Fault Studies Semester5

7/31/2019 Fault Studies Semester5

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OBSERVATION SHEET

Name : M.K.B.S.Munasinghe

Index No. : 090337U

Group : G8

Instructed by : Ms.Chenuka

Date : 01/03/2012

Line-Ground (L-G) faultVao = - 27.28 V Iao = 12 mA

Va1 = 38.34 V Ia1 = 12 mA

Va2 = - 10.75 V Ia2 = 12 mA

Line-Line (L-L) faultVa1 = 23.95 V Ia1 = + 26.5 mA

Va2 = 23.84 V Ia2 = - 26.5 mA

Line-Line-Ground (L-L-G) faultVao = 19.85 V Iao = - 8.4 mA

Va1 = 19.86 V Ia1 = + 31 mA

Va2 = 19.87 V Ia2 = - 22 mA

Impedance readingsZ1 = 1084

Z2 = 897

Z3 = 2471


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THEORY1. Importance of a fault study.

Power systems are frequently subjected to faulty conditions which may be due to insulation

failure or some natural disturbances or even human error. Under faulty conditions high currents

usually flow in the system which may lead to excessive heating of equipment and lines causing

further damage at the point of fault and throughout the system. Also these high currents maycause large voltage drops which would disrupt the quality of power supply to other consumers.

Because of this faults must be isolated from the rest of the system as quickly as possible. Fault

analysis supply required information for this.

Also fault analysis of a power system is required in order to provide information for the selection

of protection equipment, switchgear, setting of relays and stability of system operation.

2.Analogue methods of studying the fault flow in a system Symmetrical component methodIn this method phase components are decomposed in to symmetrical sequence components zero

sequence, positive sequence and negative sequence. Usually used for symmetrical fault analysis.

Bus impedance or admittance methodThis method is used for analysis of both symmetrical and asymmetrical faults. Here the bus bar

admittances are used for the analysis.

3. Using DC network analyzerA network analyzer is an instrument that can be used to measures the network parameters of

electrical networks. It can be used to model and study symmetric and asymmetric faults in a

small scale by using positive, negative and zero sequence components. Alternators, transformers

with bus bars, transmission line sections and load sections can be separately modeled using this

and metering is possible using external meters.A multiplication factors is used to convert the

values in to real scale.

Asymmetrical faults study can be performed by symmetrical component method using the DC

network analyzer. For this first the three sequence networks have to be modeled according to

the given system. Per unit values multiplied by appropriate scale factor are used for modelingin the range of DC network analyzer. DC power supply represents the generators in the network.

According to type of the fault the three independent sequence networks can be connected in

different configurations. Using metering equipments the values of current and voltage of

sequence components are measured and then by calculations the actual values can be obtained.

4. Importance of using sequence componentsUnbalanced three phase systems can be decomposed into three balanced components, Positive

Sequence (balanced and having the same phase sequence as the unbalanced supply), NegativeSequence (balanced and having the opposite phase sequence to the unbalanced supply) and Zero

Sequence (balanced but having the same phase and hence no phase sequence). These are

known as the Symmetrical Components or the Sequence Components.


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+ +a

b

c

a1b1

c1

a2

b2

c2 a0c0b0

Unbalanced system Positive Sequence Negative Sequence Zero Sequence

1 1 11 1

Where, = 1120As knowledge of one sequence component directly gives the other related two components this

decomposing makes the analysis much easier.

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

Power systemcomponent

Sequence Impedance values

ReasoningPositivesequence

Negativesequence

Zerosequence

Transmission lines No inherent directionTransformers

Depend upon thewinding connection

and earthlingarrangement

No inherent direction butzero sequence impedance

depends on earth path

Generators

Has a inherent direction

due to rotation in onedirection

CALCULATIONS

1)Calculation of phase voltages and currents for the three types of faults using observed sequencecomponents of current and voltage.

