theory lectures

1

Hands-On Relay School

Jon F. DaumeBonneville Power Administration

March 14-15, 2011

Theory TrackTransmission Protection Theory

Symmetrical Components & Fault Calculations

2

Class Outline

Power system troublesSymmetrical componentsPer unit system Electrical equipment impedancesSequence networksFault calculations

3

Power System ProblemsFaultsEquipment troubleSystem disturbances

4

Fault CausesLightningWind and iceVandalismContaminationExternal forces

Cars, tractors, balloons, airplanes, trees, critters, flying saucers, etc.

Equipment failuresSystem disturbances

Overloads, system swings

6

Fault TypesOne line to ground (most common)Three phase (rare but most severe)Phase to phasePhase to phase to ground

7

Symmetrical Components

8

Balanced & Unbalanced Systems

Balanced System:3 Phase load3 Phase fault

Unbalanced System:Phase to phase faultOne line to ground

faultPhase to phase to

ground faultOpen pole or

conductorUnbalanced load

9

Balanced & Unbalanced Systems

A

C

BBalancedSystem

A

C

BUnbalanced System

10

Sequence Currents for Unbalanced Network

Ia2

Ic2Ib2

Negative Sequence

Ic0Ib0

Ia0

Zero Sequence

Ia1

Ic1

Ib1

Positive Sequence

11

Sequence Quantities

Condition + - 03 Phase load - -3 Phase fault - -Phase to phase fault -One line to ground faultTwo phase to ground faultOpen pole or conductorUnbalanced 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 Operatora = -0.5 + j √3= 1 ∠ 120°

2a2 = -0.5 – j √3= 1 ∠ 240°

2

1 + a + a2 = 0

1

a

a 2

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)/3Ia1 = (IA + aIB + a2IC)/3Ia2 = (IA + a2IB + aIC)/3

Voltages:Va0 = (VA + VB + VC)/3Va1 = (VA + aVB + a2VC)/3Va2 = (VA + a2VB + aVC)/3

16

Zero Sequence Filter3Ia0 = Ig = Ir = IA + IB + ICand: 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 FilterSome 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

ExampleIA = 3 + j4IB = -7 - j2IC = -2 + j7

+j

-j

IA = 3+j4

IB = -7-j2

IC = -2+j7

21

Zero SequenceIa0 = (IA + IB + IC)/3= [(3+j4)+(-7-j2)+(-2+j7)]/3= -2 + j3 = 3.61 ∠ 124°

Ia0 = Ib0 = Ic0

Ic0Ib0

Ia0

Zero Sequence

22

Positive SequenceIa1 = (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 SequenceIa2 = (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

Ic2Ib2

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 TypeFault Type + - 03 Phase fault - -Phase to phase fault -One line to ground faultTwo phase to ground fault

30

Per Unit

31

Per UnitPer unit values are commonly used for fault

calculations and fault study programsPer unit values convert real quantities to

values based upon number 1Per unit values include voltages, currents and

impedances Calculations are easier

Ignore voltage changes due to transformers Ohms law still works

32

Per UnitConvert equipment impedances into per unit

valuesTransformer and generator impedances are

given in per cent (%)Line impedances are calculated in ohmsThese impedances are converted to per unit

ohms impedance

33

Base kVA or MVAArbitrarily selectedAll values converted to common KVA or MVA

Base100 MVA base is most often usedGenerator or transformer MVA rating may be

used for the base

34

Base kVUse nominal equipment or line voltages

765 kV 525 kV345 kV 230 kV169 kV 138 kV115 kV 69 kV34.5 kV 13.8 kV12.5 kV etc.

35

Base Ohms, AmpsBase 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 Amps525 2756.3 110.0345 1190.3 167.3230 529.0 251.0115 132.3 502.069 47.6 836.7

34.5 11.9 1673.513.8 1.9 4183.712.5 1.6 4618.8

37

ConversionsPercent 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 Ω3φ MVA rating

38

Ohms to Per Unitpu Ohms = ohms / base ohmsbase MVA x ohms = pu Ω @ base MVA

kV2LL

39

Per Unit to Real StuffAmps = pu amps x base amps

kV = pu kV x base kVOhms = pu ohms x base ohms

40

Converting Between Bases

Znew = Zold x base MVAnew x kV2old

base MVAold kV2new

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 MachinesMachine values:

Subtransient reactance (X"d) Transient reactance (X'd) Synchronous reactance (Xd) Negative sequence reactance (X2)Zero sequence reactance (X0)

44

Synchronous MachinesMachine neutral ground impedance: Usually

expressed in ohmsUse 3R or 3X for fault calculations

Calculations generally ignores resistance values for generators

Calculations generally uses X”d for all impedance values

45

Generator ExampleMachine nameplate values:

250 MVA, 13.8 kVX"d = 25% @ 250 MVAX'd = 30% @ 250 MVAXd = 185% @ 250 MVAX2 = 25% @ 250 MVAX0 = 10% @ 250 MVA

46

Generator ExampleConvert machine reactances to per unit @

common MVA base, (100):X"d = 25% / 250 = 0.1 puX'd = 30% / 250 = 0.12 puXd = 185% / 250 = 0.74 puX2 = 25% / 250 = 0.1 puX0 = 10% / 250 = 0.04 pu

base MVA x % Z of equipment = Z pu Ω @ base MVA3φ MVA rating 100

47

Generator Example

~R1 jX1” = 0.1

R0 jX0 = 0.04

R2 jX2 = 0.1

48

TransformersZx X

Ze

ZhH 1:N

Vh

Ih

ZhxH X

Equivalent Transformer - Impedance in %

Zhx Ω = Vh /Ih = Zh + Zx /N2

Zhx % = Vh /Ih x MVA/kV2 x 100

49

TransformersImpedances in % of the transformer MVA

ratingConvert from circuit voltage to tap voltage:

