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Theory Track Transmission Protection Theory Symmetrical Components & Fault Calculations Hands-On Relay School
Jon F. Daume Bonneville Power Administration March 14-15, 2011 1
Class Outline Power system troubles Symmetrical components Per unit system Electrical equipment impedances Sequence networks Fault calculations 2
Power System Problems Faults Equipment trouble System disturbances
3
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 4
5
Fault Types One line to ground (most common) Three phase (rare but most severe) Phase to phase Phase to phase to ground
6
Symmetrical Components
7
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 8
Balanced & Unbalanced Systems C C
A A
B
Balanced System
B Unbalanced System
9
Sequence Currents for Unbalanced Network
Ib0 Ia0
Ic1
Ic0
Zero Sequence Ia1 Ia2 Ib2
Positive Sequence
Ic2
Negative Sequence Ib1 10
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
D D D D D D D
D D D D D
D D D D 11
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 12
a Operator a
a = -0.5 + j √3= 1 ∠ 120° 2 a2 = -0.5 – j √3= 1 ∠ 240° 2 1
1 + a + a2 = 0 a2 13
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 14
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 15
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 16
Zero Sequence Current Filter Ia
Ib
Ic
3I0 = Ia + Ib + Ic 17
Zero Sequence Voltage Filter Ea
Eb
Ec
3 VO Polarizing Potential
3V0 18
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 19
Example IA = 3 + j4 IB = -7 - j2 IC = -2 + j7
+j IC = -2+j7 IA = 3+j4
IB = -7-j2 -j 20
Zero Sequence Ia0 = (IA + IB + IC)/3 = [(3+j4)+(-7-j2)+(-2+j7)]/3 = -2 + j3 = 3.61 ∠ 124°
Ib0 Ia0
Ic0
Ia0 = Ib0 = Ic0 Zero Sequence
21
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 22
Positive Sequence Ic1
Ia1
Ib1 23
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 24
Negative Sequence Ia2 Ib2
Ic2
25
Reconstruct Phase Currents Ic
Ic0
Ic2 Ic1
Ia Ia0
Ia2 Ia1
Ib
Ib0 Ib1 Ib2
26
Positive, Negative, and Zero Sequence Impedance Network Calculations for a Fault Study
27
+, -, 0 Sequence Networks Fault Simple 2 Source Power System Example
1PU
+
+
+
V1
V2
V0
I1
I2
Z1
I0
Z2
-
Z0
-
-
28
Impedance Networks & Fault Type Fault Type
+
-
0
3 Phase fault Phase to phase fault One line to ground fault Two phase to ground fault
D D D D
D D D
D D
29
Per Unit 30
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 31
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
32
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
33
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. 34
Base Ohms, Amps Base ohms: kV2 kV2 1000 = base kVA base MVA Base amps: base kVA = 1000 base MVA √3 kV √3 kV 35
Base Ohms, Amps (100 MVA Base) kV 525 345 230 115 69 34.5 13.8 12.5
Base Ohms 2756.3 1190.3 529.0 132.3 47.6 11.9 1.9 1.6
Base Amps 110.0 167.3 251.0 502.0 836.7 1673.5 4183.7 4618.8
36
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 37
Ohms to Per Unit pu Ohms = ohms / base ohms base MVA x ohms = pu Ω @ base MVA kV2LL
38
Per Unit to Real Stuff Amps = pu amps x base amps kV = pu kV x base kV Ohms = pu ohms x base ohms
39
Converting Between Bases Znew = Zold x base MVAnew x kV2old base MVAold kV2new
40
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 41
Synchronous Machines ~ Machine values: Machine reactances given in % of the machine KVA or MVA rating Ground impedances given in ohms 42
Synchronous Machines Machine values: Subtransient reactance (X"d) Transient reactance (X'd) Synchronous reactance (Xd) Negative sequence reactance (X2) Zero sequence reactance (X0)
43
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
44
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 45
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 46
Generator Example
~
R1
jX1” = 0.1
R2
jX2 = 0.1
R0
jX0 = 0.04
47
Transformers Ih H
Vh
Zh
1:N
Zx
X
Ze
Zhx Ω = Vh /Ih = Zh + Zx /N2 Zhx % = Vh /Ih x MVA/kV2 x 100 H
Zhx
X
Equivalent Transformer - Impedance in % 48
Transformers Impedances in % of the transformer MVA rating Convert from circuit voltage to tap voltage: %Xtap = %Xcircuit kV2circuit kV2tap
49
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 50
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 51
Transformer Example R1
jX1 = 0.04
R2
jX2 = 0.04
R0
jX0 = 0.04
Zero sequence connection depends upon winding configuration.
52
Transformer Connections Winding Connection
Sequence Network Connections Z0 Z1, Z2
Z1, Z2
Z0
53
Transformer Connections Winding Connection
Sequence Network Connections Z1, Z2
Z0
Z1, Z2
Z0
Z1, Z2
Z0
Z1, Z2
Z0
54
Delta Wye Transformer
A
a Ia
nIA
IA B
b Ib
nIB
IB C
Ic
nIC
IC 3I0 = IA+IB+IC 55
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 56
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 57
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 58
“T” Network Calculate corresponding “T” network impedances: ZH = (ZHL+ ZHM - ZML)/2 ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2 ZH ZHL= ZH + ZL ZHM = ZH + ZM ZML= ZM+ ZL
ZM ZL
59
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
60
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 61
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 62
Transformer Example
H , 230 kV H
0.03
-0.007
L, 13.8 kV M
H
M , 115 kV 0.03
0.07 L
+, - Sequence
-0.007 M 0.07
L
0 Sequence
63
Problem Calculate pu impedances for generators and transformers Use 100 MVA base Ignore all resistances
Problem 13.8 kV
230 kV
230 kV
13.8 kV
115 kV Fault
65
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 66
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) 67
Transmission Lines R
jX
68
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 69
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 70
Zero Sequence Line Impedance Z 0 = Ra + Re + j 0.01397 f log
De _______ 3√(GMR 2) GMD cond sep
or Z0 = Ra + Re + j (Xa + Xe - 2Xd) Ω/mile
71
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.
72
Transmission Lines Ra
j(Xa+Xd)
Z1 Ra
j(Xa+Xd)
Z2 Ra+Re j(Xa+Xe-2Xd)
Z0 73
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 74
Transmission Line Example Structure “H” frame: A
B
23 Feet
J6 Configuration
C
23 Feet
GMD = 3√(dabdbcdac) = 3√(23x23x46) = 28.978 feet Xd = 0.2794 f log GMD 60 = 0.2794 log 28.978 = 0.4085 Ω /mile
75
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 kV2LL
Z1 Line = (4.52 + j 39.03)100/2302 = 0.0085 + j 0.0743 pu
76
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 77
Transmission Line Example 0.0085
j0.0743
Z1 0.0085
0.0356
j0.0743
j0.2311
Z2
Z0 78
Long Parallel Lines Mutual impedance between lines
79
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 3I0, Line #1
Line #1
3I0, Line #2 Line #2
80
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 81
Mutual Impedance GMDcircuits is the ninth root of all possible distances between the six conductors, approximately equal to center to center spacing GMDcircuits = 9√(d a1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) Xd circuits = 0.2794 log GMDcircuits 82
Mutual Impedance Example Circuit #1 A
Circuit #2
B
C
23 Feet 23 Feet 92 Feet
A
46 Feet
B
23 Feet
C
23 Feet
115 Feet 138 Feet 69 Feet 92 Feet 115 Feet 46 Feet 69 Feet 92 Feet 83
Mutual Impedance Example GMDcircuits = 9√(d a1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) = 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) 84
Mutual Impedance Model Z0 Line 1 ZM Bus 1
Z0 Line 2
Bus 2
Z01- ZM ZM
Bus 1 Z02 - ZM
Bus 2
85
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 86
Mutual Impedance Fault Example Taft
Garrison
Taft
Garrison
645 Amps
1980 Amps
645 Amps
1315 Amps
1LG Faults With Mutual Impedances Taft
Garrison 920 Amps
920 Amps
Taft
Garrison 1370 Amps
260 Amps
87
1LG Faults Without Mutual Impedances
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 88
Problem 13.8 kV
230 kV
230 kV
13.8 kV
115 kV Fault
89
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 90
Fault Calculations and Impedance Network Connections
91
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
92
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
93
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 94
Positive Sequence Network
Fault
+
Vl = 1
Z1sl
Z1tl
Z1Ll
V1=1-I1Z1 Z1l
Z1Lr I1
Z1h
Z1sr
Vr = 1
Z1m
95
Negative Sequence Network
Fault
+
Z2sl
Z2tl
Z2Ll
V2= -I2Z2 Z2l
Z2Lr I2
Z2h
Z2sr
Z2m
96
Zero Sequence Network
Fault
+
Z0sl
Z0tl
V0= -I0Z0
Z0Ll
Z0l
Z0Lr I0
Z0h
Z0sr
Z0m
97
Network Reduction Fault Simple 2 Source Power System Example
1PU
+
+
+
V1
V2
V0
I1
I2
Z1
I0
Z2
-
Z0
-
-
98
Three Phase Fault Fault Simple 2 Source Power System Example
Only positive sequence impedance network used No negative or zero sequence currents or voltages
99
Three Phase Fault
1PU
+
+
+
V1
V2
V0
I1=11.9
Z1 0.084
-
I2=0
I0=0
Z2 0.084
Z0 0.081
-
100
Three Phase Fault Fault Simple 2 Source Power System Example
+
Vl = 1
Z1sl
Z1tl
Z1Ll
0.1
0.04
0.037
V1=1-I1Z1 Z1Lr
0.07 Z1l
0.037 Z1h Z1m I1 0.03 -0.007
Z1sr
Vr = 1
0.1
Sequence Network Connection for 3 Phase Fault 101
Three Phase Fault Fault Simple 2 Source Power System Example
+
Vl = 1
V1=1-I1Z1
0.177
0.160
Vr = 1
I1
Positive Sequence Network Reduced 102
Three Phase Fault Vectors Ic Vc
Ib
Va
Vb
Ia 103
Three Phase Fault MVAFault = MVABase ZFault pu or I pu Fault current = 1 pu ESource ZFault pu
104
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 = 0 105
