# 04-Presentation Analysis of Faults

September 23, 2017 | Author: Rajesh Pillai | Category: Electric Power System, Transformer, Electrical Impedance, Engineering, Electromagnetism

#### Description

Fault Analysis

Power System Fault Analysis (1) All Protection Engineers should have an understanding TO :z

z

z

z

z z

z z z 3

Calculate Power System Currents and Voltages during Fault Conditions Check that Breaking Capacity of Switchgear is Not Exceeded Determine the Quantities which can be used by Relays to Distinguish Between Healthy (i.e. Loaded) and Fault Conditions Appreciate the Effect of the Method of Earthing on the Detection of Earth Faults Select the Best Relay Characteristics for Fault Detection Ensure that Load and Short Circuit Ratings of Plant are Not Exceeded Select Relay Settings for Fault Detection and Discrimination Understand Principles of Relay Operation Conduct Post Fault Analysis

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Power System Fault Analysis (2)

Power System Fault Analysis also used to :-

X Consider Stability Conditions

 Required fault clearance times  Need for 1 phase or 3 phase auto-reclose

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Vectors

Vector notation can be used to represent phase relationship between electrical quantities. Z

I

V

θ

V = Vsinwt = V ∠0° I = I ∠-θ° = Isin(wt-θ)

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j Operator Rotates vectors by 90° anticlockwise : j = 1 ∠90°

90° j2 = 1 ∠180° = -1

90° 1

90°

90°

j3 = 1 ∠270° = -j

Used to express vectors in terms of “real” and “imaginary” parts. 6

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a = 1 ∠120 ° Rotates vectors by 120° anticlockwise Used extensively in “Symmetrical Component Analysis”

1 3 a = 1∠120° = - + j 2 2 120°

120°

1 120°

3 1 a = 1∠240° = − − j 2 2 2

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a = 1 ∠120 ° Balanced 3Ø voltages :VC = aVA

a2 + a + 1 = 0

VA

VB = a2VA

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Representation of Plant

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Generator Short Circuit Current The AC Symmetrical component of the short circuit current varies with time due to effect of armature reaction.

i TIME

Magnitude (RMS) of current at any time t after instant of short circuit :

Ι ac = (Ι" - Ι' )e- t/Td" + (Ι' - Ι )e- t/Td' + Ι where : I" =

10

I'

=

I

=

Initial Symmetrical S/C Current or Subtransient Current = E/Xd" ≈ 50ms Symmetrical Current a Few Cycles Later ≈ 0.5s or Transient Current = E/Xd' Symmetrical Steady State Current = E/Xd

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Simple Generator Models

Generator model X will vary with time. Xd" - Xd' - Xd

X

E

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Parallel Generators 11kV

11kV XG=0.2pu

j0.05

j0.1

11kV

20MVA

XG=0.2pu

20MVA

If both generator EMF’s are equal ∴ they can be thought of as resulting from the same ideal source - thus the circuit can be simplified.

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P.U. Diagram

j0.05

j0.2

j0.1

j0.05

j0.2

j0.2 IF

1.0

13

1.0

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j0.1

j0.2 IF

1.0

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Positive Sequence Impedances of Transformers 2 Winding Transformers P

P1

S

ZS

ZP

S1

ZM N1

P1

ZT1 = ZP + ZS

ZP

=

Primary Leakage Reactance

ZS

=

Secondary Leakage Reactance

ZM

= =

Magnetising impedance Large compared with ZP and ZS

ZM

Æ Infinity ∴ Represented by an Open Circuit

ZT1 =

S1

N1 14

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ZP + ZS = Positive Sequence Impedance

ZP and ZS both expressed on same voltage base. 14

Motors X Fault current contribution decays with time X Decay rate of the current depends on the system. From tests, typical decay rate is 100 - 150mS. X Typically modelled as a voltage behind an impedance

Xd"

M

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1.0

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Induction Motors – IEEE Recommendations Small Motors Motor load 35kW SCM = 4 x sum of FLCM

Large Motors SCM ≈ motor full load amps Xd"

Approximation :

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SCM =

locked rotor amps

SCM =

5 x FLCM ≈ assumes motor impedance 20%

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Synchronous Motors – IEEE Recommendations