%Xtap = %Xcircuit kV2circuit

kV2tap

50

TransformersConvert to common base MVA:%X @ base MVA =base MVA x %X of Transformer

MVA of Measurement

%X of Transformer = pu X @ 100 MVAMVA of MeasurementX1 = X2 = X0 unless a special value is given for

X0

51

Transformer Example250 MVA Transformer13.8 kV Δ- 230 kV Yg10% Impedance @ 250 MVAX = 10% = 0.04 pu @ 100 MVA

250X1 = X2 = X0 = XAssume R1, R2, R0 = 0

52

Transformer ExampleR1 jX1 = 0.04

R0 jX0 = 0.04

R2 jX2 = 0.04

Zero sequence connection depends upon winding configuration.

53

Transformer Connections

Winding Connection Sequence NetworkConnections

Z1, Z2 Z0

Z1, Z2 Z0

54

Transformer ConnectionsWinding 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 TransformerIa = 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 ConnectionsA YG / YG connection provides a series

connection for zero sequence currentA Δ / YG connection provides a zero sequence

(I0) current source for the YG windingAuto transformer provides same connection as

YG / YG connectionUse 3R or 3X if a Y is connected to ground

with a resistor or reactor

58

Three Winding TransformerImpedances ZHL, ZHM, & ZML given in % at

corresponding winding ratingConvert impedances to common base MVACalculate corresponding “T” network

impedances:ZH = (ZHL+ ZHM - ZML)/2

ZM = (- ZHL+ ZHM + ZML)/2 ZL = (ZHL- ZHM + ZML)/2

59

“T” NetworkCalculate 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 Example230 kV YG/115 kV YG/13.2 kV ΔNameplate Impedances ZHL= 5.0% @ 50 MVAZHM = 5.75% @ 250 MVAZML = 3.15% @ 50 MVA

61

Transformer ExampleConvert impedances to per unit @ common

MVA Base (100)ZHL= 5.0% @ 50 MVA = 5.0 / 50

= 0.10 puZHM = 5.75% @ 250 MVA = 5.75 / 250

= 0.023 puZML = 3.15% @ 50 MVA = 3.15 / 50

= 0.063 pu

62

Transformer ExampleConvert impedances to “T” network equivalentZH = (ZHL+ ZHM - ZML)/2

= (0.1 + 0.023 - 0.063)/2 = 0.03 puZM = (- ZHL+ ZHM + ZML)/2

= (-0.1 + 0.023 + 0.063)/2 = - 0.007 puZL = (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 DataMachine nameplate values:

300 MVA Nameplate rating X"d = 25% @ 300 MVAX'd = 30% @ 300 MVAXd = 200% @ 300 MVAX2 = 25% @ 300 MVAX0 = 10% @ 300 MVALeft generator: 13.8 kVRight generator: 115 kV

67

Problem - Transformer DataTwo winding transformer nameplate values

300 MVA Transformer13.8 kV Δ- 230 kV Yg10% Impedance @ 300 MVA

Three winding transformer nameplate values230 kV Yg/115 kV Yg/13.8 kV ΔZHL= 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/mileRa and Xa from conductor tables

Xd = 0.2794 f log GMD60

70

Positive & Negative Sequence Line Impedance

f = system frequencyGMDsep = 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 ImpedanceZ0 = Ra + Re +j 0.01397 f log De _______

3√(GMRcond GMDsep2)

or

Z0 = Ra + Re + j (Xa + Xe - 2Xd) Ω/mile

72

Zero Sequence Line ImpedanceRe = 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 LinesRa j(Xa+Xd)

Ra+Re j(Xa+Xe-2Xd)

Ra j(Xa+Xd)

Z1

Z2

Z0

74

Transmission Line Example230 kV Line50 Miles long1272 kcmil ACSR Pheasant Conductor

Ra = 0.0903 Ω /mile @ 80° CXa = 0.37201 Ω /mileGMR = 0.0466 feet

Structure: horizontal “H” frame

75

Transmission Line ExampleStructure “H” frame:

GMD = 3√(dabdbcdac) = 3√(23x23x46) = 28.978 feet

Xd = 0.2794 f log GMD60

= 0.2794 log 28.978 = 0.4085 Ω /mile

A CB

23 Feet 23 FeetJ6 Configuration

76

Transmission Line ExampleZ1 = 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 Basebase MVA x ohms = pu Ω @ base MVA

kV2LL

Z1 Line = (4.52 + j 39.03)100/2302

= 0.0085 + j 0.0743 pu

77

Transmission Line ExampleZ0 = 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 BaseZ0 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 LinesMutual impedance between lines

80

Mutual ImpedanceResult of coupling between parallel linesOnly affects Zero sequence networkWill affect ground fault magnitudesWill affect ground current flow in lines

Line #1

Line #2

3I0, Line #1

3I0, Line #2

81

Mutual ImpedanceZM = Re + j 0.838 log De Ω/mile

GMDcircuits

or ZM = Re + j (Xe − 3Xd circuits) Ω/mile

Re = 0.2862 @ 60 HzDe = 2160 √(ρ /f) = 2788 @ 60 HzXe = 2.89 for 100 meter ohms average

ground resistivity

82

Mutual ImpedanceGMDcircuits 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 Feet69 Feet

