57
Learning Outline Learning Outline Symmetrical Components Theory Symmetrical Components Theory Sequence Impedance: Load, Line, Generator, T f T ransf ormer Fault Analysis: LineGround, LineLine, LineLineGround. Examples and Class Exercises 1

Unsymmetrical Fault

Embed Size (px)

Citation preview

Page 1: Unsymmetrical Fault

Learning OutlineLearning Outline

Symmetrical Components TheorySymmetrical Components Theory

Sequence Impedance: Load, Line, Generator, T fTransformer

Fault Analysis: Line‐Ground, Line‐Line, Line‐Line‐Ground.

Examples and Class Exercisesp

1

Page 2: Unsymmetrical Fault

Fault Analysisy ____________________

• Fault types:– balanced faults (<5%)

• three‐phase to ground

• Three‐phase

– unbalanced faults• single‐line to ground (60%‐75%)

d bl li t d (15% 25%)• double‐line to ground (15%‐25%)

• line‐to‐line faults (5%‐15%)line to line faults (5% 15%)

2

Page 3: Unsymmetrical Fault

Example impact of faultExample impact of faultThe second largest blackout in the history of TEPCO(The Tokyo Electric Power Company Inc ) hit central(The Tokyo Electric Power Company, Inc.) hit centralTokyo area at about 7:38 a.m. on August 14, 2006. Itwas caused by a floating crane on a barge goingupstream on a river on the eastern edge of the city.

The workers on the boat did not realize that the 33meter crane was raised too high, so it hit TEPCO's275 kV double circuit transmission lines that runacross the riveracross the river.

As a result of the accident the transmission lines were short‐circuited and the wires damaged. The relay protection operated and tripped both lines

3

Page 4: Unsymmetrical Fault

Symmetrical Componentsy p __________

• Three phase voltage or current is in a balance condition if it has the following characteristic:the following characteristic:

– Magnitude of phase a,b, and c is all the same

– The system has sequence of a,b,cThe system has sequence of a,b,c

– The angle between phase is displace by 120 degree 

• If one of the above is character is not satisfied, unbalanced occur.  Example:

4

Page 5: Unsymmetrical Fault

Symmetrical Componentsy p __________

• For unbalanced system, power system analysis cannot be analyzed using per phase as in Load Flow analysis oranalyzed using per phase as in Load Flow analysis or Symmetrical fault ‐>Symmetrical components need to be used.

• Symmetrical component allow unbalanced phase quantities such as current and voltages to be replaced by three separate balanced symmetrical componentsbalanced symmetrical components. 

5

Page 6: Unsymmetrical Fault

Symmetrical Componentsy p __________

6

Page 7: Unsymmetrical Fault

Symmetrical Componentsy p __________

By convention, the direction of rotation of the phasors is taken to be counterclock wiseto be counterclock‐wise.

Positive sequence:111 0 aaa III =°∠=

1211 240 aab IaII =°∠=

111 120 III °∠

(10.1)

111 120 aac aIII =°∠=

Where we defined an operator a that causes a counterclockwise rotation of 120 degree such that:degree, such that:

866.05.0120sin120cos1201 jja +−=°+°=°∠=

866.05.02401)1201()1201(2 ja −−=°∠=°∠×°∠= 01 2 =++ aa(10.2)

0136013 ja +=°∠= (10.3)

7

Page 8: Unsymmetrical Fault

Symmetrical Componentsy p __________

Negative sequence:

°∠= 022aa II

222 120 aab aIII =°∠=2222 240 III °∠

(10.4)

2222 240 aac IaII =°∠=

000cba III ==

Zero sequence:

(10.5)

8

Page 9: Unsymmetrical Fault

Symmetrical Componentsy p __________

Consider the three‐phase unbalanced current of   cba III ,,

210

210

bbbb

aaaa

IIII

IIII

++=

++=(10.6)

210cccc IIII ++=

Based on (10.1), (10.4) and (10.5), (10.6) can be rewrite all in terms of phase acomponents

2120

210aaaa IIII ++=

⎥⎤

⎢⎡

⎥⎤

⎢⎡

⎥⎤

⎢⎡

1

0

2

111 aa II

2210

2120

aaac

aaab

IaaIII

aIIaII

++=

++= (10.7)

⎥⎥⎥

⎦⎢⎢⎢

⎣⎥⎥⎥

⎦⎢⎢⎢

=⎥⎥⎥

⎦⎢⎢⎢

⎣2

1

2

2

11

a

a

c

b

II

aaaa

II (10.8)

9

Page 10: Unsymmetrical Fault

Symmetrical Componentsy p __________

Equation 10.8 can be written as: 

012a

abc AII =

Where A is known as symmetrical components transformation matrix, 

(10.9)

which transforms phasor currents          into components currents  and 

abcI 012aI

⎥⎤

⎢⎡

2

111(10 10)