Large Synchronous Motors SCM ≈ 6.7 x FLCM for 1200 rpm

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Assumes X"d = 15%

≈ 5 x FLCM for 514 - 900 rpm

Assumes X"d = 20%

≈ 3.6 x FLCM for 450 rpm or less

Assumes X"d = 28%

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Balanced Faults

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Balanced (3Ø) Faults (1) X RARE :- Majority of Faults are Unbalanced X CAUSES :1. System Energisation with Maintenance Earthing Clamps still connected. 2. 1Ø Faults developing into 3Ø Faults

X 3Ø FAULTS MAY BE REPRESENTED BY 1Ø CIRCUIT Valid because system is maintained in a BALANCED state during the fault Voltages equal and 120° apart Currents equal and 120° apart Power System Plant Symmetrical Phase Impedances Equal Mutual Impedances Equal Shunt Admittances Equal 19

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Balanced (3Ø) Faults (2)

TRANSFORMER LINE ‘X’

GENERATOR

Ea

ZG

ZT

ZLX

IaF

Eb

IbF

Ec

IcF

ZLY

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Balanced (3Ø) Faults (3) IcF

Ea

IaF

Eb

Ec

IbF Positive Sequence (Single Phase) Circuit :Ea ZG1 ZT1 ZLX1 Ia1 = IaF

F1

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Analysis of Balanced Faults

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Different Voltages – How Do We Analyse?

11/132kV 50MVA

11kV 20MVA ZG=0.3pu

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ZT=10%

O/H Line ZL=40Ω

132/33kV 50MVA

ZT=10%

Feeder ZL=8Ω

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Referring Impedances X1

R1

N : 1

X2

R2

Ideal Transformer

Consider the equivalent CCT referred to :Primary R1 +

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N2R2

X1 + N2X2

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Secondary R1/N2

+ R2

X1/N2 + X2

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Per Unit System

Used to simplify calculations on systems with more than 2 voltages.

Definition :

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P.U. Value = Actual Value of a Quantity Base Value in the Same Units

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Base Quantities and Per Unit Values

11/132 kV 50 MVA

11 kV 20 MVA ZG = 0.3 p.u.

ZT = 10%

O/H LINE ZL = 40Ω

132/33 kV 50 MVA

ZT = 10%

FEEDER ZL = 8Ω

X Particularly useful when analysing large systems with several voltage levels X All system parameters referred to common base quantities X Base quantities fixed in one part of system X Base quantities at other parts at different voltage levels depend on ratio of intervening transformers

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Base Quantities and Per Unit Values (1)

Base quantities normally used :BASE MVA

= MVAb = 3∅ MVA Constant at all voltage levels Value ~ MVA rating of largest item of plant or 100MVA

BASE VOLTAGE = KVb

=

∅/∅ voltage in kV Fixed in one part of system This value is referred through transformers to obtain base voltages on other parts of system. Base voltages on each side of transformer are in same ratio as voltage ratio.

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Base Quantities and Per Unit Values (2)

Other base quantities :-

(kVb )2 Base Impedance = Zb = in Ohms MVAb Base Current

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= Ιb =

MVAb in kA 3 . kVb

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Base Quantities and Per Unit Values (3)

Per Unit Values = Actual Value Base Value

MVA a Per Unit MVA = MVAp.u. = MVAb KVa Per Unit Voltage = kVp.u. = KVb Per Unit Impedance = Zp.u. = Per Unit Current = Ιp.u. =

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Za MVAb = Za . Zb (kVb )2

Ιa Ιb 29

Transformer Percentage Impedance X If ZT = 5% with Secondary S/C 5% V (RATED) produces I (RATED) in Secondary. ∴ V (RATED) produces 100 x I (RATED) 5 = 20 x I (RATED) X If Source Impedance ZS = 0 Fault current = 20 x I (RATED) Fault Power = 20 x kVA (RATED) X ZT is based on I (RATED) & V (RATED) i.e. Based on MVA (RATED) & kV (RATED) ∴ is same value viewed from either side of transformer. 30

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Example (1) Per unit impedance of transformer is same on each side of the transformer. Consider transformer of ratio kV1 / kV2 1

2 MVA

kVb / kV1

kVb / kV2

Actual impedance of transformer viewed from side 1 = Za1 Actual impedance of transformer viewed from side 2 = Za2

