Power System Protective Relaying-Part Two

Power System Protective Power System Protective Relaying-Part TwoRelaying-Part Two

Wei-Jen Lee, Ph.D., PE

Professor of Electrical Engineering Dept.The Univ. of Texas at Arlington

DefinitionDefinition

• Quantity in per unit =

• Quantity in percent = (Quantity in per unit)*100

QuantityofValueBase

QuantityActual

AdvantagesAdvantages

• More meaningful when comparing different voltage levels

• The per unit equivalent impedance of the transformer remains the same when referred to either the primary or the secondary side

• The per unit impedance of a transformer in a three-phase system is the same, regardless the winding connection

• The per unit method is independent of voltage changes and phase shifts through transformers

AdvantagesAdvantages

• Manufacturers usually specify the impedance of the equipment in per unit or percent on the base of its nameplate ratings

• The per unit impedance values of various ratings of equipment lie in a narrow range

General Relations Between General Relations Between Circuit QuantitiesCircuit Quantities

LL

L

oLNLL

LLL

V

SI

VV

IVS

3

303

3

3

3



3

2

3

3

3

2

3

3033*303

303

3

30

3

30

303*

3

30

S

V

S

VV

I

V

I

VZ

V

S

Z

VII

Connection

S

V

S

VV

I

VZ

ConnectionY

oLLLLo

LLL

oLL

D

LLD

LL

o

D

LLo

LD

oLLLL

oLL

L

LNY

Base Quantity SelectionsBase Quantity Selections

• VA, V, I, and Z are four power quantities• One has to select two base quantities and derive

the other two

Base ConversionBase Conversion

2)(

2)(

)(

)()()( **

newbase

oldbase

oldbase

newbaseoldpunewpu KV

KV

MVA

MVAZZ

Example One: Base ConversionExample One: Base Conversion

• A 50-MVA, 34.5:161 kV transformer with 10% reactance is connected to a power system where all the other impedance values are on a 100 MVA, 34.5 or 161 kV base. The reactance of the transformer under new base is:

2.0*50

100*1.0

2)(

2)(

)( newbase

oldbasenewpu KV

KVZ

Example Two: Base ConversionExample Two: Base Conversion

• A generator and transformer, as shown below, are to be combined into a single equivalent reactance on a 100 MVA, 110 kV (high voltage side) base.

Example Two: Base ConversionExample Two: Base Conversion

• The transformer is operated at 3.9 kV tap. • New base voltage at high side is 110 kV.• The base voltage at low side is:

110*3.9/115 = 3.73 kV

514.1364.015.1

364.073.3

9.3*

30

100*1.0

15.173.3

4*

25

100*25.0

)()(

2

2

)(

2

2

)(

newXfernewgeneq

newXfer

newgen

ZZZ

Z

Z

Transformer PolarityTransformer Polarity

Transformer PolarityTransformer Polarity

• The ANSI/IEEE standard for transformers states that the high voltage should lead the low voltage by 30o with Y- or -Y banks.

Relay PolarityRelay Polarity

• Relays that sense the direction of current (or power) flow at a specific location and, thereby, indicate the direction of the fault, provide a good example of relay polarity.

• A directional-sensing unit requires a reference quantity that is reasonably constant against which the current in the protected circuit can be compared.


• Definition of maximum torque line and zero torque line

• Solid state units can have adjustments for (1) the maximum torque angle and (2) the angle limits of the operating zone



• In Fig. (A), the maximum operating torque or energy occurs when the current flow from polarity to non-polarity (Ipq) leads by 30o the voltage drop from polarity to non-polarity (Vrs). The minimum pick up of the directional unit is specified at the maximum torque.

• Higher current will be required when Ipq deviates from the maximum torque line.


• For ground fault protection, the 60o unit of Fig. (B) is used with a 3Vo reference and the zero unit of Fig. (C) with a 3Io current reference.

• The Fig. (C) is also used for power or var applications.


