Symmetrical Componet Fauly calculation

8/16/2019 Symmetrical Componet Fauly calculation

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Hands-On Relay School

Jon F. Daume

Bonneville Power Administration

March 14-15, 2011

Theory TrackTransmission Protection Theory

Symmetrical Components &Fault Calculations


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Class Outline

Power system troublesSymmetrical components

Per unit system

Electrical equipment impedances

Sequence networks

Fault calculations


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Power System ProblemsFaults

Equipment troubleSystem disturbances


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Fault CausesLightning

Wind and ice

VandalismContamination

External forces

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

Equipment failures

System disturbances

Overloads, system swings


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Fault TypesOne line to ground (most common)

Three phase (rare but most severe)

Phase to phase

Phase to phase to ground


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Symmetrical

Components


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Balanced & Unbalanced SystemsBalanced System:

3 Phase load3 Phase fault

Unbalanced System:

Phase to phase faultOne line to ground

fault

Phase to phase toground fault

Open pole orconductor

Unbalanced load


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Balanced & Unbalanced Systems

A

C

B

Balanced

System

A

C

BUnbalanced System


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Sequence Currents for

Unbalanced Network

Ia2

Ic2Ib2

Negative Sequence

Ic0Ib0

Ia0

Zero Sequence

Ia1

Ic1

Ib1

Positive Sequence


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Sequence QuantitiesCondition + - 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


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Phase Values From Sequence

ValuesCurrents:

I A = Ia0 + Ia1 + Ia2IB = Ib0 + Ib1 + Ib2

IC = Ic0 + Ic1 + Ic2

Voltages:

V A = Va0 + Va1 + Va2

VB = Vb0 + Vb1 + Vb2

VC = Vc0 + Vc1 + Vc2


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a Operator a = -0.5 + j √3= 1 ∠ 120°

2a2 = -0.5 – j √3= 1 ∠ 240°

2

1 + a + a2 = 0

1

a

a 2


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Phase Values From Sequence

ValuesCurrents:

I A = Ia0 + Ia1 + Ia2IB = Ia0 + a

2Ia1 + aIa2

IC = Ia0 + aIa1 + a2Ia2

Voltages:

V A = Va0 + Va1 + Va2

VB = Va0 + a2Va1 + aVa2

VC = Va0 + aVa1 + a2Va2


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Sequence Values From Phase

ValuesCurrents:

Ia0 = (I A + IB + IC)/3Ia1 = (I A + aIB + a

2IC)/3

Ia2 = (I A + a2IB + aIC)/3

Voltages:

Va0 = (V A + VB + VC)/3

Va1 = (V A + aVB + a2VC)/3

Va2 = (V A + a2VB + aVC)/3


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Zero Sequence Filter 3Ia0 = Ig = Ir = I A + IB + IC

and: 1 + a + a2

= 0

I A = Ia0 + Ia1 + Ia2

+IB = Ia0 + a2Ia1 + aIa2

+IC = Ia0 + aIa1 + a2Ia2

= Ig = 3Iao + 0 + 0


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Ia

Ic

Ib

3I0 = Ia + Ib + Ic

Zero Sequence Current Filter


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Zero Sequence Voltage Filter

3V0

3 VO Polarizing Potential

Ea Eb Ec


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Negative Sequence Filter Some protective relays are designed to

sense negative sequence currents and/or

voltages

Much more complicated than detecting zero

sequence valuesMost modern numerical relays have negative

sequence elements for fault detection

and/or directional control


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ExampleI A = 3 + j 4

IB = -7 - j 2IC = -2 + j 7

+ j

- j

IA = 3+ j 4

IB = -7- j 2

IC = -2+ j 7


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Zero SequenceIa0 = (I A + IB + IC)/3

= [(3+ j 4)+(-7- j 2)+(-2+ j 7)]/3= -2 + j 3 = 3.61 ∠ 124°

Ia0 = Ib0 = Ic0

Ic0Ib0

Ia0

Zero Sequence


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Positive SequenceIa1 = (I A + aIB + a

2IC)/3

= [(3+ j 4)+(-0.5+ j √3/2)(-7- j 2)+(-0.5- j √3/2)(-2+ j 7)]/3

= [(3+ j 4)+(5.23- j5 .06)+(7.06- j 1.77)]/3

= 5.10 - j 0.94 = 5.19 ∠ -10.5°

Ib1 is rotated -120º Ic1 is rotated +120º


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Positive Sequence

Ia1

Ic1

Ib1


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Negative SequenceIa2 = (I A + a

2IB + aIC)/3

= [(3+ j 4)+(-0.5- j √3/2)(-7- j 2)+(-0.5+ j √3/2)(-2+ j 7)]/3

= [(3+ j 4)+(1.77+ j 7.06)+(-5.06- j 5.23)]/3

= -0.1 + j 1.94 = 1.95 ∠ 92.9°

Ib2 is rotated +120º Ic2 is rotated -120º


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Negative Sequence

Ia2

Ic2Ib2


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Reconstruct Phase Currents

Ia

Ic

Ib

Ic1

Ib1

Ia1

Ib0

Ia0

Ic0

Ia2

Ib2

Ic2


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Positive, Negative, and ZeroSequence Impedance

