5.205.1)0083(. =++= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 20083.035. 0083. eetIAC pueIDC 5.99.9 2. 0083. == − kApuIRMS 2159.145.943.6 22 ==+= d
34 Turbine Gen. Energy
35 Superposition for Fault Analysis
36 Superposition for Fault Analysis New representation: IF1 IF2=0 Bus 1 Bus 1 Bus 2 IG = IG! + IG2 = IG1+ IL IM = IM1 – IL IF = IG1 + IM1 Example 7.3 IG1 IG2 ILIM1 IG IF IM
37 Example 7.3 For the system of Slide 35 and 36 the generator is operating at 100 MVA, .95 PF Lagging 5% over rated voltage Part a: Find Subtransient fault current magnitude. From Slide 36 puj j j Z V I TH F F 08.9 116. 05.1 655. )505)(.15(. 05.1 "1 −==== Part b: Neglecting load current, find Generator and motor fault current. pujjIG 7 655. 505. 08.9"1 −=−= pujjjIM 08.2)7(08.9"1 −=−−−=
38 Example 7.3 Part c: Including load current, find Generator and motor current during the fault period. 22* * 18/952. 05.1 18/1 0/05.1 95.cos/1 MG o o oLoad II V S I −==−= − = − == pujI oo G 83/35.718/953.7" −=−+−= pujI oo M 243/00.218/952.08.2" =−−−= c c
39 Z Bus Method For Z bus method of fault studies the following approximations are made: • Neglect load current • Model series impedance only • Model generators and synchronous motors by voltage behind a reactance for the positive sequence system
40 AC AC AC + Eg” - + E m - J 0 . 2 J 0 . 305 J 0 .15 1 2 -VFIF
41 Z Bus Method For the circuit of Figure 7.4d (Slide 36 & 40) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 22 12 21 11 2 1 E E Y Y Y Y I I Injected node currents [matrix Y-bus] nodal admittance Node voltages Premultiplying both sides by the inverse of [Y-bus} Pre-fault node Voltage [Z-Bus] =[Y-Bus]-1 Injected node Current -IF1 0 For a fault at Bus 1 )( 1111 FIZE −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −= − = 1111 1 1 Z V Z E I F F ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2221 1211 2 1 I I ZZ ZZ E E
42 Z-Bus Method ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −= − = 1111 1 1 Z V Z E I F F )( 1111 FIZE −= 11 1 Z V I F F = where: For a fault at Bus 1 IF1 = Fault current at bus 1 VF = Prefault voltage of the faulted bus (Bus 1)
43 Z-Bus Method For N bus system, fault on Bus n ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 . 0 0 0 . ..,... . . . . . 321 321 33333231 22232221 11131211 3 2 1 Fn NNNnNNN nNnnnnn Nn Nn Nn N N I ZZZZZ ZZZZZ ZZZZZ ZZZZZ ZZZZZ E E E E E -VF nn F Fn Z V I = Where: VF = Pre-fault voltage at faulted bus Znn = Thevinen impedance
44 Z-Bus Method After IFn is found the voltage at any bus can be found from: E1 = Z1n (-Ifn) E2 = Z2n(-Ifn) Etc. If voltage at each bus is found, current through any branch can be found: I12 = (E1 - E2) / Ž12 Etc/ Note: Ž12 is series impedance between Bus1 and Bus 2, not from Z-Bus. (Example 7.4)
45 Example 7.4 For the system of Figure 7.3 (Slide 40) using the Z-bus method find: a) Z bus b) IF and I contribution from Line for Bus 1 fault c) IF and I contribution from Line for Bus 2 fault Y20 = -j5Y10 = -j6.67 Y12 = -j3.28 1 2 IF
46 Example 7.4 [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = 95.928.3 28.395.9 jj jj YBus [ ] [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ == − 139.046. 046.1156.1 jj jj YZ busBus ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E 0 -IF 11 )1156.( IjE = -VF -IF 08.9 1156. " j j V I F F −== a b
47 Example 7.4 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E For fault at Bus 1: E1 = E1 1+ E1 2 = 0 E2 = E2 1 + E2 2 = VF + (j.046)I1 E2 = 1.05 + (j.046)(j9.08) = .632 /0o 07.2 305. 0632. 21 12 21 j jZ EE I −= − = − = Find: Line current b
48 Example 7.4 Y20 = -j5Y10 = -j6.67 Y12 = -j3.28 1 2 IF Find IF and I contribution from Line for Bus 2 fault ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E -VF puj j IF 55.7 139. 05.1 2 −== - I F2 o FF jjIjVE 0/703.)55.7)(046.(05.1))(046.(1 =+=−+= puj jZ EE I 3.2 305. 0703. 12 21 12 −= − = − = c
49 Z-Bus Method [Z-Bus] = [Y-Bus]-1 Will not cover formation of [Z-Bus] or [Y-Bus] [Z-Bus] can be considered a fictitious circuit which has the appearance of a rake. See Figure 7.6 on Page 371.
50 nn F nF Z V II == Example: Fault at Bus n ))(( 11 nnF IZVE −= Etc. Z-Bus Rake equivalent
51 Class Problem 1 pujZbus ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 08.06.04. 06.12.08. 04.08.12. For the given Bus Impedance matrix(where subtransient reactances were used) and a pre-fault voltage of 1 p.u.: a. Draw the rake equivalent circuit b. A three-phase short circuit occurs at bus 2. Determine the subtransient fault current and the voltages at buses 1, 2, and 3 during the fault.
52 Symmetrical Components
53 Symmetrical Components Symmetrical Components is often referred to as the language of the Relay Engineer but it is important for all engineers that are involved in power. The terminology is used extensively in the power engineering field and it is important to understand the basic concepts and terminology.
54 Symmetrical Components • Used to be more important as a calculating technique before the advanced computer age. • Is still useful and important to make sanity checks and back-of-an-envelope calculation. • We will be studying 3-phase systems in general. Previously you have only considered balanced voltage sources, balanced impedance and balanced currents.
55 Symmetrical Components n a a b b c Va Vb Vc Va Vb Vc Balanced load supplied by balanced voltages results in balanced currents This is a positive sequence system, In Symmetrical Components we will be studying unbalanced systems with one or more dissymmetry. ZY ZY ZY Ib Ia Ic
56 Symmetrical Components For the General Case of 3 unbalanced voltages VA VB VC 6 degrees of freedom Can define 3 sets of voltages designated as positive sequence, negative sequence and zero sequence
57 Symmetrical Components Common a operator identities a =1/120o a2 = 1/240o a3 = 1/0o a4 = 1/120o 1+a+a2 = 0 (a)(a2) = 1
58 Symmetrical Components Positive Sequence 120o 120o120o VA1 VB1 VC1 2 degrees of freedom VA1 = VA1 VB1 = a2 VA1 VC1 = a VA1 a is operator 1/120o
59 Symmetrical Components Negative Sequence 120o 120o120o VA2 VC2 VB2 2 degrees of freedom a is operator 1/120o VA2 = VA2 VB2 = aVA2 VC2 = a2 VA2
60 Symmetrical Components Zero Sequence 2 degrees of freedom VA0 VB0 VC0 VA0 = VB0 = VC0
61 Symmetrical Components Reforming the phase voltages in terms of the symmetrical component voltages: VA = VA0 + VA1 + VA2 VB = VB0 + VB1 + VB2 VC = VC0 + VC1 + VC2 What have we gained? We started with 3 phase voltages and now have 9 sequence voltages. The answer is that the 9 sequence voltages are not independent and can be defined in terms of other voltages.
62 Symmetrical Components Rewriting the sequence voltages in term of the Phase A sequence voltages: VA = VA0 + VA1 +VA2 VB = VA0 + a2 VA1 + aVA2 VC = VA0 + aVA1 +a2 VA2 VA = V0 + V1 +V2 VB = V0 + a2 V1 + aV2 VC = V0 + aV1 +a2 V2 Drop A Suggests matrix notation: VA 1 1 1 V0 VB 1 a2 a V1 VC 1 a a2 V2 = [VP] = [A] [VS]
63 Symmetrical Components We shall consistently apply: [VP] = Phase Voltages [VS] = Sequence Voltages 1 1 1 [A] = 1 a2 a 1 a a2 [VP] = [A][VS] Pre-multiplying by [A]-1 [A]-1[VP] = [A]-1[A][VS]= [I][VS] [VS] = [A]-1 [VP]
64 Operator a a = 1 /120o = - .5 + j .866 a2 = 1 / 240o = - .5 - j.866 a3 = 1 / 360o = 1 a4 = 1 / 480o = 1 / 120o = a a5 = a2 etc. 1 + a + a2 = 0 a - a2 = j 3 1 - a2 = /30o 1/a = a2 3 Relationships of a can greatly expedite calculations ( Find [A]-1)
65 Inverse of A [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA Step 1: Transpose [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA T Step 2: Replace each element by its minor ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3
66 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3 Inverse of A Step 3: Replace each element by its cofactor ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3
67 Inverse of A ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3 Step 4: Divide by Determinant [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA )(3)(1)(1)(1 2222 aaaaaaaaD −=−+−+−= a a a a aa a a a aa a = − − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − 1 111 2 2 2 2 2 2 2 2 1 1 111 a aa a aaa a == − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − −
68 Inverse of A [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = − aa aaA 2 21 1 1 111 3 1
69 Symmetrical Components Previous relationships were developed for voltages. Same could be developed for currents such that: IA IB IC [IP] = I0 I1 I2 [IS] = [IP] = [A] [IS] [IS] = [A]-1 [IP] 1 1 1 [A] = 1 a2 a 1 a a2 1 1 1 [A]-1 = 1/3 1 a a2 1 a2 a
70 Significance of I0 IA IB IC I0 I1 I2 1 1 1 = 1/3 1 a a2 1 a2 a I0 = 1/3 ( IA + IB + IC) n IA IB IC In In = IA + IB + IC = 3 I0 For a balanced system I0 = 0 For a delta system I0 = 0 (Examples 8.1, 8.2 and 8.3)
71 Example 8.1 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= a aV o o o P 277 277 277 120/277 120/277 0/277 2 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = − 0 0/277 01 1 1 111 3 277 2 2 21 2 1 0 o PS a a aa aaVA V V V V 0 1 2 Find [VS] (Sequence voltages) a b c
72 Example 8.2 Y connected load with reverse sequence [ ] ( ) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = 2 1 10 120/10 120/10 0/10 a aI o o o P a b c Find IS (Sequence Currents) [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − o PS a a aa aaIAI 0/10 0 01 1 1 111 3 10 22 21 0 1 2
73 Example 8.3 Ia = 10 / 0o Ic = 10 /120o Ib = o In [ ] [ ] [ ]PS IA I I I I 1 2 1 0 − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = aaa aaIS 0 1 1 1 111 3 10 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + = o o o S a a a a I 60/33.3 0/67.6 60/33.3 2 3 10 1 2 1 3 10 2 2 0 1 2 o n II 60/103 0 == a b c
74 Sequence Impedance for Shunt Elements Sequence Networks of balanced Y elements( Loads, Reactors, capacitor banks, etc.) VA = IAZy + (IA + IB +IC) Zn = (ZY + Zn)IA + ZnIB + ZnIC VB = ZnIA + (ZY + Zn)IB + ZnIC VC = ZnIA + ZnIB +(ZY + Zn)IC n IB IC . IA VB VA VC ZY ZY ZY Zn
75 Sequence Impedance for Shunt Elements ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ C B A nYnn nnYn nnnY C B A I I I ZZZZ ZZZZ ZZZZ V V V [VP] = [ZP] [IP] (1) Transform to sequence reference frame. We know: [VP] = [A] [VS] and [IP] = [A] [IS], Substitute in(1) [A][VS] = [ZP][A][IS] premultiply both sides by [A]-1 [VS] = [A]-1[ZP][A][IS] = [ZS][IS] where: [ZS] = [A]-1[ZP][A]
76 Sequence Impedance for Shunt Elements [ZS] = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + + ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 2 2 2 222120 121110 020100 1 1 111 1 1 111 3 1 aa aa ZZZZ ZZZZ ZZZZ aa aa ZZZ ZZZ ZZZ nYnn nnYn nnnY [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + = Y Y nY S Z Z ZZ Z 0 00 003 0 1 2 0 1 2
77 Sequence Impedance for Shunt Elements ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 1 0 2 1 0 00 00 003 I I I Z Z ZZ V V V Y Y nY V0 = Z00 I0 where: Z00 = ZY +3 Zn V1 = Z11 I1 V2 = Z22 I2 where Z11 = Z22 = ZY Systems are uncoupled: Zero sequence currents only produce zero sequence voltages. Positive sequence currents only produce positive sequence voltages, etc.
78 Sequence Impedance for Shunt Elements We can form sequence circuits which represent the equations: ZY 3 Zn ZY ZY V0 V1 V2 I0 I1 I2 Zero sequence circuit Zn only in zero Sequence No neutral: Zn = infinity Solid ground: Zn = 0 Positive sequence circuit Negative sequence circuit
79 Sequence Impedance for Shunt Elements Delta connected shunt element ZY V0 V1 V2 I0 I1 I2 open ZΔ/3 ZΔ/3 Sequence circuits .A B C IA IB IC ZΔ ZΔ ZΔ
80 Sequence Impedance for Shunt Elements For the general case: [ZS] = [A]-1[ZP][A] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 2 2 2 222120 121110 020100 1 1 111 1 1 111 3 1 aa aa ZZZ ZZZ ZZZ aa aa ZZZ ZZZ ZZZ CCCBCA BCBBBA ACABAA If there is symmetry: ZAA = ZBB = ZCC and ZAB = ZBC = ZCA we could perform multiplication and get: [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ABAA ABAA ABAA S ZZ ZZ ZZ Z 00 00 002 We see that: Z11 = Z22 and Z00 > Z11
81 ZAB ZBC ZAA ZCC VAA’ ZBB VAA’ VBB’ IA IB IC VA VB VC ZCA VA’ VB’ VC ’ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − C B A CCCBCA BCBBBA ACABAA CC BB AA I I I ZZZ ZZZ ZZZ VV VV VV ' ' ' n n Series Element Sequence Impedance
82 Series Element Sequence Impedance Matrices in compact form [VP]-[VP’] = [ZP] [IP] We can transform to the symmetrical component reference frame: [VS] - [VS’] = [ZS] [IS] where: [ZS] = [A]-1[ZP][A] If ZAA = ZBB = ZCC and ZAB = ZBC = ZCA , [ZS] will be the diagonal matrix: [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 1 0 Z Z Z ZS
83 Series Element Sequence Impedance The sequence circuits for series elements are: Z0 V0 V0’ I0 o n0 Z1 V1 V1’ I1 o n1 Z2 V2 V2’ I2 o n2
84 Series Element Sequence Impedance We have quickly covered the calculation of Positive and Negative sequence parameters for 3-phase lines. To determine the zero sequence impedance we need to take the effect of the earth into account. This is done by using Carson’s Method which treats the earth as an equivalent conductor.
