master of science in eee project by greg clawson and mira lopez spring 2012.pdf
TRANSCRIPT
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DESIGN AND FAULT ANALYSIS OF
A 345KV 220 MILE OVERHEAD TRANSMISSION LINE
A Project
Presented to the faculty of the Department of Electrical and Electronic Engineering
California State University, Sacramento
Submitted in partial satisfaction of
the requirements for the degree of
MASTER OF SCIENCEin
Electrical and Electronic Engineering
and
MASTER OF SCIENCE
inElectrical and Electronic Engineering
by
Greg Clawson
Mira Lopez
SPRING2012
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2012
Greg Clawson
Mira Lopez
ALL RIGHTS RESERVED
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DESIGN AND FAULT ANALYSIS OF
A 345KV 220 MILE OVERHEAD TRANSMISSION LINE
A Project
by
Greg Clawson
Mira Lopez
Approved by:
_____________________________________, Committee Chair
Turan Gnen, Ph.D.
_____________________________________, Second Reader
Salah Yousif, Ph.D.
____________________
Date
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Student: Greg Clawson
Mira Lopez
I certify that these students have met the requirements for the format contained in the
University format manual and that this project is suitable for shelving in the Library and
that credit is to be awarded for the project.
_____________________________, Graduate Coordinator _________________B. Preetham. Kumar, Ph.D. Date
Department of Electrical and Electronic Engineering
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Abstract
of
DESIGN AND FAULT ANALYSIS OF
A 345KV 220 MILE OVERHEAD TRANSMISSION LINE
by
Greg Clawson
Mira Lopez
Efficient and reliable transmission of bulk power economically benefits both the power
company and consumer. This report gives clarification to concept and procedure in
design of an overhead 345 kV long transmission line. The project will find an optimum
design alternative which meets certain criteria including transmission efficiency, voltage
regulation, power loss, line sag and tension. A MATLAB script will be developed to
assess which alternative solutions can fulfill the criteria.
Integration of protective devices is a fundamental part of achieving power system
reliability. To determine the sizing and setting of protective devices, analysis of potential
fault conditions provide the necessary current and voltage data. A fault analysis for the
final transmission line design will be simulated two ways: 1) by using a MATLAB script
that was developed for this project and 2) by using an available Aspen One Liner
program.
_____________________________________, Committee ChairTuran Gnen, Ph.D.
_____________________Date
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DEDICATION
I dedicate my work to my sister, Mandica Konjevod for inspiring me.
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TABLE OF CONTENTS
Page
Dedication...vi
List of Tables .xii
List of Figures...xiii
Chapter
1.INTRODUCTION.......12.LITERATURE SURVEY ....3
2.1.Introduction ....32.2.Support Structure ....32.3.Line Spacing and Transposition.5
2.3.1.Symmetrical Spacing.....62.3.2.Asymmetrical Spacing.......72.3.3.Transposed Line. ..10
2.4.Line Constants..112.5.Conductor Type and Size..122.6.Extra-High Voltage Limiting Factors...16
2.6.1.Corona ..162.6.2.Line Design Based on Corona.....192.6.3.Advantages of Corona.192.6.4.Disadvantages of Corona .....20
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2.6.5.Prevention of Corona...202.6.6.Radio Noise.. 202.6.7.Audible Noise ..21
2.7.Line Modeling...212.8.Line Loadability242.9.Fault Events..252.10.Fault Analysis.262.11.Single Line-to-Ground (SLG) Fault...272.12.Line-to-Line (L-L) Fault.....272.13.Double Line-to-Ground (DLG) Fault.282.14.Three-Phase Fault...292.15.The Per-Unit System...30
3.MATHEMATICAL MODEL....313.1.Introduction...313.2.Geometric Mean Distance (GMD)...313.3.Geometric Mean Radius (GMR)..333.4.Inductance and Inductive Reactance.....343.5.Capacitance and Capacitive Reactance.353.6.Long Transmission Line Model....353.7.Sending-End Voltage and Current....403.8.Power Loss....42
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3.9.Transmission Line Efficiency...443.10.Percent Voltage Regulation....443.11.Surge Impedance Loading (SIL)453.12.Sag and Tension .46
3.12.1.Catenary Method.... 463.12.2.Parabolic Method...50
3.13.Corona Power Loss.513.13.1.Critical Corona Disruptive Voltage...513.13.2.Visual Corona Disruptive Voltage.533.13.3.Corona Power Loss at AC Voltage54
3.14.Method of Symmetrical Components.....553.14.1.Sequence Impedance of Transposed Lines59
3.15.Fault Analysis.....613.16.Per Unit...623.17.Single Line-to-Ground (SLG) Fault...633.18.Line-to-Line (L-L) Fault....663.19.Double Line-to-Ground (DLG) Fault.....693.20.Three-Phase Fault...72
4.APPLICATION OF MATHEMATICAL MODEL....764.1.Introduction..764.2.Design Criteria......77
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4.3.Geometric Mean Distance (GMD).......774.4.Geometric Mean Radius (GMR)......784.5.Inductance and Inductive Reactance....784.6.Capacitance and Capacitive Reactance........794.7.Long Line Characteristics.........804.8.ABCD Constants......814.9.Sending-End Voltage and Current....834.10.Power Loss......844.11.Percent Voltage Regulation.. ..............864.12.Transmission Line Efficiency.....864.13.Surge Impedance Loading (SIL)864.14.Sag and Tension ......87
4.14.1.Catenary Method874.14.2.Parabolic Method...89
4.15.Corona Power Loss ....894.15.1.Critical Corona Disruptive Voltage.......894.15.2.Visual Corona Disruptive Voltage.....914.15.3.Corona Power Loss at AC Voltage....924.15.4.Corona Power Loss forFoul Weather Conditions.94
4.16.Per Unit.......974.17.Fault Analysis Outline....98
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4.18.Procedure Using Symmetrical Components.......994.19.Fault Analysis at the End of Transmission Line.......100
4.19.1.Single Line-to-Ground (SLG) Fault........1014.19.2.Line-to-Line (L-L) Fault..1044.19.3.Double Line-to-Ground (DLG) Fault..1094.19.4.Three Line-to-Ground (3LG) Fault......113
5.CONCLUSIONS.....117Appendix A. Conductor and TowerCharacteristics.......119
Appendix B. Aspen Simulation Model and Analysis.....120
Appendix C. Aspen Fault Analysis Summary....123
Appendix D. MATLABAspen Fault Analysis Results.........131
Appendix E. MATLAB Code.........137
Bibliography182
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LIST OF TABLES
Tables Page
1. Table 2.1 Typical conductor separation.52. Table 2.2 Aluminum vs. copper conductor type..133. Table 3.1 Corona Factor..554. Table 3.2 Power and functions of operatora...565. Table 4.1 Design parameters776. Table 4.2 System data for power system model......997. Table 4.3 Fault analysis of SLG fault at receiving end of line..1048. Table 4.4 Fault analysis of L-L fault at receiving end of line...1089. Table 4.5 Fault analysis of DLG fault at receiving end of line11210.Table 4.6 Fault analysis of 3LG fault at receiving end of line..116
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LIST OF FIGURES
Figures Page
1. Figure 2.1 Three-phase line with symmetrical spacing.....62. Figure 2.2 Cross section of three-phase line with horizontal tower configuration....63. Figure 2.3Three-phase line with asymmetrical spacing84. Figure 2.4A transposed three-phase line.105. Figure 2.5 Equivalent circuit of short transmission line..226. Figure 2.6 Nominal-T circuit of medium transmission line....227. Figure 2.7 Nominal- circuit of medium transmission line.238. Figure 2.8 Segment of 1-phase and neutral connection for long transmission line.239. Figure 2.9 Practical Loadability for Line Length2510.Figure 2.10 General representation for single line-to-ground fault.2711.Figure 2.11 General representation for line-to-line fault.2812.Figure 2.12 General representation of double line-to-ground fault.2913.Figure 2.13 General representation for three-phase fault....3014.Figure 3.1 Bundled conductors configurations....3215.Figure 3.2 Cross section of three-phase horizontal bundled-conductor..3216.Figure 3.3 Segment of 1-phase and neutral connection for long transmission line.3617.Figure 3.4 Parameters of catenary...4818.Figure 3.5 Parameters of parabola...5019.Figure 3.6 Sequence components....56
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20.Figure 3.7 Single line-to-ground faultsequence network connection.6421.Figure 3.8 Line-to-line fault sequence network connection....6622.Figure 3.9 Double line-to-ground fault sequence network connection7023.Figure 3.10 Three-phase fault sequence network connection..7224.Figure 4.1 3H1 wood H-frame type structure......7625.Figure 4.2 One line diagram of power system model......9826.Figure 4.3 Power system model with fault at end of line......10027.Figure 4.4 Equivalent sequence networks.10128.Figure 4.5 Sequence network connection for SLG fault...10129.Figure 4.6 Sequence network connection for L-L fault.10530.Figure 4.7 Sequence network connection for DLG fault...10931.Figure 4.8 Sequence network connection for 3LG fault ...113
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Chapter 1
INTRODUCTION
The purpose of this project is to design an overhead long transmission line that operates
at an extra-high voltage (EHV), and effectively supplies power to a specified load. The
line will have a length of 220 miles, and operate at 345 kV. The receiving end of the line
will be connected to a load of 100 MVA with a lagging power factor of 0.9.
