Unbalnced faults

8/10/2019 Unbalnced faults

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Unbalanced Faults

Faults at the Generator Terminals

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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.

Well derive these relationships for several common

faults.

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Single Line-to-Ground (SLG) Faults

With a SLG fault only one phase has non-zero fault current --

well assume it is phase A.

?

0

0

fa

fb

fc

I

I

I

3


4/20

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

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SLG Equivalent Circuit

)( 0ZZZIV f

3)( 0aa

fI

ZZZVI

0

)(

3

0

cb

a

a

II

ZZZ

VI

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Example

Consider a system with sequence impedances given byZ+ = 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.

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

The sequence currents are given by;

Therefore,

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Sequence Voltages

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Phase Voltages

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Double Line to Ground Fault

Fault conditions

Ia= 0

Vb= Vc=00//ZZZ

VI

ZIEV

VaVVV3

0

Z

VI

o

oo

Z

VI

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10/20

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


11/20


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

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12/20


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

2Z 0Z

- -

+ +

1aI 2aI 0aI

-

+

1aV 2aV aoV

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

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

0

0 upj

jZ

VI a

a

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037.268.005.3021 jjjIIII aaaa

37.2)68.0)(866.05.0()05.3)(866.05.0(0212

jjjjjIaIIaI aaab

..3.13280.4555.3230.3 0 upj

37.2)866.0)(866.05.0()05.3)(866.05.0(0221 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

0bc

V

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..711.0 upVVV acca

Expressed in amperes and volts,

0aI

AIb00 3.13240173.13280.4837

AIc00 7.4740177.4780.4837

AIn00 9059519011.7837

kVVab0066.5

3

8.13711.0

0

bcV

kVVca018066.5

3

8.13711.0

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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

VV

aa

aa

V

VV

2

2

2

1

0

1

1111

3

1

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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

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18/20

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

j

Ia 667.135.025.0

00.11

unitperjII aa 667.112 00 aI

0667.1667.1021 jjIIII aaaa

)866.05.0(667.1)866.05.0(667.10212 jjjjIaIIaI aaab unitperjjj 0886.2443.1833.0443.1833.0

unitperjII bc 0886.2 18


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Line-to-line fault

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

0aI

01802416837886.2 bI002416837886.2 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 0021 upVVVV aaaa

0212

aaab VaVVaV

)866.05.0(583.0)866.05.0(583.0 jjVV bc

..583.0 up 19


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Line-to-line fault

Line-to-line voltages are

unitperVVVbaab

749.1583.0166.1

unitperVVVcbbc

0583.0583.0

unitperVVV acca 749.1166.1583.0 Expressed in volts, the line-to-line voltages are

kVVab0094.13

3

8.13749.1

kVVbc 0

kVVca018094.13

3

8.13749.1

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Unbalnced faults

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