85 Rotating Machine Sequence Networks A B C ZK ZK ZK - - - + + + EB EAEC IC IA IB Z n ZAB ZBC ZCA ZCB ZBA ZAC eA = Em Cos ωt eB = Em Cos(ωt – 120o) eC = Em Cos(ωt + 120o) In phasor form: EA= ERMS / 0 = E EB = ERMS /-120o = a2 E EC = ERMS /120o = a E
86 Rotating Machine Sequence Networks [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = aE Ea E EPg 2 EA= ERMS / 0 = E EB = ERMS /-120o = a2 E EC = ERMS /120o = a E or [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − 0 0 1 1 111 3 1 2 2 21 E aE Ea E aa aaEAE PgSg Therefore, only the positive sequence system has a generator voltage source. 0 1 2 a b c
87 Rotating Machine Sequence Networks A B C ZK ZK ZK - - - + + + EB EAEC IC IA IB Z n ZAB ZBC ZCA ZCB ZBA ZAC Machine is not passive: Mutual Reactances: ZAB ≠ ZBA , etc. ZAB = ZBC = ZCA = ZR ZBA = ZCB = ZAC = ZQ
88 Rotating Machine Sequence Networks ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ +++ +++ +++ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ C B A NKNQNR NRNKNQ NQNRNK C B A I I I ZZZZZZ ZZZZZZ ZZZZZZ E E E [ ] [ ][ ]PPGPG IZE = [ ] [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − 2 1 0 1 00 00 00 G G G PGSG Z Z Z AZAZ From the machine diagram we can write: Where: ZG0 = ZK + ZR + ZQ ZG1 = ZK + a2 ZR + a ZQ ZG2 = ZK + a ZR + a2 ZQ uncoupled 0 1 2 0 1 2
89 Rotating Machine Sequence Networks Generator sequence circuits are uncoupled 3Zn ZG0 I0 V0 EG1 - + ZG1 I1 V1 ZG2 I2 V2 Generator Terminal Voltages
90 Rotating Machine Sequence Networks Sequence impedances are unequal ZG1 varies depending on the application a) Steady state, power flow studies: ZG1 = ZS(synchronous) b) Stability studies ZG1 = Z’ (transient) c) Short circuit and transient studies: ZG1 = Z” (subtransient) Motor circuits are similar but there is no voltage source for an induction motor. (Example 8.6)
91 Example 8.6 - [ EP ] + [ IP ] Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ Unbalanced Source [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o o PE 115/295 120/260 0/277 a b c Find phase Currents [ IP ] Ω+=Ω== Δ 43.666.740/10 3 j Z Z o Y Ω+=Ω= 996.087.85/1 jZ o L Ω=+=+=== o LY jZZZZZ 7.43/72.10426.7747.7210
92 Example 8.6 - [ EP ] + [ IP ] Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o o PE 115/295 120/260 0/277 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − o o o o o o PS aa aaEAE 6.216/22.9 77.1/1.277 1.62/91.15 115/295 120/260 0/277 1 1 111 3 1 2 21 0 1 2
93 Example 8.6 10.72 /43.7o Ώ - + 15.91 /62.1o I 0 10.72 /43.7o Ώ - + 277.1 /- 1.77o I1 10.72 /43.7o Ώ - + 9.22 /216.6o I2 00 =I o o I 7.43/72.10 77.1/277 1 − = AI o 5.45/84.251 −= o o I 7.43/72.10 6.216/22.9 2 = AI o 9.172/86.02 =
94 Example 8.6 [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − == o o o SP IAI 8.73/64.26 4.196/72.25 7.46/17.25 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o SI 9.172/86.0 5.45/84.25 0 0 1 2 a b c Amps Amps How would you do problem without Symmetrical Components?
95 Transformer Connections for Zero Sequence P Q Ic Ia Ib I C IA IB P Q Ia + Ib + Ic is not necessarily 0 if we only look at P circuit but Ia = nIA Ib = nIB and Ic = nIC Therefore since IA + IB + IC = 0 , Ia + Ib + Ic = 0 and I0 = 0 P0 Q0 Z0 n0 No zero sequence current flow through transformer
96 Transformer Connections for Zero Sequence P Q Ic Ia Ib IC IA IB P Q Ia + Ib + Ic is not necessarily 0 and IA + IB + IC is not necessarily. P0 Q0Z0 n0 I0 can flow through the transformer. Therefore I0 is not necessarily 0, I0
97 Transformer Connections for Zero Sequence P Q Ic Ia Ib IC IA IB P Q Ia + Ib + Ic is not necessarily 0 and Ia/n + Ib/n + Ic/n is not necessarily 0 P0 Q0 Z0 n0 Provides a zero sequence current source Ib/n Ic/n Ic/n I0 but IA + IB + IC = 0
98 Transformer Connections for Zero Sequence P Q Ic Ia Ib I C IA IB P Q Ia + Ib + Ic = 0 Ia/n + Ib/n + Ic/n is not necessarily 0, but IA + IB + IC = 0 P0 Q0 Z0 n0 No zero sequence current flow Ib/n Ic/n Ic/ n
99 Transformer Connections for Zero Sequence P Q IC IA IB P Q Ia + Ib + Ic = 0 IA + IB + IC = 0 P0 Q0 Z0 n0 No zero sequence current flow ∆ ∆ Ia Ib Ic
100 Power In Sequence NetworksFor a single phase circuit we know that: S = EI* = P + jQ In a 3-phase system we can add the power in each phase such that: SP = EAIA* + EBIB* + ECIC* Written in matrix form [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = * * * C B A CBAP I I I EEES
101 Power in Sequence Networks If we want the apparent power in the symmetrical component reference frame, we can substitute the following: [EP] = [A][ES] [IP] = [A][IS] [EP]T =[ES]T [A]T [IP]* = [A]*[IS]* Into (1) resulting in [SP] = [ES]T [A]T[A]*[IS]* which results in: [SP] = 3[ES]T [IS]* = 3[SS] Where: [SS] = E0I0 * + E1I1 * + E2I2 * From our previous definitions: [SP] = [EP]T [IP]* (1)
102 Class Problem 2 One line of a three-phase generator is open circuited, while the other two are short- circuited to ground. The line currents are: Ia=0, Ib= 1500/90 and Ic=1500/-30 a. Find the symmetrical components of these currents b. Find the ground current
103 Class Problem 3 The currents in a delta load are: Iab=10/0, Ibc= 20/-90 and Ica=15/90 Calculate: a. The sequence components of the delta load currents b. The line currents Ia, Ib and Ic which feed the delta load c. The sequence components of the line currents
104 Class Problem 4 The source voltages given below are applied to the balanced-Y connected load of 6+j8 ohms per phase: Vag=280/0, Vbg= 290/-130 and Vcg=260/110 The load neutral is solidly grounded. a. Draw the sequence networks b. Calculate I0, I1 and I2, the sequence components of the line currents. c. Calculate the line currents Ia, Ib and Ic
105 Unsymmetrical Faults
106 Phase and Symmetrical Component Relationship Phase Reference Frame IA IB IC n V C VB V A Symmetrical Components Reference Frame I0 I1 I2 V0 V1 V2 n0 n1 n2
107 Unsymmetrical Fault Analysis For the study of unsymmetrical faults some, or all, of the following assumptions are made: • Power system balanced prior to fault • Load current neglected • Transformers represented by leakage reactance • Transmission lines represented by series reactance
108 Assumptions Continued • Synchronous machines represented by constant voltage behind reactance(x0, x1. x2) • Non-rotating loads neglected • Small machines neglected • Effect of Δ – Y transformers may be included
109 Faulted 3-Phase Systems Sequence networks are uncoupled for normal system conditions and for the total system we can represent 3 uncoupled systems: positive, negative and zero. When a dissymmetry is applied to the system in the form of a fault, we can connect the sequence networks together to yield the correct sequence currents and voltages in each sequence network. From the sequence currents and voltages we can find the corresponding phase currents and voltages by transformation with the [A] matrix
110 Faulted 3-Phase Systems To represent the dissymmetry we only need to identify 2 points in the system: fault point and neutral point: Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 The sequence networks are connected together from knowledge of the type of fault and fault impedance Example 9.1
111 . AC Bus 1 AC Bus 2 X1=X2 =20Ώ . . ∆ ∆ 100MVA 13.8kV X”=0.15pu X2 = 0.17pu X0 =0.05pu 100MVA 13.8:138kV X = 0.1pu 100MVA 138:13.8kV X = 0.1pu 100MVA 13.8kV X”=0.20pu X2 = 0.21pu X0 =0.05pu Xn = 0.05pu Example 9.1 G M Prefault Voltage = 1.05 pu Draw the positive, negative and zero sequence diagrams for the system on 100MVA, 13.8 kV base in the zone of the generator Line Model: X0 = 60Ώ ( ) Ω= 4.190 100 138 2 BZ puj j ZZ 105.0 4.190 20 21 === puj j Z 315.0 4.190 60 0 ==
112 AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 AC AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 AC AC . AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.15 n0 Example 9.1
113 Example 9.1 Reduce the sequence networks to their thevenin equivalents as viewed from Bus 2 AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.15 n 0 Zero Sequence Thevenin Equivalent from Bus 2 f0 n0 J0.25
114 Example 9.1 AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 Positive Sequence Thevenin Equivalent from Bus 2 139. 655. )2)(.455(. j j Zthev == f1 n1 J0.139 + - 1.05 / 0 o
115 Example 9.1 Negative Sequence Thevenin Equivalent from Bus 2 146. 685. )21)(.475(. j j Zthev == f2 n2 J0.146 AC AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 AC AC .
116 Single Line-to-Ground Fault [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 0 0 FA FP I I [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − FA FA FAFA FPFS I I II aa aaIAI 3 1 0 0 1 1 111 3 1 2 21 IF0 = IF1 = IF2 EFA = IFA ZF EF0 + EF1 + EF2 = (IF0 + IF1 + IF2) ZF EF0 + EF1 + EF2 = 3IF0 ZF A B C IF A EF A IF B IFC n Z F
117 Single Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF 0 EF1 EF2 3ZF
118 Single Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 3 ZF
119 Example 9.3 For the system of Example 9.1 there is a bolted Single- Line-to-Ground fault at Bus 2. Find the fault currents in each phase and the phase voltages at the fault point. f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 IF2 IF1 96.1 146.139.25. 0/05.1 210 j jjj III o FFF −= ++ ===
120 Example 9.3 f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 = IF1 = IF2 = -j1.96 EF0 EF2 EF1 pujjVF 491.)25.)(96.1(0 −=−−= pujVF 777.)139.)(96.1(05.11 =−−= pujjVF 286.)146.)(96.1(2 −=−−=
121 Example 9.3 [ ] [ ][ ]FSFP IAI = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 88.5 96.1 96.1 96.1 1 1 111 2 2 puj j j j aa aa I I I FC FB FA [ ] [ ][ ]FSFP EAE = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ pu pu aa aa E E E o o FC FB FA 7.128/179.1 231/179.1 0 286. 777. 491. 1 1 111 2 2 Note: Unfaulted phase voltages are higher than the source voltage. a b c a b c
122 . Example 9.3a Find fault current in the transmission line, I L 1) Find ILS 2) Find ILP ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 88.5 puj I I I FC FB FA ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 96.1 96.1 96.1 2 1 0 j j j I I I F F F . . AC Bus 1 AC Bus 2 ∆ ∆ G M SLG Fault IF I L
123 Zero Sequence AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2(f0) j.15 n 0 -j1.96 I L0 = 0 I L0 =0
124 Positive Sequence AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 e j30 : 1 SLG e j30 : 1 n1 -j1.96 I T1I L1 6. 655. 2. )96.1(1 jjIT −=−= o LI 60/6.01 −= n1 j.455 j .22 I T1 -j1.96 f1 f1
125 Negative Sequence e -j30 : 1 n2 -j1.96 I T2I L2 n2 j.475 j .22 I T2 -j1.96 6. 685. 21. )96.1(2 jjIT −=−= o LI 120/6.02 −= AC AC . AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 e -j30 : 1 SLG f2 f2
126 Example 9.3a [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == puj puj aa aaIAI o o PSPL 039.1 0 039.1 120/6. 60/6. 0 1 1 111 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − −= o o LSI 120/6. 60/6. 0 0 1 2 a b c . . AC Bus 1 AC Bus 2 ∆ ∆ G M SLG Fault IF I L
127 Line to Line Fault [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = FB FBFP I II 0 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − FB FB FB FBFPFS Ij Ij I I aa aaIAI 3 3 0 3 1 0 1 1 111 3 1 2 21 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = FFBFB FB FA FP ZIE E E E n A B C IFA EF A IFB IFC EF B EF C ZF IF0 = 0 IF1 = IF2 ( ) FFFFBFFBFFBFF ZIZIjZIjZI aa EE 1 2 21 33 3 ==−−= − −=− FBF IjI 31 = so 3 1 j I I F FB = EF1 = EF2 + IF1ZF 0 1 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −+ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFBFBFA FFBFBFA FFBFBFA FFBFB FB FA FS ZaIEE ZIaEE ZIEE ZIE E E aa aaE 2 2 2 2 3 1 1 1 111 3 1
128 Line to Line Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF 2 ZF
129 Example 9.4 For the system of Example 9.1 there is a bolted Line-to-Line fault at Bus 2. Find the fault currents in each phase and the phase voltages at the fault point. f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF1 IF1 IF0 puj jj II o FF 69.3 146.139. 0/05.1 21 −= + =−=00 =FI ( ) ( )( ) pujjjIEE FFF 537.0146.69.3146.221 =−=−== EF1=EF2EF0 00 =FE
130 Example 9.4 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ pu pu jj jj j j aa aa I I I FC FB FA 39.6 39.6 0 )69.3(3 )69.3(3 0 69.3 69.3 0 1 1 111 2 2 a b c ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − −= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ pu pu pu aa aa E E E FC FB FA 537. 537. 07.1 537. 537. 0 1 1 111 2 2 a b c
131 2 Line to Ground Fault [ ] ( ) ( ) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + += ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFCFB FFCFB FA FC FB FA FP ZII ZII E E E E E A B C IFA EF A IFB IFC n EF B EF C ZF IFA = 0 = IF0 + IF1 + IF2 Since IFA = 0, IFB + IFC = 3IF0 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFFA FFFA FFFA FF FF FA FS ZIE ZIE ZIE ZI ZI E aa aaE 0 0 0 0 0 2 2 3/ 3/ 23/ 3 3 1 1 111 3 1 EF0 – EF1 = 3 IF0 ZF so EF0 = EF1 + 3IF0 ZF and EF1 = EF2 0 1 2
132 2 Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 3ZF
133 For the system of Example 9.1 there is a 2-line-to- ground bolted fault at Bus 2. a) Find the fault currents in each phase b) Find the neutral current c) Fault current contribution from motor and generator Neglect delta-wye transformers Example 9.5 . . AC Bus 1 AC Bus 2 ∆ ∆ G M 2LG Fault IF I L
134 Example 9.5 f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 IF2 IF1 puj jj IF 547.4 25.146. )25)(.146(. 139. 05.1 1 −= + + = pujII FF 674.1 25.146. 146. )( 10 = + −= pujjjIII FFF 873.2)547.4(674.1102 =−−−=−−=
135 Example 9.5 This image cannot currently be displayed. [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu j j j aa aaI o o FP 3.21/9.6 7.158/9.6 0 873.2 547.4 674.1 1 1 111 2 2 a b c pujjII FFn 02.5)674.1)(3(3 0 ===
136 Example 9.5 n1 j.455 j .22 I T1 -j4.547 n2 j.475 j .22 I T2 J2.87 3 00 =GFI pujjIII GFFMFO 674.10674.100 =−=−= 39.1 655. 2. )547.4(! jjIGF −=−= pujjjIII GFFMF 16.3)39.1(547.4111 −=−−−=−= 88. 685. 21. )8773.2(2 jjIGF == pujjjIII GFFMF 993.188.873.2222 =−=−= f1 f2
137 Example 9.5 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j aa aaI o o GFP 4.7/98.1 6.172/98.1 51. 88. 39.1 0 1 1 111 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j j aa aaI o o MFP 9.26/0.5 1.153/0.5 504. 99.1 16.3 674.1 1 1 111 2 2
138 Example 9.5 results AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.1 5 n 0 I L0 = 0 2LG J1.674 X AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2 . - + 1 2 1.05 / 0o 1.05 / 0o n1 e -j30 : 1 2LGe j30 : 1 -j3.16X -j1.39 Find the fault current contribution from the generator considering the delta-wye transformer phase shift. Example 9.6 1.39/ -60o -j1.39
139 Example 9.6 Example 9.5 results J1.99 AC AC . AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 e -j30 : 1 2LGe j30 : 1 X j.88.88/ 60o j.88 . . AC Bus 1 AC Bus 2 ∆ ∆ G M 2LG Fault I L X [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j aa aaI o o GP 7/98.1 173/98.1 51. 88. 39.1 0 1 1 111 2 2 a b c IGP
140 Class Problem 5 The system data in p.u. based on SB = 100MVA, VB = 765kV for the lines are: G1: X1=X2=.18, X0=.07 T1: X=.1 LINE 1-3: X1=X2=.4 X0=.17 G2: X1=X2=.2, X0=.10 T2: X=.1 LINE 1-2: X1=X2=.085 X0=.256 G3: X1=X2=.25, X0=.085 T3: X=.24 LINE 2-3: X1=X2=.4 X0=.17 G4: X1=.34, X2=.45, X0=.085 T4: X=.15 a) From the perspective of Bus 1, draw the zero, positive and negative sequence networks. b) Determine the fault current for a 1 L-G bolted fault on Bus 1. AC Bus 1 AC Bus 3 ∆ G1 G3 G4 G2 Bus 2 LINE 1-3 LINE 1-2 LINE 2-3∆ T1 T2 T3 T4
141 Modern Fault Analysis Methods
142 Modern Fault Analysis Tools • Power Quality Meters (Power Quality Alerts) • Operations Event Recorder (ELV, Electronic Log Viewer) • Schweitzer Relay Event Capture • Schweitzer Relay SER (Sequential Events Record)
143 Modern Fault Analysis Example: Line current diff with step distance • First indication of an event - Power Quality alert email notifying On-Call Engineer that there was a voltage sag in the area. This event was a crane contacting a 69kv line. Time of event identified.
