Modeling and Analysisof

Modern Power Systems:

The Fundamentals ofGenCo

TransCoDistCo

&PoolCo

1996 Dr. Mo-Shing Chen

TABLE OF CONTENTS

Pref ace ............ : ........................................................................ .

Otaptcr 1 Introduction ...................................................................... 1-1

Oiaptcr 2 Ca1culation of Une Constants and an Introduction to Symmetrical-Component Modeling ................ ........................... 2-1

Oiaptcr 3 Equivalent Circuit and Operating Clw'acteristics of a Synchronous Machine ........................................................ .. 3-1

Olapter 4 Mcxleling of Th.rec-Phase Transformcn .................................... . 4-1

Olapter 5 Modeling of Power System Loads ........................................... . 5-1

Oiapter 6 Reactive Power ................................................................ .. 6-1

Chapter 7 Transmission Power Flow Analysis ......................................... .

Olapter 8 Distribution NctWork. Analysis ............................................... .

7-1

-1 8-1

Oiaptcr 9 Power System Transient Stability Analysis ................................ .. 9-1

Olaptcr 10 Povrer System Dynamic Stability Analysis .................................. 10-1

Olaptcr 11 Power System Voltage Stability ................ ................. .............. 11-1

Olaptcr 12 Power System Fault Calculation ... ... . . .. . . . . .. . . . . . . . . . . . .. . . . .. . . .. . . .. . . 12-1

I

PREFACE

This text was developed from a set of course material used for a course tiLled Modeling and

Analysis of Modern Power Systems which was conducted by the Energy Systems Research Cen~ (ESRC) at the University of Texas at Arlington since 1969. It includes the professional staffs' scientific and technical experience in electric power system analysis, design and operations along

with up-to-date and first hand material related to the deregulation of power systems. The power

industry worldwide is going through significant change due to the advent of deregulation. This

course, Modeling and Analysis of Modern Power Systems: The Fundamentals for GenCo, Transco, DistCo, and PoolCo, is designed for the power industry from the point of view of

unbundling the generation, transmission, and distribution functions. 1be material covered in this

course includes the basic responsibilities of GenCo, TransCo, DistCo, and PoolCo along with the

other fundamental theoretical background required for power systems engineers. Of course, some

e of the responsibilities are not very clear because many changes will occur depending on the social and political conditions within various countries. It is believed that this is the first text designed to

discuss power industry deregulation issues. However, it is not the intent to cover everything in

great detail, nevertheless it includes discussion on the state-of-the-art of today's power industry.

It is strongly believed that deregulation will have profound and important implications on

technology within the electric power industry and its institutional structure. Not only must the

power industry as whole re-evaluate its planning for the future, but the delivery systems for power

engineering education need to up-date curriculum and teaching sttategies in order to prepare

engineers for the challenges presented by deregulation .

CHAPTER l

1.0 INTRODUCTION

Electric power systems in many countries are structured in a single vertically

integrated company, which provide electric power based on cost-of-service. However in recent years deregulation has became accepted by many countries in the world. Hence, utilities in many countries are changing from monopolies to a free and competitive market

type industry. The first primary requirement for deregulation is the separation of the entire

power system into a Generating Company (Genco.), a Transmission Company (Transco.), a Distribution Company (DistCo.) and possibly a Power Pool Company (PoolCo.). The role and responsibility of each company must be clearly defined.

Considering the generating systems, many new, small and competitive generation units may be

introduced into utility systems with varying measures of reliability, cost, and environmental

. impacts. The major immediate effect of this sector will be the development of more advanced power generation units with much higher efficiency than that presently exist, along with

efficient operation of generating units. Obviously deregulation will provide an opportunity for

non-utility ownership of GenCo, and the end results will be an open door for competition.

Since the price of electricity fluctuates from day-time-rate to night-time-rate and seasonally

from summer to winter, it is expected that electricity will become as a commodity; and free

competition will make the consumer feel that they are paying a getting fair price.

1-l

---~--- - ~ . - --~- ------- ----- -----

The transmission company will provide the wheeling of electric energy from the GenCo. to the

DistCo. The Transco will be responsible for security of the electric power system and will be

motivated to use advanced technology in their power delivery in order to ensure secure and economic operation.