Phase to Phase Fault Fault Simple 2 Source Power System Example
Positive and negative sequence impedance networks connected in parallel No zero sequence currents or voltages
106
Phase to Phase Fault
1PU
+
+
+
V1
V2
V0
I1
I2
Z1
I0
Z2
-
Z0
-
107
Phase to Phase Fault Fault
+
Vl = 1
Z1sl
Z1tl
Z1Ll
V1=1-I1Z1
I1
+
Z2sl
Z2tl
Z2Ll
Z1l
Z1Lr Z1h
Z1sr
Vr = 1
Z1m
V2= -I2Z2 Z2l
Z2Lr I2 = -I1
Z2h
Z2sr
Z2m 108
Sequence Network Connection for Phase to Phase Fault
Phase to Phase Fault Vectors Vc
Ic
Va Ib
Vb
109
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) 110
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
111
Single Line to Ground Fault Fault Simple 2 Source Power System Example
Positive, negative and zero sequence impedance networks connected in series
112
Single Line to Ground Fault +
+
+
V1
V2
V0
I1=4.02
I2=4.02
Z1 .084
Z2 .084
1PU
-
I0=4.02
Z0 .081
-
113
Single Line to Ground Fault +
Vl = 1
V1=1-I1Z1
Z1sl
Z1tl
Z1Ll
Z1Lr
0.1
0.04
0.037
I1 0.037
+
V2= -I2Z2
Z2sl
Z2tl
Z2Ll
Z2Lr
0.1
0.04
0.037
I2 0.037
+
V0= -I0Z0
Z0sl
Z0tl
Z0Ll
Z0Lr
0.04
0.04
0.116
I0 0.116
0.07 Z1l Z1h Z1m 0.03 -0.007
Z1sr
Vr = 1
0.1
0.07 Z2l Z2h Z2m 0.03 -0.007
Z2sr 0.1
0.07 Z 0l
Z0sr
Z0h Z0m 0.04 0.03 -0.007 I1 = I2 = I0
Sequence Network Connection for One Line to Ground Fault
114
Single Line to Ground Fault Vectors Vc
Va
Vb
Ia 115
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 116
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) 117
Two Phase to Ground Fault Fault Simple 2 Source Power System Example
Positive, negative and zero sequence impedance networks connected in parallel
118
Two Phase to Ground Fault
1PU
+
+
+
V1
V2
V0
I1
I2
Z1
I0
Z2
-
Z0
-
119
Two Phase to Ground Fault +
Vl = 1
Z1sl
Z1tl
Z1Ll
V1=1-I1Z1
I1
+
Z2sl
Z2tl
Z2Ll
Z0Ll
Vr = 1
Z1m
Z 2l
Z2Lr
+
Z0tl
Z1h
Z1sr
V2= -I2Z2
I2
Z0sl
Z 1l
Z1Lr
Z2h
Z2sr
Z2m
V0= -I0Z0 Z 0l
Z0Lr I0
Z0h
Z0sr
Z 0m
Sequence Network Connection for Phase to Phase to Ground Fault
120
Two Phase to Ground Fault Vectors Ic
Vc Ib Va Vb 121
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. 122
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. 123
The end
Jon F. Daume Bonneville Power Administration Retired! 124
Theory Track Transmission Protection Theory Transmission System Protection Hands-On Relay School
Jon F. Daume Bonneville Power Administration March 14-15, 2011 1
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
2
Power Transfer Vs
X
Vr
P = Vs Vr sin δ / X
Transmitted Power
Power Transfer
1
0.5
0 0
30
60
90
120
150
180
Angle Delta
3
Increase Power Transfer • Increase transmission system operating voltage • Increase angle δ • Decrease X – Add additional transmission lines – Add series capacitors to existing lines
4
5
Power Transfer During Faults Power Transfer
Transmitted Power
1.2
Normal
1
1LG
0.8
LL
0.6
LLG
0.4
3 Phase
0.2 0 0
30
60
90
120
150
180
Angle Delta 6
Vs
Vr
Power Transfer
1.2 1 5
Power
0.8 4
0.6
P1
0.4
P2
B
A 6
3
0.2 1
0 0
30
2
60
90 Angle Delta
120
150
180 7
System Stability • • • • • •
Relay operating speed Circuit breaker opening speed Pilot tripping High speed, automatic reclosing Single pole switching Special protection or remedial action schemes
8
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)
9
IEEE Device Numbers 51 52 59 62 63 67 79 81 86 87
AC time overcurrent relay AC circuit breaker Overvoltage relay Time delay stopping or opening relay Pressure switch AC directional overcurrent relay AC reclosing relay Frequency relay Lock out relay Differential relay (IEEE C37.2)
10
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
11
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 12
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
13
Transmission Line Protection 14
Western Transmission System 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
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.
15
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)
16
Line Angles vs. Voltage Z = √[Ra2 + 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.
17
Typical Line Protection
18
Distance Relays (21, 21G) 19
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 20
CT & PT Connections V Phase
I Phase 21 3V0
3I0 = Ia + Ib + Ic 67N
I Polarizing 21
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 22
Saturated CT Current 150 100 50 0 -50 -100 -0.017
0.000
0.017
0.033
0.050
0.06 23
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 24
Impedance Characteristic X
R
Directional 25
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
26
3 Zone mho Characteristic X
Zone 3
Zone 2
Zone 1
R
3 Zone Distance Elements Mho Characteristic
27
Typical Reaches
21 Zone 1 85-90% 21 Zone 2 125-180%, Time Delay Trip 21 Zone 3 150-200%, Time Delay Trip 67 Ground Instantaneous Overcurrent 67 Ground Time Overcurrent 67 Ground Time Permissive Transfer Trip Overcurrent Typical Relay Protection Zones
28
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
29
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 30
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 31
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 32
Problems for Distance Relays • • • • • •
Fault in front of relay Apparent Impedance Load encroachment Fault resistance Series compensated lines Power swings
33
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
34
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 35
Apparent Impedance Bus A
Zb = 1 Bus B
Za = 1 ohm Ia = 1 Ic = Ia + Ib = 2
Z apparent @ Bus A = Za + ZcIc/Ia = 3 Ohms
Zc = 1
Ib = 1
Bus C
Apparent Impedance
36
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 37
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
38
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
39
Load Encroachment X
Zone 3
Zone 2
Zone 1 Load Area R Load Consideration with Distance Relays
40
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
41
Lens Characteristic
42
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
43
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 44
Quadrilateral Characteristic X
R
Quadrilateral Characteristic 45
Special Load Encroachment X
Zone 4
Zone 2
Zone 1
R
46
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
47
Fault Resistance X
Zone 3
Zone 2
Rf Zone 1
R Fault Resistance Effect on a Mho Characteristic
48
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
49
Series Caps
Zl 21
Zc
Zl > Zc
21
50
Series Caps Bypass MOD
Isolating MOD
Isolating MOD
Bypass Breaker
Discharge Reactor
Triggered Gap Capacitor (Fuseless)
Damping Circuit
Metal-Oxide Varistor (MOV)
Platform
Main Power Components for EWRP Series Capacitors
51
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 52
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 53
Out Of Step Blocking X OOSB Outer Zone
OOSB Inner Zone
Zone 2 t = 30 ms?
Zone 1
R
Typical Out Of Step Block Characteristic
54
Ground Distance Protection and Kn (21G) 55
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 56
Sequence Networks
57
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)
58
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 59
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 60
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
61
Ground Fault Protection (67N) 62
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 63
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. 64
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 65
Current Polarizing H1 CT
X1
Y1
H2 X2
Y2
X3
Y3
H3
H0X0 I Polarizing
Auto Transformer Polarizing Current Source
66
Voltage Polarizing Ea
Eb
Ec
3 VO Polarizing Potential
67
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
68
Mutual Coupling 3I0, Line #1
Line #1
3I0, Line #2 Line #2
69
Mutual Coupling vs. Ground Relays Taft
Garrison
Taft
Garrison
645 Amps
1980 Amps
645 Amps
1315 Amps
1LG Faults With Mutual Impedances Taft
Garrison 920 Amps
920 Amps
Garrison
Taft 1370 Amps
260 Amps
1LG Faults Without Mutual Impedances
70
Other Line Protection Relays 71
Line Differential
87
87
72
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 73
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
74
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 75
Pilot Wire
76
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 77
Current Differential • Single pole capability • 3 Terminal line capability • May include an external, direct transfer trip feature • Immune to swings, load, series caps
78
Transfer Trip
79
Direct Transfer Trip • Line protection • Equipment protection – Transformer terminated lines – Line reactors – Breaker failure
• 2 or more signals available – Analog or digital tone equipment
80
Direct Transfer Trip Protective Relay Protective Relay Tone 1 Xmit
Tone 1 Rcvd
Tone 2 Xmit
Tone 2 Rcvd
PCB Trip Coil
PCB Trip Coil
Direct Transfer Trip
81
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 82
Permissive Transfer Trip Permissive Relay
Permissive Relay
Tone 2 Xmit
Tone 2 Xmit
Tone 2 Rcvd
Tone 2 Rcvd
PCB Trip Coil
PCB Trip Coil Permissive Transfer Trip
83
Permissive Keying • Zone 2 instantaneous • Permissive overcurrent ground (very sensitive setting) • PCB 52/b switch • Current reversal can cause problems
84
PRT Current Reversal Ia
I Fault, Line AB
A
Ic
C
Ib
B
I Fault, Line CD
Id
D
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
85
PRT Current Reversal Ia
I Fault, Line AB
A
Ic
C
B
I Fault, Line CD
Id
D
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.