144 Modern Fault Analysis Example • Event Log Viewer stores breaker operation events. Search done in ELV using time from PQ Alert and breakers identified where trip occurred. Ferris and Miller breakers operated.
145 Modern Fault Analysis Example • Next the line relays (SEL-311L) at the two substations are interrogated for a possible event at this time. • Use command EVE C 1 to capture the event you desire. The C gives you the digital elements as well as the analog quantities. Ferris and Miller triggered an event record at this time (HIS command used in SEL relay) Reclosing enabled at Miller, additional record is the uncleared fault after reclosing.
146 Modern Fault Analysis Example • If the fault distance is not reasonable from the relays, i.e. the fault distances from each end is longer then the line length, the fault magnitude can be modeled in Aspen to determine fault distance by running interim faults. This discrepancy in distance can result from tapped load or large infeed sources.
147 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller initial fault:
148 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Ferris initial fault: Unknown source voltage
149 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller reclose operation:
150 Modern Fault Analysis Example • This SEL-311L setup is a current differential with step distance protection. • Analysis from line relay SER to ensure proper relaying operation: • Question, why didn’t Z1G pickup?
151 Questions

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Symmetrical Components Fault Calculations

  • 1. 1 Hands-On Relay School Jon F. Daume Bonneville Power Administration March 14-15, 2011 Theory Track Transmission Protection Theory Symmetrical Components & Fault Calculations
  • 2. 2 Class Outline Power system troubles Symmetrical components Per unit system Electrical equipment impedances Sequence networks Fault calculations
  • 3. 3 Power System Problems Faults Equipment trouble System disturbances
  • 4. 4 Fault Causes Lightning Wind and ice Vandalism Contamination External forces Cars, tractors, balloons, airplanes, trees, critters, flying saucers, etc. Equipment failures System disturbances Overloads, system swings
  • 5. 5
  • 6. 6 Fault Types One line to ground (most common) Three phase (rare but most severe) Phase to phase Phase to phase to ground
  • 8. 8 Balanced & Unbalanced Systems Balanced System: 3 Phase load 3 Phase fault Unbalanced System: Phase to phase fault One line to ground fault Phase to phase to ground fault Open pole or conductor Unbalanced load
  • 9. 9 Balanced & Unbalanced Systems A C B Balanced System A C B Unbalanced System
  • 10. 10 Sequence Currents for Unbalanced Network Ia2 Ic2Ib2 Negative Sequence Ic0 Ib0 Ia0 Zero Sequence Ia1 Ic1 Ib1 Positive Sequence
  • 11. 11 Sequence Quantities Condition + - 0 3 Phase load - - 3 Phase fault - - Phase to phase fault - One line to ground fault Two phase to ground fault Open pole or conductor Unbalanced load
  • 12. 12 Phase Values From Sequence Values Currents: IA = Ia0 + Ia1 + Ia2 IB = Ib0 + Ib1 + Ib2 IC = Ic0 + Ic1 + Ic2 Voltages: VA = Va0 + Va1 + Va2 VB = Vb0 + Vb1 + Vb2 VC = Vc0 + Vc1 + Vc2
  • 13. 13 a Operator a = -0.5 + j √3= 1 ∠ 120° 2 a2 = -0.5 – j √3= 1 ∠ 240° 2 1 + a + a2 = 0 1 a a2
  • 14. 14 Phase Values From Sequence Values Currents: IA = Ia0 + Ia1 + Ia2 IB = Ia0 + a2Ia1 + aIa2 IC = Ia0 + aIa1 + a2Ia2 Voltages: VA = Va0 + Va1 + Va2 VB = Va0 + a2Va1 + aVa2 VC = Va0 + aVa1 + a2Va2
  • 15. 15 Sequence Values From Phase Values Currents: Ia0 = (IA + IB + IC)/3 Ia1 = (IA + aIB + a2IC)/3 Ia2 = (IA + a2IB + aIC)/3 Voltages: Va0 = (VA + VB + VC)/3 Va1 = (VA + aVB + a2VC)/3 Va2 = (VA + a2VB + aVC)/3
  • 16. 16 Zero Sequence Filter 3Ia0 = Ig = Ir = IA + IB + IC and: 1 + a + a2 = 0 IA = Ia0 + Ia1 + Ia2 +IB = Ia0 + a2Ia1 + aIa2 +IC = Ia0 + aIa1 + a2Ia2 = Ig = 3Iao + 0 + 0
  • 17. 17 Ia Ic Ib 3I0 = Ia + Ib + Ic Zero Sequence Current Filter
  • 18. 18 Zero Sequence Voltage Filter 3V0 3 VO Polarizing Potential Ea Eb Ec
  • 19. 19 Negative Sequence Filter Some protective relays are designed to sense negative sequence currents and/or voltages Much more complicated than detecting zero sequence values Most modern numerical relays have negative sequence elements for fault detection and/or directional control
  • 20. 20 Example IA = 3 + j4 IB = -7 - j2 IC = -2 + j7 +j -j IA = 3+j4 IB = -7-j2 IC = -2+j7
  • 21. 21 Zero Sequence Ia0 = (IA + IB + IC)/3 = [(3+j4)+(-7-j2)+(-2+j7)]/3 = -2 + j3 = 3.61 ∠ 124° Ia0 = Ib0 = Ic0 Ic0 Ib0 Ia0 Zero Sequence
  • 22. 22 Positive Sequence Ia1 = (IA + aIB + a2IC)/3 = [(3+j4)+(-0.5+j√3/2)(-7-j2) +(-0.5-j√3/2)(-2+j7)]/3 = [(3+j4)+(5.23-j5.06)+(7.06-j1.77)]/3 = 5.10 - j 0.94 = 5.19 ∠ -10.5° Ib1 is rotated -120º Ic1 is rotated +120º
  • 24. 24 Negative Sequence Ia2 = (IA + a2IB + aIC)/3 = [(3+j4)+(-0.5-j√3/2)(-7-j2) +(-0.5+j√3/2)(-2+j7)]/3 = [(3+j4)+(1.77+j7.06)+(-5.06-j5.23)]/3 = -0.1 + j 1.94 = 1.95 ∠ 92.9° Ib2 is rotated +120º Ic2 is rotated -120º
  • 27. 27 Positive, Negative, and Zero Sequence Impedance Network Calculations for a Fault Study
  • 28. 28 +, -, 0 Sequence Networks Simple 2 Source Power System Example Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
  • 29. 29 Impedance Networks & Fault Type Fault Type + - 0 3 Phase fault - - Phase to phase fault - One line to ground fault Two phase to ground fault
  • 31. 31 Per Unit Per unit values are commonly used for fault calculations and fault study programs Per unit values convert real quantities to values based upon number 1 Per unit values include voltages, currents and impedances Calculations are easier Ignore voltage changes due to transformers Ohms law still works
  • 32. 32 Per Unit Convert equipment impedances into per unit values Transformer and generator impedances are given in per cent (%) Line impedances are calculated in ohms These impedances are converted to per unit ohms impedance
  • 33. 33 Base kVA or MVA Arbitrarily selected All values converted to common KVA or MVA Base 100 MVA base is most often used Generator or transformer MVA rating may be used for the base
  • 34. 34 Base kV Use nominal equipment or line voltages 765 kV 525 kV 345 kV 230 kV 169 kV 138 kV 115 kV 69 kV 34.5 kV 13.8 kV 12.5 kV etc.
  • 35. 35 Base Ohms, Amps Base ohms: kV2 1000 = kV2 base kVA base MVA Base amps: base kVA = 1000 base MVA √3 kV √3 kV
  • 36. 36 Base Ohms, Amps (100 MVA Base) kV Base Ohms Base Amps 525 2756.3 110.0 345 1190.3 167.3 230 529.0 251.0 115 132.3 502.0 69 47.6 836.7 34.5 11.9 1673.5 13.8 1.9 4183.7 12.5 1.6 4618.8
  • 37. 37 Conversions Percent to Per Unit: base MVA x % Z of equipment 3φ MVA rating 100 = Z pu Ω @ base MVA If 100 MVA base is used: % Z of equipment = Z pu Ω 3φ MVA rating
  • 38. 38 Ohms to Per Unit pu Ohms = ohms / base ohms base MVA x ohms = pu Ω @ base MVA kV2 LL
  • 39. 39 Per Unit to Real Stuff Amps = pu amps x base amps kV = pu kV x base kV Ohms = pu ohms x base ohms
  • 40. 40 Converting Between Bases Znew = Zold x base MVAnew x kV2 old base MVAold kV2 new
  • 41. 41 Evaluation of System Components Determine positive, negative, and zero sequence impedances of various devices (Z1, Z2, Z0) Only machines will act as a voltage source in the positive sequence network Connect the various impedances into networks according to topography of the system Connect impedance networks for various fault types or other system conditions
  • 42. 42 Synchronous Machines ~ Machine values: Machine reactances given in % of the machine KVA or MVA rating Ground impedances given in ohms
  • 43. 43 Synchronous Machines Machine values: Subtransient reactance (X"d) Transient reactance (X'd) Synchronous reactance (Xd) Negative sequence reactance (X2) Zero sequence reactance (X0)
  • 44. 44 Synchronous Machines Machine neutral ground impedance: Usually expressed in ohms Use 3R or 3X for fault calculations Calculations generally ignores resistance values for generators Calculations generally uses X”d for all impedance values
  • 45. 45 Generator Example Machine nameplate values: 250 MVA, 13.8 kV X"d = 25% @ 250 MVA X'd = 30% @ 250 MVA Xd = 185% @ 250 MVA X2 = 25% @ 250 MVA X0 = 10% @ 250 MVA
  • 46. 46 Generator Example Convert machine reactances to per unit @ common MVA base, (100): X"d = 25% / 250 = 0.1 pu X'd = 30% / 250 = 0.12 pu Xd = 185% / 250 = 0.74 pu X2 = 25% / 250 = 0.1 pu X0 = 10% / 250 = 0.04 pu base MVA x % Z of equipment = Z pu Ω @ base MVA 3φ MVA rating 100
  • 47. 47 Generator Example ~ R1 jX1” = 0.1 R0 jX0 = 0.04 R2 jX2 = 0.1
  • 48. 48 Transformers Zx X Ze Zh H 1:N Vh Ih Zhx H X Equivalent Transformer - Impedance in % Zhx Ω = Vh /Ih = Zh + Zx /N2 Zhx % = Vh /Ih x MVA/kV2 x 100
  • 49. 49 Transformers Impedances in % of the transformer MVA rating Convert from circuit voltage to tap voltage: %Xtap = %Xcircuit kV2 circuit kV2 tap
  • 50. 50 Transformers Convert to common base MVA: %X @ base MVA = base MVA x %X of Transformer MVA of Measurement %X of Transformer = pu X @ 100 MVA MVA of Measurement X1 = X2 = X0 unless a special value is given for X0
  • 51. 51 Transformer Example 250 MVA Transformer 13.8 kV Δ- 230 kV Yg 10% Impedance @ 250 MVA X = 10% = 0.04 pu @ 100 MVA 250 X1 = X2 = X0 = X Assume R1, R2, R0 = 0
  • 52. 52 Transformer Example R1 jX1 = 0.04 R0 jX0 = 0.04 R2 jX2 = 0.04 Zero sequence connection depends upon winding configuration.
  • 53. 53 Transformer Connections Winding Connection Sequence Network Connections Z1, Z2 Z0 Z1, Z2 Z0
  • 54. 54 Transformer Connections Winding Connection Sequence Network Connections Z1, Z2 Z0 Z1, Z2 Z0 Z1, Z2 Z0 Z1, Z2 Z0
  • 56. 56 Delta Wye Transformer Ia = nIA - nIC = n(Ia0+Ia1+Ia2- Ia0-aIa1-a2Ia2 ) = n(Ia1 - aIa1 + Ia2 - a2Ia2 ) Ib = nIB - nIA = n(Ia0+a2Ia1+aIa2 -Ia0-Ia1-Ia2 ) = n(a2Ia1 - Ia1 + aIa2 - Ia2 ) Ic = nIC - nIB = n(Ia0+aIa1+a2Ia2 -Ia0-a2Ia1-aIa2 ) = n(aIa1 - a2Ia1 + a2Ia2 - aIa2 ) No zero sequence current outside delta
  • 57. 57 Transformer Connections A YG / YG connection provides a series connection for zero sequence current A Δ / YG connection provides a zero sequence (I0) current source for the YG winding Auto transformer provides same connection as YG / YG connection Use 3R or 3X if a Y is connected to ground with a resistor or reactor
  • 58. 58 Three Winding Transformer Impedances ZHL, ZHM, & ZML given in % at corresponding winding rating Convert impedances to common base MVA Calculate corresponding “T” network impedances: ZH = (ZHL+ ZHM - ZML)/2 ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2
  • 59. 59 “T” Network Calculate corresponding “T” network impedances: ZH = (ZHL+ ZHM - ZML)/2 ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2 ZHL= ZH + ZL ZHM = ZH + ZM ZML= ZM+ ZL ZH ZM ZL
  • 60. 60 Transformer Example 230 kV YG/115 kV YG/13.2 kV Δ Nameplate Impedances ZHL= 5.0% @ 50 MVA ZHM = 5.75% @ 250 MVA ZML = 3.15% @ 50 MVA
  • 61. 61 Transformer Example Convert impedances to per unit @ common MVA Base (100) ZHL= 5.0% @ 50 MVA = 5.0 / 50 = 0.10 pu ZHM = 5.75% @ 250 MVA = 5.75 / 250 = 0.023 pu ZML = 3.15% @ 50 MVA = 3.15 / 50 = 0.063 pu
  • 62. 62 Transformer Example Convert impedances to “T” network equivalent ZH = (ZHL+ ZHM - ZML)/2 = (0.1 + 0.023 - 0.063)/2 = 0.03 pu ZM = (- ZHL+ ZHM + ZML)/2 = (-0.1 + 0.023 + 0.063)/2 = - 0.007 pu ZL = (ZHL- ZHM + ZML)/2 = (0.1 - 0.023 + 0.063)/2 = 0.07 pu
  • 63. 63 Transformer Example 0.03 -0.007 0.07 H M 0.03 -0.007 0.07 H M LL H, 230 kV L, 13.8 kV M, 115 kV +, - Sequence 0 Sequence
  • 64. Problem Calculate pu impedances for generators and transformers Use 100 MVA base Ignore all resistances
  • 65. 65 Problem Fault 13.8 kV 13.8 kV230 kV230 kV 115 kV
  • 66. 66 Problem - Generator Data Machine nameplate values: 300 MVA Nameplate rating X"d = 25% @ 300 MVA X'd = 30% @ 300 MVA Xd = 200% @ 300 MVA X2 = 25% @ 300 MVA X0 = 10% @ 300 MVA Left generator: 13.8 kV Right generator: 115 kV
  • 67. 67 Problem - Transformer Data Two winding transformer nameplate values 300 MVA Transformer 13.8 kV Δ- 230 kV Yg 10% Impedance @ 300 MVA Three winding transformer nameplate values 230 kV Yg/115 kV Yg/13.8 kV Δ ZHL= 5.0% @ 50 MVA (230 kV – 13.8 kV) ZHM = 6.0% @ 300 MVA (230 kV –115 kV) ZML = 3.2% @ 50 MVA (115 kV – 13.8 kV)
  • 69. 69 Positive & Negative Sequence Line Impedance Z1 = Z2 = Ra + j 0.2794 f log GMDsep 60 GMRcond or Z1 = Ra + j (Xa + Xd) Ω/mile Ra and Xa from conductor tables Xd = 0.2794 f log GMD 60
  • 70. 70 Positive & Negative Sequence Line Impedance f = system frequency GMDsep = Geometric mean distance between conductors = 3√(dabdbcdac) where dab, dac, dbc = spacing between conductors in feet GMRcond = Geometric mean radius of conductor in feet Ra = conductor resistance, Ω/mile
  • 71. 71 Zero Sequence Line Impedance Z0 = Ra + Re + j 0.01397 f log De _______ 3√(GMRcond GMDsep 2) or Z0 = Ra + Re + j (Xa + Xe - 2Xd) Ω/mile
  • 72. 72 Zero Sequence Line Impedance Re = 0.2862 for a 60 Hz. system. Re does not vary with ρ. De = 2160 √(ρ /f) = 2788 @ 60 Hz. ρ = Ground resistivity, generally assumed to be 100 meter ohms. Xe = 2.89 for 100 meter ohms average ground resistivity.
