48572 PCT Lecture Notdfes.pdf

7/27/2019 48572 PCT Lecture Notdfes.pdf

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SCHOOL OF ELECTRICAL, MECHANICAL AND

MECHATRONIC SYSTEMS

LECTURE NOTES

48572 - Power Circuit Theory

Notes Prepared by: Dr Germane Athanasius


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2


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Contents

1 Power System - An introduction 7

1.1 Power system divisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.2 Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.1.3 Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.4 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Power in single phase AC circuits . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Complex power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Three phase balanced circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.1 Star or Y connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4.2 Delta or connection . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Power in three phase balanced circuit . . . . . . . . . . . . . . . . . . . . . . . . 12

1.6 Three phase power measurement using two wattmeter method . . . . . . . . . . 14

1.7 Power transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.7.1 Equivalent circuit of a transformer . . . . . . . . . . . . . . . . . . . . . 161.7.2 Three phase transformers . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.8 Auto transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.8.1 Tap changing transformers . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Per unit system and load modeling 23

2.1 Per unit system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.1 Changing the base ofpu quantities . . . . . . . . . . . . . . . . . . . . . 24

2.2 Power flow between two nodes . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2.1 The static stability limit . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2.2 Effect of reactive power . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3 Load modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.1 Variation with voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.2 Variation with frequency . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3.3 Models of loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Transmission line Impedance 29

3.1 Resistance of the transmission line . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Inductance of single conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3


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4 CONTENTS

3.2.1 Internal inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2.2 Inductance due to external flux . . . . . . . . . . . . . . . . . . . . . . . 31

3.3 Inductance of single phase two wire system . . . . . . . . . . . . . . . . . . . . 32

3.4 Flux linkages of one conductor in a group . . . . . . . . . . . . . . . . . . . . . 34

3.5 Inductance of 3 phase systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.5.1 Conductors with symmetrical spacing . . . . . . . . . . . . . . . . . . . 35

3.5.2 Conductors with unsymmetrical spacing . . . . . . . . . . . . . . . . . . 36

3.6 Inductance of stranded conductors . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.7 Bundled conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.8 Double circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Transmission line capacitance 41

4.1 Capacitance of two wire conductor . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2 Potential difference in a multi conductor system . . . . . . . . . . . . . . . . . . 434.3 Capacitance of three phase line . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.4 The effect of earth on line capacitance . . . . . . . . . . . . . . . . . . . . . . . 45

4.5 Capacitance of bundled conductors . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.6 Capacitance of double circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Transmission line model 47

5.1 Short transmission line model . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5.2 Medium length line model or nominal model . . . . . . . . . . . . . . . . . . 485.3 Long transmission line model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.4 Voltage and current waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.5 Surge impedance loading of the line . . . . . . . . . . . . . . . . . . . . . . . . 54

6 Symmetrical faults 55

6.1 Numerical example -1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.2 Determination of short circuit capacity (SCC) . . . . . . . . . . . . . . . . . . . 59

6.3 Fault analysis using Zbus matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 606.4 Numerical example - 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

7 Symmetrical components 65

7.1 Basics of symmetrical components . . . . . . . . . . . . . . . . . . . . . . . . . 65

7.2 Sequence impedance of a star connected load . . . . . . . . . . . . . . . . . . . 68

7.3 Sequence impedance of a transmission line . . . . . . . . . . . . . . . . . . . . 69

7.4 Sequence impedance of Synchronous generator . . . . . . . . . . . . . . . . . . 71

7.5 Sequence network of a loaded Synchronous generator . . . . . . . . . . . . . . . 71

8 Unsymmetrical Faults 73

8.1 Single line to ground fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

8.2 Line to line fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

8.3 Double line to ground fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76


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CONTENTS 5


8.4.1 Case 1: Balanced 3 phase fault . . . . . . . . . . . . . . . . . . . . . . . 79

8.4.2 Case 2: Single line to ground fault . . . . . . . . . . . . . . . . . . . . . 80

8.4.3 Case 3: Line to line fault . . . . . . . . . . . . . . . . . . . . . . . . . . 808.4.4 Case 4: Double line to ground fault . . . . . . . . . . . . . . . . . . . . 81

8.5 Zbus matrix using symmetrical components . . . . . . . . . . . . . . . . . . . . 818.5.1 Single line to ground fault . . . . . . . . . . . . . . . . . . . . . . . . . 81

8.5.2 Line to line fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

8.5.3 Double line to ground fault . . . . . . . . . . . . . . . . . . . . . . . . . 82

8.5.4 Line currents and bus voltages during fault . . . . . . . . . . . . . . . . 82


8.6.1 Case 1: Balanced three phase fault . . . . . . . . . . . . . . . . . . . . . 84

8.6.2 Case 2: Single line to ground fault . . . . . . . . . . . . . . . . . . . . . 85

8.6.3 Case 3: Line to line fault . . . . . . . . . . . . . . . . . . . . . . . . . . 868.6.4 Case 4: Double line to ground fault . . . . . . . . . . . . . . . . . . . . 88

9 Power system transients 91

9.1 Transients with AC source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

9.2 Re-striking voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

9.3 Double frequency transient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

9.4 Traveling waves on transmission lines . . . . . . . . . . . . . . . . . . . . . . . 98

9.5 Traveling waves - open end line . . . . . . . . . . . . . . . . . . . . . . . . . . 100

9.6 Traveling waves - short circuited line . . . . . . . . . . . . . . . . . . . . . . . . 101

9.7 Line terminated through a resistance . . . . . . . . . . . . . . . . . . . . . . . . 102

10 Transient Stability 105

10.1 Swing equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

10.2 Single machine on infinite bus (SMIB) model . . . . . . . . . . . . . . . . . . . 106

10.2.1 Rotor angle response to sudden change in Pm . . . . . . . . . . . . . . . 10810.2.2 Equivalent single machine system . . . . . . . . . . . . . . . . . . . . . 109

10.3 Stability based on equal area criterion . . . . . . . . . . . . . . . . . . . . . . . 109

10.3.1 Stability during sudden input power change . . . . . . . . . . . . . . . . 111

10.3.2 Stability during 3 phase fault . . . . . . . . . . . . . . . . . . . . . . . . 112

10.4 Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113


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6 CONTENTS


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Chapter 1

Power System - An introduction

Electrical energy is the most convenient and widely used form of energy. Electrical energy is

more efficient to generate, transport and to use compared with other forms of energy. With the

invention of power transformers, electrical energy can be transformed from one voltage level to

other without much difficulty. This enabled long distance transmission of electrical energy at

high voltages with reduced transmission losses.

The generating stations are usually located near to the source of power required to drive the

prime movers. These locations often geographically well separated from the load centers or

consumers. The power generated from the generating centers are transmitted at high voltages to

the load centers where it is stepped down to low voltages suitable for use.

1.1 Power system divisionsModern power system can be broadly divided into four areas:

Generation Transmission Distribution Loads

A basic simple power system layout is shown in Figure 1.1.

1.1.1 Generation

Major power generation uses hyro, nuclear and thermal (coal and petro fuels). Turbines are

driven by these energy sources are known as prime movers and are used to drive the generators.

Generators generate 3 phase electrical power at 11 or 25 kV voltage level. Transmission of

power at the generation voltage for long distance in not feasible due to high transmission losses

and several other reasons. Therefore the generated power is stepped up to transmission voltage

usually 230 or 400 kV.

7


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1.2. POWER IN SINGLE PHASE AC CIRCUITS 9

1.2 Power in single phase AC circuits

Consider a single phase AC circuit shown in Figure 1.2. The instantaneous power is given by,

Figure 1.2: Power in single phase AC circuits.

p(t) = v(t)i(t) = Vmax cos(t + v) Imax cos(t + i) (1.1)

Using the trigonometrical identity,

cos A cos B =1

2[cos (A B) + cos (A + B)] (1.2)

we get

p(t) =1

2VmaxImax [cos (v i) + cos (2t + v + i)]

=1

2VmaxImax {cos(v i) + cos [2 (t + v) (v i)]}

=1

2VmaxImax [cos (v i) + cos2 (t + v)cos(v i) + sin 2 (t + v)sin(v i)]

= V Icos [1 + cos 2 (t + v)] + V Isin sin2(t + v) (1.3)

where V = Vmax2

and I = Imax2

are the rms values of voltage and current and = v i. In(1.3) the first term represents the power flow into the circuit and the second term represents the

power absorbed during charging and discharging of the reactive elements in the circuit. Now let

us consider the first term,

pr = V Icos + V Icos cos2(t + v) (1.4)

If we take the average power over a cycle the second term which is the oscillating power in the

circuit disappears leaving the average power as,

P = V Icos (1.5)


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10 CHAPTER 1. POWER SYSTEM - AN INTRODUCTION

The term V Icos is the active or real power, cos is the power factorV I is the apparent power.The second term of (1.3) is,

pq = V Isin sin2(t + v) (1.6)

This is the power consumed during charging and discharging of the reactive elements in the

circuit. If we take the average value of this power it will be zero. The amplitude of this oscillating

power is called the reactive power and is given by,

Q = V Isin (1.7)

Even though P and Q have the same units, to distinguish P is expressed in watts and Q in var.

1.3 Complex power

Consider the phasor diagram shown in Figure 1.3 in which current lags the voltage by an angle

since the load is inductive. If we take the complex product of the vectors,

Figure 1.3: Complex power in AC circuits.

