Fault Study - 100255K

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EE 2192 Laboratory Practice IV

FAULT STUDY

Name

Index No.

Field

Group

Date of Performance

Date of Submission

: W.M.C.N.S.Kirinde

: 100255K

: Electrical Engineering

: G6

: 2012/11/22

: 2012/12/06

Instructed by : H.G.D.M.Somarathne

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Fault Point: ANURADHAPURA BUS-BAR

Single Line to Ground Fault (L-G)

Positive Sequence

(Va1, Ia1)Negative Sequence

(Va2, Ia2)Zero Sequence

(Va0, Ia0)

Fault Voltage (V) 40.19 -17.98 -22.17

Fault Current (mA) 13.5 13.5 13.5

Line to Line Fault (L-L)

Positive Sequence



(Va0, Ia0)

Fault Voltage (V) 32.07 32.06 0

Fault Current (mA) 24.5 -24.5 0

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

Positive Sequence



(Va0, Ia0)

Fault Voltage (V) 24.81 24.81 24.81

Fault Current (mA) 34 -19 -15.5

OBSERVATION SHEET:

PRACTICAL

NAME

INDEX NO.

FIELD

GROUP

DATE OF PERFORMANCE

DATE OF SUBMISSION

INSTRUCTED BY

: Fault Study

: W.M.C.N.S.Kirinde

: 100255K: Electrical Engineering

: G6

: 2012/11/22

: 2012/12/06

: H.G.D.M.Somarathne

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PROCEDURE:

X1 = 0.06

X2 = 0.04

X0 = 0.01

XP = 0.048

XT = 0.055

XS = 0.051

POLPITIYAANURADHAPURA

LAXAPANA KOLONNAWA BOLAWATTA

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 = 0.04

X1 = 0.3

X2= 0.2

X0 = 0.05

X0 = 0.1X0 = 0.09

X0 = 0.08

X0 = 0.02

All data is in pu according to following bases,

132kV, 40MVA

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Positive sequence equivalent circuit

1 pu

Reference

0.005

0.102

0.08

0.04

0.44

0.341

0.13

0.8

0.4

ANURADHAPURA

POLPITIYA


0.048

0.051

0.055

0.08

0.09

F

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Negative sequence equivalent circuit

0.09

0.048

0.0510.055

0.08

0.005

0.102

0.06

0.04

0.44

0.341

0.13

0.8

0.3

ANURADHAPURA

POLPITIYA

LAXAPANA KOLONNAWABOLAWATT

Reference

F

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Zero sequence equivalent circuit

0.09

0.048

0.051

0.055

0.08

0.005

0.102

0.06

0.04

0.44

0.341

0.13

0.8

0.3

ANURADHAPURA

POLPITIYA


Reference

F

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THEORY:

1. Importance of a fault study.Power systems are always dynamic systems due to the changes occur during switching

operations. A fault analysis is essential to determine the short circuit rating of newswitchgear and other substation infrastructure equipment that will be procured and installed.

For design, operation and protection purposes, fault studies should be done routinely. Fault

analysis plays a huge role in the safety factor. Power system equipment such as circuit

breakers can fail catastrophically if they are subjected to fault duties that exceed their

ratings. Other equipment like busbars, transformers and cables likely to fail thermally or

mechanically if subjected to fault currents in excess of ratings. Hence, to ensure the safety of

workers and the general public fault study is a major factor.

2. What are the analogue methods of studying the fault flow in a system? Symmetrical components method

All three main sequences positive, negative and zero sequences are used in this

symmetrical components method. An asymmetric fault is represented by those symmetric

components. Hence, the analysis is very easy to perform.

Bus impedance methodIn this method of fault study, bus-bar impedances are used. Both the symmetrical and

asymmetrical faults can be analyzed using this method. Bus impedance method has a higher

accuracy level though this method is hard in analyzing.

3. How DC network analyzer may be used for this purpose. (Explain for bothsymmetrical and asymmetrical faults).

DC network analyzer is used to simulate a network in its three sequence components. In

a symmetrical fault it is just the positive sequence required and in asymmetrical faults all

three sequences are used to represent the fault. Thus, this enables analyzing the sequence

components of a network. The sequential networks can be connected independently and the

values of sequential components can read using the metering equipment. The generators of

the given power system are represented by a DC power supply.

