Unsymmetrical Short Circuit

Unsymmetrical Short Circuit

Submitted By:Santosh Kumar GuptaAssistant ProfessorEE DepartmentSIT Sitamarhi

FaultAnalysis

2

• Fault types:

– balanced faults (<5%)

• three‐phase toground

• Three‐phase

– unbalanced faults

• single‐line to ground(60%‐75%)

• double‐line to ground(15%‐25%)

• line‐to‐line faults (5%‐15%)

Example impact offault

3

The second largest blackout in the history of TEPCO

(The Tokyo Electric Power Company, Inc.) hit central

Tokyo area at about 7:38 a.m. on August 14, 2006. It

was caused by a floating crane on a barge going

upstream on a river on the eastern edge of the city.

The workers on the boat did not realize that the 33

meter crane was raised too high, so it hit TEPCO's

275 kV double circuit transmission lines that run

across theriver.

As a result of the accident the transmission

lines were short‐circuited and the wires

damaged. The relay protection operatedand

tripped both lines

SymmetricalComponents

4

• Three phase voltage or current is in a balance condition if it has

the following characteristic:

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

– The system has sequence ofa,b,c

– The angle between phase is displace by 120 degree

• If one of the above is character is not satisfied, unbalanced

occur. Example:


5

• For unbalanced system, power system analysis cannot be

analyzed 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 components.


6


7

By convention, the direction of rotation of the phasors is taken

to becounterclock‐wise.

Positive sequence:

I1 = I 10 = I 1

a a a

I1 = I1240 = a2 I1

b a a

(10.1)

I1 = I 1120 = aI 1

c a a

Where we defined an operator a that causes a counterclockwise rotation of 120

degree, such that:

a = 1120 = cos120 + j sin 120 = −0.5 + j0.866

a2 = (1120) (1120) = 1240 = −0.5 − j0.866 1 + a + a2 =0

(10.2)

a3 = 1360 = 1 + j0(10.3)


8

Negative sequence:

I 2 = I 20a a

I 2 = I 2120 = aI 2

b a a

(10.4)

I 2 = I 2240 = a2 I 2

c a a

a b cI 0 = I 0 = I 0

Zero sequence:

(10.5)


9

Consider the three‐phase unbalancedcurrent of Ia , Ib , Ic

I = I 0 + I 1 + I 2

a a a a

I = I 0 + I 1 + I2

b b b b

(10.6)

I = I 0 + I1 + I 2

c c c c

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

components

I = I 0 + I 1 + I 2

a a a a

a

a I 0

I = I 0 + aI 1 + a2 I 2

c a a a

I = I 0 + a2 I 1 + aI 2

b a a a(10.7)

22

1 a

a b Ia a

I 1 1 1

a 2 a I 1

Ic

I = 1 (10.8)


10

Equation 10.8 can be writtenas:

aIabc = AI012

WhereA is known as symmetrical components transformation matrix,

(10.9)

anda

which transforms phasor currents Iabc into components currents I012

1

a2 a

1 1 1

a2 a A = 1 (10.10)

Solving (10.9) for the symmetrical components of currents:

aI012 = A−Iabc (10.11)

The inverse of A is givenby: 1 1 1 1

a

a

a2

a2

1

13

A− =(10.12)


11

From (10.10) and (10.12), we concludethat

(10.13)A− =1

A*

3

Substituting for A‐1 in (10.11), wehave:

b

a

a

a

Ia2I 1

I 0

= 13

(10.14)

c a

1 1 1 I1 a

a 2 I 2 1 a I

or in component form, the symmetrical components are:

3

1

3

1

1

0

a b ca

a b ca

I = (I + aI + a2 I )

I = (I + I + I )

(10.15)

3

12

a b caI = (I + a2 I + aI )


12

Similar expressions exists forvoltage:

(10.16)

V = V 0 + aV 1 + a2V 2

c a a a

20 2 1

V = V 0 +V 1 +V 2

a a a a

ab a aV = V + a V + aV a

V abc = AV 012 (10.17)

