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THE ELECTRICAL TRANSMISSION
SYSTEM OF A 345KV WITH 170 MILES MODEL
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 SCIENCE
in
Electrical and Electronic Engineering
by
Farshad Tavatli
Babak Kaviani Joupari
SPRING
2013
Sample Copyright page
[l]
© 2013[Year of graduation]
Farshad Tavatli
Babak Kaviani Joupari
ALL RIGHTS RESERVED
[Thesis/Project Approval Page]
ii
THE ELECTRICAL TRANSMISSION
SYSTEM OF A 345KV WITH 170 MILES MODEL
A Project
by
Farshad Tavatli
Babak Kaviani Joupari
Approved by:
__________________________________, Committee Chair
Turan Gönen, Ph.D.
__________________________________, Second Reader
Salah Yousif, Ph.D.
____________________________
Date
[Thesis/Project Format Approval Page]
iii
Student: Farshad Tavatli
Babak Kaviani Joupari
I certify that these students have met the requirements for format contained in the
University format manual, and that this project is suitable for shelving in the Library and
credit is to be awarded for the project.
__________________________, Graduate Coordinator
B. Preetham. Kumar, Ph.D.
r
Department of Electrical and Electronic Engineering
iv
___________________
Date
Abstract
of
THE ELECTRICAL TRANSMISSION
SYSTEM OF A 345KV WITH 170 MILES MODEL
by
Farshad Tavatli
Babak Kaviani Joupari
A reliable and efficient transmission system benefits not only the power utility
companies, but the consumer as well. This report provides clarification to concepts and
calculations in the analysis of an electrical transmission system involving a 345KV with
170 miles of distance. The project will analyze the efficiency, voltage regulation, power
quality, power losses by comparing different sizes of cables.
A fundamental part of achieving a reliable transmission system is by considering all
necessary factors such as; careful design and complicated calculations “with help of
MATLAB”. Different compensator devices will be discussed and used in no load and full
load situations in this report in order to design an efficient transmission circuit.
_______________________, Committee Chair
Turan Gönen, Ph.D.
_______________________
Date
v
DEDICATION
[Optional]
I dedicate this work to my parents and a great instructor “DR Gonen”.
vi
[This Table of Contents covers many possible headings. Pages are optional.
Use only the headings t
TABLE OF CONTENTS
Page
Dedication .................................................................................................................... vi
List of Tables .................................................................................................................x
List of Figures .............................................................................................................. xi
Chapter
1. INTRODUCTION ...................................................................................................1
2. LITERATURE SURVEY ........................................................................................2
2.1. Introduction ....................................................................................................2
2.2. Line Routing ..................................................................................................4
2.3. Ground Clearance ..........................................................................................4
2.4. Separation Between Conductors ....................................................................6
2.5. Insulation and Insulators ................................................................................6
2.6. Tower Structure .............................................................................................7
2.7. Sag and Tension .............................................................................................9
2.8. Thermal Ampacity .......................................................................................10
2.9. Line Parameters ...........................................................................................10
2.10. Inductance ..................................................................................................11
2.11. Capacitance ................................................................................................11
2.12. Transposition with Ground Wire ...............................................................12
vii
2.13. Long Line Model .......................................................................................12
2.14. Line Compensation ....................................................................................13
3. MATHEMATICAL MODEL ................................................................................14
3.1.
Thermal Ampacity rating ....................................................................14
3.2.
Series Impedance and shunt Admittance ............................................14
3.2.1.
Series Impedance .................................................................................15
3.2.1a Self and Mutual Impedances with effects of Ground Wire.............17
3.2.2 Shunt Admittance..............................................................................23
3.3.
Long Line Model .................................................................................29
3.4.
ABCD Constants .................................................................................32
3.5.
Voltage Regulation, Power Loss and Efficiency ................................32
3.6.
Line Compensation .............................................................................33
3.6.1 Full Load Compensation ...................................................................34
3.6.2 No Load Compensation ....................................................................35
4. APPLICATION OF THE MATHEMATICAL MODEL ......................................39
4.1.
Introduction ..........................................................................................39
4.2.
Thermal Ampacity Rating...................................................................41
4.3.
Line Characteristics and ABCD Constants ..........................................42
4.3.1 Series Impedance ..............................................................................43
4.3.2 Shunt Admittance..............................................................................46
4.3.3 ABCD Constants ...............................................................................50
viii
4.4.
Uncompensated Line ...........................................................................51
4.4.1 Full Load ...........................................................................................52
4.4.2 No Load ............................................................................................53
4.4.3 Voltage Regulation, Power Loss and Efficiency ..............................53
4.4.4 MATLAB ..........................................................................................54
4.5.
Compensation ......................................................................................55
4.5.1 Full Load Compensation ...................................................................55
4.5.2 No Load Compensation ....................................................................56
5. SUMMARY ...........................................................................................................60
Appendix A. Overhead Ground Wire Characteristics .................................................62
Appendix B. ACRS Table............................................................................................64
Appendix C. Wire Compensation of Uncompensated Line data .................................65
Appendix D. MATLAB Code of Uncompensated Line ..............................................67
Appendix E. MATLAB Code, Full Load Compensation ............................................71
Appendix F. MATLAB Code, No Load Compensation ..............................................76
Bibliography ................................................................................................................79
ix
LIST OF TABLES
Tables
Page
1.
Table 2.1 Vertical Clearance of Conductors ....................................................... 5
2.
Table 4.1 MATLAB VS Theoretical Comparison for 1/0 ACSR .................... 54
x
LIST OF FIGURES
Figures
Page
1.
Figure 2.1 Suspension Type Insulators, 345 KV Tower Structures ................... 8
2.
Figure 2.2 Pole Type, 345 KV Tower Structures ............................................... 9
3.
Figure 3.1 Carson’s Return Theory ................................................................. 16
4.
Figure 3.2 Effects of the Ground Wire ............................................................. 19
5.
Figure 3.3 Sectioning of Completely Transposed line...................................... 19
6.
Figure 3.4 Image Method .................................................................................. 24
7.
Figure 3.5 Model of a Long Transmission Line ............................................... 30
8.
Figure 3.6 Series Compensation of a Long Line .............................................. 35
9.
Figure 3.7 No Load Compensation with a Shunt Reactor ................................ 36
10.
Figure 4.1 Tower configuration ........................................................................ 40
11.
Figure 4.2 Series Capacitor Compensation ....................................................... 56
12.
Figure 4.3 Shunt Reactor Compensation .......................................................... 57
xi
1
Chapter 1
INTRODUCTION
The purpose of this project is to find a means of choosing a line based on its voltage
regulation, power loss, and efficiency. A voltage regulation between 3% and 5% will be
ideal, while minimizing power loss and maximizing efficiency. Since ACSR (Aluminum
Conductor cable Steel Reinforced) conductors are widespread used for transmission line
design, therefore in this project we will be using ACSR as well. A line will be selected in
the beginning just to show the hand calculations then MATLAB will be used to execute
the same calculations but for different line sizes. A static receiving end load with a
lagging power factor will be analyzed for a 170 miles long line at 345kv. An overhead
ground wire will be used for lightning protection and will be taken into consideration
when computing the long line parameters. The tower configuration to be used will be
arbitrarily selected as well as the overhead ground wire type. The computational method
is the most significant portion of this analysis; therefore all structural and mechanical
aspects will be ignored. From this analysis the data can then be used for an economic
analysis which would involve line initial cost, construction, maintenance, operation,
efficiency, compensation Vars, and reliability.[1]
Equation Chapter (Next) Section 1
Equation Chapter (Next) Section 1
2
Chapter 2
LITERATURE SURVEY
2.1 INTRODUCTION
Designing Transmission lines is definitely not a simple task. There are so many
considerations that have to be factored in to make an efficient and worth building
transmission line. In addition, there are certain regulations that have to be met for the
safety and wellbeing not just for the people but also for the environment. Such
considerations are as following; Line Routing, Cost Estimating, Station Interfacing,
Right-of Way Acquisition, Structure Design, Sag and Tension, Conductor Analysis,
clearances, strengths and loading, and many more which we cannot explain all of them in
this research. On top of all regulations also have to be considered from the National
Electric Code (NEC) and the National Electric Safety Code (NESC). The noticeable
difference between the two is that the NESC is not intended as a design specification or
an instruction manual. The NEC covers the proper installation guidelines of electric
conductors on buildings or structures that need to be considered to protect the people and
property from hazards that occur from the usage of electricity in building and structures.
The NESC addresses the practical safety and the basic provisions that are considered
essential for the safety of the workers and the public. [2]
Transmission lines were not always generated by alternating current (AC). In fact, Direct
Current (DC) was first used as generation in 1882. However, it was quickly found
3
inefficient because voltage could not be increased for a long distance. The classes of
loads are variable so it required different voltages and would result in different generators
and circuits. Nikola Tesla proposed the use of alternating current which allowed efficient
generation. With the use of rotary converters, an electrical machine used to convert one
form of electrical power to another, networks connecting different generating plants with
loads having different frequencies could be connected. Also, AC could be converted to
DC with the use of rotary converter to be served to the loads that required DC. By
permitting multiple interconnections between several generating plants, electricity
production was more efficient resulting in the decrease in cost. Reliability immensely
improved and also resulted in lower production cost. [2]
AC Power can be transported by two ways. One is overhead transmission and the other is
underground transmission. In overhead transmission, the electrical conductors are not
covered by insulation. The conductors that are used are mainly Aluminum Conductor
Steel Reinforced (ACSR). They are used as bare overhead transmission cable or primary
and secondary distribution cable. The underground transmission is mainly used in
densely congested areas and also where there are natural obstacles that do not permit the
use of overhead transmission. The use of underground transmission is a more expensive
option, actually about two to four times the cost of an overhead line.
4
2.2 LINE ROUTING
One major consideration that has to be taken into account for transmission line design is
the route of the lines. Where will these lines carrying so much voltage go? Selecting a
route may not be an easy task. There are many considerations that need to be thought
taken. For instance, there are physical (highways, railroads, other transmission lines),
biological (wetlands, wildlife, woodlands), or other (federal, state controlled lands). Not
only do these locations have to be examined, the cost of clearing, ease of maintaining,