V I = V I(v i) = V I

= V Icos +V Isin (1.8)

Using (1.5) and (1.7) we can write (1.8) as,

S = V I = P +Q (1.9)

Equation (1.9) is a complex equation and the power represented by it is called complex power.

For leading power factors Q is negative and for lagging power factors Q is positive. Equation(1.9) can be diagrammatically represented using power triangle as shown in Figure 1.3.


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Figure 1.5: Three phase star connected system.

From (1.11), for star connected system the line voltage is

3 times the phase voltage and leadsthe set of phase voltages by 300 for positive sequence system.

1.4.2 Delta or connection

A three phase delta connected system is shown in Figure 1.7. The currents IA, IB and IC are line

currents and the currents IAB, IBC and ICA are phase currents. In delta connected system linevoltage and the phase voltage are equal. Using the phasor diagram shown in Figure 1.8 the line

currents in terms of phase currents are given by,

IA = IAB ICA = Iphase00 Iphase1200 =

3Iphase300IB = IBC IAB = Iphase2400 Iphase00 =

3Iphase210

0

IC = ICA IBC = Iphase1200 Iphase2400 =

3Iphase900 (1.12)

From (1.12), for delta connected system the line current is

3 times the phase voltage and lagsthe set of phase currents by 300 for positive sequence system.

1.5 Power in three phase balanced circuit

Consider a balanced three phase system, the instantaneous phase voltages are given by,

vAN =

2Vphase cos(t)

vBN =

2Vphase cos

t 1200vCN =

2Vphase cos

t 2400 (1.13)


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1.5. POWER IN THREE PHASE BALANCED CIRCUIT 13

Figure 1.6: Three phase line and phase voltages.

Figure 1.7: Three phase delta connected system.

If the respective phase currents lag the phase voltages by , we can write the instantaneous phasecurrents as,

iA =

2Iphase cos(t )iB =

2Iphase cos t

1200

iC = 2Iphase cos

t 2400 (1.14)The net instantaneous three phase power is given by,

p3phase = vANiA + vBNiB + vCNiC

= 2VphaseIphase cos(t)cos(t )+ 2VphaseIphase cos

t 1200 cos t 1200

+ 2VphaseIphase cos

t 2400 cos t 2400


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Figure 1.8: Three phase line and phase voltages.

Using the trigonometrical relation (1.2)

p3phase = VphaseIphase [cos + cos (2t )]+ VphaseIphase

cos + cos

2t 2400

+ VphaseIphase

cos + cos

2t 4800three 2t terms will add up to zero. The instantaneous three phase power becomes

p3phase = 3VphaseIphase cos = P3phase (1.15)

From (1.15) we can see that even though the individual single phase power is pulsating the three

phase power is constant. The three phase reactive and apparent power is given by,

Q3phase = 3VphaseIphase sin

S3phase = 3VphaseIphase = P3phase +Q3phase (1.16)

Three phase power in terms of line quantities is given by,

P3phase =

3VlineIline cos

Q3phase =

3VlineIline sin (1.17)

1.6 Three phase power measurement using two wattmeter method

Three phase power can be measured by three single phase wattmeters having current coils in

each line and potential coils connected across the given line and any common junction. Since

this common junction is completely arbitrary, it may be placed on one of the three lines, in which

case the wattmeter connected in that line will indicate zero power because its potential coil has no

voltage across it. Hence that wattmeter can be dispensed with, and three phase power by means


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1.6. THREE PHASE POWER MEASUREMENT USING TWO WATTMETER METHOD 15

of only two single phase wattmeters having a common potential junction on any of the three lines

in which there is no current coil. This method is valid for both balanced and unbalanced circuits

with either the load or the source unbalanced.

Consider the circuit arrangement for the power measurement as shown in Figure 1.9

Figure 1.9: Two-wattmeter method circuit diagram.

Consider the phasor diagram shown in Figure 1.10 The wattmeter readings are given by,

Figure 1.10: Two-wattmeter method phasor diagram.


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W1 = VABIA cos

+ 300

=

3VphaseIphase cos

+ 300

W2 = VCBICcos 300

=

3VphaseIphase cos 300

W1 + W2 = 3VphaseIphase cos = P3phase

W2 W1 =

3VphaseIphase sin =Q3phase

3

P3phase = W1 + W2 and Q =

3 (W2 W1) (1.18)

1.7 Power transformers

Transformers are used at different stages in a power system. The generated voltage usually at 11

or 25 kV is raised to transmission voltage of 230 - 750 kV and at the receiving end voltage is

reduced to distribution level 11 - 33 kV and further reduced to 400-220 V for house hold use.

1.7.1 Equivalent circuit of a transformer

The equivalent circuit of a transformer with primary voltage Vp and secondary voltage Vs is givenin Figure 1.11 In the figure, Rp and Rs are the resistances of primary and secondary windings

Figure 1.11: Transformer equivalent circuit

and Xp and Xs is the leakage reactances. Rc and Xm indicate the no-load loss and magnetis-ing components of the transformer. Ep and Es are the induced emfs in primary and secondary

windings. The voltage transformation ratio of the transformer is given by k = EpEs

. The equiv-

alent circuit can be simplified by moving the parallel branch to the supply side and moving the

impedances to primary or secondary side. While moving the voltage, current and impedances

the voltage transformation need to be included by properly transforming the quantities. Figure

1.12(a) shows the equivalent circuit referred to primary side. Further simplification in the equiv-

alent circuit is made by neglecting the no-load branch as shown in Figure 1.12(b). Figure 1.13


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1.7. POWER TRANSFORMERS 17

Figure 1.12: Transformer equivalent circuit referred to primary.

shows the equivalent circuit referred to secondary side.

Figure 1.13: Transformer equivalent circuit referred to secondary.

1.7.2 Three phase transformersThree phase power transformers can be either connected in star or delta connection. Generally

transformer windings are connected internally in Y or formation. The large transformers areusually constitute of three single phase transformer banks and connected externally in Y or configuration. The different combinations are Y Y, , Y and Y. Depending onthe voltage level and other requirements a particular configuration is chosen.

Three phase transformers are represented using per phase equivalent circuits. The no-load

shunt path is neglected and only the winding resistance and leakage reactances are considered


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while deriving the per phase equivalent. When working with Y configuration, the side isrepresented by the equivalent Y formation and the neutrals are joined and the per phase equiva-lent is worked out. The equivalent Y connected impedance is given by ZY = Z/3. The per

phase equivalent circuits referred to primary and secondary sides are given in 1.14(a) and (b).

Figure 1.14: Per phase equivalent circuit.

1.8 Auto transformers

A two winding transformer can be connected as an autotransformer as shown in Figure 1.15. Theworking principle of the auto transformer is the same as the two winging transformer except that

the former is electrically connected. The autotransformer has an increased power rating when

compared with two winding transformer. The voltage, current and the power rating advantage

over two wing transformer are given by,

VpauVsau

= 1 +N1N2

= 1 + k

IpauIsau

=1

1 + k

Power advantage Padv

=V Apau

V A2wind= 1 +

1

k(1.19)

The equivalent per unit impedance of the two winding and the auto transformer are related by,

Zau =Z2wind

Padv(1.20)

The advantage of using the autotransformer is that for the same power rating the autotransformer

requires less copper. But the insulation cost will be more because both windings need same level

of insulation.


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1.8. AUTO TRANSFORMERS 19

Figure 1.15: Auto-transformer.

1.8.1 Tap changing transformers

In order to control the voltage at different levels of the power system and also to control the

reactive power flow, transformers are equipped with tap changing facility. Generally the tap

changing facility will provide a voltage variation

5 to

10 % of the rated voltage. The

primary or the secondary windings are provided with taps to alter the transformation ratio kthereby altering the voltage. Based on the tap changing operation transformers are classified as

off load tap changing transformers and on load tap changing transformers (OLTCs). Off load tap

changing transformers require the transformer to be switched off while changing the taps and are

used where frequent voltage adjustments are not required. OLTCs permit tap changing without

switching off the load. Generally OLTCs operate under closed loop conditions which help to

maintain the voltage irrespective of changes in load variations.