  • 73. 73 Transmission Lines Ra j(Xa+Xd) Ra+Re j(Xa+Xe-2Xd) Ra j(Xa+Xd) Z1 Z2 Z0
  • 74. 74 Transmission Line Example 230 kV Line 50 Miles long 1272 kcmil ACSR Pheasant Conductor Ra = 0.0903 Ω /mile @ 80° C Xa = 0.37201 Ω /mile GMR = 0.0466 feet Structure: horizontal “H” frame
  • 75. 75 Transmission Line Example Structure “H” frame: GMD = 3√(dabdbcdac) = 3√(23x23x46) = 28.978 feet Xd = 0.2794 f log GMD 60 = 0.2794 log 28.978 = 0.4085 Ω /mile A CB 23 Feet 23 Feet J6 Configuration
  • 76. 76 Transmission Line Example Z1 = Z2 = Ra + j (Xa + Xd) = 0.0903 + j (0.372 + 0.4085) = 0.0903 + j 0.781 Ω /mile Z1 Line = 50(0.0903 + j 0.781) = 4.52 + j 39.03 Ω = 39.29 Ω ∠ 83.4 ° Per unit @ 230 kV, 100 MVA Base base MVA x ohms = pu Ω @ base MVA kV2 LL Z1 Line = (4.52 + j 39.03)100/2302 = 0.0085 + j 0.0743 pu
  • 77. 77 Transmission Line Example Z0 = Ra + Re + j (Xa + Xe – 2Xd) = 0.0903 + 0.286+ j (0.372 + 2.89 - 2 x0.4085) = 0.377 + j 2.445 Ω /mile Z0 Line = 50(0.377 + j 2.445) = 18.83 + j 122.25 Ω = 123.69 Ω ∠ 81.2 ° Per unit @ 230 kV, 100 MVA Base Z0 Line = (18.83 + j 122.25)100/2302 = 0.0356 + j 0.2311 pu
  • 78. 78 Transmission Line Example Z1 Z2 Z0 0.0085 j0.0743 0.0356 j0.2311 0.0085 j0.0743
  • 79. 79 Long Parallel Lines Mutual impedance between lines
  • 80. 80 Mutual Impedance Result of coupling between parallel lines Only affects Zero sequence network Will affect ground fault magnitudes Will affect ground current flow in lines Line #1 Line #2 3I0, Line #1 3I0, Line #2
  • 81. 81 Mutual Impedance ZM = Re + j 0.838 log De Ω/mile GMDcircuits or ZM = Re + j (Xe − 3Xd circuits) Ω/mile Re = 0.2862 @ 60 Hz De = 2160 √(ρ /f) = 2788 @ 60 Hz Xe = 2.89 for 100 meter ohms average ground resistivity
  • 82. 82 Mutual Impedance GMDcircuits is the ninth root of all possible distances between the six conductors, approximately equal to center to center spacing GMDcircuits = 9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) Xd circuits = 0.2794 log GMDcircuits
  • 83. 83 Mutual Impedance Example A CB 23 Feet 23 Feet A CB 23 Feet 23 Feet Circuit #1 Circuit #2 46 Feet 46 Feet 92 Feet 69 Feet 69 Feet 92 Feet 115 Feet 138 Feet 115 Feet 92 Feet
  • 84. 84 Mutual Impedance Example GMDcircuits = 9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) = 9√(92x115x138x69x92x115x46x69x92) = 87.84 feet Xd circuits = 0.2794 log GMDcircuits = 0.2794 log 87.84 = 0.5431 Ω/mile ZM = Re + j (Xe − 3Xd circuits) = 0.2862 + j (2.89 - 3x0.5431) = 0.2862 + j 1.261 Ω/mile (Z0 = 0.377 + j 2.445 Ω /mile)
  • 85. 85 Mutual Impedance Model Bus 1 Bus 2 Z0 Line 1 Z0 Line 2 ZM Bus 1 Bus 2Z02 - ZM Z01- ZM ZM
  • 86. 86 Mutual Impedance Model Model works with at least 1 common bus ZM Affects zero sequence network only ZM For different line voltages: pu Ohms = ohms x base MVA kV1 x kV2 Mutual impedance calculations and modeling become much more complicated with larger systems
  • 87. 87 Mutual Impedance Fault Example Taft Taft 645 Amps 1315 Amps645 Amps 1980 Amps 920 Amps 260 Amps920 Amps 1370 Amps 1LG Faults With Mutual Impedances 1LG Faults Without Mutual Impedances Garrison Garrison Taft Garrison Taft Garrison
  • 88. 88 Problem Calculate Z1 and Z0 pu impedances for a transmission line Calculate R1, Z1, R0 and Z0 Calculate Z1 and Z0 and the angles for Z1 and Z0 Calculate Z0 mutual impedance between transmission lines Use 100 MVA base and 230 kV base
  • 89. 89 Problem Fault 13.8 kV 13.8 kV230 kV230 kV 115 kV
  • 90. 90 Transmission Line Data 2 Parallel 230 kV Lines 60 Miles long 1272 kcmil ACSR Pheasant conductor Ra = 0.0903 Ω /mile @ 80° C Xa = 0.37201 Ω /mile GMR = 0.0466 feet H frame structure - flat, 23 feet between conductors Spacing between circuits = 92 feet centerline to centerline
  • 91. 91 Fault Calculations and Impedance Network Connections
  • 92. 92 Why We Need Fault Studies Relay coordination and settings Determine equipment ratings Determine effective grounding of system Substation ground mat design Substation telephone protection requirements Locating faults
  • 93. 93 Fault Studies Fault Types: 3 Phase One line to ground Phase to phase Phase to phase to ground Fault Locations: Bus fault Line end Line out fault (bus fault with line open) Intermediate faults on transmission line
  • 94. 94 Fault Study Assumptions Ignore loads Use generator X”d Generator X2 equal X”d Ignore generator resistance Ignore transformer resistance 0 Ω Fault resistance assumed Negative sequence impedance = positive sequence impedance
  • 95. 95 Positive Sequence Network Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 Fault
  • 96. 96 Negative Sequence Network Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + I2 Fault
  • 97. 97 Zero Sequence Network Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + I0 Fault
  • 98. 98 Network Reduction Simple 2 Source Power System Example Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
  • 99. 99 Three Phase Fault Only positive sequence impedance network used No negative or zero sequence currents or voltages Simple 2 Source Power System Example Fault
  • 100. 100 Three Phase Fault 1PU Z1 0.084 I1=11.9 I2=0 I0=0 V0 - + V2 - + V1 - + Z0 0.081 Z2 0.084
  • 101. 101 Three Phase Fault Simple 2 Source Power System Example Fault Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 Sequence Network Connection for 3 Phase Fault I1 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007
  • 102. 102 Three Phase Fault Positive Sequence Network Reduced Simple 2 Source Power System Example Fault V1=1-I1Z1 + Vl = 1 Vr = 1 I1 0.177 0.160
  • 103. 103 Three Phase Fault Vectors Va Vc Vb Ia Ic Ib
  • 104. 104 Three Phase Fault MVAFault = MVABase ZFault pu or I pu Fault current = 1 pu ESource ZFault pu
  • 105. 105 Three Phase Fault I1 = E / Z1 = 1 / Z1 I2 = I0 = 0 IA = I1 + I2 + I0 = I1 IB = a2I1 IC = aI1 V1 = 1 – I1Z1 = 0 V2 = 0, V0 = 0 VA = VB = VC = 0
  • 106. 106 Phase to Phase Fault Positive and negative sequence impedance networks connected in parallel No zero sequence currents or voltages Simple 2 Source Power System Example Fault
  • 107. 107 Phase to Phase Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
  • 108. 108 Phase to Phase Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + I2 = -I1 Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 I1 + Vl = 1 Vr = 1 Sequence Network Connection for Phase to Phase Fault Fault
  • 109. 109 Phase to Phase Fault Vectors Va Vc Vb Ic Ib
  • 110. 110 Phase to Phase Fault I1 = - I2 = E = ___1___ I0 = 0 (Z1 + Z2) (Z1 + Z2) IA = I0 + I1 + I2 = 0 IB = I0 + a2I1 + aI2 = a2I1 - aI1 IB = (a2 - a) E = _-j √3 E_ = -j 0.866 E (Z1 + Z2) (Z1 + Z2) Z1 IC = - IB (assume Z1 = Z2)
  • 111. 111 Phase to Phase Fault V1 = E - I1Z1 = 1 - I1Z1 V2 = - I2Z2 = V1 V0 = 0 VA = V1 + V2 + V0 = 2 V1 VB = V0 + a2V1 + aV2 = a2V1 + aV1 = -V1 VC = -V1 Phase to phase fault = 86.6% 3 phase fault
  • 112. 112 Single Line to Ground Fault Positive, negative and zero sequence impedance networks connected in series Simple 2 Source Power System Example Fault
  • 113. 113 Single Line to Ground Fault 1PU Z1 .084 I0=4.02 V0 - + V2 - + V1 - + Z2 .084 Z0 .081 I2=4.02I1=4.02
  • 114. 114 Single Line to Ground Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 I2 I0 Sequence Network Connection for One Line to Ground Fault I1 = I2 = I0 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007 0.04 0.1160.04 0.116 0.04 0.03 0.07 -0.007 0.1 0.0370.04 0.037 0.1 0.03 0.07 -0.007
  • 115. 115 Single Line to Ground Fault Vectors Va Vc Vb Ia
  • 116. 116 Single Line to Ground Fault I1 = I2 = I0 = ____E_____ = ____1_____ (Z1 + Z2 + Z0) (Z1 + Z2 + Z0) IA = I1 + I2 + I0 = 3 I0 IB = I0 + a2I1 + aI2 = I0 + a2I0 + aI0 = 0 IC = 0 I Ground = I Residual = 3I0
  • 117. 117 Single Line to Ground Fault V1 = E - I1Z1 = 1 - I1Z1 V2 = - I2Z2 V0 = - I0Z0 VA = V1 + V2 + V0 = 0 VB = V0 + a2V1 + aV2 = (Z1 - Z0 ) + a2 (Z0+Z1+Z1) VC = V0 + aV1 + a2V2 = (Z1 - Z0 ) + a (assumes Z1 = Z2) (Z0+Z1+Z1)
  • 118. 118 Two Phase to Ground Fault Positive, negative and zero sequence impedance networks connected in parallel Simple 2 Source Power System Example Fault
  • 119. 119 Two Phase to Ground Fault 1PU Z1 I1 Z2 I2 Z0 I0 V0 - + V2 - + V1 - +
  • 120. 120 Two Phase to Ground Fault Z2sl Z2tl Z2Ll Z2Lr Z2sr Z2h Z2m Z2l V2= -I2Z2 + Z0sl Z0tl Z0Ll Z0Lr Z0sr Z0h Z0m Z0l V0= -I0Z0 + Z1sl Z1tl Z1Ll Z1Lr Z1sr Z1h Z1m Z1l V1=1-I1Z1 + Vl = 1 Vr = 1 I1 I2 I0 Sequence Network Connection for Phase to Phase to Ground Fault
  • 121. 121 Two Phase to Ground Fault Vectors Va Vc Vb Ic Ib
  • 122. 122 Other Conditions Fault calculations and symmetrical components can also be used to evaluate: Open pole or broken conductor Unbalanced loads Load included in fault analysis Transmission line fault location For these other network conditions, refer to references.
  • 123. 123 References Circuit Analysis of AC Power Systems, Vol. 1 & 2, Edith Clarke Electrical Transmission and Distribution Reference Book, Westinghouse Electric Co., East Pittsburgh, Pa. Symmetrical Components, Wagner and Evans, McGraw-Hill Publishing Co. Symmetrical Components for Power Systems Engineering, J. Lewis Blackburn, Marcel Dekker, Inc.
  • 124. 124 The end Jon F. Daume Bonneville Power Administration Retired!
  • 125. 1 Hands-On Relay School Jon F. Daume Bonneville Power Administration March 14-15, 2011 Theory Track Transmission Protection Theory Transmission System Protection
  • 126. 2 Discussion Topics • Protection overview • Transmission line protection – Phase and ground fault protection – Line differentials – Pilot schemes – Relay communications – Automatic reclosing • Breaker failure relays • Special protection or remedial action schemes
  • 127. 3 Power Transfer Vs VrX Power Transfer 0 0.5 1 0 30 60 90 120 150 180 Angle Delta TransmittedPower P = Vs Vr sin δ / X
  • 128. 4 Increase Power Transfer • Increase transmission system operating voltage • Increase angle δ • Decrease X – Add additional transmission lines – Add series capacitors to existing lines
  • 129. 5
  • 130. 6 Power Transfer During Faults Power Transfer 0 0.2 0.4 0.6 0.8 1 1.2 0 30 60 90 120 150 180 Angle Delta TransmittedPower Normal 1LG LL LLG 3 Phase
  • 131. 7 Vs Vr Power Transfer 0 0.2 0.4 0.6 0.8 1 1.2 0 30 60 90 120 150 180 Angle Delta Power B P1 3 21 P2 6 4 5 A
  • 132. 8 System Stability • Relay operating speed • Circuit breaker opening speed • Pilot tripping • High speed, automatic reclosing • Single pole switching • Special protection or remedial action schemes
  • 133. 9 IEEE Device Numbers Numbers 1 - 97 used 21 Distance relay 25 Synchronizing or synchronism check device 27 Undervoltage relay 32 Directional power relay 43 Manual transfer or selector device 46 Reverse or phase balance current relay 50 Instantaneous overcurrent or rate of rise relay (fixed time overcurrent) (IEEE C37.2)
  • 134. 10 51 AC time overcurrent relay 52 AC circuit breaker 59 Overvoltage relay 62 Time delay stopping or opening relay 63 Pressure switch 67 AC directional overcurrent relay 79 AC reclosing relay 81 Frequency relay 86 Lock out relay 87 Differential relay (IEEE C37.2) IEEE Device Numbers
  • 135. 11 Relay Reliability • Overlapping protection – Relay systems are designed with a high level of dependability – This includes redundant relays – Overlapping protection zones • We will trip no line before its time – Relay system security is also very important – Every effort is made to avoid false trips
  • 136. 12 Relay Reliability • Relay dependability (trip when required) – Redundant relays – Remote backup – Dual trip coils in circuit breaker – Dual batteries – Digital relay self testing – Thorough installation testing – Routine testing and maintenance – Review of relay operations
  • 137. 13 Relay Reliability • Relay security (no false trip) – Careful evaluation before purchase – Right relay for right application – Voting • 2 of 3 relays must agree before a trip – Thorough installation testing – Routine testing and maintenance – Review of relay operations
  • 139. 15 Western Transmission System Northwest includes Oregon, Washington, Idaho, Montana, northern Nevada, Utah, British Columbia and Alberta. WECC is Western Electricity Coordinating Council which includes states and provinces west of Rocky Mountains. Voltage, kV Northwest WECC 115 - 161 27400 miles 48030 miles 230 20850 miles 41950 miles 287 - 345 4360 miles 9800 miles 500 9750 miles 16290 miles 260 - 500 DC 300 miles 1370 miles
  • 140. 16 Transmission Line Impedance • Z ohms/mile = Ra + j (Xa + Xd) • Ra, Xa function of conductor type, length • Xd function of conductor spacing, length Ra j(Xa+Xd)
  • 141. 17 Line Angles vs. Voltage Z = √[Ra 2 + j(Xa+Xd)2] ∠θ ° = tan-1 (X/R) Voltage Level Line Angle (∠θ °) 7.2 - 23 kV 20 - 45 deg. 23 - 69 kV 45 - 75 deg. 69 - 230 kV 60 - 80 deg. 230 - 765 kV 75 - 89 deg.