and the effect of the line may have on the environment. After the route has been selected,
the surveyors conduct a field examination and prepare a universal drawing for the whole
route. Also, there are many permits and authorizations that need to be obtained. For
instance, federal permits or licenses are required, the use of private property, permit from
state/county/city, permission of utility. There are others that need to be obtained but for
the limited space of this report, they shall not be listed. [1]
2.3 GROUND CLEARANCE
This is where the NESC come into play. There has to be enough clearances under the line
so it will not pose any danger to anything underneath the line. Such clearances to
consider are clearances over water surface, ground, roadways, or rails. The NESC
specifies all of these under different types of nominal line voltages. For example, for a
230 KV line, there has to be at least 32.9 ft of clearance for track rails, 24.9 ft for roads
and streets, and 20.9 for spaces and ways for accessible to pedestrians only. Not only is
there vertical clearance, there is also horizontal clearance in order to provide more
5
cushion of safety. Often the conductors that are displaced by wind are given more
clearance than those that are at rest. Some horizontal clearances may include traffic signal
support or structure, buildings, or bridges. Again, all of these rules come from the NESC.
In Table 2.1, a table from RUS Bulletin 1724E-200: Rural Utilities Service, U.S.
Department of Agriculture is provided to give a sense of the clearances for different kinds
roadways, rails, or water surfaces. [2]
Table 2.1 Vertical Clearance of Conductors
6
2.4 SEPARATION BETWEEN CONDUCTORS
Conductors are always assumed to move. On that note, there should always be a
minimum separation between each phase of the conductors. If insulators are free to
swing, that means the conductors are also going to swing since the conductor hangs from
the insulator. The separation between the lines will depend on the spans and sags of the
lines as well as how structures match up with another. The standard separation value
should be on a worst case analysis.
Galloping or “dancing” occurs where the conductors move with very high
amplitudes. This is bad because it can cause contact between phase conductors or the
overhead ground wire, which will cause a major electrical outage. Also, the conductor
themselves may break due to the intense stress, or maybe the structure or towers
themselves may have possible damage. Galloping is caused by stable wind blowing over
the conductors covered by a layer of ice deposited from the freezing rain. There are
measures to take to reduce the possible contact between the conductor phases that are
caused by galloping. For instance, shorter spans, which are the distance from one tower
to another, or increased, phase separation between the conductors. [1]
2.5 INSULATION AND INSULATORS
Insulators are objects that have to be considered when designing transmission lines. An
object made of a material like glass, porcelain or composite polymer that is a poor
conductor of electricity. All of the materials for insulators are handled with epoxy resins,
which acts like reinforcement glue for the materials of the insulators. Insulators are used
7
to attach conductors to the transmission structure and to prevent a short circuit from
happening between the conductor and the structure. For this design suspension type
insulators will be used, since they are used for very high voltage systems is not a practical
or safe to use other types of insulators. Suspension type insulators are usually made of
porcelain that can be stacked in a string and hangs from a cross arm on a tower or pole
that supports the line conductor. Fig 2.2 illustrates suspension type insulators hanging
from a steel lattice tower.
There are also pin-typed insulators to secure the conductor to the insulator. Another type
of insulator is the strain insulator. These types of insulators are used on when the line is
dead-ended. It provides adequate mechanical strength to counterbalance the forces due to
the tension of the conductors and as well as provide insulation. [1][3]
2.6 TOWER STRUCTURE
The type of towers that needs to be used for such transmission lines has to be well
thought of. It has to be just right to meet the requirements of the project. For example, the
height has to be tall enough so it provides that ground clearance that was mentioned
earlier. The towers have to be mechanically strong enough to withstand the weight of the
insulators, the tension of the conductors, or the other possible hazards that may arise from
nature such as a hurricane if installed in the east coast line. For this project a 345 KV line
will be used. A typical structure for this line can be a single pole tower or an H-frame
type tower. Fig 2.3 shows a typical representation of a pole tower and H-Frame.
8
Figure 2.1 Suspension Type Insulators, 345 KV Tower Structure
9
Figure 2.2 Pole Type, 345 KV Tower Structure
Another type of tower structure is the steel lattice tower. These towers are for extra high
voltages. [1] [3]
2.7 SAG AND TENSION
In transmission lines, the behaviors of the hanging conductors are a key variable for the
design. For instance, some factors that will contribute to the sag are the temperature of
the environment, temperature of the conductor itself, the current running through the line,
and the weight of the conductor. These are just the simple factors that will depend on
10
how much the line will sag at each span (distance between each tower). In addition, all
have to be met within the desired NESC regulations. Tension is correlated with the sag.
Tension should not be too high or too low because that will also contribute to the sag.
Having a higher tension will increase the sag and vice versa.
2.8 THERMAL AMPACITY
Transmission lines have the capability of transmitting power, which is limited to the
thermal loading and stability limits. The temperature of the conductor reflects the
thermal loading limit of the line and the real power loss of the line is increased as the
temperature of the conductors rise. The thermal limit is specified to be at 75% of the
current carrying capacity given for a conductor at a temperature of 50° Celsius. This
temperature of 50° Celsius is based on a 25° Celsius air temperature with a 25° Celsius
rise in conductor temperature. In order to find the current carrying capacity, the power of
the line and the rated phase voltage of the selected line should be known. [3]
2.9 LINE PARAMETERS
In modeling a transmission line, the resistance, inductance and capacitance of the line
must be taken into consideration. These parameters will be used to find the series
impedance and shunt admittance of the line which will then be used to find the ABCD
constants that will relate the sending and receiving end of the line.
11
2.10 INDUCTANCE
The inductance of a conductor is caused by the current being carried by the conductor
and the magnetic field around the conductor. As the current changes, the flux changes
and a voltage is induced in the circuit. There is also inductance inside a conductor. For
this design, the skin effect, which is the tendency of the alternating current to dispense
itself into the surface of the conductor making the current greater at the surface than the
center, will be neglected and assume that there is uniform current density throughout the
conductor cross section. The total inductance of the line will be the sum of the internal
and external flux linkages of the conductor.
2.11 CAPACITANCE
The capacitance of a transmission line is determined by the potential differences between
the conductors. The capacitance is a ratio of the charge of the conductor to the potential
difference of the conductor. The bigger the potential difference between the conductors
will result in smaller capacitance of the line and it is conversely the same if the potential
difference was very small. In contrast with the inductor, capacitance is associated with
electric field. The charge on a conductor increases the electric field with radial flux lines.
Overall the total capacitance of the line in a three phase system is the sum of the self and
mutual capacitances.
12
2.12 TRANSPOSITION WITH GROUND WIRE
The transposition of a line allows for a line to have approximately equal parameters. This
allows for the line to be modeled by one single phase to neutral line, with the addition of
an Overhead Ground Wire. Since all three phases occupy the same position for the same
amount of time,
2.13 LONG LINE MODEL
For a long transmission line the magnitude of the voltage over the entire line is not
constant. For relatively small loads or an open circuit condition, the voltage increases
from the sending end to the receiving end. For larger loads or the short circuit condition,
the voltage decreases depending on the amount of load. These increases and decreases in
the line voltage are unwanted. A decrease in voltage means an increase in current. This
would increase the power losses in the line and decrease the lines loading capability. For
the open circuit or small load situation, an increase in voltage would cause the line to be
more prone to arcing.
There are many ways of compensating a line. In this report, full load and no load
compensation will be our primary focus. The open circuit situation has its voltage
compensated with a single shunt reactor applied at the receiving end. For the full load
situation, compensation is achieved with the placement of a series capacitor in the middle
of the transmission line. Again, the idea is to understand how the line can be
compensated for the worst case scenarios up to but not including the thermal ampacity of
the line.
13
2.14 LINE COMPENSATION
For a long transmission line the magnitude of the voltage over the entire line is not
constant. For relatively small loads or an open circuit condition, the voltage increases
from the sending end to the receiving end. For larger loads or the short circuit condition,
the voltage decreases depending on the amount of load. These increases and decreases in
the line voltage are unwanted. A decrease in voltage means an increase in current. This
would increase the power losses in the line and decrease the lines loading capability. For
the open circuit or small load situation, an increase in voltage would cause the line to be
more prone to arcing.
There are many ways of compensating a line. In this report, full load and no load
compensation will be our primary focus. The open circuit situation has its voltage
compensated with a single shunt reactor applied at the receiving end. For the full load
situation, compensation is achieved with the placement of a series capacitor in the
middle of the transmission line. Again, the idea is to understand how the line can be
compensated for the worst case scenarios up to but not including the thermal ampacity
of the line.
14
Chapter 3
MATHEMATICAL MODEL
3.1 THERMAL AMPACITY RATING
The first step in our search is to figure out which lines meet the thermal ampacity rating.
This ampacity is given in Appendix A, which is 75% of the current carrying capacity of
the line given for a conductor temperature of 50° Celsius. From the equation for three
phase apparent power with the receiving end voltage at 345kV,
IR =
SR(3Φ)
3VR(L-L)
=
200 MVA
=334.7 Amps
3×345 KV
(3-1)
3.2 SERIES IMPEDANCE AND SHUNT ADMITTANCE
Next, the series impedance and shunt admittance of the line will be found. These
parameters will then be used to model the line. The calculations will take the effects of
the earth into consideration and the effect of the overhead ground wire. This will greatly
increase the amount of calculations for the compensation of the line.
15
3.2.1 SERIES IMPEDANCE
Series impedance exists on a line due to inductance and resistance. The current through
any conductor develops a magnetic field of proportional magnitude. The energy is then
stored in the magnetic field, and there is opposition to change in currents, meaning the
current through the inductors cannot change rapidly. Each conductor in a three phase
system develops a magnetic field as it carries charging currents for the capacitance
between the wires. Since there is always a return current path present, the magnetic field
created by a changing magnetic field current in the circuit itself induces a voltage in the
same circuit. Since the system is three phase, there will be induced voltage on a line
from a current adjacent to another line.
From Carson’s Earth Return Theory, there flows a current through a given conductor "a"
above the earth and another given conductor "d" in the earth.
By using Carson’s line formula, [5]
VA =(z A +z DD  2z AD )IA V/unit length.
(3-2)
Also knowing that,
z AA =(z A +z DD  2z AD )
(3-3)
The self-impedance of conductor a can be established, by using Newman’s formula
z AA =rA + jωL.
(3-4)
z AA =rA + jωkln
2s
Ω/unit length
DSA
(3-5)
16
“s” is the length of the line. For a 3 phase line, there would be conductors a, b, and c
above the earth. The mutual and self-impedance can be found in a similar way from the
single conductor.
Figure 3.1 Carson’s Return Theory
Self Impedance
z AA =(rA + rD ) + jωkln
De
Ω/unit length
Ds
z BB =(rB + rD ) + jωkln
De
Ω/unit length
Ds
z CC =(rC + rD ) + jωkln
De
Ω/unit length
Ds
z WW =(rW + rD ) + jωkln
De
Ω/unit length
Dsw
(3-6)
17
Mutual Impedance
z AB = z BA = rD + jωkln
De
Ω/unit length
D AB
z BC = z CB = rD + jωkln
De
Ω/unit length
D BC
z CA =z AC = rD + jωkln
De
Ω/unit length
DCA
z AW = z WA = rD + jωkln
De
Ω/unit length
D AW
z BW = z WB = rD + jωkln
De
Ω/unit length
D BW
z CW =z WC =rD +jωkln
De
Ω/unit length
DCW
(3-7)
3.2.1a SELF AND MUTUAL IMPEDANCES WITH EFFECTS OF GROUND
WIRE
The self and mutual impedances are a result of the self and mutual inductance.
The impedance is the sum of the inductance and the resistance. A ground wire “W” will
be added to our 3 phase system.
18
where:
ω= 2πf @ f = 60 hz
k= 0.3219x10-3 mile
De = 2160
ρ
= 2790 ft. @ ρ=100 Ω×m for Avg. soil
f
DS , DSW = GMR of each conductor
D AB , D BC , DCA , D AW , D BW & DCW = distance between conductors
rD = 1.588x10-3f Ω/mile
rA = rB = rC = resistance of lines
Ω/mile
rW = resistance of ground wire Ω/mile
For a three phase line with 1 Overhead Ground Wire, the V=IZ equation becomes,
 VA   z AA
 V  z
 B  =  BA
 VC   z CA

