CHAPTER 8

UNSYMMETRICAL FAULTS

The sequence circuits and the sequence networks developed in the previous chapter will now be used for finding out fault current during unsymmetrical faults. Specifically we are going to discuss the following three types of faults:

• Single-line-to-ground (1LG) fault • Line-to-line (LL) fault • Double-line-to-ground (2LG) fault

For the calculation of fault currents, we shall make the following assumptions:

• The power system is balanced before the fault occurs such that of the three sequence

networks only the positive sequence network is active. Also as the fault occurs, the sequence networks are connected only through the fault location.

• The fault current is negligible such that the pre-fault positive sequence voltages are same at all nodes and at the fault location.

• All the network resistances and line charging capacitances are negligible. • All loads are passive except the rotating loads which are represented by synchronous

machines. Based on the assumptions stated above, the faulted network will be as shown in Fig. 8.1 where the voltage at the faulted point will be denoted by Vf and current in the three faulted phases are Ifa, Ifb and Ifc. We shall now discuss how the three sequence networks are connected when the three types of faults discussed above occur.

Fig. 8.1 Representation of a faulted segment.

8.1 SINGLE-LINE-TO-GROUND FAULT

Let a 1LG fault has occurred at node k of a network. The faulted segment is then as shown in Fig. 8.2 where it is assumed that phase-a has touched the ground through an impedance Zf. Since the system is unloaded before the occurrence of the fault we have

0== fcfb II (8.1)

3.35

Fig. 8.2 Representation of 1LG fault. Also the phase-a voltage at the fault point is given by

fafka IZV = (8.2)

From (8.1) we can write

=

00

11

111

31

2

2012

fa

fa

I

aaaaI (8.3)

Solving (8.3) we get

3210fa

fafafa

IIII === (8.4)

This implies that the three sequence currents are in series for the 1LG fault. Let us denote the zero, positive and negative sequence Thevenin impedance at the faulted point as Zkk0, Zkk1 and Zkk2 respectively. Also since the Thevenin voltage at the faulted phase is Vf we get three sequence circuits that are similar to the ones shown in Fig. 7.7. We can then write

222

111

000

fakkka

fakkfka

fakkka

IZV

IZVV

IZV

−=

−=

−=

(8.5)

Then from (8.4) and (8.5) we can write

( ) 0210

210

fakkkkkkf

kakakaka

IZZZVVVVV

++−=++=

(8.6)

Again since

( ) 0210 3 faffafafaffafka IZIIIZIZV =++== we get from (8.6)

3.36

fkkkkkk

ffa ZZZZ

VI

32100 +++

= (8.7)

The Thevenin equivalent of the sequence network is shown in Fig. 8.3.

Fig. 8.3 Thevenin equivalent of a 1LG fault.

Example 8.1: A three-phase Y-connected synchronous generator is running unloaded with rated voltage when a 1LG fault occurs at its terminals. The generator is rated 20 kV, 220 MVA, with subsynchronous reactance of 0.2 per unit. Assume that the subtransient mutual reactance between the windings is 0.025 per unit. The neutral of the generator is grounded through a 0.05 per unit reactance. The equivalent circuit of the generator is shown in Fig. 8.4. We have to find out the negative and zero sequence reactances.

Fig. 8.4 Unloaded generator of Example 8.1.

Since the generator is unloaded the internal emfs are

°∠=°−∠== 1200.11200.10.1 cnbnan EEE Since no current flows in phases b and c, once the fault occurs, we have from Fig. 8.4

3.37

( ) 0.405.02.0

1 jj

I fa −=+

=

Then we also have

2.0−=−= fann IXV From Fig. 8.4 and (7.34) we get

°∠=+−=++=

°−∠=−−=++==

72.1240536.1866.06.0025.0

72.1240536.1866.06.0025.00

jIjVEV

jIjVEVV

fancnc

fanbnb

a

Therefore

−

−=

°∠°−∠=

3.07.04.0

72.1240536.172.1240536.1

0

012 CVa

From (7.38) we can write Z1 = jω(Ls + Ms) = j0.225. Then from Fig. 7.7 we have

3333.1225.0

7.011

11 j

jZVEI aan

fa −=−

=−

=

Also note from (8.4) that

210 fafafa III == Therefore from Fig. 7.7 we get

( ) 15.015.03.030

00 jjZ

IVZ n

a

ag =−=−−=

225.02

22 j

IVZ

a

a =−=

Comparing the above two values with (7.37) and (7.39) we find that Z0 indeed is equal to jω(Ls − 2Ms) and Z2 is equal to jω(Ls + Ms). Note that we can also calculate the fault current from (8.7) as