  • 144. 20 Distance Relays • Common protective relay for non radial transmission lines • Fast and consistent trip times – Instantaneous trip for faults within zone 1 – Operating speed little affected by changes in source impedance • Detect multiphase faults • Ground distance relays detect ground faults • Directional capability
  • 145. 21 CT & PT Connections 21 67N I Phase 3I0 = Ia + Ib + Ic 3V0 V Phase I Polarizing
  • 146. 22 Instrument Transformers • Zsecondary = Zprimary x CTR / VTR • The PT location determines the point from which impedance is measured • The CT location determines the fault direction – Very important consideration for • Transformer terminated lines • Series capacitors • Use highest CT ratio that will work to minimize CT saturation problems
  • 148. 24 Original Distance Relay • True impedance characteristic – Circular characteristic concentric to RX axis • Required separate directional element • Balance beam construction – Similar to teeter totter – Voltage coil offered restraint – Current coil offered operation • Westinghouse HZ – Later variation allowed for an offset circle
  • 150. 26 mho Characteristic • Most common distance element in use • Circular characteristic – Passes through RX origin – No extra directional element required • Maximum torque angle, MTA, usually set at line angle, ∠θ ° – MTA is diameter of circle • Different techniques used to provide full fault detection depending on relay type – Relay may also provide some or full protection for ground faults
  • 151. 27 3 Zone mho Characteristic X R Zone 1 Zone 2 Zone 3 3 Zone Distance Elements Mho Characteristic
  • 152. 28 Typical Reaches 21 Zone 1 85-90% 21 Zone 2 125-180%, Time Delay Trip 21 Zone 3 150-200%, Time Delay Trip Typical Relay Protection Zones 67 Ground Instantaneous Overcurrent 67 Ground Time Overcurrent 67 Ground Time Permissive Transfer Trip Overcurrent
  • 153. 29 Coordination Considerations, Zone 1 • Zone 1 – 80 to 90% of Line impedance – Account for possible errors • Line impedance calculations • CT and PT Errors • Relay inaccuracy – Instantaneous trip
  • 154. 30 Coordination Considerations • Zone 2 – 125% or more of line impedance • Consider strong line out of service • Consider lengths of lines at next substation – Time Delay Trip • > 0.25 seconds (15 cycles) • Greater than BFR clearing time at remote bus • Must be slower if relay overreaches remote zone 2’s. – Also consider load encroachment – Zone 2 may be used with permissive overreach transfer trip w/o time delay
  • 155. 31 Coordination Considerations • Zone 3 – Greater than zone 2 • Consider strong line out of service • Consider lengths of lines at next substation – Time Delay Trip • > 1 second • Greater than BFR clearing time at remote bus • Must be longer if relay overreaches remote zone 3’s. – Must consider load encroachment
  • 156. 32 Coordination Considerations • Zone 3 Special Applications – Starter element for zones 1 and 2 – Provides current reversal logic for permissive transfer trip (reversed) – May be reversed to provide breaker failure protection – Characteristic may include origin for current only tripping – May not be used
  • 157. 33 Problems for Distance Relays • Fault in front of relay • Apparent Impedance • Load encroachment • Fault resistance • Series compensated lines • Power swings
  • 158. 34 3 Phase Fault in Front of Relay • No voltage to make impedance measurement-use a potential memory circuit in distance relay • Use a non-directional, instantaneous overcurrent relay (50-Dead line fault relay) • Utilize switch into fault logic – Allow zone 2 instantaneous trip
  • 159. 35 Apparent Impedance • 3 Terminal lines with apparent impedance • Fault resistance also looks like an apparent impedance • Most critical with very short or unbalanced legs • Results in – Short zone 1 reaches – Long zone 2 reaches and time delays • Pilot protection may be required
  • 160. 36 Apparent Impedance Bus A Bus BZa = 1 ohm Ia = 1 Zb = 1 Ib = 1 Z apparent @ Bus A = Za + ZcIc/Ia = 3 Ohms Apparent Impedance Ic = Ia + Ib = 2 Zc = 1 Bus C
  • 161. 37 Coordination Considerations • Zone 1 – Set to 85 % of actual impedance to nearest terminal • Zone 2 – Set to 125 + % of apparent impedance to most distant terminal – Zone 2 time delay must coordinate with all downstream relays • Zone 3 – Back up for zone 2
  • 162. 38 Load Encroachment • Z Load = kV2 / MVA – Long lines present biggest challenge – Heavy load may enter relay characteristic • Serious problem in August, 2003 East Coast Disturbance • NERC Loading Criteria – 150 % of emergency line load rating – Use reduced voltage (85 %) – 30° Line Angle • Z @ 30° = Z @ MTA cos (∠MTA° -∠30° ) for mho characteristic
  • 163. 39 Load Encroachment • NERC Loading Criteria – Applies to zone 2 and zone 3 phase distance • Other overreaching phase distance elements – All transmission lines > 200 kV – Many transmission lines > 100 kV • Solutions – Don’t use conventional zone 3 element – Use lens characteristic – Use blinders or quadrilateral characteristic – Tilt mho characteristic toward X axis – Utilize special relay load encroachment characteristic
  • 164. 40 Load Encroachment X R Zone 1 Zone 2 Zone 3 Load Consideration with Distance Relays Load Area
  • 165. 41 Lens Characteristic • Ideal for longer transmission lines • More immunity to load encroachment • Less fault resistance coverage • Generated by merging the common area between two mho elements
  • 167. 43 Tomato Characteristic • May be used as an external out of step blocking characteristic • Reaches set greater than the tripping elements • Generated by combining the total area of two mho elements
  • 168. 44 Quadrilateral Characteristic • High level of freedom in settings • Blinders on left and right can be moved in or out – More immunity to load encroachment (in) – More fault resistance coverage (out) • Generated by the common area between – Left and right blinders – Below reactance element – Above directional element
  • 171. 47 Fault Resistance • Most severe on short lines • Difficult for ground distance elements to detect • Solutions: – Tilt characteristic toward R axis – Use wide quadrilateral characteristic – Use overcurrent relays for ground faults
  • 172. 48 Fault Resistance X R Zone 1 Zone 2 Zone 3 Fault Resistance Effect on a Mho Characteristic Rf
  • 173. 49 Series Compensated Lines • Series caps added to increase load transfers – Electrically shorten line • Negative inductance • Difficult problem for distance relays • Application depends upon location of capacitors
  • 175. 51 Series Caps Bypass MOD Bypass Breaker Discharge Reactor Damping Circuit Metal-Oxide Varistor (MOV) Capacitor (Fuseless) Triggered Gap Isolating MOD Isolating MOD Platform Main Power Components for EWRP Series Capacitors
  • 176. 52 Coordination Considerations • Zone 1 – 80 to 90% of compensated line impedance – Must not overreach remote bus with caps in service • Zone 2 – 125% + of uncompensated apparent line impedance – Must provide direct tripping for any line fault with caps bypassed – May require longer time delays
  • 177. 53 Power Swing • Power swings can cause false trip of 3 phase distance elements • Option to – Block on swing (Out of step block) – Trip on swing (Out of step trip) • Out of step tripping may require special breaker • Allows for controlled separation • Some WECC criteria to follow if OOSB implemented
  • 178. 54 Out Of Step Blocking X R Zone 1 Zone 2 Typical Out Of Step Block Characteristic OOSB Outer Zone OOSB Inner Zone t = 30 ms?
  • 180. 56 Fault Types • 3 Phase fault – Positive sequence impedance network only • Phase to phase fault – Positive and negative sequence impedance networks in parallel • One line to ground fault – Positive, negative, and zero sequence impedance networks in series • Phase to phase to ground fault – Positive, negative, and zero sequence impedance networks in parallel
  • 182. 58 What Does A Distance Relay Measure? • Phase current and phase to ground voltage Zrelay = VLG/IL (Ok for 3 phase faults only) • Phase to phase current and phase to phase voltage Zrelay = VLL/ILL (Ok for 3 phase, PP, PPG faults) • Phase current + compensated ground current and phase to ground voltage Zrelay = VLG/(IL + 3KnI0) (Ok for 3 phase, 1LG, PPG faults)
  • 183. 59 Kn - Why? • Using phase/phase or phase/ground quantities does not give proper reach measurement for 1LG fault • Using zero sequence quantities gives the zero sequence source impedance, not the line impedance • Current compensation (Kn) does work for ground faults • Voltage compensation could also be used but is less common
  • 184. 60 Current Compensation, Kn Kn = (Z0L - Z1L)/3Z1L Z0L = Zero sequence transmission line impedance Z1L = Positive sequence transmission line impedance IRelay = IA + 3I0(Z0L- Z1L)/3Z1L = IA + 3KnI0 ZRelay = VA Relay/IRelay = VA/(IA + 3KnI0) = Z1L Reach of ground distance relay with current compensation is based on positive sequence line impedance, Z1L
  • 185. 61 Current Compensation, Kn • Current compensation (Kn) does work for ground faults. • Kn = (Z0L – Z1L)/3Z1 – Kn may be a scalar quantity or a vector quantity with both magnitude and angle • Mutual impedance coupling from parallel lines can cause a ground distance relay to overreach or underreach, depending upon ground fault location • Mutual impedance coupling can provide incorrect fault location values for ground faults
  • 187. 63 Ground Faults • Directional ground overcurrent relays (67N) • Ground overcurrent relays – Time overcurrent ground (51) – Instantaneous overcurrent (50) • Measure zero sequence currents • Use zero sequence or negative sequence for directionality
  • 188. 64 Typical Ground Overcurrent Settings • 51 Time overcurrent Select TOC curve, usually very inverse Pickup, usually minimum Time delay >0.25 sec. for remote bus fault • 50 Instantaneous overcurrent >125% Remote bus fault • Must consider affects of mutual coupling from parallel transmission lines.
  • 189. 65 Polarizing for Directional Ground Overcurrent Relays • I Residual and I polarizing – I Polarizing: An autotransformer neutral CT may not provide reliable current polarizing • I Residual and V polarizing – I Residual 3I0 = Ia + Ib + Ic – V Polarizing 3V0 = Va + Vb + Vc • Negative sequence – Requires 3 phase voltages and currents – More immune to mutual coupling problems
  • 190. 66 Current Polarizing I Polarizing Auto Transformer Polarizing Current Source CT H1 X1 H3 X3 H2 X2 Y1 Y2 Y3 H0X0
  • 191. 67 Voltage Polarizing 3 VO Polarizing Potential Ea Eb Ec
  • 192. 68 Mutual Coupling • Transformer affect between parallel lines – Inversely proportional to distance between lines • Only affects zero sequence current • Will affect magnitude of ground currents • Will affect reach of ground distance relays
  • 193. 69 Mutual Coupling Line #1 Line #2 3I0, Line #1 3I0, Line #2
  • 194. 70 Mutual Coupling vs. Ground Relays Taft Taft 645 Amps 1315 Amps645 Amps 1980 Amps 920 Amps 260 Amps920 Amps 1370 Amps 1LG Faults With Mutual Impedances 1LG Faults Without Mutual Impedances Garrison Garrison Taft Garrison Taft Garrison
  • 197. 73 Line Differential Relays • Compare current magnitudes, phase, etc. at each line terminal • Communicate information between relays • Internal/external fault? Trip/no trip? • Communications dependant! • Changes in communications paths or channel delays can cause potential problems
  • 198. 74 Phase Comparison • Compares phase relationship at terminals • 100% Channel dependant – Looped channels can cause false trips • Nondirectional overcurrent on channel failure • Immune to swings, load, series caps • Single pole capability
  • 199. 75 Pilot Wire • Common on power house lines • Uses metallic twisted pair – Problems if commercial line used – Requires isolation transformers and protection on pilot wire • Nondirectional overcurrent on pilot failure • Newer versions use fiber or radio • Generally limited to short lines if metallic twisted pair is used
  • 201. 77 Current Differential • Similar to phase comparison • Channel failure? – Distance relay backup or – Non directional overcurrent backup or – No backup – must add separate back up relay • Many channel options – Changes in channel delays may cause problems – Care required in setting up digital channels
  • 202. 78 Current Differential • Single pole capability • 3 Terminal line capability • May include an external, direct transfer trip feature • Immune to swings, load, series caps
  • 204. 80 Direct Transfer Trip • Line protection • Equipment protection – Transformer terminated lines – Line reactors – Breaker failure • 2 or more signals available – Analog or digital tone equipment
  • 205. 81 Tone 1 Xmit Tone 2 Xmit PCB Trip Coil PCB Trip Coil Tone 1 Rcvd Tone 2 Rcvd Direct Transfer Trip Protective Relay Protective Relay Direct Transfer Trip
  • 206. 82 Direct Transfer Trip Initiation • Zone 1 distance • Zone 2 distance time delay trip • Zone 3 distance time delay trip • Instantaneous ground trip • Time overcurrent ground trip • BFR-Ring bus, breaker & half scheme • Transformer relays on transformer terminated lines • Line reactor relays
  • 207. 83 Tone 2 Xmit Tone 2 Rcvd Permissive Relay PCB Trip Coil PCB Trip Coil Tone 2 Xmit Tone 2 Rcvd Permissive Transfer Trip Permissive Relay Permissive Transfer Trip
  • 208. 84 Permissive Keying • Zone 2 instantaneous • Permissive overcurrent ground (very sensitive setting) • PCB 52/b switch • Current reversal can cause problems
  • 209. 85 PRT Current Reversal A C D B Ib Id Ia Ic Fault near breaker B. Relays at B pick up Relays at B key permissive signal to A, trip breaker B instantaneously Relays at A pick up and key permissive signal to B. Relays at C pick up and key permissive signal to C. Relays at D block I Fault, Line AB I Fault, Line CD
  • 210. 86 PRT Current Reversal A C D B Id Ia Ic Breaker B opens instantaneously. Relays at B drop out. Fault current on line CD changes direction. Relays at A remain picked up and trip by permissive signal from B. Relays at C drop out and stop keying permissive signal to C. Relays at D pick up and key permissive signal to D. I Fault, Line AB I Fault, Line CD
  • 211. 87 Directional Comparison Blocking • Overreaching relays • Delay for channel time • Channel failure can allow overtrip • Often used with “On/Off” carrier
  • 212. 88 Block Xmit Block Rcvd PCB Trip Coil PCB Trip Coil Block Xmit Block Rcvd Directional Comparison Blocking Scheme Time DelayTime Delay Forward Relay Reverse Relay Reverse Relay Forward Relay TDTD Directional Comparison
  • 213. 89 Directional Comparison Relays • Forward relays must overreach remote bus • Forward relays must not overreach remote reverse relays • Time delay (TD) set for channel delay • Scheme will trip for fault if channel lost – Scheme may overtrip for external fault on channel loss
  • 214. 90 Tone Equipment • Interface between relays and communications channel • Analog tone equipment • Digital tone equipment • Security features – Guard before trip – Alternate shifting of tones – Parity checks on digital
  • 215. 91 Tone Equipment • Newer equipment has 4 or more channels – 2 for direct transfer trip – 1 for permissive transfer trip – 1 for drive to lock out (block reclose)
  • 216. 92 Relay to Relay Communications • Available on many new digital relays • Eliminates need for separate tone gear • 8 or more unique bits of data sent from one relay to other • Programmable functions – Each transmitted bit programmed for specific relay function – Each received bit programmed for specific purpose
  • 217. 93 Telecommunications Channels • Microwave radio – Analog (no longer available) – Digital • Other radio systems • Dedicated fiber between relays – Short runs • Multiplexed fiber – Long runs • SONET Rings
  • 218. 94 Telecommunications Channels • Power line carrier current – On/Off Carrier often used with directional comparison • Hard wire – Concern with ground mat interconnections – Limited to short runs • Leased line – Rent from phone company – Considered less reliable