 
 VW =0  z WA
z AB
z AC
z BB
z CB
z WB
z BC
z CC
z WC
z AW   I A 
z BW   I B  Volts
×
mile
z CW   IC 
  
z WW   I W 
(3-8)
19
Figure 3.2 Effects of the Ground Wire
Figure 3.3 Sectioning of Completely Transposed line
20
Since our line is completely transposed, the line can be broken up into three equal
sections as in figure 3.3. Positions 1, 2, and 3 will show the transposing of the line for
the different sections
If the impedances of the line are broken up into the different transposed sections S1, S2,
and S3, as shown above, they can be written as,
Evaluating the impedances separately, the impedance matrices with respect to the
different positions become,
 Z11
Z
ZS1 =  21
 Z31

 Z W1
 Z22
Z
ZS2 =  32
 Z12

 Z W2
 Z33
Z
ZS3 =  13
 Z23

 Z W3
Z12
Z13
Z22
Z23
Z32
Z33
Z W2
Z W3
Z23
Z33
Z21
Z31
Z13
Z11
Z W3
Z W1
Z31
Z32
Z11
Z12
Z21
Z22
Z W1
Z W2
Z1W 
Z2W 
Z3W 

Z WW 
Z2W 
Z3W 
Z1W 

Z WW 
Z3W 
Z1W 
Z2W 

Z WW 
Using kron’s reduction for the different sections of the line to only include the phase
voltages,
21
 Z11
ZS1'=  Z21
 Z31
Z12
Z22
Z32
Z13   Z1W 
 1 
Z23    Z2W  
  Z W1
Z
Z33   Z3W   WW 
Z W2
Z W3 


Z1W Z W1
Z Z
Z Z
) (Z12  1W W2 ) (Z13  1W W3 ) 
 (Z11 
ZWW
ZWW
ZWW 



Z Z
Z Z
Z Z
= (Z21  2W W1 ) (Z22  2W W2 ) (Z23  2W W3 ) 
ZWW
Z WW
Z WW 



Z Z
Z Z
Z Z
 (Z31  3W W1 ) (Z32  3W W2 ) (Z33  3W W3 ) 
ZWW
ZWW
ZWW 

 Z22
ZS2 '=  Z32
 Z12
Z23
Z33
Z13
Z21   Z2W 
 1 
Z31    Z3W  
  Z W2
Z
WW

Z11   Z1W  
Z W3
Z W1 


Z2W Z W2
Z Z
Z Z
) (Z23  2W W3 ) (Z21  2W W1 ) 
(Z22 
Z WW
Z WW
Z WW 



Z Z
Z Z
Z Z
=  (Z32  3W W2 ) (Z33  3W W3 ) (Z31  3W W1 ) 
Z WW
Z WW
Z WW 



Z Z
Z Z
Z Z
 (Z12  1W W2 ) (Z13  1W W3 ) (Z11  1W W1 ) 
Z WW
Z WW
ZWW 

 Z33
ZS3'=  Z13
 Z23
Z31
Z11
Z21
Z32   Z3W 
 1 
Z12    Z1W  
  Z W3
ZWW 

Z22   Z2W 
Z W1
(3-9)
(3-10)
Z W2 


Z3W Z W3
Z Z
Z Z
) (Z31  3W W1 ) (Z32  3W W2 ) 
 (Z33 
Z WW
Z WW
Z WW 



Z Z
Z Z
Z Z
=  (Z13  1W W3 ) (Z11  1W W1 ) (Z12  1W W2 ) 
Z WW
Z WW
Z WW 



Z Z
Z Z
Z Z
(Z23  2W W3 ) (Z21  2W W1 ) (Z22  2W W2 ) 
Z WW
ZWW
ZWW 

(3-11)
22
From,
1
ZABC = (ZS1 +ZS2 +ZS3 )
3
Since the line is completely transposed the diagonal terms of the impedance matrix are
equal. Similarly, the off diagonal elements will be equal to each other, and both can be
represented as,
ZS =
Z Z
Z Z 
1
1 Z Z
 Z11 +Z22 +Z33    1W W1 + 2W W2 + 3W W3 
3
3  ZWW
ZWW
ZWW 
ZM =
Z Z
Z Z 
1
1 Z Z
 Z12 +Z23 +Z31    1W W2 + 2W W3 + 3W W1 
3
3  ZWW
ZWW
ZWW 
(3-12)
Now the impedance matrix of the line can be written as,
 ZS
ZABC =  ZM
 ZM
ZM
ZS
ZM
ZM   zS
ZM  =  z M
ZS   z M
zM
zS
zM
zM 
z M  ×total length of line
zS 
(3-13)
Now that the impedance matrix of the line is found, a single phase must be represented to
continue our analysis.
 VA   ZS
V  = Z
 B  M
 VC   ZM
ZM
ZS
ZM
ZM   I A 
ZM   IB 
ZS   IC 
(3-14)
Writing the phase currents of phases B and C in terms of the current in phase A we have,
IB = (1120°)IA = aIA
IC = (1240°)IA = a 2 IA (3-15)
23
 VA   ZS
V  = Z
 B  M
 VC   ZM
ZM
ZS
ZM
ZM   I A 
ZM   aI A 
ZS  a 2 I A 
(3-16)
Using only phase A to represent the series impedance of the line,
V=I A ZS +aI A ZM +a 2 I A ZM


=I A  ZS + a+a 2 ZM 
=I A  ZS -ZM 
(3-17)
Therefore, the series impedance of the line can be found as,
Z=ZS  ZM
Z Z
Z Z  1
Z Z
Z Z 
1 Z Z
1 Z Z
=Z11   1W W1 + 2W W2 + 3W W3    Z12 +Z23 +Z31    1W W2 + 2W W3 + 3W W1 
3  ZWW
ZWW
ZWW  3
3  ZWW
ZWW
ZWW 
(3-18)
3.2.2 SHUNT ADMITTANCE
The shunt admittance of a line is of the form Y = g +jωC. Where g is the
conductance of the line and is considered negligible. Therefore Y = jωC. For a 3 phase
line, the capacitance is found first by its potential coefficients where,
V
q
 Pq volts
C
and
P  C 1 unit length
Farad
24
In addition to that, the capacitance of the 3 phase line can also be found by using the
image method. This entails that the earth affects the capacitance of a transmission line
because its presence alters the line’s electric field. The presence of the earth’s surface
forces the electric field of the charged conductors to conform. Figure 3.5 below is a
representation of the method of images that is utilized in this analysis.
Variables for Shunt Admittance calculations [5]
q  charge
c  capacitance
RA  RB  RC  RW radius of conductor
H A  H B  H C  HW  distance between conductor and it's image
H AB  distance between conductor A and conductor B's image
DAB  distance between conductor A and conductor B
Figure 3.4 Image Method
25
With this in place, it is desired to obtain the voltages of all the conductors to ground but
first consider that,
1
VA  VAA'
2
1
VB  VBB'
2
1
VC  VCC '
2
Then, the voltages can be determined by,
1
VA  VAA'
2
H AC
H AW
HA
H AB

 q A ln r  qB ln D  qC ln D  qW ln D
1 
A
AB
AC
AW


D
D
r
D
4
 q A ln A  qB ln AB  qC ln AC  qW ln AW

HA
H AB
H AC
H AW






 (3-19)
Then, now by combining the terms,
VA 
H
H 
HA
H
1 
 qB ln AB  qC ln AC  qW ln AW 
 q A ln
2 
rA
DAB
DAC
DAW 
H
H 
H AB
H
1 
 qB ln B  qC ln BC  qW ln BW 
 q A ln
2 
DAB
rB
DBC
DBW 
H AC
H
H
H 
1 
VC 
 qB ln BC  qC ln C  qW ln CW 
 q A ln
2 
DAC
DBC
rC
DCW 
VB 
VW 
H AW
H
H
H 
1 
 qB ln BW  qC ln CW  qW ln W 
 q A ln
2 
DAW
DWB
DCW
rW 
(3-20)
Continuing the process to find the capacitance, these voltages must be related to the
potential coefficients by,
V  Pq
26
Then, the potential coefficient are calculated by,
Pii 
Pij 
1
2
1
2
ln
ln
H i 1
F m Self Potential Coefficient 
ri
H ij
Dij
F 1m  Mutual Potential Coefficient 
Here, the i and j subscripts can represent phases a, b and c. Epsilon, , is the permeability
of air and its value is . Applying that value in for the potential coefficients,
The
1
2
term becomes 18 109 F1m
Then converting this number from meters to mile calls for,
18 109 F1m
 11.185 F1mile
m
1609
mile
Using this to in the following to find the self-potential coefficients of the line with the
OHGW results in,
p AA =11.185ln
HA
MF-1mile
RA
p BB =11.185ln
HB
MF-1mile
RB
pCC =11.185ln
HC
MF-1mile
RC
p WW =11.185ln
HW
MF-1mile
RW
27
Then the mutual potential coefficients between the phase conductors are,
p AB =p BA =11.185ln
H AB
MF-1mile
D AB
p BC =pCB =11.185ln
H BC
MF-1mile
D BC
pCA =p AC =11.185ln
H CA
MF-1mile
DCA
Also the mutual charge coefficients between the OHGW and phases,
p AW =p WA =11.185ln
H AW
MF-1mile
D AW
p BW =p WB =11.185ln
H BW
MF-1mile
D BW
pCW =p WC =11.185ln
H CW
MF-1mile
DCW
The line can then be represented by its potential coefficients with the overhead ground
wire.
 VA   p AA
 V  p
 B  =  BA
 VC   pCA

