fault analysis final

Exp.No:5

Date:

FAULT ANALYSIS

AIM

To compute the fault level, post-fault voltages and currents for both

symmetrical and unsymmetrical faults.

Software Required

MIPOWER

Theory

Need for short circuit analysis:

The system must be protected against heavy flow of short circuit current

during the occurrence of the fault. This is done by disconnecting the faulty section

from the healthy section by means of circuit breaker. To estimate the magnitude of

fault current for the proper choice of circuit breaker and protective relays, short

circuit study is essential.

Causes of short circuit fault:

Insulation failure of equipment.

Flash over of lines initiated by lightning stroke or through accidental faulty

operation.

Salt spray or pollution on insulators.

Animals or plants coming in contact with the wires.

Fault types:

There are two main types of faults:

Symmetrical faults: System remains balanced; these faults are relatively

rare, but easiest to analyze.

Unsymmetrical faults: System is no longer balanced; very common, but

relatively difficult to analyze.

The most common type of fault on a three phase system by far is the line-to-

ground (L-G), followed by the line-to line faults (L-L), double line-to-ground (L-

L-G) faults, and balanced three phase faults. The probability of two or more

simultaneous faults on a power system is removed and is therefore ignored in

system design for abnormal conditions.

In power system, loads are specified and the load currents are unknown. The

effects of load currents in the fault analysis are to express the loads by constant

impedance evaluated at the prefault bus voltages.

Prefault fault bus voltages are obtained from the results of power flow

solution.

Loads are expressed by constant admittance using prefault bus voltages.

Replace reactances of synchronous machines by their sub-transient /

transient values.

Draw reactance diagram for the short circuit.

Draw Thevenin’s equivalent viewed from faulted bus (qth bus) and find the

fault current using the following formula

Fqq

q

fZZ

VI

0

(5.1)

Where 0

qV = Prefault voltage

qqZ = Thevenin’s resistance

FZ = Fault impedance

Determine current contributed by each generator in the system using current

division technique.

Determine post fault bus voltage using formula.

fNqN

fqqq

fq

f

N

f

q

f

IZV

IZV

IZV

V

V

V

0

0

1

0

11

(5.2)

Determine post fault line flows using the following formula (5.3).

series

ij

f

j

f

if

ijZ

VVI

(5.3)

Where f

ijI= the post fault line current connecting the buses ‘i’ and ‘j’

series

ijZ= the series impedance of the line connecting the buses ‘i’ and ‘j’.

Short circuit capacity (SCC) or fault level of a bus is defined as the product of

the prefault voltage and the post fault current and is given by

MVAupZ

VIVSCC

qq

q

fq ...

20

0 (5.4)

If the prefault voltages are assumed to be 1.0 p.u then,

qqZSCC 1 (5.5)

Unsymmetrical fault analysis:

Most of the systems have unsymmetrical faults. It consists of unsymmetrical

short circuit fault or unsymmetrical faults through impedance, or open conductor

faults. If the insulation of the system fails at any point or if a conducting object

comes in contact with a bare conductor, an unsymmetrical short circuit fault is said

to occur. Due to this unbalanced currents flow in the system.

Shunt type faults:

Line to ground fault (L-G)

Line to line fault(L-L)

Double line to ground fault (L-L-G)

Series type faults:

Open conductor fault

Three phase (3L) fault being the most severe must be used to calculate the

rupturing capacity of circuit breakers, even though this type of fault has a low

frequency of occurrence, when compared to the unsymmetrical faults listed above.

There are, however, situations when LG fault can cause greater fault current than a

three phase fault.

Causes of unsymmetrical faults:

Lightning

Wind damage, tree falling across lines, vehicles colliding with towers,

breaks due to excessive ice loading.

Salt spray.

Braking of one or more conductors.

Symmetrical Component Analysis of Unsymmetrical Faults

The key idea of symmetrical component analysis is to decompose the system

into three sequence networks. The networks are then coupled only at the point of

the unbalance (i.e., the fault).

The three sequence networks are known as the

Positive sequence

Negative sequence

Zero sequence

Positive sequence:

The positive sequence sets have three phase currents/voltages with equal

magnitude, with phase ‘b’ lagging phase ‘a’ by 120°, and phase ‘c’ lagging phase

‘b’ by 120° as shown in figure (5.1).

Fig (5.1)

Negative Sequence:

The negative sequence have three phase currents/voltages with equal

magnitude, with phase ‘b’ leading phase ‘a’ by 120°, and phase ‘c’ leading phase

‘b’ by 120°. Negative sequence sets are similar to positive sequence, except that

Ic

Ib

Ia

the phase order is reversed as shown in figure (5.2).

Fig (5.2)

Zero sequence:

Zero sequence sets have three values with equal magnitude and angle. Zero

sequence sets have neutral current as shown in figure (5.3).