Design of an overhead transmission line is an intricate process that essentially involves a
complete study of conductors, structure, and equipment [1]. The study determines the
potential effectiveness of a proposed system of components in satisfying design criteria.
The design criteria for this project are primarily focused on electrical performance
requirements. The criteria include transmission line efficiency, power loss, voltage
regulation, line sag and tension. To simplify the design process for this project, the same
support structure will be used for all design options, and for the final solution. The
options in conductor size with the predetermined structure will provide alternative
solutions. A MATLAB program will be used to determine the performance of all
alternative solutions with respect to each design criteria. Amongst the options, the ones
that meet all design criteria will be considered and compared for selecting the optimal
final solution.
A fault analysis will be completed for the final solution in order to demonstrate the
electrical behavior and performance of the transmission line system, when subjected to
fault conditions. The system model will interconnect the transmission line to a typical
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generator source, via a step-up transformer, in order to supply power to the line. On the
receiving end of the line, the load will be connected. This fault study will be completed
twice, one time via a MATLAB program, and another time via the ASPEN One-Liner
software. Current and voltage conditions will be found during the different fault events.
The study will cover fault events occurring at the following three locations: 1) beginning
of the transmission line, 2) midpoint on the transmission line and 3) end of the
transmission line. At each location, the four classical fault types will be considered.
The last analysis for this project will use the ASPEN One-Liner software to simulate the
load flow for the final line design using the same system model as described for the fault
study. The results will indicate performance of the final line design under normal
operating conditions.
Equation Chapter (Next) Section 1
Equation Chapter (Next) Section 1
Equation Section (Next)
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Chapter 2
LITERATURE SURVEY
2.1 INTRODUCTION
This chapter succinctly introduces and explains important fundamental concepts and
terminology involved with transmission line design and fault analysis. Some basic theory
is provided as circumstantial information that leads to general questions and issues that
must be addressed during the design and analysis processes.
2.2 SUPPORT STRUCTURE
A line design usually has structure support requirements that are very similar to
requirements of some existing lines [1]. Thus, an existing structure design can likely be
found and leveraged to accommodate the support requirements. For this reason, most of
the work associated with the structure involves defining the configuration and mechanical
load requirements that the structure must support in order to select the appropriate
existing structure design.
Many factors must be considered when defining the configuration and mechanical load of
an overhead transmission line. First, data about the environmental conditions and climate
must be gathered and reviewed. Parameters such as air temperature, wind velocity,
rainfall, snow, ice, relative humidity and solar radiation must be studied [9].
Subsequently, other factors are assessed, including conductor weight, ground shielding
needs, clearance to ground, right of way, equipment mounting needs, material
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availability, terrain to be crossed, cost of procurement, and lifetime upgrading and
maintenance [1].
Conductor load is found by calculating sag/tension on the conductor. The amount of
tension depends on the conductor's weight, sag, and span. In addition, wind and ice
loading increases the tension and must be included in the load specifications [9]. For safe
operation of conductors, the structure must have a margin of strength under all expected
load/tension conditions. For all conditions, the structure must also provide adequate
clearance between conductors.
The three main types of structures are pole, lattice, and H-frame. The lattice and H-frame
types are stronger than the pole type, and provide more clearance between conductors.
Common materials used for structure fabrication are wood, steel, aluminum and concrete
[1]. For an extra-high or ultra-high voltage line, conductors are larger and heavier, so the
structure must be stronger than ones that are used for lower voltage lines. Steel lattice
type structures are the most reliable, having advantages in strength of structure type and
material and in additional clearance between conductors. In comparison, wood and
concrete pole type structures are suited more for lower load stresses. Wood has
advantages of less procurement cost and natural insulating qualities [1].
Since the scope of this project is primarily focused on the electrical design criteria of
transmission lines, a structure for this design will be selected from a group of existing
345kV support structures without defining specific load support requirements.
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2.3 LINE SPACING AND TRANSPOSITION
When designing a transmission line the spacing between conductors should be taken into
consideration. There are two aspects of spacing analysis: mechanical and electrical.
Mechanical Aspect:Wing conductors usually swing synchronously. However, in cases
of small size conductors and long spans there is the likelihood that conductors might
swing non-synchronously.In order to determine correct conductor spacing the following
factors should be included into analysis: the material, the diameter and the size of the
conductor, in addition to maximum sag at the center of the span. A conductor with
smaller cross-section will swing out further than a conductor of large cross-section. There
are several formulas in use to determine right spacing [8].This is NESC, USA formula
3.6812
LD A S (2.1)
D = horizontal spacing in cm
A = 0.762 cm per kV line voltage
S = sag in cm
L = length of insulator string in cm
Voltage between
conductors
Minimum horizontal
spacing
Minimum vertical
spacing
Up to 8700V 12in 16in
8701 to 50,000V12in, plus 0.4in for each1000 V above 8700V*
40in
Above 50,000V 12in, plus 0.4in for each1000 V above 8700V
40in, plus 0.4in for each1000 V above 50,000V
*This is approximate.
Table 2.1 Typical conductor separation [11].
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Electrical Aspect: When increasing spacing (GMD = geometric mean distance between
the phase conductors in ft) Z1 (positive-sequence impedance) increases and Z0 (zero-
sequence impedance) decreases. If the neutral is placed closer to the phase conductors it
will reduce Z0 but may increase the resistive component of Z0. A small neutral with high
resistance increases the resistance part of Z0 [8].
2.3.1 SYMMETRICAL SPACING
Three-phase line with symmetrical spacing forms an equilateral triangle with a distance D
between conductors. Assuming that the currents are balanced:
0a b c
I I I (2.2)
(a) (b)
Figure 2.1 Three-phase line with symmetrical spacing: a) geometry; b) phase inductance [8].
b ca
D12 D23
D31
Figure 2.2 Cross section of three-phase line with horizontal tower configuration.
La
neutral
D
D
D
Ia
IbIcr
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The total flux linkage of phase conductor is:
7
'
1 1 1
2 10 ( )a a b cI ln I ln I lnDr D
(2.3)
b c aI I I (2.4)
7
'
1 12 10 a a aI ln l
rI n
D
(2.5)
72 10'
a aln
r
DI (2.6)
Because of symmetry:
a b c (2.7)
and the three inductances are identical.
The inductance per phase per kilometer length:
0.2s
D mHL ln
D km (2.8)
r= the geometric mean radius, GMR, and is shown by Ds
For a solid round conductor:
1
4s
D r e
(2.9)
Inductance per phase for a three-phase circuit with equilateral spacing is the same as for
one conductor of a single-phase circuit.
2.3.2 ASYMMETRICAL SPACING
While constructing a transmission line it is necessary to take into account the practical
problem of how to maintain symmetrical spacing. With asymmetrical spacing between
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the phases, the voltage drop due to line inductance will be unbalanced even when the line
currents are balanced. The distances between the phases are denoted byD12,D32 andD13.
The following flux linkages for the three phases are obtained:
D13
D23
D12
a
b
c
Figure 2.3 Three-phase line with asymmetrical spacing [8].