 
 VW =0  p WA
p AB
p AC
p BB
pCB
p WB
pBC
pCC
p WC
p AW   q A 
pBW   q B 
pCW   q C 
 
p WW  q W 
(3-21)
However, just like the series impedance, the phase values are only needed. Therefore by
Kron’s reduction,

p AW pW A 

 p AA 
pW W 


V A 
V    p  p BW pW A 
 B   BA
pW W 
VC  

pCW pW A 

 pCA 
pW W 


p p 
 p AB  AW W B 
pW W 


p p 
 p BB  BW W B 
pW W 


pCW pW B 
 pCB 

pW W 


p p 
 p AC  AW W C 
pW W 

q A 

p p 
 p BC  BW W C   q B 
pW W    

q 

pCW pW C    C 
 pCC 

pW W  

(3-22)
28
Converting the charge coefficients to capacitances,
CAA
C=P =  CBA
 CCA
-1
CAB
CBB
CCB
CAC 
CBC 
CCC 
The capacitances to ground for the different phases are,
CAG = CAA  CAB  CAC
CBG = CBBCAB  CBC
CCG = CCC  CAC  CBC (3-23)
Since the line is transposed, the capacitance to ground in each phase is the average of the
capacitance to ground of the three phases,
CG =
1
 CAG  CBG  CCG 
3
Now to find the capacitance to neutral, the mutual capacitances of the line has to be
calculated. Since the mutual capacitances are delta connected, they must be converted to
wye,
 Cmutual  (1/ 3)(C AB  CBC  CCA )
CY  3  Cmutual
And the capacitance to neutral,
C  CG  CY
Therfore, Shunt Admittance, Y=jωC
29
3.3
LONG LINE MODEL
Since the lines are so long (150 mi or more), it is not an accurate way to calculate for the
voltage and currents assuming the parameters to be lumped. Various points should be
considered anywhere along the line to fully have the most precise values that are being
desired. Figure 3.6 is used to find the equations for the voltage and current equations.
From Kirchoff’s voltage and current law, Voltage and current can be expressed with
relation between sending and receiving as:
V(x+Δx)=V(x) + zΔxI(x)
V(x+Δx)  V(x)
=zI(x)
Δx
(1)
(3-24)
I(x+Δx)=I(x)+yΔxV(x+Δx)
I(x+Δx)-I(x)
=yV(x+Δx)
Δx
(2)
Taking the limit as x  0 of eqn. (1) and (2),
dV(x) V(x)  V(x) 0
=
=  L'hopital=zI(x)
dx
Δx
0
(3)
dI(x) I(x)  I(x) 0
=
=  L'hopital=yV(x)
dx
Δx
0
(4)
d 2 V(x) dI(x)
=z
dx 2
dx
(5)
30
Figure 3.5 Model of a Long Transmission Line
Differentiating eqn. (3) and plugging in eqn. (5) to (4) gives (6)
d 2 V(x)
= zyV(x)
dx 2
Let γ 2 =zy
d 2 V(x) 2
 γ V(x)
dx 2
(6)
31
Long Line Paramenters:
γ = zy =α + jβ = propogation constant
where z = impedance / unit length
y = admittance / unit length
α = attenuation constant rad/unit length
β = phase constant
rad/unit length
Characteristics impedance:
zC =
z
y
The voltage and current equations are the following.
Therefore: V(x)=A1e γx +A 2e-γx
I(x)=
1
(A1e γx  A 2e-γx )
ZC
Let,
VR +ZC I R
2
V  ZC I R
A2 = R
2
A1 =
e γx +e  γx
e γx  e  γx
V(x)=
VR +ZC
IR
2
2
1 e γx  e  γx
e γx +e  γx
I(x)=
VR +ZC
IR
ZC
2
2
(7)
(8)
(3-25)
32
3.4 ABCD CONSTANTS
From the voltage and current equations, ABCD constants were created and the voltage
and currents were rewritten into a matrix form by,
Rewriting eqns. (7) and (8)
V(x) = coshγxVR +ZCsinhγxI R
I(x) =
1
sinhγxVR +coshγxI R
ZC
Considering in the length of the line, x  l , γ= ZY and ZC =
Z
Y
ABCD Constants:
 VS   A B   VR 
I  =  C D I 
 R 
S  
where:
A=cosh ZY
B=
Z
sinh ZY
Y
1
sinh ZY
D=cosh ZY
Z
Y
A=D and AD-BC=1
C=
3.5
VOLTAGE REGULATION, POWER LOSS AND EFFICIENCY
The voltage regulation can be found using the magnitudes of the receiving end voltages
calculated for the no load and full load conditions. The percent voltage regulation is
defined by the rise in voltage when full load is removed, that is given by,
%Voltage Regulation=
V
R L-L  NO LOAD 
 VR L-L  FULL LOAD
VR L-L  FULL LOAD
 ×100%
(3-26)
33
For calculate of the power loss and efficiency of the line, the receiving and sending end
three phase apparent powers and their respective power factors must be found for the full
load condition by the following, [1]
SS,3  3VSI*S MVA
SR,3  3VR I*R MVA
PFSECD  cos ( Vs   Is )
PFREC  cos (VR  IR )
(3-27)
Therefore the power loss in the line is given as the difference between the 3 phase
powers of the sending end and receiving end,
PLOSS =PS,3  PR,3 MW
The efficiency of the line is the ratio of the receiving end and sending end 3 phase
powers,
η=
PR,3
PS,3
×100%
3.6 LINE COMPENSATION
Compensating for the voltage in a line allows for an improved voltage regulation. This
is why the line must be compensated for its full and no load situations.
34
3.6.1 FULL LOAD COMPENSATION
For the full load condition, a single capacitor was used at the middle of the line to
improve the voltage regulation. Because of the complexity of the calculations for the full
load condition, MATLAB iteration will be used to calculate the full load compensation
vars and then the line variables. Splitting the line down the middle gives the new equation
relating the sending end and receiving end of the line.
 VS   A1
 I  = C
 S  1
B1  1 ZCAP   A 2
D1  0
1   C 2
B2   VR 
D 2   I R  (3-28)
Since the line is split down the middle, the lengths of both ABCD matrices are equal.
Therefore, both matrices are equal and can be represented by the same ABCD constants.
 VS   A
 I  = C
 S 
B  1
D   0
ZCAP   A
1   C
B   VR 
D   I R 
(3-29)
This matrix will then be used in MATLAB plugging in different values for Z.
35
Figure 3.6 Series Compensation of a Long Line
3.6.2 NO LOAD COMPENSATION
When compensating a transmission line for the no load condition, the situation
calls for an inductive compensation using a shunt reactor. To keep the analysis short, a
single inductive reactor was used at the receiving end of the line.
From the equation of the line relating the sending end and the receiving end the first
equation is found.
 VS   A B   VR 
 I  =  C D  I 
 R 
 S 
Vs=AVR +BIR
Relating the shunt reactance to the voltage and current at the receiving end then
substituting into the previous equation we have,
36
IR =
VR
jX Lsh
Vs=AVR +B
VR
jX Lsh
(3-30)
Solving for the reactance,
jX Lsh =
BVR
BVR
=
VS -AVR VS δ-AVR
(3-31)
where δ is the angle of VS
Figure 3.7 No Load Compensation with a Shunt Reactor
Setting the receiving end voltage as the reference, the only other variable other than the
reactance is the angle of the sending end voltage. Since the compensation of the line is
purely reactive, the real power at the receiving end is zero. From the ABCD constants of
the line we solve for the receiving end current.
37
VS =AVR +BI R
VS  AVR
B
changing into its phasor notation,
IR =
(3-32)
IR =
IR =
VS δ  A θ A VR 0
B θ B
VS
B
δ  θ B 
A VR
B
θ A  θ B
From the equation of three phase power the current is substituted,
SR,3 =3VR I*R
*
V

A VR
SR,3 =3 VR 0  S δ  θ B 
θ A  θ B 
 B

B


°
SR,3 =
3 VR VS
B
θ B  δ 
3 A VR
B
(3-33)
2
θ B  θ A
Now only the real power equation is needed to find the angle of the sending end voltage,
38
Therfore, PR,3 =0
PR,3 =
3 VR VS
B
3 VR VS
B
cos  θ B  δ  
cos  θ B  δ  =
3 A VR
3 A VR
B
2
B
2
cos  θ B  θ A  =0
(3-34)
cos  θ B  θ A 
Solving for the sending end voltage angle,
 A VR

δ=θ B  cos -1 
cos  θ B  θ A   (3-35)
 V

S


Now the two equations can be put together and the shunt inductive reactance can be
found by,
jX Lsh =
BVR

 A VR

VS   θ B  cos-1 
cos  θ B  θ A     AVR


 VS


Ohms (3-36)
Also the shunt reactor to be installed must have a Mvar rating that corresponds to the
reactance found where,
3 VR
2
Q =
var
3 X
Lsh
39
Chapter 4
APPLICATION OF THE MATHEMATICAL MODEL
4.1 INTRODUCTION
A load of 200MVA at a lagging power factor of 0.95 will be located 170 miles out from
the point of origin.
A 7 No. 8 Alumoweld Overhead Ground Wire for protection and an
earth resistivity of 100 Ohm meters will be used. Also, a minimum line size of 1/0
ACSR will be understood for structural limitations. Assuming the pre-design
specifications call for a 345kV transposed line, a comparison of the different lines will be
made for their no and full load scenarios. These comparisons will be done based on their
uncompensated & compensated values. In the case of the uncompensated line values, the
voltage rise for the no load situation and the voltage drop for the full load situation will
be calculated by hand.
For the analysis, many assumptions were made. For one, the tower configuration of the
line was randomly chosen based on a standard 345kV tower design. The height of the
lowest conductor was set to 53.83 feet at the towers. Two, the load was assumed to be a
worst case scenario and future growth was already considered. Three, all the mechanical
and structural considerations were not considered in the calculations. Other than the
minimum line size of 1/0 ACSR all else will be neglected. And four, different line types
will be compared from among the different sizes of ACSR conductors. Other types of
conductors will be omitted from our analysis.
40
Figure 4.1 Tower configuration
In preparation of the analysis, first the thermal ampacities must be compared to the
estimated ampacity found for the receiving end voltage at 345 kV and load at 200MVA.
Then the lines that meet or exceed the ampacity rating at 345kV will be compared by
directly connecting their sending end to a 345kV source and comparing their no load and
full load values. From these values, voltage regulation and efficiency will be found for
all the lines in the appropriate range.
41
For the application, only the two extremes of the spectrum will be analyzed and
compensated. There are many ways of compensating a line, but the very basic
methodology will be used. The receiving end voltage will be compensated to a 100% for
both situations which sets the magnitudes of the receiving and sending end voltage equal
to each other. For the open circuit situation, a single shunt reactor will be used at the
receiving end of the line to compensate for the voltage. For the full load situation, a
series capacitor compensation will be used midway of the transmission line. A
MATLAB program will be used to verify the hand calculations that will be done and it
will also, instantly calculate the same parameters obtained from the hand calculations but
for all other lines in the range.
4.2 THERMAL AMPACITY RATING
The first step in our search is to figure out which lines meet the thermal ampacity rating.
From the equation for three phase apparent power with the receiving end voltage at
345kV,
IR =
SR(3 )
3VR(L-L)
200MVA
=
=334.69 Amps
3×345kV
(2.1)
From the ACSR table in appendix B, our current of 334.696 Amps could be handled by a
#1 AWG. But since we are limited to 1/0, we will start our analysis from there.
42
4.3 LINE CHARACTERISTICS AND ABCD CONSTANTS
The next step in our analysis is to model the line. Figure 4.1 shows the tower
configuration of our line where the distance from the ground to the lowest conductor is at
53.83 feet. In considering the effects of ground on the line impedance the earth condition
is of “average damp earth” with a resistivity of 100 Ohm meters.
For our long line, the series impedance and shunt admittance per unit length must be
found first before calculating the ABCD constants of the line. The effects of the earth
will be taken into consideration and the overhead ground wire. The GMD’s between the
conductors and their images are first found.
GMD Between conductors
DAB  DAB  11.52  182  21.36 Feet
D BC  DCB  11.52  182  21.36 Feet
DCA  DAC  23 Feet
D AW  DWA  92  23.52  25.1645 Feet
D BW  DWB  92  452  45.8912 Feet
DCW  DCW  92  56.52  57.2123Feet
GMD Between conductors & their images
H A  2  76.83  153.66 Feet
H B  2  65.03  130.06 Feet
HC  2  53.83  107.66 Feet
HW  2 110.03  220.06 Feet
43
GMD Between conductors & images of other conductors
H AB  H BA  182   76.83  65.03  142.997 FeeT
2
H BC  H CB  182   65.03  53.83  120.215 Feet
2
H CA  H AC  (76.83  53.83)  130.66 Feet
H AW  HWA  92  110.03  76.83  187.077 Feet
2
H BW  HWB  92  110.03  65.03  175.291 Feet
2
H CW  HWC  92  103.03  53.83  164.107 Feet
2
4.3.1 SERIES IMPEDANCE
For a 60Hz line, with a return earth condition for “average damp earth”,
rD  earth resist ance  1.588  10 3 (60 Hz )  0.09528 
and DE  2160