Network Calculations for aFault Study


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+, -, 0 Sequence Networks

Simple 2 Source Power System Example

Fault

1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

+

V2

-

+

V1

-

+


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Impedance Networks & Fault TypeFault Type + - 0

3 Phase fault

- -Phase to phase fault -

One line to ground fault

Two phase to ground fault


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Per Unit


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Per UnitPer 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 andimpedances

Calculations are easier

Ignore voltage changes due to transformers

Ohms law still works


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Per Unit

Convert equipment impedances into per unit

values

Transformer and generator impedances are

given in per cent (%)

Line impedances are calculated in ohmsThese impedances are converted to per unit

ohms impedance


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Base kVA or MVA

Arbitrarily selected

All values converted to common KVA or MVABase

100 MVA base is most often used

Generator or transformer MVA rating may beused for the base


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Base kV

Use nominal equipment or line voltages

765 kV 525 kV345 kV 230 kV

169 kV 138 kV

115 kV 69 kV

34.5 kV 13.8 kV

12.5 kV etc.


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Base Ohms, Amps

Base ohms:

kV

2

1000 = kV

2

base kVA base MVA

Base amps:

base kVA = 1000 base MVA

√3 kV √3 kV


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Base Ohms, Amps (100 MVA Base)

kV Base Ohms Base Amps

525 2756.3 110.0

345 1190.3 167.3

230 529.0 251.0

115 132.3 502.0

69 47.6 836.7

34.5 11.9 1673.513.8 1.9 4183.7

12.5 1.6 4618.8


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Conversions

Percent to Per Unit:

base MVA x % Z of equipment3φ MVA rating 100

= Z pu Ω @ base MVA

If 100 MVA base is used:

% Z of equipment = Z pu Ω3φ MVA rating


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Ohms to Per Unit

pu Ohms = ohms / base ohms

base MVA x ohms = puΩ

@ base MVAkV2LL


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Per Unit to Real Stuff

Amps = pu amps x base amps

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


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Converting Between Bases

Znew = Zold x base MVAnew x kV2old

base MVAold kV2new

Evaluation of System


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Evaluation of System

ComponentsDetermine positive, negative, and zero

sequence impedances of various devices

(Z1, Z2, Z0)

Only machines will act as a voltage source in

the positive sequence networkConnect the various impedances into networks

according to topography of the system

Connect impedance networks for various fault

types or other system conditions


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Synchronous Machines

~

Machine values:Machine reactances given in % of the

machine KVA or MVA rating

Ground impedances given in ohms


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Machine values:

Subtransient reactance (X"d)

Transient reactance (X'd)

Synchronous reactance (Xd)

Negative sequence reactance (X2)

Zero sequence reactance (X0)


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Machine neutral ground impedance: Usually

expressed in ohms

Use 3R or 3X for fault calculations

Calculations generally ignores resistance

values for generatorsCalculations generally uses X”d for all

impedance values


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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


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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 puX2 = 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


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Generator Example

~R1 jX1” = 0.1

R0 jX0 = 0.04

R2 jX2 = 0.1


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Transformers

Zx X

Ze

ZhH 1:N

Vh

Ih

ZhxH

X

Equivalent Transformer - Impedance in %

ZhxΩ = Vh /Ih = Zh + Zx /N2

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


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Transformers

Impedances in % of the transformer MVA

rating

Convert from circuit voltage to tap voltage:

%Xtap = %Xcircuit kV2circuit

kV2tap

f


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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 MeasurementX1 = X2 = X0 unless a special value is given for

X0

T f E l


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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

T f E l


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Transformer ExampleR1 jX1 = 0.04

R0 jX0 = 0.04

R2 jX2 = 0.04

Zero sequence connection depends upon

winding configuration.


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T f C ti


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Transformer ConnectionsWinding Connection Sequence Network Connections

Z1, Z2 Z0

Z1, Z2 Z0

Z1, Z2 Z0

Z1, Z2 Z0

D lt W T f


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Delta Wye Transformer

A

B

C

Ia

Ic

Ib

IA

IB

IC

nIA

nIC

3I0 = IA+IB+IC

nIB


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Transformer Connections


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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 asYG / YG connection

Use 3R or 3X if a Y is connected to ground

with a resistor or reactor

Three Winding Transformer


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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

“T” Network


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T Network

Calculate corresponding “T” network

impedances:

ZH = (ZHL+ ZHM - ZML)/2

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

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

ZHM = ZH + ZMZML= ZM+ ZL

ZH ZM

ZL

Transformer Example


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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

Transformer Example


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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


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Transformer Example


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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


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Problem

Calculate pu impedances for generators and

transformers

Use 100 MVA baseIgnore all resistances

Problem


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Problem

Fault

13.8 kV 13.8 kV230 kV230 kV

115 kV

Problem - Generator Data


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Problem Generator Data

Machine nameplate values:

300 MVA Nameplate rating

X"d = 25% @ 300 MVA

X'd = 30% @ 300 MVA

Xd = 200% @ 300 MVAX2 = 25% @ 300 MVA

X0 = 10% @ 300 MVA

Left generator: 13.8 kVRight generator: 115 kV

Problem - Transformer Data


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Problem Transformer Data

Two winding transformer nameplate values

300 MVA Transformer

13.8 kV Δ- 230 kV Yg10% Impedance @ 300 MVA

Three winding transformer nameplate values

230 kV Yg/115 kV Yg/13.8 kV ΔZ

HL

= 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)

Transmission Lines


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Transmission Lines

R jX

Positive & Negative Sequence


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Line Impedance

Z1 = Z2 = Ra + j 0.2794 f log GMDsep60 GMRcond

or

Z1 = Ra + j (Xa + Xd) Ω/mile

Ra and Xa from conductor tables

Xd = 0.2794 f log GMD

60


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Positive & Negative SequenceLine Impedancef = system frequency

GMDsep = Geometric mean distance

between conductors = 3√(dabdbcdac) where

dab, dac, dbc = spacing between conductorsin feet

GMRcond = Geometric mean radius of

conductor in feetRa = conductor resistance, Ω/mile

Zero Sequence Line Impedance


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q p

Z0 = Ra + Re +

j 0.01397 f log De _______ 3√(GMRcond GMDsep2)

or

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

Zero Sequence Line Impedance


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q p

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.

Transmission Lines


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Ra j(Xa+Xd)

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

Ra j(Xa+Xd)

Z1

Z2

Z0

Transmission Line Example


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p

230 kV Line

50 Miles long

1272 kcmil ACSR Pheasant Conductor

Ra = 0.0903 Ω /mile @ 80° C

Xa = 0.37201 Ω /mileGMR = 0.0466 feet

Structure: horizontal “H” frame



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pStructure “H” frame:

GMD = 3√(dabdbcdac) =3√(23x23x46)

= 28.978 feet

Xd = 0.2794 f log GMD60

= 0.2794 log 28.978 = 0.4085Ω

/mile

A CB

23 Feet 23 FeetJ6 Configuration



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p

Z1 = Z2 = Ra + j (Xa + Xd)

= 0.0903 + j (0.372 + 0.4085)

= 0.0903 + j 0.781 Ω /mileZ1 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



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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 /mileZ0 Line = 50(0.377 + j 2.445)

= 18.83 + j 122.25Ω

= 123.69Ω

∠ 81.2 °

Per unit @ 230 kV, 100 MVA Base

Z0 Line = (18.83 + j 122.25)100/2302

= 0.0356 + j 0.2311 pu



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Z1

Z2

Z0

0.0085 j0.0743

0.0356 j0.2311

0.0085 j0.0743

Long Parallel Lines


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Mutual impedance between lines

Mutual Impedance


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Result of coupling between parallel lines

Only affects Zero sequence network

Will affect ground fault magnitudes

Will affect ground current flow in lines

Line #1

Line #2

3I0, Line #1

3I0, Line #2

Mutual Impedance


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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 averageground resistivity

Mutual Impedance


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GMDcircuits is the ninth root of all possible

distances between the six conductors,

approximately equal to center to centerspacing

GMDcircuits =

9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2)

Xd circuits = 0.2794 log GMDcircuits

Mutual Impedance Example


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A CB

23 Feet 23 Feet

A CB

23 Feet 23 Feet

Circuit #1 Circuit #2

46 Feet

46 Feet

92 Feet

69 Feet

69 Feet

92 Feet

115 Feet

138 Feet

115 Feet

92 Feet

Mutual Impedance Example


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GMDcircuits =9√(da1a2da1b2da1c2db1a2db1b2db1c2dc1a1dc1b2dc1c2) =

9√(92x115x138x69x92x115x46x69x92)= 87.84 feet

Xd circuits = 0.2794 log GMDcircuits

= 0.2794 log 87.84 = 0.5431 Ω/mileZM = Re + j (Xe − 3Xd circuits)

= 0.2862 + j (2.89 - 3x0.5431)

= 0.2862 + j 1.261 Ω/mile(Z 0 = 0.377 + j 2.445 Ω /mile)

Mutual Impedance Model


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Bus 1 Bus 2

Z0 Line 1

Z0 Line 2

ZM

Bus 1 Bus 2Z02 - ZM

Z01- ZM

ZM

Mutual Impedance Model


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Model works with at least 1 common bus

ZM Affects zero sequence network only

ZM For different line voltages:

pu Ohms = ohms x base MVAkV1 x kV2

Mutual impedance calculations and modelingbecome much more complicated with larger

systems


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Problem


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Calculate Z1 and Z0 pu impedances for a

transmission line

Calculate R1, Z1, R0 and Z0

Calculate Z1 and Z0 and the angles for Z1and Z0

Calculate Z0 mutual impedance between

transmission linesUse 100 MVA base and 230 kV base

Problem


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Fault

13.8 kV 13.8 kV230 kV230 kV

115 kV

Transmission Line Data


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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 feetH frame structure - flat, 23 feet between

conductors

Spacing between circuits = 92 feet centerline tocenterline


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Fault Calculations andImpedance Network

Connections

Why We Need Fault Studies


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Relay coordination and settings