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Example (2) Base voltage on each side of a transformer must be in the same ratio as voltage ratio of transformer. 11.8kV

Incorrect selection of kVb Correct selection of kVb

Alternative correct selection of kVb

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132/11kV 11.8/141kV OHL

11.8kV

Distribution System

132kV

11kV

132x11.8 141 = 11.05kV

132kV

11kV

11.8kV

141kV

141x11 = 11.75kV 132

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Conversion of Per Unit Values from One Set of Quantities to Another

Z p.u. 2

Z p.u.1

Zb1

Zb2

MVAb1 MVAb2 kVb1

kVb2

Zp.u.1 =

Za Zb1

Zp.u.2 =

Za Z = Zp.u.1 x b1 Zb2 Zb2

(kVb1)2 MVAb2 x = Zp.u.1 x MVAb1 (kVb2 )2 MVAb2 (kVb1)2 = Zp.u.1 x x MVAb1 (kVb2 )2

Actual Z = Za

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Example 132/33 kV 50 MVA

11/132 kV 50 MVA

11 kV 20 MVA

0.3p.u.

10%

40Ω

10%

8Ω 3∅ FAULT

kVb

11

132

33

MVAb

50

50

50

349 Ω

21.8 Ω

Zb = kVb2 MVAb Ib = MVAb √3kV b Zp.u.

2.42Ω

∴ I11 kV = 0.698 x Ib = 219 A

2625 A

874 A

0.698 x 2625 = 1833A I132 kV = 0.698 x 219 = 153A

0.3 x 50 20 0.1p.u.

8 = 0.367 40 = 0.115 p.u. 0.1 p.u. p.u. 21.8 349

I33 kV = 0.698 x 874 = 610A

= 0.75p.u. 1.432p.u.

V 1p.u.

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> Fault Analysis – January 2004

IF =

1 = 0.698p.u. 1.432

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Fault Types

Line - Ground (65 - 70%) Line - Line - Ground (10 - 20%) Line - Line (10 - 15%) Line - Line - Line (5%) Statistics published in 1967 CEGB Report, but are similar today all over the world.

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Unbalanced Faults

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Unbalanced Faults (1) In three phase fault calculations, a single phase representation is adopted. 3 phase faults are rare. Majority of faults are unbalanced faults. UNBALANCED FAULTS may be classified into SHUNT FAULTS and SERIES FAULTS. SHUNT FAULTS: Line to Ground Line to Line Line to Line to Ground SERIES FAULTS: Single Phase Open Circuit Double Phase Open Circuit 37

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Unbalanced Faults (2) LINE TO GROUND LINE TO LINE LINE TO LINE TO GROUND Causes : 1) Insulation Breakdown 2) Lightning Discharges and other Overvoltages 3) Mechanical Damage

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Unbalanced Faults (3)

OPEN CIRCUIT OR SERIES FAULTS Causes : 1) Broken Conductor 2) Operation of Fuses 3) Maloperation of Single Phase Circuit Breakers

DURING UNBALANCED FAULTS, SYMMETRY OF SYSTEM IS LOST

∴ SINGLE PHASE REPRESENTATION IS NO LONGER VALID

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Unbalanced Faults (4)

Analysed using :X Symmetrical Components X Equivalent Sequence Networks of Power System X Connection of Sequence Networks appropriate to Type of Fault

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Symmetrical Components

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Symmetrical Components Fortescue discovered a property of unbalanced phasors ‘n’ phasors may be resolved into :X (n-1) sets of balanced n-phase systems of phasors, each set having a different phase sequence plus X 1 set of zero phase sequence or unidirectional phasors VA = VA1 + VA2 + VA3 + VA4 - - - - - VA(n-1) + VAn VB = VB1 + VB2 + VB3 + VB4 - - - - - VB(n-1) + VBn VC = VC1 + VC2 + VC3 + VC4 - - - - - VC(n-1) + VCn VD = VD1 + VD2 + VD3 + VD4 - - - - - VD(n-1) + VDn -----------------------------------------Vn = Vn1 + Vn2 + Vn3 + Vn4 - - - - - Vn(n-1) + Vnn (n-1) x Balanced

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1 x Zero Sequence 42

Unbalanced 3-Phase System VA = VA1 + VA2 + VA0 VB = VB1 + VB2 + VB0 VC = VC1 + VC2 + VC0 VA2