86
• • • •
Directional Comparison Blocking
Overreaching relays Delay for channel time Channel failure can allow overtrip Often used with “On/Off” carrier
87
Directional Comparison Forward Relay
TD
Reverse Relay
Reverse Relay
Forward Relay
Block Xmit
Block Xmit
Block Rcvd
Block Rcvd
Time Delay
PCB Trip Coil
Time Delay
TD
PCB Trip Coil
Directional Comparison Blocking Scheme
88
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
89
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 90
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)
91
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 92
Telecommunications Channels • Microwave radio – Analog (no longer available) – Digital
• Other radio systems • Dedicated fiber between relays – Short runs
• Multiplexed fiber – Long runs
• SONET Rings 93
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 94
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 95
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 96
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 97
Breaker Failure Relay (50BF) 98
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
99
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) 100
Breaker Failure Relay Protective Relay
PCB Trip Coil #1
BFR Fault BFR Retrip Detector TD
PCB Trip Coil #2
BFR Time Delay, 8~
TD
Trip
86
Block Close Typical Breaker Failure Scheme with Retrip 101
Typical BFR Clearing Times Proper Clearing: 0 Fault occurs +1~ Relays PU, Key TT +2~ PCB trips +1~ Remote terminal clears
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
3-4 Cycles local clearing time 4-5 Cycles remote clearing time
12-13 Cycles local back up clearing time 13-14 Cycles remote backup clearing 102
Remedial Action Schemes (RAS) aka: Special Protection Schemes 103
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 104
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 105
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 106
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
107
Generator Dropping • Trip generators for loss of load • Trip generators for loss of transmission lines or paths – Prevent overloading
108
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 109
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
110
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 111
Typical RAS Controller
112
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
113
Chief Jo Brake
1400 Megawatts @ 230 kV
114
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
115
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
116
UFOs vs. Power Outages
117
the end
Jon F. Daume Bonneville Power Administration retired March 15, 2011 118
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
Symmetrical Faults
3
Faults Shunt faults: Three phase Line to line Line to ground 2 Line to ground
a b c a b c c b a a b c
4
Faults Series faults One open phase:
a b c a b c
2 open phases Increased phase impedance
Z
a b c 5
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 6
Symmetrical Faults L
R
i(t) AC
2V sin( ω t + α )
e(t ) =
Fault at t = 0
2V α t=0 7
Symmetrical Faults For a short circuit at generator terminals at t=0 and generator initially open circuited: di e ( t ) = Ri + L dt
(L is considered constant)
di 2VSin(ωt + α ) = Ri + L dt by using Laplace transforms i(t) can be found
8
Symmetrical Faults 2V [ Sin(ωt + α − θ ) − Sin(α − θ )e − t / T ] i (t ) = Z Where: Z = R 2 + (ωL) 2 = R 2 + X 2
θ = Tan
T=
i(t ) =
ωL
−1
2I
R
= Tan
L X = R ωR
ac
Where: Iac
−1
X R
Note that for a 3phase system α will be different for each phase. Therefore, DC offset will be different for each phase
Time Constant
[ Sin ( ω t + α − θ ) − Sin (α − θ ) e − t / T ]
= ac RMS fault current at t=0
(Examples)
9
i(t ) =
2I
ac
[ Sin ( ω t + α − θ ) − Sin (α − θ ) e − t / T ]
α = θ = 90 e(t)
2V
α = 90 o
t=0
2 I ac
Idc = 0
iac
10
o
i(t ) =
2I
2V
ac
[ Sin ( ω t + α − θ ) − Sin (α − θ ) e − t / T ]
θ = 90 o α =0
e(t)
t=0
iac
2 I ac 0
2 I ac 0
idc 11
i(t ) =
θ = 90 o α =0
2I
ac
[ Sin ( ω t + α − θ ) − Sin ( α − θ ) e − t / T ]
iac
2 I ac 0
2 I ac 0
2 2 I ac 0
idc i (t ) t 12
Symmetrical Faults Iac and Idc are independent after t = 0
2V [ Sin(ωt + −π / 2) + e − t / T ] i (t ) = Z I
RMS
I
=
dc
I
2
ac
= 2I
+I
aco
2
dc −t T e
Substituting:
t − − 2t T 2 2 T I (max)= (I + 2I )e = I 1+ 2e RMS ac ac ac 13
Asymmetry Factor IRMS(max) = K(τ) Iac Asymmetry Factor = K(τ)
−4π xτ r K (τ ) = 1 + 2e Where: τ = number of cycles T = X / 2 π fR (Example 7.1)
14
Example 7.1 R = 0.8 Ώ
-
CB
I
+ AC
XL = 8 Ώ
Fault at t = 0
V = 20 kVLN
•Fault at a time to produce maximum DC offset •Circuit Breaker opens 3 cycles after fault inception Find: 1. Iac at t = 0 2. IRMS Momentary at
τ = 0.5 cycles
3. IRMS Interrupting Current 15
Example 7.1 a.
b.
I AC (0) =
20 0.8 + 8 2
K ( 0 .5 ) = 1 + 2 e
2
= 2.488kARMS
− 4 Π ( .5
10
)
= 1.438
I momentary = (1.438)( 2.488) = 3.577 KA c.
K ( 3 ) = 1 + 2e
− 4Π ( 3
10
)
= 1.023
I Interrupting = (1.023)( 2.488) = 2.545 KA 16
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 17 (See Phasor Diagram)
AC Decrement R
XL
EI
XAR Load
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”. 18
I=IL + EI
XL
XAR
-
Armature Reaction
Field Flux
Load
Loaded Generator EI
Resultant Field
jILXAR(t)
ET IL
jILXL
19
I=0 + E” = E’ = E = ET0
XL
XAR=0
-
ET0
Unloaded Generator Armature Reaction = 0
Field Flux
Resultant Field
t=0E” ET0
20
I=0 + E” = E’ = E = ET0
XL
XAR
Faulted Generator
t=0
Armature Reaction = 0
Field Flux
Resultant Field
E”
ET0 = 0 21
I = I” XL
+ E” = E’ = E = ET0
XAR=0
Armature Reaction = 0
Field Flux
Resultant Field
t=0+ E” = jI”XL
ET = 0 I” 22
I = I’ XL
+ E” = E’ = E = ET0
XAR’
-
Armature Reaction = 0
t ≈ 3Cyc. Field Flux
Resultant Field
E’ = jI’(XL + XAR’)
ET = 0 I’ 23
I=I XL
+ E” = E’ = E = ET0
XAR
-
Armature Reaction = 0
t =∞ Field Flux
Resultant Field
E’ = jI(XL + XAR)
ET = 0 I’ 24
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.
2I2I"
2I '
2I
25
AC Decrement For a round rotor machine we only need to consider the direct axis reactance. 2I"=
2I '=
2I =
2E" X "d 2E' X 'd
2E X d
Subtransient Transient Synchronous (steadystate)
26
AC Decrement Can write the ac decrement equation
[
i (t ) = 2 ( I "− I ' )e ac
−t
T "d
+ ( I '− I )e T 'd + I ][Sin(ω t + α − θ )]
E " = E '= E = ET T”d:
−t
For an unloaded generator (special case):
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
27
AC Decrement
[
i (t ) = 2 ( I "− I ' )e ac
−t
T "d
−t
+ ( I '− I )e T 'd + I ][Sin(ω t + α − θ )]
For t = 0
iac (max) = 2[( I "− I ' ) + ( I '− I ) + I ] = 2 I " For t = ∞
iac (max) = 2[0 + 0 + I ] = 2 I 28
ac and dc Decrement Transform ac decrement equation to phasor form _ ⎡ − t − t I ac = ⎢( I "− I ' )e T " d + ( I '− I )e T ' d + I ] /α − θ ⎢⎣ dc decrement equation: −
I dc =
t
2 I " Sin ( α − θ ) e
T
A
Where TA = Armature circuit time constant (Example 7.2)
29
Example 7.2
S
R=0
-
CB
I
+ AC
x”d =.15pu T”d = .035 Sec. x’d = .24pu T’d = 2.0 Sec. xd = 1.1pu TA = 0.2 Sec.