Determine equipment ratings

Determine effective grounding of system

Substation ground mat design

Substation telephone protectionrequirements

Locating faults

Fault Studies


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Fault Types:

3 Phase

One line to ground

Phase to phase

Phase to phase to ground

Fault Locations:

Bus fault

Line endLine out fault (bus fault with line open)

Intermediate faults on transmission line

Fault Study Assumptions


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Ignore loads

Use generator X”d

Generator X2 equal X”d

Ignore generator resistance

Ignore transformer resistance0 Ω Fault resistance assumed

Negative sequence impedance = positivesequence impedance

Positive Sequence Network


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Z1sl Z1tl Z1Ll Z1Lr Z1sr

Z1h Z1m

Z1l

V1=1-I1Z1+

Vl = 1 Vr = 1

I1

Fault

Negative Sequence Network


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Z2h Z2m

Z2lV2= -I2Z2

+

I2

Fault

Zero Sequence Network


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Z0h Z0m

Z0l

V0= -I0Z0+

I0

Fault

Network Reduction


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Fault

1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

+

V2

-

+

V1

-

+

Three Phase Fault


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Only positive sequence impedance network

usedNo negative or zero sequence currents or

voltages


Fault

Three Phase Fault

+++


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1PU

Z1

0.084

I1=11.9 I2=0 I0=0

V0

-

+

V2

-

+

V1

-

+

Z0

0.081

Z2

0.084

Three Phase Fault


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Fault


Z1h Z1m

Z1lV1=1-I1Z1

+

Vl = 1 Vr = 1

Sequence Network Connection for 3 Phase Fault

I1

0.1 0.0370.04 0.037 0.1

0.03

0.07

-0.007

Three Phase Fault


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Positive Sequence Network Reduced


Fault

V1=1-I1Z1+

Vl = 1 Vr = 1

I1

0.177 0.160

Three Phase Fault VectorsIc


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Va

Vc

Vb

Ia

Ib

Three Phase Fault

MVA = MVA


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MVAFault = MVABase

ZFault pu

or

I pu Fault current = 1 pu ESource

ZFault pu

Three Phase Fault

I = E / Z = 1 / Z


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I1 = E / Z1 = 1 / Z1

I2 = I0 = 0

I A = I1 + I2 + I0 = I1

IB = a2I1

IC = aI1V1 = 1 – I1Z1 = 0

V2

= 0, V0

= 0

V A = VB = VC = 0

Phase to Phase Fault


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Positive and negative sequence impedance

networks connected in parallelNo zero sequence currents or voltages


Fault



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1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

+

V2

-

+

V1

-

+


Fault


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Z2h Z2m

Z2l

V2= -I2Z2+

I2 = -I1


Z1h Z1m

Z1l

V1=1-I1Z1

I1

+

Vl = 1 Vr = 1

Sequence Network Connection for Phase to Phase Fault

Fault

Phase to Phase Fault Vectors


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Va

Vc

Vb

Ic

Ib

Phase to Phase FaultI1 = - I2 = E = 1 I0 = 0


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I1 I2 E ___1___ I0 0

(Z1 + Z2) (Z1 + Z2)

I A = I0 + I1 + I2 = 0

IB = I0 + a2I1 + aI2 = a

2I1 - aI1

IB = (a2 - a) E = _- j √3 E_ = - j 0.866 E(Z1 + Z2) (Z1 + Z2) Z1

IC

= - IB

(assume Z 1 = Z 2 )

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


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V1 E I1Z1 1 I1Z1

V2 = - I2Z2 = V1

V0 = 0

V A = V1 + V2 + V0 = 2 V1

VB = V0 + a2

V1 + aV2 = a2

V1 + aV1 = -V1VC = -V1

Phase to phase fault = 86.6%3 phase fault

Single Line to Ground Fault


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Positive, negative and zero sequence

impedance networks connected in series


Fault


+++


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113

1PU

Z1

.084

I0=4.02

V0

-

+

V2

-

+

V1

-

+

Z2

.084

Z0

.081

I2=4.02I1=4.02


Z l Z tl Z Ll Z L ZZ1l

V1=1-I1Z1+

Vl = 1 Vr = 10.07


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Z2h Z2m

Z2l

V2= -I2Z2+


Z0h Z0m

Z0l

V0= -I0Z0+


Z1h Z1mI1

I2

I0

Sequence Network Connection for One Line to Ground Fault

I1 = I2 = I0

0.1 0.0370.04 0.037 0.1

0.03 -0.007

0.04 0.1160.04 0.116 0.04

0.03

0.07

-0.007

0.1 0.0370.04 0.037 0.10.03

0.07

-0.007

Single Line to Ground Fault VectorsVc


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Va

Vb Ia


I1 = I2 = I0 = E = 1


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1 2 0 ____ _____ ____ _____

(Z1 + Z2 + Z0) (Z1 + Z2 + Z0)

I A = I1 + I2 + I0 = 3 I0

IB = I0 + a2I1 + aI2 = I0 + a

2I0 + aI0 = 0

IC = 0

I Ground = I Residual = 3I0


V1 = E - I1Z1 = 1 - I1Z1


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1 1 1 1 1

V2 = - I2Z2

V0 = - I0Z0

V A = V1 + V2 + V0 = 0

VB = V0 + a2

V1 + aV2 = (Z1 - Z0 ) + a2

(Z0+Z1+Z1)