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Then current & voltage sequence components can be measured using metering

equipment. Measurements are taken in per unit values. The values of the phase voltage and

phase current can then be calculated by applying to the matrix equation. To obtain actual

values, calculated values have to be multiplied by the corresponding scale factors and base

factors.

4. Importance of using sequence components.The model we used to analyze a power system in this practical uses the three main

symmetrical components the positive, the negative and the zero sequences. And the

asymmetric fault is represented by those symmetric components after being done some

calculations. This method is easy in analyzing and calculations. Because the circuit converts

to the symmetrical components, the equipment in the circuit such as generators, transformers

and transmission lines replace by impedances and voltage sources. Apart from that the

sequence matrix calculation is much easier than that of complex asymmetric calculations.

Another advantage is that using the per unit values the high values of currents, voltages

under fault conditions can be replaced by the measurable quantities.

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5. The relationships between the sequence impedances for generators, transformers andtransmission lines with suitable reasons.

Generator

The impedance for the currents of the three sequences will generally be different for

each sequence due to the direction of the rotation of generator as its a rotating machine in

operation. The rotational emf generated for positive and negative sequences would also tend

to change because of the rotating direction of the generator.

Transformer

Transformer, in the gross is a passive and stationary device. It does not have any

specific direction in rotation as we discussed in the generator. Thus, all the impedances of

positive sequence, negative sequence and zero sequence are almost the same in value.

However, the zero sequence paths across the windings of a transformer depend on the

winding connection and even grounding impedance. The winding connection configuration

may either be star or delta. In the delta winding the current can circulate within the winding.

Transmission Lines

Much similar to the transformer as they are passive, stationary and do not have specific

direction in rotation. Thus, they have the same positive sequence and negative sequence

impedances. Because the zero sequence impedance comes across the grounding path, those

earth wire impedances might be considerably large and needed to be added. Hence, zero

sequence impedance will be somewhat different from the other impedances.

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CALCULATIONS:

i. Calculations using practical valuesSingle Line to Earth Fault (L-G)

Va= 0 (fault impedance is zero),

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

Voltage

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

a2

a1

a0

2

2

c

b

a

VV

V

11

111

VV

0V

V04.0V

98.1719.40-22.17V

VVVV

a

0

a

a2a1a0a

V446.12337.60V

12098.1724019.40-22.17V

VVVV

0b

00

b

a2a1

2

a0b

V446.12337.60V

24098.1712019.40-22.17V

VVVV

0

c

00

c

a2

2

a1a0c

Actual voltage values,

kV0609.0V

)503/132(04.0V

acta,

acta,

kV

FSupply

Side

a

b

c

Ia

Ib

Ic

Va = 0

Vb

Vc

EfZ1 Z2 Z0Ia1 Ia2 Ia0

Va1 Va2 Va0

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kV6123.44-016.29V

)503/132(446.12337.60V

0

actb,

0

actb,

kV

kV446.123016.29V)503/132(446.12337.60V

0

actc,

0

actc,

kV

Fault Current

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

I

3I

0I

0I

I

1

1

111

3

1

I

I

I

21a0

c

b

a

2

2

a2

a1

a0

a

aa

III

Fault current, If= Ia

= 3Ia0

= 3 x 13.5 = 40.5 mA

mA40.5I

5.135.135.31I

IIII

a

0

a

a2a1a0a

Actual fault current = Ia,act x

=

= 566.85 A

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00

b

a2a1

2

a0b

1205.132405.135.31I

IIII

A0I

b

00

c

a22

a1a0c

2405.131205.135.31I

IIII

A0Ic

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

Vb = Vc

Ib = - Ic

Voltage

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

a2

a1

a0

2

2

bc

b

a

V

V

V

1

1

111

VV

V

V

V13.64V

06.3207.320V

VVVV a2a1a0

a

a

a

V98.179065.32V12006.3224007.320V

VVVV

0

b

00

b

a2a1

2

a0b


kVkV 74.97)503/132(13.64V acta,

kV98.17987.48V

)503/132(98.179065.32V

actb,

0

actb,

kV

kV98.179-87.48VV actb,actc,

Supply

Side

a

b

c

Ia

Ib

Ic

Va

Vb

Vc

Va0

Z0

Ia0

Va2

Z1 Ia1

Ef

Va1

Z2Ia2

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Fault Current

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

I

a2

a1

a0

2

2

bc

b

a

I

I

I

1

1

111

II

I

0I

A0I

5.245.240I

IIII

a

a

a2a1a0a

mA90435.42I

1205.242405.240I

IIII

0

b

00

b

a2a1

2

a0b

Ic = - Ib = mA

Actual current values,

A

A

90936.593I

(50/4000)