⎥⎥⎥

⎦⎢⎢⎢

=2

2

11

aaaaA (10.10)

Solving (10.9) for the symmetrical components of currents:

abca IAI −=012 (10.11)

The inverse of A is given by:⎥⎤

⎢⎡ 111

1 (10 12)

⎥⎥⎥

⎦⎢⎢⎢

=−

aaaa

2

2

11

31A (10.12)

10

Page 11: Unsymmetrical Fault

Symmetrical Componentsy p __________

From (10.10) and (10.12), we conclude that

(10.13)*

31 AA =−

Substituting for A‐1 in (10.11), we have:Substituting for A in (10.11), we have: 

⎥⎥⎥⎤

⎢⎢⎢⎡

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥⎤

⎢⎢⎢⎡

b

a

a

a

III

aaIII

2

2

2

1

0

11

111

31 (10.14)

⎥⎦⎢⎣⎥⎦⎢⎣⎥⎦⎢⎣ ca IaaI 22 1

or in component form, the symmetrical components are:  

)(31

)(31

21

0

cbaa

cbaa

IaaIII

IIII

++=

++=

(10.15)

)(31 22

cbaa aIIaII ++=

11

Page 12: Unsymmetrical Fault

Symmetrical Componentsy p __________

Similar expressions exists for voltage:

(10.16)2210

2120

210

aaac

aaab

aaaa

VaaVVV

aVVaVV

VVVV

++=

++=

++=012a

abc AVV = (10.17)

aaac

1

The symmetrical components in terms of unbalanced voltages are:

)(31

)(31

21

0

cbaa

cbaa

VaaVVV

VVVV

++=

++=

(10.18) abca VAV −=012 (10.19)

)(31 22

cbaa aVVaVV ++=

12

Page 13: Unsymmetrical Fault

Symmetrical Componentsy p __________

The apparent power may also be expressed in terms of the symmetrical  componentscomponents.

*

)3(abcabcT

S IV=φ (10.20)

*

*012012)3( )()( a

Ta

T

S AIAV=φ

Substituting (10.9) and (10.17) in (10.20), we obtain: 

(10.21)*012*012a

Ta

T

IAAV=

3, * == AAAA TTSince                                      complex power becomes

(10.21)

*012012T

*** 221100

012012)3(

333

)(3

aaaaaa

T

S

IVIVIV

IV

++=

Total power for unbalance 3 phase system can be obtained from the sum of

(10.22)

Total power for unbalance 3‐phase system can be obtained from the sum of symmetrical components powers. 

13

Page 14: Unsymmetrical Fault

Example 1p _______________________

One conductor of a three‐phase line is open. The current flowing to delta‐connected load through line a is 10 A With the current in line a asconnected load through line a is 10 A. With the current in line a asreference and assuming that line c is open, find the symmetricalcomponents of the line currents.

a AIa °∠= 010

)(31

)(31

21

0

cbaa

cbaa

IaaIII

IIII

++=

++=

b AIb °∠= 18010

AI 0

)(31

)(3

22cbaa

cbaa

aIIaII ++=

c AIc 0=

14

Page 15: Unsymmetrical Fault

Solution________________________

The line current are:

°∠= 010aI °∠= 18010bI 0=cI

1From (10.15): )(

31

)(31

21

0

cbaa

cbaa

IaaIII

IIII

++=

++=

0)018010010(31)0( =+°∠+°∠=aI

Ia +°+°∠+°∠= )0)120180(10010(31)1(

0 Sequence

+ Sequence

)(313

22cbaa aIIaII ++=

Aj °−∠=−= 3078.589.25

Aj

Ia

°∠=+=

+°+°∠+°∠=

307858925

)0)240180(10010(31)2(

‐ SequenceAj ∠=+= 3078.589.25

From (10.4) 0)0( =bI 0)0( =cI

AIb °−∠= 15078.5)1( AIc °∠= 9078.5)1(

AIb °∠= 15078.5)2( AIc °−∠= 9078.5)2(

15

Page 16: Unsymmetrical Fault

Example 2p _______________________

16

Page 17: Unsymmetrical Fault

Exercise 1_______________________

: that Show

°∠=++

)(1

1201)a(1

a)(1 (a)

2

2

°−∠=+− 1803

a)(1a)(1(b) 2

2

17

Page 18: Unsymmetrical Fault

Exercise 2_______________________Obtain the symmetrical components for the set of unbalanced voltages 

°−∠=°∠=°−∠= 30100,90200,120300 cba VVV

)(31

)(31

21

0

cbaa

cbaa

VaaVVV

VVVV

++=

++=

°−∠=

1202650.42012V

The symmetrical components of a set of unbalanced three‐phase currents are

)(313

22cbaa aVVaVV ++=°−∠

°−∠8961.849473.86

1351852.193

The symmetrical components of a set of unbalanced three phase currents are

Obtain the original unbalanced phasors.