f
mile
 2788.55 feet
Using appendix B for a 1/0 ACSR conductor, the resistances per phase conductor,
rA  rB  rC  r  1.12 
and their self GMR' s
mile
DSA  DSB  DSC  DS  0.00446 feet
For the 7 No. 8 Alumoweld Overhead ground wire,
The self & mutual impedances are than calculated as,
rW  2.44  /mile
D SW  0.002085 Feet  Self GMR 
D AW  DWA  92  23.52  25.1645 Feet
D BW  DWB  92  452  45.8912 Feet
DCW  DWC  92  56.52  57.2123 Feet
44
D 
where Z S  self impedance  (r  rD )  j 0.12134 ln  E 
 DS 
 2788.55 
 (1.12  0.09528)  j 0.12134 ln 

 0.00446 
 1.2153  j1.6194 
mile
and
 Z ABC
 D 
Z M  mutual impedance  rD  j 0.12134 ln  E 
 DEQ 


 2788.55 
 0.09528  j 0.12134 ln 

 21.8932 
 0.09528  j 0.5881 
mile
 1.2153  j1.6194 0.09528  j 0.5881 0.09528  j 0.5881
 0.09528  j 0.5881 1.2153  j1.6194 0.09528  j 0.5881
0.09528  j 0.5881 0.09528  j 0.5881 1.2153  j1.6194 
Now the self and mutual impedances for the ground wire,
 2788.55 
Z SW  self impedance  (2.44  0.09528)  j 0.12134 ln 

 0.002085 
 2.5353  j1.7117
 2788.55 
Z AW  ZWA  mutual impedance  (0.09528)  j 0.12134 ln 

 25.1645 
 0.09528  j 0.57125 
mile
 2788.55 
Z BW  ZWB  mutual impedance  (0.09528)  j 0.12134 ln 

 45.8912 
 0.09528  j 0.49834 
mile
 2788.55 
Z CW  ZWC  mutual impedance  (0.09528)  j 0.12134 ln 

 57.2123 
 0.09528  j 0.47159 
mile
45
Kron’s reduction will be used to find the new self and mutual impedances (zs’ & zm’) in
order to determine the series impedance of the line with the ground wire,
z z
z z 
1
1 z z
 z AA  zBB  zCC    AW WA  BW WB  CW WC 
3
3  zWW
zWW
zWW 
zs ' 
z AA  zBB  zCC , z AW  zWA , zBW  zWB , zCW  zWC
1  z AW 2 zBW 2 zCW 2 
 zs  z AA  



3  zWW zWW zWW 
'
1   0.09528  j 0.57125   0.09528  j 0.49834   0.09528  j 0.47159 
 zs  1.2153  j1.6194   


3  2.5353  j1.7117
2.5353  j1.7117
2.5353  j1.7117

zs '  1.280  j1.533 /mile
2
'
zm ' 
2
2




z z
z z 
1
1 z z
 z AB  zBC  zCA    AW WA  BW WB  CW WC 
3
3  zWW
zWW
zWW 
z AB  zBC  zCA & z AW  zWA , z BW  zWB , zCW  zWC
1 z 2 z 2 z 2 
 zm '  z AB   AW  BW  CW 
3  zWW zWW zWW 
2
2
2
1   0.09528  j 0.6289   0.09528  j 0.5535  0.09528  j 0.5074  
'


 zm   0.09528  j 0.582  


3  2.5353  j1.7117
2.5353  j1.7117
2.5353  j1.7117 


zm '  0.160  j 0.496  /mile
Z s '  zs '  l
Z s '  1.280  j1.533 170
Z s '  217.6  j 260.61 
Z s '  339.5150.14 
Z m '  zm '  l
Z m '   0.160  j 0.496  170
Z m '  27.2  j84.32 
Z m '  88.59972.12 
46
Z series  Z s '  Z m '
Z series   217.6  j 260.61   27.2  j84.32 
Z series  190.4  j176.29 
Z series  259.48142.80 
4.3.2 SHUNT ADMITANCE
The shunt admittance will be found by the image method. Figure 4.1 shows the
configuration of the line with the overhead ground wire being at the top of the tower 14
feet above the highest phase conductor.
The self-potential coefficients of the line with the OHGW are,
153.66
MF 1mile
0.016583
 102.165 MF 1mile
p AA  11.185ln
p AA
130.06
MF 1mile
0.016583
1
 100.3MF mile
pBB  11.185ln
pBB
107.66
MF 1mile
0.016583
 98.1859 MF 1mile
pCC  11.185ln
pCC
220.06
MF 1mile
0.016042
 106.553 MF 1mile
pWW  11.185ln
pWW
The mutual potential coefficients between the phase conductors are,
47
142.997
MF 1mile
21.36
 21.2661 MF 1mile
p AB  pBA  11.185ln
p AB  pBA
120.215
MF 1mile
21.36
 19.325 MF 1mile
pBC  pCB  11.185ln
pBC  pCB
130.66
MF 1mile
23
 19.4295 MF 1mile
pCA  p AC  11.185ln
pCA  p AC
The mutual potential coefficients between the OHGW and phases,
187.077
MF 1mile
25.1645
 22.4381 MF 1mile
p AW  pWA  11.185ln
p AW  pWA
175.291
MF 1mile
45.8912
 14.9898 MF 1mile
pBW  pWB  11.185ln
pBW  pWB
164.107
MF 1mile
57.2123
 11.7862 MF 1mile
,
pCW  pWC  11.185ln
pCW  pWC
48
PAB '  PBA'  PAB 
PAW PWB
PWW
22.438114.9898
106.553
1
 18.1095 MF mile
PAB '  PBA'  21.2661 
PAB '  PBA'
PAC '  PCA'  PAC 
PAW PWC
PWW
22.438111.7862
106.553
1
 16.9475 MF mile
PAC '  PCA'  19.4295 
PAC '  PCA'
PBC '  PCB '  PBC 
PBW PWC
PWW
14.9898 11.7862
106.553
1
 17.6669 MF mile
PBC '  PCB '  19.325 
PBC '  PCB '
From Kron’s Reduction
PAA'  PAA 
PAW PWA
PWW
PAW  PWA & PBW = PWB & PCW  PWC
 PAA'  PAA 
PAW 2
P 2
P 2
& PBB '  PBB  BW & PCC '  PCC  CW
PWW
PWW
PWW
22.43812
106.553
 97.4399 MF 1mile
PAA'  102.165 
PAA'
14.98982
106.553
 98.1912 MF 1mile
PBB '  100.3 
PBB '
11.7862 2
106.553
 96.8822 MF 1mile
PCC '  98.1859 
PCC '
49
Converting the charge coefficients to capacitances
PABC
'
97.4399 18.1095 16.9475 
 18.1095 98.1912 17.6669  MF 1mile
16.9475 17.6669 96.8822 
c  P'
c ABC
1
10.858 1.717 1.586 
  1.717 10.801 1.669  nF
mile
 1.586 1.669 10.904 
c AG  c AA - c AB - c AC  10.858 -  -1.717  -1.586 
c AG  14.161 nF
mile
cBG  cBB - c AB - cBC  10.801-  -1.717  -1.669 
cBG  14.187 nF
mile
cCG  cCC - c AC - cBC  10.904 -  -1.586  -1.669 
cCG  14.159 nF
mile
1
1
 cAG  cBG  cCG   14.161  14.187  14.159 
3
3
cG  14.169 nF
mile
cG 
cmutual  1
3
c AB  cBC  cCA   1
3
1.717  1.669  1.586   1.6573 nF
c  3cmutual  cG  4.972  14.169  9.197 nF
mile
mile
50
Finally the shunt admittance of the line is given as,
y  jc  j (2 )  60  (9.197 109 )  j 3.467 
6
mile
4
Y  j 3.467 10 170  j 5.8939 10 
4.3.3 ABCD CONSTANTS
After calculating the series impedance and shunt admittance from the previous
sections, those quantities will now be used to determine the ABCD constants for the line.
As calculated previously,
Z  190.4  j176.29 
Y  j 5.8939 104 
Using the series impedance and shunt admittance for the 170 mile long line,
A  cosh YZ  cosh
 j5.8939 10   190.4  j176.29 
4
A  0.9480  j 0.0550  0.916.52
B
Z
190.4  j176.29
sinh YZ 
sinh
Y
j 5.8939 104
 j5.8939 10   190.4  j176.29 
4
B  183.49  j176.75  254.77343.9
C
Y
j 5.8939 104
sinh YZ 
sinh
Z
190.4  j176.29
 j5.8939 10   190.4  j176.29 
C  1.0889 105  j5.7918 10 4  5.79 10 4 91.1
D  A  0.9480  j 0.0550  0.916.52
4
51
Therefore, the line can be represented in terms of its ABCD constants with the following
equation:
VS   A B  VR 
 I   C D    I 
  R
 S 
In rectangular form,
0.9480  j 0.0550
VS  
 I    1.0889 105  j 5.7918 104
 S 
183.49  j176.75  VR 