)]3*[40/(13290435.42I

0

actb,

0

actb,

A

A

90936.593I

(50/4000))]3*[40/(13290435.42I

0

actc,

0actc,

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Supply

Side

a

b

c

VaIa

Ib

Ic

Vb=0

Vc=0

When fault impedance is zero,

Vb =Vc =0

Ia= 0

Voltage

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

V

V

V

a2

a1

a0

2

2

c

b

a

V

V

V

1

1

111

0V

0V

V

V43.74V

81.2481.2481.24V

VVVV

a

a

a2a1a0a


kV44.113V

)503/132(43.74V

a

acta,

kV

V0V

12081.2424081.2481.24V

VVVV

b

00

b

a2a1

2

a0b

V0V

24081.2412081.2481.24V

VVVV

c

00

c

a2

2

a1a0c

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Fault Current

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

I

a2

a1

a0

2

2

c

b

a

I

I

I

1

1

111

I

I

0I

mA0

a

a

a2a1a0a

1805.0I

19345.51I

IIII

mA0

b

00

b

a2a1

2

a0b

62.11634.51I

12019240345.51I

IIII

mA0

c

00

c

a2

2

a1a0c

62.11634.51I

24019120345.51I

IIII

Actual current values,

A

A

180998.6I

(50/4000)

)]3*[40/(1321805.0I

0

acta,

0

acta,

A

A

62.11657.718I

(50/4000)

)]3*[40/(13262.11634.51I

0

actb,

0

actb,

A

A

62.11657.718I

(50/4000)

)]3*[40/(13262.11634.51I

0actc,

0

actc,

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ii. Theoretical calculations using given dataConsidering that the fault has occurred at Anuradhapura bus bar.

Single Line to Earth Fault (L-G)