69 Feet

92 Feet115 Feet

138 Feet115 Feet

92 Feet

84

Mutual Impedance ExampleGMDcircuits = 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 ModelModel works with at least 1 common busZM 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 ExampleTaft

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

ProblemCalculate Z1 and Z0 pu impedances for a

transmission lineCalculate R1, Z1, R0 and Z0

Calculate Z1 and Z0 and the angles for Z1and 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 Data2 Parallel 230 kV Lines

60 Miles long1272 kcmil ACSR Pheasant conductorRa = 0.0903 Ω /mile @ 80° CXa = 0.37201 Ω /mileGMR = 0.0466 feetH frame structure - flat, 23 feet between

conductorsSpacing between circuits = 92 feet centerline to

centerline

91

Fault Calculations and Impedance Network

Connections

92

Why We Need Fault StudiesRelay coordination and settingsDetermine equipment ratingsDetermine effective grounding of systemSubstation ground mat designSubstation telephone protection

requirementsLocating faults

93

Fault StudiesFault Types:

3 PhaseOne line to groundPhase to phasePhase to phase to ground

Fault Locations:Bus faultLine end Line out fault (bus fault with line open)Intermediate faults on transmission line

94

Fault Study AssumptionsIgnore loadsUse generator X”d

Generator X2 equal X”d

Ignore generator resistanceIgnore transformer resistance0 Ω Fault resistance assumedNegative sequence impedance = positive

sequence impedance

95

Positive Sequence Network

Z1sl Z1tl Z1Ll Z1Lr Z1sr

Z1h Z1m

Z1lV1=1-I1Z1

+

Vl = 1 Vr = 1

I1

Fault

96

Negative Sequence Network


Z2h Z2m

Z2lV2= -I2Z2

+

I2

Fault

97

Zero Sequence Network


Z0h Z0m

Z0lV0= -I0Z0

+

I0

Fault

98

Network Reduction


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


Fault

100

Three Phase Fault

1PU

Z10.084

I1=11.9 I2=0 I0=0

V0

-

+

V2

-

+

V1

-

+

Z00.081

Z20.084

101

Three Phase Fault


Fault


Z1h Z1m

Z1lV1=1-I1Z1

+

Vl = 1 Vr = 1

Sequence Network Connection for 3 Phase Fault

I10.1 0.0370.04 0.037 0.1

0.03

0.07

-0.007

102

Three Phase Fault

Positive Sequence Network Reduced


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 FaultMVAFault = MVABase

ZFault puorI pu Fault current = 1 pu ESource

ZFault pu

105

Three Phase FaultI1 = E / Z1 = 1 / Z1

I2 = I0 = 0IA = I1 + I2 + I0 = I1IB = a2I1IC = aI1V1 = 1 – I1Z1 = 0V2 = 0, V0 = 0VA = VB = VC = 0

106

Phase to Phase Fault

Positive and negative sequence impedance networks connected in parallel

No zero sequence currents or voltages


Fault

107


1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

+

V2

-

+

V1

-

+

108



Z2h Z2m

Z2lV2= -I2Z2

+

I2 = -I1


Z1h Z1m

Z1lV1=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 FaultI1 = - I2 = E = ___1___ I0 = 0

(Z1 + Z2) (Z1 + Z2)IA = I0 + I1 + I2 = 0IB = I0 + a2I1 + aI2 = a2I1 - aI1IB = (a2 - a) E = _-j √3 E_ = -j 0.866 E

(Z1 + Z2) (Z1 + Z2) Z1

IC = - IB(assume Z1 = Z2)

111

Phase to Phase FaultV1 = E - I1Z1 = 1 - I1Z1

V2 = - I2Z2 = V1

V0 = 0VA = 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


Fault

113


1PU

Z1.084

I0=4.02

V0

-

+

V2

-

+

V1

-

+

Z2.084

Z0.081

I2=4.02I1=4.02

114



Z2h Z2m

Z2l

V2= -I2Z2+


Z0h Z0m

Z0lV0= -I0Z0

+


Z1h Z1m

Z1lV1=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 0.10.03

0.07

-0.007

0.04 0.1160.04 0.116 0.040.03

0.07

-0.007

0.1 0.0370.04 0.037 0.10.03

0.07

-0.007

115

Single Line to Ground Fault Vectors

Va

Vc

Vb Ia

116

Single Line to Ground FaultI1 = I2 = I0 = ____E_____ = ____1_____

(Z1 + Z2 + Z0) (Z1 + Z2 + Z0)IA = I1 + I2 + I0 = 3 I0IB = I0 + a2I1 + aI2 = I0 + a2I0 + aI0 = 0IC = 0

I Ground = I Residual = 3I0

117

Single Line to Ground FaultV1 = E - I1Z1 = 1 - I1Z1

V2 = - I2Z2

V0 = - I0Z0

VA = V1 + V2 + V0 = 0VB = 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


Fault

119


1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

+

V2

-

+

V1

-

+

120



Z2h Z2m

Z2lV2= -I2Z2

+


Z0h Z0m

Z0lV0= -I0Z0

+


Z1h Z1m

Z1lV1=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 ConditionsFault calculations and symmetrical

components can also be used to evaluate:Open pole or broken conductorUnbalanced loadsLoad included in fault analysisTransmission line fault location

For these other network conditions, refer to references.

123

ReferencesCircuit Analysis of AC Power Systems, Vol. 1 &

2, Edith ClarkeElectrical 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


Retired!