Connection Unit Type Phase A Phase B Phase C Maximum torque occurs when

30o Fig. 3.7C Ia, Vac Ib, Vba Ic, Vcb I lags 30o

60o Delta Fig. 3.7C Ia-Ib, Vac Ib-Ic, Vba Ic-Ia, Vcb I lags 60o

60o Wye Fig. 3.7C Ia, -Vc Ib, -Va Ic, -Vb I lags 60o

90o-45o Fig. 3.7A(max. torque: 45o)

Ia, Vbc Ib, Vca Ic, Vab I lags 45o

90o-60o Fig. 3.7A Ia, Vbc Ib, Vca Ic, Vab I lags 60o

The 90The 90oo-60-60oo Connection for Connection for Phase-Fault ProtectionPhase-Fault Protection

The 90The 90oo-60-60oo Connection for Connection for Phase-Fault ProtectionPhase-Fault Protection

Directional Sensing for Ground Directional Sensing for Ground Faults: Voltage PolarizationFaults: Voltage Polarization

Directional Sensing for Ground Directional Sensing for Ground Faults: Voltage PolarizationFaults: Voltage Polarization

Directional Sensing for Ground Directional Sensing for Ground Faults: Current PolarizationFaults: Current Polarization

Symmetrical ComponentsSymmetrical Components

2

1

0

2

2

1

1

111

I

I

I

aa

aa

I

I

I

c

b

a

c

b

a

I

I

I

aa

aa

I

I

I

2

2

2

1

0

1

1

111

3

1

Zero Sequence Current and Voltage Zero Sequence Current and Voltage for Ground Fault Protectionfor Ground Fault Protection

Sequence NetworksSequence Networks

• Single-Line Diagram


• Positive Sequence Network


• Negative Sequence Network


• Zero Sequence Network

Sequence Network ReductionSequence Network Reduction

• Consider faults at bus H for the positive sequence network of the sample system. The Z1 is equal to the parallel of (Xd”+XTG+X1GH) and (Z1S+XHM)

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Three-phase fault studies for applying and setting

phase relays• Single-phase-to-ground fault studies for applying

and setting ground protection relays• Fault Impedance: Faults are seldom solid, but

involve varying amount of resistance. • It is generally ignored in protective relaying and

faults studies for high voltage transmission or sub-transmission system.

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• In distribution systems, very large or basically

infinite impedance can exist. High impedance fault detection relay may be necessary for distribution system.

• For arcing fault, the arc resistance varies a lot. However, a commonly accepted value for currents between 70 and 20,000 A has been an arc drop of 440 V per foot, essentially independent of current magnitude. Therefore,

I

lZ arc

440

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• In low voltage (480 V) switchboard-type

enclosures, typical arc voltages of about 150 V can be experienced.

• Substation and Tower-Footing Impedance is another highly variable factor. Several technical papers have been written and computer programs have been developed in this area with many variables and assumptions. The general practice is to neglect these in most fault studies and relay applications and settings.

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination

2

1

0

2

2

2

2

2

1

0

1

1

111

1

1

111

3

1

I

I

I

aa

aa

I

I

I

I

I

I

aa

aa

I

I

I

c

b

a

c

b

a

2

1

0

2

1

0

2

1

0

00

00

00

0

0

I

I

I

Z

Z

Z

V

V

V

V

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Sequence Interconnections for Three-Phase Faults

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Sequence Interconnections for Three-Phase Faults

– Three-Phase faults are assumed to be symmetrical.

– The positive-sequence network can be used to calculate the fault current.

– Since Ia, Ib, and Ic are balanced, only I1 appears in the

circuit. If there is fault impedance ZF among phases, Z1

should be changed to Z1 + ZF. V-( Z1 + ZF) I1=0

or1

1 Z

VII aF

FaF ZZ

VII

11

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Sequence Interconnections for Single Phase-to-

Ground Faults


Ground Faults – A phase-a-to-ground fault is represented by

connecting the three sequence networks together (either with or without fault impedance).

