Unbalanced Faults

Faults at the Generator Terminals

1

Unbalanced Faults

• Common unbalanced faults are SLG, LL and LLG faults.

• Unbalanced faults unbalance the network, but only at

the fault location.

• This causes a coupling of the sequence networks.

• How the sequence networks are coupled depends upon

the fault type.

• We’ll derive these relationships for several common

faults.

2

Single Line-to-Ground (SLG) Faults

• With a SLG fault only one phase has non-zero fault current -- we’ll assume it is phase A.

?

0

0

fa

fb

fc

I

I

I

3

SLG

0

2 0

2

Then since

1 1 1 ?1 1

1 03 3

01

f

ff f f f a

f

I

I I I I I

I

Va = 0 for a single line to ground fault

V+ + V- +V0 = 0

With the generators normally producing balanced three-phase voltages,

which are positive sequence only, we can write

4

SLG Equivalent Circuit

)( 0ZZZIV f

3)( 0aa

f

I

ZZZ

VI

0

)(

30

cb

aa

II

ZZZ

VI

5

Example

Consider a system with sequence impedances given by

Z+ = j0.2577, Z- = j0.2085, and Z0 = j0.14; find the voltages

and currents at the fault point for a single line-to-ground fault.

Solution

The sequence networks are connected in series for a SLG fault.

The sequence currents are given by;

Therefore,

6

Sequence Voltages

7

Phase Voltages

8

Double Line to Ground Fault

• Fault conditions

Ia = 0

Vb= Vc=0 0// ZZZ

VI

ZIEV

VaVVV

3

0

Z

VI

o

oo

Z

VI

9

Double line-to-ground fault

bI

cI

a

b

c

nI

0 cb VV 0aI

0

0

1

1

111

3

1

2

2

2

1

0 a

a

a

a V

aa

aa

V

V

V

aaaa VVVV3

1021

2

1

0

2

1

0

2

1

0

00

00

00

0

0

a

a

a

f

a

a

a

I

I

I

Z

Z

Z

V

V

V

V

10

Double line-to-ground fault

2

1

0

2

1

0

11

11

11

00

00

00

0

0

a

a

a

f

af

af

af

I

I

I

Z

Z

Z

V

ZIV

ZIV

ZIV

2

1

0

1

2

1

0

11

11

11

00

00

00

0

0

a

a

a

f

af

af

af

I

I

I

Z

Z

Z

V

ZIV

ZIV

ZIV

11

Double line-to-ground fault

2

1

0

2

1

0

11

11

11

100

010

001

a

a

a

af

faf

af

I

I

I

Z

Z

Z

ZIV

VZIV

ZIV

0210 aaaa IIII)( 02

021

1

ZZZZ

Z

VI

f

a

fV

-

+

1Z

2Z0Z

- -

+ +

1aI 2aI0aI

-

+

1aV2aV aoV

12

Double line-to-ground fault

Example : Find the line currents and the line-to-line voltages when a

double line-to-ground fault occurs at the terminals of the generator

described in single line-to-ground fault example. Assume that the generator

was unloaded and operating at rated voltage when the fault occurred. The

generator has positive sequence reactance of 0.25 , negative sequence

reactance of 0.35 and zero sequence reactance of 0.10 per unit. The neutral

of the generator is solidly grounded. Neglect resistance.

..05.33278.0

0.1

)10.035.0/()10.035.0(25.0

00.1

)/( 020211 upj

jjjjjj

j

ZZZZZ

EI aa

)25.0)(05.3(111021 jjZIEVVV aaaaa

..237.0763.00.1 up

..68.035.0

237.0

2

22 upj

jZ

VI aa

..37.2.010.0

237.0

0

00 upj

jZ

VI aa

13

Double line-to-ground fault

037.268.005.3021 jjjIIII aaaa

37.2)68.0)(866.05.0()05.3)(866.05.0(021

2 jjjjjIaIIaI aaab

..3.13280.4555.3230.3 0 upj

37.2)866.0)(866.05.0()05.3)(866.05.0(02

2

1 jjjjjIIaaII aaac

..7.4780.4555.3230.3 0 upj

..11.737.233 0 upjjII an

..11.7555.3230.3555.3230.3 upjjjIII cbn

..711.0237.033 1021 upVVVVV aaaaa

0 cb VV

..711.0 upVVV baab

0bcV

14

Double line-to-ground fault

..711.0 upVVV acca

Expressed in amperes and volts,

0aI

AIb

00 3.13240173.13280.4837

AIc

00 7.4740177.4780.4837

AIn

00 9059519011.7837

kVVab

0066.53

8.13711.0

0bcV

kVVca

018066.53

8.13711.0

15

Line-to-line fault

bI

cI

a

b

c

cb VV 0aI cb II

c

c

a

a

a

I

I

aa

aa

I

I

I 0

1

1

111

3

1

2

2

2

1

0

00 aI 12 aa II

c

c

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

16

Line-to-line fault

2

1

0

2

1

0

2

1

0

00

00

00

0

0

a

a

a

f

a

a

a

I

I

I

Z

Z

Z

V

V

V

V

21 aa VV

22110 ZIZIV aaf

21

1ZZ

VI

f

a

fV

-

+

1Z

2Z

- -

+ +

1aV2aV

1aI 2aI

17

Line-to-line fault

Example : Find the subtransient currents and the line-to-line voltages at the fault under

subtransient conditions when a line-to-line fault occurs at the terminals of the generator

described in the sing line-to-ground fault example. Assume that the generator is unloaded

and operating at rated terminal voltage when the fault occurs. Neglect resistance.

unitperjjj

jIa 667.1

35.025.0

00.11

unitperjII aa 667.112 00 aI

0667.1667.1021 jjIIII aaaa

)866.05.0(667.1)866.05.0(667.1021

2 jjjjIaIIaI aaab

unitperjjj 0886.2443.1833.0443.1833.0

unitperjII bc 0886.2 18

Line-to-line fault

in the sing line-to-ground fault example, base current is 837A, and so

0aI

01802416837886.2 bI

002416837886.2 cI

unitperjjVV aa 583.0)25.0)(667.1(121

00 aV (Neutral of generator grounded)

Line-to-ground voltages are

..0166.1583.0583.0 0

021 upVVVV aaaa

021

2

aaab VaVVaV

)866.05.0(583.0)866.05.0(583.0 jjVV bc

..583.0 up 19

Line-to-line fault

Line-to-line voltages are

unitperVVV baab 749.1583.0166.1

unitperVVV cbbc 0583.0583.0

unitperVVV acca 749.1166.1583.0

Expressed in volts, the line-to-line voltages are

kVVab

0094.133

8.13749.1

kVVbc 0

kVVca

018094.133

8.13749.1

20