VC

= V0

+ aV1

+ a2V2

= (Z1

- Z0

) + a

(assumes Z 1 = Z 2 ) (Z0+Z1+Z1)

Two Phase to Ground Fault


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Positive, negative and zero sequence

impedance networks connected in parallel


Fault

Two Phase to Ground Fault

+++


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1PU

Z1

I1

Z2

I2

Z0

I0

V0

-

V2

-

V1

-


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Two Phase to Ground Fault Vectors

Ic


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Va

Vc

Vb

Ib

Other Conditions

Fault calculations and symmetrical


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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.

References

Circuit Analysis of AC Power Systems, Vol. 1 &2, Edith Clarke


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, d C a e

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, MarcelDekker, Inc.


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The end

Jon F. DaumeBonneville Power Administration

Retired!

Theory Track

Transmission Protection TheoryTransmission System


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Hands-On Relay School

Jon F. DaumeBonneville Power Administration

March 14-15, 2011

Transmission System

Protection

Discussion Topics• Protection overview

• Transmission line protectionPh d d f lt t ti


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– Phase and ground fault protection

– Line differentials

– Pilot schemes

– Relay communications

– Automatic reclosing

• Breaker failure relays• Special protection or remedial action schemes

Power Transfer

Vs VrX


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Power Transfer

0

0.5

1

0 30 60 90 120 150 180

Angle Delta

T r a n s m i t t e d P o w

e r

P = Vs Vr sin / X

Increase Power Transfer

• Increase transmission system operating

voltage


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voltage

• Increase angle δ• Decrease X

– Add additional transmission lines

– Add series capacitors to existing lines


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Power Transfer During Faults

Power Transfer

1.2


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0

0.2

0.4

0.6

0.8

1

1.2

0 30 60 90 120 150 180

Angle Del ta

T r a n s

m i t t e d P o w e

r

Normal

1LG

LL

LLG

3 Phase

Vs Vr


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Power Transfer

0

0.2

0.4

0.6

0.8

1

1.2

0 30 60 90 120 150 180

Angle Delta

P o w e

r

BP1

3

21

P2

6

4

5

A

System Stability

• Relay operating speed

• Circuit breaker opening speed


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• Circuit breaker opening speed

• Pilot tripping• High speed, automatic reclosing

• Single pole switching

• Special protection or remedial action

schemes

IEEE Device Numbers

Numbers 1 - 97 used

21 Distance relay


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21 Distance relay

25 Synchronizing or synchronism checkdevice

27 Undervoltage relay

32 Directional power relay43 Manual transfer or selector device

46 Reverse or phase balance current relay

50 Instantaneous overcurrent or rate of riserelay (fixed time overcurrent)

(IEEE C37.2)

51 AC time overcurrent relay52 AC circuit breaker

59 O lt l

IEEE Device Numbers


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59 Overvoltage relay

62 Time delay stopping or opening relay63 Pressure switch

67 AC directional overcurrent relay

79 AC reclosing relay

81 Frequency relay

86 Lock out relay87 Differential relay

(IEEE C37.2)

Relay Reliability

• Overlapping protection

– Relay systems are designed with a high level


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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

Relay Reliability

• Relay dependability (trip when required)

– Redundant relays


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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

Relay Reliability

• Relay security (no false trip)

– Careful evaluation before purchase


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p

– 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


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Transmission

Line Protection

Western Transmission System

Voltage, kV Northwest WECC

115 - 161 27400 miles 48030 miles


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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.

230 20850 miles 41950 miles

287 - 345 4360 miles 9800 miles

500 9750 miles 16290 miles

260 - 500 DC 300 miles 1370 miles

Transmission Line Impedance

• Z ohms/mile = Ra + j (Xa + Xd)

• R X function of conductor type length


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Ra, Xa function of conductor type, length

• Xd function of conductor spacing, length

Ra j(Xa+Xd)

Line Angles vs. Voltage

Z = √[Ra2 + j(Xa+Xd)2]

∠θ ° = tan-1 (X/R)


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( )

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.

Typical Line Protection


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Distance Relays

(21, 21G)


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CT & PT Connections

V Phase


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21

21

67N

I Phase

3I0 = Ia + Ib + Ic 3V0

I Polarizing

Instrument Transformers

• Zsecondary = Zprimary x CTR / VTR• The PT location determines the point from

which impedance is measured


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which impedance is measured

• The CT location determines the faultdirection

– Very important consideration for • Transformer terminated lines

• Series capacitors

• Use highest CT ratio that will work tominimize CT saturation problems

Saturated CT Current

100

150


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-100

-50

0

50

-0.017 0.000 0.017 0.033 0.050 0.0

Original Distance Relay

• True impedance characteristic – Circular characteristic concentric to RX axis


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• 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

Impedance Characteristic

X


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R

Directional

mho Characteristic

• Most common distance element in use• Circular characteristic

Passes through RX origin


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– Passes through RX origin

– No extra directional element required

• Maximum torque angle, MTA, usually setat 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