0.9480  j 0.0550   I R 
In polar form,
VS   0.916.52
I   
4

 S  5.79 10 91.1
254.77343.9  VR 
 
0.916.52   I R 
This solution will be used in the upcoming sections to calculate the uncompensated and
compensated line parameters such as voltage regulation, power loss, and efficiency.
However, for the full load compensation, the ABCD constant matrix will not be used
because of the series capacitor installed at the center of the line. A new ABCD constant
matrix will have to be found.
4.4 UNCOMPENSATED LINE
For the uncompensated line, the voltage regulation, efficiency and power loss was found
using the 1/0 ACSR conductor to show the hand calculations [6]. For voltage regulation,
the no load and full load values will be found by setting the receiving end as the
reference and calculations will only be done for the full load situation. The calculations
will be followed by a MATLAB program that will calculate the same values for all the
lines within our range.
52
4.4.1 FULL LOAD
For the full load condition, the receiving end voltage was fixed to 230kV with an
associated angle of 0°. Using the specified apparent power at the receiving end given in
section 4.1, all other values were found.
From,
VRLL  FULL LOAD   3450 kV
S R ,3  3VR I R*
IR 
S R ,3 *
3VR
*
 200  cos

1
0.95  MVA
345kV
3
3
 334.696  18.19 A
Now the sending end values can be calculated using the ABCD matrix of the line.
VS   0.916.52
I   
4

 S  5.79 10 91.1
345000

254.77343.9  
3




0.916.52  334.696  18.19


VS   26328912.6311 V 
 I   300.89410.2805 A

 S 
VSLL  3  VS  456030 kV
53
4.4.2 NO LOAD
For the no load condition, the sending end voltage calculated above was used to
determine the no load receiving end voltage.
VS   A B  VR 
 I   C D    I 
  R
 S 
and
IR  0
VR 
VS
A
Plugging in the values for the 1/0 ACSR line,
VR NO LOAD  
263.28912.6311
 2893296.1111V
0.916.52
and the line to line voltage,
VRLL  NO LOAD   3  2893296.1111   5011336.1111 kV
4.4.3 VOLTAGE REGULATION, POWER LOSS AND EFFICIENCY
Now the voltage regulation can be found using the receiving end voltages calculated for
the no load and full load conditions.
%Voltage Regulation 
V
RL L  NO LOAD 
%Voltage Regulation  45.26%
 VRLL  FULL LOAD 
VRLL  FULL LOAD 
 100%  501.133kV  345kV  100%
345kV
54
To calculate the power loss and efficiency of the line the receiving end three phase
powers are found for the full load condition.
S S ,3  3VS I S*  3  26328912.6311  300.894  10.2805   237.466  j 9.748 MVA

    cos 12.6311  10.2805   0.9991
PFSEND  cos VS   I S
S R ,3  190  j 62.45 MVA
Therefore the power loss in the line,
PLOSS  PS ,3  PR ,3  237.466  190  47.466 MW
And the efficiency of the line is,

PR ,3
PS ,3

190
 100%  80.01%
237.466
4.4.4 MATLAB
A MATLAB code was generated for sections 4.3 and 4.4 which calculate all the voltages,
currents, powers, and line constants. Appendix E,F and G, shows the table for all the
ACSR lines within range. This table was made to compare voltage regulation, power
loss, and efficiency of all the lines. In addition to that, it will be used to compare to the
compensated values calculated in section 4.5. For the 1/0 ACSR line used in the hand
calculations the values are compared to MATLAB in the table 4.2.
Stranding
Is
(Amps)
Sending
End PF
Vs(L-L)
kV
Ps
(MW)
Qs
(Mvar)
Vreg
Ploss
(MW)
Efficiency
MATLAB
1-Jun
306.97
0.998
467.59
248.47
16.68
42
58.047
76.6
THEORETICAL
1-Jun
300.89
0.99
456.03
237.47
9.748
45
47.466
80.01
Kcmil / Awg
Table 4.1 MATLAB VS Theoretical comparison for 1/0 ACSR
55
4.5 COMPENSATION
Now to minimize the power loss and voltage regulation and to maximize the line
efficiency, the no load and full load voltages were compensated for. As with section 4.4,
the calculated sending end voltage will be used to calculate the no load values. However,
the full load calculations were not straight forward so MATLAB was used to generate the
variables.
4.5.1 FULL LOAD COMPENSATION
For the full load condition a series capacitor will be applied halfway through the line to
correct for the voltage drop at the receiving end of the line. There are many different
arrangements using multiple series and/or shunt capacitors, but for our analysis only one
midway will be used.
Since the line is divided in half, the ABCD constants must be calculated for half the
distance. Therefore, the ABCD constants for the two sections must be recalculated with
the line length being,
l
170
 85 miles
2
Y  y  l  j 5.249 106  85
Y  j 4.462 104 
Z  z  l  1.120  j1.037   85
Z=95.195+j88.151=129.74142.80 
56
Figure 4.2 Series Capacitor Compensation
And the new ABCD constant matrix of the line with the series capacitor is,
VS   A B  1 ZCAP   A B  VR 


 I   C D    0
1  C D   I R 
 
 S 
Lines 1/0, 266.8kcmil, 336.4kcmil, 500.0kcmil, and 636.0 were chosen to further
analyze. For these lines, there was an impedance found that minimized the sending end
voltage while fixing the receiving end voltage at 230kV. Appendix D shows the values
found for these lines.
4.5.2
NO LOAD COMPENSATION
For the no load condition a shunt reactor will be used to correct for the voltage rise at
the receiving end of the line. From the analysis done for the full load condition, a
different sending end voltage magnitude was found for the different lines. This voltage
will be used to calculate the receiving end voltage for the no load condition while fixing
57
the voltage regulation to 3%. Many different methods could be used to do the analysis,
but to simplify the calculations further the voltage regulation was held constant.
Voltage Regulation  3%
0.05 
V
RL L  NO LOAD 
 VRLL  FULL LOAD 
VRLL  FULL LOAD 
  V
RL L  NO LOAD 
 345000

345000
VRLL  NO LOAD   1.05  230000  362.250kV
therefore
VR ( NO LOAD )  362.250
3
 209.145kV
Figure 4.3 Shunt Reactor Compensation
This voltage will then be set as the reference and the angle for Vs will be found. Again
using the magnitude of the sending end voltage calculated for the 1/0 line to show
illustrate the procedure,
VR  209.1450 kV
VS  275.452kV
58
Therefore the power angle of the sending end voltage is,
 A VR

δ=θ B -cos -1 
cos  θ B -θ A  
 V

 S

  0.91 209145 
  43.9  cos 1 

275452

cos  43.9-6.52  

  12.7986
The shunt reactance is found by,
jX Shunt 
jX Shunt 
BVR
VS   AVR
254.77343.9   209145 
 275452  12.7986    0.916.52    209145 
jX Shunt  j 464.649 
The Reactive Power is calculated as so,
QR ,3 
3 VR
2
X Shunt