Consider the power system shown in Figure 1.16, the generation voltage VG is stepped upthrough OLTC and transmitted and at the receiving end it is stepped down by OLTC to the load

voltage level VL. The secondary voltage of the OLTC at the sending end is Vs and the primary

voltage at the receiving end is Vr. The tap positions of the OLTCs are Ts and Tr. The voltagesand impedances are referred to the high voltage side. The phasor diagram of the system is shownin Figure 1.17. From the phasor diagram we can write,

|Vs| |Vr| + xy + yz|Vs| = |Vr| + |I|Req cos + |I|Xeq sin (1.21)

We can write real and reactive power as,

Pr = |Vr||I| cos and Qr = |Vr||I| sin (1.22)


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Substituting in (1.21) we get,

|Vs

|=

|Vr

|+

ReqPr + XeqQr

|Vr|(1.23)

We have Vs = TsVG and Vr = TrVL and substituting this in (1.23) we get,

Ts|VG| = Tr|VL| + ReqPr + XeqQrTr|VL|

Ts =1

|VG|

Tr|VL| + ReqPr + XeqQrTr|VL|

(1.24)

Assuming the product TsTr 1 and substituting in (1.23) we get,

Ts = 1|VG| |VL|

Ts+ ReqPr + XeqQr|VL|

Ts

=

|VL| + T2s (ReqPr + XeqQr)|VG||VL|Ts

T2s [|VG||VL| (ReqPr + XeqQr)] = |VL|

Ts =

|VL|

|VG||VL| (ReqPr + XeqQr) (1.25)

Figure 1.16: Single line diagram of OLTC power system


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1.8. AUTO TRANSFORMERS 21

Figure 1.17: Phasor diagram of OLTC power system


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Chapter 2

Per unit system and load modeling

2.1 Per unit systemPower system operates at different voltage levels from generation to the distribution end. It

will be difficult to analyse the power system with different voltages especially when dealing

with impedances. To overcome this difficulty per unit (pu) system is used. The power system

variables are expressed as a fraction or multiples of base quantities. While working with pu

system the voltage differences are eliminated. The per unit quantity of any variable is defined as,

Quantity in pu =actual quantity

base value of the quantity

%pu = Quantity in pu 100% (2.1)For example,

power pu =actual power

base power

voltage pu =actual voltage

base voltage

To arrive the pu values of voltage, current, volt-amperes and impedance, it is only necessary

to choose any of the two values as base values and other base values will get fixed accordingly.

Generally the power of the largest generating unit in MVA and maximum operating voltage in

kV are selected as base values. It is not a hard and fast rule, depending on convenience the base

values can be selected. If we select Base kVA and Base kV, other base quantities are given by,

base current IBase(A) =Base kVA

Base kV

base impedance ZBase() =Base kV

IBase

=Base kV2 1000

Base kVA

=Base kV2

Base MVA

23


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24 CHAPTER 2. PER UNIT SYSTEM AND LOAD MODELING

Generally in power system analysis we work with per-phase models and equivalent circuits.

But for the power system line voltages and 3 phase power will be specified. The following

expressions relate these two.

pu phase voltage(in line-neutral Base) = pu line voltage(in line-line Base)

pu phase kVA(in per phase Base kVA) = pu 3 phase kVA(in 3 phase Base kVA)

2.1.1 Changing the base ofpu quantities

Generally the manufacturers specify the equipments pu impedance, but these pu impedances of

all equipments need not be on the same base values. We need to convert them to the common

selected base for the analysis. The pu impedance is given by,

pu impedance =

actual impedance

base kVA

(base kV)2 1000To convert to the new base,

Znew base = Zold base

base kVold

base kVnew

2base kVAnew

base kVAold

2.2 Power flow between two nodes

To analyse the power flow between two nodes, consider the nodes 1 and 2 as shown in Figure 1.1

If12 = V1V2, then power leaving from 1 to 2 is given by,

Figure 2.1: Power flow between two nodes

S12 = V1I = V1Y (V1

V2)

= Y|V1|2 V1V2

= Y|V1|2 |V1||V2| (cos 12 + sin 12)

= (g b) |V1|2 |V1||V2| cos 12 (|V1||V2| sin 12) (2.2)The real and reactive power are given by,

P12 = g|V1|2 |V1||V2| cos 12 b (|V1||V2| sin 12) (2.3)

Q12 = g (|V1||V2| sin 12) b|V1|2 |V1||V2| cos 12 (2.4)


25/115

2.2. POWER FLOW BETWEEN TWO NODES 25

Power arriving at node 2 from 1 is given by,

S21 = V2I = V2Y (V1 V2)

= Y

V2V1 |V2|2= Y

|V1||V2| (cos 12 sin 12) |V2|2= (g b) |V1||V2| cos 12 |V2|2 (|V1||V2| sin 12) (2.5)

The real and reactive power are given by,

P21 = g|V1||V2| cos 12 |V2|2 b (|V1||V2| sin 12) (2.6)

Q12 = g (|V1||V2| sin 12) b|V1||V2| cos 12 |V2|2 (2.7)

The complex power lost between the nodes is given by,

Sloss = S12 S21 = (g b) |V1|2 + |V2|2 2|V1||V2| cos 12 (2.8)If we neglect the resistance between the nodes and the reactance is given by X then g = 0 andb = 1

X. The real power transferred is given by substituting in (1.3) and (1.6).

P12 = P21 =|V1||V2| sin 12

X(2.9)

The reactive power transferred is given by substituting in (1.4) and (1.7).

Q12 =|V1|2 |V1||V2| cos 12

X

Q21 = |V1

||V2

|cos 12

|V2

|2

X (2.10)

From (1.9) the direction of real power flow depends on the value of 12. If V1 leads V2 powerflows from node 1 to 2 otherwise from 2 to 1. 12 is known as power angle. The average reactivepower transferred is given by,

Qave =Q12 + Q21

2=

|V1|2 |V2|22X

(2.11)

Equation (1.11) shows that the reactive power flows node with high voltage magnitude to the low

voltage node and it is independent of12. When V1 = V2, there will be no average reactive powertransferred and the reactive power losses are supplied equally from both ends.

2.2.1 The static stability limit

Differentiating (1.3) with respect to 12 and equating to zero we get,

dP12d12

= g|V1||V2| sin 12 b|V1||V2| cos 12 = 0

tan 12 =b

g=

X

R(2.12)


26/115


IfR = 0, the stability limit occurs at 12 =2

and the maximum power transferred is given by

P12max =

|V1||V2|X

if|V1| |V2| = V, then P12max = |V|2

X(2.13)

Equation (1.13) shows that high voltage is required to transmit power over long distances with

high X.

2.2.2 Effect of reactive power

Because of inductive nature of transformers, transmission lines, induction motors and other in-

ductive loads large amount of inductive power is consumed. Even though reactive power does

not produce any useful work, still it has to be generated and transmitted which produces lossesand other economic difficulties. The best way to get rid of this is to cancel the reactive power

by generating capacitive reactive power at the point where negative inductive reactive power is

generated. For this purpose capacitor banks and other equipments like synchronous condensers

and static var compensators are used through out the network.

2.3 Load modeling

Power system loads consists of motors, heaters, lighting and electronic equipments. Most system

loads are a mixture of different types. The loads vary in size (watts and vars), symmetry, daily

and seasonal variations, short term fluctuations (e.g, welding machines, wood chip mills). Someare nonlinear loads (e.g, rectifiers and other power conversion equipments) which may produce

significant harmonic currents. Most loads vary with voltage and frequency.

2.3.1 Variation with voltage

Let the variation in power be described by

P = k|V|n (2.14)Consider a small change in voltage magnitude d|V|. The relative change is given by d|V|

V. A first

order Taylors series approximation gives the change in power as,

dP dPd|V|d|V| = kn|V|

n1 d|V| = n P|V| d|V| (2.15)

Hence the relative change in power is given by,

dP

P nd|V||V| (2.16)

A similar expression can be arrived for reactive power Q the value of exponent n may differ.


27/115

2.3. LOAD MODELING 27

2.3.2 Variation with frequency

Let the power P be a function of frequency f, P = g(f). Then for a small change in frequency

df using first order Taylors series approximation we have,

dP dPdf

d|f| (2.17)

and the relative change in power is given by,

dP

P

f

P

dP

df

df

f(2.18)

2.3.3 Models of loads

Constant impedance load

Admittance Y = 1Z

= g + b = constant (b is negative for inductive load). S = Y |V|2 andP = g|V|2 and Q = b|V|2, therefore the value ofn in (1.16) is 2, ie., a 1% drop in voltageresults in 2% drop in both P and Q.

Incandescent lighting load

The resistance of the light globes increases significantly with increasing operating temperature,

therefore the exponent n < 2 in (1.14) and (1.16). A value ofn = 1.6 is typical. Hence a dropof1% drop in voltage results in a 1.6% drop in P (drop in Q is negligible).

Fluorescent lighting load

The P, |V| relationship is more complicated than for the incandescent lamps but in the absenceof better information, n = 1.6 may be assumed.

Synchronous motor load

The speed of the motor is not affected by small changes of voltage but is proportional to fre-

quency. However, when the voltage drops too low the motor loses synchronism. As the me-

chanical load is unaffected by voltage, the electrical power P may also be assumed to remain

constant. Therefore the exponent n 0 in (1.14) and (1.16). Variation of P with frequencydepends on how the mechanical load varies with speed.

Induction motor load

Induction motor torque-slip characteristics give torque T s|V|2 where s is the slip, but thetorque is actually a characteristic of the mechanical load. For example, if T is constant, thes |V|2. In practice s < 0.05 at rated conditions. For s = 0.05, and we increase the voltageby 1%. Then the new slip is given by 0.05 1.012 = 0.049. Speed of the motor is proportional


28/115


to (1 s), therefore with constant T the mechanical power is proportional to (1 s). Hencethe mechanical power increases by the ratio 10.049

10.05 , i.e., the mechanical power increases by only0.105% for 1% voltage increase.

For an ideal motor (1s) is equal to the theoretical efficiency, hence the improved efficiencyat lower slip exactly compensates for the increased mechanical power, so that the electrical power

P remains constant in this ideal case. This is not quite so far a practical motor, particularly asthe torque varies with speed but we would generally be justified in assuming that the real power

P is independent of voltage for an induction motor. So n = 0 in (1.14) and (1.16) for P. This isnot true for the reactive power Q which increases with voltage (n > 0)


29/115

Chapter 3

Transmission line Impedance

The transmission line is characterised by its resistance, inductance and capacitance. The induc-

tance of the transmission line is the effect related to the emf induced on the conductor due to

changing flux caused by changing current. The capacitance is due to the charge created on the

conductors due to per unit potential difference between them.