1

Hands-On Relay School


March 14-15, 2011

Theory TrackTransmission 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 TransferVs VrX

Power Transfer

0

0.5

1

0 30 60 90 120 150 180

Angle Delta

Tran

smitt

ed P

ower

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

6

Power Transfer During FaultsPower Transfer

0

0.2

0.4

0.6

0.8

1

1.2

0 30 60 90 120 150 180

Angle Delta

Tran

smitt

ed P

ower

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

Pow

er BP1

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 NumbersNumbers 1 - 97 used21 Distance relay25 Synchronizing or synchronism check

device27 Undervoltage relay32 Directional power relay43 Manual transfer or selector device46 Reverse or phase balance current relay50 Instantaneous overcurrent or rate of rise

relay (fixed time overcurrent)(IEEE C37.2)

10

51 AC time overcurrent relay52 AC circuit breaker59 Overvoltage relay62 Time delay stopping or opening relay63 Pressure switch67 AC directional overcurrent relay 79 AC reclosing relay81 Frequency relay86 Lock out relay87 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 WECC115 - 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. VoltageZ = √[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 CharacteristicX

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


– 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 ImpedanceBus 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


– 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 EncroachmentX

R

Zone 1

Zone 2

Zone 3

Load Consideration with Distance Relays

LoadArea

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 ResistanceX

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 CapsBypass 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 BlockingX

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 voltageZrelay = VLG/IL (Ok for 3 phase faults only)

• Phase to phase current and phase to phase voltageZrelay = VLL/ILL (Ok for 3 phase, PP, PPG

faults)• Phase current + compensated ground

current and phase to ground voltageZrelay = 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, KnKn = (Z0L - Z1L)/3Z1L

Z0L = Zero sequence transmission line impedanceZ1L = Positive sequence transmission line

impedanceIRelay = 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 overcurrentSelect TOC curve, usually very inversePickup, usually minimumTime 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 RelayProtective 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


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 upRelays 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


Block Xmit

Block Rcvd

Directional Comparison Blocking Scheme

Time DelayTime Delay

ForwardRelay

ReverseRelay

ReverseRelay

ForwardRelay

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

deionizet = 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 Relays1. Fault on line2. 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 FaultDetectorPCB Trip

Coil #1TD

Protective Relay

86

Trip

Block Close

TD

PCB TripCoil #2

BFR Retrip

BFR TimeDelay, 8~

102

Typical BFR Clearing TimesProper 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


retired

March 15, 2011

Transmission System Faults and Event AnalysisFault Analysis Theory

andModern Fault Analysis Methods

Presented by:Matthew Rhodes

Electrical Engineer, SRP1

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

FaultsShunt faults:Three phase a

bc

Line to line

Line to ground

2 Line to ground

ba

c

abc

abc

5

Faults

Series faultsOne open phase:

abc

2 open phasesabc

Increased phase impedance

Z abc

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=0and generator initially open circuited:

dtdiLRite +=)(

dtdiLRitVSin +=+ )(2 αω

by using Laplace transforms i(t) can be found

(L is considered constant)

9

Symmetrical Faults]/)()([2)( TteSintSin

ZVti −−−−+= θαθαω

2222 )( XRLRZ +=+= ω

RXTan

RLTan 11 −− ==

ωθ

Where:

RX

RLT

ω== Time Constant

]/)()([2)( TteSintSinacIti −−−−+= 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


o90== θαV2 e(t)

o90=α

11


0=αo90=θ

V2 e(t)

t = 0

iac02 acI

02 acI idc

12

iac02 acI

02 acI idc

022 acI

t

0=αo90=θ


)(ti

13

Symmetrical FaultsIac and Idc are independent after t = 0

22

dcIacIRMSI +=

TteacoIdcI −= 2

Substituting:

Tteac

ITt

eac

Iac

IRMS

I 221)222((max) −+=−

+=

]/)2/([2)( TtetSinZVti −+−+= πω

14

Asymmetry FactorIRMS(max) = K(τ) Iac

Asymmetry Factor = K(τ)

rxeK

τπτ

421)(−

+=

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

IFault 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 7.1a. RMSAC kAI 488.2

88.020)0(

22=

+=

b.438.121)5.0( )10

5.(4=+=

Π−eKKAImomentary 577.3)488.2)(438.1( ==

c.

023.121)3( )103(4

=+=Π−eK

KAI ngInterrupti 545.2)488.2)(023.1( ==

17

AC DecrementIn 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 generatorXL 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=0ET0

-

+E” = E’ = E = ET0

I=0

Unloaded Generator

21

XL XAR

-

+E” = E’ = E = ET0

I=0

t=0

E”Field Flux


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”


23

XL XAR’

-

+E” = E’ = E = ET0

I = I’

E’ = jI’(XL + XAR’)

t ≈ 3Cyc.

Field Flux

Resultant Field

ET = 0

I’


24

XL XAR

-

+E” = E’ = E = ET0

I = I

E’ = jI(XL + XAR)

t =∞

Field Flux

Resultant Field

ET = 0

I’


25

AC DecrementAs 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 DecrementFor a round rotor machine we only need to

consider the direct axis reactance.

dXEI

""2"2 = Subtransient

dXEI

''2'2 =

dXEI 22 =

Transient

Synchronous(steadystate)

27

AC Decrement

Can write the ac decrement equation[ ] ([ )])'()'"(2)( '" θαω −++−+−=

−− tSinIeIIeIItaci dTtdTt

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 −++−+−=−− tSinIeIIeIItaci dT

tdTt

For t = ∞

IIiac 2]00[2(max) =++=

"2])'()'"[(2(max) IIIIIIiac =+−+−=

29

ac and dc DecrementTransform ac decrement equation to phasor form

] θα −+−

−+−

−=⎢⎢⎣

⎡/')'(")'"(

_IdT

teIIdT

teIIacI

dc decrement equation:

AT

t

eSinIdcI−

−= )("2 θα

Where TA = Armature circuit time constant

(Example 7.2)

30

Example 7.2

IFault 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⎥⎦⎤

⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −= −−

1.11

1.11

24.1

24.1

15.105.1)( 2035. tt

AC eetI

2.max "2)(

t

DC eItI −= KAIBase 434.14

320500

==

kApudx

EI 1010.715.05.1

""" ==== a

DCI

2.2.max 9.9)7(2)(

tt

DC eetI −−== b

32

Example 7.2Part c: Find IRMS at interruption (3 cycles)

.sec05.0603==t

⎥⎦⎤

⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −= −−

1.11

1.11

24.1

24.1

15.105.1)( 205.035.

05.eetI AC

( )[ ] puI AC 92.4909.)975)(.258.3()24(.5.205.1)05(. =++=

pueI DC 71.79.9)05(. 2.05.

==−

kApuI RMS 132146.971.792.4)05(. 22 ==+= c

33

Example 7.2Part d: Find IMomentary(max) at t = ½ cycle

sec0083.605.==t

( )[ ] puI AC 43.6909.)996)(.258.3()79(.5.205.1)0083(. =++=

⎥⎦⎤

⎢⎣⎡ +⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −= −−

1.11

1.11

24.1

24.1

15.105.1)( 20083.035.

0083.eetI AC

pueI DC 5.99.9 2.0083.

==−

kApuI RMS 2159.145.943.6 22 ==+= d

34

TurbineGen.

Energy

35

Superposition for Fault Analysis

36

Superposition for Fault AnalysisNew representation:

IF1

IF2=0

Bus 1

Bus 1 Bus 2

IG = IG! + IG2 = IG1+ IL IM = IM1 – IL IF = IG1 + IM1Example 7.3

IG1IG2

ILIM1

IGIF IM

37

Example 7.3For 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

pujjjZ

VITH

FF 08.9

116.05.1

655.)505)(.15(.

05.1"1 −====

Part b: Neglecting load current, find Generator and motor fault current.

pujjIG 7655.505.08.9"1 −=−=

pujjjI M 08.2)7(08.9"1 −=−−−=

38

Example 7.3Part c: Including load current, find Generator and motor current during the fault period.

22*

*

18/952.05.118/1

0/05.195.cos/1

MGo

o

oLoad IIVSI −==−=

−=

−==

pujI ooG 83/35.718/953.7" −=−+−=

pujI ooM 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 . 305J 0 . 15

1 2

-VFIF

41

Z Bus MethodFor the circuit of Figure 7.4d (Slide 36 & 40)

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

22

12

21

11

2

1

EE

YY

YY

II

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

0For a fault at Bus 1

)( 1111 FIZE −=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

−=

1111

11 Z

VZEI F

F

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2221

1211

2

1

II

ZZZZ

EE

42

Z-Bus Method

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

−=

1111

11 Z

VZEI F

F

)( 1111 FIZE −=

111 Z

VI FF =

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 MethodFor N bus system, fault on Bus n

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

0.

000

...,...

.

.

.

.

.

321

321

33333231

22232221

11131211

3

2

1

Fn

NNNnNNN

nNnnnnn

Nn

Nn

Nn

N

N I

ZZZZZ

ZZZZZZZZZZZZZZZZZZZZ

E

EEEE

-VF

nn

FFn Z

VI = Where: VF = Pre-fault voltage at faulted bus Znn = Thevinen impedance

44

Z-Bus MethodAfter 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 7.4For the system of Figure 7.3 (Slide 40) using the Z-bus method find:a) Z busb) IF and I contribution from Line for Bus 1

faultc) IF and I contribution from Line for Bus 2

fault

Y20 = -j5Y10 = -j6.67

Y12 = -j3.28

1 2

IF

46

Example 7.4[ ] ⎥

⎦

⎤⎢⎣

⎡−

−=

95.928.328.395.9

jjjj

YBus

[ ] [ ] ⎥⎦

⎤⎢⎣

⎡== −

139.046.046.1156.1

jjjj

YZ busBus

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2

1

139.046.046.1156.

II

jjjj

EE

0

-IF

11 )1156.( IjE =

-VF -IF

08.91156.

" jjVI F

F −==

a

b

47

Example 7.4

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2

1

139.046.046.1156.

II

jjjj

EE

For fault at Bus 1: E1 = E11+ E1

2 = 0

E2 = E21 + E2

2 = VF + (j.046)I1E2 = 1.05 + (j.046)(j9.08) = .632 /0o

07.2305.

0632.21

1221 j

jZEEI −=

−=

−=

Find: Line current

b

48

Example 7.4

Y20 = -j5Y10 = -j6.67

Y12 = -j3.281 2

IFFind IF and I contribution from Line for Bus 2 fault

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2

1

139.046.046.1156.

II

jjjj

EE

-VF

pujj

IF 55.7139.05.1

2 −==

- I F2

oFF jjIjVE 0/703.)55.7)(046.(05.1))(046.(1 =+=−+=

pujjZ

EEI 3.2305.