3 209.145kV
2
464.649
QR ,3  282.417 MVar
A three phase shunt reactor rating of up to 282.417 Mvar would be needed to compensate
the line for a 5% voltage regulation. This means the voltage between the magnitudes of
the sending end and receiving end will be equal, but it does not mean that the voltage is
constant throughout the line. In fact the voltage rises from the sending end to
approximately the midway point of the line, and then it begins to decline as it approaches
the receiving end. The rise in the voltage is due to the capacitance of the line therefore,
59
the reactor is used increase the current and the impedance of the line which in turn
decreases the voltage.
60
Chapter 5
SUMMARY
The basis of our analysis was to determine a method of selecting a line based on its
electrical characteristics. These electrical characteristics include but are not limited to
power loss, efficiency, and voltage regulation. However, those three characteristics are
the focus of our analysis. This investigation was made by fixing the full load apparent
power and receiving end voltage to 200MVA at 0.95 lagging power factor and 345kV
respectively. Other parameters that remained constant throughout the analysis were the
tower configuration, line length, and overhead ground wire type. This enabled us keep
the mathematical model as simple as possible without leaving anything out. Once the
lines were modeled, a table was generated for the uncompensated condition to show the
difference in line characteristics as size varies from 1/0 up to 1590kcmil. Then out of the
table, 5 different lines were selected to further investigate its electrical properties.
In solving for the uncompensated condition the full load receiving end voltage was fixed
to 345kV. Then the full load sending end voltage was applied to the no load condition to
calculate the no load receiving end voltage. The analysis showed that for the no load
condition, the voltage drastically increases and for full load condition it decreases. Also,
by varying the line sizes from 1/0 to 1590.0kcmil, as the line size goes up, the full load
sending end voltage goes down while power loss, efficiency, and voltage regulation
improve.
Compensating for the voltage of a transmission line increases its voltage capability. It
also allows for a line to stay within a specified range which is determined by the tower
61
configuration, insulator other hardware ratings. In the compensation section in our
analysis, five different lines were selected to further analyze. These lines were the 1/0,
266.8kcmil 6/7 strand, 336.4kcmil 26/7 strand, 500.0kcmil 30/7 strand, and 636.0kcmil
26/7 strand ACSR conductors. A comparison was first done between the uncompensated
and compensated values found for the different lines. Then a comparison was done
between the 5 lines for the compensated condition.
Now that a basis has been established, the electrical characteristics can then be used to
perform an economic analysis to determine which line to choose. Power loss, power
supplied, and Mvar corrected would be converted to its dollar value and compared
between the different wires and the most economical line will be chosen.
62
Appendix A
OVERHEAD GROUND WIRE CHARACTERISTICS
A.1 ACSR CHARACTERISTICS
63
A.2 ACSR PHYSICAL AND ELECTRICAL CHARACTERISTICS
64
Appendix B
ACSR TABLE
65
Appendix C
WIRE COMPENSATION OF UNCOMPENSATED LINE DATA
MATLAB PRINTOUT
Comparison of ACSR conductors from 1/0 to 1590.0 kcmil for Vr=345kV
Sr=200MVA, PF=0.95 lagging,
Uncompensated Line
66
67
Appendix D
MATLAB CODE OF UNCOMPENSATED LINE
% ACSR Table from 1/0 to 1,590kcmil table B.8
%
GMR/feet
Diameter/inch
Resistance
table=[
0.00394
0.198
3.98
0.00416
0.223
3.18
0.00452
0.257
2.55
0.00401
0.250
2.57
0.00403
0.281
2.07
0.00504
0.325
1.65
0.00418
0.316
1.69
0.00418
0.355
1.38
0.00446
0.398
1.12
0.00510
0.447
0.895
0.00600
0.502
0.723
0.00814
0.563
0.592
0.00684
0.633
0.552
0.02170
0.642
0.385
0.02410
0.700
0.342
0.02300
0.680
0.342
0.02550
0.741
0.306
0.02440
0.721
0.306
0.02780
0.806
0.259
0.02650
0.783
0.259
0.03040
0.883
0.216
0.02900
0.858
0.216
0.03110
0.904
0.206
0.03280
0.953
0.1859
0.03130
0.927
0.1859
0.03270
0.966
0.1720
0.03210
0.953
0.1775
0.03510
1.019
0.1618
0.03350
0.990
0.1618
0.03290
0.977
0.1688
0.03370
1.000
0.1601
0.03720
1.081
0.1442
0.03550
1.051
0.1442
0.03490
1.036
0.1482
0.03930
1.140
0.1288
0.03750
1.108
0.1288
0.03680
1.093
0.1378
0.03860
1.146
0.1228
0.03910
1.162
0.1185
0.04030
1.196
0.1128
68
0.04200
1.246
0.1035
0.04350
1.293
0.0969
0.04500
1.338
0.0906
0.04650
1.382
0.0851
0.04790
1.424
0.0803
0.04930
1.465
0.0760
0.05070
1.506
0.0720
0.05200
1.545
0.0684
];
%CONSTANTS FOR ALL LINES
Length = 170;
%miles
Srmag = 200;
%in MVA
PFrec = 0.95;
%receiving end power factor
rd = 0.09528;
%earth resistance ohms per mile
De = 2788.55;
%in feet
%%% OHGW %%%
rw
= 2.44;
%ohms per mile
Dsw
= 0.002085; %GMR feet
radiusw = 0.016042; %feeT
%%% Distances %%%
Dab = sqrt(11.5^2+18^2);
Dbc = sqrt(11.5^2+18^2);
Dca = 23;
Daw = sqrt(9^2+23.5^2);
Dbw = sqrt(9^2+45^2);
Dcw = sqrt(9^2+56.5^2);
Deq = (Dab*Dbc*Dca)^(1/3);
Ha = 2*76.83;
Hb = 2*65.03;
Hc = 2*53.83;
Hw = 2*110.03;
Hab = sqrt(18^2+(76.83+65.03)^2);
Hbc = sqrt(18^2+(65.03+53.83)^2);
Hca = 76.83+53.83;
Haw = sqrt(9^2+(110.03+76.83)^2);
Hbw = sqrt(9^2+(110.03+65.03)^2);
Hcw = sqrt(9^2+(110.03+53.83)^2);
fprintf('
Is
reactive(Mvar)
PF
Vreg
Vs(L-L)
Ploss
power(MW)
Efficiency \n');
n=1;
for row=1:n
Ds
=table(row,1);
diameter
=table(row,2)/12;
r
=table(row,3);
radius
= diameter/2;
%put in feet
69
%%%%%%%
Calculating Z and Y
%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% series impedance %%%
zaa = r+rd+j*0.12134*log(De/Ds);
zbb = zaa;
zcc = zaa;
zww = rw+rd+j*0.12134*log(De/Dsw);
zab = rd+j*0.12134*log(De/Deq);
zbc = rd+j*0.12134*log(De/Deq);
zca = rd+j*0.12134*log(De/Deq);
zaw = rd+j*0.12134*log(De/Daw);
zbw = rd+j*0.12134*log(De/Dbw);
zcw = rd+j*0.12134*log(De/Dcw);
Zs = (1/3)*(zaa+zbb+zcc)-(1/3)*((zaw*zaw+zbw*zbw+zcw*zcw)/zww);
Zm = (1/3)*(zab+zbc+zca)-(1/3)*((zaw*zaw+zbw*zbw+zcw*zcw)/zww);
z =Zs-Zm;
paa = 11.185*log(Ha/radius);
%%mile/MF
pbb = 11.185*log(Hb/radius);
pcc = 11.185*log(Hc/radius);
pww = 11.185*log(Hw/radiusw);
pab = 11.185*log(Hab/Dab);
pbc = 11.185*log(Hbc/Dbc);
pca = 11.185*log(Hca/Dca);
paw = 11.185*log(Haw/Daw);
pbw = 11.185*log(Hbw/Dbw);
pcw = 11.185*log(Hcw/Dcw);
potential = [
(paa-(paw*paw)/pww) (pab-(paw*pbw)/pww) (pca(pcw*paw)/pww) ;
(pab-(paw*pbw)/pww) (pbb-(pbw*pbw)/pww) (pbc-(pbw*pcw)/pww)
(pca-(pcw*paw)/pww) (pbc-(pbw*pcw)/pww) (pcc(pcw*pcw)/pww)]*1000000;
capacitan = inv(potential); %% F/mile
Ca = capacitan(1,1)-capacitan(1,2)-capacitan(1,3);
Cb = capacitan(2,2)-capacitan(2,1)-capacitan(2,3);
Cc = capacitan(3,3)-capacitan(3,2)-capacitan(3,1);
Cmutual=(1/3)*(capacitan(1,2)+capacitan(2,3)+capacitan(3,1));
Cap = (1/3)*(Ca + Cb + Cc)+(3*Cmutual);
y = j*2*pi*60*Cap;
%%%% In ohms/mile
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Z = z * Length;
Y = y * Length;
%calculating ABCD constants
Zc = sqrt(Z/Y);
A = cosh(sqrt(Z*Y));
B = Zc*sinh(sqrt(Z*Y));
C = (1/Zc)*sinh(sqrt(Z*Y));
D = A;
ABCD = [A B;C D];
70
%%%%%% FULL LOAD VALUES %%%%%
Vr_full = (345000 + j*0)/sqrt(3);
angleS = acos(PFrec);
Sr_3ph = 200000000*cos(angleS)+j*200000000*sin(angleS);
Ir_full = conj(Sr_3ph)/(3*Vr_full);
VsIs_full = ABCD*[Vr_full ; Ir_full];
Vs_full = VsIs_full(1);
Is_full = VsIs_full(2);
%%%%%%% NO LOAD VALUES %%%%%%
Ir_noload = 0;
Vr_noload = Vs_full/A;
%%%%%%% POWER CALCULATIONS %%%%%%
Ss = 3*Vs_full*conj(Is_full);
Sr = 3*Vr_full*conj(Ir_full);
PFsend = cos(angle(Vs_full)-angle(Is_full));
%%%%%% VOLTAGE REGULATION %%%%%
V_reg
=
(Vr_noload - Vr_full)/Vr_full;
Ploss
= real(Ss) - real(Sr);
Efficiency = real(Sr)/real(Ss);
fprintf('%6.3f
', abs(Is_full))
%Amps Sending end
current
fprintf('%4.3f
', PFsend)
%Sending end PF
fprintf('%6.3f
', sqrt(3)*abs(Vs_full)/1000)
%kV Sending end
line to line volatge
fprintf('%6.3f
', real(Ss)/1000000)
%Sending end power in
watts
fprintf('%6.3f
', imag(Ss)/1000000)
%Sending end reactive
power
fprintf('%6.3f
', V_reg)
%Voltage
regulation
fprintf('%6.3f
', Ploss/1000000)
%Power Loss
fprintf('%6.3f
\n', Efficiency)
%Efficiency
end
71
Appendix E
MATLAB CODE, FULL LOAD COMPENSATION
% ACSR Table from 1/0 to 1,590kcmil table B.8
%
GMR/feet
Diameter/inch
Resistance
table=[
0.00394
0.00416
0.00452
0.00401
0.00403
0.00504
0.00418
0.00418