VA1

120°

VC1

VB1

Positive Sequence

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240°

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VC2

VB2

Negative Sequence

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Unbalanced 3-Phase System

VA0 VB0 VC0

Zero Sequence

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Symmetrical Components Phase ≡ Positive + Negative + Zero VA VA = VA1 + VA2 + VA0 VB = VB1 + VB2 + VB0 VC = VC1 + VC2 + VC0 VC VA1

VB VA0VB0

VA2 + VC1 VB1

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VC2

+

VC0

VB2

VB1 = a2VA1

VB2 = a VA2

VB0 = VA0

VC1 = a VA1

VC2 = a2VA2

VC0 = VA0

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Converting from Sequence Components to Phase Values VA = VA1 + VA2 + VA0 VB = VB1 + VB2 + VB0 = a2VA1 + a VA2 + VA0 VC = VC1 + VC2 + VC0 = a VA1 + a2VA2 + VA0 VA0

VA

VA2 VA1

VC0

VC

VC1 VC2 VB1

VB VB0

VB2 46

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Converting from Phase Values to Sequence Components VA1 = 1/3 {VA + a VB + a2VC} VA2 = 1/3 {VA + a2VB + a VC} VA0 = 1/3 {VA + VB + VC} VA

VB 3VA0

VC

VA0

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Summary VA = VA1 VB = ∝2VA1 VC = ∝VA1

+ VA2 + VA0 + ∝VA2 + VA0 + ∝2VA2 + VA0

IA = IA1 IB = ∝2IA1 IC = ∝IA1

+ IA2 + ∝A2 + ∝2IA2

VA1 = 1/3 {VA + VA2 = 1/3 {VA + VA0 = 1/3 {VA +

∝VB + ∝2VB + VB +

IA1 = 1/3 {IA + ∝IB IA2 = 1/3 {IA + ∝2IB IA0 + 1/3 {IA + IB 48

> Fault Analysis – January 2004

+ IA0 + IA0 + IA0

∝2VC} ∝VC } VC }

+ ∝2IC } + ∝IC } + IC } 48

Residual Current Used to detect earth faults

IA IB IC IRESIDUAL = IA + IB + IC = 3I0 E/F IRESIDUAL is zero for :-

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Balanced Load 3∅ Faults Ø/∅ Faults

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IRESIDUAL is ∅/E Faults present for :- ∅/Ø/E Faults Open circuits (with current in remaining phases)

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Residual Voltage Used to detect earth faults Residual voltage is measured from “Open Delta” or “Broken Delta” VT secondary windings. VRESIDUAL is zero for:Healthy unfaulted systems 3∅ Faults ∅/∅ Faults VRESIDUAL is present for:VRESIDUAL = VA + VB + VC = 3V0

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∅/E Faults ∅/∅/E Faults Open Circuits (on supply side of VT)

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Example Evaluate the positive, negative and zero sequence components for the unbalanced phase vectors : VA = 1 ∠0°

VC

VB = 1.5 ∠-90°

VA

VC = 0.5 ∠120°

VB 51

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Solution

VA1

=

1/3 (VA + aVB + a2VC)

=

1/3 [ 1 + (1 ∠120) (1.5 ∠-90) + (1 ∠240) (0.5 ∠120) ]

VA2

=

0.965 ∠15

=

1/3 (VA + a2VB + aVC)

=

1/3 [ 1 + (1 ∠240) (1.5 ∠-90) + (1 ∠120) (0.5 ∠120) ]

VA0

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> Fault Analysis – January 2004

=

0.211 ∠150

=

1/3 (VA + VB + VC)

=

1/3 (1

=

0.434 ∠-55

+ 1.5 ∠-90 + 0.5 ∠120)

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Positive Sequence Voltages VC1 = aVA1

VA1 = 0.965∠15º 15º

VB1 = a2VA1 53

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VC2 = a2VA2

VA2 = 0.211∠150°

-55º

150º

VA0 = 0.434∠-55º VB0 = VC0 = VB2 = aVA2

Zero Sequence Voltages

Negative Sequence Voltages

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Example Evaluate the phase quantities Ia, Ib and Ic from the sequence components IA1