The DistCo will offer different kinds of quality of service to its customers. The customers will

have a choice in purchasing electricity with different reliability levels. If a customer requires a

high degree of reliability for electric power service, pay a premium price will be charged, on .

the other hand if a customer requires a low degree of reliability, a discounted price will be

charged. Therefore based on the customers need, the level of reliability requirements in the power system will be determined. Because of this condition, the amount of the required

reserve will be defined by the distribution system rather than the generation or transmission

company. Thus, the amount of required reserve will be different for each area. Furthermore,

distribution management assistance must be provided to make the efficiency of the total

distribution system much better than at present.

The emphasis of the distribution company will be on demand side management. If a distribution company manages well it will be able to purchase electricity from a generation

company at a very attractive and fair price. For instance if a distribution company has a very sophisticated metering system, its tariff structure may vary a great deal from off-peak to on-

peak time. If a distribution company can achieve an overall flat power demand both day and night then it may negotiate a much lower rate from the generating company. Distribution

companies with a sophisticated demand side management program along with its customers'

desire to purchase a lower degree of reliability, may use its own load as a reserve.

Hence, after deregulation it is expected that the DistCo will negotiate directly with the

Genco. to purchase electric energy and with the Transco. for wheeling electric energy.

1-2

Electric power companies even today are willing to offer their consumers many incentives so

the whole power system could be operated much more efficient1y than previous1y. For

instance if consumers desire to use more electricity during off-peak time, they will not only pay a lower rate for electricity but also will receive credit similar to a frequent advantage mile

'

offered by the air line industry. If the power factors of the consumer are maintained near unity during the entire year, they may qualify as a preferred customer and may receive other benefits

from the utility company.

It is clear that, without traditional protection some utility companies in the world will either

go bankrupt or merge. Also, it is expected that without protection the low efficiency power

plant will soon be closed down. Obviously the Nuclear power industry must compete with

other resources of power on an equal basis. Therefore a base load power plant will not

receive the same rate as the load following units. It is expected that power storage systems

may soon be very attractive options. Further more, power brokers will serve as a cushion and

provide a convenient linkage between the power producer and consumer. The transition from

a regulated industry to a deregulated industry will produce uncertainty and confusion. It is

not expected that the consumers will see and enjoy the immediate benefits. However, in the long term it is believed the consumer will be able to purchase electricity at lower rates.

Today utilities in Japan charge their customers 25 cents per KW, the highest in the world.

This gives US companies an opportunity to compete in Japan and export electricity to Japan.

It could help in balancing the huge trade deficit to the US, and enable the US electric

companies to stay in the market strongly and deal with the challenge. However, in many

countries in the world electricity is produced by government owned companies. Therefore,

these nationalized systems are even more of a monopoly then a regulated monopo1y in the private utility sector. It is clear that there exists a definite opportunity for the utility industry to

explore both the domestic and international markets. Actually this trend has already begun by

major utilities in the US and they are already doing utility business outside of the US or at least considering doing business outside of the US.

1-3

This text is developed from a set of course material used in the short course titled "Modeling

and Analysis of Modern Power Systems" conducted by Energy Systems Research Center

(ESRC) at the University of Texas at Arlington since 1969. The material covered in this text '

book provide fundamental theoretical and technical background for understanding the changes

required for the future structuring of GenCo, Transco, and DistCo.

The Fundamentals for GenCo., Transco .. DistCo., and poolCo., are designed for the power

industry from the unbundling of generation, transmission, and distribution point of view. The

material covered in this text will include the basic responsibilities of a GenCo, Transco,

DistCo, and a PoolCo., along with the other fundamental theoretical background required for

power systems engineers. Of course, some of the responsibilities are not very clear because

many changes occur depending on the social and political conditions within the various

countries. It is believed that this is the first text designed to address to the changes related to

deregulation of the power industry. It is not the intent to cover everything in great detail,

however, it is planned to discuss the state-of-the art in today's power industry and the

expected power industry requirements to deal with the challenge.

It is strongly believed that deregulation will have profound and important implications on

technology within the electric power industry and the institutional structure. Therefor not

only must the power industry as whole re-evaluate its future planning, but the power

engineering educational institutions need to up date the educational curriculum and strategy in

order to prepare engineers for the challenge presented by deregulation.