Fig (5.3)

Single Line - to - Ground fault (L- G)

The single line to ground fault, the most common type, is caused by

lightning or by conductors making contact with grounded structures.

Suppose a line to ground fault occurs on phase ‘a’ connected to ground through

impedance Zf as shown in figure (5.4).

Fig (5.4)

Ib

Ia

Ic

Ia

Ib

Ic

a

b

c

F

Ic=0

ZF

Ib=0

Ia

Fault current is calculated by

(5.6) (6.4) (6.4)

Where 0

aI = zero sequence current

aI = positive sequence current

aI = negative sequence current

KKZ = positive sequence impedance

KKZ =negative sequence impedance 0

KKZ =zero sequence impedance

Fault phase current:

0 cb II (5.7) aaf III 3 (5.8)

a

a

a

c

b

a

I

I

I

aa

aa

I

I

I 0

22

22

1

1

111

(5.9)

Symmetrical voltages are

(5.10)

aKKqa IZVV 0

(5.11)

aKKa IZV (5.12)

FKKKKKK

q

aaaZZZZ

VIII

30

0

0

aKKa IZV 00

Post fault positive sequence bus voltages

f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

.

.

1.1

(5.13)

Post fault negative sequence bus voltages

f

KNK

f

KKK

f

KK

f

N

f

k

f

IZ

IZ

IZ

V

V

V

11

(5.14)

Post fault zero sequence bus voltages

00

.

00

.

00

1.

0

0

0

1

f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

(5.15)

Positive sequence line current

ij

f

j

f

if

ijZ

VVI

(5.16)

Negative sequence line current

ij

f

j

f

if

ijZ

VVI

(5.17)

Zero sequence line current 0

00

0

ij

f

j

f

if

ijZ

VVI

(5.18)

LINE –TO-LINE FAULT (L-L)

A three phase generator with a fault through impedance Zf between phases

‘b’ and ‘c’ is shown in fig (5.5) Assume the generator is unloaded (no load), the

conditions at the fault bus ‘K’ are expressed by the following relations.

Fig (5.5)

Fault current:

00 aI (5.19)

FKKKK

q

aaZZZ

VII

0

(5.20)

Fault phase currents:

0aI (5.21)

acb IjII 3 (5.22)

FKKKK

q

bfZZZ

VjII

0)3(

(5.23)

Symmetrical voltages:

0aV (5.24) aKKqa IZVV 0

(5.25)

KKaaKKa ZIIZV (5.26)

ZF

a

b

c F

Ia=0

Ib Ic


f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

.

.

1.1

(5.27)


f

KNK

f

KKK

f

KK

f

N

f

k

f

IZ

IZ

IZ

V

V

V

11

(5.28)


00

.

00

.

00

1.

0

0

0

1

f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

(5.29)


ij

f

j

f

if

ijZ

VVI

(5.30)


ij

f

j

f

if

ijZ

VVI

(5.31)


00

0

ij

f

j

f

if

ijZ

VVI

(5.32)

DOUBLE LINE – TO – GROUND FAULT (L-L-G)

A three phase generator with a fault on phase ‘b’ to ‘c’ through impedance

Zf. Assuming the generator is initially on no-load, the condition at the fault K is

expressed by the following relations.

Fig (5.6)

Fault currents:

(5.33)

(5.34)

FKKKK

KKaa

ZZZ

ZII

30

00

(5.35)

Fault phase currents:

03 af II (5.36)

0aI (5.37)

fcb III (5.38)

Symmetrical voltage faults:

000

aKKa IZV (5.39)

aKKqaa IZVVV 0

(5.40)

a

b

c F

Ia=0

Ib Ic

ZF 3Ig0

FKKKK

FKKKKKK

q

a

ZZZ

ZZZZ

VI

3

)(0

0

0

FKKKK

FKKaa

ZZZ

ZZII

3

)3(0

0


f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

.

.

1.1

(5.41)


f

KNK

f

KKK

f

KK

f

N

f

k

f

IZ

IZ

IZ

V

V

V

11

(5.42)


00

.

00

.

00

1.

0

0

0

1

f

KNKfp

f

KKKfp

f

KKfp

f

N

f

k

f

IZV

IZV

IZV

V

V

V

(5.43)


ij

f

j

f

if

ijZ

VVI

(5.44)


ij

f

j

f

if

ijZ

VVI

(5.45)


00

0

ij

f

j

f

if

ijZ

VVI

(5.46)

FLOW CHART (Symmetrical Fault Analysis)

Draw the Thevenin’s equivalent circuit and obtain the fault

current using equation (5.1)

Start

Read line data, bus data, fault bus sub-transient reactance of each machine

Assume prefault load currents, shunt elements in transformer, transformer taps, shunt

capacitances, series resistances

Draw the prefault per phase network (positive sequence network) and

then obtain busZ matrix using bus building algorithm

Compute the post fault bus voltage using equation (5.2)

Calculate Post fault line currents using equation (5.3)

Calculate the fault level using the equation (5.4)

Print fI , post fault voltages, post fault currents, SCC

Stop

Flow chart (Unsymmetrical Fault Analysis):

NO

START

Read the bus data, line data, fault bus, reactance of each

machine, Assume prefault voltages.