°∠=°∠=°−∠= 304,905,303 210aaa III

210aaaa IIII ++=

2210

2120

aaac

aaab

aaaa

IaaIII

aIIaII

++=

++=

°−∠°∠

=

3042163.421854.8

Iabc

°−∠ 2163.1021854.8

18

Page 19: Unsymmetrical Fault

Exercise 3_______________________The line‐to‐line voltages in an unbalanced three‐phase supply are 

°∠=°−∠=°∠= 120500,1500254.866,01000 cabcab VVV

Determine the symmetrical components for line and phase voltages, then find the phase voltages Van, Vbn, and Vcn.

°∠°∠

=

8934107626763300.0

VL 012

°∠=

00.0Va 012

°−∠=

1066.199586.440Vabc

°∠°−∠

306751.2888934.107626.763

°∠°−∠

606667.1668934.409586.440

°∠°−∠

603333.3331021.1669252.600

19

Page 20: Unsymmetrical Fault

Sequence Impedanceq p ______________

• The impedance of an equipment or component to theThe impedance of an equipment or component to thecurrent of different sequences.

iti i d (Z1) I d th t• positive‐sequence impedance (Z1): Impedance thatcauses a positive‐sequence current to flow

• negative‐sequence impedance (Z2): Impedance thatcauses a negative‐sequence current to flow

• zero‐sequence impedance (Z0): Impedance that causesa zero‐sequence current to flow

20

Page 21: Unsymmetrical Fault

Sequence Impedance of Y‐Connected Loadq p

Line to ground voltages are:

(10.23)

nncsbmamc

nncmbsamb

nncmbmasa

IZIZIZIZVIZIZIZIZVIZIZIZIZV

++++=+++=+++=

(10.24)cban IIII ++=

Kirchhoff’ current law:

(10 25)

Substituting In into (10.23):

⎥⎤

⎢⎡

⎥⎤

⎢⎡ +++

⎥⎤

⎢⎡ anmnmnsa IZZZZZZV )(

(10.25)

(10.26)

⎥⎥⎥

⎦⎢⎢⎢

⎣⎥⎥⎥

⎦⎢⎢⎢

⎣ ++++++=

⎥⎥⎥

⎦⎢⎢⎢

⎣ c

b

nsnmnm

nmnsnm

c

b

II

ZZZZZZZZZZZZ

VV

)()(

abcabcabc IZV = (10.26)IZV =

21

Page 22: Unsymmetrical Fault

Sequence Impedance of Y‐Connected Loadq p

(10 27)⎥⎥⎤

⎢⎢⎡

++++++

)()( nmnmns

abc ZZZZZZZZZZZZ

Z (10.27)⎥⎥⎥

⎦⎢⎢⎢

⎣ ++++++=

)()(

nsnmnm

nmnsnm

ZZZZZZZZZZZZZ

Writing Vabc and Iabc in terms of their symmetrical components: 

(10.28)012012a

abca AIZAV =

Multiplying (10.28) by A‐1 : 

(10.29)012012

012012 )(

a

aabc

a

IZ

IAZAV

=

= −

AZAZ abc−=012where  (10.30)

(10.31)⎥⎥⎤

⎢⎢⎡

⎥⎥⎤

⎢⎢⎡

++++++

⎥⎥⎤

⎢⎢⎡

22012 1111

)()(

1111

1 aaZZZZZZZZZZZZ

aaZnmnmns

Substituting for Zabc, A and A‐1 from (10.27), (10.10) and (10.12): 

⎥⎥⎦⎢

⎢⎣⎥⎥⎦⎢

⎢⎣ +++

+++⎥⎥⎦⎢

⎢⎣

=22 1

1)(

)(11

3aaaa

ZZZZZZZZZZZZ

aaaaZ

nsnmnm

nmnsnm

22

Page 23: Unsymmetrical Fault

Sequence Impedance of Y‐Connected Loadq p

⎤⎡

Performing the multiplication in (10.31): 

⎥⎥⎥

⎢⎢⎢

−−

++=

)(000)(000)23(

012

ms

ms

mns

ZZZZ

ZZZZ (10.32)

⎥⎤

⎢⎡ +

0)(000)3(

012ns

ZZZ

Z

When there is no mutual coupling, Zm = 0, and the impedance matrix becomes

(10.33)

⎥⎥⎥

⎦⎢⎢⎢

=)(00

0)(0012

s

s

ZZZ (10.33)

23

Page 24: Unsymmetrical Fault

Sequence Impedance of Transmission Linesq p

For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different and larger approximately 3 times than positive and negative sequence.