0Va , 0Ib , 0Ic

3

IIII

0I

0I

I

1

1

111

3

1

I

I

I

a2a1aa0

c

b

a

2

2

a2

a1

a0

Fault Current

Since,

Z1 =0.239

Z2 = 0.223

Z0 = 0.609

Ef= 1 pu

AI

I

II

II

II

II

Z

Z

Z

E

V

V

V

actf

actf

af

aa

aa

aa

f

a

a

a

07.490

101323

1040

609.0223.0239.0

13

ZZZ

E3

3

3

3

00

00

00

0

0

,

3

6

,

021

f

2

1

0

2

1

0

2

1

0

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Voltage

AII aa 36.1633

490.07I 21a0

6.43540MVA

(132kV)Z

2

base

kV868.1536.1636.4350.223IZV

kV59.20336.1636.4350.2393

10132IZEV

kV336.4336.1636.4350.609IZV

a22a2

3

a11fa1

a00a0

0VVVVV a2a1a0a

kV134.99-93.91V

120868.15240203.59336.43V

VVVV

0

b

00

b

a2a1

2

a0b

kV134.9993.91V

042868.15021203.59336.43V

VVVV

0

c

00

c

a2

2

a1a0c

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Ia = 0, Vb = Vc ,Ib = -Ic

Fault Current

Since,

Z1 =0.239

Z2 = 0.223

Z0 = 0.608

Ef= 1 pu

AI

AII

AI

ZZ

EI

a

aa

a

f

a

0

69.378

69.378

101323

1040

)223.0(0.239

1

0

12

1

3

6

21

1

A90-91.655I

12069.37824069.3780I

IIII

0

b

00

b

a2a1

2

a0b

A

A9091.655I

A90-91.655I

II

0

c

0

c

bc

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Voltage

VZIV

kVVV

ZIVV

aa

aa

aaa

0

79.36

6.435223.069.378

000

21

2221

kV80179.36VV

kV80179.36V

12079.3624079.360V

VVVV

58.73V

79.3679.360V

VVVV

0

bc

0b

00

b

a2a1

2

a0b

a

a

a2a1a0a

kV

kV

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Ia = 0

Vb = Vc = 0

Fault Current

Since,

Z1 =0.239

Z2 = 0.223

Z0 = 0.608

Ef= 1 pu

AI

ZZ

ZII

AI

ZZ

ZII

AIZZZ

EI

a

aa

a

aa

a

f

a

75.116

223.0608.0

223.004.435

29.318

223.0608.0

608.004.435

04.435101323

1040

))608.0//223.0(239.0(

1

)//(

0

20

210

2

20

012

1

3

6

021

1

A105.03-49.675I

120318.29240435.04116.75I

IIII

0

b

00

b

a2a1

2

a0b

A

A105.0349.675I

240318.29120435.04116.75I

IIII

0

c

00c

a2

2

a1a0c

A

Voltage

kVV

IZEVV

a

afaa

76.92

04.4356.435239.03

101323)(333

111

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RESULTS:

L-G fault Practical Theoretical

Fault

Currents

Ia 566.85 A 490.07A

Ib 0 0

Ic 0 0

Fault

Voltages

Va 0.0609 kV 0

Vb 92.016 -123.4460

kV 91.93 -134.990

kV

Vc 92.016123.4460

kV 91.93134.990

kV

L-L fault Practical Theoretical

Fault

Currents

Ia 0 0

Ib 593.936 -900

A 655.91 -900

A

Ic 593.936900

A 655.91900

A

Fault

Voltages

Va 97.74 kV 73.58 kV

Vb 48.87 -179.980 kV 36.79180

0 kV

Vc 48.87 -179.980

kV 36.791800

kV

L-L-G fault Practical Theoretical

Fault

Currents

Ia 6.9981800 A 0

Ib 718.57 -116.620

A 675.49 -105.030

A

Ic 718.57116.620

A 675.49105.030

A

Fault

Voltages

Va 113.44 kV 92.76 kV

Vb 0 0

Vc 0 0

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DISCUSSION:

1. What are the assumptions made in your fault study and how valid are they?

a. Pre-fault voltages are at 1 pu (All generators are running at their rated voltages andrated frequency with their emfs in phase).Even though we assume, the sources to be balanced in magnitude and phase, they can be

slightly different from each other. However, those wont affect in the calculations in a

large scale.

b.Negligible line resistance compared to reactance.When considering the impedances in transmission lines, reactance is much higher than

the resistance. Hence, it is not much an error to neglect the resistance.

c. Negligible load currents compared to fault currents.When comparing with the fault current, the load currents are very small. While the load

currents are in Amperes range, the fault currents are at the range of kilo Amperes at the

fault. Hence, the effect from the load current to the system can be neglected when

comparing with the effect from the fault current.

d.Shunt capacitances are negligible.

e. Large systems may be represented by infinite bus bars.In larger systems, we can assume that the voltage and the frequency are constant. Hence,by using infinite bus bars, we can consider the voltage, the frequency to be constants,

phase sequence and phase voltage.

f. Transformers are on nominal tap position.Transformer impedances can be changed with the tap changings. By this assumption, we

can use the nominal voltages of the primary and secondary windings of the transformer as

well as the given impedances.

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REFERENCES

Electrical Power Systems by Debapriya Das Principles of Power Systems by V.K.Mehta Power Systems II-Lec 1, Lec 2, Lec 3Lecture Slides & Notes http://www.elect.mrt.ac.lk/EE423_%20Fault_Analysis_Notes.pdf http://en.wikipedia.org/wiki/Fault_(power_engineering)
http://www.elect.mrt.ac.lk/EE423_%20Fault_Analysis_Notes.pdfhttp://www.elect.mrt.ac.lk/EE423_%20Fault_Analysis_Notes.pdfhttp://en.wikipedia.org/wiki/Fault_(power_engineering)http://en.wikipedia.org/wiki/Fault_(power_engineering)http://en.wikipedia.org/wiki/Fault_(power_engineering)http://www.elect.mrt.ac.lk/EE423_%20Fault_Analysis_Notes.pdf

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