V = 1.05 pu
Fault at t = 0
500 MVA, 20kV, 60 Hz Synchronous Generator 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)
30
Example 7.2 1 ⎞ − t .035 ⎛ 1 1 ⎞ −t 2 1 ⎤ ⎡⎛ 1 + ⎥ − +⎜ − I AC ( t ) = 1.05 ⎢⎜ ⎟e ⎟e 1.1 ⎦ ⎝ .24 1.1 ⎠ ⎣⎝ .15 .24 ⎠
I DC ( t ) max = 2 I " e
−t
.2
I Base
500 = = 14.434 KA 20 3
E " 1.05 I"= = = 7.0 puI DC = 101kA x" d .15 I DC ( t ) max = 2 (7)e
−t
.2
= 9.9e
−t
.2
a b
31
Example 7.2 Part c: Find IRMS at interruption (3 cycles) 3 t= = 0.05 sec . 60
1 ⎞ − .05 .035 ⎛ 1 1 ⎞ − .05 2 1 ⎤ ⎡⎛ 1 I AC ( t ) = 1.05 ⎢⎜ + ⎥ +⎜ − − ⎟e ⎟e 1.1 ⎦ ⎝ .24 1.1 ⎠ ⎣⎝ .15 .24 ⎠
I AC (.05) = 1.05[(2.5 )(.24) + ( 3.258)(.975) + .909] = 4.92 pu I DC (.05) = 9.9e
− .05
.2
= 7.71 pu
I RMS (.05) = 4.92 2 + 7.712 = 9.146 pu = 132kA
c
32
Example 7.2 Part d: Find IMomentary(max) at t = ½ cycle
.5 t= = .0083 sec 60 1 ⎞ − .0083 .035 ⎛ 1 1 ⎞ − .0083 2 1 ⎤ ⎡⎛ 1 + ⎥ I AC ( t ) = 1.05 ⎢⎜ +⎜ − − ⎟e ⎟e 1.1 ⎦ ⎝ .24 1.1 ⎠ ⎣⎝ .15 .24 ⎠
I AC (.0083) = 1.05[(2.5 )(.79) + ( 3.258)(.996) + .909] = 6.43 pu
I DC = 9.9e
− .0083
.2
= 9.5 pu
I RMS = 6.43 + 9.5 = 14.9 pu = 215kA 2
2
d 33
Turbine
Gen.
Energy
34
Superposition for Fault Analysis
35
Superposition for Fault Analysis Bus 1
New representation:
IF
IG
Bus 2
IM
Bus 1
IG1
IF1
IM1
IG = IG! + IG2 = IG1+ IL IM = IM1 – IL Example 7.3
IG2
IF = IG1 + IM1
IL IF2=0
36
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
VF 1.05 1.05 I F1"= = = = − j 9.08 pu (.15)(.505) j .116 Z TH j .655 Part b: Neglecting load current, find Generator and motor fault current.
.505 I G 1 " = − j 9.08 = − j 7 pu .655
I M 1 " = − j 9.08 − ( − j 7 ) = − j 2.08 pu
37
Example 7.3 Part c: Including load current, find Generator and motor current during the fault period.
I Load
S * 1/ − cos .95 1/ − 18 o o = * = = = . 952 / − 18 = IG2 = − I M 2 o 1.05 1.05 / 0 V
I G " = − j 7 + .953 / − 18 = 7.35 / − 83 pu o
o
c
I M " = − j 2.08 − .952 / − 18 = 2.00 / 243 pu o
o
c
38
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 39
1
2 J 0 . 305
J 0 . 15 AC
J 0 .2
IF
-VF
+ AC
Eg” -
+ AC
E
m
-
40
Z Bus Method For the circuit of Figure 7.4d (Slide 36 & 40)
⎡ Y 11 Y 12 ⎤ ⎡ E ⎡ I1 ⎤ ⎢ I ⎥ = ⎢Y Y ⎥ ⎢ E ⎣ 21 22 ⎦ ⎣ ⎣ 2 ⎦
Injected node currents
[matrix
1 2
⎤ ⎥ ⎦
Node voltages
Y-bus] nodal admittance Premultiplying both sides by the inverse of [Y-bus}
⎡ E1 ⎤ ⎡ Z 11 ⎢E ⎥ = ⎢Z ⎣ 2 ⎦ ⎣ 21
Z 12 ⎤ ⎡ I 1 ⎤ Z 22 ⎥⎦ ⎣⎢ I 2 ⎥⎦
Pre-fault [Z-Bus] Injected node =[Y-Bus]-1 node Current Voltage
-IF1 0
For a fault at Bus 1
E1 = Z11 (− I F 1 )
⎛ − VF − E1 = −⎜⎜ I F1 = Z11 ⎝ Z11
⎞ ⎟⎟ ⎠
41
Z-Bus Method For a fault at Bus 1 E 1 = Z 11 ( − I IF1 I
F 1
⎛ − VF − E1 = = −⎜⎜ Z 11 ⎝ Z 11 =
V Z
F 1
)
⎞ ⎟⎟ ⎠
F 11
where:
IF1 = Fault current at bus 1 VF = Prefault voltage of the faulted bus (Bus 1) 42
Z-Bus Method For N bus system, fault on Bus n
-VF
⎡ E1 ⎤ ⎡ Z 11 ⎢E ⎥ ⎢Z ⎢ 2 ⎥ ⎢ 21 ⎢ E3 ⎥ ⎢ Z 31 ⎢ ⎥=⎢ ⎢ E N ⎥ ⎢ Z n1 ⎢ . ⎥ ⎢ . ⎢ ⎥ ⎢ ⎢⎣ E N ⎥⎦ ⎢⎣ Z N 1
I Fn
VF = Z nn
Z 12 Z 22
Z 13 Z 23
Z 1n Z 2n
Z 32
Z 33
Z 3n
Z n2 .
Z n3 .
Z nn ,
ZN2
Z N3
Z Nn
. Z 1N ⎤ ⎡ 0 ⎤ . Z 2 N ⎥⎥ ⎢ 0 ⎥ ⎢ ⎥ . Z 3N ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ . Z nN ⎥ ⎢− I Fn ⎥ . . ⎥⎢ . ⎥ ⎥⎢ ⎥ . Z NN ⎥⎦ ⎢⎣ 0 ⎥⎦
Where: VF = Pre-fault voltage at faulted bus Znn = Thevinen impedance 43
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) 44
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 1 2 Y12 = -j3.28 Y10 = -j6.67
Y20 = -j5 IF 45
Example 7.4 ⎡ − j 9.95 j 3.28 ⎤ [YBus ] = ⎢ ⎥ j 3 . 28 − j 9 . 95 ⎣ ⎦
[Z Bus ] = [Ybus ]
−1
⎡ j .1156 =⎢ ⎣ j .046
⎡ E1 ⎤ ⎡ j .1156 ⎢ E ⎥ = ⎢ j .046 ⎣ 2⎦ ⎣
j .046⎤ j .139⎥⎦
a
j .046⎤ ⎡ I 1 ⎤ -IF j .139⎥⎦ ⎣⎢ I 2 ⎥⎦ 0
E1 = ( j .1156) I 1 -VF
-IF
VF IF "= = − j 9.08 j .1156
b
46
Example 7.4 Find: Line current
⎡ E1 ⎤ ⎡ j .1156 ⎢ E ⎥ = ⎢ j .046 ⎣ 2⎦ ⎣
j .046⎤ ⎡ I 1 ⎤ j .139⎥⎦ ⎣⎢ I 2 ⎥⎦
For fault at Bus 1: E1 = E11+ E12 = 0 E2 = E21 + E22 = VF + (j.046)I1 E2 = 1.05 + (j.046)(j9.08) = .632 /0o
I 21
E 2 − E1 .632 − 0 = = = − j 2.07 Z 21 j .305
b 47
Example 7.4 1
2 Y12 = -j3.28
Y10 = -j6.67
Y20 = -j5
IF Find IF and I contribution from Line for Bus 2 fault
⎡ E1 ⎤ ⎡ j .1156 =⎢ ⎢ ⎥ -VF ⎣ E 2 ⎦ ⎣ j .046 IF2
j .046⎤ ⎡ I 1 ⎤ j .139⎥⎦ ⎣⎢ I 2 ⎥⎦
1.05 = = − j 7.55 pu j .139
- I F2
E1 = VF + ( j .046)( − I F ) = 1.05 + ( j .046)( j 7.55) = .703 / 0 o E1 − E 2 .703 − 0 = = − j 2.3 pu I 12 = c 48 Z 12 j .305
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.
49
Z-Bus Rake equivalent
Example: Fault at Bus n IF = In =
VF Z nn
E1 = VF − ( Z 1n )( I n )
Etc.
50
Class Problem 1 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.
⎡.12 .08 .04⎤ Zbus = j ⎢⎢.08 .12 .06⎥⎥ pu ⎢⎣.04 .06 .08⎥⎦
51
Symmetrical Components
52
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. 53
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.
54
Symmetrical Components Ib bb Ia a
a
Vb Va ZY
ZY n
Vc
Vb ZY
Va
c
Vc Ic Balanced load supplied by balanced voltages results in balanced currents This is a positive sequence system, In Symmetrical Components we will be studying 55 unbalanced systems with one or more dissymmetry.