7
12 13
1 1 12 10 ( )
'a a b cI ln I ln I ln
r D D (2.10)
7
12 23
1 1 12 10 ( )
'b a b cI ln I ln I ln
D r D (2.11)
7
13 23
1 1 12 10 (
' )c a b cI ln I ln I ln
D D r (2.12)
In matrix form:
L I (2.13)
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The symmetrical inductance matrix:
12 13
7
12 23
13 23
1 1 1
'
'
'
1 1 12 10
1 1 1
ln ln lnr D D
L ln ln lnD
r
r D
ln ln lnD D
(2.14)
With Ia as a reference for balanced three-phase currents:
2240b a aI I a I (2.15)
120c a aI I a I (2.16)
The operatora:
120aa I (2.17)
2240aa I
(2.18)
The phase inductances are not equal and they contain an imaginary term due to the
mutual inductance:
7
12 13
1 1 1 2 10 ( )
'
aa a b c
a
L I ln I ln I lnI r D D
(2.19)
7
12 23
1 1 1 2 10 (
')
bb a b c
b
L I ln I ln I lnI D r D
(2.20)
7
13 23
1 1 1 2 1'
0
cc a b c
c
L I ln I ln I lnI D D r
(2.21)
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2.3.3 TRANSPOSED LINE
In most power system analysis a per-phase model of the transmission line is required.
The previously above stated inductances are unwanted because they result in an
unbalanced circuit configuration. The balanced nature of the circuit can be restored by
exchanging the positions of the conductors at consistent intervals. This is known as
transposition of line and is shown in Figure 2.4. In this example each segment of the line
is divided into three equal sub-segments. Transposition involves interchanging of the
phase configuration every one-third the length so that each conductor is moved to occupy
the next physical position in a regular sequence.
a
a
a
b
b
b
c
c
c
1
2
3
D23
D12
S/3
Length of line, S
S/3 S/3
1
2
3
1
2
3
Section IISection I Section III
Figure 2.4 A Transposed three-phase line [7].
In a transposed line, each phase takes all the three positions. The inductance per phase
can be found as the average value of the three inductances (La,Lb andLc) previously
calculated in (2.19) to (2.21). Consequently,
3a b cL L LL
(2.22)
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Since, 2 1 120 1 240 1o oa a (2.23)
The average of a b cL L L
come to be
7
'
12 23 13
2 10 1 1 1 1 3
3L ln ln ln ln
r D D D
(2.24)
1
37 12 23 31( )2 10
'
D D DL ln
r
(2.25)
The inductance per phase per kilometer length:
0.2s
GMD mHL ln
D km (2.26)
Ds is the geometric mean radius, (GMR). For stranded conductorDs is obtained from the
manufactures data. However, for solid conductor:
1
4'sD r r e
(2.27)
GMD (geometric mean distance) is the equivalent conductor spacing:
312 23 31GMD D D D (2.28)
For the modeling purposes it is convenient to treat the circuit as transposed.
2.4 LINE CONSTANTS
Transmission lines have four basic constants: series resistance, series inductance, shunt
capacitance, and shunt conductance [8].
Series resistance is the most important cause of power loss in a transmission line. The ac
resistance or effective resistance of a conductor is
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2 Lac
PR
I (2.29)
where the real power loss (PL) in the conductor is in watts, and the conductor's rms
current (I) is in amperes [8]. The amount of resistance in the line depends mostly upon
conductor material resistivity, conductor length, and conductor cross-sectional area.
The inductance of a transmission line is calculated as flux linkages per ampere. An
accurate measure of inductance in the line must include both flux internal to each
conductor and the external flux that is produced by the current in each conductor [5].
Both series resistance and series inductance, i.e. series impedance, bring about series
voltage drops along the line.
Shunt capacitance produces line-charging currents. Shunt capacitance in a transmission
line is due to the potential difference between conductors [1].
Shunt conductance causes, to a much lesser degree, real power losses as a result of
leakage currents between conductors or between conductors and ground. The current
leaks at insulators or to corona [8]. Shunt conductance of overhead lines is usually
ignored.
2.5 CONDUCTOR TYPE AND SIZE
A conductor consists of one or more wires appropriate for carrying electric current. Most
conductors are made of either aluminum or copper.
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Aluminum (Al) Copper (Cu) ObservationMelting Point 660C 1083C
Annealing
starts
Most rapidly
above 100C 100C200C and 325C
Both soften and lose
tensile strength.
Resistance to
corrosionGood Very Good
Al corrodes quicklythrough electricalcontact with Cu or
steel. This galvaniccorrosionaccelerates in the
presence of salt.
OxidationWhen exposed to the
atmosphere
Al thin invisibleoxidation film
protects against
most chemicals,weather and evenacids.
Resistivity Very low
Cu conductor hasequivalent
ampacity of analuminumconductor that istwo AWG sizes
larger. A larger Alcross-sectional areais required to obtainthe same loss as in aCu conductor
Usage
Al is lighter, lessexpensive and so it has
been used for almost all
new overheadinstallations
Cu is widely used as apower conductor, butrarely as an overheadconductor. Cu isheavier and more
expensive than Al
The supply of Al isabundant, whereas
that of Cu is limited.
Table 2.2 Aluminum vs. copper conductor type.
Since aluminum is lighter and less expensive for a given current-carrying capability it has
been used by utilities for almost all new overhead installations. Aluminum for power
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conductors is alloy 1350, which is 99.5% pure and has a minimum conductivity of 61.0%
IACS [10].
Different types of aluminum conductors are available:
AACall-aluminum conductor
Aluminum grade 1350-H19 AAC has the highest conductivity-to-weight ratio of all
overhead conductors [10].
ACSRaluminum conductor, steel reinforced
Because of its highmechanical strength-to-weight ratio, ACSR has equivalent or higher
ampacity for the same size conductor. The steel adds extra weight, normally 11 to 18% of
the weight of the conductor. Several different strandings are available to provide different
strength levels. Common distribution sizes of ACSR have twice the breaking strength of
AAC. High strength means the conductor can withstand higher ice and wind loads.
Also, trees are less likely to break this conductor [10]. Stranded conductors are easier to
manufacture, since larger conductor sizes can be obtained by simply adding successive
layers of strands. Stranded conductors are also easier to handle and more flexible than
solid conductors, especially in larger sizes. The use of steel strands gives ACSR
conductors a high strength-to-weight ratio. For purposes of heat dissipation, overhead
transmission-line conductors are bare (no insulating cover) [8].
AAACall-aluminum alloy conductor
This alloy of aluminum, the 6201-T81 alloy, has high strength and equivalent ampacities
of AAC or ACSR. AAAC finds good use in coastal areas where use of ACSR is
prohibited because of excessive corrosion [10].
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ACARaluminum conductor, alloy reinforced
Strands of aluminum 6201-T81 alloy are used along with standard 1350 aluminum. The
alloy strands increase the strength of the conductor. The strands of both are the same
diameter, so they can be arranged in a variety of configurations. For most urban and
suburban applications, AAC has sufficient strength and has good thermal characteristics
for a given weight. In rural areas, utilities can use smaller conductors and longer pole
spans, so ACSR or another of the higher-strength conductors is more appropriate [10].
Conductor Sizes
The American Wire Gauge (AWG) is the standard generally employed in this country
and where American practices prevail. The circular mil(cmil) is usually used as the unit
of measurement for conductors. It is the area of a circle having a diameter of 0.001 in,
which works out to be 0.7854 106 in2. In the metric system, these figures are a
diameter of 0.0254 mm and an area of 506.71 106 mm2
[11]. Wire sizes are given in
gauge numbers, which, for distribution system purposes, range from a minimum of no. 12
to a maximum of no.0000 (or 4/0) for solid type conductors. Solid wire is not usually
made in sizes larger than 4/0, and stranded wire for sizes larger than no. 2 is generally
used. Above the 4/0 size, conductors are generally given in circular mils (cmil) or in
thousands of circular mils (cmil 103); stranded conductors for distribution purposes
usually range from a minimum of no. 6 to a maximum of 1,000,000 cmil (or 1000 cmil
103) and may consist of two classes of strandings. Gauge numbers may be determined
from the formula:
0.3249
1.123nDiameter in (2.30)
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105,500Cross sectionalarea
1,261ncmil (2.31)
where n is the gauge number (no. 0 = 0; no. 00 =1; no. 000 =2; no.0000 =3) [11].