Aluminium is used as conductor material because of lower weight, cheapness and large di-

ameter for the same resistance when compared with copper. Higher diameter reduces conduc-

tor surface voltage gradient which helps to reduce corona loss. Generally stranded aluminium

conductors are reinforced by steel is used. Aluminium alloy helps to improve the mechanical

properties of the conductor.

3.1 Resistance of the transmission lineThe transmission line resistance causes power loss Ploss during transmission. The effective re-sistance R of the conductor is given by,

R =Ploss|I|2 (3.1)

where I is the rms value of the current. The effective resistance will be equal to the dc resistanceof the conductor, if the current distribution is uniform across the conductor. With alternating

current the current distribution will not be uniform due to skin effectand this will increase the ac

resistance of the conductor. The dc resistance of the conductor is given by,

Rdc =lA

(3.2)

The resistance of the conductor varies with the temperature as,

R2 = R1T + t2T + t1

(3.3)

where R1 is the resistance at a temperature of t01 C, R2 is the resistance at a temperature of t

02 C

and T is the temperature constant whose value depends on the material of the conductor (T

29


30/115


31/115


32/115

32 CHAPTER 3. TRANSMISSION LINE IMPEDANCE

Figure 3.2: Inductance due to external flux

dx where the flux linkage dx is equal to flux dx since dx links the entire current. Now theflux over an area of width dx and unit length is given by,

dx = dx = Bxdx 1 = I2x

Now the flux linkage between the points D1 and D2 is given by,

ext =D2D1

I

2x dx

=I

2ln

D1D2

wb/m (3.9)

Now the inductance between the points D1 and D2 is

Lext =ext

I=

2ln

D1D2

H/m (3.10)

If the relative permeability is 1,

Lext = 2 107

ln

D1D2 H/m (3.11)

3.3 Inductance of single phase two wire system

Consider a two wire single phase system with radius r1 and r2 and separated through a distanceD as shown in Figure 1.3. The current in conductor A is equal and opposite to that of B. Letus consider the flux line produced by A. The flux lines beyond D need not be considered sincethose flux lines links a net current of zero.


33/115

3.3. INDUCTANCE OF SINGLE PHASE TWO WIRE SYSTEM 33

Figure 3.3: Inductance of single phase two wire system

The net inductance due to internal and external flux using (1.8) and (1.11) is given by,

L1 = Lint + Lext =1

2 107 + 2 107 ln D

r1

= 2 107

1

4+ ln

D

r1

(3.12)

Similarly for the conductor B,

L2 = 2 107

1

4+ ln

D

r2

(3.13)

The net inductance is given by,

L = L1 + L2 = 2 107

1

4+ ln

D

r1+

1

4+ ln

D

r2

= 2 107

ln e1

4 + lnD

r1+ ln e

1

4 + lnD

r2

(3.14)

= 2 107lnD

r1

+ lnD

r2 (3.15)

where r1 = r1e 1

4 = 0.7788r1 and r2 = r2e

14 = 0.7788r2 and are known as self-

geometric mean distance or geometric mean radius Ds (GMR), the circle with radii r1 or r

2 and

can be considered as the radius of fictitious conductor with no internal flux but with the same

inductance with a conductor of radius r1 or r2.Ifr1 = r2, substituting in (1.14) we get,

L = 4 107 ln Dr


34/115


3.4 Flux linkages of one conductor in a group

In the two wire system considered in 1.3, the inductance for conductors A and B can be expressed

using equation (1.14),

L1 = 2 107 ln 1r1

+ 2 107 ln D1

L2 = 2 107 ln 1r2

+ 2 107 ln D1

(3.16)

Now the flux linkages is given by,

1 = 2 107 ln 1r1

I1 + 2 107 ln D1

I1

2 = 2 107 ln 1r2

I2 + 2 107 ln D1

I2 (3.17)

Using the fact, I1 = I2 in (1.17),

1 = 2 107 ln 1r1

I1 + 2 107 ln 1D

I2

2 = 2 107 ln 1D

I1 + 2 107 ln 1r2

I2 (3.18)

Equation (1.18) can expressed in short as,

1 = L11I1 + L12I2

2 = L21I1 + L22I2 (3.19)

Equation (1.19) can be thought to be as an expression of flux linkages in terms of self and mutual

inductances and can be figuratively expressed as shown in Figure 1.4. Equation (1.19) can be ex-

Figure 3.4: Self and mutual Inductance in two wire system

tended to a system with n conductors with currents, I1, I2, , In so that I1 + I2 + + In = 0.


35/115

3.5. INDUCTANCE OF 3 PHASE SYSTEMS 35

Then the flux linkage of the ith conductor is given by,

i = LiiI1 +n

j=1

LijIj j

= i

= 2 107

ln1

riIi +

nj=1

ln1

DijIj

j = i (3.20)

3.5 Inductance of 3 phase systems

3.5.1 Conductors with symmetrical spacing

We shall consider a three phase transmission system of unit length with conductors of equal

radius r and spaced symmetrically in the form of an equilateral triangle as shown in Figure 1.5.Using (1.20) we can write the flux linkage as,

Figure 3.5: Conductors with symmetrical spacing

a = 2 107

ln1

rIa + ln

1

DIb + ln

1

DIc

(3.21)

For a balanced three phase system, Ia + Ib + Ic = 0 and Ia = (Ib + Ic). Substituting in(1.21),

a = 2

107Ia ln 1r

ln1

D

= 2 107Ia ln Dr

(3.22)

Because of symmetry a = b = c = and the inductance per phase is given by,

La = 2 107 ln Dr

= 0.2 lnD

rmH/km (3.23)


36/115


3.5.2 Conductors with unsymmetrical spacing

We shall consider a three phase transmission system of unit length with conductors of equal

radius r and spaced unsymmetrically as shown in Figure 1.6. Using (1.20) we can write the flux

Figure 3.6: Conductors with unsymmetrical spacing

linkage as,

a = 2 107

ln1

rIa + ln

1

D12Ib + ln

1

D31Ic

b = 2 107ln1

rIb + ln

1

D12Ia + ln

1

D23Ic

c = 2 107

ln1

rIc + ln

1

D31Ia + ln

1

D23Ib

(3.24)

Expressing the currents in terms of symmetrical components we get,

La =aIa

= 2 107

ln1

r+ ln

1

D12a2 + ln

1

D31a

Lb =b

Ib

= 2

107ln

1

r+ ln

1

D12

a + ln1

D23

a2Lc =

cIc

= 2 107

ln1

r+ ln

1

D31a2 + ln

1

D23a

(3.25)

From (1.25), it can be seen that the inductance of the three phases will be unequal. To make

the inductance of the phases equal the conductors are transposed at equal intervals as shown

in Figure 1.7. Because of the transposition each conductors take all the three positions over

the entire transmission length. For the transposed line the inductance per phase is given by the


37/115

3.6. INDUCTANCE OF STRANDED CONDUCTORS 37

Figure 3.7: Transposition of conductors with unsymmetrical spacing

average inductance,

L =La + Lb + Lc

3

=2 107

3

3 ln

1

r ln 1

D12 ln 1

D23 ln 1

D31

a + a2 = 11200 + 12400 = 1

= 2 107

ln1

r ln 1

(D12D23D31)1

3

= 2 107 ln (D12D23D31)1

3

r= 2 107 ln GMD

Ds(3.26)

where GMD = (D12D23D31)1

3 is the geometrical mean distance and Ds = r = re

1

4 is the

geometric mean radius.

3.6 Inductance of stranded conductors

We have derived expression for inductance of the solid conductors but in actual transmission

lines conductors are made up of strands. Stranded conductors offer better mechanical properties

and handling qualities. Let us consider a single phase system with stranded conductors p andq. Let the conductor p have n strands and the conductor qhave m strands and each strand havean equal diameter as shown in Figure 1.8. If the total current is I, then current per strand in p

conductor is

I

n and in qconductor isI

m . Using (1.20) we can write the flux linkage in strand 1 as,

1 = 2 107 In

ln

1

rp+ ln

1

D12+ ln

1

D13+ + ln 1

D1n

2 107 Im

ln

1

D11+ ln

1

D12+ + ln 1

D1m

= 2 107Ilnm

D11D12 D1mn

rpD12D13 D1n(3.27)


38/115


Figure 3.8: Stranded conductors

The inductance of strand 1 is given by,

L1 = 1In

= 2n 107 lnm

D11D12 D1mn

rpD12D13 D1n(3.28)

Similarly the inductance of all individual strands in the conductor p can be found. The averageinductance of a strand in the conductor p is given by,

Lave =L1 + L2 + + Ln

n(3.29)

The net inductance due all the strands in the conductor p is given by,

Lp = Laven = L1 + L2 + + Lnn2

= 2 107 lnmn

(D11D12 D1m) (Dn1Dn2 Dnm)n2

(D11D12D13 D1n) (Dn1Dn2Dn3 Dnn)= 2 107 ln GMD

GMR1(3.30)

where D11 = D22 = = rp. Using the similar procedure the inductance of the conductorqcan be obtained.