S I d f S h M hiSequence Impedance of Synchronous MachineThe positive-sequence generator impedance is the value found when positive-sequence current flows from the action of an imposed positive-sequence set of

"2

q p p qvoltages.

The negative-sequence reactance is close to the positive-sequencesubstransient reactance i e : "2

dXX ≈substransient reactance, i.e :

Zero-sequence reactance is approximated to the leakage reactance, i.e :

lXX ≈0

24

Page 25: Unsymmetrical Fault

Sequence Impedances of Transformerq p

• Series Leakage Impedance.– the magnetization current and core losses represented by the shunt branch t e ag et at o cu e t a d co e osses ep ese ted by t e s u t b a c

are neglected (they represent only 1% of the total load current)– the transformer is modeled with the equivalent series leakage impedance

• Since transformer is a static device, the leakage impedance will not change if the phase sequence is changedif the phase sequence is changed. 

• Therefore, the positive and negative sequence impedance are the same; Z0 = Z1 = Z2 = Zl

• Wiring connection always cause a phase shift. In Y‐Delta or Delta‐Y transformer:– Positive Sequence rotates by a +30 degrees from HV to LV side – Negative Sequence rotates by a ‐30 degrees from HV to LV side – Zero Sequence does not rotateZero Sequence does not rotate

• The equivalent circuit for zero‐sequence impedance depends on the winding connections and also upon whether or not the neutrals are grounded.

25

Page 26: Unsymmetrical Fault

Sequence Impedances of Transformerq pConnection diagram Zero‐sequence circuit

Figure (a)

Figure (b)g ( )

Figure (c)Figure (c)

Figure (d)Figure (d)

Figure (e)

26

Figure (e)

Page 27: Unsymmetrical Fault

Sequence Impedances of Transformerq p

Description of Zero sequence Equivalent Circuit

(a) Y‐Y connections with both neutrals grounded – We know that the zero sequence currentequals the sum of phase currents. Since both neutrals are grounded, there is a path for the zerosequence current to flow in the primary and secondary, and the transformer exhibits theequivalent leakage impedance per phase as shown in Fig. (a).

(b) Y‐Y connections with primary the neutral grounded – The primary neutral is grounded, but( ) p y g p y g ,since the secondary neutral is isolated, the secondary phase current must sum up to zero. Thismeans that the zero‐sequence current in the secondary is zero. Consequently, the zerosequence current in the primary is zero, reflecting infinite impedance or an open circuit asshown in Fig. (b).g ( )

27

Page 28: Unsymmetrical Fault

Sequence Impedances of Transformerq pc) Y‐Δ with grounded neutral – in this configuration, the primary currents canflow because the zero‐sequence circulating current in the Δ‐connectedsecondary and a ground return path for the Y‐connected primary. Note that noy g p p yzero‐sequence current can leave the Δ terminals, thus there is an isolationbetween the primary and secondary sides as shown in figure (c)

d) Y Δ ti ith i l t d t l i thi fi ti b thd) Y‐Δ connection with isolated neutral – in this configuration, because theneutral is isolated, zero sequence current cannot flow and the equivalentcircuit reflects an infinite impedance or an open as shown in figure (d)

e) Δ‐Δ connection – in this configuration, zero‐sequence currents circulate inthe Δ‐connected windings, but no currents can leave the Δ terminals, and theequivalent circuit is as shown in figure (e)

Notice that the neutral impedance plays an important part in the equivalentcircuit. When the neutral is grounded through an impedance Zn, becauseIn=3Io, in the equivalent circuit, the neutral impedance appears as 3Zn in the

28

path of Io.

Page 29: Unsymmetrical Fault

Sequence Impedances of a Loaded Generatorq p

A synchronous machine generates balanced three‐phase internal voltages and is represented as a positive‐sequence set of phasors

aabc E

aa⎥⎥⎥

⎢⎢⎢

⎡= 2

1E

(10.44)

a ⎥⎦⎢⎣

29

Page 30: Unsymmetrical Fault

Sequence Impedances of a Loaded Generatorq p

The machine is supplying a three‐phase balanced load. Applying kirchhoff’s voltage law to each phase we obtain:

(10.45)nnbsbb

nnasaa

IZIZEV

IZIZEV

−−=

−−=

nncscc IZIZEV −−=

Substituting for In = Ia + Ib + Ic into (10.45):

⎤⎡⎤⎡⎤⎡⎤⎡ IZZZZEV )(

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

++

+−

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

c

b

a

nsnn

nnsn

nnns

c

b

a

c

b

a

III

ZZZZZZZZZZZZ

EEE

VVV

)()(

)(

(10.46)

In compact form: abcabcabcabc IZEV −= (10.47)