  • 219. 95 Automatic Reclosing (79) • First reclose ~ 80% success rate • Second reclose ~ 5% success rate • Must delay long enough for arc to deionize t = 10.5 + kV/34.5 cycles 14 cycles for 115 kV; 25 cycles for 500 kV • Must delay long enough for remote terminal to clear • 1LG Faults have a higher success rate than 3 phase faults
  • 220. 96 Automatic Reclosing (79) • Most often single shot • Delay of 30 to 60 cycles following line trip is common • Checking: – Hot bus & dead line – Hot line & dead bus – Sync check • Utilities have many different criteria for transmission line reclosing
  • 221. 97 More on Reclosing • Only reclose for one line to ground faults • Block reclose for time delay trip (pilot schemes) • Never reclose on power house lines • Block reclosing for transformer fault on transformer terminated lines • Block reclosing for bus faults • Block reclosing for BFR • Do not use them
  • 223. 99 Breaker Failure • Stuck breaker is a severe impact to system stability on transmission systems • Breaker failure relays are recommended by NERC for transmission systems operated above 100 kV • BFRs are not required to be redundant by NERC
  • 224. 100 Breaker Failure Relays 1. Fault on line 2. Normal protective relays detect fault and send trip to breaker. 3. Breaker does not trip. 4. BFR Fault detectors picked up. 5. BFR Time delay times out (8 cycles) 6. Clear house (open everything to isolate failed breaker)
  • 225. 101 Breaker Failure Relay Typical Breaker Failure Scheme with Retrip BFR Fault Detector PCB Trip Coil #1 TD Protective Relay 86 Trip Block Close TD PCB Trip Coil #2 BFR Retrip BFR Time Delay, 8~
  • 226. 102 Typical BFR Clearing Times Proper Clearing: 0 Fault occurs +1~ Relays PU, Key TT +2~ PCB trips +1~ Remote terminal clears 3-4 Cycles local clearing time 4-5 Cycles remote clearing time Failed Breaker: 0 Fault occurs +1~ BFR FD PU +8~ BFR Time Delay +1~ BFR Trips 86 LOR +2~ BU PCBs trip +1~ Remote terminal clears 12-13 Cycles local back up clearing time 13-14 Cycles remote backup clearing
  • 227. 103 Remedial Action Schemes (RAS) aka: Special Protection Schemes
  • 228. 104 Remedial Action Schemes • Balance generation and loads • Maintain system stability • Prevent major problems (blackouts) • Prevent equipment damage • Allow system to be operated at higher levels • Provide controlled islanding • Protect equipment and lines from thermal overloads • Many WECC & NERC Requirements
  • 229. 105 Remedial Action Schemes • WECC Compliant RAS – Fully redundant – Annual functional test – Changes, modifications and additions must be approved by WECC • Non WECC RAS – Does not need full redundancy – Local impacts only – Primarily to solve thermal overload problems
  • 230. 106 Underfrequency Load Shedding • Reduce load to match available generation • Undervoltage (27) supervised (V > 0.8 pu) • 14 Cycle total clearing time required • Must conform to WECC guidelines • 4 Steps starting at 59.4 Hz. • Restoration must be controlled • Must coordinate with generator 81 relays • Responsibility of control areas
  • 231. 107 Undervoltage Load Shedding • Detect 3 Phase undervoltage • Prevent voltage collapse • Sufficient time delay before tripping to ride through minor disturbances • Must Conform to WECC Guidelines • Primarily installed West of Cascades
  • 232. 108 Generator Dropping • Trip generators for loss of load • Trip generators for loss of transmission lines or paths – Prevent overloading
  • 233. 109 Reactive Switching • On loss of transmission lines – Trip shunt reactors to increase voltage – Close shunt capacitors to compensate for loss of reactive supplied by transmission lines – Close series capacitors to increase load transfers – Utilize generator var output if possible – Static Var Compensators (SVC) provide high speed adjustments
  • 234. 110 Direct Load Tripping • Provide high speed trip to shed load – May use transfer trip – May use sensitive, fast underfrequency (81) relay • Trip large industrial loads
  • 235. 111 Other RAS Schemes • Controlled islanding – Force separation at know locations • Load brake resistor insertion – Provide a resistive load to slow down acceleration of generators • Out of step tripping – Force separation on swing • Phase shifting transformers – Control load flows
  • 237. 113 Typical RAS Controller Outputs • Generator tripping • Load tripping • Controlled islanding and separation (Four Corners) • Insert series caps on AC Intertie • Shunt capacitor insertion • Shunt reactor tripping • Chief Jo Load Brake Resister insertion • Interutility signaling • AGC Off
  • 238. 114 Chief Jo Brake 1400 Megawatts @ 230 kV
  • 239. 115 RAS Enabling Criteria • Power transfer levels • Direction of power flow • System configuration • Some utilities are considering automatic enabling/disabling based on SCADA data • Phasor measurement capability in relays can be used to enable RAS actions
  • 240. 116 RAS Design Criteria • Generally fully redundant • Generally use alternate route on telecommunications • Extensive use of transfer trip for signaling between substations, power plants, control centers, and RAS controllers
  • 241. 117 UFOs vs. Power Outages
  • 242. 118 the end Jon F. Daume Bonneville Power Administration retired March 15, 2011
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  • 372. Transmission System Faults and Event Analysis Fault Analysis Theory and Modern Fault Analysis Methods Presented by: Matthew Rhodes Electrical Engineer, SRP 1
  • 373. Transmission System Fault Theory • Symmetrical Fault Analysis • Symmetrical Components • Unsymmetrical Fault Analysis using sequence networks • Lecture material originally developed by Dr. Richard Farmer, ASU Research Professor 2
  • 375. 4 Faults Shunt faults: Three phase a b c Line to line Line to ground 2 Line to ground b a c a b c a b c
  • 376. 5 Faults Series faults One open phase: a b c 2 open phases a b c Increased phase impedance Z a b c
  • 377. 6 Why Study Faults? • Determine currents and voltages in the system under fault conditions • Use information to set protective devices • Determine withstand capability that system equipment must have: – Insulating level – Fault current capability of circuit breakers: • Maximum momentary current • Interrupting current
  • 378. 7 Symmetrical Faults α t=0 2 V i(t) Fault at t = 0AC R L )sin(2)( αω += tVte
  • 379. 8 Symmetrical Faults For a short circuit at generator terminals at t=0 and generator initially open circuited: dt di LRite +=)( dt di LRitVSin +=+ )(2 αω by using Laplace transforms i(t) can be found (L is considered constant)
  • 380. 9 Symmetrical Faults ]/)()([ 2 )( TteSintSin Z V ti −−−−+= θαθαω 2222 )( XRLRZ +=+= ω R X Tan R L Tan 11 −− == ω θ Where: R X R L T ω == Time Constant ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω Where: Iac = ac RMS fault current at t=0 (Examples) Note that for a 3- phase system α will be different for each phase. Therefore, DC offset will be different for each phase
  • 381. 10 t = 0 acI2 iac Idc = 0 ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω o 90== θα V2 e(t) o 90=α
  • 382. 11 ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω 0=α o 90=θ V2 e(t) t = 0 iac02 acI 02 acI idc
  • 383. 12 iac02 acI 02 acI idc 022 acI t 0=α o 90=θ ]/)()([2)( TteSintSin ac Iti −−−−+= θαθαω )(ti
  • 384. 13 Symmetrical Faults Iac and Idc are independent after t = 0 22 dc I ac I RMS I += Tt e aco I dc I − = 2 Substituting: Tt e ac IT t e ac I ac I RMS I 2 21)222((max) − += − += ]/)2/([ 2 )( TtetSin Z V ti −+−+= πω
  • 385. 14 Asymmetry Factor IRMS(max) = K(τ) Iac Asymmetry Factor = K(τ) rx eK τπ τ 4 21)( − += Where: τ = number of cycles (Example 7.1) fRXT π2/=
  • 386. 15 Example 7.1 •Fault at a time to produce maximum DC offset •Circuit Breaker opens 3 cycles after fault inception I Fault at t = 0AC R = 0.8 Ώ XL = 8 Ώ V = 20 kVLN - + CB Find: 1. Iac at t = 0 2. IRMS Momentary at = 0.5 cycles 3. IRMS Interrupting Current τ
  • 387. 16 Example 7.1 a. RMSAC kAI 488.2 88.0 20 )0( 22 = + = b. 438.121)5.0( ) 10 5.(4 =+= Π− eK KAImomentary 577.3)488.2)(438.1( == c. 023.121)3( ) 10 3(4 =+= Π− eK KAI ngInterrupti 545.2)488.2)(023.1( ==
  • 388. 17 AC Decrement In the previous analysis we treated the generator as a constant voltage behind a constant impedance for each phase. The constant inductance is valid for steady state conditions but for transient conditions, the generator inductance is not constant. The equivalent machine reactance is made up of 2 parts: a) Armature leakage reactance b) Armature reaction (See Phasor Diagram)
  • 389. 18 AC Decrement Steady state model of generator XL is leakage reactance XAR is a fictitious reactance and XAR>> XL XAR is due to flux linkages of armature current with the field circuit. Flux linkages can not change instantaneously. Therefore, if the generator is initially unloaded when a fault occurs the effective reactance is XL which is referred to as Subtransient Reactance, x”. EI R XL XAR Load
  • 391. 20 E”Field Flux Armature Reaction = 0 Resultant Field ET0 t = 0 - XL XAR=0 ET0 - + E” = E’ = E = ET0 I=0 Unloaded Generator
  • 392. 21 XL XAR - + E” = E’ = E = ET0 I=0 t=0 E”Field Flux Armature Reaction = 0 Resultant Field ET0 = 0 Faulted Generator
  • 393. 22 XL XAR=0 - + E” = E’ = E = ET0 I = I” E” = jI”XL t=0+ Field Flux Resultant Field ET = 0 I” Armature Reaction = 0
  • 394. 23 XL XAR’ - + E” = E’ = E = ET0 I = I’ E’ = jI’(XL + XAR’) t ≈ 3Cyc. Field Flux Resultant Field ET = 0 I’ Armature Reaction = 0
  • 395. 24 XL XAR - + E” = E’ = E = ET0 I = I E’ = jI(XL + XAR) t =∞ Field Flux Resultant Field ET = 0 I’ Armature Reaction = 0
  • 396. 25 AC Decrement As fault current begins to flow, armature reaction will increase with time thereby increasing the apparent reactance. Therefore, the ac component of the fault current will decrease with time to a steady state condition as shown in the figure below. "2I '2I I2 "2I
  • 397. 26 AC Decrement For a round rotor machine we only need to consider the direct axis reactance. dX E I " "2 "2 = Subtransient dX E I ' '2 '2 = d X E I 2 2 = Transient Synchronous (steadystate)
  • 398. 27 AC Decrement Can write the ac decrement equation [ ] ([ )])'()'"(2)( '" θαω −++−+−= − − tSinIeIIeIIt ac i dT t dTt For an unloaded generator (special case):TEEEE === '" T”d: Subtransient time constant (function of amortisseur winding X/R) T’d: Transient time constant (function of field winding X/R) Look at equation for t=0 and t=infinity
  • 399. 28 AC Decrement For t = 0 [ ] ([ )])'()'"(2)( '" θαω −++−+−= − − tSinIeIIeIIt ac i dT t dTt For t = ∞ IIiac 2]00[2(max) =++= "2])'()'"[(2(max) IIIIIIiac =+−+−=
  • 400. 29 ac and dc Decrement Transform ac decrement equation to phasor form ] θα −+ − −+ − −= ⎢ ⎢ ⎣ ⎡ /')'(")'"( _ IdT t eIIdT t eIIacI dc decrement equation: A T t eSinIdcI − −= )("2 θα Where TA = Armature circuit time constant (Example 7.2)
  • 401. 30 Example 7.2 I Fault at t = 0AC R = 0 V = 1.05 pu - + CB x”d =.15pu T”d = .035 Sec. x’d = .24pu T’d = 2.0 Sec. xd = 1.1pu TA = 0.2 Sec. No load when 3-phase fault occurs Breaker clears fault in 3 cycles. Find: a) I”, b) IDC(t) c) IRMS at interruption d) Imomentry (max) S 500 MVA, 20kV, 60 Hz Synchronous Generator
  • 402. 31 Example 7.2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 2035. t t AC eetI 2. max "2)( t DC eItI − = KAIBase 434.14 320 500 == kApu dx E I 1010.7 15. 05.1 " " " ==== a DCI 2.2. max 9.9)7(2)( tt DC eetI −− == b
  • 403. 32 Example 7.2 Part c: Find IRMS at interruption (3 cycles) .sec05.0 60 3 ==t ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 205.035. 05. eetIAC ( )[ ] puIAC 92.4909.)975)(.258.3()24(.5.205.1)05(. =++= pueIDC 71.79.9)05(. 2. 05. == − kApuIRMS 132146.971.792.4)05(. 22 ==+= c
  • 404. 33 Example 7.2 Part d: Find IMomentary(max) at t = ½ cycle sec0083. 60 5. ==t ( )[ ] puIAC 43.6909.)996)(.258.3()79(.5.205.1)0083(. =++= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= −− 1.1 1 1.1 1 24. 1 24. 1 15. 1 05.1)( 20083.035. 0083. eetIAC pueIDC 5.99.9 2. 0083. == − kApuIRMS 2159.145.943.6 22 ==+= d
  • 407. 36 Superposition for Fault Analysis New representation: IF1 IF2=0 Bus 1 Bus 1 Bus 2 IG = IG! + IG2 = IG1+ IL IM = IM1 – IL IF = IG1 + IM1 Example 7.3 IG1 IG2 ILIM1 IG IF IM
  • 408. 37 Example 7.3 For the system of Slide 35 and 36 the generator is operating at 100 MVA, .95 PF Lagging 5% over rated voltage Part a: Find Subtransient fault current magnitude. From Slide 36 puj j j Z V I TH F F 08.9 116. 05.1 655. )505)(.15(. 05.1 "1 −==== Part b: Neglecting load current, find Generator and motor fault current. pujjIG 7 655. 505. 08.9"1 −=−= pujjjIM 08.2)7(08.9"1 −=−−−=
  • 409. 38 Example 7.3 Part c: Including load current, find Generator and motor current during the fault period. 22* * 18/952. 05.1 18/1 0/05.1 95.cos/1 MG o o oLoad II V S I −==−= − = − == pujI oo G 83/35.718/953.7" −=−+−= pujI oo M 243/00.218/952.08.2" =−−−= c c
  • 410. 39 Z Bus Method For Z bus method of fault studies the following approximations are made: • Neglect load current • Model series impedance only • Model generators and synchronous motors by voltage behind a reactance for the positive sequence system
  • 411. 40 AC AC AC + Eg” - + E m - J 0 . 2 J 0 . 305 J 0 .15 1 2 -VFIF
  • 412. 41 Z Bus Method For the circuit of Figure 7.4d (Slide 36 & 40) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 22 12 21 11 2 1 E E Y Y Y Y I I Injected node currents [matrix Y-bus] nodal admittance Node voltages Premultiplying both sides by the inverse of [Y-bus} Pre-fault node Voltage [Z-Bus] =[Y-Bus]-1 Injected node Current -IF1 0 For a fault at Bus 1 )( 1111 FIZE −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −= − = 1111 1 1 Z V Z E I F F ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2221 1211 2 1 I I ZZ ZZ E E
  • 413. 42 Z-Bus Method ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −= − = 1111 1 1 Z V Z E I F F )( 1111 FIZE −= 11 1 Z V I F F = where: For a fault at Bus 1 IF1 = Fault current at bus 1 VF = Prefault voltage of the faulted bus (Bus 1)
  • 414. 43 Z-Bus Method For N bus system, fault on Bus n ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 . 0 0 0 . ..,... . . . . . 321 321 33333231 22232221 11131211 3 2 1 Fn NNNnNNN nNnnnnn Nn Nn Nn N N I ZZZZZ ZZZZZ ZZZZZ ZZZZZ ZZZZZ E E E E E -VF nn F Fn Z V I = Where: VF = Pre-fault voltage at faulted bus Znn = Thevinen impedance
  • 415. 44 Z-Bus Method After IFn is found the voltage at any bus can be found from: E1 = Z1n (-Ifn) E2 = Z2n(-Ifn) Etc. If voltage at each bus is found, current through any branch can be found: I12 = (E1 - E2) / Ž12 Etc/ Note: Ž12 is series impedance between Bus1 and Bus 2, not from Z-Bus. (Example 7.4)
  • 416. 45 Example 7.4 For the system of Figure 7.3 (Slide 40) using the Z-bus method find: a) Z bus b) IF and I contribution from Line for Bus 1 fault c) IF and I contribution from Line for Bus 2 fault Y20 = -j5Y10 = -j6.67 Y12 = -j3.28 1 2 IF
  • 417. 46 Example 7.4 [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = 95.928.3 28.395.9 jj jj YBus [ ] [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ == − 139.046. 046.1156.1 jj jj YZ busBus ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E 0 -IF 11 )1156.( IjE = -VF -IF 08.9 1156. " j j V I F F −== a b
  • 418. 47 Example 7.4 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E For fault at Bus 1: E1 = E1 1+ E1 2 = 0 E2 = E2 1 + E2 2 = VF + (j.046)I1 E2 = 1.05 + (j.046)(j9.08) = .632 /0o 07.2 305. 0632. 21 12 21 j jZ EE I −= − = − = Find: Line current b
  • 419. 48 Example 7.4 Y20 = -j5Y10 = -j6.67 Y12 = -j3.28 1 2 IF Find IF and I contribution from Line for Bus 2 fault ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 1 2 1 139.046. 046.1156. I I jj jj E E -VF puj j IF 55.7 139. 05.1 2 −== - I F2 o FF jjIjVE 0/703.)55.7)(046.(05.1))(046.(1 =+=−+= puj jZ EE I 3.2 305. 0703. 12 21 12 −= − = − = c
  • 420. 49 Z-Bus Method [Z-Bus] = [Y-Bus]-1 Will not cover formation of [Z-Bus] or [Y-Bus] [Z-Bus] can be considered a fictitious circuit which has the appearance of a rake. See Figure 7.6 on Page 371.