2.6 EXTRA HIGH VOLTAGE LIMITING FACTORS
Limiting factors for extra high voltage are:
a) Coronab)
Radio noise (RN)
c) Audible Noise (AN)
2.6.1 CORONA
Air surrounding conductors act as an insulator between them. Under certain conditions
air gets ionized and its partial breakdown occurs. Disruption of air dielectrics when the
electrical field reaches the critical surface gradient is known as corona. Corona effect
causes significant power loss and a high frequency current. Corona comes in different
forms: visual corona as violet or blue glows, audible corona as high pitched sound and
gaseous corona as ozone gas which can be identified by its specific odor. In addition
high conductor surface gradient causes the emission of radio and television interference
(RI and TVI) to the surrounding antennas known as radio corona. In order to design
corona free lines it is necessary to take into consideration following factors:
1) Electrical2) Atmospheric
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3) Conductor1) Electrical Factors:
a) Frequency and waveform of the supply: Corona loss is a function of frequency.For that reason the higher the frequency of the supply voltage the higher is corona
loss. This means that corona loss at 60 Hz is greater than at 50 Hz. As a result
direct current (DC) corona loss is less than the alternate current (AC).
b) Line Voltage: Line voltage factor is significant for voltages higher than disruptivevoltage. Corona and line voltage are directly proportional.
c) Conductor electrical field: Conductor electrical field depends on the voltage andconductor configuration i.e., vertical, horizontal, delta etc. In horizontal
configuration the middle conductor has a larger electrical field than the outsides
ones. This means that the critical disruptive voltage is lower for the middle
conductor and therefore corona loss is larger.
2) Atmospheric Factors: Air density, humidity, wind, temperature and pressure have aneffect on the corona loss. In addition rain, snow, hail and dust can reduce the critical
disruptive voltage and hence increase the corona loss. Rain has more effect on the
corona loss than any other weather conditions. The most influential are temperature
and pressure. Atmospheric condition such as air density is directly proportional to the
air strength breakdown.
3) Conductor Factors: Several different conductor factors affect the corona loss:
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a) Radius or size of the conductor: The larger the size of the conductor (radius) thelarger the power lower loss. For a certain voltages the larger the conductor size,
the larger the critical disruptive voltage and therefore the smaller the power loss.
2( )loss ln cP V V (2.32)
Vln = line-to-neutral (phase) operating voltage in kV
Vc = disruptive (inception) critical voltage kV (rms)
b) Spacing between conductors: The larger the spacing between conductors thesmaller the power loss. This can be observed from power loss approximation:
lossr
PD
(2.33)
r = conductor radius
D = distance (spacing) between conductors
c) Number of conductors / Phases: In case of a single conductor per phase for highervoltages there is a significant corona loss. In order to reduce corona loss two or
more conductors are bundled together. By bundling conductors the self-
geometric mean distance (GMD) and the critical disruptive voltage are greater
than in case of a single conductor per phase which leads to reducing corona loss.
d) Profile or shape of the conductor: Conductors can have different shapes orprofiles. The profile of the conductor (cylindrical, oval, flat, etc.,) affects the
corona loss. Cylindrical shape has better field uniformity than any other shape and
hence less corona loss.
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e) Surface conditions of the conductors:The disruptive voltage is higher for smoothcylindrical conductors. Conductors with uneven surface have more deposit (dust,
dirt, grease, etc.,) which lowers the disruptive voltage and increases corona.
f) Clearance from ground: Electrical field is affected by the height of the conductorfrom the ground. Corona loss is greater for smaller clearances.
g) Heating of the conductor by load current:Load currentcauses heating of theconductor which accelerates the drying of the conductor surface after rain. This
helps to minimize the time of the wet conductor and indirectly reduces the corona
loss [12].
2.6.2 LINE DESIGN BASED ON CORONA
When designing a long transmission line (TL) it is desirable to have corona-free lines for
fair weather conditions and to minimize corona loss under wet weather conditions. The
average corona value is calculated by finding out corona loss per kilometer at various
points at long transmission line and averaging them out. For typical transmission line in
fair weather condition corona loss of 1kW per three-phase mile and foul weather loss of
20 kW per three-phase mile is acceptable [7].
2.6.3 ADVANTAGES OF CORONA
Corona reduces the magnitude of high voltage waves due to lightning by partially
dissipating as a corona loss. In this case it has a purpose of a safety valve.
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2.6.4 DISADVANTAGES OF CORONA
a) Loss of powerb) The effective capacitance of the conductor is increased which increases the
flow of charging current.
c) Due to electromagnetic and electrostatic induction field corona interferes withthe communication lines which usually run along the same route as the power
lines [7].
2.6.5 PREVENTION OF CORONA
Corona loss can be prevented by:
a) increasing the radius of conductorb) increasing spacing of the conductorsc) selecting proper type of the conductord) using bundled conductors [7].
2.6.6 RADIO NOISE
Radio noise (RN) happens due to corona and gap discharges (sparking). It is unwanted
interference within radio frequency band. RN includes radio interference (RI) and
television interference (TVI).
Radio interference(RI): It affects amplitude modulated (AM) radio waves within the
standard broadcast band (0.5 to 1.6 MHz). Frequency modulated (FM) waves are less
affected.
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Television interference (TVI): In general TVI is caused by sparking within VHF (30-
300MHz) and UHF (300-3000MHz) bands. Two types of TVI are recognized due to
weather conditions: fair and foul [1].
2.6.7 AUDIBLE NOISE
Audible noise (AN) takes place predominantly during foul weather conditions due to
corona. AN sounds like a hiss or sizzle. In addition corona produces low-frequency
humming tones (120 - 240Hz) [1].
2.7 LINE MODELING
To understand the electrical performance of a transmission line, electrical parameters at
both ends of a line must be evaluated. When voltage and current is given at one end of a
line, an accurate calculation of voltage and current at the other end, or at some point
along the line, requires a sufficiently accurate model of a line. How a transmission line is
modeled depends on the line length. There are three classes of line lengths. For line
lengths that are classified as short, up to 50 miles, the model is simplified because shunt
capacitance and shunt admittance can be omitted because they have little effect on the
accuracy of the model. Because the line impedance is constant throughout the line, the
current will be the same from the sending end to the receiving end, so the model can be a
simple, lumped impedance value, as shown in Figure 2.5 [1].
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Figure 2.5 Equivalent circuit of short transmission line [1].
For line lengths that are classified as medium, between 50 and 150 miles, there is enough
current leaking through the shunt capacitance that shunt admittance must be included in
order for the model to be an acceptable representation. However, a medium line is still
short enough that lumping the shunt admittance at some points along the line is a
sufficiently accurate model [1]. Typically, a medium line is modeled either as a T or
network, as shown in Figures 2.6 and 2.7.
Figure 2.6 Nominal-T circuit of medium transmission line [1].
I S Z = R + jX L I R
+
V S-
+
V R-
a a
N N
S e n d i n g
e n d
( s o u r c e )
R e c e i v i n g
e n d
( s o u r c e )
l
I S R / 2 + j ( X L / 2 ) R / 2 + j( X L / 2 ) I R
+
V S-
+
V R-
C G
I Ya a
N N
V Y
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Figure 2.7 Nominal- circuit of medium transmission line [1].
For line lengths that are classified as long, above 150 miles, the needed accuracy from the
model requires that the series impedance and shunt admittance be represented by a
uniform distribution of the line parameters [1]. Each differential length is infinitely small
and defined as a unit length. The series impedance and shunt admittance is represented
for each unit length of line, as shown in Figure 2.8.
Figure 2.8 Segment of one phase and neutral connection for long transmission line [5].
I S R + jX L I R
+
V S-
+
V R-
C / 2 G / 2
IC 1
a a
N N
C / 2 G / 2
I C 2
I I
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This model accounts for the changes in voltage and current throughout the line exactly as
the series impedance and shunt admittance affect them. In this way, the difference
between voltage and current at the sending end and receiving end can be analyzed
accurately [5]. The scope of this project will only cover the mathematical model used for
designing long line lengths. For details of the long transmission line mathematical model,
see Chapter 3.
2.8 LINE LOADABILITY
The characteristic impedance of a line, also known as surge impedance, is a function of
line inductance and capacitance. Surge impedance loading (SIL) is a measure of the
amount of power the line delivers to a purely resistive load equal to its surge impedance.
SIL provides a comparison of the capabilities of lines to carry load, and permissible
loading of a line can be expressed as a fraction of SIL.
The theoretical maximum power that can be transmitted over a line is when the angular
displacement across the line is = 90, for the terminal voltages. However, for reasons
of system stability, the angular displacement across the line is typically between 30 and
45 [8]. Figure 2.9 illustrates the differences in curve plots for the theoretical steady-
state stability limit and a practical line loadability. The practical line loadability is
derived from a typical voltage-drop limit of and a maximum angular
displacement of 30 to 35 across the line [8]. The loadability curve is generally
applicable to overhead 60-Hz lines with no compensation.
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Figure 2.9 Practical Loadability for Line Length [8].
