my080544r-fault anls -root(3) - copy

7/31/2019 My080544R-Fault Anls -Root(3) - Copy

1/18

FAULT STUDY

Instructed by :- Ms. Chenuka Perera

Name : D.M.A.Wijerathne

Index No : 080544R

Field : Electrical

Group : G-14

Date of sub : 20-03-2012


2/18

OBSERVATION SHEET

Experiment : Fault Study

Name : D.M.A.Wijerathne

Index No : 080544R

Field : Electrical Engineering

Year : Level 4

1) Single line to earth fault

Positive Sequence Negative Sequence Zero Sequence

Fault voltage ( V ) 38.72 -10.5 -28.41

Fault current ( mA ) 11.5 11.5 11.5

2) Line to line fault

Positive Sequence Negative Sequence

Fault voltage ( V ) 24.08 24.08

Fault current ( mA ) 27 -27

3) Double line to earth fault

Positive Sequence Negative Sequence Zero Sequence

Fault voltage ( V ) 20.21 20.21 20.21

Fault current ( mA ) 31 -23 -8


3/18

Calculationsi. Practical Calculations :

All data is in pu according to following bases,

Voltage base = 132 kVPower base = 40 MVA

F

Figure 1Simplified Power System

X= 0.08

X= 0.09

X1 = 0.06

X2 = 0.04

X0 = 0.01

XP = 0.048

XT = 0.055

XS = 0.051

Z = 0.002 + j0.005

X0 = 0.02

Z = 0.017 + j0.04

X0 = 0.15

Z = 0.19 + j0.44

X0 = 2.0

Z = 0.156 + j0.341

X0 = 1.1

Z = 0.347 + j0.8

X0 = 2.5

Z = 0.057 + j0.13

X0 = 0.45

Z = 0.058 + j0.102

X0 = 0.4

X1 = 0.3

X2 = 0.2

X0 = 0.05

X= 0.1

POLPITIYA ANURADHAPURA

KOLONNAWA BOLAWATTALAXAPANA

X = 0.02


4/18

Resistive part is neglected in the impedance since the reactive part is very much higher than

resistive part.

Positive Sequence Network

F

Figure 2

0.06

0.005 0.04

0.44

0.341 0.8

0.130.102

0.1

0.02

0.3

N1

N1

POLPITIYAANURADHAPURA


0.09

0.055

0.051

0.048 0.08


5/18

Negative Sequence Network

F

Figure 3

N2

N2

0.04

0.005 0.04

0.44

0.341 0.8

0.130.102

0.1

0.02

0.2

POLPITIYAANURADHAPURA


0.09

0.051

0.055

0.048 0.08


6/18

Zero Sequence Network

F

Figure 3

0.45

0.1

0.05

BOLAWATTA

0.08

N0 Refference

LAXAPANA

0.02

KOLONNAWA

2.0

0.4

0.02 0.15 1.1 2.5

POLPITIYA

0.01

ANURADHAPURA

0.051

0.055

0.048

0.09


7/18

In the DC network analyzer 50V DC supply was used instead of 132kV supply and resistances

are multiplied by a factor of 4000. So the results have to be transformed to actual values.

actual current in the network = Current from the observations

400050

)3(13240MVA

V

kV

= Current from the observations 13.996kA

Actual voltage in the network = Voltage from the observations

350

132

V

kV

= Voltage from the observations 2.64/ 3

Single Line to Earth Fault (L-G)

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

I

11.5mA

11.5mA

11.5mA

1

1

111

I

I

I

2

2

c

b

a

Actual values ;If= Ia = 311.5

Ia = 482.87A

A0I

87.48210))5.11()5.11(5.11(I

b

32

b

A0I

87.48210))5.11()5.11(5.11(I

b

32

c

Ef Z1 Z2 Z0Ia1 Ia2 Ia0

Va1 Va2 Va0

Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc


8/18

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

10.5V-

38.72V

28.41V-

1

1

111

V

V

V

2

2

c

b

a

Actual values

0.29kVkV32.64/10.5)38.7228.41(Va

kV93.13491.77V

V3/64.2)1205.102408.72341.28(V

V3/64.2))5.10()72.38(41.28(V

0

b

00

b

2

b

kV93.13477.91V

V32.64/)24010.512038.7241.28(V

V3/64.2))5.10()72.38(28.41(V

0

c

00

c

2

c

Line to Line Fault (L-L)