30

Page 31: Unsymmetrical Fault

Sequence Impedances of a Loaded Generatorq pTransforming the terminal voltages and currents phasors into their symmetrical components:

(10.48)012012012a

abcaa AIZAEAV −=

Multiplying (10.48) by A‐1:

(10.49)012012012

012012012 )(

aa

aabc

aa

IZE

IAZAEV

−=

−= −

h ⎤⎡⎤⎡ +⎤⎡ 111)(111 ZZZZ(10.50)

Where:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

++

+

⎥⎥⎥

⎢⎢⎢

⎡=

2

2

2

2012

11

111

)()(

)(

11

111

31

aaaa

ZZZZZZZZZZZZ

aaaa

nsnn

nnsn

nnns

Z

Performing the above multiplication:

⎥⎥⎤

⎢⎢⎡

=⎥⎥⎤

⎢⎢⎡ +

= 1

0

012 0000

0000)3(

ZZ

ZZZ

s

ns

Z (10.51)

⎥⎥

⎦⎢⎢

⎣⎥⎥⎦⎢

⎢⎣

20000 ZZs

s

31

Page 32: Unsymmetrical Fault

Sequence Impedances of a Loaded Generatorq pSince the generated emf is balanced, there is only positive‐sequence voltage, i.e:

(10 52)⎥⎤

⎢⎡ 0

(10.52)⎥⎥⎥

⎦⎢⎢⎢

=0

012aa EE

( )⎥⎤

⎢⎡

⎥⎤

⎢⎡

⎥⎤

⎢⎡

⎥⎤

⎢⎡ 000 000 aa IZV

Substituting for         and           in (10.49):012aE 012Z

111

000 0 aa IZV −=(10.54)

⎥⎥⎥

⎦⎢⎢⎢

⎣⎥⎥⎥

⎦⎢⎢⎢

−⎥⎥⎥

⎦⎢⎢⎢

=⎥⎥⎥

⎦⎢⎢⎢

⎣2

1

2

1

2

1

0000

0 a

aa

a

a

II

ZZE

VV (10.53) or

222

111

0 aa

aaa

IZV

IZEV

−=

−=

32

Page 33: Unsymmetrical Fault

Sequence Impedances of a Loaded Generatorq pThe three equations in (10.54) can be represented by the three equivalent sequence networks: 

• Important observations:– The three sequences are independent. 

Th i i k i h h li di d i– The positive‐sequence network is the same as the one‐line diagram used in studying balance three‐phase currents and voltages.

– Only the positive‐sequence network has a source and no voltage source for other sequences.h l f h h f f d– The neutral of the system is the reference for positive‐ and negative‐sequence networks, but ground is the reference for zero‐sequence networks. Thus, zero sequence current can only flow if the circuit from the system neutrals to ground is complete.The grounding impedance is reflected in the zero sequence network as 3Zn– The grounding impedance is reflected in the zero sequence network as 3Zn

– The three‐sequence systems can be solved separately on a per phase basis. The phase currents and voltages can then be determined by superposing their symmetrical components of current and voltage respectively.  33

Page 34: Unsymmetrical Fault

Single Line‐To‐Ground Faultg

Three‐phase generator with neutral grounded through impedance Zn and SLGF occurs at phase a through impedance Zf .

Assuming the generator is initially on no‐load, the boundary conditions at the fault point are:

afa IZV = (10.55)

0== cb II (10.56)34

Page 35: Unsymmetrical Fault

Single Line‐To‐Ground Faultg

Substituting for Ib = Ic = 0, the symmetrical components of currents from (10.14) are:are:

⎥⎥⎥⎤

⎢⎢⎢⎡

⎥⎥⎥⎤

⎢⎢⎢⎡

=⎥⎥⎥⎤

⎢⎢⎢⎡

01111

31

2

2

2

1

0a

a

a IaaI

I(10.57)

⎥⎦⎢⎣⎥⎦⎢⎣⎥⎦⎢⎣ 013 22

a aaI

From the above equation, we find that:

1 (10 58) 111

000 0 aa IZV −=

aaaa IIII31210 ===

Phase a voltage in terms of symmetrical components is :

(10.58)

222

111

0 aa

aaa

IZV

IZEV

−=

−=

210aaaa VVVV ++=

: noting and (10.54) from and, ngSubstituti 210210aaaaaa IIIVVV ==

(10.59)

0210 )( aaa IZZZEV ++−= (10. 60)

35

Page 36: Unsymmetrical Fault

Single Line‐To‐Ground Faultg

:get we,3 noting and (10.55), fromfor ngSubstituti .3 Where 0a

0aans IIVZZZ =+=

(10 61)02100 )(3 aaaf IZZZEIZ ++−=

or

(10.61)

f

aa ZZZZ

EI

32100

+++=

h f l

(10.62)

The fault current is

f

aaa ZZZZ

EII

33

3 2100

+++== (10.63)

f

In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is substituted into Eq. (10.54) 

36

Page 37: Unsymmetrical Fault

Single Line‐To‐Ground Faultg

Eq. (10.58) and (10.62) can be represented by connecting the sequence networks in series as shown in the following figure. 

aaaa IIII31210 ===

f

aa ZZZZ

EI

32100

+++=(10.58) (10.62)

37

Page 38: Unsymmetrical Fault

Line‐To‐Line Fault

Three‐phase generator with a fault through an impedance Zf between phase band c.