As indicated by the chart, for lines classified as short, the power transfer capability is
determined by the thermal loading limit. For medium and long lines, maximum power
transfer is determined by the stability limit.
2.9 FAULT EVENTS
A fault event in an electric power transmission system is any abnormal change in the
physical state of a transmission system that impairs normal current flow. Typically, a
fault in a transmission line occurs when an external object or force causes a short circuit.
Examples of external objects that intrude upon an overhead transmission line are
lightning strikes, tree limbs, animals, high winds, earthquakes, and local structures.
Other faults occur when components or devices in a transmission system fail. During a
fault, the network can experience either an open circuit or a short circuit. Short-circuit
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faults impose the most risk of damaging elements in a power system. Open circuit faults
are typically not a threat for causing damage to other network elements.
2.10 FAULT ANALYSIS
Important part of TL designing includes fault analysis. In order to have well protected
network typically faults are simulated at different points throughout the transmission
system. It is crucial to have precise analysis of the designed system to prevent faults
interruption. In general the three phase faults can be classified as:
1. Shunt faults (short circuits)1.1.Unsymmetrical faults (Unbalanced)
1.1.1. Single line-to-ground (SLG) fault1.1.2. Line-to-line (L-L) fault1.1.3. Double line-to-ground (DLG) fault
1.2.Symmetrical fault (Balanced)1.2.1. Three-phase-fault
2. Series Faults (open conductor)2.1.Unbalanced faults
2.1.1. One line open (OLO)2.1.2. Two lines open (TLO)
3. Simultaneous faults
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2.11 SINGLE LINE-TO-GROUND (SLG) FAULT
Seventy percent of all transmission line faults are attributable to when a single conductor
is physically damaged and either lands a connection to the ground or makes contact with
the neutral wire [1]. This fault type makes the system unbalanced and is called a single
line-to-ground (SLG) fault. The failed phase conductor, generally defined as phase a, is
connected to ground by an impedance value Zf. Figure 2.10 shows the general
representation of an SLG fault.
Zf
a
b
c
n
F
+
Vaf-
Ibf= 0af Icf= 0
Figure 2.10 General representation for single line-to-ground fault [1].
2.12 LINE-TO-LINE (L-L) FAULT
A line-to-line fault is unsymmetrical (unbalanced) fault and it takes place when two
conductors are short-circuited. This can happen for various reasons i.e., ionization of air,
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flashover, or bad insulation. Figure 2.11 shows the general representation of an LL fault
[1].
a
b
c
F
Ibfaf=0 Icf= -Ibf
Zf
Figure 2.11 General representation for line-to-line fault [1].
2.13 DOUBLE LINE-TO-GROUND (DLG) FAULT
Ten percent of all transmission line faults are attributable to when two conductors are
physically damaged and both of them land a connection through the ground or both
contact the neutral wire [1]. This fault type makes the system unbalanced and is called a
double line-to-ground (DLG) fault. The failed phase conductors, generally defined as
phases b and c, are each connected to ground by their own separate fault impedance value
Zfand a common ground impedance value Zg.
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Figure 2.12 shows the general representation of a DLG fault.
Zf
a
b
c
n
F
Ibfaf= 0 Icf
Zf
Zg
N
Ibf+Icf
Figure 2.12 General representation of double line-to-ground fault [1].
2.14 THREE-PHASE FAULT
A three-phase (3) fault occurs when all three phases of a TL are short-circuited to each
other or earthed. It is a symmetrical (balanced) fault and the most severe one. Since 3
fault is balanced it is sufficient to identify the positive sequence network. As all three
phases carry 120 displaced equal currents the single line diagram can be used for the
analysis. Three-phase faults make 5% of the initial faults in a power system [1].
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Figure 2.13 shows the general representation of a 3 fault.
Zf
a
b
c
n
F
Ibfaf Icf
Zf
ZgN
Iaf+Ibf+ Icf= 3Ia0
Zf
Figure 2.13 General representation for three-phase (3) fault [1].
2.15 THE PER-UNIT SYSTEM
In power system analysis it is beneficial to normalize or scale quantities because of
different ratings of the equipment used. Usually the impedances of machines and
transformers are specified in per-unit or percent of nameplate rating. Using per-unit
system has more than a few advantages such as simplifying hand calculations,
elimination of ideal transformers as circuit component, bringing voltage from beginning
to end of the system close to unity, and simplifies analysis of the system overall.
Particular disadvantages are that sometimes phase shifts are eliminated and equivalent
circuits look more abstract. In spite of this per-unit system is widely used in industry.
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Chapter 3
MATHEMATICAL MODEL
Equation Chapter (Next) Section 1
3.1 INTRODUCTION
This chapter steps through the mathematical approach which is used for design and
analysis of an overhead extra-high voltage long transmission line. Some information
about the physical solution is included to relate the mathematical model and physical
solution.
After design requirements are established, the first step in preliminary design is to choose
a standardized support structure that can be adapted to provide the best solution for the
given job. The selection should be taken from a group of structures that have been
categorized as standard designs for the transmission voltage level that matches the design
requirement. A selected structure will define the spacing between conductor phases and
the limits on conductor size that can be supported.
The next step in preliminary design is to choose a conductor type and size that has
adequate capacity to handle the load current. With a preliminary selection of support
structure and conductor type and size, a detailed design analysis can be undertaken, as
shown in the mathematical approach from the following sections.
3.2 GEOMETRIC MEAN DISTANCE (GMD)
Bundling of conductors is used for extra-high voltage (EHV) lines instead of one large
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conductor per phase. The bundles used at the EHV range usually have two, three, or four
subconductors [1].
(a) (b) (c)
d
d
d d
d
d d
d
Figure 3.1 Bundled conductors configurations: (a) two-conductor bundle; (b) three-conductorbundle; (c) four-conductor bundle [1].
b'b
d
c'c
d
a'a
d
D12 D23
D31
Figure 3.2 Cross section of bundled-conductor three-phase line with horizontal towerconfiguration [1].
The three-conductor bundle has its conductors on the vertices of an equilateral triangle,
and the four-conductor bundle has its conductors on the corners of a square.
For balanced three-phase operation of a completely transposed three-phase line only one
phase needs to be considered.Deq, the cube root of the product of the three-phase
spacings, is the geometric mean distance (GMD) between phases:
312 23 31 eq mD D D D D ft (3.1)
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3.3 GEOMETRIC MEAN RADIUS (GMR)
Geometric Mean Radius (GMR) of bundled conductors for
Two-conductor bundle:
bS SD D d ft (3.2)
Three-conductor bundle:
23 bS SD D d ft (3.3)
Four-conductor bundle:
34 bS SD D d ft (3.4)
where:
= GMR of subconductors
distance between two subconductors
If the phase spacings are large compared to the bundle spacing, then sufficient accuracy
forDeqis obtained by using the distances between bundle centers. If the conductors are
stranded and the bundle spacing dis large compared to the conductor outside radius, each
stranded conductor is replaced by an equivalent solid cylindrical conductor with
GMR= .
The modified GMR of bundled conductors used in capacitance calculations for
Two-conductor bundle:
bSCD r d ft (3.5)
Three-conductor bundle:
3 2 bSC
D r d ft (3.6)
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Four-conductor bundle:
34
1.09
b
SCD r d ft
(3.7)
where:
= outside radius of subconductors
= distance between two subconductors.
3.4 INDUCTANCE AND INDUCTIVE REACTANCE
For three-phase transmission lines that are completely transposed, Equation (3.1) can be
used to find the equivalent equilateral spacing for the line. Thus, the average inductance
per phase is
72 10 ln eq
a
s
D HL
D m
(3.8)
or
100.7411 log eq
a
s
D mHL
D mi
(3.9)
and the inductive reactance is found by
2L a
X f L per phase (3.10)
or
0.1213 ln eq
L
s
DX
D mi
per phase (3.11)
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3.5 CAPACITANCE AND CAPACITIVE REACTANCE
The average line-to-neutral capacitance per phase is
10
0.0388 F toneutral
log
N
eq
CD mi
r
(3.12)
where
1
3 eq m ab bc caD D D D D ft (3.13)
radius of cylindrical conductor in feet.
The capacitive reactance is calculated by
1
2C
N
Xf C
(3.14)
or
100.06836 log .eq
C
DX M mi
r
(3.15)
3.6 LONG TRANSMISSION LINE MODEL
For lines 150 miles and longer, i.e. long lines, modeling with lumped parameters is not
sufficiently accurate for representing the effects of the parameters uniform distribution
throughout the length of the line. An acceptable model provides mathematical
expressions for voltage and current at any point along the line [1]. Figure 3.3 depicts a
segment of one phase of a three-phase transmission line of length l.