Symmetrical Components For the General Case of 3 unbalanced voltages VC
6 degrees of freedom VA
VB Can define 3 sets of voltages designated as positive sequence, negative sequence and zero sequence
56
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
57
Symmetrical Components Positive Sequence 2 degrees of freedom VC1
VA1
120o 120o
120o
VA1 = VA1 VB1 = a2 VA1 VC1 = a VA1
VB1
a is operator 1/120o 58
Symmetrical Components Negative Sequence 2 degrees of freedom VB2
VA2
120o 120o
120o
VC2
VA2 = VA2 VB2 = aVA2 VC2 = a2 VA2 a is operator 1/120o 59
Symmetrical Components Zero Sequence 2 degrees of freedom VA0 VB0 VC0
VA0 = VB0 = VC0
60
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. 61
Symmetrical Components Rewriting the sequence voltages in term of the Phase A sequence voltages: VA = VA0 + VA1 +VA2 Drop A VB = VA0 + a2 VA1 + aVA2 VC = VA0 + aVA1 +a2 VA2
VA = V0 + V1 +V2 VB = V0 + a2 V1 + aV2 VC = V0 + aV1 +a2 V2
Suggests matrix notation: VA
1
1
1
V0
VB = 1
a2
a
V1
VC
a
a2
V2
[VP] =
1
[A]
[VS]
62
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]
63
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 = 3 /30o 1/a = a2 Relationships of a can greatly expedite calculations ( Find [A]-1)
64
Inverse of A ⎡1 1 [A] = ⎢⎢1 a 2 ⎢⎣1 a
1⎤ ⎥ a⎥ 2 a ⎥⎦
⎡1 1 T Step 1: [ A] = ⎢⎢1 a 2 Transpose ⎢⎣1 a
1⎤ a ⎥⎥ a 2 ⎥⎦
Step 2: Replace each element by its minor 1
2
3
2 2 2 ⎡ a −a a−a ⎤ 1 a−a ⎢ 2 ⎥ 2 a −1 ⎥ 2 ⎢a − a a − 1 2 2 ⎢ a − 1 a − 1 ⎥⎦ 3 ⎣a − a 65
Inverse of A 1
2
3
2 2 2 ⎡ a −a a−a ⎤ 1 a−a ⎢ 2 ⎥ 2 a −1 ⎥ 2 ⎢a − a a − 1 2 2 ⎢ ⎥ − 1 − 1 a a 3 a−a ⎣ ⎦
Step 3: Replace each element by its cofactor 1 2 3
⎡a − a a − a a − a ⎤ ⎢ ⎥ 2 2 a −1 1− a ⎥ 2 ⎢a − a 2 2 ⎢ ⎥ 1 − − 1 a a 3 a−a ⎣ ⎦ 1
2
2
2
66
1
Inverse of A
2
3
⎡a − a a − a a − a ⎤ ⎢ ⎥ 2 2 a −1 1− a ⎥ 2 ⎢a − a 2 2 ⎢ ⎥ 1 − − 1 a a 3 a−a ⎣ ⎦ 2
1
2
2
⎡1 1 [A] = ⎢⎢1 a 2 ⎢⎣1 a
1⎤ ⎥ a⎥ 2 a ⎥⎦
Step 4: Divide by Determinant
D = 1(a − a ) + 1(a − a ) + 1(a − a ) = 3(a − a ) 2
2
2
2
a −1 a −1 ⎛ a ⎞ a −1 =⎜ ⎟ =a 2 =a 2 2 a −1 a−a ⎝a⎠a−a 2
2
2
1− a ⎛ 1 ⎞ 1− a 1 2 = =a =⎜ ⎟ 2 a−a ⎝ a ⎠ 1− a a
67
Inverse of A
[A]
−1
⎡1 1 1⎢ = ⎢1 a 3 2 ⎢⎣1 a
1⎤ 2⎥ a ⎥ a ⎥⎦ 68
Symmetrical Components Previous relationships were developed for voltages. Same could be developed for currents such that: IA [IP] = IB IC
I0 [IS] = I1 I2
[IP] = [A] [IS]
[IS] = [A]-1 [IP]
1 1 1 [A] = 1 a2 a 1 a a2
[A]-1
1 1 1 = 1/3 1 a a2 1 a2 a 69
Significance of I0 I0 I1 I2
1 1 1 = 1/3 1 a a2 1 a2 a
IA IB IC
I0 = 1/3 ( IA + IB + IC) IA In = IA + IB + IC = 3 I0
IC In IB
n
For a balanced system I0 = 0 For a delta system I0 = 0 (Examples 8.1, 8.2 and 8.3)
70
Example 8.1 ⎡ 277 / 0 o ⎤ ⎡ 277 ⎤ a ⎢ o⎥ [V P ] = ⎢ 277 / − 120 ⎥ = ⎢⎢ 277a 2 ⎥⎥ b ⎢⎣ 277 / 120 o ⎥⎦ ⎢⎣ 277a ⎥⎦ c Find [VS] (Sequence voltages)
⎡1 1 ⎡V0 ⎤ 277 ⎢ −1 ⎥ ⎢ [VS ] = ⎢V1 ⎥ = [ A] [VP ] = ⎢1 a 3 ⎢⎣V2 ⎥⎦ ⎢⎣1 a 2
1 ⎤⎡ 1 ⎤ ⎡ 0 ⎤ 0 o⎥ 2 ⎥⎢ 2 ⎥ ⎢ a a = 277/ 0 1 ⎥ ⎥⎢ ⎥ ⎢ a ⎥⎦ ⎣⎢ a ⎥⎦ ⎢⎣ 0 ⎥⎦ 2
71
Example 8.2 Y connected load with reverse sequence
⎡ 10 / 0 o ⎤ ⎡ 1 ⎤a ⎢ o ⎥ [I P ] = ⎢ 10/ 120 ⎥ = (10)⎢⎢ a ⎥⎥ b ⎢⎣10 / − 120 o ⎥⎦ ⎢⎣a 2 ⎥⎦ c Find IS
(Sequence Currents)
⎡1 1 10 ⎢ −1 [I S ] = [ A] [I P ] = ⎢1 a 3 ⎢⎣1 a 2
1 ⎤⎡ 1 ⎤ ⎡ 0 ⎤ 0 2 ⎥⎢ ⎥ ⎢ ⎥ a a = 0 ⎥1 ⎥⎢ ⎥ ⎢ a ⎥⎦ ⎢⎣a 2 ⎥⎦ ⎢⎣10 / 0 o ⎥⎦ 2 72
Ia = 10 /
0o
Example 8.3 ⎡I0 ⎤ [I S ] = ⎢⎢ I 1 ⎥⎥ = [ A]−1 [I P ] ⎢⎣ I 2 ⎥⎦
Ic = 10 /120o In Ib = o
⎡1 1 10 ⎢ [I S ] = ⎢1 a 3 ⎢⎣1 a 2
1 ⎤ ⎡1⎤ a a 2 ⎥ ⎢0⎥ b ⎥⎢ ⎥ a ⎥⎦ ⎣⎢a ⎥⎦ c
2 o ⎡ ⎡ ⎤ −a 3.33 / 60 ⎤ 0 ⎡ 1+ a ⎤ 10 ⎢ 10 ⎢ ⎥ ⎢ o ⎥ ⎥ [I S ] = ⎢ 2 ⎥ = ⎢ 2 ⎥ = ⎢ 6.67 / 0 ⎥ 1 3 3 2 o ⎥⎦ 2 ⎢ ⎥ ⎢ − 1 a a 3 . 33 / 60 + − ⎥⎦ ⎢⎣ ⎣ ⎦ ⎣
I n = 3 I 0 = 10 / 60 o
73
Sequence Impedance for Shunt Elements Sequence Networks of balanced Y elements( Loads, Reactors, IA capacitor banks, etc.) IC VA
ZY
VC
VB
IB
ZY n ZY
Zn
. VA = IAZy + (IA + IB +IC) Zn = (ZY + Zn)IA + ZnIB + ZnIC VB = ZnIA + (ZY + Zn)IB + ZnIC VC = ZnIA + ZnIB +(ZY + Zn)IC
74
Sequence Impedance for Shunt Elements Zn Zn ⎤⎡IA⎤ ⎡VA⎤ ⎡ZY +Zn ⎥ ⎢ ⎥ ⎢V ⎥ = ⎢ Z Z Z Z I + n Y n n B ⎥⎢ B ⎥ ⎢ ⎥ ⎢ ⎢⎣VC ⎥⎦ ⎢⎣ Zn Zn ZY +Zn ⎥⎦⎢⎣IC ⎥⎦ [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]
75
Sequence Impedance for Shunt Elements Zn ⎤⎡1 1 1 ⎤ ⎡Z00 Z01 Z02⎤ ⎡1 1 1 ⎤⎡ZY +Zn Zn ⎢Z Z Z ⎥ = 1⎢1 a a2⎥⎢ Z ⎥⎢1 a2 a ⎥ Z Z Z + [ZS] = ⎢ 10 11 12⎥ ⎢ n Y n n ⎥⎢ ⎢ ⎥ ⎥ 3 ⎢⎣Z20 Z21 Z22⎥⎦ ⎢⎣1 a2 a ⎥⎦⎢⎣ Zn Zn ZY +Zn⎥⎦⎢⎣1 a a2⎥⎦ 0
⎡ ZY + 3 Z n [Z S ] = ⎢⎢ 0 0 ⎣⎢
1
0 ZY
2
0 ⎤ 0 ⎥ ⎥ Z Y ⎥⎦
0 1 2 76
Sequence Impedance for Shunt Elements ⎡V 0 ⎤ ⎡ Z Y + 3 Z n ⎢V ⎥ = ⎢ 0 1 ⎢ ⎥ ⎢ ⎢⎣V 2 ⎥⎦ ⎢⎣ 0
0 ZY 0
0 ⎤⎡I0 ⎤ 0 ⎥⎥ ⎢⎢ I 1 ⎥⎥ Z Y ⎥⎦ ⎢⎣ I 2 ⎥⎦
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.