3.7 Bundled conductors

Conductors are bundled together as two or three or four depending on the requirement as shown

in Figure 1.9 inEHVtransmission. Bundling of conductors offers have many advantages. Mainly

it reduces the inductance thereby the reactance of the transmission line which will increase the

power transfer capacity of the line. Also, bundling reduces the surface voltage gradient which

will in turn reduce corona loss, radio interference etc.,. The distance between the conductors is

maintained by using spacer-dampers.


39/115

3.8. DOUBLE CIRCUITS 39

Figure 3.9: Bundled conductors

IfDs is the GMR of each conductor in a bundle and d is the distance between the conductorsthen GMR equivalent for the bundled conductor is given by,

for double conductor bundle,

Dsbundle =

Dsd (3.31)

for three

Dsbundle =3

Dsd2 (3.32)

for four

Dsbundle = 1.094

Dsd3 (3.33)

3.8 Double circuits

To increase the power transfer capacity double circuits are used. The phases are paralleled as

aa, b b and cc. To balance reactance and thereby voltage drop conductors are transposedwithin phases and also between phases at regular intervals. To find the inductance we use (1.30).

Identical phases are grouped together and GMD between such phase group is given by,

DAB =4

DabDabDabDab

DBC =4

DbcDbcDbcDbc

DAC =4

DacDacDacDac (3.34)

The equivalent GMD of a phase is given by,

GMD = 3

DABDBCDAC (3.35)

GMR of each phase group is given by,

DSA =

DsbundleDaa

DSB =

DsbundleDbb

DSC =

DsbundleDcc


40/115


Figure 3.10: Double circuit

where Dsbundle is the GMR of each bundled conductor given by (1.31) to (1.33). Now GMRper phase is given by,

GMRP =3

DSADSBDSC (3.36)

Inductance of the line per meter is given by,

L = 2

107 lnGMD

GMRP(3.37)


41/115

Chapter 4

Transmission line capacitance

Capacitance of a transmission line is due to the potential difference between two conductors

separated by a dielectric medium usually air. Because of the alternating nature of the voltage

applied, there will be charging and discharging currents through the line capacitance. We can

compute the capacitance of the transmission line using Gausss law for electric field which states

the total electric charge within a closed surface equals the total electric flux emerging from the

surface.

Let us consider a conductor of radius r carrying a charge ofqcoulombs. To find the electricfield intensity, we shall consider a cylindrical space of 1 m length at a distance x from the centerof the conductor as shown in Figure 1.1 The electric flux is assumed to be uniformly distributed

Figure 4.1: Potential difference between two points

on the surface of the cylinder and radially outward. The electric flux density is given by,

Df =q

surface area=

q

2x(4.1)

The electric field intensity is given by,

E =Df

permittivity of the medium=

q

20x(4.2)

41


42/115

42 CHAPTER 4. TRANSMISSION LINE CAPACITANCE

where 0 = 8.85 1012 F/m is the permittivity of free space. The potential difference betweenthe cylinders of radius D1 and D2 is given by the work done in moving a unit charge from P2 toP1 through the electric field produced by the conductor and is given by,

v12 =

D2D1

Edx =

D2D1

q

20xdx

=q

20ln

D2D1

(4.3)

4.1 Capacitance of two wire conductor

Consider a two wire system shown in Figure 1.2. Let the charge on conductor a be qa and on onconductor b be qb. Now let us consider the effect of charge qa alone and the interaction due to qb

Figure 4.2: Capacitance of two wire conductor

and ground are neglected. The voltage between a and b is given by,

vab(qa) =qa

20ln

D

ra(4.4)

Now consider with charge qb,

vba(qb) =qb

20ln

D

rb(4.5)

since vba(qb) = vab(qb) we get,

vab(qb) =qb

20ln

rbD

(4.6)

Using superposition the potential difference between a and b due to both charges is given by,

vab = vab(qa) + vab(qb) =qa

20ln

D

ra+

qb20

lnrbD

(4.7)

Ifqb = qa = qand ra = rb = r, then we get,

vab =q

0ln

D

r(4.8)


43/115

4.2. POTENTIAL DIFFERENCE IN A MULTI CONDUCTOR SYSTEM 43

The capacitance between a and b is given by,

Cab =q

vab=

0

lnDr

F/m (4.9)

If we assume the line voltage is twice the voltage between phase and neutral (if power is supplied

by a center tapped neutral earthed transformer). Then the capacitance to neutral or ground is

given by,

Cn = Can = Cbn = 2Cab =20

ln Dr

F/m (4.10)

4.2 Potential difference in a multi conductor system

Let us consider a system with n conductors with charges q1, , qn. If the system is balancedthe net charge is given by,

q1 + q2 + + qn = 0 (4.11)

Using (1.7) we can write the potential difference between conductors i and j as,

vij =1

20

nk=1

qk lnDkjDki

for k = i, Dii = ri (4.12)

4.3 Capacitance of three phase line

Consider a balanced 3 phase system with unsymmetrical spacing as shown in Figure 1.3. The

conductors are transposed at regular intervals. Using (1.12) the voltage difference between con-

Figure 4.3: Capacitance of three phase line

ductor ab in Section 1 of the transmission line can be found as,

vab(1) =1

20

qa ln

D12r

+ qb lnr

D12+ qc ln

D23D13

(4.13)


44/115


for Section 2

vab(2) =1

20qa ln D23

r+ qb ln

r

D23+ qc ln

D13

D12 (4.14)

for Section 3

vab(3) =1

20

qa ln

D13r

+ qb lnr

D13+ qc ln

D12D23

(4.15)

The average voltage is given by,

vab =1

3 20

qa ln

D12D23D13

r3

+ qb ln

r3

D12D23D13

+ qc ln

D12D23D13D12D23D13

= 120

qa ln

(D12D23D13)

1

3

r

+ qb ln

r

(D12D23D13)1

3

(4.16)

Similarly,

vac =1

20

qa ln

(D12D23D13)

1

3

r

+ qc ln

r

(D12D23D13)1

3

(4.17)

adding (1.16) and (1.17) and substituting qb + qc = qa since the net charge for the balancedsystem is zero, we get,

vab + vac =1

20

2qa ln

(D12D23D13)

13

r

qa ln

r

(D12D23D13)1

3

=3qa

20ln

(D12D23D13)

1

3

r

=

3qa20

lnGMD

r(4.18)

where GMD = (D12D23D13)1

3 is the geometrical mean distance. For a balanced 3 phase

system,

vab + vac = 3van (4.19)

Combining (1.18) and (1.19) gives the capacitance to neutral as,

Cn =qa

van=

20

ln GMDr

F/m (4.20)

The charging current is given by

Icharging = Cnvan A/m (4.21)


45/115

4.4. THE EFFECT OF EARTH ON LINE CAPACITANCE 45

4.4 The effect of earth on line capacitance

The presence of earth affects the electric field of the conductor. Let us consider a single conduc-

tor, when it is charged the earth acquires equal and opposite charge. The electric flux from theconductor to ground will be perpendicular to the earths equipotential surface. In order to model

this, the effect of earth can be replaced by a fictitious conductor of the same size and shape of the

over head conductor lying below the surface of the earth at a depth equal to the height of the con-

ductor above the earth. Now if the earth is removed, the electric flux between the fictitious and

actual conductor represents the effect of the earth with earths surface as equipotential surface.

The fictitious conductor is referred as image conductor.

Let us calculate the capacitance of a 3 phase unsymmetrical system as shown in Figure 1.4.

If the charges on the conductors be qa, qb and qc then the charge on the image conductors be

Figure 4.4: Effect of earth on line capacitance

qa, qb and qc. As explained in Section 1.3, the voltage equation vab can be written for threesections of the transposed system and averaged over three sections. Now the equation for the

voltage vab for section 1 can be written using (1.12) as,

vab =1

20

qa

ln

D12r

ln H12H1

+ qb

ln

r

D12 ln H2

H12

+ qc

ln

D23D31

ln H23H31

(4.22)


46/115


Similarly after getting the equations for other two sections of the line, average vab can be foundout. van can be obtained as explained in Section 1.3 through vab and vac. Now the capacitance toneutral is given by,

Cn =20

ln GMDr

ln

3H12H23H313H1H2H3

F/m (4.23)With reference to (1.20) and (1.23), we can see that due to the effect of earth the capacitance

increases. Generally the conductors are situated at a height well above the ground, so the effect

of earth can be neglected. But in the case of unbalanced system calculations the earth effect will

be important.

4.5 Capacitance of bundled conductors

As in the case of inductance calculations for bundled conductors described in Section 1.7, we

can find the capacitance as,

Cn =20

ln GMDrbundle

F/m (4.24)

The equivalent radius rbundle for different configurations as shown in Figure 1.9 are given by, fordouble conductor bundle,

rbundle =

rd (4.25)

for three

rbundle =3

rd2 (4.26)

for four

rbundle = 1.094

rd3 (4.27)

4.6 Capacitance of double circuit

Let us consider a double circuit line with transposed section as shown in Figure 1.10. We can

work out the capacitance similar to the inductance calculations in Section 1.8 as,

Cn =20

ln GMDGMRbundle

F/m (4.28)

where GMRbundle =3

RaRbRc and Ra =

rbundleDaa , Rb =

rbundleDbb and Rc =

rbundleDcc.


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Chapter 5

Transmission line model

The transmission line parameters evaluated for unit length of the line is used to arrive at the

transmission line model between sending and receiving ends. The line models are worked out

based on the length of the transmission line.