0703.12

2112 −=

−=

−= 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

FnF Z

VII ==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


53


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


naa

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

ZYZY

Ib

Ia

Ic

56

Symmetrical ComponentsFor the General Case of 3 unbalanced voltages

VA

VB

VC6 degrees of freedom

Can define 3 sets of voltages designated as positive sequence, negative sequence and zero sequence

57

Symmetrical ComponentsCommon a operator identities

a =1/120o

a2 = 1/240o

a3 = 1/0o

a4 = 1/120o

1+a+a2 = 0

(a)(a2) = 1

58

Symmetrical ComponentsPositive 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 ComponentsNegative Sequence

120o

120o120o

VA2

VC2

VB2


a is operator 1/120o

VA2 = VA2 VB2 = aVA2 VC2 = a2 VA2

60

Symmetrical ComponentsZero Sequence


VA0VB0VC0

VA0 = VB0 = VC0

61

Symmetrical ComponentsReforming 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 ComponentsRewriting the sequence voltages in term of the Phase A sequence voltages:

VA = VA0 + VA1 +VA2VB = VA0 + a2 VA1 + aVA2VC = VA0 + aVA1 +a2 VA2

VA = V0 + V1 +V2VB = V0 + a2 V1 + aV2VC = 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 ComponentsWe 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 aa = 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

11

111

aaaaA

Step 1: Transpose

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2

2

11

111

aaaaA T

Step 2: Replace each element by its minor

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−−−

1111

22

22

222

aaaaaaaa

aaaaaa1

1

2

3

2 3

66

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−−−

1111

22

22

222

aaaaaaaa

aaaaaa1

1

2

3

2 3

Inverse of A

Step 3: Replace each element by its cofactor

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−−−

1111

22

22

222

aaaaaaaaaaaaaa1

1

2

3

2 3

67

Inverse of A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−−−

1111

22

22

222

aaaaaaaaaaaaaa1

1

2

3

2 3

Step 4: Divide by Determinant

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2

2

11

111

aaaaA

)(3)(1)(1)(1 2222 aaaaaaaaD −=−+−+−=

aaaa

aaa

aa

aaa

=−−

=−−

⎟⎠⎞

⎜⎝⎛=

−−

1111

2

2

2

2

2

2

22

11111 a

aaa

aaaa

==−−

⎟⎠⎞

⎜⎝⎛=

−−

68

Inverse of A

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

aaaaA

2

21

11

111

31

69

Symmetrical ComponentsPrevious relationships were developed for voltages. Same could be developed for currents such that:

IAIBIC

[IP] =I0I1I2

[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

IAIBIC

I0I1I2

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

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=aaV

o

o

o

P

277277

277

120/277120/2770/277

2

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= −

00/277

01

11

111

3277 2

2

21

2

1

0o

PS

aa

aaaaVA

VVV

V0

12

Find [VS] (Sequence voltages)

a

bc

72

Example 8.2Y connected load with reverse sequence

[ ] ( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

2

110

120/10120/10

0/10

aaI

o

o

o

P

a

bc

Find IS (Sequence Currents)

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

oPS

aa

aaaaIAI

0/10001

11

111

310

22

210

1

2

73

Example 8.3Ia = 10 / 0o

Ic = 10 /120o

Ib = o

In

[ ] [ ] [ ]PS IAIII

I 1

2

1

0−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

aaaaaI S 0

1

11

111

310

2

2

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+

+=

o

o

o

S

a

a

a

aI

60/33.30/67.6

60/33.32

310

12

1

310

2

2

0

1

2

on II 60/103 0 ==

a

bc

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

ZYZY

ZYZn

75


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

C

B

A

nYnn

nnYn

nnnY

C

B

A

III

ZZZZZZZZZZZZ

VVV

[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 forShunt Elements

[ZS] =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

2

2

2

222120

121110

020100

11

111

11

111

31

aaaa

ZZZZZZZZZZZZ

aaaa

ZZZZZZZZZ

nYnn

nnYn

nnnY

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +=

Y

Y

nY

S

ZZ

ZZZ

000003 0

1

2

0 1 2

77


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

1

0

0000003

III

ZZ

ZZ

VVV

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


We can form sequence circuits which represent the equations:

ZY

3 Zn

ZY

ZY

V0

V1

V2

I0

I1

I2

Zero sequence circuit Znonly in zero Sequence No neutral: Zn = infinity Solid ground: Zn = 0

Positive sequence circuit

Negative sequence circuit

79


Delta connected shunt element

ZYV0

V1

V2

I0

I1

I2

open

ZΔ/3

ZΔ/3

Sequence circuits.A

B

C

IA

IB

IC

ZΔ

ZΔ ZΔ

80


For the general case: [ZS] = [A]-1[ZP][A]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

2

2

2

222120

121110

020100

11

111

11

111

31

aaaa

ZZZZZZZZZ

aaaa

ZZZZZZZZZ

CCCBCA

BCBBBA

ACABAA

If there is symmetry: ZAA = ZBB = ZCC and ZAB = ZBC = ZCA we could perform multiplication and get:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

+=

ABAA

ABAA

ABAA

S

ZZZZ

ZZZ

0000002

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

III

ZZZZZZZZZ

VVVVVV

'

'

'

n n

Series Element Sequence Impedance

82

Series Element Sequence ImpedanceMatrices 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

ZZ

ZZS

83


The sequence circuits for series elements are:

Z0V0 V0’I0

o n0

Z1V1 V1’I1

o n1

Z2V2 V2’I2

o n2

84


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

ICIA

IB

Zn

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


[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

aEEa

EEPg

2EA= ERMS / 0 = E EB = ERMS /-120o = a2 E EC = ERMS /120o = a E

or

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

0

0

11

111

31 2

2

21 EaE

EaE

aaaaEAE PgSg

Therefore, only the positive sequence system has a generator voltage source.