1-4

CHAPTER2

CALCULATION OF LINE CONSTANTS AND AN INTRODUCTION TO SYMMETRICAL-COMPONENT MODELING

2.0 Introduction

Electricity is one of the most commonly used forms of energy in industry, residences, commerce and transportation, because of its flexibility, convenience and efficiency of delivery and utilization. Due to some limitations, transferring electric power over a transmission line is more complicated than any other form of energy. These limitations include thermal limits, transient stability, dynanuc stability and voltage stability that are determined by transmission line parameters. Flow of electric energy from one point to another is controlled by Transco.

Since load demands are varied hourly, daily, weekly, and even seasonally, the most challenging problem for Transco is how to coordinate and wheel energy flow in the most efficient, secure and competitive manner.

Since transmission line characteristics play an important role in the analytical simulation of electric power systems, before studying any other fonnal power system analysis, this chapter will discuss the fundamental characteristics of three phase transmission lines.

Actually one can use the analogy that transmission lines are similar to interstate highways and distribution lines are similar to country roads. Where bulk power is delivered over the transmission system and then through distribution lines to each consumer. The engineers at the TransCO are expected to predict and estimate the real and reactive power that is to be carried on each segment of the transmission line in the network. Then by utilization of suitable hardware and software they should be able to control and wheel the energy flow in the transmission line in a most desired manner.

2.1 The Power Transmission Line

Transmission lines are one of the major components in every electric power system that are used to deliver the bulk power from power generating plants to various load centers. Generally, both overhead and underground transmission lines are currently in use to

2-1

transmit electricity. The underground line offers the advantages of safety and aesthetic appeal. Offsetting these advantages are its highly specialized skills' requirement for both installation and repair as well as higher costs of installation compared to overhead transmission lines. In addition, the capacities of underground transmission lines are limited by their initial installation, and in order to increase their power carrying capability additional lines are built either in parallel with the original one or to replace the old ope. Therefo' .. undi'' "round transmission lines usually are installed only in high density populated urban areas. In this chapter only overhead transmission lines' parameters, particularly inductance and capacitance are discussed and their associated models are developed. Since there is no single model that can accurately represent various types of underground cables, models are customarily obtained from the cable manufacturers.

The basic overhead transmission line consists of a group of phase conductors that transmit the electric energy, the earth return, and usually one or more neutral conductors which are nominally at zero (or ground) potential.

In electric power networks, the transmission and distribution systems from both an operational and functional point of view is actually one interconnected system, but based upon operating voltage level, the high voltage lines are called transmission lines (interstate powe1 highway system) and low voltage lines are called distribution lines. The line-to-line voltage level between pairs of phase conductors may range from 12.5 KV or less in the distribution system to 138 KV, 345 KV, 500 KV or even higher in the transmission network. In Texas, the backbone of the transmission system is 345 KV.

Transmission and primary distribution circuits invariably involve all three of the phase conductors, while secondary distribution circuits may carry only one or two of the phases. In this chapter primary concern is with the analysis of three-phase systems; in Chapter 8 will deal with distribution network that may consist of only one or two of the phases.

Compared to any other component in electric power systems, like generators or transformers, the three-phase transmission lines seem to be simpler, but the analysis is complicated due to the existence of inductive and capacitive coupling between conductors and between conductors and the ground. The calculation of the inductance, capacitance, and resistance associated with a particular line geometry (e.g., wire size and configuration) involves the solution of field-theory problems. Equations for the distributed parameters of the line may be readily derived and suitably modified to account for the presence of the earth in the practical line. Once the series and shunt parameters have been calculated, it is ---......._

2-2

..

more meaningful to consider the line as a circuit element whose electrical behavior can be described in terms of a lumped-parameter model such as the 7t-model of the form shown in Figure 2.1. l. The development and analysis of an equivalent lumped-parameter model to represent an electric power transmission line is the main objective of the present chapter.

R L

IC/2 C/2I

Figure 2.1. l Lumped-parameter n:-model of line

As mentioned, the conductors of the three-phase line with earth return are inductively and capacitively coupled to each other and to the ground. The typical lumped-parameter 7t-model of a segment of three phase lines with coupling included is shown in Figure 2.1.2, where the fictitious conductor nn' is included to account for the resistive and inductive effects of the earth. Obviously, perfonning any circuit analysis on this model would require considerable effort. Where, if the series and shunt parameters are examined independently and the results are superimposed, relatively little error will incur. In many applications. specially for short transmission lines it is even permissible to neglect the shunt capacitance entirely. ln the following, we are concerned only with the series impedances of the lines and their representations, then later on the effect of shunt capacitance will be discussed.