MVA 3 base = 40 MVAResistance multiplication factor = 4000Simulation voltage/real voltage = 50V/132KVTherefore,

1524.20 Zbase= &kVLL base/MVA3 base

/

435.6

7 7 8 9://;< :;


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Practical calculations

a)Single Line to Earth Fault (L-G Fault)

From observations, Ia1=Ia2=Ia0= 12 mA therefore,

@@B@C 1 1 11 1 121212@D @ 3 @EF 3 12 GTherefore the actual fault currents are,

@ 3 12 10H 136.3 503. A@B &1 J J 12 0

@C &1 J J 12 0

From the observations,

3.34, 10.5, 2.2 BC

1 1 11 1 2.2J3.3410.5

Therefore the actual voltages are,

&2.2 J 3.34 10.5 1524.20 0.425 kVB &2.2 J 3.34 10.5 2640 &2.20 J 3.34240 10.5120 1524.20 0.102 134.01 OC &2.2 J 3.34 10.5 2640 &2.20 J 3.34120 10.5240 1524.20 0.102134.01 O


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

From observations, Ia0= -8.4 mA, Ia1= 31 mA, Ia2= - 22 mA therefore,

@@B@C

1 1 11 1

.4J3122

Therefore the actual fault currents are,

@ &.4 J 31 22 10H 136.3 .3 A@B &.4J 31 22 10H 24242.4 &.40 J 31240 22120 10H 136.3 66.31 105.6

@C &.4 J 31 22 10H 41.11 &.40 J 31120 22240 10H 136.3

66.31105.6 From the observations,

1.5 1.6 1. Therefore,BC

1 1 11 1 1.51.61.


&1.5 J 1.6 J 1. 1524.20 0.11 kVB &1.5 J 1.6 J 1. 2640 &1.50 J 1.6240 J 1.120 1524.20 0.0264150 OC &1.5J 1.6 J 1. 2640 &1.50 J 1.6120 J 1.240 1524.20 0.0264 150 O


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c) Line to Line Fault (L-L Fault)

From observations, Ia0= 0, Ia1= 26.5 mA, Ia2= - 26.5 mA therefore,

@@B@C

1 1 11 1

0J26.5

26.5

Therefore the actual fault currents are,

@ &1 0 J 1 26.5 J 1 26.5 10H 136.3 0 A@B &0 J 26.5 26.5 10H 24242.4 &00 J 26.5240 26.5120 10H 136.3 642.42 0 @C &0 J 26.5 26.5 10H 24242.4

&00 J 26.5120 26.5240 10H

136.3

642.420 From the observations,

0 23.5 23.4 Therefore,BC

1 1 11 1 023.523.4


&0 J 23.5 J 23.4 1524.20 2.42 kVB &0 J 23.56 J 23.4 2640 &00 J 23.5240 J 23.4120 1524.20 36.421. OC &0 J 23.5 J 23.4 2640 &00 J 23.5120 J 23.4240 1524.20 36.421. O


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2) Theoretical calculation of phase voltages and currents for the three types of faultsThe given network is shown below. All values are in per unit with Vbase=132kV and

Sbase=40MVA.


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Usually in large power systems the resistances of the transmission lines and machines are negligible

compared to their inductance values. Therefore calculations carried out below are done using only

the inductance components.

Positive sequence equivalent circuit

Negative sequence equivalent circuit


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Zero sequence equivalent circuit

From the above equivalent circuit diagrams,Z1 = 0.2391 pu Z2 = 0.2225 pu Z0 = 0.6092 pu Ef= 1 pu

Base current @BPQ RSTUWRTUW XRSYS 0.14 OBase impedance BPQ &RTUW/RSTUW /XRS 435.6 Z

a) Single Line to Earth Fault (L-G Fault)