0

12

abc

87


A

B

C

ZK

ZK

ZK

-

- -

+

+ +

EB

EAEC

ICIA

IB

Zn

ZAB

ZBC

ZCA

ZCB

ZBA

ZAC

Machine is not passive: Mutual Reactances: ZAB ≠ ZBA , etc.

ZAB = ZBC = ZCA = ZR

ZBA = ZCB = ZAC = ZQ

88


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++++++++

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

C

B

A

NKNQNR

NRNKNQ

NQNRNK

C

B

A

III

ZZZZZZZZZZZZZZZZZZ

EEE

[ ] [ ][ ]PPGPG IZE =

[ ] [ ] [ ][ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

2

1

01

000000

G

G

G

PGSG

ZZ

ZAZAZ

From the machine diagram we can write:

Where: ZG0 = ZK + ZR + ZQ

ZG1 = ZK + a2 ZR + a ZQ

ZG2 = ZK + a ZR + a2 ZQ

uncoupled0

12

0 1 2

89

Rotating Machine Sequence NetworksGenerator sequence circuits are uncoupled

3Zn

ZG0

I0

V0

EG1-

+ ZG1 I1V1

ZG2 I2V2

Generator Terminal Voltages

90


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

LoadZ∆ = 30 / 40o Ώ

Unbalanced Source

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=o

o

o

PE115/295120/2600/277 a

bc

Find phase Currents [ IP ]

Ω+=Ω== Δ 43.666.740/103 jZZ oY

Ω+=Ω= 996.087.85/1 jZ oL

Ω=+=+=== oLY jZZZZZ 7.43/72.10426.7747.7210

92

Example 8.6- [ EP ] +

[ IP ] Z L = 1.0 / 85o Ώ

LoadZ∆ = 30 / 40o Ώ[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=o

o

o

PE115/295120/2600/277

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

o

o

o

o

o

o

PS

aaaaEAE

6.216/22.977.1/1.2771.62/91.15

115/295120/2600/277

11

111

31

2

21012

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

I7.43/72.10

77.1/2771

−=

AI o5.45/84.251 −=

o

o

I7.43/72.106.216/22.9

2 =

AI o9.172/86.02 =

94

Example 8.6

[ ] [ ][ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −==

o

o

o

SP IAI8.73/64.264.196/72.257.46/17.25

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

o

oSI

9.172/86.05.45/84.25

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 IaIb

IC

IAIB

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

P0Q0Z0

n0

No zero sequence current flow through transformer

96


P Q

Ic IaIb

IC IAIB

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


P Q

Ic IaIb IC

IA

IB

P Q

Ia + Ib + Ic is not necessarily 0 and Ia/n + Ib/n + Ic/n is not necessarily 0

P0 Q0Z0

n0

Provides a zero sequence current source

Ib/nIc/n

Ic/n

I0but IA + IB + IC = 0

98


P Q

Ic IaIb 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 Q0Z0

n0

No zero sequence current flow

Ib/nIc/n

Ic/n

99


P Q

IC

IA

IB

P Q

Ia + Ib + Ic = 0 IA + IB + IC = 0

P0 Q0Z0

n0

No zero sequence current flow

∆ ∆Ia

IbIc

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

III

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 2One 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 3The 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 4The 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 FrameIAIBIC

nVC

VBVA

Symmetrical Components Reference Frame

I0I1

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 SystemsTo 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 n1n2

IF0IF1 IF2

EF0 EF1 EF2

The sequence networks are connected together from knowledge of the type of fault and fault impedanceExample 9.1

111

.

AC

Bus 1

AC

Bus 2X1=X2 =20Ώ

. .

∆ ∆

100MVA 13.8kV

X”=0.15puX2 = 0.17puX0 =0.05pu

100MVA 13.8:138kV X = 0.1pu

100MVA 138:13.8kV X = 0.1pu

100MVA 13.8kV

X”=0.20puX2 = 0.21puX0 =0.05puXn = 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

100138 2

BZ pujjZZ 105.04.190

2021 === pujjZ 315.0

4.19060

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.05J0.1

J0.315

J0.1J0.1.

1 2

j.15

n0

Example 9.1

113

Example 9.1Reduce the sequence networks to their

thevenin equivalents as viewed from Bus 2

AC AC

.AC

j.05J0.1

J0.315

J0.1J0.1.

1 2

j.15

n0Zero 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

n1Positive Sequence Thevenin Equivalent

from Bus 2

139.655.

)2)(.455(. jjZthev ==

f1

n1

J0.139+

-1.05 / 0 o

115

Example 9.1

Negative Sequence Thevenin Equivalent from Bus 2

146.685.

)21)(.475(. jjZthev ==

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

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

00FA

FP

II

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

FA

FA

FAFA

FPFS

IIII

aaaaIAI

31

00

11

111

31

2

21

IF0 = IF1 = IF2EFA = IFA ZF

EF0 + EF1 + EF2 = (IF0 + IF1 + IF2) ZF EF0 + EF1 + EF2 = 3IF0 ZF

ABCIFA

EF

A

IFB

IFC

nZF

117


Zero System

Positive System

Negative System

f0 f1 f2

n0n1 n2

IF0 IF1 IF2

EF

0

EF1 EF2

3ZF

118


Zero System

Positive System

Negative System

f0

f1

f2

n0

n1

n2

IF0

IF1

IF2

EF0

EF1

EF2

3 ZF

119

Example 9.3For 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

J0.146IF0IF2

IF1

96.1146.139.25.