I 0

Zs

Ia=0

NZsZs

EaEb

Ec

Z

Ib

Va

ZfIcVb

Vc

bfb IZVV =− I = 0

Assuming the generator is initially on no‐load, the boundary conditions at the fault point are:

(10.64) (10.66)bfcb IZVV Ia 00=+ cb II

(10.64)

(10.65)

(10.66)

38

Page 39: Unsymmetrical Fault

Line‐To‐Line Fault

Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents from (10.14) are:from (10.14) are:

⎥⎥⎤

⎢⎢⎡

⎥⎥⎤

⎢⎢⎡

=⎥⎥⎤

⎢⎢⎡

ba

a

IaaII 0

1111

1 21

0

(10.67)

⎥⎥⎦⎢

⎢⎣−⎥⎥⎦⎢

⎢⎣⎥

⎦⎢⎢

⎣ b

b

a

a

IaaI 13 22

From the above equation we find that:From the above equation, we find that:

00 =aI

1

(10.68)

(10 69)ba IaaI )(

31 21 −=

ba IaaI )(31 22 −=

(10.69)

(10.70)

ba 3

39

Page 40: Unsymmetrical Fault

Line‐To‐Line Fault

Also, from (10.69) and (10.70), we note that:21 (10 71)21aa II −=

From (10.16), we have:

VVVVVVVV )()( 22102120

(10.71)

2120

210

aaab

aaaa

aVVaVV

VVVV

++=

++=

(10.16)

bf

aa

aaaaaacb

IZVVaa

VaaVVaVVaVVV

=−−=

++−++=−

))((

)()(212

22102120

(10.72)

2210aaac VaaVVV ++=

000 0 aa IZV −=:get we, noting and (10.54) from and for ngSubstituti 1221

a aaa IIVV −=

bfaa IZIZZEaa =+−− ])()[( 1212 (10.73)(10.54)

222

111

0

0

aa

aaa

aa

IZV

IZEV

V

−=

−=

:get we(10.69), from for ngSubstituti bI

))((3

)( 22

1121

aaaaI

ZIZZE afaa −−

=+− (10.74) ba IaaI )(31 21 −= (10.69)

))((

40

Page 41: Unsymmetrical Fault

Line‐To‐Line Fault

:in results for solving,3 Since 1aIa))(aa(a 22 =−−

)( 211

f

aa ZZZ

EI

++= (10.75)

The phase currents are

⎥⎤

⎢⎡⎥⎤

⎢⎡

⎥⎤

⎢⎡ 0111aI

(10 76)

⎥⎥⎥

⎦⎢⎢⎢

⎣−⎥⎥⎥

⎦⎢⎢⎢

=⎥⎥⎥

⎦⎢⎢⎢

⎣1

1

2

2

11

a

a

c

b

II

aaaa

II

The fault current is

(10.76)

The fault current is

12 )( acb IaaII −=−= or 13 ab IjI −=(10.77) (10.78)

41

Page 42: Unsymmetrical Fault

Line‐To‐Line Fault

Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –sequence networks as shown in the following figure. 

21aa II −= 1 aE

I =

42

aa)( 21

fa ZZZ

I++

=

Page 43: Unsymmetrical Fault

Double Line‐To‐Ground FaultFigure 10.14 shows a three‐phase generator with a fault on phases b  and c through an impedance Zf to ground. Assuming the generator is initially on no‐load the boundary conditions at the fault point areload, the boundary conditions at the fault point are

0

)(210 =++=

+==

aaaa

cbfcb

IIII

IIZVV (10.79)(10.80)

From (10.16), the phase voltages Vb and Vc are

Figure 10 14Figure 10.14Double line‐to‐ground fault

43

Page 44: Unsymmetrical Fault

Double Line‐To‐Ground Fault

(10.81)2210

2120aaab

VaaVVV

aVVaVV

++=

++=

(10 82)

21 ( )

aaac VaaVVV ++ (10.82)

thatnote weabove from ,VSinceVb c=

21aa VV =

Substituting for the symmetrical components of current in (10.79), we get

(10.83)