Actual values ;

a2

a1

a0

2

2

c

b

a

I

II

1

1

111

I

II

27mA-

27mA

0

1

1

111

I

I

I

2

2

c

b

a

0AI

)27270(I

a

a

Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc

Va0

Z0 Ia0

Va2

Ia2

Z1 Ia1

Va1

Z2

Ef


9/18

Va2

Va1

kA90-6531.0I

13.96610)1202724027(I

996.3110))27()27(0(I

0

b

300

b

32

b

kA906531.0I

13.96610)2402712027(I

13.96610))27()27(0(I

0c

300

c

32

c

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

24.08V

24.08V

0

1

1

111

V

V

V

2

2

c

b

a

73.4kVV

kV32.640/24.08)(24.08V

a

a

kV18036.7V

32.640/)12024.08240(24.08V

32.640/24.08))(24.08)((0V

0

b

00

b

2

b

kV

kV

kV18036.7V

32.64/)24024.08120(24.08V

32.64/24.08))(24.08)((0V

0

c

00

c

2

c

Double Line to Earth Fault (L-L-G)

Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc

Va0

EfZ1 Ia1

Z2 Ia2

Ia0Z0


10/18

Actual values ;

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

I

23mA-

31mA

8mA-

1

1

111

I

I

I

2

2

c

b

a

0AI

23)-31-8(I

a

a

kA104.39-674.0I

13.96610)12023240318(I

996.3110))23()31(8-(I

0

b

300

b

32

b

kA104.39674.0I

13.96610)24023120318(I

13.96610))23()31(8-(I

0

c

300

c

32

c

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

20.21V

20.21V

20.21V

1

1

111

V

V

V

2

2

c

b

a

92.41kVV

kV32.64/20.21)20.21(20.21V

a

a

kV0V

3/64.2)12021.2024020.2120.21(V3/64.2))21.20

()21.20

(21.20(V

b

00

b

2

b

kV0V

3/64.2)24021.2012020.2121.20(V

3/64.2))21.20()21.20(21.20(V

c

00

c

2

c


11/18

ii. Theoretical Calculations :Single Line to Earth Fault (L-G)

Va= 0 (fault impedance is zero)

Ib=Ic = 0 ( load currents are negligible compared to fault current)

3

IIII

0I

0I

I

1

1

111

31

I

I

I

aa2a1a0

c

b

a

2

2

a2

a1

a0

021

faf

a2a

a1a

a0a

2

1

0

f

2a

1a

0a

ZZZ

E3II

3II

3II

3II

Z00

0Z0

00Z

0

E

0

V

V

V

Since Z1 = 0.238pu , Z2 = 0.2225pu, = 0.6083 pu , Ef= 1 pu

Fault Current

021

f

fZZZ

E3)pu(I

baseI)pu(

fI)actual(

fI

kA

kV

MVAI

base

base

base175.0

3132

40

3

kA175.0)6083.02225.0(0.238

13(actual)If

163.69A3

491.07III

491.07AIa(actual)I

a2a1a0

f

Ef Z1 Z2 Z0Ia1 Ia2 Ia0

Va1 Va2 Va0Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc


12/18

Fault Voltages from the diagram

15.86kV10163.676.4350.2225IZV

kV59.2410163.696.4350.2383/132IZEV

kV373.3410163.696.4350.6083IZV

6.43540MVA

(132kV)Z

3-

a22a2

3-

a11fa1

3-

a00a0

2

base

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

0VVVVV a2a1a0a

kV135.01-91.98V

12015.8624059.2443.37V86.1524.5943.37V

VVVV

0

b

00

b

2

b

a2a1

2

a0b

kV.0135191.98V

24015.8612059.2373.43V

86.1523.5943.37V

VVVV

0

c

00

c

2

c

a2

2

a1a0c


Ia = 0 ( load currents are negligible compared to fault current)

Vb = Vc

Ib = - Ic

Z1 = 0.238pu , Z2 = 0.2225pu, = 0.6083 pu , Ef= 1 pu

Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc

Va0

Z0 Ia0

Va2

Ia2

Z1 Ia1

Va1Z2

Ef


13/18

0AI

0.379kAII

0.379kA435.60.2225)(0.238

kV3132/1

ZZ

EI

a0

a1a2

21

fa1

0VV

36.73kV435.60.2225379.0ZIVV

a0

2a2a2a1

kV18036.73V

kV18036.73V

kV12036.7324036.73V

36.73kV36.730V

VVVV

73.46kVV

73.3636.730V

VVVV

0

c

0

b

00

b

2

b

a2a1

2

a0b

a

a

a2a1a0a

A90-656.45I

A120379240379I

379A3790I

IIII

0b

00

b

2

b

a2a1

2

a0b

A9045.656I

II

0

c

bc


Ia = 0 ( load currents are negligible compared to fault current)