( ) 3333.105.0315.0225.0225.0

10 j

jI fa −=

×+++=

∆∆∆

3.38

8.2 LINE-TO-LINE FAULT

The faulted segment for an L-L fault is shown in Fig. 8.5 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf. Since the system is unloaded before the occurrence of the fault we have

0=faI (8.8)

Fig. 8.5 Representation of L-L fault. Also since phases b and c are shorted we have

fcfb II −= (8.9) Therefore from (8.8) and (8.9) we have

( )( )

−−=

−=

fb

fb

fb

fbfa

IaaIaa

IICI

2

2012

0

31

0 (8.10)

We can then summarize from (8.10)

21

0 0

fafa

fa

III

−=

= (8.11)

Therefore no zero sequence current is injected into the network at bus k and hence the zero sequence remains a dead network for an L-L fault. The positive and negative sequence currents are negative of each other.

Now from Fig. 8.5 we get the following expression for the voltage at the faulted point

fbfkckb IZVV =− (8.12) Again

( ) ( )( ) ( )( )( )21

22

21

22211

210210

kaka

kaka

kckbkckb

kckckckbkbkbkckb

VVaa

VaaVaa

VVVVVVVVVVVV

−−=

−+−=

−+−=−−−++=−

(8.13)

3.39

Moreover since Ifa0 = Ifb0 = 0 and Ifa1 = − Ifb2, we can write

( ) 12

212

21 fafbfafbfbfb IaaaIIaIII −=+=+= (8.14) Therefore combining (8.12)-(8.14) we get

121 fafkaka IZVV =− (8.15) Equations (8.12) and (8.15) indicate that the positive and negative sequence networks are in parallel. The sequence network is then as shown in Fig. 8.6. From this network we get

fkkkk

ffafa ZZZ

VII

++=−=

2121 (8.16)

Fig. 8.6 Thevenin equivalent of an LL fault.

Example 8.2: Let us consider the same generator as given in Example 8.1. Assume that the generator is unloaded when a bolted (Zf = 0) short circuit occurs between phases b and c. Then we get from (8.9) Ifb = − Ifc. Also since the generator is unloaded, we have Ifa = 0. Therefore from (7.34) we get

0.1== anan EV

fbfbbnbn IjIjEV 225.01201225.0 −°−∠=−=

fbfccncn IjIjEV 225.01201225.0 +°∠=−= Also since Vbn = Vcn, we can combine the above two equations to get

849.345.0

12011201−=

°∠−°−∠=−=

jII fcfb

Then

−=

−=

2222.22222.2

0

849.3849.30

012

jjCI fa

We can also obtain the above equation from (8.16) as

2222.2225.0225.0

121 j

jjII fafa −=

+=−=

3.40

Also since the neutral current In is zero, we can write Va = 1.0 and

5.0−=== bncb VVV Hence the sequence components of the line voltages are

=

−−=

5.05.0

0

5.05.0

0.1

012 CVa

Also note that

5.0225.00.1 11 =−= faa IjV 5.0225.0 22 =−= faa IjV

which are the same as obtained before.

∆∆∆

8.3 DOUBLE-LINE-TO-GROUND FAULT

The faulted segment for a 2LG fault is shown in Fig. 8.7 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Since the system is unloaded before the occurrence of the fault we have the same condition as (8.8) for the phase-a current. Therefore

( ) ( )fcfbfa

fcfbfcfbfafa

III

IIIIII

+=⇒

+=++=

0

0

331

31

(8.17)

Fig. 8.7 Representation of 2LG fault. Also the voltages of phases b and c are given by

( ) 03 fafcbfkckb IZIIZVV =+== (8.18)

3.41

Therefore

( )( )

++++

+=

=

kbka

kbka

kbka

kb

kb

ka

ka

VaaVVaaV

VV

VVV

CV2

2012

2

31 (8.19)

We thus get the following two equations from (8.19)

21 kaka VV = (8.20)

kbkakakakbkaka VVVVVVV 223 2100 +++=+= (8.21) Substituting (8.18) and (8.20) in (8.21) and rearranging we get

0021 3 fafkakaka IZVVV −== (8.22) Also since Ifa = 0 we have

0210 =++ fafafa III (8.23)

The Thevenin equivalent circuit for 2LG fault is shown in Fig. 8.8. From this figure we get

( ) ( )fkkkk

fkkkkkk

f

fkkl

kkkk

ffa

ZZZZZZ

Z

VZZZZ

VI

333

02

021

0211

+++

+

=++

= (8.24)

The zero and negative sequence currents can be obtained using the current divider principle as

++−=

fkkkk

kkfafa ZZZ

ZII302

210 (8.25)

+++

−=fkkkk

fkkfafa ZZZ

ZZII

33

02

012 (8.26)

Fig. 8.8 Thevenin equivalent of a 2LG fault.