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Figure 3.3 Segment of one phase and neutral connection for long transmission line. [5]
The following derivation is given by Saadat [5]. The series impedance per unit length is
z, and the shunt admittance per phase is y, wherez = r + jL andy = g + jC. Consider
one small segment of line x at a distance x from the receiving end of the line. The
phasor voltages and currents on both sides of this segment are shown as a function of
distance. From Kirchhoffs voltage law
( )V x x V x z xI x (3.16)
or
( )( )
V x x V xzI x
x
(3.17)
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Taking the limit as , we have
( )( )
dV x
zI xdx (3.18)
Also, from Kirchhoffs current law
I x x I x y xV x x (3.19)
or
( )( )
I x x I xyV x x
x
(3.20)
Taking the limit as , we have
( )( )
dI xyV x
dx (3.21)
Differentiating (3.18) and substituting from (3.21), we get
2
2
( ) ( )( )
d x dI xz z x
dx
VyV
dx (3.22)
Let
2zy (3.23)
The following second-order differential equation will result.
2
2
2
( )0
d V xV x
dx (3.24)
The solution of the above equation is
1 2x x
V x Ae A e (3.25)
where , known as thepropagation constant, is a complex expression given by (3.23) or
( ) ( )j zy r j L g j C (3.26)
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The real part is known as the attenuation constant, and the imaginary component is
known as thephase constant. is measured in radian per unit length. From (3.18), the
current is
1 2 1 21 ( ) x x x xdV x yI x Ae A e Ae A ez dx z z
(3.27)
or
1 21 x x
C
I x A e A eZ
(3.28)
whereZc is known as the characteristic impedance, given by
C
zZ
y (3.29)
To find the constants and, we note that when , () , and () .
From (3.25) and (3.28) these constants are found to be
12
R C RV Z IA (3.30)
22
R C RV Z IA
(3.31)
Upon substitution in (3.25) and (3.28), the general expressions for voltage and current
along a long transmission line become
2 2x xR C R R C R
V Z I V Z I V x e e
(3.32)
2 2
R RR R
x xC C
V VI I
Z ZI x e e
(3.33)
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The equations for voltage and currents can be rearranged as follows:
2 2
x x x x
R C R
e e e eV x V Z I
(3.34)
1
2 2
x x x x
R R
C
e e e eI x V I
Z
(3.35)
Recognizing the hyperbolic functions sinh, and cosh, the above equations are written as
follows:
cosh sinhR C RV x x V Z x I
(3.36)
1
sinh coshR RC
I x x V x IZ
(3.37)
We are particularly interested in the relation between the sending-end and the receiving-
end on the line. Setting , () and () , the result is
cosh sinhS R C RV l V Z l I (3.38)
1
sinh coshS R RC
I l V l IZ
(3.39)
Rewriting the above equations in terms of ABCD constants, we have
S R
S R
V VA B
I IC D
(3.40)
where
cosh cosh coshA l YZ (3.41)
sinh sinh sinhC CZ
B Z l YZ ZY
(3.42)
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sinh sinh sinhC CY
C Y l YZ Y Z
(3.43)
cosh cosh coshD A l YZ (3.44)
where
LZ r jx l (3.45)
is total line series impedance per phase
SY g jb l (3.46)
is total line shunt admittance per phase.
Note that A D (3.47)
and 1AD BC . (3.48)
For a long transmission line, conductance is very small compared to susceptance, and can
be omitted for simplicity. Thus, can be reduced to the following equation:
1
C
Y jb l j l X
(3.49)
3.7 SENDING-END VOLTAGE AND CURRENT
One step in line design is analyzing what power input, i.e. voltage and current, is needed
at the sending-end in order to deliver the load power requirements. If the resulting power
input needs are within parameters that are acceptable to the overall power system, the
design is viable. However, the line design may still be adjusted to match preferred input
parameters. After the ABCDconstants are determined, as shown in the previous section,
the following steps can be used to find the sending-end voltage and current.
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Using the receiving-end design requirements for load power, voltage, and power factor,
the receiving-end line-to-neutral voltage and current magnitude are determined by the
following equations:
( )
3
R L L
R L N
VV (3.50)
and
( )3
R
R L L
S
IV
(3.51)
where:
() receiving-end line-to-neutral voltage (kV),
magnitude of receiving-end line current (A).
The receiving-end current phasor can be found by
(cos sin )R R R Rj I I (3.52)
where:
angle difference between () and
and can be found by taking the inverse cosine of the power factor.
Using the calculated values for ABCD constants and receiving-end voltage and current,
we can use Equation (3.40) to determine the corresponding sending-end voltage and
current.
The sending-end voltage and current can be equated by
( )( ) ( )S L N RR L N V A V B I (3.53)
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and
( ) ( )
S RR L N
I C V D I (3.54)
where:
sending-end line current (A),
() sending-end line-to-neutral voltage (kV).
The sending-end line-to-line voltage is
( ) 3 1 30S L L S L N V V (3.55)
where:
() sending-end line-to-line voltage (kV).
Note that an additional is added to the angle since the line-to-line voltage is
ahead of its line-to-neutral voltage.
3.8 POWER LOSS
Typically, referring to power loss in a transmission line means the difference in real
power between the sending- and receiving- ends. To calculate the power loss, the first
step is to determine the power factor at each end. For the receiving-end, the power factor
is normally specified per design criteria. For the sending-end, the power factor is found
by determining the angle S between the sending-end current and voltage phasors. The
expression for sending-end power factor is
cos( os) c
SSS IV L N s
pf
(3.56)
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where:
sending-end power factor,
() = angle of sending-end line-to-neutral voltage phasor,
angle of sending-end current phasor,
angle difference between () and .
Then, using the value of, the equation for calculating real power at the sending-end is
( )3 3 cosS L L S S SP V I (3.57)
where:
() sending-end real power in the line (MW).
A similar equation for calculating the receiving-end real power is
( )3 3 cosR L L R RRP V I (3.58)
where:
() receiving-end real power in the line (MW).
Using the calculated values from the above equations, real power loss in the line is found
by
(3 ) (3 ) (3 )L S RP P P (3.59)
where:
() total real power loss in the line (MW).
The majority of power loss in a transmission line is a result of real power loss due to the
resistance of the line. A good design will minimize the total real power loss in the line.
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3.9 TRANSMISSION LINE EFFICIENCY
Performance of transmission lines is determined by efficiency and regulation of lines.
Transmission line efficiency is:
100R
S
P
P (3.60)
where:
= transmission line efficiency
= receiving-end power
= sending end power
% 100
Powerdeliverd at receiving endtransmissionlineefficiency
Powersent fromthe sendingend (3.61)
% 100R
S
Ptransmissionlineefficiency
P (3.62)
The end of the line where source of supply is connected is called the sending end and
where load is connected is called the receiving end [1].
3.10 PERCENT VOLTAGE REGULATION
Voltage regulation of the line is a measure of the decrease in receiving-end voltage as
line current increases. In mathematical terms, percent voltage regulation is defined as the
percent change in receiving-end voltage from the no-load to the full-load condition at a
specified power factor with sending-end voltage VS held constant, that is,
, ,
,
- | |
| |
R NL R FL
R FL
V VPercentVR 100
V (3.63)
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where:
|VR,NL|=magnitude of receiving-end voltage at no-load,
|VR,FL|=magnitude of receiving-end voltage at full-load with constant |Vs|,
|VS|=magnitude of sending-end phase (line-to-neutral) voltage at no load.
3.11 SURGE IMPEDANCE LOADING (SIL)
In power system analysis of high frequencies or surges caused by lightning, losses are
typically ignored and surge impedance becomes important. A line is lossless when its
series resistance and shunt conductance are zero [6]. The surge impedance of a lossless
line, also known as characteristic impedance, is a function of line inductance and
capacitance, and can be expressed as
CL
ZC
(3.64)
or
C C LZ X X (3.65)
where:
shunt capacitive reactance (),
series inductive reactance (),
characteristic impedance ().
Surge impedance loading (SIL), a measure of the amount of power the line delivers to a
purely resistive load equal to its surge impedance [6], is found for a three-phase line by
the following equation:
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2
( )| |Lr L
c
kVSIL MW
Z
(3.66)
where: SIL = surge impedance loading (MW).