77
Sequence Impedance for Shunt Elements We can form sequence circuits which represent the equations:
V0
I0
3 Zn
Zero sequence circuit Zn only in zero Sequence No neutral: Zn = infinity Solid ground: Zn = 0 Positive sequence circuit
ZY
V1
I1
ZY
V2
I2
ZY
Negative sequence circuit 78
Sequence Impedance for Shunt Elements Delta connected shunt element IA . A ZΔ ZΔ IB B ZΔ C IC
Sequence circuits
V0
I0
ZY open
V1
I1
V2
I2
ZΔ/3
ZΔ/3 79
Sequence Impedance for Shunt Elements For the general case: [ZS] = [A]-1[ZP][A]
⎡ Z 00 Z 01 Z 02 ⎤ ⎡1 1 1 ⎤ ⎡ Z AA Z AB Z AC ⎤ ⎡1 1 1 ⎤ 1⎢ ⎢Z ⎥ ⎥ ⎢1 a 2 a ⎥ 2 ⎥⎢ = Z Z 1 a a Z Z Z 11 12 ⎥ BB BC ⎥ ⎢ ⎢ 10 ⎥ ⎢ BA ⎥ 3⎢ ⎢⎣ Z 20 Z 21 Z 22 ⎥⎦ ⎢⎣1 a 2 a ⎥⎦ ⎢⎣ Z CA Z CB Z CC ⎥⎦ ⎢⎣1 a a 2 ⎥⎦ If there is symmetry: ZAA = ZBB = ZCC and ZAB = ZBC = ZCA we could perform multiplication and get: 0 0 ⎤ ⎡ Z AA + 2 Z AB ⎥ [Z S ] = ⎢⎢ 0 Z AA − Z AB 0 ⎥ ⎢⎣ 0 0 Z AA − Z AB ⎥⎦ We see that: Z11 = Z22 and Z00 > Z11
80
Series Element Sequence Impedance IA ZAB
IB VA VB n
IC
ZBC VC
VAA’ ZAA VBB’
ZCA
ZBB VAA’
VB’
ZCC
⎡V A − V A' ⎤ ⎡ Z AA ⎢V − V ⎥ = ⎢ Z B' ⎥ ⎢ BA ⎢ B ⎢⎣VC − VC ' ⎥⎦ ⎢⎣ Z CA
VC ’
Z AB Z BB Z CB
VA’
n
Z AC ⎤ ⎡ I A ⎤ ⎥ ⎢ ⎥ Z BC ⎥ ⎢ I B ⎥ Z CC ⎥⎦ ⎢⎣ I C ⎥⎦ 81
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:
⎡ Z0 [Z S ] = ⎢⎢ ⎢⎣
Z1
⎤ ⎥ ⎥ Z 2 ⎥⎦
82
Series Element Sequence Impedance The sequence circuits for series elements are: V0
I0
Z0
V 0’
o n0 V1
I1
Z1
V 1’
o n1 V2
I2
Z2 o n2
V 2’ 83
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.
84
Rotating Machine Sequence Networks ZK
ZAC ZCA + EC -
IC
ZBC n
ZCB
A
EA + ZAB
Z
C
+
EB
ZK
ZK ZBA
IA 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 o = a E E = E /120 C RMS IB B
85
Rotating Machine Sequence Networks EA= ERMS / 0 = E EB = ERMS /-120o = a2 E or EC = ERMS /120o = a E
[E ] = [ A] [E ] −1
Sg
Pg
[E ] Pg
⎡1 1 1⎢ = 1 a 3⎢ 2 ⎣⎢1 a
⎡ E ⎤ a = ⎢a 2 E ⎥ b ⎢ ⎥ ⎢⎣ aE ⎥⎦ c
1 ⎤⎡ E ⎤ a 2 ⎥ ⎢a 2 E ⎥ = ⎥⎢ ⎥ a ⎥⎦ ⎢⎣ aE ⎥⎦
⎡0⎤ 0 ⎢E⎥ ⎢ ⎥ 1 ⎣⎢ 0 ⎥⎦ 2
Therefore, only the positive sequence system has a generator voltage source. 86
Rotating Machine Sequence Networks ZK
ZAC ZCA + EC -
IC EA
ZBC n
ZCB
+
A
IA
ZK
ZAB Z BA
Z
C
+
ZAB = ZBC = ZCA = ZR ZBA = ZCB = ZAC = ZQ
EB
ZK
IB
Machine is not passive: Mutual Reactances: ZAB ≠ ZBA , etc.
B
87
Rotating Machine Sequence Networks From the machine diagram we can write: ⎡EA ⎤ ⎡Z K + Z N ⎢E ⎥ = ⎢ Z + Z N ⎢ B⎥ ⎢ Q ⎢⎣ E C ⎥⎦ ⎢⎣ Z R + Z N
ZR + ZN ZK + ZN ZQ + Z N
ZQ + Z N ⎤⎡I A ⎤ ⎥ Z R + Z N ⎥⎢IB ⎥ ⎢ ⎥ Z K + Z N ⎥⎦ ⎢⎣ I C ⎥⎦
[E PG ] = [Z PG ][I P ]
0
⎡ ZG0 [Z SG ] = [ A]−1 [Z PG ][ A] = ⎢⎢ 0 ⎣⎢ 0
1
0
Z G1 0
2
0 ⎤0 ⎥ 0 1 uncoupled ⎥ Z G 2 ⎥⎦ 2
Where: ZG0 = ZK + ZR + ZQ ZG1 = ZK + a2 ZR + a ZQ ZG2 = ZK + a ZR + a2 ZQ
88
Rotating Machine Sequence Networks Generator sequence circuits are uncoupled I0 ZG0 3Zn
V0
+ EG1
ZG1
I1 V1
ZG2
I2
Generator Terminal Voltages
V2 89
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)
90
Example 8.6 - [ EP ] +
[ IP ]
Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ
Unbalanced Source
⎡ 277 / 0o ⎤ a ⎢ o⎥ [EP ] = ⎢260/ − 120 ⎥ b ⎢ 295/ 115o ⎥ c ⎣ ⎦
Find phase Currents [ IP ]
ZY =
ZΔ
3
= 10 / 40 o Ω = 7.66 + j 6.43Ω
Z L = 1/ 85 o Ω = .087 + j .996Ω Z 0 = Z 1 = Z 2 = Z Y + Z L = 7.747 + j 7.426 = 10.72 / 43.7 o Ω
91
Example 8.6 - [ EP ] + [ IP ] ⎡ 277 / 0o ⎤ ⎢ o⎥ [EP ] = ⎢260/ − 120 ⎥ ⎢ 295/ 115o ⎥ ⎣ ⎦ ⎡1 1 [E S ] = [ A]−1 [E P ] = 1 ⎢⎢1 a 3 ⎢⎣1 a 2
Z L = 1.0 / 85o Ώ Load Z∆ = 30 / 40o Ώ
1 ⎤ ⎡ 277 / 0 o ⎤ ⎡ 15.91/ 62.1o ⎤ 0 ⎢ 2 ⎥⎢ o⎥ o⎥ a ⎢ 260 / − 120 ⎥ = ⎢ 277.1/ − 1.77 ⎥ 1 ⎥ a ⎥⎦ ⎢⎣ 295 / 115 o ⎥⎦ ⎢⎣ 9.22 / 216.6 o ⎥⎦ 2
92
I0
10.72 /43.7o Ώ
+ 15.91 /62.1o
Example 8.6
I1 10.72 /43.7o Ώ
+ 277.1 /1.77o
I2 +
9.22 /216.6o
I0 = 0
-
10.72 /43.7o Ώ
277 / − 1.77 o I1 = 10.72 / 43.7 o
I 1 = 25.84 / − 45.5 o A
9.22 / 216.6 o I2 = 10.72 / 43.7 o I 2 = 0.86 / 172.9 o A 93
Example 8.6
0 ⎡ ⎤0 o⎥ ⎢ [I S ] = ⎢25.84/ − 45.5 ⎥ 1 o ⎢⎣ 0.86 / 172.9 ⎥⎦ 2
Amps
⎡ 25.17 / − 46.7 o ⎤ a ⎢ o ⎥ [I P ] = [ A][I S ] = ⎢ 25.72/ 196.4 ⎥ b ⎢⎣ 26.64 / 73.8 o ⎥⎦ c
Amps
How would you do problem without Symmetrical Components? 94
Transformer Connections for Zero Sequence P
Q
P
Q
Ic
Ia
Ib
I C
IB
IA
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 No zero sequence Q0 P0 Z0 current flow through 95 transformer n 0
Transformer Connections for Zero Sequence P
Q
P
Q
Ic
Ia
Ib
IC
IB
IA
Ia + Ib + Ic is not necessarily 0 and IA + IB + IC is not necessarily. Therefore I0 is not necessarily 0, P0
I0
I0 can flow through the transformer. Q0 Z0 n0
96
Transformer Connections for Zero Sequence P
Q
P
IA Q Ic
P0
Ib
Ic/n
IB
Ic/n
IC
Ia Ib/n
Ia + Ib + Ic is not necessarily 0 and Ia/n + Ib/n + Ic/n is not necessarily 0 but IA + IB + IC = 0 I0 Q0 Provides a zero sequence Z0 current source n0 97
Transformer Connections for Zero Sequence P
Q
P
IA Q Ic/n Ic
Ib
Ia
Ib/n
IB
I Ic/ n C Ia + Ib + Ic = 0 Ia/n + Ib/n + Ic/n is not necessarily 0, but IA + IB + IC = 0 No zero sequence current flow P0 Q0 Z0 n0
98
Transformer Connections for Zero Sequence P P
∆
Ia Ib Ic
IA
Q
IB IC
Ia + Ib + Ic = 0 P0
∆
Q
IA + IB + IC = 0 No zero sequence current flow Q0
Z0 n0
99
Power In Sequence Networks For 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
[S P ] = [E A
EB
⎡ I A *⎤ E C ]⎢ I B *⎥ ⎢ ⎥ ⎣⎢ I C *⎥⎦ 100
Power in Sequence Networks From our previous definitions: [SP] = [EP]T [IP]* (1) If we want the apparent power in the symmetrical component reference frame, we can substitute the following: [EP] = [A][ES] [EP]T =[ES]T [A]T
[IP] = [A][IS] [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* 101
Class Problem 2 One line of a three-phase generator is open circuited, while the other two are shortcircuited 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
102
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
103
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
104
Unsymmetrical Faults
105
Phase and Symmetrical Component Relationship Phase Reference Frame IA IB IC V
VB
V A
n C Symmetrical Components Reference Frame n0 n1 n2
I0 I1 I2
V0 V1 V2
106
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 107
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
108
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 109
Faulted 3-Phase Systems To represent the dissymmetry we only need to identify 2 points in the system: fault point and neutral point: f0 EF0
IF0 Zero System
f1
f2
IF1 EF1
Positive System
IF2 EF2
Negative System
n2 n0 n1 The sequence networks are connected together from knowledge of the type of fault and fault impedance Example 9.1
110
Example 9.1 Bus 1
∆
Bus 2
∆
X1=X2 =20Ώ
G
M
AC
100MVA 13.8kV X”=0.15pu X2 = 0.17pu X0 =0.05pu
100MVA 13.8:138kV X = 0.1pu
X0 = 60Ώ
.