Assuming that the fault impedance is zero and load currents are negligible,

0 @B @C 0@@@ 1 1 11 1 @

@B 0@C 0 This yield @ @ @ [U


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0\D0

0 00 00 0 ]@ @ 3@ @ 3@ @ 3 _ ` @D @

c


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fgh fgi fgj@ \D& J 132 10

3l0.231 J &0.6020.2225m 435.6 435.13

@ &\D @ &132 103 0.231 435.6 435.130.2225 435.6 31.2 @ &\D @ &132 10

3 0.231 435.6 435.130.602 435.6 116.41 & J J 3 3 n\D @o

3 n132 10

3 0.231 435.6 53.66o

2.62 Opgh J pgi J pgj j@B &IE J IE J IE &116.410 J 435.13240 31.2120 65.0 104. @r &IE J @ J IE

&116.410 J 435.2120 31.2240

65.0104. c) Line to Line Fault (L-L Fault)

Assuming zero impedance and negligible load currents, B C @ 0&@ @Using the equivalent circuit diagram for the fault,

@ 0

@ \D J 132 103l0.231 J 0.2225m 435.6 3.02 @ @ 3.02


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@ 3.02 435.6 0.2225 36.35 O @ 0, 0 & J J 0J 36.35 J 36.35 3.4 OB & J J 00 J 36.35240 J 36.35120 36.35 10OC B 36.3510O@B &@ J @ J @ 00 J 3.02240 3.02120 656.4 0@C @B 656.4 0 656.4 0

RESULTS

1. Comparison of theoretical results with results obtained through the practical using DC networkanalyzer.

Type offault

ValuesFault current(A) Fault voltage(kV)

Ia Ib Ic Va Vb Vc

L-G

Practical1511.61

00 0

0.4725

0

90.102

-134.01

90.102

134.01

Theoretical503.87

00 0

0.009

0

91.931

-135.03

91.931

135.03

L-L-G

Practical8.398

0

667.31

-105.70

667.31

105.70

90.811

0

0.0264

150

0.0264

-150

Theoretical 0675.80

-104.97

675.80

90

92.672

00 0

L-L

Practical 0642.42

-90

642.42

90

72.842

0

36.42

179.77

36.42

179.77

Theoretical 0

656.48

-90

656.48

90

73.47

0

36.735

180

36.735

180


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DISCUSSION

1. Assumptions made in a fault study and their validity.

All the sources in the system under consideration are assumed to be balanced and equal inmagnitude and phase. Although slight differences magnitude and phase may be present

between different source voltages that does nothave a major effect on the results of the

calculations.

When analyzing the fault,Sources that are connected to the system are represented by theThevenins equivalent voltage prior to fault at the fault point.

Large systems are usually represented by infinite bus bars.The effect from a small system toa much larger scale system, on to which the smaller system is connected, is very small.

Therefore this assumption is valid for all practical purposes

Transformers are assumed to be on the nominal tap position. But if any tap changingtransformers are present, the actual transformer impedance values of those transformers may

be different from the values used for calculations. This may have a effect on the accuracy of

the calculations.

Resistances are usually assumed to be negligible compared to reactance.But when resistancevalues too are significant, like in the system used for this experiment, neglecting those

values may have negative effect on the accuracy and usefulness of the calculations.

Transmission lines are assumed to be fully transposed so that all three phase have the sameimpedance. But three can be small differences if the transmission lines are not transposed or

if the spacing between conductors is uneven.This would change the line inductances which

would eventually lead to errors in the calculations. Loads currents are negligible compared to fault currents. Depending upon the fault level of

the point the fault current may vary in kA range. But compared with load currents (in

ampere range) this is hundreds of times larger. So there is no significant effect to the final

result by ignoring load currents.

Line charging currents can be completely neglected. As line charging currents are smallcompared to load current, ignoring line charging currents will not have a significant effect

on the final results.

2.

State the reasons for the deviation of practical results and theoretical results.

The resistance values which were set on the DC network analyzer were not exactly the sameas the values used for calculations.

Some assumptions made above may not hold true for the experiment. Resistance values of connecting wires and joints were not taken in to consideration. Observer errors

3. Practical problems encountered during the experiment and state the steps you have taken toavoid them

Not many problems were encountered in the experiment. Only the resistor values of the DCanalyzer posed a problem as a result of un availability of fine tunable resistors. This was


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overcome by selecting suitable the resistor combinations to approximate the needed values

as close as possible.

Fault Studies Semester5

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