  • 421. 50 nn F nF Z V II == Example: Fault at Bus n ))(( 11 nnF IZVE −= Etc. Z-Bus Rake equivalent
  • 422. 51 Class Problem 1 pujZbus ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 08.06.04. 06.12.08. 04.08.12. For the given Bus Impedance matrix(where subtransient reactances were used) and a pre-fault voltage of 1 p.u.: a. Draw the rake equivalent circuit b. A three-phase short circuit occurs at bus 2. Determine the subtransient fault current and the voltages at buses 1, 2, and 3 during the fault.
  • 424. 53 Symmetrical Components Symmetrical Components is often referred to as the language of the Relay Engineer but it is important for all engineers that are involved in power. The terminology is used extensively in the power engineering field and it is important to understand the basic concepts and terminology.
  • 425. 54 Symmetrical Components • Used to be more important as a calculating technique before the advanced computer age. • Is still useful and important to make sanity checks and back-of-an-envelope calculation. • We will be studying 3-phase systems in general. Previously you have only considered balanced voltage sources, balanced impedance and balanced currents.
  • 426. 55 Symmetrical Components n a a b b c Va Vb Vc Va Vb Vc Balanced load supplied by balanced voltages results in balanced currents This is a positive sequence system, In Symmetrical Components we will be studying unbalanced systems with one or more dissymmetry. ZY ZY ZY Ib Ia Ic
  • 427. 56 Symmetrical Components For the General Case of 3 unbalanced voltages VA VB VC 6 degrees of freedom Can define 3 sets of voltages designated as positive sequence, negative sequence and zero sequence
  • 428. 57 Symmetrical Components Common a operator identities a =1/120o a2 = 1/240o a3 = 1/0o a4 = 1/120o 1+a+a2 = 0 (a)(a2) = 1
  • 429. 58 Symmetrical Components Positive Sequence 120o 120o120o VA1 VB1 VC1 2 degrees of freedom VA1 = VA1 VB1 = a2 VA1 VC1 = a VA1 a is operator 1/120o
  • 430. 59 Symmetrical Components Negative Sequence 120o 120o120o VA2 VC2 VB2 2 degrees of freedom a is operator 1/120o VA2 = VA2 VB2 = aVA2 VC2 = a2 VA2
  • 431. 60 Symmetrical Components Zero Sequence 2 degrees of freedom VA0 VB0 VC0 VA0 = VB0 = VC0
  • 432. 61 Symmetrical Components Reforming the phase voltages in terms of the symmetrical component voltages: VA = VA0 + VA1 + VA2 VB = VB0 + VB1 + VB2 VC = VC0 + VC1 + VC2 What have we gained? We started with 3 phase voltages and now have 9 sequence voltages. The answer is that the 9 sequence voltages are not independent and can be defined in terms of other voltages.
  • 433. 62 Symmetrical Components Rewriting the sequence voltages in term of the Phase A sequence voltages: VA = VA0 + VA1 +VA2 VB = VA0 + a2 VA1 + aVA2 VC = VA0 + aVA1 +a2 VA2 VA = V0 + V1 +V2 VB = V0 + a2 V1 + aV2 VC = V0 + aV1 +a2 V2 Drop A Suggests matrix notation: VA 1 1 1 V0 VB 1 a2 a V1 VC 1 a a2 V2 = [VP] = [A] [VS]
  • 434. 63 Symmetrical Components We shall consistently apply: [VP] = Phase Voltages [VS] = Sequence Voltages 1 1 1 [A] = 1 a2 a 1 a a2 [VP] = [A][VS] Pre-multiplying by [A]-1 [A]-1[VP] = [A]-1[A][VS]= [I][VS] [VS] = [A]-1 [VP]
  • 435. 64 Operator a a = 1 /120o = - .5 + j .866 a2 = 1 / 240o = - .5 - j.866 a3 = 1 / 360o = 1 a4 = 1 / 480o = 1 / 120o = a a5 = a2 etc. 1 + a + a2 = 0 a - a2 = j 3 1 - a2 = /30o 1/a = a2 3 Relationships of a can greatly expedite calculations ( Find [A]-1)
  • 436. 65 Inverse of A [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA Step 1: Transpose [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA T Step 2: Replace each element by its minor ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3
  • 437. 66 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3 Inverse of A Step 3: Replace each element by its cofactor ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3
  • 438. 67 Inverse of A ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− 11 11 22 22 222 aaaa aaaa aaaaaa1 1 2 3 2 3 Step 4: Divide by Determinant [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 2 1 1 111 aa aaA )(3)(1)(1)(1 2222 aaaaaaaaD −=−+−+−= a a a a aa a a a aa a = − − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − 1 111 2 2 2 2 2 2 2 2 1 1 111 a aa a aaa a == − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − −
  • 439. 68 Inverse of A [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = − aa aaA 2 21 1 1 111 3 1
  • 440. 69 Symmetrical Components Previous relationships were developed for voltages. Same could be developed for currents such that: IA IB IC [IP] = I0 I1 I2 [IS] = [IP] = [A] [IS] [IS] = [A]-1 [IP] 1 1 1 [A] = 1 a2 a 1 a a2 1 1 1 [A]-1 = 1/3 1 a a2 1 a2 a
  • 441. 70 Significance of I0 IA IB IC I0 I1 I2 1 1 1 = 1/3 1 a a2 1 a2 a I0 = 1/3 ( IA + IB + IC) n IA IB IC In In = IA + IB + IC = 3 I0 For a balanced system I0 = 0 For a delta system I0 = 0 (Examples 8.1, 8.2 and 8.3)
  • 442. 71 Example 8.1 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= a aV o o o P 277 277 277 120/277 120/277 0/277 2 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = − 0 0/277 01 1 1 111 3 277 2 2 21 2 1 0 o PS a a aa aaVA V V V V 0 1 2 Find [VS] (Sequence voltages) a b c
  • 443. 72 Example 8.2 Y connected load with reverse sequence [ ] ( ) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = 2 1 10 120/10 120/10 0/10 a aI o o o P a b c Find IS (Sequence Currents) [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − o PS a a aa aaIAI 0/10 0 01 1 1 111 3 10 22 21 0 1 2
  • 444. 73 Example 8.3 Ia = 10 / 0o Ic = 10 /120o Ib = o In [ ] [ ] [ ]PS IA I I I I 1 2 1 0 − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = aaa aaIS 0 1 1 1 111 3 10 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + = o o o S a a a a I 60/33.3 0/67.6 60/33.3 2 3 10 1 2 1 3 10 2 2 0 1 2 o n II 60/103 0 == a b c
  • 445. 74 Sequence Impedance for Shunt Elements Sequence Networks of balanced Y elements( Loads, Reactors, capacitor banks, etc.) VA = IAZy + (IA + IB +IC) Zn = (ZY + Zn)IA + ZnIB + ZnIC VB = ZnIA + (ZY + Zn)IB + ZnIC VC = ZnIA + ZnIB +(ZY + Zn)IC n IB IC . IA VB VA VC ZY ZY ZY Zn
  • 446. 75 Sequence Impedance for Shunt Elements ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ C B A nYnn nnYn nnnY C B A I I I ZZZZ ZZZZ ZZZZ V V V [VP] = [ZP] [IP] (1) Transform to sequence reference frame. We know: [VP] = [A] [VS] and [IP] = [A] [IS], Substitute in(1) [A][VS] = [ZP][A][IS] premultiply both sides by [A]-1 [VS] = [A]-1[ZP][A][IS] = [ZS][IS] where: [ZS] = [A]-1[ZP][A]
  • 447. 76 Sequence Impedance for Shunt Elements [ZS] = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + + ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 2 2 2 222120 121110 020100 1 1 111 1 1 111 3 1 aa aa ZZZZ ZZZZ ZZZZ aa aa ZZZ ZZZ ZZZ nYnn nnYn nnnY [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + = Y Y nY S Z Z ZZ Z 0 00 003 0 1 2 0 1 2
  • 448. 77 Sequence Impedance for Shunt Elements ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 1 0 2 1 0 00 00 003 I I I Z Z ZZ V V V Y Y nY V0 = Z00 I0 where: Z00 = ZY +3 Zn V1 = Z11 I1 V2 = Z22 I2 where Z11 = Z22 = ZY Systems are uncoupled: Zero sequence currents only produce zero sequence voltages. Positive sequence currents only produce positive sequence voltages, etc.
  • 449. 78 Sequence Impedance for Shunt Elements We can form sequence circuits which represent the equations: ZY 3 Zn ZY ZY V0 V1 V2 I0 I1 I2 Zero sequence circuit Zn only in zero Sequence No neutral: Zn = infinity Solid ground: Zn = 0 Positive sequence circuit Negative sequence circuit
  • 450. 79 Sequence Impedance for Shunt Elements Delta connected shunt element ZY V0 V1 V2 I0 I1 I2 open ZΔ/3 ZΔ/3 Sequence circuits .A B C IA IB IC ZΔ ZΔ ZΔ
  • 451. 80 Sequence Impedance for Shunt Elements For the general case: [ZS] = [A]-1[ZP][A] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 2 2 2 2 222120 121110 020100 1 1 111 1 1 111 3 1 aa aa ZZZ ZZZ ZZZ aa aa ZZZ ZZZ ZZZ CCCBCA BCBBBA ACABAA If there is symmetry: ZAA = ZBB = ZCC and ZAB = ZBC = ZCA we could perform multiplication and get: [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ABAA ABAA ABAA S ZZ ZZ ZZ Z 00 00 002 We see that: Z11 = Z22 and Z00 > Z11
  • 453. 82 Series Element Sequence Impedance Matrices in compact form [VP]-[VP’] = [ZP] [IP] We can transform to the symmetrical component reference frame: [VS] - [VS’] = [ZS] [IS] where: [ZS] = [A]-1[ZP][A] If ZAA = ZBB = ZCC and ZAB = ZBC = ZCA , [ZS] will be the diagonal matrix: [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 2 1 0 Z Z Z ZS
  • 454. 83 Series Element Sequence Impedance The sequence circuits for series elements are: Z0 V0 V0’ I0 o n0 Z1 V1 V1’ I1 o n1 Z2 V2 V2’ I2 o n2
  • 455. 84 Series Element Sequence Impedance We have quickly covered the calculation of Positive and Negative sequence parameters for 3-phase lines. To determine the zero sequence impedance we need to take the effect of the earth into account. This is done by using Carson’s Method which treats the earth as an equivalent conductor.
  • 456. 85 Rotating Machine Sequence Networks A B C ZK ZK ZK - - - + + + EB EAEC IC IA IB Z n ZAB ZBC ZCA ZCB ZBA ZAC eA = Em Cos ωt eB = Em Cos(ωt – 120o) eC = Em Cos(ωt + 120o) In phasor form: EA= ERMS / 0 = E EB = ERMS /-120o = a2 E EC = ERMS /120o = a E
  • 457. 86 Rotating Machine Sequence Networks [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = aE Ea E EPg 2 EA= ERMS / 0 = E EB = ERMS /-120o = a2 E EC = ERMS /120o = a E or [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − 0 0 1 1 111 3 1 2 2 21 E aE Ea E aa aaEAE PgSg Therefore, only the positive sequence system has a generator voltage source. 0 1 2 a b c
  • 458. 87 Rotating Machine Sequence Networks A B C ZK ZK ZK - - - + + + EB EAEC IC IA IB Z n ZAB ZBC ZCA ZCB ZBA ZAC Machine is not passive: Mutual Reactances: ZAB ≠ ZBA , etc. ZAB = ZBC = ZCA = ZR ZBA = ZCB = ZAC = ZQ
  • 459. 88 Rotating Machine Sequence Networks ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ +++ +++ +++ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ C B A NKNQNR NRNKNQ NQNRNK C B A I I I ZZZZZZ ZZZZZZ ZZZZZZ E E E [ ] [ ][ ]PPGPG IZE = [ ] [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − 2 1 0 1 00 00 00 G G G PGSG Z Z Z AZAZ From the machine diagram we can write: Where: ZG0 = ZK + ZR + ZQ ZG1 = ZK + a2 ZR + a ZQ ZG2 = ZK + a ZR + a2 ZQ uncoupled 0 1 2 0 1 2
  • 460. 89 Rotating Machine Sequence Networks Generator sequence circuits are uncoupled 3Zn ZG0 I0 V0 EG1 - + ZG1 I1 V1 ZG2 I2 V2 Generator Terminal Voltages
  • 461. 90 Rotating Machine Sequence Networks Sequence impedances are unequal ZG1 varies depending on the application a) Steady state, power flow studies: ZG1 = ZS(synchronous) b) Stability studies ZG1 = Z’ (transient) c) Short circuit and transient studies: ZG1 = Z” (subtransient) Motor circuits are similar but there is no voltage source for an induction motor. (Example 8.6)
  • 462. 91 Example 8.6 - [ EP ] + [ IP ] Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ Unbalanced Source [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o o PE 115/295 120/260 0/277 a b c Find phase Currents [ IP ] Ω+=Ω== Δ 43.666.740/10 3 j Z Z o Y Ω+=Ω= 996.087.85/1 jZ o L Ω=+=+=== o LY jZZZZZ 7.43/72.10426.7747.7210
  • 463. 92 Example 8.6 - [ EP ] + [ IP ] Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o o PE 115/295 120/260 0/277 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − o o o o o o PS aa aaEAE 6.216/22.9 77.1/1.277 1.62/91.15 115/295 120/260 0/277 1 1 111 3 1 2 21 0 1 2
  • 464. 93 Example 8.6 10.72 /43.7o Ώ - + 15.91 /62.1o I 0 10.72 /43.7o Ώ - + 277.1 /- 1.77o I1 10.72 /43.7o Ώ - + 9.22 /216.6o I2 00 =I o o I 7.43/72.10 77.1/277 1 − = AI o 5.45/84.251 −= o o I 7.43/72.10 6.216/22.9 2 = AI o 9.172/86.02 =
  • 465. 94 Example 8.6 [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − == o o o SP IAI 8.73/64.26 4.196/72.25 7.46/17.25 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −= o o SI 9.172/86.0 5.45/84.25 0 0 1 2 a b c Amps Amps How would you do problem without Symmetrical Components?