AC
100MVA 138:13.8kV X = 0.1pu
Prefault Voltage = 1.05 pu
.
100MVA 13.8kV X”=0.20pu X2 = 0.21pu X0 =0.05pu Xn = 0.05pu
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: ZB =
(138)2 100
190.4Ω
Z1 = Z 2 =
j 60 j 20 = j 0.315 pu = j 0.105 pu Z 0 = 190.4 190.4 111
Example 9.1 1 j.05
2
J0.315
AC
J0.1
.
J0.1 . AC
J0.1
j.15
AC
1 j.15
J0.1
J0.105
2
J0.1
AC
+ AC
-
+
.
1.05 / 1
j.17
n0
1.05 / 0o
0o J0.1
n1
J0.105
J0.1
. J0.2 AC
AC
-
2 . J0.21
AC
AC
. AC
AC
112
n2
Example 9.1 Reduce the sequence networks to their thevenin equivalents as viewed from Bus 2 1 j.05
AC
.
2
J0.315 J0.1
J0.1 . AC
J0.1 AC
j.15
n 0
Zero Sequence Thevenin Equivalent from Bus 2 f0 J0.25 n0
113
Example 9.1 1 j.15
J0.1
J0.105
J0.1
2
AC
+ AC
-
+
.
1.05 /
0o
1.05 / 0o
. J0.2 AC
AC
-
n1 Positive Sequence Thevenin Equivalent from Bus 2 f1
Z thev
j (.455)(.2) = = j .139 .655
J0.139 + 1.05 / 0 o n1 114
Example 9.1 1 j.17
J0.1
J0.105
J0.1
2 . J0.21
AC
AC
. AC
AC
n2 Negative Sequence Thevenin Equivalent from Bus 2 f2
Z thev
j (.475)(.21) = = j .146 .685
J0.146 n2 115
Single Line-to-Ground Fault A B C IF Z EF An F A
⎡ I FA ⎤ [I FP ] = ⎢⎢ 0 ⎥⎥ ⎢⎣ 0 ⎥⎦
IF B
IFC ⎡1 1 [I FS ] = [ A]−1 [I FP ] = 1 ⎢⎢1 a 3 ⎢⎣1 a 2
EFA = IFA ZF EF0 + EF1 + EF2 = (IF0 + IF1 + IF2) ZF
1 ⎤ ⎡ I FA ⎤ ⎡ I FA ⎤ 1 a 2 ⎥ ⎢ 0 ⎥ = ⎢ I FA ⎥ ⎥⎢ ⎥ 3⎢ ⎥ a ⎥⎦ ⎣⎢ 0 ⎥⎦ ⎣⎢ I FA ⎥⎦
IF0 = IF1 = IF2 EF0 + EF1 + EF2 = 3IF0 ZF
116
Single Line to Ground Fault 3ZF f0 IF0
f1 IF1
EF Zero 0 System n0
f2 IF2
EF1 Positive System n1
EF2 Negative System n2
117
Single Line to Ground Fault f0
EF0
EF1
EF2
IF0
Zero System n0 IF1 f1 Positive System n1 IF2 f2 Negative System
3 ZF
n2 118
Example 9.3 For the system of Example 9.1 there is a bolted SingleLine-to-Ground fault at Bus 2. Find the fault currents in each phase and the phase voltages at the fault point. f1 f0 IF0
IF1
J0.25 n0
IF 0 = IF1 = IF 2
J0.139 + 1.05 / 0 o n1
f2 IF2
J0.146 n2
1.05 / 0 o = = − j1.96 j .25 + j .139 + j .146
119
Example 9.3 IF0 = IF1 = IF2 = -j1.96 f1 f0 EF0
J0.25 n0
EF1
J0.139 + 1.05 / 0 o n1
f2 EF2
J0.146 n2
VF 0 = − ( − j1.96)( j .25) = −.491 pu VF 1 = 1.05 − ( −1.96)( j .139) = .777 pu
VF 2 = − ( − j1.96)( j .146) = −.286 pu
120
Example 9.3
[I FP ] = [ A][I FS ] ⎡ I FA ⎤ ⎡1 1 ⎢ I ⎥ = ⎢1 a 2 ⎢ FB ⎥ ⎢ ⎢⎣ I FC ⎥⎦ ⎢⎣1 a
1 ⎤ ⎡ − j1.96⎤ ⎡ − j 5.88 pu⎤ a ⎥ 0 a ⎥ ⎢ − j1.96⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥b 0 a 2 ⎥⎦ ⎢⎣ − j1.96⎥⎦ ⎢⎣ ⎥⎦ c
[E FP ] = [ A][E FS ] ⎡ E FA ⎤ ⎡1 1 ⎢ E ⎥ = ⎢1 a 2 ⎢ FB ⎥ ⎢ ⎢⎣ E FC ⎥⎦ ⎢⎣1 a
0 1 ⎤ ⎡ − .491⎤ ⎡ ⎤a a ⎥ ⎢ .777 ⎥ = ⎢ 1.179 / 231o pu ⎥ b ⎥⎢ ⎥ ⎢ ⎥ 2 o a ⎥⎦ ⎢⎣ − .286⎥⎦ ⎢⎣1.179 / 128.7 pu⎥⎦ c
Note: Unfaulted phase voltages are higher than the source voltage. 121
Example 9.3a Bus 1
∆
G
IL
Bus 2
∆
M
AC
AC
IF ⎡ I F 0 ⎤ ⎡ − j1.96⎤ ⎢ I ⎥ = ⎢ − j1.96⎥ ⎢ F1 ⎥ ⎢ ⎥ ⎢⎣ I F 2 ⎥⎦ ⎢⎣ − j1.96⎥⎦
.
⎡ I FA ⎤ ⎡ − j 5.88 pu⎤ ⎢I ⎥ = ⎢ ⎥ 0 ⎢ FB ⎥ ⎢ ⎥ 0 ⎢⎣ I FC ⎥⎦ ⎢⎣ ⎥⎦
.
SLG Fault .
Find fault current in the transmission line, I L 1) Find ILS 2) Find ILP 122
Zero Sequence 1 j.05
AC
.
J0.315 J0.1
I L0 = 0 J0.1
2(f0) -j1.96 AC
J0.1 . AC
j.15
n 0
I L0 =0
123
Positive Sequence 1 j.15
J0.1
J0.105
e j30 : 1
AC
+ AC
-
.
1.05 /
j.455 I T1
2 f1
j .2 -j1.96
n1
.2 = ( − j1.96) = − j .6 .655
I L1 e j30 : 1
2
SLG
1.05 / 0o
0o n1
IT 1
J0.1
+
. J0.2 AC
AC
-
I T1 f1 -j1.96 n1
I L1 = 0.6 / − 60
o
124
Negative Sequence 1 j.17
J0.1
J0.105
J0.1
e -j30 : 1
AC
2
SLG
. J0.21 AC
. AC
n2 j.475 2 I T2
j .2
f2 -j1.96 n2
IT 2
.21 = ( − j1.96) = − j .6 .685
I L2
I T2
e -j30 : 1
f2 -j1.96 n2
I L 2 = 0.6/ − 120o 125
Example 9.3a Bus 1
∆
G
IL
Bus 2
∆
M
AC
AC
IF
0 . ⎤0 ⎡ [I LS ] = ⎢⎢ .6/ − 60 o ⎥⎥ 1 ⎢⎣.6 / − 120 o ⎥⎦ 2
⎡1 1 [I PL ] = [ A][I PS ] = ⎢⎢1 a 2 ⎢⎣1 a
.