3.42

Example 8.3: Let us consider the same generator as given in Examples 8.1 and 8.2. Let us assume that the generator is operating without any load when a bolted 2LG fault occurs in phases b and c. The equivalent circuit for this fault is shown in Fig. 8.9. From this figure we can write

fcfbnnbn IjIjVVE 025.02.01201 −=+°−∠=+

fbfcnncn IjIjVVE 025.02.01201 −=+°∠=+ ( )fcfbn IIjV +−= 05.0

Fig. 8.9 Equivalent circuit of the generator in Fig. 8.4 for a 2LG fault in phases b and c. Combining the above three equations we can write the following vector-matrix form

°∠°−∠

=

12011201

25.0025.0025.025.0

fc

fb

II

j

Solving the above equation we get

8182.1849.38182.1849.3

jIjI

c

b

+=+−=

Hence

−=

+−+−=

6162.18283.2

2121.1

8182.1849.38182.1849.3

0

012

jj

j

jjCi fa

We can also obtain the above values using (8.24)-(8.26). Note from Example 8.1 that

( ) 0 and 3.005.0315.0,225.0 021 ==×+=== fZjjZjZZ Then

3.43

8283.2

525.03.0225.0225.0

0.11 j

jjjj

I fa −=

×+

=

6162.1525.0

3.012 j

jjII fafa =−=

2121.1525.0225.0

10 jjjII fafa =−=

Now the sequence components of the voltages are

3636.0225.00.1 11 =×−= faa IjV

3636.0225.0 22 =×−= faa IjV 3636.03.0 00 =×−= faa IjV

Also note from Fig. 8.9 that

( ) 0909.10225.0 =+++= fcfbnana IIjVEV and Vb = Vc = 0. Therefore

=

=

3636.03636.03636.0

00

0909.1

012 CVa

which are the same as obtained before.

∆∆∆

8.4 FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS

In this section we shall demonstrate the use of sequence networks in the calculation of fault currents using sequence network through some examples.

Example 8.4: Consider the network shown in Fig. 8.10. The system parameters are

given below:

Generator G: 50 MVA, 20 kV, X′′ = X1 = X2 = 20%, X0 = 7.5% Motor M: 40 MVA, 20 kV, X′′ = X1 = X2 = 20%, X0 = 10%, Xn = 5% Transformer T1: 50 MVA, 20 kV∆/110 kVY, X = 10% Transformer T2: 50 MVA, 20 kV∆/110 kVY, X = 10% Transmission line: X1 = X2 = 24.2 Ω, X0 = 60.5 Ω

We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-2.

3.44

Fig. 8.10 Radial power system of Example 8.4.

Let us choose a base in the circuit of the generator. Then the per unit impedances of the generator are:

075.0,2.0 021 === GGG XXX The per unit impedances of the two transformers are

1.021 == TT XX The MVA base of the motor is 40, while the base MVA of the total circuit is 50. Therefore the per unit impedances of the motor are

0625.0405005.0,125.0

40501.0,25.0

40502.0 021 =×==×==×== nMMM XXXX

For the transmission line

Ω== 24250

1102

baseZ

Therefore

25.0242

5.60,1.0242

2.24021 ===== LLL XXX

Let us neglect the phase shift associated with the Y/∆ transformers. Then the positive, negative and zero sequence networks are as shown in Figs. 8.11-8.13.

Fig. 8.11 Positive sequence network of the power system of Fig. 8.10.

3.45

Fig. 8.12 Negative sequence network of the power system of Fig. 8.10.

Fig. 8.13 Zero sequence network of the power system of Fig. 8.10.

From Figs. 8.11 and 8.12 we get the following Ybus matrix for both positive and negative sequences

−−

−

==

14100010201000102010001015

21 jYY busbus

Inverting the above matrix we get the following Zbus matrix

==

1667.01333.01000.00667.01333.01867.01400.00933.01000.01400.01800.01200.00667.00933.01200.01467.0

21 jZZ busbus

Again from Fig. 8.13 we get the following Ybus matrix for the zero sequence

−−

−−

=

2.300001440041400003333.13

0 jYbus

Inverting the above matrix we get

=

3125.000000778.00222.0000222.00778.00000075.0

0 jZbus

3.46

Hence for a fault in bus-2, we have the following Thevenin impedances

0778.0,18.0 021 jZjZZ === Alternatively we find from Figs. 8.11 and 8.12 that

18.045.03.021 jjjZZ l ===

0778.035.01.00 jjjZ l ==

(a) Single-Line-to-Ground Fault: Let a bolted 1LG fault occurs at bus-2 when the system is unloaded with bus voltages being 1.0 per unit. Then from (8.7) we get