)2(

)(210

22102120)(

aaaf

aaaaaafb

IIIZ

IaaIIaIIaIZV

−−=

+++++=

03 af IZ= (10.84)

44

Page 45: Unsymmetrical Fault

:havewe(10.81), into (10.83)fromfor and(10.84)from for ngSubstituti 2b aVV

10

1200 )(3

aa

aaaf

VV

VaaVIZ

−=

++=

( ),( )( )g b a

(10.85)

110 aa IZE

I−

:get we,for solving and(10.85)into(10.54)from voltageofcomponentslsymmetrica for the ngSubstituti

0aI

(10 86))3( 0

0

f

aaa ZZ

I+

−=

Also, substituting for the symmetrical components of voltage in (10.83), we obtain11IZE − (10 87)

(10.86)

22

ZIZE

I aaa

−−=

E

:get we,for solving and (10.80) into I and for ngSubstituti 120a aa II

(10.87)

f

f

aa

ZZZZZZ

Z

EI

3)3(

02

021

1

++

++

= (10.88)

45

Page 46: Unsymmetrical Fault

sequence-negative theofn combinatio paralel with theseriesin impedancesequence-positive theconnectingby drepresentebecan (10.88)-(10.86)Equation

(10 8)fromfoundthenarecurrentphaseThefoundareIandIand (10.87), and (10.86)in dsubstitute is (10.86) from found I of valueThe

10.15. figure ofcircuit equivalent in theshown as networks sequence-zero and

20

1a

from obtained iscurrent fault theFinally,(10.8).fromfoundthen arecurrent phaseThefound. are I and I and aa

03bf IIII =+= (10.89)3 acbf IIII + ( )

Figure 10.15 Sequence network connection for double line‐to‐ground fault

46

Page 47: Unsymmetrical Fault

The one-line diagram of a simple power system is show in Figure 10 16 The neutral of each

EXAMPLE

The one-line diagram of a simple power system is show in Figure 10.16. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-mva base tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phasephase.Determine the fault current for the following faultsa. A balaced three-phase fault at bus 3 through a fault impedance = j 0.1 per unitb. A single line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit c A line-to-line fault at bus 3 through a fault impedance = j 0 1 per unit

fZfZ

fZc. A line-to-line fault at bus 3 through a fault impedance j 0.1 per unit d. A double line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit

fZfZ

Item Base Rated X1 X2 X0

MVA VoltageG1 100 20-kV 0.15 0.15 0.05G2 100 20 kV 0.15 0.15 0.05T1 100 20/220 kV 0.10 0.10 0.10T1 100 20/220 kV 0.10 0.10 0.10T2 100 20/220 kV 0.10 0.10 0.10L12 100 220 kV 0.125 0.125 0.30L13 100 220 kV 0.15 0,15 0.35L23 100 220 kV 0 25 0 25 0 7125L23 100 220 kV 0.25 0.25 0.7125

47

Page 48: Unsymmetrical Fault

Figure 10 16Figure 10.16

Fault

Item BaseMVA

RatedVoltage

X1 X2 X0

G1 100 20-kV 0.15 0.15 0.05G2 100 20 kV 0 15 0 15 0 05G2 100 20 kV 0.15 0.15 0.05T1 100 20/220 kV 0.10 0.10 0.10T2 100 20/220 kV 0.10 0.10 0.10L12 100 220 kV 0.125 0.125 0.30

48

L13 100 220 kV 0.15 0,15 0.35L23 100 220 kV 0.25 0.25 0.7125

Page 49: Unsymmetrical Fault

To find Thevenin impedance viewed from the faulted bus (bus 3), we convert the delta formed by buses 123 to an equivalent Y as shown belowformed by buses 123 to an equivalent Y as shown below

03571430)15.0)(125.0( jjjZ )250)(1250( jj

Fig. 10.17Positive-sequence impedance

0357143.0525.01 j

jZ S == 0595238.0

525.0)25.0)(125.0(

2 jj

jjZ S ==

0714286.05250

)25.0)(15.0(3 j

jjjZ S ==

525.0j

49

Page 50: Unsymmetrical Fault

0714286.05952381.0

)3095238.0)(2857143.0(133 j

jjjZ +=

22.0jj

=

50

Page 51: Unsymmetrical Fault

To find thevenin impedance viewed from the faulted bus (bus 3), we convert the delta formed by buses 123 to an equivalent Y as shown in figure 10.19(b)