The symmetrical components in terms of unbalanced voltages are:

+ aV + a2V )b c

3

1

3

1

1

0

aa

aa

V = (V

+ Vb + Vc )V = (V

(10.18) aV012 =A− Vabc (10.19)

+ a2V + aV )b c3

12

aaV = (V


13

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

components.

abcT abc*

S(3 ) = V I (10.20)

S = (AV012 )T (AI012)*

(3 ) a a

Substituting (10.9) and (10.17) in (10.20), weobtain:

(10.21)012

a

T * 012

a

T *

A A I= V

= A,ATSince AT A* = 3 complex powerbecomes

*** 2 20 0 1 1

a aa aa a= 3V I + 3V I + 3V I

012T 012*

S(3 ) = 3(V I )

(10.22)

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

symmetrical componentspowers.

Example1

14

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 as

reference and assuming that line c is open, find the symmetrical

components of the linecurrents.

aIa =100A

1

3

1

1

0

a b ca

aa

I = (I + aI + a2 I )

I = (I + Ib + Ic )

b Ib = 10180A 3

1

3

2

a b caI = (I + a2 I + aI )

c Ic = 0A

Solution

15

Ic = 0

1

The line currentare:

Ia = 100 Ib =10180

From (10.15):3

1

3

1

1

a

0

a

a b c

a b c

= (I + aI + a2 I )

= (I + I + I )

I

I

(100 + 10180 + 0) = 0a

I (0) =

a (100 + 10(180 +120) + 0)

3

1

3I (1) =

0 Sequence

+ Sequence

3

1a b c

2

aI = (I + a2 I + aI )

=5 − j2.89 = 5.78 − 30A

a ) + 0)0 + 10(180+ 240I = (103

1(2)

‐Sequence

=5 + j2.89 =5.7830A

From (10.4) bI (0) = 0 c

I (0) = 0

bI (1) = 5.78 −150 A c

I (1) = 5.7890A

bI = 5.78150A(2) cI (2) = 5.78 − 90A

Example2

16

Exercise1

17

Show that :

(a)(1+ a)

=1120(1+ a2 )

= 3−180(1+ a)2

(1− a)2

(b)

Exercise2

18

Obtain the symmetrical components for the set of unbalanced voltages

Va = 300 −120,Vb = 20090,Vc = 100 − 30

1

3

1

1

0

a b ca

a b ca

V = (V + aV + a2V )

V = (V +V + V )V 012 =

42.2650 − 120

193 .1852 − 135

86.9473 − 84.8961

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

3

1

3

2

a b caV = (V + a2V + aV )

I 0 = 3 − 30, I 1 = 590, I 2 = 430a a a

Obtain the original unbalancedphasors. I = I 0 + I 1 + I 2

a a a a

I = I 0 + aI 1 + a2 I 2

c a a a

I = I 0 + a2 I 1 + aI 2

b a a a

Iabc =

8.1854 42.2163

4 − 30

8.1854 − 102 .2163

Exercise3

19

The line‐to‐line voltages in an unbalanced three‐phase supply areVab = 1000 0,Vbc = 866 .0254 −150,Vca = 500120

Determine the symmetrical components for line and phase voltages, then find the

phase voltages Van, Vbn, andVcn.

VL 012 =

0.030 0.00

Va 012 = Vabc =

440 .9586 − 19.1066

600 .9252 − 166 .1021

333 .3333 60288 .675130

763 .7626 − 10.8934 440 .9586 − 40.8934

166.6667 60

SequenceImpedance

20

• The impedance of an equipment or component to the current of differentsequences.