SIL provides a comparison of the capabilities of lines to carry load, and permissible
loading of a line can be expressed as a fraction of SIL. SIL, or natural loading, is a
function of the line-to-line voltage, line inductance and line capacitance. Since the
characteristic impedance is based on the ratio of inductance and capacitance, SIL is
independent of line length. The relationship between SIL and voltage explains why an
extra-high voltage line has more power transfer capability than lower voltage lines.
3.12 SAG AND TENSION
3.12.1 CATENARY METHOD
Sag-tension calculations predict the behavior of conductors based on recommended
tension limits under varying loading conditions. These tension limits specify certain
percentages of the conductors rated breaking strength that are not to be exceeded upon
installation or during the life of the line. These conditions, along with the elastic and
permanent elongation properties of the conductor, provide the basis for defining the
amount of resulting sag during installation and long-term operation of the line.
Accurately determined initial sag limits are essential in the line design process. Final sags
and tensions depend on initial installed sags and tensions and on proper handling during
installation. The final sag shape of conductors is used to select support point heights and
span lengths so that the minimum clearances will be maintained over the life of the line.
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If the conductor is damaged or the initial sags are incorrect, the line clearances may be
violated or the conductor may break during heavy ice or wind loadings [1].
( )max
T w c d (3.67)
( )max
T w c d (3.68)
minT w c (3.69)
H w c (3.70)
Hcw
(3.71)
minT H (3.72)
T= the tension of the conductor at any point P in the direction of the curve
w = the weight of the conductor per unit length
H= the tension at origin 0
c = catenary constant
s = the length of the curve between points 0 and P
v = that the weight of the portions is ws
L = horizontal distance.
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d
y
AB
L
0H
2
ls
2
ls
Hc
w
V ws
0' (Directrix) x
y
T=wyTy=ws
Tx=wc
Figure 3.4 Parameters of catenary [1].
An increase in the catenary constant, having the units of length, causes the catenary curve
to become shallower and the sag to decrease. Although it varies with conductor
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temperature, ice and wind loading, and time, the catenary constant typically has a value
in the range of several thousand feet for most transmission-line catenaries.
For equilibrium
x
T H (3.73)
yT w s (3.74)
Tx = the horizontal component
Ty= the vertical component.
The total tension in the conductor at any pointx:
w xT H cos
H
(3.75)
The total tension in the conductor at the support:
2
w LT H cos
H
(3.76)
The sag or deflection of the conductor for a span of length L between supports on the
same level:
cosh 12
H w Ld
w H
(3.77)
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3.12.2 PARABOLIC METHOD
The conductor curve can be observed as a parabola for short spans with small sags.
x
y
A B
L
0 Hwx
d
Ty
H
T Ty
Tx
P
Figure 3.5 Parameters of parabola [1].
The following assumptions can be taken into consideration when using parabolic method:
1. The tension is considered uniform throughout the span.2. The change in length of the conductor due to stretch or temperature is the same as
the change of the length due to the horizontal distance between the towers [1].
Approximate value of tension by using parabolic method can be calculated as
2
8
w LT
d
(3.78)
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or
2
8
w Ld T
(3.79)
when
1
2x L (3.80)
y d . (3.81)
3.13 CORONA POWER LOSS
3.13.1 CRITICAL CORONA DISRUPTIVE VOLTAGE
The maximum stress on the surface of the conductor is given by:
ln
LNmax
V kVE
D cmm r
r
(3.82)
VLN = the phase or line-to-neutral voltage in kV
D = is equivalent spacing in cm
r = radius of the conductor in cm
mc = surface irregularity factor ( )
mc = 1 for smooth, solid, polished round conductor
mc = 0.930.98 for roughened or weathered conductor
mc = 0.800.87 for up to seven strands conductor
mc = approx. 0.90 for large conductor with more than seven strands [12]
Mean voltage gradient can be calculated from:
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ln 3
LNmean
V kVE
D cmm r
r
(3.83)
The air density correction factor is defined as:
3.9211
273
p
t
(3.84)
where:
p = the barometric pressure in cm Hg
t = temperature inC
The critical disruptive voltage (corona inception voltage) Vc is voltage at which complete
disruption of dielectric occurs. The dielectric stress is 30 kV/cm peak or 21.1 rms at
NTP i.e., 25C and 76 mmHg. Vc is minimum conductor voltage with respect to earth at
which the corona is expected to start. At Vc corona is not visible [7].
30 c cD
V m r ln kV peak r
(3.85)
21.1 c cD
V m r ln kV rmsr
(3.86)
Vc = the critical disruptive voltage in kV
The critical disruptive voltage Vc line-to-line is
( ) ( )3
c L L c rmsV V kV
(3.87)
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3.13.2 VISUAL CORONA DISRUPTIVE VOLTAGE
In order to observe corona visually the inception voltage has to exceed the critical
disruptive voltage Vc. The visual critical voltage Vv is given by:
0.301
30 1 vD
V m r ln kV peak rr
(3.88)
0.301
21.1 1 vD
V m r ln kV rmsrr
(3.89)
Vv = the visual critical voltage in kV
D = equivalent spacing of conductors in cm
r = radius of the conductor in cm [7]
mv = surface irregularity factor ( )
m = 1 for smooth, solid, polished round conductor
For local and general visual corona:
m = 0.930.98 for roughened or weathered conductor
For local visual corona:
m = 0.700.75 for weathered stranded conductor
For general visual corona:
m = 0.800.85 for weathered stranded conductor
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3.13.3 CORONA POWER LOSS AT AC VOLTAGE
For AC transmission lines empirical equations are used to determine corona loss.
According to Peek corona power loss can be determined from:
2 5241 25 10 / LN c
r kWP f V V phase peak
D km
(3.90)
P = corona loss in kW/km/phase
= density correction factor
VLN = the phase or line-to-neutral voltage in kV
Vc = the critical disruptive voltage in kV
f = frequency
r = radius of the conductor in cm
D = equivalent spacing of conductors in cm.
It is desirable to design transmission line with corona loss between 0.10 and 0.21
kW/km/phase for fair weather conditions. For lower loss range i.e., when
V 1.8LN
cV (3.91)
Peeks formula is not accurate [7].
According to Petersons corona power loss formula:
2
5
10
2.1 10 /cV kW
P f F phaseD km
logr
(3.92)
F = corona loss function
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VLN / Vc 1.0 1.2 1.4 1.5 1.6 1.8 2.0
F 0.037 0.082 0.3 0.9 2.2 4.95 7.0
Table 3.1 Corona factor [7].
Above stated formulas are used for fair weather conditions. For wet weather conditions
critical disruptive voltage is approximately 0.80 of the fair weather calculated value. The
calculated disruptive critical voltage for three-phase horizontal conductor configuration
can be determined as:
3 0.96c c fair V V (3.93)
for the middle conductor and
3 1.06
c c fair V V
(3.94)
for the two outer conductors [1].
3.14 METHOD OF SYMMETRICAL COMPONENTS
According to Charles Fortescue, a set of three-phase voltages are resolved into the
following three sets of sequence components:
1. Zero-sequence components: consisting of three phasors with equal magnitudes and
with zero phase displacement
2. Positive-sequence components, consisting of three phasors with equal magnitudes,
120 phase displacement
3. Negative-sequence components, consisting of three phasors with equal magnitudes,
120 phase displacement[1].
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Vb0 Vc0 = V0Va0
Vc2
Vb2 Va2 = V2
Va1 = V1
Vb1
Vc1
(a) (b) (c)
Figure 3.6 Sequence components: ( a) zero ( b) positive ( c) negative [8 ]
0 1 2
a a a a V V V V (3.95)
0 1 2
b b b b V V V V (3.96)
0 1 2
c c c c V V V V (3.97)
Operatora is a complex number with unit magnitude and a 120 phase angle. When any
phasor is multiplied by a, that phasor rotates by 120 (counterclockwise).
A list of some common powers, functions and identities involving a:
Power or Function In Polar Form In Rectangular Form
a 1120 -0.5+j0.866
a2 1240=1-120 -0.5-j0.866
a3
1360=10 1.0+j0.0
a4 1120 -0.5+j0.866
1+a= -a2 160 0.5+j0.866
1- a -30 1.5-j0.8661+ a
2= -a 1-60 0.5-j0.866
1- a2 30 1.5+j0.866
a -1 150 -1.5+j0.866a+ a2 1180 -1.0+j0.0
a - a2 90 0.0+j1.732a
2- a -90 0.0-j1.732
a2- 1 -150 -1.5-j0.866
1 + a+ a2 0 0.0+j0.0ja 1210 -0.884+j0.468
Table 3.2 Power and functions of operatora[1].