3 Zone mho Characteristic

X

Zone 3


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R

Zone 1

Zone 2

3 Zone Distance Elements Mho Characteristic

Typical Reaches

%


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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

Coordination Considerations,

Zone 1• Zone 1

– 80 to 90% of Line impedance

Account for possible errors


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– Account for possible errors

• Line impedance calculations

• CT and PT Errors

• Relay inaccuracy

– Instantaneous trip

Coordination Considerations

• Zone 2 – 125% or more of line impedance

• Consider strong line out of service


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g

• 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 permissiveoverreach transfer trip w/o time delay


• Zone 3 – Greater than zone 2



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• 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


• Zone 3 Special Applications – Starter element for zones 1 and 2

Provides current reversal logic for permissive


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– 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

Problems for Distance Relays

• Fault in front of relay

• Apparent Impedance


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• Load encroachment• Fault resistance

• Series compensated lines

• Power swings

3 Phase Fault in Front of Relay

• No voltage to make impedancemeasurement-use a potential memory

circuit in distance relay


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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

Apparent Impedance• 3 Terminal lines with apparent

impedance

• Fault resistance also looks like anapparent impedance


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apparent impedance

• Most critical with very short orunbalanced legs

• Results in – Short zone 1 reaches

– Long zone 2 reaches and time delays

• Pilot protection may be required

Apparent Impedance

Bus A Bus BZa = 1 ohm

Ia = 1

Zb = 1

Ib = 1


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Z apparent @

Bus A = Za +ZcIc/Ia

= 3 Ohms

Apparent Impedance

Ic = Ia + Ib = 2 Zc = 1

Bus C


• Zone 1 – Set to 85 % of actual impedance to nearest

terminal


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• 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

Load Encroachment

• Z Load = kV2

/ MVA – Long lines present biggest challenge

– Heavy load may enter relay characteristic


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• 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


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Load Encroachment

X

Zone 3


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R

Zone 1

Zone 2

Load Consideration with Distance Relays

Load

Area

Lens Characteristic

• Ideal for longer transmission lines• More immunity to load encroachment


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• Less fault resistance coverage• Generated by merging the common area

between two mho elements

Lens Characteristic


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Tomato Characteristic

• May be used as an external out of stepblocking characteristic

• Reaches set greater than the tripping


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g pp g

elements

• Generated by combining the total area of

two mho elements

Quadrilateral Characteristic

• High level of freedom in settings• Blinders on left and right can be moved in

or out


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– 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


X


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R


Special Load Encroachment

X

Zone 4


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R

Zone 1

Zone 2


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Fault ResistanceX

Zone 3


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R

Zone 1

Zone 2

Fault Resistance Effect on a Mho Characteristic

Rf

Series Compensated Lines

• Series caps added to increase loadtransfers

– Electrically shorten line


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• Negative inductance

• Difficult problem for distance relays

• Application depends upon location ofcapacitors

Series Caps

Zl Zc

Zl > Zc


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50

21

21 Zl > Zc


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• Zone 1 – 80 to 90% of compensated line impedance

– Must not overreach remote bus with caps in

service


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52

service

• Zone 2

– 125% + of uncompensated apparent lineimpedance

– Must provide direct tripping for any line fault

with caps bypassed – May require longer time delays

Power Swing

• Power swings can cause false trip of 3phase distance elements

• Option to


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– 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

Out Of Step Blocking

X

OOSB Outer Zone

OOS

B Inner Zone


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R

Zone 1

Zone 2

Typical Out Of Step Block Characteristic

t = 30 ms?

Ground Distance

P t ti


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Protection

and Kn(21G)

Fault Types

• 3 Phase fault – Positive sequence impedance network only

• Phase to phase fault

– Positive and negative sequence


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– Positive and negative sequenceimpedance networks in parallel

• One line to ground fault

– Positive, negative, and zero sequenceimpedance networks in series

• Phase to phase to ground fault

– Positive, negative, and zero sequenceimpedance networks in parallel

Sequence Networks


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Kn - Why?• Using phase/phase or phase/ground

quantities does not give proper reachmeasurement for 1LG fault

• Using zero sequence quantities gives thezero sequence source impedance not the


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zero sequence source impedance, not theline impedance

• Current compensation (Kn) does work forground faults

• Voltage compensation could also be used

but is less common

Current Compensation, KnK

n= (Z

0L- Z

1L)/3Z

1LZ0L = Zero sequence transmission line impedance

Z1L = Positive sequence transmission line

impedance


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impedanceIRelay = I A + 3I0(Z0L- Z1L)/3Z1L = I A + 3KnI0

ZRelay = V A Relay/IRelay = V A/(I A + 3KnI0) = Z1L

Reach of ground distance relay with current

compensation is based on positive sequence line

impedance, Z1L

Current Compensation, Kn• Current compensation (Kn) does work for

ground faults.