A section of this chapter is devoted to the special representation of the transmission lines by symmetrical components, where particular emphases are given to the syni.metrical-component representation of the completely transposed three-phase line. Both the mathematical background and the physical significance of the symmetrical component method are discussed in detail; however an extensive treatment of symmetrical-component modeling will cover in a later chapter.

2-3

C~n Ctn Can 2 2 2

.JII Lan

Rn Ln '\/V'v,--~~~-----.~ III~.

Figure 2.1.2 Lumped-parameter 1t-model of a three-phase line

2.2 Self -inductance of a Long .filamentary Conductor

Before discussing the inductance associated with an actual' three phase transmission line, it is helpful to examine the electromagnetic behavior of a single conductor and pairs of conductors in space. Consider a thin circular conductor of length t and radius r, where r

Figure 2.2. l Pertaining to the derivation of the inductance of a single filamentary conductor

With reference to Figure 2.2.1, the Biot-Savart law may be written as

dH = lsin0 dy p 4itc2

2.2. l

But in Figure 2.2.1 it can be seen that

and so

=--.......,..--

4it(a2 +(y-b)2 r2 2.2.2

The total magnetic field intensity produced at point P as a result of the current I flowing in the line may then be found by integrating dHp over the entire length of line t, or

2.2.3

=- [ l-b + b ] 4na ~a2+(l-b)2 ~a2+b2 The differential magnetic flux d{j> passing through the differential area tdx is found by integrating the expression for H along this differential strip and multiplying by the

2-5

' ,' ,~ .

differential length dx of the strip. Since the medium surrounding the line is as~ ,med to be air, we can assume that B = 0H, and so

d= 0dxfHpdy=.&!.dxf[ e-b ,+ b ,]db o 4 1ta o ~a2 +(l-b)" ~a 2 +b- 2.2.4 = 2oldx [~a2 + e2 -a]= ol [[T;if -l]dx

41ta 21t a

The total flux outside the wire may then be found by integrating the differential flux crossing this infinitesimal strip between the limits of r and oo. Since this integration is performed in the x-direction, the variable a is replaced by x, resulting in

2.2.5 OQ

= 1::2!. [~ x 2 + e2 - x - e In e+,,J;.2;li] 21t x = ol [t In e+..['(2;;2 - ~ 2 + r2 + r] 2 tt r 2.2.6

Since r

Where is the magnetic permeability of the conductor material. Since most practical power transmission lines consist predominantly of nonferrous conductors, ~ 0 for the interiors of the conductors as well as for the medium surrounding them, and therefore

2.2.10

The total inductance associated with this single conductor in space may then be written as

L = L + L =of [1nlf - 1]+ ot = oe [tnlf - .J.] ext mt 2n r 81t 21t r 4 2.2.l l

2.3 Mutual Inductance of Two Parallel Filamentary Conductors

Having developed expressions for the magnetic flux resulting from the flow of current in a single long conductor, it is easy to find an expression for the mutual inductance between two such conductors. The mutual inductance M between two coupled circuits i and j in regions of constant permeability is defined as

2.3.1

i.e., the mutual inductance between two conductors is the ratio of the resulting change in the flux linkage of one conductor to the change in excitation current in the other. The flux associated with this change may be found by integrating Equation 2.2.4 between the limits of d and oo, where d is the spacing between conductors, to obtain

4> = i~ I[ ~x2x+f2 - l]dx 2.3.2

= ol [ e In e+~ e2+d2 -21t d ~e2 +d2 +ct]

from which

M = ~[ e1n l+Je2+d2 - +ct] ~e2 +d2 21t d 2.3.3

Recalling that r

2.3.4

Equation 2.3.5 is sufficiently accurate for most practical transmission line problems and is therefore used in the derivation of other equations within this chapter.