0/05.1210 j

jjjIII

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 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0088.5

96.196.196.1

11

111

2

2

puj

jjj

aaaa

III

FC

FB

FA

[ ] [ ][ ]FSFP EAE =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

pupu

aaaa

EEE

o

o

FC

FB

FA

7.128/179.1231/179.10

286.777.491.

11

111

2

2

Note: Unfaulted phase voltages are higher than the source voltage.

abc

abc

122

.

Example 9.3a

Find fault current in the transmission line, I L

1) Find ILS

2) Find ILP

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0088.5 puj

III

FC

FB

FA

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

96.196.196.1

2

1

0

jjj

III

F

F

F. .

AC

Bus 1

AC

Bus 2∆ ∆G M

SLG Fault

IF

I L

123

Zero Sequence

AC AC

.AC

j.05J0.1

J0.315

J0.1J0.1.

1 2(f0)

j.15

n0

-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 -j1.96f1 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

aaaaIAI

o

oPSPL

039.10039.1

120/6.60/6.

0

11

111

2

2

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

o

oLSI

120/6.60/6.

0 012

abc

. .

AC

Bus 1

AC

Bus 2∆ ∆G M

SLG Fault

IF

I L

127

Line to Line Fault

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

FB

FBFP

III0

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== −

FB

FB

FB

FBFPFS

IjIj

II

aaaaIAI

330

31

0

11

111

31

2

21

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

FFBFB

FB

FA

FP

ZIEEE

E

n

ABC

IFA

EF

A

IFB IFC

EF

B

EF

CZF

IF0 = 0 IF1 = IF2

( ) FFFFBFFBFFBFF ZIZIjZIjZIaaEE 1

2

21 333

==−−=−

−=−

FBF IjI 31 = so 31

jII F

FB =

EF1 = EF2 + IF1ZF

012

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

FFBFBFA

FFBFBFA

FFBFBFA

FFBFB

FB

FA

FS

ZaIEEZIaEEZIEE

ZIEEE

aaaaE 2

2

2

2

31

11

111

31

128

Line to Line Fault

Zero System

Positive System

Negative System

f0 f1 f2

n0 n1 n2

IF0IF1 IF2

EF0 EF1 EF

2

ZF

129

Example 9.4For 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

IF1IF1

IF0

pujjj

IIo

FF 69.3146.139.

0/05.121 −=

+=−=00 =FI

( ) ( )( ) pujjjIEE FFF 537.0146.69.3146.221 =−=−==

EF1=EF2EF0

00 =FE

130

Example 9.4

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

pupu

jjjj

jj

aaaa

III

FC

FB

FA

39.639.60

)69.3(3)69.3(3

0

69.369.3

0

11

111

2

2abc

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

pupu

pu

aaaa

EEE

FC

FB

FA

537.537.07.1

537.537.0

11

111

2

2abc

131

2 Line to Ground Fault

[ ] ( )( ) ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

FFCFB

FFCFB

FA

FC

FB

FA

FP

ZIIZII

E

EEE

E

ABC

IFA

EF

A

IFB IFC

nEF

B

EF

CZF

IFA = 0 = IF0 + IF1 + IF2

Since IFA = 0, IFB + IFC = 3IF0

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

FFFA

FFFA

FFFA

FF

FF

FA

FS

ZIEZIEZIE

ZIZI

E

aaaaE

0

0

0

0

02

2

3/3/

23/

33

11

111

31

EF0 – EF1 = 3 IF0 ZF so EF0 = EF1 + 3IF0 ZF and EF1 = EF2

012

132

2 Line to Ground Fault

Zero System

Positive System

Negative System

f0 f1 f2

n0 n1 n2

IF0IF1 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

J0.146IF0IF2

IF1

pujjj

IF 547.4

25.146.)25)(.146(.139.

05.11 −=

++

=

pujII FF 674.125.146.

146.)( 10 =+

−=

pujjjIII FFF 873.2)547.4(674.1102 =−−−=−−=

135

Example 9.5

This image cannot currently be displayed.[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

pupu

jj

j

aaaaI

o

oFP

3.21/9.67.158/9.6

0

873.2547.4

674.1

11

111

2

2abc

pujjII FFn 02.5)674.1)(3(3 0 ===

136

Example 9.5

n1

j.455 j .22I T1 -j4.547

n2

j.475 j .22I T2 J2.87

3

00 =GFI pujjIII GFFMFO 674.10674.100 =−=−=

39.1655.2.)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

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

pupu

puj

jj

aaaaI

o

oGFP

4.7/98.16.172/98.1

51.

88.39.1

0

11

111

2

2

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

pupu

puj

jj

j

aaaaI

o

oMFP

9.26/0.51.153/0.5

504.

99.116.3

674.1

11

111

2

2

138

Example 9.5 results

AC AC

.AC

j.05J0.1

J0.315

J0.1J0.1.

1 2

j.15n

0

I L0 = 0

2LGJ1.674

X

AC AC

.

AC AC

j.15

-

+

J0.1J0.105 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 9.6Example 9.5 results

J1.99

AC AC

.

AC

j.17

J0.1J0.105 J0.1

J0.21.

1 2

n2

e -j30 : 1 2LGe j30 : 1 X

j.88.88/ 60oj.88

. .

AC

Bus 1

AC

Bus 2∆ ∆G M

2LG Fault

I L

X

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

pupu

puj

jj

aaaaI

o

oGP

7/98.1173/98.151.

88.39.1

0

11

111

2

2abc

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


view waveforms and digital elements of event. Ferris initial fault: Unknown

source voltage

149


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

theory lectures

Documents