=

0.6 ∠0

IA2

=

-0.4 ∠0

IA0

=

-0.2 ∠0

IA

=

IA1 + IA2 + IA0 = 0

IB

=

∝2IA1 + ∝IA2 + IA0

=

0.6∠240 - 0.4∠120 - 0.2∠0 = 0.91∠-109

=

∝IA1 + ∝2IA2 + IA0

=

0.6∠120 - 0.4∠240 - 0.2∠0 = 0.91∠-109

Solution

IC

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Unbalanced Voltages and Currents acting on Balanced Impedances (1) Va

Vb Vc

VA VB VC

Ia

ZS

Ib

ZS

Zm

Ic

ZS

Zm

IAZS IAZM IAZM

= = =

+ + +

IBZM IBZS IBZM

+ + +

Zm

ICZM ICZM ICZS

In matrix form VA VB VC

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=

ZS

ZM

ZM

IA

ZM

ZS

ZM

IB

ZM

ZM

ZS

IC

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Unbalanced Voltages and Currents acting on Balanced Impedances (2) Resolve V & I phasors into symmetrical components 1 1 1

1 a2 a

1 a a2

V0 V1 V2

=

ZS ZM ZM

ZM ZS ZM

ZM ZM ZS

1 1 1

1 a2 a

1 a a2

I0 I1 I2

Multiply by [A]-1 -1

1 1 1

1 a2 a

1 a a2

ZS ZM ZM

ZM ZS ZM

ZM ZM ZS

1 1 1

1 a2 a

1 a a2

I0 I1 I2

V0 1 V1 = 1/3 1 V2 1

1 a a2

1 a2 a

ZS ZM ZM

ZM ZS ZM

ZM ZM ZS

1 1 1

1 a2 a

1 a a2

I0 I1 I2

V0 V1 V2

=

V0 ZS + 2ZM V1 = 1/3 ZS - ZM V2 ZS - ZM 1 1 1 57

> Fault Analysis – January 2004

1 a2 a

1 a a2

ZS + 2ZM ZM + aZS + a2ZM ZM + a2ZS + aZM

ZS + 2ZM ZM + aZM + a2ZS ZM + a2ZM + aZS

I0 I1 I2 57

Unbalanced Voltages and Currents acting on Balanced Impedances (3) V0 V1

=

V2 V0 V1 V2

0

0

I0

0

ZS - ZM

0

I1

0

0

ZS - ZM

I2

ZS + 2ZM

=

Z0

0

0

I0

0

Z1

0

I1

0

0

Z2

I2

The symmetrical component impedance matrix is a diagonal matrix if the system is symmetrical. The sequence networks are independent of each other. The three isolated sequence networks are interconnected when an unbalance such as a fault or unbalanced loading is introduced. 58

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Representation of Plant Cont…

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Transformer Zero Sequence Impedance

P

Q

ZT0

a

a Q

P

b

b

N0

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General Zero Sequence Equivalent Circuit for Two Winding Transformer Primary Terminal

Z T0

'a'

'b'

'a'

Secondary Terminal

'b'

N0

On appropriate side of transformer :

61

Earthed Star Winding

-

Delta Winding

-

Unearthed Star Winding

-

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Zero Sequence Equivalent Circuits (1)

P

P0

S

ZT0

a

b

a

S0

b

N0

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Zero Sequence Equivalent Circuits (2)

P

P0

S

ZT0

a

b

a

S0

b

N0

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Zero Sequence Equivalent Circuits (3)

P

P0

S

ZT0

a

b

a

S0

b

N0

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Zero Sequence Equivalent Circuits (4)

P

P0

S

ZT0

a

b

a

S0

b

N0

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3 Winding Transformers P

S

T P ZP ZM

ZS

ZP, ZS, ZT = Leakage reactances of Primary, Secondary and Tertiary Windings

S

ZM = Magnetising Impedance = Large ∴ Ignored

ZT T N1

P

ZS

ZP ZT

T

S ZP-S = ZP + ZS = Impedance between Primary (P) and Secondary (S) where ZP & ZS are both expressed on same voltage base N1 Similarly ZP-T = ZP + ZT and ZS-T = ZS + ZT

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Auto Transformers

H

L

H

ZL

ZH1

L

1

ZM1

ZT1 T

T

N1

Equivalent circuit is similar to that of a 3 winding transformer.