2.4 Self-Inductance of a Two-Line Return Circuit

For a simple example of the application of the equations developed in Sections 2.2 and 2.3, consider the two-line return circuit, or loop, shown in Figure 2.4. l. The circuit consists of two parallel conductors, each of length l and radius r which are separated by a distance d and are joined at one end to form a loop. For this circuit r

bilaterally symmetrical, L1 = L2 = L and M 12 = M21 = M, and so the total inductance of the loop becomes

L =2L-2M 2.4.2

Substitution of the approximate expressions for self- and mutual-inductance (Equations 2.2. l l and 2.3.S, respectively) into Equation 2.4.2 yields

L = 2{i~ [in 2/-:t]}-2{i~ [in 2J - 1]} 2.4.3

= ol [tnlf-tnlf-1+ 1] 1t r d 4

= ol [1n .4. + .l] 1t r 4

Equation 2.4.3 is more directly applicable to paired conductors such as telephone loops and control cable than to power transmission lines. Nevertheless, the results of this section have several important practical meanings for the power system engineers. First, by developing Equation 2.4.3 we have shown that the distributed inductance of a transmission line configuration, as shown in Figure 2.4.1. may be mo

2.5 Calculation of the Inductance of a Three-Phase Transmission Line with a Return Conductor

The inductance associated with a long three-phase transmission line may now be calculated from the expressions for self- and mutual-inductance derived in Sections 2.2 and 2.3. Consider a long transmission line of length t that consists of parallel ph~e conductors a, b, and c and the neutral or return conductor n, each of radius r. An end view of this line is shown in Figure 2.4.3(a} and a lateral view in Figure 2.4.3(b). It is assumed that the line is not in close proximity to any ground plane.

b Ia-a

Dab Doc lb-Dbn b

a Dea c 1,-c

In-Dan Den n ~ ~I

n

(a} End view of a three-phase line (b) Lateral view of the three-phase line

Figure 2.4.3 Three-phase line with return conductor

An expression for the total flux linking any conductor may be written in terms of the currents in each line and a group of self- and mutual-inductances. A set of such equations is expressed concisely in matrix form as

n Lna Lnb Lne Lnn In

or [

Where

[ :: J ~ ~~ In E.m, + .l In~ In~ [::J r 4 Dab Dae I Dbn In Dbn +1- I Dbn n- n-Dab r 4 Dbc c I Den I Den In Den +.l n- n-Dae Doc r 4

2.5.10

which may be rewritten as l $, HL,._,. Lab-n L~_,, ][1' l b - Lba-n Lbb-n Lbcn lb c Lca-n Lcb-n Lcc-n I_.

2.5. l l

or (abc] = (Labc-n] (Iabc] 2.5.12

In Equations 2.5.l and 2.5.2 the flux vector (

develop expressions and notations for the inductance, tlux linkages, currents, and impedance associaLcd with the three-phase line having an earth return.

A lumped-parameter model of the series elements of a length of three-phase line with neutral return is shown in Figure 2.6. l. This is simply a circuit model of the line in Figure 2.4.3, we have not yet stipulated that the neutral must be an actual earth return.

Assuming steady-state sinusoidal excitation, an equation for the voltage drop around the loop which includes conductors aa' and n'n may be written as

where

Va = (Ra + jwLa )I a + jwL3bI b + jwLaclc + jwLanI n + V~ -(Rn + jwLn )In - jwLanla - jwLbnlb - jwLcnlc

=(Ra + jWLa - jwLan )la + (jwLab - jwLbn )lb + (jwLac - jwLcn )le -(Rn+ jwLn - jwLan )In+ V~

Ra is resistance of phase a, Rn is resistance of neutral return, La is self inductance of phase a, L

0 is self inductance of the ground return, and

2.6.l

Lab, Lbc , Lac , Lan, Lbn, and Len are the mutual inductances between two phases, and between the phase and neutral.

Since only one return path is specified, it must be true that

2.6.2

2-13

- --------------------------

and so

Ya= (Ra+ jroLa - jroLanHa +(jWLab - jwLbn)Ib +(jWLac - jroLcnHc -(Rn + jroL 11 - jroLan )(-I a - I b - I c) + V ~ =(Ra+ Rn+ jroLa -2jwLan + jroLn)la +(Rn+ jroLab - jroLbn + jroLn - jroLan )lb+ (R 0 + jroLac - jroLcn + jroLn - jroLan )le+ V~

2.6.3

Equation 2.6.3 is expressed explicitly in terms of the phase currents, while the effect of the neutral is implicitly accounted for. This Otun's law equation, which is in terms of impedance rather than inductance alone, may be written in an analogous manner to Equation 2.5. l l as

where

Zaa-n =Ra+ Rn+ jro(La -2Lan + Ln) zab-n =Rn+ jro(Lab - Lbn + Ln - Lan) Zac-n =Rn+ jro(Lac - Len+ Ln - Lan)

2.6.4

The subscript -n denotes, as bef

where the subscript -g indicates the presence of the earth return. Observe that the expressions for the terms of ( Zabc-g] are the same as those for the corresponding terms of [zabc-n] in the equations of the form of 2.6.3; the numerical values of these terms, however, are in general different.