)350)(300( jj 0770642.03625.1

)35.0)(30.0(1 j

jjjZ S ==

15688070)7125.0)(30.0( jjjZ ==

j0.077064

1568807.03625.12 j

jZ S ==

1830257.03625.1

)7125.0)(35.0(3 j

jjjZ S ==

51Fig. 10.19Zero-sequence impedance

j

Page 52: Unsymmetrical Fault

Combining the parallel branches, the zero‐sequence thevenin impedance isg p q p

1830275.073394490

)2568807.0)(4770642.0(033 j

jjjZ +=

35.07339449.033

j

jj

=

j0.077064

So, the zero‐sequence impedance diagram is show in fig. 10.20

52

Fig. 10.20 Zero‐sequence network

Page 53: Unsymmetrical Fault

(a) Balanced three‐phase fault at bus 3Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault current is

A90820 1j3 1250.1(F)I )0(3 °∠V a

A90-820.1pu-j3.1250.1 j0.22Z Z

(F)If

133

)0(33 °∠==

+=

+=

j

(b) Single line‐to ground fault at bus 3F (10 62) th t f th f lt tFrom (10.62), the sequence component of the fault current are

puj

V a

-j0.91743(j0.1)0.35 j0.22 j0.22

0.1 3ZZZ Z

II If

033

233

133

)0(323

13

03 =

+++=

+++===

The fault current is :

jIII a ⎤⎡−⎤⎡⎤⎡⎤⎡⎤⎡ 752323111 03

033

pujI

III

aaaa

III

c

b

⎥⎥⎥

⎢⎢⎢

⎡=⎥⎥⎥

⎢⎢⎢

=⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

007523.2

00

3

11

111 3

03

03

3

2

2

3

3

3

53

Page 54: Unsymmetrical Fault

(c) Line‐to Line fault at bus 3The zero‐sequence component of current is zero, i.e.,

0 I03 =

The positive‐and negative‐sequence components of the fault current are

V a

puV a

-j1.8519j0.1 j0.22 j0.22

0.1 ZZZ Z

I If

033

233

133

)0(323

13 =

++=

+++=−=

The fault current is

pujj

aaaa

III

c

b

a

⎥⎥⎥

⎢⎢⎢

−−=

⎥⎥⎥

⎢⎢⎢

−−

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

2075.32075.30

8519.18519.10

11

111

2

2

3

3

3

j ⎦⎣⎦⎣⎦⎣⎦⎣ 3

(d) Double Line‐to Line‐fault at bus 3The positive‐sequence component of the fault current is

puV a

-j2.6017

j0.3j0.35j0.22j0.3)(j0.35 j0.22 j0.22

0.1

)Z3ZZ)Z3Z(Z

Z I

f033

233

f033

2331

33

)0(313 =

+++

+=

+++

+=

)f3333

54

Page 55: Unsymmetrical Fault

The negative‐sequence component of current is :

pujjV a

j1.9438 j0.22

)6017.2)(22.0(0.1 Z

IZ I 2

33

133

133)0(32

3 =−−

−=−

−=

Th t f t iThe zero‐sequence component of current is:

pujjV a

j0.6579j0 3j0 35

)6017.2)(22.0(0.1 3ZZ

IZ I 0

133

133)0(30

3 =+−−

−=+

−−=

j0.3j0.353ZZ f33 ++

And the phase currents are :

jI a ⎤⎡⎤⎡⎤⎡⎥⎤

⎢⎡ 06579.01113

pujj

j

aaaa

II

c

b

⎥⎥⎥

⎢⎢⎢

°∠°∠=

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

07.14058.493.165058.4

0

9438.16017.2

6579.0

11

2

2

3

3

3

And the fault currents is:

°∠=+= 909732.1)( 333cb IIFI

55

Page 56: Unsymmetrical Fault

UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX

Single Line‐to‐Ground Fault Using Zbus

kkkk

VIII )0(021

210 === (10.90)

fkkkkkkkkk ZZZZ 3021 +++

k. busat oltageprefault v theis)0(V andmatrix impedance bus

ingcorrespond theof axisk in the elements diagonal theare Zand Z, ZWhere 2kk

1kk

k

okk

: iscurrent phasefault The

=abckI A 012

kI (10.91)

Line‐to‐Line Fault Using Zbus

00 =kIV )0(

(10.92)

fkkkk

kkk ZZZ

VII++

=−= 2121 )0(

(10.93)

56

Page 57: Unsymmetrical Fault

Double Line‐to‐Ground Fault Using Zbus

fkkkk

fkkkkkk

kk

ZZZZZZ

Z

VI

3)3(

)0(

02

021

1

++

++

= (10.94)

fkkkk

2

112 )0(

kk

kkkkk Z

IZVI −−= (10.95)

kk

kkkkk ZZ

IZVI3

)0(0

110

+−

−= (10.96)fkk ZZ 3+

iscurrent result theand(10.91), fromobtainedarecurrentsphaseThe matrix. impedance bus ingcorrespond theof axisk in the elements diagonal theare Zand , Zand , ZWhere 2

kk1kk

okk

( ),pp

Ck

bkk IIFI +=)( (10.97)

57