• Kn = (Z0L – Z1L)/3Z1

– Kn may be a scalar quantity or a vector quantitywith both magnitude and angle


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K may be a scalar quantity or a vector quantitywith both magnitude and angle

• Mutual impedance coupling from parallel

lines can cause a ground distance relay tooverreach or underreach, depending uponground fault location

• Mutual impedance coupling can provideincorrect fault location values for groundfaults

Ground Fault

Protection


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Protection

(67N)

Ground Faults

• Directional ground overcurrent relays(67N)

• Ground overcurrent relays

– Time overcurrent ground (51)


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Time overcurrent ground (51)

– Instantaneous overcurrent (50)

• Measure zero sequence currents• Use zero sequence or negative sequence

for directionality

Typical Ground OvercurrentSettings

• 51 Time overcurrentSelect TOC curve, usually very inverse

Pickup, usually minimum

Time delay >0.25 sec. for remote bus fault


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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.

Polarizing for DirectionalGround Overcurrent Relays

• I Residual and I polarizing – I Polarizing: An autotransformer neutral CT

may not provide reliable current polarizing

• I Residual and V polarizing


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p g

– 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

Current Polarizing

CT

H1

X1

H2

Y1


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I Polarizing

Auto Transformer Polarizing Current Source

H3

X3

X2 Y2

Y3

H0X0

Voltage PolarizingEa Eb Ec


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3 VO Polarizing Potential

Mutual Coupling

• Transformer affect between parallel lines – Inversely proportional to distance between

lines

• Only affects zero sequence current


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y q

• Will affect magnitude of ground currents

• Will affect reach of ground distance relays

Mutual Coupling

Line #13I0, Line #1


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Line #2

3I0, Line #2

Mutual Coupling vs. GroundRelays

Taft

645 Amps

1315 Amps645 Amps

1980 Amps

GarrisonTaft Garrison


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Taft

920 Amps

260 Amps920 Amps

1370 Amps

1LG Faults With Mutual Impedances

1LG Faults Without Mutual Impedances

GarrisonTaft Garrison

Other Line


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Other Line

Protection Relays

Line Differential


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87 87

Line Differential Relays

• Compare current magnitudes, phase, etc.at each line terminal

• Communicate information between relays

• Internal/external fault? Trip/no trip?


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p p

• Communications dependant!

• Changes in communications paths orchannel delays can cause potential

problems

Phase Comparison

• Compares phase relationship at terminals• 100% Channel dependant

– Looped channels can cause false trips

• Nondirectional overcurrent on channel


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Nondirectional overcurrent on channel

failure

• Immune to swings, load, series caps• Single pole capability

Pilot Wire

• Common on power house lines• Uses metallic twisted pair

– Problems if commercial line used

– Requires isolation transformers and protection


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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


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Current Differential

• Single pole capability• 3 Terminal line capability

• May include an external, direct transfer trip

feature


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• Immune to swings, load, series caps

Transfer Trip


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Transfer Trip


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Tone 1 Xmit

Tone 2 Xmit

Tone 1 Rcvd

Tone 2 Rcvd

Protective Relay

Protective Relay

Direct Transfer Trip


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PCB Trip Coil PCB Trip Coil

Direct Transfer Trip

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


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• Time overcurrent ground trip

• BFR-Ring bus, breaker & half scheme• Transformer relays on transformer

terminated lines

• Line reactor relays

Tone 2 Xmit

Tone 2 Rcvd

Permissive Relay

Tone 2 Xmit

Tone 2 Rcvd

Permissive Relay

Permissive Transfer Trip


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Permissive Transfer Trip

Permissive Keying

• Zone 2 instantaneous• Permissive overcurrent ground (very

sensitive setting)

• PCB 52/b switch


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• Current reversal can cause problems


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PRT Current Reversal

A B

Id

Ia

Ic

I Fault, Line AB

I Fault, Line CD


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C 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.

Directional ComparisonBlocking

• Overreaching relays• Delay for channel time

• Channel failure can allow overtrip

• Often used with “On/Off” carrier


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Block Xmit

Block Rcvd

Block Xmit

Block Rcvd

Forward

Relay

Reverse

Relay

Reverse

Relay

Forward

Relay

Directional Comparison


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Directional Comparison Blocking Scheme

Time DelayTime Delay TDTD

Directional Comparison Relays

• Forward relays must overreach remotebus

• Forward relays must not overreach remote

reverse relays

• Time delay (TD) set for channel delay


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• Time delay (TD) set for channel delay

• Scheme will trip for fault if channel lost – Scheme may overtrip for external fault on

channel loss

Tone Equipment

• Interface between relays andcommunications channel

• Analog tone equipment

• Digital tone equipment


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• Security features

– Guard before trip

– Alternate shifting of tones

– Parity checks on digital

Tone Equipment

• Newer equipment has 4 or morechannels

– 2 for direct transfer trip

– 1 for permissive transfer trip

1 for drive to lock out (block reclose)


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– 1 for drive to lock out (block reclose)

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


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• Programmable functions

– Each transmitted bit programmed for specificrelay function

– Each received bit programmed for specific

purpose

TelecommunicationsChannels

• Microwave radio – Analog (no longer available)

– Digital

• Other radio systems• Dedicated fiber between relays


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Dedicated fiber between relays