0 cb II

a

a

aa

I

I

II

aa

aa

I

I

I

3

1

0

0

1

1

111

3

1

2

2

2

1

0

03* IZV Fa


Ground Faults

12

11

10

1

1

1

2

1

0

2

1

0

00

00

00

0

0

IZ

IZV

IZ

I

I

I

Z

Z

Z

V

V

V

V

11210210 3)( IZIZZZVVVVV Fa

1021

021021

3

)3(

IIIII

ZZZZ

VIII

aF

F

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Sequence Interconnections for Phase-to-Phase

Faults - It is convenient to show the fault between phases b and

c with fault impedance of ZF.


Faults

cbbFcba IIandIZVVI ,,0

bb

bb

b

b

aIIa

IaaI

I

I

aa

aa

I

I

I

2

2

2

2

2

1

00

3

1

0

1

1

111

3

1


Faults

12

11

1

1

2

1

0

2

1

0 00

00

00

00

0

0

IZ

IZV

I

I

Z

Z

Z

V

V

V

V

2

1

0

2

2

1

1

111

V

V

V

aa

aa

V

V

V

c

b

a


Faults

22

12 )()( VaaVaaVV cb

))(( 212 VVaa

21

1212

3

))()((

aa

ZIZI

IZZVaa

FFb

FF

Fb ZIaaaa

ZIZIIZZV 122

1121 ))((

3))((


Faults

• Assume Z1 = Z2, then I1 = V/2Z1. Just considering the magnitude,

122

1

1212

21

2121

3

3

0

)(

IjIaaII

IjaIIaI

III

ZZZ

VII

cF

bF

aF

F

311

866.0866.02

3I

Z

V

Z

VII cFbF

Fault Studies for Relay Settings Fault Studies for Relay Settings and Coordinationand Coordination• Sequence Interconnections for Double Phase-to-

Ground Faults • The connection for this type of fault is similar to the

phase-to-phase fault with the addition of the zero sequence impedance in parallel with the negative sequence impedance.


Ground Faults

0

)(

210

IIII

IIZVV

a

cbFcb

FFcbb

c

b

ZIZIIV

VV

VaaVVV

aVVaVV

0

21

22

10

212

0

3)(

222

111

000

2

1

0

2

1

0

2

1

0

00

00

00

0

0

IZV

IZVV

IZV

I

I

I

Z

Z

Z

V

V

V

V


Ground Faults

1012

00 )(3 VVVaaVZIV Fb

F

F

ZZ

IZVI

IZVIZVVZI

3

3

0

110

1100100

)( 021

2

112

III

Z

IZVI


Ground Faults (Solid faults)

02

210

02

012

02

021

1

ZZ

ZII

ZZ

ZII

ZZ

ZZZ

VI


Ground Faults (Line-to-line fault impedance: ZF and phase-to-ground fault impedance: ZFG)

FGF

F

FGF

FGF

FGF

FGFFF

ZZZZ

ZZII

ZZZZ

ZZZII

ZZZZ

ZZZZZZZ

VI

3

))2/((

3

)3)2/((

3

)3)2/())(2/(()

2(

02

210

02

012

02

021

1

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • Step One: Transfer all the constants to a common

base - VG:

- VS: No change

- Two winding Transformer:

puXX Gd 2.080

100*16.02

"

1375.080

100*11.0 TGX


base

- Three winding transformer:

puX

puX

puX

ML

HL

HM

18667.0150

100*280.0

2400.0150

100*360.0

03667.0150

100*055.0

puX

puX

puX

L

M

H

1950.0)18667.02400.003667.0(2

1

00833.0)18667.02400.003667.0(2

1

0450.0)18667.02400.003667.0(2

1


base

- Line Impedance:

puXX

puX

Zbase

18147.0

6200.025.132

82

25.13210*100

10*115

21

0

6

62

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • Develop sequence network for different fault

conditions (Fault at point “G”).

Positive & negative sequence network

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • Develop sequence network for different fault

conditions (Fault at point “G”).