• positive‐sequence impedance (Z1): Impedance that causes a positive‐sequence current toflow

• negative‐sequence impedance (Z2): Impedance that causes a negative‐sequence current toflow

• zero‐sequence impedance (Z0): Impedance that causes a zero‐sequence current toflow

Sequence Impedance of Y‐ConnectedLoad

21

(10.23)

Line to ground voltagesare:

Va = Zs Ia + Zm Ib + Zm Ic + Zn In

(10.24)I n = I a + I b + I c

Vb = Zm Ia + Zs Ib + Zm Ic + Zn In

Vc = Zm Ia + Zm Ib + +Zs Ic + Zn In

Kirchhoff’ current law:

Substituting In into (10.23):

Zm + Zn IaVa (Zs + Zn ) Zm +Zn

(10.25)

(10.26)

m n b mb

I

Zm + Zn Zm +ZnVc

V = Z

(Zs + Zn )Ic

+ Zn (Zs +Zn ) Z + Z

Vabc = Zabc Iabc


22

Zm + Zn Zm + Zn

(Zs + Zn )

(10.27) m n m n

Zm +Zn

+ Z (Zs +Zn ) Z + Z

Zm + Zn (Zs + Zn )

Zabc = Z

Writing Vabc and Iabc in terms of their symmetrical components:

(10.28)AV012 = Zabc AI012

a a

Multiplying (10.28) by A‐1:

(10.29)a

a a

= Z012I012

V012 = (A− Zabc A)I012

Z012 = A− Zabc Awhere (10.30)

(10.31)

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

1

13

a a2 Z + Z (Z + Z )1

+ Z (Z + Z ) Z + Z

a Z + Za2

a2 Z 1

1 1 1 (Zs + Zn ) Zm + Zn Zm + Zn 1 1 1 1

a a2 a Z 012 =

n sm n m n

n ms nm n


23

Performing the multiplication in(10.31):

0

0s m

0 (Zs − Zm )

0 (Z − Z )

0

(Zs + 3Zn +2Zm ) 0

Z012 = (10.32)

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

(Zs +3Zn ) 0 0 (10.33)

0

s 0 (Zs )

0 (Z )Z012 = 0

Sequence Impedance of TransmissionLines

24

For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different

and larger approximately 3 times than positive and negative sequence.

Sequence Impedance of SynchronousMachineThe positive-sequence generator impedance is the value found when positive-

sequence current flows from the action of an imposed positive-sequence set of

voltages.

The negative-sequence reactance is close to the positive-sequence

dX 2

X "substransient reactance, i.e :

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

X lX 0

Sequence Impedances ofTransformer

25

• Series Leakage Impedance.– the magnetization current and core losses represented by the shunt branch

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 notchange if the phase sequence ischanged.

• Therefore, the positive andnegative 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 notrotate

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


26

Connection diagram Zero‐sequence circuit

Figure (a)

Figure (b)

Figure (c)

Figure (d)

Figure (e)


27

Description of Zero sequence EquivalentCircuit

(a)Y‐Y connections with both neutrals grounded – We know that the zero sequence current

equals the sum of phase currents. Since both neutrals are grounded, there is a path for the zero

sequence current to flow in the primary and secondary, and the transformer exhibits the

equivalent leakage impedance per phase as shown in Fig. (a).

(b)Y‐Y connections with primary the neutral grounded – The primary neutral is grounded, but

since the secondary neutral is isolated, the secondary phase current must sum up to zero. This

means that the zero‐sequence current in the secondary is zero. Consequently, the zero

sequence current in the primary is zero, reflecting infinite impedance or an open circuit as

shown in Fig.(b).


28

c)Y‐Δ with grounded neutral – in this configuration, the primary currents can

flow because the zero‐sequence circulating current in the Δ‐connected

secondary and a ground return path for the Y‐connected primary. Note that no

zero‐sequence current can leave the Δ terminals, thus there is an isolation

between the primary and secondary sides as shown in figure (c)

d)Y‐Δ connection with isolated neutral – in this configuration, because the

neutral is isolated, zero sequence current cannot flow and the equivalent

circuit reflects an infinite impedance or an open as shown in figure (d)

e)Δ‐Δ connection – in this configuration, zero‐sequence currents circulate in

the Δ‐connected windings, but no currents can leave the Δ terminals, and the

equivalent circuit is as shown in figure (e)

Notice that the neutral impedance plays an important part in the equivalent

circuit. When the neutral is grounded through an impedance Zn, because

In=3Io, in the equivalent circuit, the neutral impedance appears as 3Zn in the

path of Io.