H

ZH1

ZL1

ZM = Magnetising Impedance = Large ∴ Ignored

L ZHL1 = ZH1 + ZL1 (both referred to same voltage base)

ZT1

ZHT1 = ZH1 + ZT1 (both referred to same voltage base) T

ZLT1 = ZL1 + ZT1 (both referred to same voltage base) N1

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Sequence Networks

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Sequence Networks (2)

X +ve, -ve and zero sequence networks are drawn for a ‘reference’ phase. This is usually taken as the ‘A’ phase. X Faults are selected to be ‘balanced’ relative to the reference ‘A’ phase. e.g. For Ø/E faults consider an A-E fault For Ø/Ø faults consider a B-C fault X Sequence network interconnection is the simplest for the reference phase.

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Positive Sequence Diagram E1 Z1

N1

1.

2.

70

Phase-neutral voltage

Impedance network -

4.

All generator and load neutrals are connected to N1

Include all source EMF’s -

3.

F1

Positive sequence impedance per phase

Diagram finishes at fault point F1

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70

Example Generator

Transformer

Line

F

N R E

N1

E1

ZG1

ZT1

ZL1

I1

F1 V1 (N1)

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V1

=

Positive sequence PH-N voltage at fault point

I1

=

Positive sequence phase current flowing into F1

V1

=

E1 - I1 (ZG1 + ZT1 + ZL1)

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Negative Sequence Diagram

Z2

N2

1.

2.

72

No negative sequence voltage is generated!

Impedance network -

4.

All generator and load neutrals are connected to N2

No EMF’s included -

3.

F2

Negative sequence impedance per phase

Diagram finishes at fault point F2

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Example Generator

Transformer

Line

F

N R

System Single Line Diagram

E

ZG2

N2

ZT2

ZL2

I2

F2 V2

Negative Sequence Diagram

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(N2)

V2

=

Negative sequence PH-N voltage at fault point

I2

=

Negative sequence phase current flowing into F2

V2

=

-I2 (ZG2 + ZT2 + ZL2)

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Zero Sequence Diagram (1) For “In Phase” (Zero Phase Sequence) currents to flow in each phase of the system, there must be a fourth connection (this is typically the neutral or earth connection). IA0

N

IB0 IC0

IA0 + IB0 + IC0 = 3IA0

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Zero Sequence Diagram (2) Resistance Earthed System :N

3ΙA0 Zero sequence voltage between N & E given by R

V0 = 3IA0.R Zero sequence impedance of neutral to earth path

E

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> Fault Analysis – January 2004

Z0 = V0 = 3R IA0

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Zero Sequence Diagram (3) Generator

Transformer

Line

F

N

RT

R

System Single Line Diagram E

ZG0

N0 3R

ZL0

I0

F0

3RT

E0

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ZT0

Zero Sequence Network

V0 (N0)

V0

=

Zero sequence PH-E voltage at fault point

I0

=

Zero sequence current flowing into F0

V0

=

-I0 (ZT0 + ZL0)

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76

Network Connections

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Interconnection of Sequence Networks (1) Consider sequence networks as blocks with fault terminals F & N for external connections. F1 POSITIVE SEQUENCE NETWORK

N1 I2 F2 NEGATIVE SEQUENCE NETWORK

V2

N2 I0 ZERO SEQUENCE NETWORK

F0 V0

N0 78

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Interconnection of Sequence Networks (2) For any given fault there are 6 quantities to be considered at the fault point i.e.

VA

VB

VC

IA

IB

IC

Relationships between these for any type of fault can be converted into an equivalent relationship between sequence components V1, V2, V0 and I1, I2 , I0 This is possible if :1) or

2)

Any 3 phase quantities are known (provided they are not all voltages or all currents) 2 are known and others are known to have a specific relationship.

From the relationship between sequence V’s and I’s, the manner in which the isolation sequence networks are connected can be determined. The connection of the sequence networks provides a single phase representation (in sequence terms) of the fault. 79

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79

To derive the system constraints at the fault terminals :-

F

IA

VA

IB

VB

IC

VC

Terminals are connected to represent the fault. 80

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80

Line to Ground Fault on Phase ‘A’

IA

VA

81

IB

VB

> Fault Analysis – January 2004

IC

VC

At fault point :VA VB VC

= = =

0 ? ?