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1 120 a (3.98)
1 31 1202 2
j a (3.99)
Similarly, when any phasor is multiplied by
2 1 120 1 240 a (3.100)
the phasor rotates by 240.
The phase voltages in terms of the sequence voltages i.e.synthesis equations:
0 1 2 0a a a a V V V V (3.101)
2
0 1 2 0b a a a V a V aV V (3.102)
2
0 1 2 0c a a a V aV a V V (3.103)
The sequence voltages in terms of phase voltages i.e. analysis equations:
01
3a a b c V V V V (3.104)
211
3a a b c
V V aV aV (3.105)
221
3a a b c V V aV aV (3.106)
In matrix form the phase voltages can be expressed as
0
2
1
2
2
1 1 1
1
1
a a
b a
c a
V V
V a a V
V a a V
(3.107)
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and the sequence voltages can be expressed as
02
1
2
2
1 1 11 13
1
a a
a b
a c
V V
V a a V
V a a V
(3.108)
or
012abc V A V (3.109)
1
012 abc
V A V (3.110)
where
22
1 1 1
1
1
A a a
a a
(3.111)
1 2
2
1 1 11
13
1
A a a
a a
(3.112)
Similarly, the phase currents in matrix form can be expressed as
0
2
1
2
2
1 1 1
1
1
a a
b a
c a
I I
I a a I
I a a I
(3.113)
and the sequence currents can be expressed as
0
2
1
2
2
1 1 11
13
1
a a
a b
a c
I I
I a a I
I a a I
(3.114)
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or
012abcI A I
(3.115)
1
012 abc
I A I (3.116)
3.14.1 SEQUENCE IMPEDANCES OF TRANSPOSED LINES
In order to attain equal mutual impedances the line should be transposed or conductors
should have equilateral spacings.
Hence, for the equal mutual impedances
ab bc ca m Z Z Z Z (3.117)
In case when the self-impedances of conductors are equal to each other
aa bb cc s Z Z Z Z . (3.118)
Therefore,
s m m
abc m s m
m m s
Z Z Z
Z Z Z Z
Z Z Z
(3.119)
where,
0.1213ln es a es
Dr r j l
D
Z (3.120)
and
0.1213ln em eeq
Dr j l
D
Z . (3.121)
ra = resistance of a single conductora
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re= resistance of Carsons equivalent earth return conductor which is a function of
frequency
31.588 10er f
. (3.122)
At 60 Hz,
0.09528e
r
(3.123)
At 60 Hz frequency and for 100
average earth resistivity
2788.55eD ft. (3.124)
The equilateral spacings of the conductors can be calculated as
3eq m ab bc caD D D D D (3.125)
TheDs is geometric mean radius (GMR) of the phase conductor.
The sequence impedance matrix of a transposed transmission line can be expressed as
012
2 0 0
0 0
0 0
s m
s m
s m
Z Z
Z Z Z
Z Z
(3.126)
where, by definition,
Z0 is zero-sequence impedance at 60Hz
0 00 2s mZ Z =Z Z (3.127)
3
0 23 0.1213ln ea e
s eq
Dr r j l
D D
Z , (3.128)
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Z1 is positive-sequence impedance at 60Hz
1 11 s m-Z Z =Z Z (3.129)
1 0.1213lneq
a
s
Dr j l
D
Z , (3.130)
Z2 is negative-sequence impedance at 60Hz
2 22 s m-Z Z =Z Z (3.131)
2 0.1213ln
eq
a
s
D
r j lD
Z . (3.132)
Therefore, the sequence impedance matrix of a transposed transmission line can be
expressed as
0
012 1
2
0 0
0
0 0
Z
Z Z
Z
(3.133)
3.15 FAULT ANALYSIS
Three-phase faults can be balanced (i.e., symmetrical) or unbalanced (i.e.,
unsymmetrical). The unbalanced faults are more common. In order to resolve an
unbalanced system the method of symmetrical components can be applied by converting
the system into positive, negative and the zero-sequence fictitious networks. After
defining positive, negative and zero-sequence currents for specific fault phase currents,
sequence and phase voltages can calculated.
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3.16 PER UNIT
Power, current, voltage, and impedance are often expressed in per-unit or percent of
specified base values.
The per-unit values are calculated as:
-
actualquantityper unit quantity
basevalueof quantity (3.134)
where actual quantity is the value of the quantity in the actual units and the base value
has the same units as the actual quantity, forcing the per-unit quantityto be
dimensionless. The actual value may be complex but the base value is always a real
number. Consequently, theangle of the per-unit value is the same as the angle of the
actual value [8] .
In a given power system two independent base values can be arbitrarily selected at one
point. Typically the base complex power Sbase1 and the base voltage VbaseLN are chosen
for either a single-phase circuit or for one phase of a three-phase circuit. In order to
preserve electrical laws in the per-unit system, the following equations must be used for
other base values:
1 1 1base base baseP Q S (3.135)
11
basebase
baseLN
SI
V
(3.136)
2
1
baseLN baseLNbase base basebase base
V VR X
I SZ
(3.137)
1 base base base
base
Y G BZ
(3.138)
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2
basebasebase
kVZ
MVA (3.139)
( )
3
basebase
base LL
MVAI
kV (3.140)
The subscriptsLNand 1represent line-to-neutraland per-phase respectively, for
three-phase circuits. Equations (2.35) and (2.36) are also effective for single-phase
circuits by omitting the subscripts.
By agreement, the following two rules for base quantities are assumed:
1) The value of Sbase1 is the same for the entire power system2) The ratio of the voltage bases on either side of a transformer is selected to be the
same as the ratio of the transformer voltage ratings.
As a result per-unit impedance remains unchanged when referred from one side of a
transformer to the other [8].
3.17 SINGLE LINE-TO-GROUND (SLG) FAULT
An SLG fault generally occurs when one phase conductor either falls to the ground or
makes contact with the neutral wire. Figure 3.7 depicts the typical representation of an
SLG fault at a fault point F with a fault impedance Zf. It is customary to show the fault
occurring on phase a. If the fault actually takes place on another phase, the phases of the
system can simply be relabeled in the appropriate sequence.
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(a)
(b)
Figure 3.7Single line-to-ground fault: (a) general representation; (b) sequence networkconnection [1]
From inspection of Figure 3.7a, the currents for phases b and c are
0bf cf
I I (3.141)
Zf
a
b
c
n
F
+
Vaf-
Ibf= 0af Icf= 0
F0 Z0 N0
+ VA0 -
F1 Z1 N1
+ VA1 -
F2 Z2 N2
+ VA2 -
1.0+ -
0o
3ZfIa1
Ia1a0 Ia2
Zero-sequence
network
Positive-sequence
network
Negative-sequence
network
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Substituting values into Equation (3.114), the symmetrical components for
the currents are given as
0
2
1
2
2
1 1 11
1 03
1 0
a af
a
a
a a
a a
I I
I
I
(3.142)
Using the above equation, the sequence currents for phase a are
0 1 2
1
3a a a af I I I I (3.143)
and can be rewritten as
0 1 23 3 3 af a a a I I I I (3.144)
By inspection of Figure 3.7b, the zero-, positive-, and negative-sequence currents are
equal and can be determined by
0 1 2
0 1 2
1.0 0
3a a a
f
I I I
Z Z Z Z(3.145)
Substituting (3.145) into Equation (3.144) gives
0 1 2
1.0 03
3af
f
IZ Z Z Z
(3.146)
With the sequence current values, Equation (3.155) can be used to find the sequence
voltages, and then Equation (3.107) can be used to find the phase voltages.
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3.18 LINE-TO-LINE (L-L) FAULT
A line-to-line (L-L) fault occurs when two conductors are short-circuited. Figure 3.8a
represents the characteristic representation of a L-L fault at a fault point F with a fault
impedance Zf. Figure 3.8b indicates the interconnection of resulting sequence networks.
It is presumed that L-L fault is between phases b and c.
(a)
(b)
Figure 3.8Line-to-line fault: (a) general representation; (b) sequence network connection [1].
a
b
c
F
Ibfaf=0 Icf= -Ibf
Zf
F0
Z0
N0
+
VA0 = 0-
F1
Z1
N1
+
VA1-
F2
Z2
N2
+
VA2-
+
1.0-0o
Ia1a0=0 Ia2
Zf