Sequence Impedances of a LoadedGenerator

29

A synchronous machine generates balanced three‐phase internal voltages and is

represented as a positive‐sequence set ofphasors

a a

1

Eabc = a2E(10.44)


30

The machine is supplying a three‐phase balanced load.Applying kirchhoff’s voltage

law to each phase weobtain:

(10.45)

Va = Ea − Zs Ia − Zn In

Vb = Eb − Zs Ib − Zn In

Vc = Ec − Zs Ic −Zn In

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

b I Ia

(Zs + Zn )Ic

nb b Vc

Ec Zn Zn

V = E −Va Ea

Zn (Zs + Zn ) Z

(Zs + Zn ) Zn Zn

(10.46)

In compactform: V abc = Eabc −Zabc Iabc (10.47)


31

Transforming the terminal voltages and currents phasors into their symmetrical

components:

(10.48)AV012 = AE012 − Zabc AI012

a a a

Multiplying (10.48) byA‐1:

(10.49)a a

a a a

= E012 −Z012I012

V012 = E012 − (A−Zabc A)I012

(10.50)Where:

a

1 1

a2

a a2 1

1

(Z + Z )1

13

Z

a Z

a2

Z

(Z + Z ) Z

s n

ZnZn

nn

ns nn

1 1 1 (Zs + Zn )1

a

a2

Z012 = 1

Performing the abovemultiplication:

00 = 0

0 Z0 0 0

Z1 0 Z

(Z + 3Z ) 0

s

s n

Z012 = (10.51)

0 00 Z2 0 Z

s


32

Since the generated emf is balanced, there is only positive‐sequence voltage, i.e:

0

(10.52)0

012 aaE = E

0 aaZ0 0 0 I 0 V 0

Substituting for and Z012 in (10.49):012

aE

V 0 = 0 − Z 0I 0

a a

(10.54)

0

1

0

a V 2

1 1

a

0 Ia Z 2 I 2

Va = Ea − 0 Z

0

(10.53) orV 2 = 0 − Z 2I 2

a a

V 1 = E − Z 1I 1

a a a


The three equations in (10.54) can be represented by the three equivalent

sequence networks:

• Important observations:– The three sequences are independent.

– The positive‐sequence network is the same as the one‐line diagram used in studying balance three‐phase currents andvoltages.

– Only the positive‐sequence network has a source and no voltage source for other sequences.

– 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 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 andvoltage respectively. 33

Single Line‐To‐GroundFault

34

Three‐phase generator with neutral grounded through impedance Zn andSLGF

occurs at phase a through impedanceZf.

Assuming the generator is initially on no‐load, the boundary conditions at the

fault point are:

Va = Z f I a(10.55)

Ib = Ic = 0 (10.56)


35

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

are:

0= 13

a

a

a

1 I

a2 I 1 (10.57)

a 1 a 0

I0 1 1 1 a

a2 I 2

From the above equation, we findthat: V 0 = 0 − Z 0I 0

a a

aaaa3

1210I = I = I = I

Phase a voltage in terms of symmetrical components is :

(10.58) 1

V 2 = 0 − Z 2I 2

a a

a a aV = E − Z 1I 1

V = V 0 + V 1 + V2

a a a a

Substituting V 0 ,V 1and V 2 from (10.54) and noting I 0 = I 1 = I 2 :a a a a a a

(10.59)

V = E − (Z 0 + Z 1 + Z 2 )I 0

a a a(10.60)