0.00446
0.00510
0.00600
0.00814
0.00684
0.02170
0.02410
0.02300
0.02550
0.02440
0.02780
0.02650
0.03040
0.02900
0.03110
0.03280
0.03130
0.03270
0.03210
0.03510
0.03350
0.03290
0.03370
0.03720
0.03550
0.03490
0.03930
0.03750
0.03680
0.198
0.223
0.257
0.250
0.281
0.325
0.316
0.355
0.398
0.447
0.502
0.563
0.633
0.642
0.700
0.680
0.741
0.721
0.806
0.783
0.883
0.858
0.904
0.953
0.927
0.966
0.953
1.019
0.990
0.977
1.000
1.081
1.051
1.036
1.140
1.108
1.093
3.98
3.18
2.55
2.57
2.07
1.65
1.69
1.38
1.12
0.895
0.723
0.592
0.552
0.385
0.342
0.342
0.306
0.306
0.259
0.259
0.216
0.216
0.206
0.1859
0.1859
0.1720
0.1775
0.1618
0.1618
0.1688
0.1601
0.1442
0.1442
0.1482
0.1288
0.1288
0.1378
72
0.03860
0.03910
0.04030
0.04200
0.04350
0.04500
0.04650
0.04790
0.04930
0.05070
0.05200
];
1.146
1.162
1.196
1.246
1.293
1.338
1.382
1.424
1.465
1.506
1.545
0.1228
0.1185
0.1128
0.1035
0.0969
0.0906
0.0851
0.0803
0.0760
0.0720
0.0684
%CONSTANTS FOR ALL LINES
Length = 170/2;
Srmag = 200;
PFrec = 0.95;
rd = 0.09528;
De = 2788.55;
%miles
%in MVA
%receiving end power factor
%earth resistance ohms per mile
%in feet
%%% OHGW %%%
rw
= 2.44;
%ohms per mile
Dsw
= 0.002085; %GMR feet
radiusw = 0.016042; %feet
%%% Distances %%%
Dab
Dbc
Dca
Daw
Dbw
Dcw
=
=
=
=
=
=
sqrt(11.5^2+18^2);
sqrt(11.5^2+18^2);
23;
sqrt(9^2+23.5^2);
sqrt(9^2+45^2);
sqrt(9^2+56.5^2);
Deq = (Dab*Dbc*Dca)^(1/3);
Ha
Hb
Hc
Hw
=
=
=
=
2*76.83;
2*65.03;
2*53.83;
2*110.03;
73
Hab
Hbc
Hca
Haw
Hbw
Hcw
=
=
=
=
=
=
sqrt(18^2+(76.83+65.03)^2);
sqrt(18^2+(65.03+53.83)^2);
76.83+53.83;
sqrt(9^2+(110.03+76.83)^2);
sqrt(9^2+(110.03+65.03)^2);
sqrt(9^2+(110.03+53.83)^2);
fprintf('
Is
reactive(Mvar)
#
Ploss
PF
Vs(L-L)
Efficiency \n');
power(MW)
n=20;
for row=20:n
Ds
=table(row,1);
diameter
=table(row,2)/12;
r
=table(row,3);
radius
= diameter/2;
%%%%%%%
Calculating Z and Y
%put in feet
%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% series impedance %%%
zaa
zbb
zcc
zww
=
=
=
=
r+rd+j*0.12134*log(De/Ds);
zaa;
zaa;
rw+rd+j*0.12134*log(De/Dsw);
zab = rd+j*0.12134*log(De/Dab);
zbc = rd+j*0.12134*log(De/Dbc);
zca = rd+j*0.12134*log(De/Dca);
zaw = rd+j*0.12134*log(De/Daw);
zbw = rd+j*0.12134*log(De/Dbw);
zcw = rd+j*0.12134*log(De/Dcw);
Zs
Zm
= (1/3)*(zaa+zbb+zcc)-(1/3)*((zaw*zaw+zbw*zbw+zcw*zcw)/zww);
= (1/3)*(zab+zbc+zca)-(1/3)*((zaw*zbw+zbw*zcw+zca*zaw)/zww);
z =Zs-Zm;
%%% shunt admittance variables %%%
74
paa
pbb
pcc
pww
=
=
=
=
11.185*log(Ha/radius);
11.185*log(Hb/radius);
11.185*log(Hc/radius);
11.185*log(Hw/radiusw);
%%mile/MF
pab = 11.185*log(Hab/Dab);
pbc = 11.185*log(Hbc/Dbc);
pca = 11.185*log(Hca/Dca);
paw = 11.185*log(Haw/Daw);
pbw = 11.185*log(Hbw/Dbw);
pcw = 11.185*log(Hcw/Dcw);
potential = [
(paa-(paw*paw)/pww) (pab-(paw*pbw)/pww) (pca(pcw*paw)/pww) ;
(pab-(paw*pbw)/pww) (pbb-(pbw*pbw)/pww) (pbc-(pbw*pcw)/pww) ;
(pca-(pcw*paw)/pww) (pbc-(pbw*pcw)/pww) (pcc(pcw*pcw)/pww)]*1000000;
capacitan = inv(potential); %% F/mile
Ca = capacitan(1,1)-capacitan(1,2)-capacitan(1,3);
Cb = capacitan(2,2)-capacitan(2,1)-capacitan(2,3);
Cc = capacitan(3,3)-capacitan(3,2)-capacitan(3,1);
Cap = (1/3)*(Ca + Cb + Cc);
y = j*2*pi*60*Cap;
%%%% In ohms/mile
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Z = z * Length;
Y = y * Length;
%calculating ABCD constants
Zc = sqrt(Z/Y);
A = cosh(sqrt(Z*Y));
B = Zc*sinh(sqrt(Z*Y));
C = (1/Zc)*sinh(sqrt(Z*Y));
D = A;
ABCD = [A
k=0;
B;C
D];
for Z_cap = 84:84
75
ABCD_new = ABCD*[1 -j*Z_cap ;0 1]*ABCD;
%%%%%% FULL LOAD VALUES %%%%%
Vr_full =
angleS =
Sr_3ph =
(345000 + j*0)/sqrt(3);
acos(PFrec);
200000000*cos(angleS)+j*200000000*sin(angleS);
Ir_full =
conj(Sr_3ph)/(3*Vr_full);
VsIs_full = ABCD_new*[Vr_full ; Ir_full];
Vs_full =
Is_full =
VsIs_full(1);
VsIs_full(2);
%%%%%%% POWER CALCULATIONS %%%%%%
Ss = 3*Vs_full*conj(Is_full);
Sr = 3*Vr_full*conj(Ir_full);
PFsend = cos(angle(Vs_full)-angle(Is_full));
Ploss
= real(Ss) - real(Sr);
Efficiency = real(Sr)/real(Ss);
fprintf('%6.3f %d
', abs(Is_full),k)
%Amps Sending end
current
fprintf('%4.3f
', PFsend)
%Sending end PF
fprintf('%8.5f
', sqrt(3)*abs(Vs_full)/1000)
%kV Sending end
line to line volatge
fprintf('%6.3f
', real(Ss)/1000000)
%Sending end power in watts
fprintf('%6.3f
', imag(Ss)/1000000)
%Sending end reactive power
fprintf('%6.3f
', Ploss/1000000)
%Power loss
fprintf('%6.3f
', Efficiency)
%Efficiency
fprintf('%8.8f
\n', 1/(Z_cap*2*pi*60)) %capacitance
k=k+1;
end
end
76
Appendix F
MATLAB Code, No load compensation
%
GMR/feet
table=[
0.00446
];
Diameter/inch
0.398
1.12
Resistance
275452
%CONSTANTS FOR ALL LINES
Length = 170;
%miles
Vrmag = 1.03*345000/sqrt(3);
%for 3% voltage regulation
rd = 0.09528;
%earth resistance ohms per mile
De = 2788.55;
%in feet
%%% OHGW %%%
rw
= 2.44;
%ohms per mile
Dsw
= 0.002085; %GMR feet
radiusw = 0.016042; %feet
%%% Distances %%%
Dab = sqrt(11.5^2+18^2);
Dbc = sqrt(11.5^2+18^2);
Dca = 23;
Daw = sqrt(9^2+23.5^2);
Dbw = sqrt(9^2+45^2);
Dcw = sqrt(9^2+56.5^2);
Deq = (Dab*Dbc*Dca)^(1/3);
Ha = 2*76.83;
Hb = 2*65.03;
Hc = 2*53.83;
Hw = 2*110.03;
Hab = sqrt(18^2+(76.83+65.03)^2);
Hbc = sqrt(18^2+(65.03+53.83)^2);
Hca = 76.83+53.83;
Haw = sqrt(9^2+(110.03+76.83)^2);
Hbw = sqrt(9^2+(110.03+65.03)^2);
Hcw = sqrt(9^2+(110.03+53.83)^2);
fprintf('Xshunt
Is
Vs(L-N)
n=1;
for row=1:n
inductance
Qshunt
sendpower(MW)
reactive(Mvar) \n');
77
Ds
diameter
r
Vsmag
radius
%%%%%%%
=table(row,1);
=table(row,2)/12;
=table(row,3);
=table(row,4);
= diameter/2;
Calculating Z and Y
%put in feet
%%%%%%%%%
%%% series impedance %%%
zaa
zbb
zcc
zww
=
=
=
=
r+rd+j*0.12134*log(De/Ds);
zaa;
zaa;
rw+rd+j*0.12134*log(De/Dsw);
zab = rd+j*0.12134*log(De/Dab);
zbc = rd+j*0.12134*log(De/Dbc);
zca = rd+j*0.12134*log(De/Dca);
zaw
zbw
zcw
Zs
Zm
=
=
=
=
=
rd+j*0.12134*log(De/Daw);
rd+j*0.12134*log(De/Dbw);
rd+j*0.12134*log(De/Dcw);
(1/3)*(zaa+zbb+zcc)-(1/3)*((zaw*zaw+zbw*zbw+zcw*zcw)/zww);
(1/3)*(zab+zbc+zca)-(1/3)*((zaw*zbw+zbw*zcw+zca*zaw)/zww);
z =Zs-Zm;
%%% shunt admittance variables %%%
paa
pbb
pcc
pww
=
=
=
=
11.185*log(Ha/radius);
11.185*log(Hb/radius);
11.185*log(Hc/radius);
11.185*log(Hw/radiusw);
%%mile/MF
pab = 11.185*log(Hab/Dab);
pbc = 11.185*log(Hbc/Dbc);
pca = 11.185*log(Hca/Dca);
paw = 11.185*log(Haw/Daw);
pbw = 11.185*log(Hbw/Dbw);
pcw = 11.185*log(Hcw/Dcw);
potential = [
(paa-(paw*paw)/pww) (pab-(paw*pbw)/pww) (pca(pcw*paw)/pww) ;
(pab-(paw*pbw)/pww) (pbb-(pbw*pbw)/pww) (pbc-(pbw*pcw)/pww) ;
(pca-(pcw*paw)/pww) (pbc-(pbw*pcw)/pww) (pcc(pcw*pcw)/pww)]*1000000;
capacitan = inv(potential); %% F/mile
78
Ca = capacitan(1,1)-capacitan(1,2)-capacitan(1,3);
Cb = capacitan(2,2)-capacitan(2,1)-capacitan(2,3);
Cc = capacitan(3,3)-capacitan(3,2)-capacitan(3,1);
Cap = (1/3)*(Ca + Cb + Cc);
y = j*2*pi*60*Cap;
%%%% In ohms/mile
%%%% Calculating ABCD Constants %%%%%
Z = z * Length;
Y = y * Length;
Zc = sqrt(Z/Y);
A = cosh(sqrt(Z*Y));
B = Zc*sinh(sqrt(Z*Y));
C = (1/Zc)*sinh(sqrt(Z*Y));
D = A;
ABCD = [A B;C D];
%%%%% SHUNT COMPENSATION %%%%%%
Vr
= Vrmag + j*0;
K
= (abs(A)*Vrmag)/Vsmag;
delta
= angle(B)-acos(K*cos(angle(B)-angle(A)));
Vs_calc = Vsmag*cos(delta)+j*Vsmag*sin(delta);
X_shunt = B*Vr/(Vs_calc-A*Vr);
%calculating the voltages and currents
Ir = Vr/X_shunt;
VsIs = ABCD * [Vr ; Ir];
Vs = VsIs(1);
Is = VsIs(2);
Ismag = abs(Is);
Sr = (3*abs(Vr)^2)/conj(X_shunt);
%Sending end PF,S,P,3-phasevoltage
PFsend = cos(angle(Vs_full)-angle(Is_full));
Ss=3*Vs*conj(Is);
Vs3ph = sqrt(3) * abs(Vs);
fprintf('%6.3f
',imag(X_shunt))
%single phase shunt
reactance
fprintf('%6.3f
',imag(X_shunt)/(2*pi))
fprintf('%6.3f
',imag(Sr)/1000000) %3-phase power compensation
fprintf('%6.3f
', Ismag)
%Amps Sending end
current
fprintf('%6.3f
', sqrt(3)*Vs/1000)
%kV Sending end line to
line voltage
fprintf('%6.3f
', real(Ss)/1000000)
%3-phase Sending end
power
fprintf('%6.3f
', imag(Ss)/1000000)
%3-phase Sending end
reactive power
fprintf('%6.3f
', Ploss/1000000)
%Power loss
fprintf('%6.3f
', Efficiency)
%Efficiency
fprintf('%4.3f
\n', PFsend)
end
79
BIBLIOGRAPHY
[1] Gonen, T. 2009.Electrical power transmission system engineering analysis and
design, 2nd ed. Florida: CRC Press.
[2] Gonen, T. 1986.Electrical power distribution system engineering. New York:
McGraw-Hill.
[3] Sadaat, H. 2002.Power system analysis, 2nded. New York: McGraw-Hill.
[4] Granger, J. J. and W. D. Stevenson.1994. Power system analysis. New York:
McGraw-Hill.
[5] SMUD. 1995. UG/OH Conductor sequence impedance study.
[6] Gonen, T. 1988. Modern power system analysis. New York: McGraw-Hill.