Zero sequence network

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • For the fault at G, the right side impedance

(j0.18147+j0.03667+j0.03=j0.2481) is parallel with left side impedance (j0.20+j0.1375=j0.3375).

1430.02481.03375.0

2482.0*3375.021 j

jj

jjZZ

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • The right side network is reduced for a fault at bus

G by first paralleling X0S + ZH with ZL and then

adding ZM and X0GH (The equivalent impedance is

equal to j0.6709). Paralleling with the left side impedance (XTG = j0.1375), the zero sequence

impedance X0 = j0.1141.

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • Three-phase fault at Bus G

• The division of current from the left (IaG) and the

right (IaH) are:

kVatAjpujj

II aF 1158.3510115*3

000,100993.6993.6

143.0

11

puI

puI

aH

aG

030.4993.6*5763.0

963.2993.6*4237.0

Example: Fault Calculations on a Example: Fault Calculations on a Sample System Sample System • Single-phase-to-ground fault at Bus G

kVatApujII

pujj

III

aF 1154.37645.73

5.2)1141.0143.0143.0(

0.1

1

021

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers • Autotransformers have become very common in

recent years.• Consider a typical autotransformer in a system, as

shown in the figure, and assume that a single-phase-to-ground fault occurs at the H or 34.5 kV terminal

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers

puX

puX

puX

ML

HL

HM

54.040

100*216.0

68.050

100*34.0

05333.0150

100*08.0

puX

puX

puX

H

M

H

58334.0)0533.054.068.0(2

1

04334.0)68.054.00533.0(2

1

09667.0)54.068.00533.0(2

1

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers • The sequence networks are shown below:

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers • When the fault happens at Bus H, the equivalent

sequence impedance for the networks are: • Positive & negative sequence network

• Zero sequence network

puXX 04637.0)08.00967.00433.0057.0(

)08.0(*)0967.00433.0057.0(21

puX

puX left

06527.028.0085177.0

28.0*085177.0

085177.00967.0)583.00433.0032.0(

583.0*)0433.0032.0(

0

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers • Single-phase-to-ground fault at H:

kVatApuII

kVatAIII

puIII

aF 34529.3177986.183287.6*33

3451.105910*345*3

10*100*3287.6

3287.6)06527.004637.004637.0(

0.1

0

3

6

210

210

Example: Fault Calculations for Example: Fault Calculations for Autotransformers Autotransformers • Fault current distribution

Example: Open-Phase Conductor Example: Open-Phase Conductor

• A blown fuse or broken conductor that opens one of the three phases results in a serious unbalance that has to be detected and resolved as soon as possible.

• The sample system (Assume phase a open at “H”)


• The positive sequence network (X-Y indicates the fault location)


• The negative sequence network (X-Y indicates the fault location)


• The zero sequence network (X-Y indicates the fault location)


• It is necessary to consider the load current in this case. (This is similar to two-phase-to-ground fault calculation)

Example: Open-Phase Falling to Example: Open-Phase Falling to Ground on One Side Ground on One Side • Same sample system as before.• Assume phase a conductor on the line at bus H

opens and falls to ground on the H side (right side).

Example: Open-Phase Falling to Example: Open-Phase Falling to Ground on One Side Ground on One Side • Same sequence network• Since this is a simultaneous fault (an open phase

fault and a phase-to-ground fault), we can insert three ideal transformers at H to isolate the open phase fault and phase-to-ground fault. (Load current can be ignored)

Example: Open-Phase Falling to Example: Open-Phase Falling to Ground on One Side Ground on One Side

Example: Open-Phase Falling to Example: Open-Phase Falling to Ground on One Side Ground on One Side

Example: Open-Phase Falling to Example: Open-Phase Falling to Ground on One Side Ground on One Side • The other possibility is that the open conductor

falls to ground on the line side. We can use the same approach as before. However, the load must be considered in the fault current calculation.

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Power System Protective Relaying-Part Two