36

Where Z 0 = Z + 3Z .Substituting for V from (10.55), and noting I = 3I 0 , we get :s n a a a

3Z I 0 = E − (Z 0 + Z 1 + Z 2 )I 0

f a a a

or

(10.61)

f

Ea

aZ 0 + Z 1 + Z 2 + 3Z

I 0=

(10.62)

The fault current is

f

3Ea

Z 0 + Z1 + Z 2 + 3ZI = 3I 0 =

a a

(10.63)

In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is

substituted into Eq.(10.54)


37

Eq. (10.58) and (10.62) can be represented by connecting the sequence

networks in series as shown in the following figure.

aaaa3

1210I = I = I = I

f

Ea

aZ 0 + Z 1 + Z 2 + 3Z

I 0=(10.58) (10.62)

Line‐To‐LineFault

38

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

Ia=0

Zs

N

ZsZs

EaEb

Ec

Ib

Va

ZfIcVb

Vc

Assuming the generator is initially on no‐load, the boundary conditions at the

fault point are:

(10.64) (10.66)Vb − Vc = Z f Ib Ia = 0

Ib + I c =0 (10.65)


39

Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents

from (10.14) are:

a

I= 1

1 0

a 2 I 1 (10.67)

b

b

a

a

a −I13

I0 1 1 1 a

a2 I 2

From the above equation, we findthat:

= 0aI 0 (10.68)

ba3

11 2I = (a − a )I

ba

12 2I = (a − a)I

(10.69)

(10.70)

3


40

Also, from (10.69) and (10.70), we notethat:

I 1 = −I 2

a a

From (10.16), wehave:

(10.71)

V = V 0 + a2V 1 + aV 2

b a a a

V = V 0 +V 1 +V 2

a a a a

(10.16)

= Z f Ib

V −V = (V 0 + a2V 1 + aV 2 ) − (V 0 + aV 1 + a2V 2 )b c a a a a a a

= (a2 − a)(V 1 −V 2 )a a (10.72)

V = V 0 + aV 1 + a2V 2

c a a a

Substituti ng for V 1 and V 2 from (10.54) and noting I 2 = −I1 , we get :a a a a

(10.73)

(10.54)

V 0 = 0 − Z 0I 0

a a

V 2 = 0 − Z 2I 2

a a

V 1 = E − Z 1I 1

a a a

(a 2 − a)[E − (Z 1 + Z 2 )I 1 ] = Z Ia a f b

Substituti ng for I b from (10.69), we get :

3I 1a

(a − a 2)(a 2 − a)f

E − (Z 1 + Z 2 )I1 = Za a

(10.74) ba3

11 2I = (a − a )I (10.69)


41

aSince (a − a2 )(a2 − a) = 3,solving for I 1 results in :

f

Ea

a(Z 1 + Z 2 + Z )

I 1 =(10.75)

The phase currentsare

1 0 I a

a

1 a

c

b I

a 2 −I 1 I 1

1 1 I = 1 a 2 a

a

(10.76)


I = −I = (a 2 − a)I 1

b c aor

abI = − j 3I 1(10.77) (10.78)


42

Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –

sequence networks as shown in the following figure.

I 1 = −I 2

a a1 Ea

I =1 2

f

a(Z + Z + Z )

Double Line‐To‐GroundFault

43

Figure 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 pointare

I = I 0 + I1 + I 2 = 0a a a a

From (10.16), the phase voltages Vb and Vcare

Vb = Vc = Z f (Ib + Ic ) (10.79)

(10.80)

Figure 10.14Double line‐to‐ground fault

Double Line‐To‐GroundFault

44

(10.81)V = V 0 + a2V 1 + aV 2

b a a a

V = V 0 + aV 1 + a2V 2

c a a a (10.82)

SinceVb = Vc , from above we note that

V 1 = V 2

a a

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

V = Z (I 0 + a 2 I 1 + aI 2 + I 0 + aI 1 + a 2 I 2 )(b ) f a a a a a a

= Z (2 I 0 − I 1 − I 2 )f a a a

(10.83)