Vb = Vc = 0

Z1 = 0.238pu , Z2 = 0.2225pu, = Z3 = 0.6083 pu , Ef= 1 pu

Supply

side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc

Va0

EfZ1 Ia1

Z2 Ia2

Ia0Z0

Va2

Va1


14/18

Fault current from the diagram

0.117kA435.60.6083

436).0435.60.2383/132(1

Z

)IZ(EI

0.32kA435.60.2225

436).0435.60.2383/132(1

Z

)IZ(EI

0.436kA435.66083)0.2225//0.(0.238

kV3132/1

//ZZZ

EI

0

a11fa0

2

a11fa2

021

fa1

93kV436).0435.60.2383/132(13)IZ(E3V3V a11fa1a

A0I

320436117I

a

a

kA104.99-0.678I

120320240436117I

320436117I

IIII

0

b

00

b

2

b

a2a12a0b

kA99.104678.0I

240320120436117I

320436117I

IIII

0

c

00

c

2

c

a2

2

a1a0c


15/18

ResultsSingle Line to Earth Fault (L-G)

Theoretical Practical

Fault

currents

Ia 491.07 A 482.87 A

Ib 0 0

Ic 0 0

Fault

voltages

Va 0 kV0.29-

Vb kV.0135191.980 kV93.13491.77 0

Vc kV.0135191.980 kV93.13491.77 0



Fault

currents

Ia 0 0

Ib A90-45.656 0 A90-1.6530

Ic A9045.6560

A901.6530

Fault

voltages

Va 73.46kV 73.4kV

Vb kV80136.73 0 kV18036.7 0

Vc kV80136.73 0 kV18036.7 0



Fault

currents

Ia 0 0

Ib A104.99-678 0 A104.39-674 0

Ic A104.99678 0 A104.396740

Fault

voltages

Va 93kV 92.41kV

Vb 0 0

Vc 0 0


16/18

THEORY1. Importance of Fault StudyThe fault study of a power system is required in order to provide information for the selection of

switch gear ,setting of relays and stability of system operation.A power system is not static but

changes during operation(switching on or off of generators and transmission lines) and during

planning(addition of generators and transmission lines).

2. Analogue methods of studying the fault flow

Symmetrical components method and Bus impedance or admittance method commonly

used in fault flow studies.

Symmetrical components method can only be used to study asymmetric faults by

decomposing the phase components in to a sequence of symmetric components. Bus

impedance or admittance method can be used for the purpose of analyzing bothsymmetrical and asymmetrical faults with the use of bus bar impedances of the system.

3. DC Network AnalyzerDC Network Analyzer is a simulation equipment used to model and study symmetrical and

asymmetric faults by using positive,Negative and zero sequence components. sequential

networks can be connected independently and the sequential components can be read with

means of metering equipments. It has variable resistances,an Ammeter and an Ohmmeter andalso variable power supplies ansd altenators ,transformers with bus bars,transmission line

sections and load sections.As real systems have small impedances ,we use multiplication

factors to have measurable and usable impedance values.DC suppies used as generators.

4. Importance of using Sequence ComponentsAnalysing balanced system is much more easier than analysing a unbalanced system.So,an

unbalanced 3 phase system can be devided in to three symmetrical sequences.They are

positive,Negative and Zero sequences. The removal of the complexity of the values makes it

easier to perform the matrix calculation.

5. Relationships between the sequence impedance for generators, transformers andtransmission lines


17/18

GeneratorThe generator has a inherent direction of rotation ,and the sequence considered may

either have the same direction(no relative motion) or the opposite direction(relative

motion at twice the speed). Thus the rotational emf developed for the positive

sequence and the negative sequence would also be different. Thus the generator has

different values for positive sequence, negative sequence and zero sequence.

TransformerTransformer being passive and stationary,do not have an inherent direction. Thus it

always has the same positive sequence and negative sequence impedance and even

zero sequence impedances. However, the zero sequence paths across the windings of a

transformer depend on the winding connection and even grounding impedance.

Transmission LinesThe conductors of the transmission lines too,being passive and stationary,do not have

inherent direction. Thus, they always have the same positive sequence and negative

sequence impedances.However,as the zero sequence path also involves the earth wire

and or the earth return path. Therefore zero sequence impedance is higher in value.

DISCUSSION1. Assumptions made in Fault Study

The following assumtions are usually made in fault analysis in three phase transmission

lines

All sources are balanced and equal in magnitude and phase. Sources represented by the Thevenins voltage prior to fault at the fault point. Large systems may be represented by infinite bus-bars. Transformers are on nominal tap position. Resistances are negligible compared to reactance. Transmission lines are assumed fully transposed and all three phase have same Z. Loads currents are negligible compared to the fault currents. Line charging currents can be completely neglected.

2. Reasons for the deviations in practical values and theoretical values .Due to the additional resistance of connectors,connecting points and wires.Errors of the measuring equipment and human errors.


18/18

3. Practical problems encountered during the experiment and steps taken to overcomethem.

There were so many wires in the three sequence networks hence it becomes complex.Therefore we numbered bus bars according to the three sequence networks and noted

all the connecting points in a paper.

Connectivity of the conductors.So we check each and every wire to make sure it wontbe a problem.

my080544r-fault anls -root(3) - copy

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