( ) 2841.20778.018.02

1210 j

jIII fafafa −=

+×=== per unit

Also from (8.4) we get

8524.63 0 jII fafa −== per unit Also Ifb = Ifc = 0. From (8.5) we get the sequence components of the voltages as

4111.018.0

5889.018.01

1777.00778.0

222

112

002

−=−=

=−=

−=−=

faa

faa

faa

IjV

IjV

IjV

Therefore the voltages at the faulted bus are

°∠°−∠=

=

−

11.1079061.011.1079061.0

0

22

12

021

a

a

a

c

b

a

VVV

CVVV

(b) Line-to-Line Fault: For a bolted LL fault, we can write from (8.16)

7778.218.02

121 j

jII fafa −=

×=−= per unit

Then the fault currents are

−=

=

−

8113.48113.400

2

11

fa

fa

fc

fb

fa

IIC

III

Finally the sequence components of bus-2 voltages are

3.47

5.018.0

5.018.010

222

112

02

=−=

=−==

faa

faa

a

IjV

IjVV

Hence faulted bus voltages are

−−=

=

−

5.05.0

0.1

22

12

021

a

a

a

c

b

a

VVV

CVVV

(c)Double-Line-to-Ground Fault: Let us assumes that a bolted 2LG fault occurs at

bus-2. Then

0543.00778.018.0 jjjZ leq ==

Hence from (8.24) we get the positive sequence current as

2676.418.0

11 j

ZjI

eqfa −=

+= per unit

The zero and negative sequence currents are then computed from (8.25) and (8.26) as

( ) 9797.20778.018.0

18.010 j

jjII fafa =+

−= per unit

( ) 2879.10778.018.0

0778.012 j

jjII fafa =

+−= per unit

Therefore the fault currents flowing in the line are

°∠°∠=

=

−

89.42657.611.137657.6

0

2

1

01

fa

fa

fa

fc

fb

fa

III

CIII

Furthermore the sequence components of bus-2 voltages are

2318.018.0

2318.018.01

2318.00778.0

222

112

002

=−=

=−=

=−=

faa

faa

faa

IjV

IjV

IjV

Therefore voltages at the faulted bus are

3.48

=

=

−

00

6954.0

22

12

021

a

a

a

c

b

a

VVV

CVVV

∆∆∆

Example 8.5: Let us now assume that a 2LG fault has occurred in bus-4 instead of the one in bus-2. Therefore

3125.0,1667.0 021 jXjXX === Also we have

1087.03125.01667.0 jjjZ leq ==

Hence

631.31667.0

11 j

ZjI

eqfa −=

+= per unit

Also

( ) 2631.13125.01667.0

1667.010 j

jjII fafa =

+−= per unit

( ) 3678.23125.01667.0

3125.012 j

jjII fafa =

+−= per unit

Therefore the fault currents flowing in the line are

°∠°∠=

=

−

04.205298.596.1595298.5

0

2

1

01

fa

fa

fa

fc

fb

fa

III

CIII

We shall now compute the currents contributed by the generator and the motor to the

fault. Let us denoted the current flowing to the fault from the generator side by Ig, while that flowing from the motor by Im. Then from Fig. 8.11 using the current divider principle, the positive sequence currents contributed by the two buses are

2103.175.025.0

11 jjjII faga −=×= per unit

4206.275.05.0

11 jjjII fama −=×= per unit

Similarly from Fig. 8.12, the negative sequence currents are given as

3.49

7893.075.025.0

22 jjjII faga =×= per unit

5786.175.05.0

22 jjjII fama =×= per unit

Finally notice from Fig. 8.13 that the zero sequence current flowing from the

generator to the fault is 0. Then we have

00 =gaI 2631.10 jIma = per unit

Therefore the fault currents flowing from the generator side are

°∠°∠

°−∠=

=

−

93.67445.107.1737445.1

904210.0

2

1

01

ga

ga

ga

gc

gb

ga

III

CIII

and those flowing from the motor are

°∠°∠

°∠=

=

−

93.258512.307.1548512.3

904210.0

2

1

01

ma

ma

ma

mc

mb

ma

III

CIII

It can be easily verified that adding Ig and Im we get If given above.