  • 466. 95 Transformer Connections for Zero Sequence P Q Ic Ia Ib I C IA IB P Q Ia + Ib + Ic is not necessarily 0 if we only look at P circuit but Ia = nIA Ib = nIB and Ic = nIC Therefore since IA + IB + IC = 0 , Ia + Ib + Ic = 0 and I0 = 0 P0 Q0 Z0 n0 No zero sequence current flow through transformer
  • 467. 96 Transformer Connections for Zero Sequence P Q Ic Ia Ib IC IA IB P Q Ia + Ib + Ic is not necessarily 0 and IA + IB + IC is not necessarily. P0 Q0Z0 n0 I0 can flow through the transformer. Therefore I0 is not necessarily 0, I0
  • 468. 97 Transformer Connections for Zero Sequence P Q Ic Ia Ib IC IA IB P Q Ia + Ib + Ic is not necessarily 0 and Ia/n + Ib/n + Ic/n is not necessarily 0 P0 Q0 Z0 n0 Provides a zero sequence current source Ib/n Ic/n Ic/n I0 but IA + IB + IC = 0
  • 469. 98 Transformer Connections for Zero Sequence P Q Ic Ia Ib I C IA IB P Q Ia + Ib + Ic = 0 Ia/n + Ib/n + Ic/n is not necessarily 0, but IA + IB + IC = 0 P0 Q0 Z0 n0 No zero sequence current flow Ib/n Ic/n Ic/ n
  • 470. 99 Transformer Connections for Zero Sequence P Q IC IA IB P Q Ia + Ib + Ic = 0 IA + IB + IC = 0 P0 Q0 Z0 n0 No zero sequence current flow ∆ ∆ Ia Ib Ic
  • 471. 100 Power In Sequence NetworksFor a single phase circuit we know that: S = EI* = P + jQ In a 3-phase system we can add the power in each phase such that: SP = EAIA* + EBIB* + ECIC* Written in matrix form [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = * * * C B A CBAP I I I EEES
  • 472. 101 Power in Sequence Networks If we want the apparent power in the symmetrical component reference frame, we can substitute the following: [EP] = [A][ES] [IP] = [A][IS] [EP]T =[ES]T [A]T [IP]* = [A]*[IS]* Into (1) resulting in [SP] = [ES]T [A]T[A]*[IS]* which results in: [SP] = 3[ES]T [IS]* = 3[SS] Where: [SS] = E0I0 * + E1I1 * + E2I2 * From our previous definitions: [SP] = [EP]T [IP]* (1)
  • 473. 102 Class Problem 2 One line of a three-phase generator is open circuited, while the other two are short- circuited to ground. The line currents are: Ia=0, Ib= 1500/90 and Ic=1500/-30 a. Find the symmetrical components of these currents b. Find the ground current
  • 474. 103 Class Problem 3 The currents in a delta load are: Iab=10/0, Ibc= 20/-90 and Ica=15/90 Calculate: a. The sequence components of the delta load currents b. The line currents Ia, Ib and Ic which feed the delta load c. The sequence components of the line currents
  • 475. 104 Class Problem 4 The source voltages given below are applied to the balanced-Y connected load of 6+j8 ohms per phase: Vag=280/0, Vbg= 290/-130 and Vcg=260/110 The load neutral is solidly grounded. a. Draw the sequence networks b. Calculate I0, I1 and I2, the sequence components of the line currents. c. Calculate the line currents Ia, Ib and Ic
  • 477. 106 Phase and Symmetrical Component Relationship Phase Reference Frame IA IB IC n V C VB V A Symmetrical Components Reference Frame I0 I1 I2 V0 V1 V2 n0 n1 n2
  • 478. 107 Unsymmetrical Fault Analysis For the study of unsymmetrical faults some, or all, of the following assumptions are made: • Power system balanced prior to fault • Load current neglected • Transformers represented by leakage reactance • Transmission lines represented by series reactance
  • 479. 108 Assumptions Continued • Synchronous machines represented by constant voltage behind reactance(x0, x1. x2) • Non-rotating loads neglected • Small machines neglected • Effect of Δ – Y transformers may be included
  • 480. 109 Faulted 3-Phase Systems Sequence networks are uncoupled for normal system conditions and for the total system we can represent 3 uncoupled systems: positive, negative and zero. When a dissymmetry is applied to the system in the form of a fault, we can connect the sequence networks together to yield the correct sequence currents and voltages in each sequence network. From the sequence currents and voltages we can find the corresponding phase currents and voltages by transformation with the [A] matrix
  • 481. 110 Faulted 3-Phase Systems To represent the dissymmetry we only need to identify 2 points in the system: fault point and neutral point: Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 The sequence networks are connected together from knowledge of the type of fault and fault impedance Example 9.1
  • 482. 111 . AC Bus 1 AC Bus 2 X1=X2 =20Ώ . . ∆ ∆ 100MVA 13.8kV X”=0.15pu X2 = 0.17pu X0 =0.05pu 100MVA 13.8:138kV X = 0.1pu 100MVA 138:13.8kV X = 0.1pu 100MVA 13.8kV X”=0.20pu X2 = 0.21pu X0 =0.05pu Xn = 0.05pu Example 9.1 G M Prefault Voltage = 1.05 pu Draw the positive, negative and zero sequence diagrams for the system on 100MVA, 13.8 kV base in the zone of the generator Line Model: X0 = 60Ώ ( ) Ω= 4.190 100 138 2 BZ puj j ZZ 105.0 4.190 20 21 === puj j Z 315.0 4.190 60 0 ==
  • 483. 112 AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 AC AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 AC AC . AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.15 n0 Example 9.1
  • 484. 113 Example 9.1 Reduce the sequence networks to their thevenin equivalents as viewed from Bus 2 AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.15 n 0 Zero Sequence Thevenin Equivalent from Bus 2 f0 n0 J0.25
  • 485. 114 Example 9.1 AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 Positive Sequence Thevenin Equivalent from Bus 2 139. 655. )2)(.455(. j j Zthev == f1 n1 J0.139 + - 1.05 / 0 o
  • 486. 115 Example 9.1 Negative Sequence Thevenin Equivalent from Bus 2 146. 685. )21)(.475(. j j Zthev == f2 n2 J0.146 AC AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 AC AC .
  • 487. 116 Single Line-to-Ground Fault [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 0 0 FA FP I I [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − FA FA FAFA FPFS I I II aa aaIAI 3 1 0 0 1 1 111 3 1 2 21 IF0 = IF1 = IF2 EFA = IFA ZF EF0 + EF1 + EF2 = (IF0 + IF1 + IF2) ZF EF0 + EF1 + EF2 = 3IF0 ZF A B C IF A EF A IF B IFC n Z F
  • 488. 117 Single Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF 0 EF1 EF2 3ZF
  • 489. 118 Single Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 3 ZF
  • 490. 119 Example 9.3 For the system of Example 9.1 there is a bolted Single- Line-to-Ground fault at Bus 2. Find the fault currents in each phase and the phase voltages at the fault point. f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 IF2 IF1 96.1 146.139.25. 0/05.1 210 j jjj III o FFF −= ++ ===
  • 491. 120 Example 9.3 f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 = IF1 = IF2 = -j1.96 EF0 EF2 EF1 pujjVF 491.)25.)(96.1(0 −=−−= pujVF 777.)139.)(96.1(05.11 =−−= pujjVF 286.)146.)(96.1(2 −=−−=
  • 492. 121 Example 9.3 [ ] [ ][ ]FSFP IAI = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 88.5 96.1 96.1 96.1 1 1 111 2 2 puj j j j aa aa I I I FC FB FA [ ] [ ][ ]FSFP EAE = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ pu pu aa aa E E E o o FC FB FA 7.128/179.1 231/179.1 0 286. 777. 491. 1 1 111 2 2 Note: Unfaulted phase voltages are higher than the source voltage. a b c a b c
  • 493. 122 . Example 9.3a Find fault current in the transmission line, I L 1) Find ILS 2) Find ILP ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 88.5 puj I I I FC FB FA ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ 96.1 96.1 96.1 2 1 0 j j j I I I F F F . . AC Bus 1 AC Bus 2 ∆ ∆ G M SLG Fault IF I L
  • 494. 123 Zero Sequence AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2(f0) j.15 n 0 -j1.96 I L0 = 0 I L0 =0
  • 495. 124 Positive Sequence AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2. - + 1 2 1.05 / 0o 1.05 / 0o n1 e j30 : 1 SLG e j30 : 1 n1 -j1.96 I T1I L1 6. 655. 2. )96.1(1 jjIT −=−= o LI 60/6.01 −= n1 j.455 j .22 I T1 -j1.96 f1 f1
  • 496. 125 Negative Sequence e -j30 : 1 n2 -j1.96 I T2I L2 n2 j.475 j .22 I T2 -j1.96 6. 685. 21. )96.1(2 jjIT −=−= o LI 120/6.02 −= AC AC . AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 e -j30 : 1 SLG f2 f2
  • 497. 126 Example 9.3a [ ] [ ][ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡− = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == puj puj aa aaIAI o o PSPL 039.1 0 039.1 120/6. 60/6. 0 1 1 111 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − −= o o LSI 120/6. 60/6. 0 0 1 2 a b c . . AC Bus 1 AC Bus 2 ∆ ∆ G M SLG Fault IF I L
  • 498. 127 Line to Line Fault [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = FB FBFP I II 0 [ ] [ ] [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ == − FB FB FB FBFPFS Ij Ij I I aa aaIAI 3 3 0 3 1 0 1 1 111 3 1 2 21 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = FFBFB FB FA FP ZIE E E E n A B C IFA EF A IFB IFC EF B EF C ZF IF0 = 0 IF1 = IF2 ( ) FFFFBFFBFFBFF ZIZIjZIjZI aa EE 1 2 21 33 3 ==−−= − −=− FBF IjI 31 = so 3 1 j I I F FB = EF1 = EF2 + IF1ZF 0 1 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −+ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFBFBFA FFBFBFA FFBFBFA FFBFB FB FA FS ZaIEE ZIaEE ZIEE ZIE E E aa aaE 2 2 2 2 3 1 1 1 111 3 1
  • 499. 128 Line to Line Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF 2 ZF
  • 500. 129 Example 9.4 For the system of Example 9.1 there is a bolted Line-to-Line fault at Bus 2. Find the fault currents in each phase and the phase voltages at the fault point. f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF1 IF1 IF0 puj jj II o FF 69.3 146.139. 0/05.1 21 −= + =−=00 =FI ( ) ( )( ) pujjjIEE FFF 537.0146.69.3146.221 =−=−== EF1=EF2EF0 00 =FE
  • 502. 131 2 Line to Ground Fault [ ] ( ) ( ) ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + += ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFCFB FFCFB FA FC FB FA FP ZII ZII E E E E E A B C IFA EF A IFB IFC n EF B EF C ZF IFA = 0 = IF0 + IF1 + IF2 Since IFA = 0, IFB + IFC = 3IF0 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = FFFA FFFA FFFA FF FF FA FS ZIE ZIE ZIE ZI ZI E aa aaE 0 0 0 0 0 2 2 3/ 3/ 23/ 3 3 1 1 111 3 1 EF0 – EF1 = 3 IF0 ZF so EF0 = EF1 + 3IF0 ZF and EF1 = EF2 0 1 2
  • 503. 132 2 Line to Ground Fault Zero System Positive System Negative System f0 f1 f2 n0 n1 n2 IF0 IF1 IF2 EF0 EF1 EF2 3ZF
  • 504. 133 For the system of Example 9.1 there is a 2-line-to- ground bolted fault at Bus 2. a) Find the fault currents in each phase b) Find the neutral current c) Fault current contribution from motor and generator Neglect delta-wye transformers Example 9.5 . . AC Bus 1 AC Bus 2 ∆ ∆ G M 2LG Fault IF I L
  • 505. 134 Example 9.5 f0 n0 J0.25 f1 n1 J0.139 + - 1.05 / 0 o f2 n2 J0.146 IF0 IF2 IF1 puj jj IF 547.4 25.146. )25)(.146(. 139. 05.1 1 −= + + = pujII FF 674.1 25.146. 146. )( 10 = + −= pujjjIII FFF 873.2)547.4(674.1102 =−−−=−−=
  • 506. 135 Example 9.5 This image cannot currently be displayed. [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu j j j aa aaI o o FP 3.21/9.6 7.158/9.6 0 873.2 547.4 674.1 1 1 111 2 2 a b c pujjII FFn 02.5)674.1)(3(3 0 ===
  • 507. 136 Example 9.5 n1 j.455 j .22 I T1 -j4.547 n2 j.475 j .22 I T2 J2.87 3 00 =GFI pujjIII GFFMFO 674.10674.100 =−=−= 39.1 655. 2. )547.4(! jjIGF −=−= pujjjIII GFFMF 16.3)39.1(547.4111 −=−−−=−= 88. 685. 21. )8773.2(2 jjIGF == pujjjIII GFFMF 993.188.873.2222 =−=−= f1 f2
  • 508. 137 Example 9.5 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j aa aaI o o GFP 4.7/98.1 6.172/98.1 51. 88. 39.1 0 1 1 111 2 2 [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j j aa aaI o o MFP 9.26/0.5 1.153/0.5 504. 99.1 16.3 674.1 1 1 111 2 2
  • 509. 138 Example 9.5 results AC AC . AC j.05 J0.1 J0.315 J0.1 J0.1. 1 2 j.1 5 n 0 I L0 = 0 2LG J1.674 X AC AC . AC AC j.15 - + J0.1 J0.105 J0.1 J0.2 . - + 1 2 1.05 / 0o 1.05 / 0o n1 e -j30 : 1 2LGe j30 : 1 -j3.16X -j1.39 Find the fault current contribution from the generator considering the delta-wye transformer phase shift. Example 9.6 1.39/ -60o -j1.39
  • 510. 139 Example 9.6 Example 9.5 results J1.99 AC AC . AC j.17 J0.1 J0.105 J0.1 J0.21. 1 2 n2 e -j30 : 1 2LGe j30 : 1 X j.88.88/ 60o j.88 . . AC Bus 1 AC Bus 2 ∆ ∆ G M 2LG Fault I L X [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = pu pu puj j j aa aaI o o GP 7/98.1 173/98.1 51. 88. 39.1 0 1 1 111 2 2 a b c IGP
  • 511. 140 Class Problem 5 The system data in p.u. based on SB = 100MVA, VB = 765kV for the lines are: G1: X1=X2=.18, X0=.07 T1: X=.1 LINE 1-3: X1=X2=.4 X0=.17 G2: X1=X2=.2, X0=.10 T2: X=.1 LINE 1-2: X1=X2=.085 X0=.256 G3: X1=X2=.25, X0=.085 T3: X=.24 LINE 2-3: X1=X2=.4 X0=.17 G4: X1=.34, X2=.45, X0=.085 T4: X=.15 a) From the perspective of Bus 1, draw the zero, positive and negative sequence networks. b) Determine the fault current for a 1 L-G bolted fault on Bus 1. AC Bus 1 AC Bus 3 ∆ G1 G3 G4 G2 Bus 2 LINE 1-3 LINE 1-2 LINE 2-3∆ T1 T2 T3 T4
  • 513. 142 Modern Fault Analysis Tools • Power Quality Meters (Power Quality Alerts) • Operations Event Recorder (ELV, Electronic Log Viewer) • Schweitzer Relay Event Capture • Schweitzer Relay SER (Sequential Events Record)
  • 514. 143 Modern Fault Analysis Example: Line current diff with step distance • First indication of an event - Power Quality alert email notifying On-Call Engineer that there was a voltage sag in the area. This event was a crane contacting a 69kv line. Time of event identified.
  • 515. 144 Modern Fault Analysis Example • Event Log Viewer stores breaker operation events. Search done in ELV using time from PQ Alert and breakers identified where trip occurred. Ferris and Miller breakers operated.
  • 516. 145 Modern Fault Analysis Example • Next the line relays (SEL-311L) at the two substations are interrogated for a possible event at this time. • Use command EVE C 1 to capture the event you desire. The C gives you the digital elements as well as the analog quantities. Ferris and Miller triggered an event record at this time (HIS command used in SEL relay) Reclosing enabled at Miller, additional record is the uncleared fault after reclosing.
  • 517. 146 Modern Fault Analysis Example • If the fault distance is not reasonable from the relays, i.e. the fault distances from each end is longer then the line length, the fault magnitude can be modeled in Aspen to determine fault distance by running interim faults. This discrepancy in distance can result from tapped load or large infeed sources.
  • 518. 147 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller initial fault:
  • 519. 148 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Ferris initial fault: Unknown source voltage
  • 520. 149 Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller reclose operation:
  • 521. 150 Modern Fault Analysis Example • This SEL-311L setup is a current differential with step distance protection. • Analysis from line relay SER to ensure proper relaying operation: • Question, why didn’t Z1G pickup?