Carson's equations, which are based upon the method of images, allow the direct calculation of the terms of the series impedance matrix (zabc-g] from the geometry and material properties of the circuit. While it is beyond the scope of this chapter to completely derive these equations, it is of interest to examine the techniques by means of which they are derived and to observe the similarities between these equations and those derived earlier in the chapter for the calculation of line inductance.

The method of images is a technique that facilitates the solution of a certain class of field-theory boundary-value problems, including the case of the long filamentary conductor parallel to an infinite conducting plane. Essentially, it allows the conducting plane to be replaced by a fictitious conductor located at the mirror image of the actual conductor.

A single line above a ground plane of infinite extent is shown in cross section in Figure 2.6.2(a). The line carries a total charge q and the ground plane, being the only other conductor in the system, carries a total charge -q. Even when current flows in the system, we see that the ground plane appears as an equipotential surface along lines transverse to the filamentary conductors, since the return current does not flow transverse to the overhead wire.

From the uniqueness theorem, we know that if a potential field can be found such that it is equipotential over the transverse section of the ground plane shown in Figure 2.6.2(a) and behaves like the field of a point source q in the region of the conducting wire, and if this potential field furthermore satisfies Laplace's equation, then it is a valid solution. These requirements can all be satisfied by a potential field that results from the combination of conductor A carrying a charge q and conductor A' carrying a charge -q located at the mirror image of A, as shown in Figure 2.6.2(b ). It is easy to show that the two systems result in an equivalent potential at any point P above the ground plane. With reference to Figure 2.6.2(b ), the potential V p at point P may be found by superposition as

2.6.9(a)

2-15

A Oq

77777777777777777777777

(a)

p

(b)

A q

-q A'

A single conductor above ground Equivalent system by method of images

Figure 2.6.2 Method of images

This function satisfies Laplace's equation, since it involves the superposition of solutions which individually satisfy Laplace's equation. In addition, the potential function is zero in

the vicinity of the ground plane where D1 = D2, and is approximately equal to 4ru::o1 in the vicinity of the overhead line, where D1

Figure 2.6.2(b) is the same as that produced by the single conducting wire in Figure 2.6.2(a). Thus Figures 2.6.2(a) and (b) possess the same solutions for the electromagnetic case as well as for the electrostatic case.

These significant results allow the techniques pertaining to the analysis of parallel lines in space to be applied to overhead lines above a conducting ground plane. The equations

' derived by Carson for the self- and mutual-impedances of lines above ground therefore contain terms similar to the inductances in Equation 2.5.10 which were derived for the return-wire circuit without an earth return.

With reference to Figure 2.6.3 (only two conductors i and j and their images are shown. for reasons of clarity), Carson's equation may be written in the form

Off-diagonal terms: Zii-g = Rii& + jXiii

where ri is the internal resistance of conductor i, xi is the internal reactance of conductor i,

2.6.10

2.6. l l

Rii-g is the external resistive component of the self-impedance Zii-g with earth

return taken into account, X 11 & is the external reactive component of the self-impedance Zii-g with earth

return taken into account, and Rij-g and Xij-g arc the resistive and reactive components of the mutual-impedance Zij-g, respectively, with earth return taken into account.

The internal components ri and xi for a particular conductor may be obtained from

transmission line handbooks. The external components are found from the equations

Rii-g = 0.2528x10-\0 + ~ii-g O/mile Rij-g = 0.2528x10-\0 + ~ij-g O/mile Xii-g = 10-3w(o.321868ln(t.JP7f)+3.4944)+L\.Xii-g 0/mile

xij-g = w-3w( 0.321868 ln(t.JP7f)+ 2.4715) + L\.Xij-g Q/mile

where d is the diameter of the conductor in inches, .