∆∆∆

In the above two examples we have neglected the phase shifts of the Y/∆ transformers. However according to the American standard, the positive sequence components of the high tension side lead those of the low tension side by 30°, while the negative sequence behavior is reverse of the positive sequence behavior. Usually the high tension side of a Y/∆ transformer is Y-connected. Therefore as we have seen in Fig. 7.16, the positive sequence component of Y side leads the positive sequence component of the ∆ side by 30° while the negative sequence component of Y side lags that of the ∆ side by 30°. We shall now use this principle to compute the fault current for an unsymmetrical fault.

Example 8.6: Let us consider the same system as given in Example 8.5. Since the phase shift does not alter the zero sequence, the circuit of Fig. 8.13 remains unchanged. The positive and the negative sequence circuits must however include the respective phase shifts. These circuits are redrawn as shown in Figs. 8.14 and 8.15.

Note from Figs. 8.14 and 8.15 that we have dropped the √3α vis-à-vis that of Fig. 7.16. This is because the per unit impedances remain unchanged when referred to the either high tension or low tension side of an ideal transformer. Therefore the per unit impedances will also not be altered.

3.50

Fig. 8.14 Positive sequence network of the power system of Fig. 8.10 including transformer phase shift.

Fig. 8.15 Negative sequence network of the power system of Fig. 8.10 including transformer phase shift.

Since the zero sequence remains unaltered, these currents will not change from those computed in Example 8.6. Thus

00 =gaI and 2631.10 jIma = per unit Now the positive sequence fault current from the generator Iga1, being on the Y-side of the Y/∆ transformer will lead Ima1 by 30°. Therefore

°−∠=°∠×−= 602103.13012103.11 jI ga per unit 4206.21 jIma −= per unit

Finally the negative sequence current Iga2 will lag Ima2 by 30°. Hence we have

°∠=°−∠×= 607893.03017893.02 jjI ga per unit 5786.12 jIma = per unit

Therefore

°∠°−∠°−∠

=

=

−

04.200642.11809996.1

04.200642.1

2

1

01

ga

ga

ga

gc

gb

ga

III

CIII

3.51

Also the fault currents flowing from the motor remain unaltered. Also note that the currents flowing into the fault remain unchanged. This implies that the phase shift of the Y/∆ transformers does not affect the fault currents.

∆∆∆

Example 8.7: Let us consider the same power system as given in Example 1.2, the sequence diagrams of which are given in Figs. 7.18 to 7.20. With respect to Fig. 7.17, let us define the system parameters as:

Generator G1: 200 MVA, 20 kV, X″ = 20%, X0 = 10% Generator G2: 300 MVA, 18 kV, X″ = 20%, X0 = 10% Generator G3: 300 MVA, 20 kV, X″ = 25%, X0 = 15% Transformer T1: 300 MVA, 220Y/22 kV, X = 10% Transformer T2: Three single-phase units each rated 100 MVA, 130Y/25 kV,

X = 10% Transformer T3: 300 MVA, 220/22 kV, X = 10% Line B-C: X1 = X2 = 75 Ω, X0 = 100 Ω Line C-D: X1 = X2 = 75 Ω, X0 = 100 Ω Line C-F: X1 = X2 = 50 Ω, X0 = 75 Ω

Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is

300. The base voltages are then same as those shown in Fig. 1.23. Per unit reactances are then computed as shown below.

Generator G1: 3.02003002.0 =×=′′X , X0 = 0.15

Generator G2: 1312.022.22

182.02

=

×=′′X , X0 = 0.0656

Generator G3: 2.0=′′X , X0 = 0.15

Transformer T1: 121.02002201.0

2

=

×=X

Transformer T2: 1266.022.22

251.02

=

×=X

Transformer T3: 121.020221.0

2

=

×=X

Line B-C: 5625.033.133

7521 === XX , 75.0

33.133100

0 ==X

Line C-D: 5625.033.133

7521 === XX , 75.0

33.133100

0 ==X

3.52

Line C-F: 375.033.133

5021 === XX , 5625.0

33.13375

0 ==X

Neglecting the phase shifts of Y/∆ connected transformers and assuming that the system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of Fig. 7.17).

From Figs. 7.18 and 7.19, we can obtain the positive and negative sequence Thevenin impedance at point C as (verify)

X1 = X2 = j0.2723 per unit Similarly from Fig. 7.20, the Thevenin equivalent of the zero sequence impedance is

X0 = j0.4369 per unit Therefore from (8.7) we get

( ) 0188.14369.02723.02

10 j

jI fa −=

+×= per unit

Then the fault current is Ifa = 3Ifa0 = 3.0565 per unit.

∆∆∆