= 3 Z f aI 0 (10.84)

Substituti ng for V from (10.84) and for V 2 from (10.83) into (10.81), we have :b a

3Z I 0 = V 0 + (a 2 + a)V 1

f a a a

= V 0 −V 1 (10.85)a a

Substituti ng for the symmetrical components of voltage from (10.54) into (10.85)

E − Z 1I 1

aand solving for I 0 , we get :

a(Z 0 + 3Z )

I 0 = − aa

f

Also, substituting for the symmetrical components of voltage in (10.83), we obtain

E − Z 1I 1

(10.86)

Z 2aI 2 = − a a

Substituti ng for I 0 and I2 into (10.80) and solving for I 1 , we get :a a a

(10.87)

f

45

f

Ea

a

Z 2 + Z 0 +3Z

Z 2 (Z 0 + 3Z )Z 1 +

I 1 = (10.88)

Equation (10.86) - (10.88) can be represented by connecting the positive -sequence

impedance in series with the paralel combination of the negative - sequence

46

The value of I1 found from (10.86) is substitute d in (10.86) and (10.87),

and zero - sequence networks as shown in the equivalent circuit of figure 10.15.

and I0 and I2 are found. The phase current are then found from (10.8).a a

Finally, the fault current is obtained from

a

= 3I 0 (10.89)acI f = Ib + I

Figure 10.15 Sequence network connection for double line‐to‐groundfault

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 phase.

Determine the fault current for the following faults

a. A balaced three-phase fault at bus 3 through a fault impedance Z f = j 0.1 per unit

47

b. A single line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unitZ f

c. A line-to-line fault at bus 3 through a fault impedance Z f = j 0.1 per unit

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

Item Base Rated X1 X2 X0

MVA Voltage

G1 100 20-kV 0.15 0.15 0.05

G2 100 20 kV 0.15 0.15 0.05

T1 100 20/220 kV 0.10 0.10 0.10

T2 100 20/220 kV 0.10 0.10 0.10

L12 100 220 kV 0.125 0.125 0.30

L13 100 220 kV 0.15 0,15 0.35

L23 100 220 kV 0.25 0.25 0.7125

Figure 10.16

Fault

MVA Voltage

G1 100 20-kV 0.15 0.15 0.05

G2 100 20 kV 0.15 0.15 0.05

T1 100 20/220 kV 0.10 0.10 0.10

T2 100 20/220 kV 0.10 0.10 0.10

L12 100 220 kV 0.125 0.125 0.30

L13 100 220 kV 0.15 0,15 0.35

L23 100 220 kV 0.25 0.25 0.7125

Item

48

Base Rated X1 X2 X0

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 below

Fig. 10.17Positive-sequence impedance

j0.525Z1S

=( j0.125)( j0.15)

= j0.0357143

j0.525Z

2S=

( j0.125)( j0.25)= j0.0595238

Z 3S=

( j0.15)( j0.25) = j0.0714286

j0.525

49

33j0.5952381

Z 1 =( j0.2857143 )( j0.3095238 )

+ j0.0714286

50

= j0.22

Tofind thevenin impedance viewed from the faulted bus (bus 3), we convert the

delta formed by buses 123 to an equivalent Yas shown in figure 10.19(b)

j1.3625

( j0.30)( j0.35)Z1S = j0.0770642=

Z

j0.077064

=( j0.30)( j0.7125)

= j0.15688072S

j1.3625

51

Z3S

j1.3625

=( j0.35)( j0.7125)