– Short runs

• Multiplexed fiber

– Long runs

• SONET Rings

TelecommunicationsChannels

• Power line carrier current – On/Off Carrier often used with directional

comparison

• Hard wire

– Concern with ground mat interconnections


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Concern with ground mat interconnections

– Limited to short runs• Leased line

– Rent from phone company

– Considered less reliable

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


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y

14 cycles for 115 kV; 25 cycles for 500 kV

• Must delay long enough for remoteterminal to clear

• 1LG Faults have a higher success ratethan 3 phase faults

Automatic Reclosing (79)

• Most often single shot• Delay of 30 to 60 cycles following line trip

is common

• Checking:

– Hot bus & dead line


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Hot bus & dead line

– Hot line & dead bus – Sync check

• Utilities have many different criteria for

transmission line reclosing

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


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Block reclosing for transformer fault on

transformer terminated lines• Block reclosing for bus faults

• Block reclosing for BFR

• Do not use them

Breaker Failure

Relay


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y

(50BF)

Breaker Failure

• Stuck breaker is a severe impact tosystem stability on transmission systems

• Breaker failure relays are recommended

by NERC for transmission systemsoperated above 100 kV


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• BFRs are not required to be redundant by

NERC

Breaker Failure Relays

1. Fault on line2. Normal protective relays detect fault and

send trip to breaker.

3. Breaker does not trip.

4. BFR Fault detectors picked up.


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p p

5. BFR Time delay times out (8 cycles)6. Clear house (open everything to isolate

failed breaker)

Breaker Failure Relay

BFR Fault

Detector PCB TripCoil #1

TD

Protective Relay

TripTD

BFR Retrip

BFR Time

Delay, 8~


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Typical Breaker Failure Scheme with Retrip

86

Block Close

PCB Trip

Coil #2

Delay, 8

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


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3-4 Cycles local clearingtime

4-5 Cycles remote clearing

time

12-13 Cycles local back upclearing time

13-14 Cycles remote

backup clearing

Remedial Action

Schemes (RAS)


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aka: SpecialProtection Schemes

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


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levels

• Provide controlled islanding

• Protect equipment and lines from thermal

overloads• Many WECC & NERC Requirements

Remedial Action Schemes

• WECC Compliant RAS – Fully redundant

– Annual functional test

– Changes, modifications and additions must beapproved by WECC

Non WECC RAS


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• Non WECC RAS

– Does not need full redundancy

– Local impacts only

– Primarily to solve thermal overload problems

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.


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• Restoration must be controlled• Must coordinate with generator 81 relays

• Responsibility of control areas

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


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• Primarily installed West of Cascades

Generator Dropping

• Trip generators for loss of load• Trip generators for loss of transmission

lines or paths

– Prevent overloading


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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


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– Utilize generator var output if possible – Static Var Compensators (SVC) provide high

speed adjustments

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


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Other RAS Schemes

• Controlled islanding – Force separation at know locations

• Load brake resistor insertion

– Provide a resistive load to slow downacceleration of generators

• Out of step tripping


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• Out of step tripping

– Force separation on swing

• Phase shifting transformers

– Control load flows

Typical RAS Controller


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Typical RAS Controller Outputs

• Generator tripping

• Load tripping

• Controlled islanding and separation (Four

Corners)• Insert series caps on AC Intertie

• Shunt capacitor insertion


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Shunt capacitor insertion

• Shunt reactor tripping

• Chief Jo Load Brake Resister insertion

• Interutility signaling• AGC Off

Chief Jo Brake


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1141400 Megawatts @ 230 kV

RAS Enabling Criteria

• Power transfer levels• Direction of power flow

• System configuration

• Some utilities are considering automatic

enabling/disabling based on SCADA data


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• Phasor measurement capability in relayscan be used to enable RAS actions

RAS Design Criteria

• Generally fully redundant• Generally use alternate route on

telecommunications

• Extensive use of transfer trip for signalingbetween substations, power plants, control

centers and RAS controllers


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centers, and RAS controllers

UFOs vs. Power Outages


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the end


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Jon F. Daume

Bonneville Power Administration

retired

March 15, 2011


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Transmission System Faults and

Event AnalysisFault Analysis Theory

andModern Fault Analysis Methods

Presented by:

Matthew Rhodes

Electrical Engineer, SRP

1


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Transmission System FaultTheory

• Symmetrical Fault Analysis• Symmetrical Components

• Unsymmetrical Fault Analysis usingsequence networks

• Lecture material originally developed byDr. Richard Farmer, ASU Research

Professor 2


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3

Symmetrical Faults


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Faults

Shunt faults:

Three phase a bc

Line to line

Line to ground

2 Line to ground

b

a

c

a

bc

a b

c


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5

Faults

Series faults

One open phase: a bc

2 open phasesa bc

Increased phaseimpedance

Za bc


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6

Why Study Faults?• Determine currents and voltages in the

system under fault conditions• Use information to set protective devices

• Determine withstand capability thatsystem equipment must have:

– Insulating level

– Fault current capability of circuit breakers:• Maximum momentary current

• Interrupting current


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7

Symmetrical Faults

α

t=0

2 V

Symmetrical Componet Fauly calculation

Documents