5.1 Short transmission line model

This model is used to study lines with length less than 80 km. Due to short length, the value of

capacitance is usually small and can be neglected without sacrificing the accuracy. We include

only line resistance and inductance in the model as shown in Figure 1.1 If l is the length of the

Figure 5.1: Short transmission line model

line, rl the resistance per unit length and Ll inductance per unit length. Then the net impedanceis given by,

Zl = Rl +Xl = l (rl +Ll) (5.1)

If Vs and Vr are the per phase voltages at sending and receiving ends respectively, the sendingend voltage is given by,

Vs = Vr + ZlIr and Is = Ir (5.2)

47


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48 CHAPTER 5. TRANSMISSION LINE MODEL

writing (1.2) in matrix form,

VsIs

= 1 Zl0 1

VrIr

= A BC D

VrIr

(5.3)where A, B, Cand D are called transmission line ABCD constants. This can be diagrammat-ically represented as two port network as shown in Figure 1.2 The percentage regulation of the

Figure 5.2: Two port network model of a transmission line

transmission line is given by,

% voltage regulation =|Vrnoload| |Vrfullload|

|Vrfullload| 100% (5.4)

and the transmission line efficiency is given by,

% =receiving end power

sending end power 100% = Pr

Ps 100% (5.5)

5.2 Medium length line model or nominal model

This model is suitable for representing transmission line of length from 80 km to 250 km. The

total capacitance of the line is divided into two equal parts and each part is lumped as single ca-

pacitance and added at sending and receiving ends as shown in Figure 1.3. If Cis the capacitanceof the line per unit length and l the line length, then the shunt admittance is given by,

Y = (g +C) l (5.6)

where g is the conductance which is usually neglected. At the receiving end,

Il = Ir +Y

2Vr (5.7)

At the sending end,

Vs = Vr + ZlIl (5.8)


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5.3. LONG TRANSMISSION LINE MODEL 49

Figure 5.3: Nominal model of a transmission line

substituting (1.7) in (1.8),

Vs = Vr + Zl

Ir + Y2 Vr

=

1 +

ZlY

2

Vr + ZlIr (5.9)

and

Is = Il +Y

2Vs (5.10)

substituting from (1.8) and (1.9) in (1.10),

Is = Ir + Y2

Vr + Y2

1 + ZlY

2

Vr + ZlIr

= Y

1 +

ZlY

4

Vr +

1 +

ZlY

2

Ir (5.11)

Expressing (1.9) and (1.11) in matrix form,VsIs

=

1 + ZlY

2Zl

Y

1 + ZlY4

1 + ZlY2

VrIr

=

A BC D

VrIr

(5.12)

We can represent (1.12) in terms of sending end quantities as,VrIr

=

D B

C A

VsIs

(5.13)

5.3 Long transmission line model

For transmission line lengths exceeding 250 km, the lumped parameter models will not give

accurate results. Line models should be worked out with distributed parameters.


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Figure 5.4: Model of a long transmission line

Let us consider a small section A B of length x of a long transmission line of length l at a

distance x from the receiving end as shown in Figure 1.4. The series impedance per unit lengthbe z = r +L and the shunt admittance be y = g +C. Voltage at A is given by,

V(x + x) = Vx + zxI(x)

V(x + x) V(x)x

= zI(x) (5.14)

When the limit x 0,dV(x)

dx= zI(x) (5.15)

At junction A,

I(x + x) = Ix + yxV(x + x)

I(x + x) I(x)x

= yV(x) (5.16)

Now taking the limits,

dI(x)

dx= yV(x) (5.17)

Differentiating (1.15) with respect to x,

d2V(x)

dx2= z

dI(x)

dx(5.18)

Substituting from (1.17) we have,

d2V(x)

dx2= zyV(x)

d2V(x)

dx2 2V(x) = 0 (5.19)


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5.3. LONG TRANSMISSION LINE MODEL 51

where 2 = zy. The solution for the differential equation (1.19) is of the form,

V(x) = A1ex + A2e

x (5.20)

where is known as the propagation constantand = + , where the real part is knownas attenuation constantand the imaginary part is known as phase constant. Now from (1.15)

I(x) =1

z

dV(x)

dx=

z

A1e

x A2ex

=

y

z

A1e

x A2ex

=1

Zc

A1e

x A2ex

(5.21)

where Zc = zy

is known as characteristic impedance. When x = 0, V(x) = Vr and

I(x) = Ir, substituting in 1.20 and 1.21,

Vr = A1 + A2

ZcIr = A1 A2A1 =

Vr + IrZc2

A2 =Vr IrZc

2(5.22)

Substituting in (1.20),

V(x) =

Vr + IrZc2

ex +

Vr IrZc

2

ex

=

ex + ex

2

Vr +

ex ex

2

ZcIr

= cosh xVr + Zc sinh xIr (5.23)

Similarly,

I(x) =1

Zc

Vr + IrZc

2

ex 1

Zc

Vr IrZc

2

ex

=

1

Zcex e

x

2

Vr +ex + ex

2

Ir

=1

Zcsinh xVr + cosh xIr (5.24)

To get Vs and Is, substitute x = l in (1.23) and (1.24),

Vs = cosh lVr + Zc sinh lIr

Is =1

Zcsinh lVr + cosh lIr (5.25)


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in matrix form

VsIs

= cosh l Zc sinh l1Zc sinh l cosh l

VrIr

= A BC D

VrIr

(5.26)From the values ofABCD constants given by (1.26) we can arrive the more accurate equivalent-model of the system. We can express (1.26) in the form of -model equation given by (1.12)as,

VsIs

=

1 +

ZlY

2Zl

Y

1 +ZlY

4

1 +

ZlY

2

VrIr

(5.27)

By equating the coefficients ofA and B in (1.26) and (1.28) we get,

Zl = Zc sinh l

1 +ZlY

2= cosh l

Y

2=

1

Zc

(cosh l 1)sinh l

=1

Zctanh

l

2(5.28)

The equivalent -model for long transmission line can be represented diagrammatically as shownin Figure 1.5.

Figure 5.5: Equivalent -model of a long transmission line

5.4 Voltage and current waves

The instantaneous value of the voltage in equation (1.20) can be expressed in time domain as,

v(t, x) =

2A1ex cos(t + x) +

2A2e

x cos(t x)= v1(t, x) + v2(t, x) (5.29)

The voltage at a point along the transmission line is the sum of two waves v1 and v2. As thedistance x increases v1 increases and it is called incident wave. v2 decreases as x increases andis called reflected wave. They behave like traveling waves along the line.


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5.4. VOLTAGE AND CURRENT WAVES 53

Let us consider the reflected wave, for the amplitude to be maximum,

t

x = 2n and x =t

2n

(5.30)

The speed or velocity of propagation of the wave is given by,

dx

dt= v =

=

2f

(5.31)

For wavelength the distance x should produce a phase change of2, using (1.30)

=2

(5.32)

For lossless line r = 0 and g = 0, using = zy we get, =

LC (5.33)

and the characteristic impedance becomes,

Zc =

L

C(5.34)

which is called surge impedance of the line. Combining (1.31) and (1.33) we get,

v =1

LCand =

1

fLC(5.35)

If we neglect internal flux linkage, and using (1.37) for L and (1.28) for C in (1.35) we get,

v =1

00and =

1

f

00

v 3 108m/s and 6000 km for 50 Hz. (5.36)

For lossless line = and cosh x = coshx = cos x and sinh x = sinhx = sin x.Using these substitutions in (1.25), for a line under noload with Ir = 0, we get

Vrnoload = Vscos l (5.37)

under noload the current is capacitive line charging current and Vrnoload Vs. The value of Vrincreases with the total length of the line. If the receiving end is shorted, Vr = 0, substitutingin (1.25) we get,

Vs = Zc sin lIr

Is = cos lIr (5.38)


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56/115

56 CHAPTER 6. SYMMETRICAL FAULTS

6.1 Numerical example -1

Consider a simple power system shown in Figure 1.1. Generators are represented using transient

reactance and other reactance are expressed in pu values. Shunt capacitances and resistances areneglected. A three phase fault is initiated with fault impedance Zfault = 0.15 pu at a) bus 1,b)bus 2 and c)bus 3. Find the fault current, bus voltages and line currents.

Figure 6.1: Network for numerical example

Solution: a) Fault at bus 1

Let consider the pre-fault bus voltages as, V10 = V20 = V30 = 1 pu. We shall insert thefault impedance at bus 1 and inject the Thevenins voltage Vth = V10 at bus 1 and shorting thegenerator sources as shown in Figure 1.2.

Figure 6.2: Circuit for Section a.