S is the spacing between conductors in feet, p is the resistivity of the earth in O/meter cube,

2-17

2.6.12

2.6.13

2.6.14

2.6.15

w is the angular frequency in rad/sec, and f is the cyclic frequency in Hz.

i' 9

j

hj

1----=m,..,-ii--l 1 j'

Figure 2.6.3 Geometric nomenclature pertaining to Carson's eq.uations

The terms 6Rii-g, 6Rij-g, 6X 1;.g and 6Xiig are correction values that account for the conductor spacing and height, the system frequency, and the conductivity of the earth. Their solution requires the evaluations of infinite series. Sufficient accuracy is achieved for all practical power system calculations when these correction terms are expressed by -the equations

6Rij-g = 10-4w[-1.299x 10-3 Jf7P Scos0 +6.785x10-7 f;( 3.661+0.43429 ln(i.JP7f))s 2 cos20

( )3/2

+2.951x10-7 fS 20sin20+6.355x10-11 f; S3 cos30 -7.084x10-15 (; )

2 S4 cos4e]

6Xii-g = u)-4u{ 2599x 10-3hi ff!P-9.211x 10-1 hr {J+s.o84x 10-whr(f,)3'2

- 3.322 x 10-13 ht (i )' ( 3541+0 43429 tn( * .[P7f))]

2-18

2.6.17

2.6.18

dXij-g = 10-4w[1.299 x 10-3 ffp S cos0-2.318 x 10-7 tS2 cos20 ( )

3/2 ( 2 +6.355x10- 11 * S3cos30-9.020xl0- 15 i) S40sin40 -20.77 x 10-15(t )2 S4 cos40(3.842 +0.43429 ln(~.JP!f))]

2.6.19

In Equations 2.6.16 through 2.6.19, the distances hi and S = Sij defined in Figure 2.~.3 are expressed in feet, and the angle 0 is given by

e -l hi+hj . -I mij =COS - 5- .. - = Stn S:-:-J IJ

(radian)

A popular set of alternate expressions of Carson's equations is given as

2.6.20

l'.u-g = (ri +6.4374x 10-4u:P)+ j( xi +32187 x I0-4wln 2~~ii +6.4374x I0-4 CJQ] Q/rnile ZiJg = (6.4374 x 10-4 roP) + { 3.2187x10-4wln ~; + 6.4374 x 10-4 ooQJ Q/mile

where p - 1l k e k2 0 28 2 ? k2 . k3 . nk4

- 8-JJicos + 16( .67 + lnk)cos-0 + 16 ~sm20 + 45Ti cos30- 1536 cos40

in which

All other quantities are the same as previously defined.

for mutual terms for self terms

Carson's equations may be used to advantage with nearly any overhead line, regardless of the number of metallic conductors present. The self- and mutual-impedance terms associated with each metallic conductor (including ground wires) must be calculated; however, the justified assumption of a symmetrical impedance matrix allows the savings of nearly half the calculation time for lines having more than three conductors. It will be shown in succeeding sections of this chapter that any single-circuit line, regardless of the number of conductors, may be reduced to an equivalent three-conductor line by means of matrix reduction techniques, after Carson's equations have been applied to the original circuit. The application of Carson's equations to a typical distribution circuit and to a typical transmission circuit is illustrated, respectively, in the following examples.

2-19

2. 7 Physical Significance of the Terms of the Impedance Matrix [Zabc-gl

The introduction of Carson's equations in the preceding section enabled us to cakulate the tenns of the coefficient matrix Zabcg required in Equation 2.6.7 in terms of line geometry and material properties. It is interesting to note that the terms of Zabcg have simple

physical meanings as well, and may be detennined from the actual circuit by measurem~nt.

Recall Equation 2. 6. 7 in the preceding section, which was expressed as

2.7.l

Each element of [Vabc] or [V'abc] represents the voltage measured from a particular conductor to earth, the primed elements denoting the voltages at one end of the line and the unprimed elements denoting voltages at the other end. The elements of [labc] represent the actual currents in the respective lines. The terms of [Zabcg] of course take into account the finite conductivity and non-zero inductance of the ground as well as the electrical properties of the metallic conductors. E4uation 2. 7. l is simply an expression of Ohm's Law for the three-phase line with earth return.