IA IB IC

= = =

? 0 0

81

Phase to Earth Fault on Phase ‘A’ At fault point VA

=

0 ; IB = 0 ; IC = 0

but

VA

=

V1 + V2 + V0

V1 I0

+ =

V2 + V0 = 0 ------------------------- (1) 1/3 (IA + IB + IC ) = 1/3 IA

I1

=

1/3 (IA + aIB + a2IC) = 1/3 IA

I2

=

1/3 (IA + a2IB + aIC) = 1/3 IA

I1 = I2 = I0 = 1/3 IA

------------------------- (2)

To comply with (1) & (2) the sequence networks must be connected in series :+ve Seq N/W

I1

F1 V1 N1

-ve Seq N/W

I2

F2

V2

N2

Zero Seq N/W

I0 F0

V0

N0 82

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82

Example : Phase to Earth Fault SOURCE

F

LINE

A-G FAULT

ZL1 = 10Ω ZL0 = 35Ω

132 kV 2000 MVA ZS1 = 8.7Ω ZS0 = 8.7Ω 8.7

10

IF

I1

F1 N1

8.7

10

I2

F2 N2

8.7

35

I0

F0 N0

Total impedance = 81.1Ω I1 = I2 = I0 = 132000 = 940 Amps √3 x 81.1 IF = IA = I1 + I2 + I0 = 3I0 = 2820 Amps 83

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83

Earth Fault with Fault Resistance

I1 POSITIVE SEQUENCE NETWORK

F1 V1

N1 I2 NEGATIVE SEQUENCE NETWORK

F2 V2

3ZF

N2 I0 ZERO SEQUENCE NETWORK

F0 V0

N0

84

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84

Phase to Phase Fault:- B-C Phase

I1 +ve Seq N/W

F1 V1 N1

85

> Fault Analysis – January 2004

I2 -ve Seq N/W

F2 V2 N2

I0 Zero Seq N/W

F0 V0 N0

85

Example : Phase to Phase Fault SOURCE 132 kV 2000 MVA ZS1 = ZS2 = 8.7Ω 132000 √3

F

LINE

B-C FAULT

ZL1 = ZL2 = 10Ω

8.7

10

I1

F1 N1

8.7

10

I2

F2 N2

Total impedance = 37.4Ω I1 = 132000 = 2037 Amps √3 x 37.4 I2 = -2037 Amps 86

> Fault Analysis – January 2004

IB = = = = =

a2I1 + aI2 a2I1 - aI1 (a2 - a) I1 (-j) . √3 x 2037 3529 Amps. 86

Phase to Phase Fault with Resistance

ZF

I1 +ve Seq N/W

F1

I2

-ve Seq N/W

V1

F2 V2

N1

N2

Zero Seq N/W

I0

F0 V0 N0

87

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87

Phase to Phase to Earth Fault:- B-C-E

I1 +ve Seq N/W

F1 V1 N1

88

> Fault Analysis – January 2004

I2 -ve Seq N/W

F2 V2 N2

I0 Zero Seq N/W

F0 V0 N0

88

Phase to Phase to Earth Fault:B-C-E with Resistance

3ZF

I1 +ve Seq N/W

89

F1 V1

> Fault Analysis – January 2004

N1

-ve Seq N/W

I2

F2 V2 N2

Zero Seq N/W

I0

F0 V0 N0

89

Maximum Fault Level

Single Phase Fault Level :

X Can be higher than 3Φ fault level on solidlyearthed systems

Check that switchgear breaking capacity > maximum fault level for all fault types.

90

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90

3Ø Versus 1Ø Fault Level (1)

E

XT

Xg

3Ø Xg

XT

ΙF = Z1 E

91

E Xg + XT

E Z1

IF

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91

3Ø Versus 1Ø Fault Level (2)

Xg

XT

Z1

E

Xg2

XT2

IF

Z2 = Z1

Xg0

3E ΙF = 2Z1 + Z0

XT0

Z0

92

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3Ø Versus 1Ø Fault Level (3)

3∅FAULTLEVEL =

3E 3E E = = 2Z1 + Z1 3Z1 Z1

3E 1∅FAULTLEVEL = 2Z1 + Z0 ∴ IF Z0 < Z1 1∅FAULTLEVEL > 3∅FAULTLEVEL

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