Reducing the parallel paths between 1-2 and 1-3-2,

Zequ1 =Z12 (Z31 + Z23)

Z12 + (Z31 + Z23)= 0.3733

Now the network reduces as shown in Figure 1.3. On further simplification of the network,

Z11 =(Zequ1 + ZG2 + ZT2) (ZG1 + ZT1)

(Zequ1 + ZG2 + ZT2) + (ZG1 + ZT1)= 0.1808


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The change in bus voltages is given by,

V1 =

0.2356 pu; V2 =

0.6020 pu; V3 =

0.4188 pu

The fault bus voltages are given by,

V1 = 0.7646 pu; V2 = 0.3980 pu; V3 = 0.5812 pu

The currents in different sections during fault are given by,

I12 = 0.5234 pu; I23 = 0.4580 pu; I13 = 0.4580 pu

Solution: c) Fault at bus 3

We shall insert the fault impedance at bus 3 and inject the Thevenins voltage Vth = V30 at bus3 and shorting the generator sources as shown in Figure 1.4. To work out the impedance viewed

Figure 6.4: Circuit for Section c.

from fault, we shall convert the - connection between buses 1,2 and 3 to equivalent Y asshown in Figure 1.5. Now solving the resultant series parallel combination we get the impedance

Z33 = 0.3463. Now the fault current is given by,

Ifault =V30

Z33 + Zfault = 2.0149 pu

By current division IG1 = 1.1317 pu and IG2 = 0.8832 pu. The change in bus voltagesis given by,

V1 = 0 IG1 (ZG1 + ZT1) = 0.2716 puV2 = 0 IG2 (ZG2 + ZT2) = 0.3180 puV3 = Ifault Zfault V30 = 0.6978 pu


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6.2. DETERMINATION OF SHORT CIRCUIT CAPACITY (SCC) 59

Figure 6.5: Circuit for Section c.

The fault bus voltages are given by,

V1 = V10 + V1 = 0.7284 pu

V2 = V20 + V2 = 0.6820 pu

V3 = V30 + V3 = 0.3022 pu

The currents in different sections during fault are given by,

I12 =V1 V2

Z12= 0.0662 pu

I23 =V2 V3

Z23= 0.9495 pu

I13 =V1 V3

Z31= 1.0654 pu

6.2 Determination of short circuit capacity (SCC)

Determination of fault current at a bus will help us to find the interrupting capacity of the circuitbreaker and bus bar current carrying capacity required.

The fault current in pu is given by

Ifaultpu =Vfb0Xfb

(6.1)

where Vfb0 is the rated pu voltage at fault bus and Xfb is the pu reactance to the point of fault.The resistances are neglected hence the computed fault current gives the worst scenario. The


60/115


base current is given by,

Ibase

=SbaseMVA

103

3VbasekV(6.2)

Now the fault current is given by,

Ifault = Ifaultpu Ibase

=

Vfb0Xfb

SbaseMVA 103

3VbasekV

(6.3)

Now SCC is given by,

SC C =

3VLfbIfault

103 MVA (6.4)

where VLfb is the line voltage at fault bus in kV. Substituting from (1.3),

SC C =Vfb0SbaseMVA

Xfb

VLfbVbasekV

MVA (6.5)

IfVLfb = VbasekV then,

SC C =Vfb0SbaseMVA

XfbMVA (6.6)

and ifVfb0 = 1

SC C =SbaseMVA

XfbMVA (6.7)

6.3 Fault analysis using Zbus matrix

The fault analysis procedure given in the previous numerical example is suitable only for small

networks. It will become more complex if we have to deal with large networks. For large

networks fault analysis is made simple and straight forward using the Zbus matrix method.To analyse the fault we need the impedance matrix of the system. We represent all the

voltages and impedances in pu values. The generator voltages are given by the voltage behindthe reactance Xd or X

d or Xd. The transmission line is represented by equivalent -model. The

load is modelled as equivalent constant impedance as given by ZLi =|Vi0|2SL

, where Vi0 is the

pre-fault voltage at load bus and SL is the complex load value.Let us consider a n bus system and a three phase fault is initiated at the kth bus. If the pre-fault

voltages of all buses are arranged as a vector given by,

Vbus0 = [V10, V20, , Vk0, , Vn0]T (6.8)


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62/115


From Thevenins circuit we have,

Vkfault = ZfaultIkfault (6.17)

Equating (1.16) and (1.20) we have,

Ikfault =Vk0

Zkk + Zfault(6.18)

To get the fault current we need only Zkk and Zfault. Now the fault voltage at the ith bus is given

by,

Vifault = Vi0 ZikIkfault= Vi0 Zik

Zkk + ZfaultVk0 (6.19)

With all fault bus voltages known we can find the current between bus i and j as,

Iijfault =Vifault Vjfault

zij(6.20)

where zij is the impedance between buses i and j.

6.4 Numerical example - 2

Consider the power system given in numerical example-1. We shall try to solve for the fault

current, bus voltages and currents using Zbus method for the fault at bus 3 case.The admittance matrix for the three bus system shown in Figure 1.6 can be written as,

Figure 6.6: Network for numerical example -2

Ybus =

1ZG1+ZT1

+ 1Z12

+ 1Z13

1

Z12 1

Z13

1Z21

1

ZG2+ZT2+ 1

Z21+ 1

Z23

1

Z23

1Z31

1Z32

1

Z31+ 1

Z32

=

0 8.0952 0 +1.4286 0 +2.50000 +1.4286 0 6.7063 0 +2.5000

0 +2.5000 0 +2.5000 0 5.0000


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65/115

Chapter 7

Symmetrical components

Method of symmetrical components helps to reduce the complexity involved in analysing un-

balanced polyphase system. In a balanced, unbalanced fault conditions can lead to unbalanced

voltages and currents.

Using Fortescues theorem, an unbalanced system of n phases can be resolved into n sys-tems of balanced phasors of equal length and these phasors are called symmetrical components

of original phasors. The analysis of unbalanced system using symmetrical components involves,

finding the responses of individual elements of the symmetrical components and applying super-

position principle to get the overall response.

7.1 Basics of symmetrical components

Using symmetrical components three unbalanced phasors of 3 phase system can be resolved into

3 balanced system of phasors.

i) Positive sequence: Displaced by 1200, equal in magnitude and has the same phase se-quence as the unbalanced system.

ii) Negative sequence: Displaced by 1200, equal in magnitude and has opposite phase se-quence as the unbalanced system.

iii) Zero sequence: Zero phase displacement and equal in magnitude.

The symmetrical components are diagrammatically represented in Figure 2.1. The superscripts

1, 2, 0 are used to represent positive, negative and zero sequences respectively. Let us define avector operator a = 11200 which performs 1200 counterclockwise rotation, so that,

a = 11200 = 0.5 +0.866a2 = 12400 = 0.5 0.866

so that 1 + a + a2 = 0 (7.1)

65


66/115

66 CHAPTER 7. SYMMETRICAL COMPONENTS

Figure 7.1: Symmetrical components

Using the above we can define positive sequence as,

V1a = V1a 0

0 = V1aV1b = V

1a 240

0 = a2V1aV1c = V

1a 120

0 = aV1a (7.2)

Similarly we can define the negative sequence as,

V2a = V2a 00 = V2aV2b = V

2a 120

0 = aV2aV2c = V

2a 240

0 = a2V2a (7.3)

The zero sequence is defined as,

V0a = V0b = V

0c (7.4)

We can get back the original unbalanced system from the symmetrical components as follows:

Va = V0

a+ V1

a+ V2

aVb = V

0b + V

1b + V

2b

Vc = V0c + V

1c + V

2c (7.5)

Using the set of equations (2.2) to (2.4) we can write (2.5) as, VaVb

Vc

=

1 1 11 a2 a

1 a a2

V0aV1a

V2a

(7.6)


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7.1. BASICS OF SYMMETRICAL COMPONENTS 67

In short form as,

Vabc = A V012 (7.7)

where A is the symmetrical component transformation matrix. From (2.7) we can write

V012 = A1 Vabc (7.8)

where

A1 =

1

3

1 1 11 a a2

1 a2 a

(7.9)

Using (2.8) and (2.9) we can write,

V0a

V1a

V2a

=

1

3 1 1 1

1 a a2

1 a2 a Va

VbVc

(7.10)

ie.,

V0a =1

3(Va + Vb + Vc)

V1a =1

3

Va + aVb + a

2Vc

V2a =1

3

Va + a

2Vb + aVc

(7.11)

Similarly we can write for currents as,

I0a =1

3(Ia + Ib + Ic)

I1a =1

3

Ia + aIb + a

2Ic

I2a =1

3

Ia + a

2Ib + aIc

(7.12)

From (2.12) for ungrounded three phase system the sum of the phase currents is zero hence no

zero sequence current. But for grounded system zero sequence current flows between neutral

and ground.