SLG Fault
0 1 ⎤⎡ ⎤ ⎡ − j1.039 pu⎤ a ⎥ a ⎥ ⎢ .6 / − 60 o ⎥ = ⎢ 0 ⎥b ⎥ ⎢ ⎥⎢ a 2 ⎥⎦ ⎢⎣.6 / − 120 o ⎥⎦ ⎢⎣ j1.039 pu ⎥⎦ c 126
Line to Line Fault A B C IFA
⎡ 0 ⎤ [I FP ] = ⎢⎢ I FB ⎥⎥ ⎣⎢ − I FB ⎦⎥
IFB
EF nA
EF B
IFC EF
ZF
⎡1 1 [I FS ] = [ A]−1 [I FP ] = 1 ⎢⎢1 a 3 ⎢⎣1 a 2
C
E FA ⎡ ⎤ ⎥ [E FP ] = ⎢⎢ E FB ⎥ ⎢⎣ E FB − I FB Z F ⎥⎦ ⎡1 1 [EFS ] = 1 ⎢⎢1 a 3 ⎢⎣1 a 2
0 1 ⎤⎡ 0 ⎤ ⎤ ⎡ 1⎢ 2 ⎥⎢ ⎥ a I FB = j 3 I FB ⎥ ⎥ ⎥⎢ ⎥ 3⎢ a ⎥⎦ ⎢⎣ − I FB ⎥⎦ ⎢⎣ − j 3 I FB ⎥⎦
IF0 = 0
IF1 = IF2
I F 1 = j 3 I FB so I FB =
EFA 1 ⎤⎡ ⎤ ⎡ EFA + 2 EFB − I FB Z F ⎤ ⎥ = 1 ⎢E − E − a2 I Z ⎥ a 2 ⎥⎥ ⎢⎢ EFB FB FB F ⎥ ⎥ 3 ⎢ FA ⎢⎣ EFA − EFB − aI FB Z F ⎥⎦ a ⎥⎦ ⎢⎣ EFB − I FB Z F ⎥⎦
a2 − a EF1 − EF 2 = − IFBZF = −(− j 3)IFBZF = j 3IFBZ F= IF1 ZF 3
EF1 = EF2 + IF1ZF
0 1 2
127
IF1 j 3
Line to Line Fault ZF
f0
EF0
IF0 Zero System n0
IF1
f1
EF1 Positive System n1
IF2
f2
EF Negative System 2 n2
128
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. f1 f2 f0 IF1 J0.139 IF1 + J0.146 E =E IF0 J0.25 E F1 F2 F0 1.05 / 0 o n2 n0 n1 1.05 / 0 o IF1 = − IF 2 = = − j 3.69 pu IF0 = 0 j .139 + j .146
E F 1 = E F 2 = − I F 2 ( j .146 ) = (− j 3.69 )( j .146 ) = 0.537 pu
EF 0 = 0
129
Example 9.4 ⎡ I FA ⎤ ⎡1 1 ⎢ I ⎥ = ⎢1 a 2 ⎢ FB ⎥ ⎢ ⎢⎣ I FC ⎥⎦ ⎢⎣1 a
1 ⎤⎡ 0 ⎤ ⎡ 0 0 ⎤a ⎤ ⎡ a ⎥ ⎢ − j 3.69⎥ = ⎢ j 3 ( j 3.69) ⎥ = ⎢ − 6.39 pu⎥ b ⎥ ⎥ ⎢ ⎥ ⎢ ⎥⎢ 2 a ⎥⎦ ⎢⎣ j 3.69 ⎥⎦ ⎢⎣ − j 3 ( j 3.69)⎥⎦ ⎢⎣ 6.39 pu ⎥⎦ c
⎡ E FA ⎤ ⎡1 1 ⎢ E ⎥ = ⎢1 a 2 ⎢ FB ⎥ ⎢ ⎢⎣ E FC ⎥⎦ ⎢⎣1 a
1 ⎤ ⎡ 0 ⎤ ⎡ 1.07 pu ⎤ a a ⎥ ⎢.537 ⎥ = ⎢ − .537 pu⎥ b ⎥ ⎥⎢ ⎥ ⎢ 2 a ⎥⎦ ⎢⎣.537 ⎥⎦ ⎢⎣ − .537 pu⎥⎦ c
130
2 Line to Ground Fault A B C IFA EF nA
IFA = 0 = IF0 + IF1 + IF2 IFB EF B
IFC ZF
EF
E FA ⎡ E FA ⎤ ⎡ ⎤ [E FP ] = ⎢⎢ E FB ⎥⎥ = ⎢⎢( I FB + I FC )Z F ⎥⎥ ⎢⎣ E FC ⎥⎦ ⎢⎣( I FB + I FC )Z F ⎥⎦
C
Since IFA = 0, IFB + IFC = 3IF0 ⎡1 1 [EFS ] = 1 ⎢⎢1 a 3 ⎢⎣1 a 2
1 ⎤ ⎡ EFA ⎤ ⎡ EFA / 3 + 2 I F 0 Z F ⎤ a 2 ⎥ ⎢3I F 0 Z F ⎥ = ⎢ EFA / 3 − I F 0 Z F ⎥ ⎥⎢ ⎥ ⎥ ⎢ a ⎥⎦ ⎢⎣3I F 0 Z F ⎥⎦ ⎢⎣ EFA / 3 − I F 0 Z F ⎥⎦
0 1 2
EF0 – EF1 = 3 IF0 ZF so EF0 = EF1 + 3IF0 ZF and EF1 = EF2 131
2 Line to Ground Fault
3ZF f0
EF0
IF0 Zero System n0
IF1
f1
EF1 Positive System n1
IF2
f2
EF2 Negative System n2
132
Example 9.5 For the system of Example 9.1 there is a 2-line-toground 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 Bus 1
∆
IL
Bus 2
∆
G
M
AC
AC
IF .
.
2LG Fault 133
Example 9.5 f1 f0 IF0
J0.25 n0
IF1 =
f2 IF1
J0.139 + 1.05 / 0 o -
n1
IF2
J0.146 n2
1.05 = − j 4.547 pu (.146)(.25) j .139 + j .146 + .25
I F 0 = (− I F 1 )
.146 = j1.674 pu .146 + .25
I F 2 = − I F 0 − I F 1 = − j1.674 − ( − j 4.547) = j 2.873 pu
134
Example 9.5 ⎡1 1 [I FP ] = ⎢⎢1 a 2 ⎢⎣1 a This image cannot currently be display ed.
1 ⎤ ⎡ j1.674 ⎤ ⎡ 0 ⎤ a a ⎥ ⎢ − j 4.547 ⎥ = ⎢6.9 / 158.7 o pu⎥ b ⎥⎢ ⎥ ⎢ ⎥ 2 o a ⎥⎦ ⎢⎣ j 2.873 ⎥⎦ ⎢⎣ 6.9 / 21.3 pu ⎥⎦ c
I Fn = 3 I F 0 = ( 3)( j1.674) = j 5.02 pu
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Example 9.5
I GF 0 = 0
I MFO = I F 0 − I GF 0 = j1.674 − 0 = j1.674 pu
j.455 2 j .2 I T1 -j4.547 f1
I GF !
.2 = ( − j 4.547 ) = − j1.39 .655
n1
I MF 1 = I F 1 − I GF 1 = − j 4.547 − ( − j1.39) = − j 3.16 pu j.475
2 j .2
I T2
J2.87 f2 3 n2
I GF 2
.21 = ( j 2.8773) = j .88 .685
I MF 2 = I F 2 − I GF 2 = j 2.873 − j .88 = j1.993 pu 136
Example 9.5 ⎡1 1 [I GFP ] = ⎢⎢1 a 2 ⎢⎣1 a
1 ⎤ ⎡ 0 ⎤ ⎡ − j .51 pu ⎤ a ⎥ ⎢ − j1.39⎥ = ⎢1.98 / 172.6 o pu⎥ ⎥⎢ ⎥ ⎢ ⎥ a 2 ⎥⎦ ⎣⎢ j .88 ⎥⎦ ⎢⎣ 1.98 / 7.4 o pu ⎥⎦
⎡1 1 [I MFP ] = ⎢⎢1 a 2 ⎢⎣1 a
1 ⎤ ⎡ j1.674 ⎤ ⎡ − j .504 pu ⎤ o ⎥ ⎢ ⎥ ⎢ ⎥ a − j 3.16 = 5.0 / 153.1 pu ⎥⎢ ⎥ ⎢ ⎥ 2 o a ⎥⎦ ⎣⎢ j1.99 ⎥⎦ ⎢⎣ 5.0 / 26.9 pu ⎥⎦
137
Example 9.6 Find the fault current contribution from the generator considering the delta-wye transformer phase shift. Example 9.5 results 1 j.05
AC
I L0 = 0
J0.315
J0.1
.
J0.1
2 X J0.1 2LG . J1.674 j.1 5 AC
AC
n 0
-j1.39 1 j.15
AC
-
J0.105
e -j30 : 1
AC
+
J0.1
1.39/ -60o
AC
.
1.05 /
-j1.39 2 J0.1 X -j3.16 . j30 e :1 2LG J0.2 +
0o
1.05 / 0o AC
n1
138
Example 9.6 Example 9.5 results j.88 j.17
1
.88/ 60o
J0.1
J0.105
e j30 : 1
AC
j.88
e -j30 : 1
J0.1
2 X 2LG
.
AC
J1.99
AC
Bus 1
G
. J0.21
IGP
AC
⎡1 1 [I GP ] = ⎢⎢1 a 2 ⎢⎣1 a
∆
n2 IL
∆
Bus 2
X 2LG Fault 1 ⎤ ⎡ 0 ⎤ ⎡ − j .51 pu ⎤ a . o. ⎢ ⎥ ⎢ ⎥ a − j1.39 = 1.98 / 173 pu⎥ b ⎥ ⎥⎢ ⎥ ⎢ 2 o a ⎥⎦ ⎢⎣ j .88 ⎥⎦ ⎣⎢ 1.98 / 7 pu ⎥⎦ c
M AC
139
T1
Bus 1
Class Problem 5
Bus 3 T3
LINE 1-3
G1
G3
AC
∆
AC
LINE 1-2
Bus 2
LINE 2-3
T4
T2
∆
G4
G2
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
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
G4: X1=.34, X2=.45, X0=.085
T4: X=.15
a)
X0=.17 X0=.17
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.
140
Modern Fault Analysis Methods
141
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)
142
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.
143
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 Ferris and Miller breakers occurred. operated.
144
Modern Fault Analysis Example • Next the line relays (SEL-311L) at the two substations are interrogated for a possible event at this time.Ferris and Miller triggered an Reclosing enabled at Miller, event record at this time (HIS additional record is the uncleared command used in SEL relay) fault after reclosing.
• Use command EVE C 1 to capture the event you desire. The C gives you the digital elements as well as the analog quantities. 145
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.
146
Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller initial fault:
147
Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Unknown source voltage Ferris initial fault:
148
Modern Fault Analysis Example • Event capture file is opened in SEL-5601 to view waveforms and digital elements of event. Miller reclose operation:
149
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?
150
Questions
151