Draw the positive, negative and zero sequence networks using the

sequence impedances of the power system. Compute positive,

negative, zero sequence using bus building algorithm

IS LG FAULT?

Symmetrical

components of fault

current at bus k for

LL-G fault are

calculated using the

equations (5.34),

(5.35) and (5.36).

Symmetrical

components of

fault current at

bus ‘k’ for L-G

fault are

calculated

using the

equation (5.6)

IS LL FAULT?

Symmetrical

components of

fault current at bus

k for L-L fault are

calculated using

the equations

(5.19) and (5.20)

E F

G

YES

YES

NO

Assume fault at phase ‘a’ and bus ‘k’

Symmetrical voltages for

L-G fault are calculated

using the equation s (5.10),

(5.11) and (5.12)

Fault phase currents

for L-L fault are

calculated using

equations (5.21),

(5.22) and (5.23)


for L-G fault are

calculated using

equations (5.7), (5.8)

and (5.9)


L-L fault are calculated

using the equation s

(5.24), (5.25) and (5.26)


for LL-G fault are

calculated using

equations (5.37),

(5.38) and (5.39)


LLG fault are calculated

using the equations

(5.40) and (5.41)

Post fault positive sequence bus voltages are calculated using the

formula (5.13)

Post fault negative sequence bus voltages are calculated using the

formula (5.14)

Post fault zero sequence bus voltages are calculated using the

formula (5.15)

E F G

Positive sequence line current is calculated using the equation (5.16)

Negative sequence line current is calculated using the equation (5.17)

Zero sequence line current is calculated using the equation (5.18)

Phase voltages

sp VTV .

Phase current

sp ITI .

Stop

Algorithm:

Symmetrical fault analysis

1. Draw the prefault per phase network (positive sequence network). Determine the

matrix using step by step bus building algorithm.

2. Obtain prefault bus voltages from power flow solution.

3. Assume prefault currents to be negligible. Represent all the components and

loads by their appropriate impedances and draw the Thevenin’s circuit. Obtain the

fault current using the equation (5.1).

4. Obtain the Thevenin’s network by inserting the Thevenin’s voltage source in

series with .

5. Obtain the post fault bus voltages using the equation (5.2).

6. Calculate the post fault line current using the equation (5.3).

7. Determine the short circuit capacity using the equation (5.4).

Unsymmetrical fault analysis:

1. Draw the positive, negative and zero sequence networks using the sequence

impedance of the power system and compute the positive, negative, zero sequence

impedance matrices using bus building algorithm.

2. Check if the fault is L-G Fault. If YES then go to next step else jump to step 6.

3. For the L-G fault calculate the symmetrical components of the fault current at

bus ‘k’ using the formula (5.6)

Where are the diagonal elements in the K axis of .

4. Determine the fault phase current using the symmetrical components obtained in

the previous step using equations (5.7), (5.8) and (5.9).

5. Symmetrical voltages can be calculated from the equations (5.10), (5.11) and

(5.12) and then jump to step 13.

6. Check if the fault is L-L fault. If YES jump to step 10 else go to next step.

7. Calculate the symmetrical components for LLG fault using the equations (5.34),

(5.35) and (5.36).

8. Determine the fault phase currents for the LLLG fault from the symmetrical

component currents obtained in the previous step using the equations (5.37), (5.38)

and (5.39).

9. Compute the symmetrical voltage components for the LLLG fault using the

equations (5.40) and (5.41) and then jump to step 13.

10. Calculate the symmetrical components for L-L fault using the equations (5.19),

and (5.20).

11. Determine the fault phase currents for the L-L fault from the symmetrical

component currents obtained in the previous step using the equations (5.21), (5.22)

and (5.23).

12. Compute the symmetrical voltage components for the LLLG fault using the

equations (5.24), (5.25) and (5.26) and then jump to step 13.

13. Calculate the post fault positive, negative and zero sequence bus voltages using

the equations (5.13), (5.14) and (5.15) respectively.

14. Determine the positive, negative and zero sequence line currents using the

equations (5.16), (5.17) and (5.18) respectively.

Problem statement

For the above circuit with X= 15% generator, Transmission line X =30% and

Transformer X = 20%, obtain

a) Fault current at fault point and post fault voltage for a bolted fault at bus 4.

b) Symmetrical fault.

c) Line to ground fault

d) Line to line fault

F

G1 G2

1 2 3 4

RESULT

Thus the fault level, post-fault voltages and currents for both symmetrical and

unsymmetrical faults were computed using MIPOWER. The obtained outputs were

also verified using hand calculations.

fault analysis final

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