Apparent 3 phase power can be expressed as symmetrical components as,

S3 = VabcTIabc

=AV012

T AI012

= V012TATAI012 (7.13)

Since AT = A and ATA = 3 substituting in (2.13),

S3 = 3

V012TI012

= 3V0a I0a + 3V

1a I

1a + 3V

2a I

2a (7.14)


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7.3. SEQUENCE IMPEDANCE OF A TRANSMISSION LINE 69

From (2.19) we can write,

V0a = (Zs + 3Zng) I0a = Z

0aI

0a

V1a = ZsI1a = Z1aI1aV2a = ZsI

2a = Z

2aI

2a (7.21)

where Z0a is the zero sequence impedance, Z1a is the positive sequence impedance and Z

2a is the

negative sequence impedance. This can be diagrammatically shown in Figure 2.3

Figure 7.3: Sequence impedance

7.3 Sequence impedance of a transmission line

Consider a section of a transmission line shown in Figure 2.4. If we assume the line as perfectly

symmetrical, the self inductance of all phase are equal and is denoted as Zs and the mutualinductance between phases as Zm and the mutual inductance between phase and neutral as Zan.We can write Kirchhoffs voltage loop equation for the section of the line as follows: Van VanVbn Vbn

Vcn Vcn

=

Zs Zan Zm Zan Zm ZanZm Zan Zs Zan Zm Zan

Zm Zan Zm Zan Zs Zan

IaIb

Ic

+

Zan ZnZan Zn

Zan Zn

In

(7.22)

But In = (Ia + Ib + Ic) substituting in (2.22) we get,Van VanVbn Vbn

Vcn Vcn

=

Zs + Zn 2Zan Zm + Zn 2Zan Zm + Zn 2ZanZm + Zn 2Zan Zs + Zn 2Zan Zm + Zn 2Zan

Zm + Zn 2Zan Zm + Zn 2Zan Zs + Zn 2Zan

IaIb

Ic

VaaVbb

Vcc

=

Van VanVbn Vbn

Vcn Vcn

=

Zss Zmm ZmmZmm Zss Zmm

Zmm Zmm Zss

IaIb

Ic

(7.23)


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7.4. SEQUENCE IMPEDANCE OF SYNCHRONOUS GENERATOR 71

7.4 Sequence impedance of Synchronous generator

Generators positive sequence reactance is given by Xd , Xd or Xd. But the negative sequence

reactance does not vary during transients and is approximately equal to Xd . The zero sequencereactance is approximately equal to the leakage reactance of the machine.

7.5 Sequence network of a loaded Synchronous generator

Consider a synchronous machine with generated voltage Ea per phase on a balanced load condi-tions as shown in Figure 2.5. Applying KVL we can get the phase voltage to neutral as,

Figure 7.5: Loaded Synchronous generator

VanVbn

Vcn

=

Ea ZphIaEb ZphIb

Ec ZphIc

In

ZnZn

Zn

(7.27)

The neutral current is given by In = Ia + Ib + Ic, substituting in (2.27) we get,

VanVbnVcn

=

EaEbEc

Zph + Zn Zn Zn

Zn Zph + Zn ZnZn Zn Zph + Zn

IaIbIc

(7.28)

In short

Vabcn = Eabc ZabcIabcTo get the symmetrical components,

AV012abcn = AE012abc ZabcAI012abc (7.29)


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72 CHAPTER 7. SYMMETRICAL COMPONENTS

multiplying (2.29) by A1,

V012abcn = E012abc

A

1ZabcAI012abc (7.30)

where

A1ZabcA = A1

Zph

1 0 00 1 0

0 0 1

+ Zn

1 1 11 1 1

1 1 1

A

=

Zph + 3Zn 0 00 Zph 0

0 0 Zph

=

Z0 0 00 Z1 0

0 0 Z2

(7.31)

Since the generated emf is balanced there will be only positive sequence voltage, therefore

E012a =

0Ea

0

(7.32)

Substituting in (2.30), V0anV1an

V2an

=

0 Z0I0aEa Z1I1a

0 Z2I2a

(7.33)


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Chapter 8

Unsymmetrical Faults

8.1 Single line to ground faultConsider a unloaded 3 phase synchronous generator shown in Figure 3.1. The neutral of the

machine is grounded through a impedance Zng. Consider a phase to ground fault between phasea and ground through fault path impedance Zfault. During fault we can write,

Figure 8.1: Single line to ground fault

Van = ZfaultIa and Ib = Ic = 0 (8.1)

For these conditions the symmetrical components are given by, I0aI1a

I2a

= 1

3

1 1 11 a a2

1 a2 a

Ia0

0

(8.2)

I0a = I1a = I

2a =

1

3Ia (8.3)

73


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74 CHAPTER 8. UNSYMMETRICAL FAULTS

Similarly we can express the phase voltage Van in terms of symmetrical components and substi-tuting from (2.30) along with (3.2) we get

Van = V0a + V1a + V2a3ZfaultI

0a = Ea

Z0 + Z1 + Z2

I0a

I0a =Ea

Z0 + Z1 + Z2 + 3Zfault(8.4)

and the fault current is given by

3I0a = Ia =3Ea

Z0 + Z1 + Z2 + 3Zfault(8.5)

Equations (3.2) and (3.5) can be diagrammatically represented as in Figure 3.2

Figure 8.2: Single line to ground fault - sequence network

8.2 Line to line fault

Let us consider a line to line fault between phase b and c on a generator operating under no loadconditions as shown in Figure 3.3. During fault we can write

Vbn Vcn = ZfaultIb, Ib + Ic = 0 and Ia = 0 (8.6)

We can write the symmetrical components of current including (3.6) as,

I0aI1a

I2a

= 1

3

1 1 11 a a2

1 a2 a

0Ib

Ib

(8.7)

(8.8)


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8.2. LINE TO LINE FAULT 75

Figure 8.3: Line to line fault

From (3.7) we get,

I0a = 0

I1a =1

3

a a2 Ib

I2a =1

3

a2 a Ib (8.9)

From (3.9) we can write

I1a = I

2a (8.10)

Using symmetrical components we have

Vbn Vcn =

a2 a V1an V2an=

a2 a Ea Z1I1a + Z2I2a

Using (3.6) and (3.10) we get,

ZfaultIb =

a2 a Ea Z1 + Z2 I1aZfault

3I1a(a

a2)

= a2 a Ea Z

1 + Z2 I1a

I1a =Ea

Z1 + Z2 + Zfault

a a2 a2 a = 3 (8.11)

The phase currents are given by, IaIb

Ic

=

1 1 11 a2 a

1 a a2

0I1a

I1a

(8.12)

(8.13)


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Fault current is given by,

Ib = Ic = a2 a I1aEquation (3.11) can be diagrammatically represented as in Figure 3.4

Figure 8.4: Line to line fault - sequence network

8.3 Double line to ground fault

For fault condition shown in Figure 3.5, during fault condition we can write,

Vbn = Vcn = Zfault (Ib + Ic) and Ia = I0a + I

1a + I

2a = 0 (8.14)

expressing the voltages in symmetrical components,

Vbn =

V0an + a2V1an + aV

2an

(8.15)

Vcn =

V0an + aV1an + a

2V2an

(8.16)

Since Vbn = Vcn, from (3.15) and (3.16) we get V1a = V

2a . Now we have from (3.14),

Vbn = Zfault (Ib + Ic)

V0an

+ a2V1an

+ aV2an

= Zfault I0a

+ a2I1a

+ aI2a

+ I0a

+ aI1a

+ a2I2a

V0an +

a2 + a

V1an = Zfault

2I0a I1a I2a

= 3ZfaultI0a

V0an V1an = 3ZfaultI0a (8.17)Substituting from (2.30),

Z0I0a Ea + Z1I1a = 3ZfaultI0aI0a =

Ea + Z1I1aZ0 + 3Zfault

(8.18)


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8.4. NUMERICAL EXAMPLE -1 77

Figure 8.5: Double line to ground fault

From (2.30) and using V1an = V2an,

Ea Z1I1a = Z2I2aI2a =

Ea Z1I1aZ2

(8.19)

Substituting (3.18) and (3.19) in (3.14) we get,

Ea + Z1I1aZ0 + 3Zfault

+ I1a Ea Z1I1a

Z2= 0

I1a =Ea

Z1 +Z2(Z0+3Zfault)Z2+Z0+3Zfault

(8.20)

We can find I2a and I0a by substituting for I

1a in (3.18) and (3.19). Using the symmetrical compo-

nents the value of fault current is found to be,

Ifault = Ib + Ic = 3I0a (8.21)

The sequence impedance for the double line to ground fault can be expressed as in Figure 3.6.

8.4 Numerical example -1

Consider the 3 bus power system shown in Figure 3.7 with two generators. The generator

and transformer configurations are indicated in the diagram. The neutrals of the generator are

grounded through 0.3 pu reactors. The generators are not loaded and running at rated frequencyand voltage with their emfs in phase. The system sequence impedance data in 100 MVA base is


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Figure 8.6: Line to line fault - sequence network

Figure 8.7: 3 bus 2 machine network

given in Table 3.1. If the following different types of fault appears in bus 3 with Zfault = 0.12pu, find the fault current Ifault.Case 1: Balanced 3 phase fault

Case 2: Single line to ground fault

Case 3: Line to line fault

Case 4: Double line to ground fault

We shall derive the sequence impedances of the equivalent Thevenins circuit. To get the

positive sequence impedance, we convert the connection between bus 123 to Y connec-tion as in Figure 3.8. Now solving the resultant series parallel combination we get the pos-

itive sequence impedance Z133 = 0.2546. The negative sequence impedance is given byZ233 = Z

133 = 0.2546. We have to create the zero sequence diagram considering the trans-


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8.4. NUMERICAL EXAMPLE -1 79

Equipment Voltage (kV) X1 X2 X0

G1 25 0.14 0.14 0.04G2 25 0.16 0.16 0.06

T1 25/220 0.15 0.15 0.15T2 25/220 0.15 0.15 0.15

Line 1-2 220 0.14 0.14 0.3

Line 2-3 220 0.22 0.22 0.65

Line 1-3 220 0.2 0.2 0.35

Table 8.1: Power system parameters.

Figure 8.8: Positive sequence circuit reduction

former configurations into account as shown in Figure 3.9 As in the positive sequence case we

shall convert the connection between bus 123 to Y connection and then solve the parallelseries combination of the resulting impedances to get the Thevenins zero sequence equivalent

Z033 = 0.3716.

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