= j0.1830257

Fig. 10.19Zero-sequence impedance

Combining the parallel branches, the zero‐sequence thevenin impedanceis

33Z 0 =

( j0.4770642)( j0.2568807) + j0.1830275

j0.7339449

= j0.35

j0.077064

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

Fig. 10.20

Zero‐sequence network

52

(a) Balanced three‐phase fault at bus3

Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault

current is

1.03(0)V a

Z1

33 f

3 = -j3.125 pu =820.1 - 90 A=I (F)=+ Z j0.22 + j0.1

(b) Single line‐to ground fault at bus3

From (10.62), the sequence component of the fault current are

V a1.0

j0.22 + j0.22 + j0.35 +3(j0.1)= = -j0.9174 pu

f33Z1

33 33+ Z2 + Z0 + 3ZI0 = I1 = I2 =

3 3 3

3(0)

The fault current is:

I a 1 1 1 I 0 3I 0

pu

53

aI b

I c

0

0

−j2.7523

0

= 0 == 1

3

I 0

3

3

3

a 2

1 a a 2 I 03

3

3

(c) Line‐to Line fault at bus3

54

The zero‐sequence component of current is zero,i.e.,

I0 = 03

The positive‐and negative‐sequence components of the fault currentare

V a1.0

fZ1

33

3(0)= = -j1.8519 pu

+ Z2 + Z0 + Z j0.22 + j0.22 + j0.133 33

I1 = −I 2 =3 3


I b

I c

a − j1.8519 = − 3.2075 pu

0 0

1

= 1

1

a 2 − j1.8519 −3.2075

Ia 1 1 a2

a3

3

3

(d) Double Line‐to Line‐fault at bus3

The positive‐sequence component of the fault currentis

V a

j0.22 + j0.35 + j0.3

1.0

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

= = -j2.6017 pu

f33Z2

33

33

3(0)

3

+ Z0 + 3Z )

Z2 (Z0 + 3Z )Z1 + 33 33 f

I1 =

The negative‐sequence component of current is:

= j1.9438 pu1.0 − ( j0.22)(− j2.6017)V a

j0.22Z2

33

33 333(0)

3

− Z1 I1

I2 = − = −

The zero‐sequence component of currentis:

= j0.6579 pu1.0 − ( j0.22)(− j2.6017)

Z0

V a I1

3(0) 33 33

3+ 3Z

− Z1

I0 = − = −j0.35 + j0.3

33 f

And the phase currents are:

I a 1

a

55

I b

I c

− j2.6017 = 4.058165.93 pu= 1

a 2 j1.9438 4.05814.07

1 1 j0.6579 0 a2

1 a3

3

3

And the fault currents is:

I (F ) = I b + I c =1.9732903 3 3

UNBALANCED FAULTANALYSIS USING BUS IMPEDANCE MATRIX

Single Line‐to‐Ground Fault UsingZbus

Vk (0)I 0 = I 1 = I 2 =k k k

(10.90)

kkZ1 + Z 2 +Z 0 + 3Z

kk kk f

Where Z1 , Z2 and Zo are the diagonal elements in the k axis of the correspondingkk kk kk

bus impedance matrix and Vk (0)is the prefault voltage at busk.

The fault phase current is :

I abc

k k= A I 012 (10.91)

Line‐to‐Line Fault Using Zbus

kI 0 = 0 (10.92)

f

56

kk kk

Vk (0)I k = −I 2

kZ + Z 2 +Z

=1

1(10.93)

Double Line‐to‐Ground Fault UsingZbus

57

kk fkk

kk kk f

kk

Vk (0)k

Z 2 + Z 0 + 3Z

Z 2 (Z 0 + 3Z )Z1

I 1 =

+

(10.94)

k Z 2

V (0) − Z 1 I 1

I 2 =− k kk k (10.95)kk

Ik0 V (0) − Z 1 I 1

= − k kk k(10.96)

fkkZ 0 + 3Z

Where Z1 , and Z 2 , and Zo are the diagonal elements in the k axis of the correspond ingkk kk kk

bus impedance matrix. The phase currents are obtained from (10.91), and the result current is

I (F ) = I b + I C k k k

(10.97)

Thank You

Date post:	05-Jan-2022
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Unsymmetrical Short Circuit

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