Suppose that we energize only conductor a at the unprimed end with a voltage Va, short

the other end of conductor a to ground, and leave conductors b and c undisturbed, as shown in Figure 2.7. l(a). AIJ current must now flow around the loop containing conductor a and the ground return. No current will be induced in conductors b or c, since these conductors are open-circuited at both ends, and so

2.7.2

In addition, since the primed end of conductor a is grounded,

2.7.3

Substitution of Equations 2.7.2 and 2.7.3 into Equation 2.7.l and subtraction of the vector [V'abc] from both sides yields

2.7.4

2-20

from which

or Z - Va aa-g - -I a

2.7.5

Thus the diagonal tenn Zaa-g may be found experimentally by taking the ratio of tenninal

voltage to line current for the circuit shown in Figure 2.7. l(a).

a la a' a la a'

b b' b b'

+

+ c c' + c c'

v,. Ya vb

(a) Arrangement for determining Zaa-g (b) Arrangement for determining Zba-g

Figure 2.7. l Arrangement for determining series impedances

Now suppose that we again energize conductor a as before but short tenninal b' to ground and measure the voltage Yb as shown in Figure 2.7.l(b). For this case

lb= le= 0 and Ya = 0 as before, and Yb. = 0

Thus Equation 2.7.1 becomes

Zacg] [Ia] ~hc-g 0 Zccg 0

2.7.6

from which

or 2.7.7

Generalizing the results of this section, we see that the tenns of the impedance matrix for a three-phase line with earth return may be found from measured values as

2-21

z .. =~ 11-g I, z - v,

ij-g -Tj

2.7.8

2.7.9

provided that the measurements are made in the sense of Figure 2. 7. l and assuming that electromagnetic coupling from adjacent sources is negligible. In practice readily available computer software is used to calculate transmission and distribution line constants. From the results one can realize that the ratio of RIX for transmission lines is greater than the RIX ratio for distribution lines.

In order to have some physical feeling for the component of the impedance matrix, als_o in order to demonstrate a direct method of measurements for the impedance matrix the following example is given.

Exan1ple 1

A three-phase transmission line is represented by Figure 2. 7 .2. Some measurements are made to determine the series impedance of th.is transmission line. The arrangement, measured data and calculated impedances are shown in the table.

a a'

b b'

c c'

I I I I I I I I I I I I I I Figure 2.7.2 Three-phase transmission line for example l

2-22

Arrangement Measurement Calculated Zabcg Apply 120 L 0 V lo lerminal a

v . a' grounded Ia =42.6L-29.l

0 A Zaa-g = T = 2.46 + jl.37 n b' and c' (~en ! a

Apply 120 L 0 V lo lerminal a Vb =24.6L4l.7V v

a' and b' grounded Zba-g = T- = 0.22 + j059 n c' (!I!_en 13 = 39.2 L -27.8 A a

Apply 120 L 0 V lo terminal a Ve= 21.9L 39.6 V v . a' and c' grounded Zca-g = T = 0.22 + J052 n h' O[!_en 13 =38.8L-27.5A a

Apply 120 L 0 V lo terminal b vb .

b' grounded i lb = 42.5 L -29.0 A zbb-g ~ = 2.47 + Jl.37 n a' and c' 0_2..en

Apply 120 L 0 V lo terminal b Ve = 25.1 L 42.8 V v b' and c' grounded Zcb-g = _, = 0.22 + j0.60 11 a' OJ!.en lb= 39.3L -27.1 A lb

Apply 120 L 0 V to terminal c v

c' grounded le= 42.6 L -29. l 0 A zcc-g _s_ = 2.46 + jl.37 n a' and b' open le

Since Zab-g = Zba-g. Zac-g = Zca-g and Zbc-g = Zcb-g, the series impedance matrix. is

[

2.46 + jl.37 0.22 + j0.59 0.22 + j0.52] zarc-g = 0.22 + j0.59 2.47 + jl.37 0.22 + j0.60 n

o.:::: + j0.52 0.2.2 + j0.60 2.46 + jl.37

2.8 Series Impedance of a Three-Phase Line with Earth Return and Ground wires

Most overhead power transmission and distribution circuits include at least one neutral wire in addition to the earth return. The effect of the return wire or earth return was accounted for in previous sections by applying Kirchhoffs current law directly to the line. However, when developing simple models for circuits containing more than one ground wire it is advantageous to employ techniques of matrix reduction through partitioning as well. In this manner the matrix equation may be condensed while the physical sense of the impedance matrix is retained.

Figure 2.8.1 shows a three-phase line with two groundwires, w and v, and an earth return. This configuration is typical of that found on wooden or metal H-frame transmission structures and on some single-circuit steel towers. The conductors w and v are typically grounded to earth at the ends of every 1pan. We wish to develop a simple circuit model

2-23

for this line which would prevent the need for considering the groundwires explicitly in later analyses.

Figure 2.8. l Three-phase line with two ground wires

A voltage equation analogous to Equation 2.6.1 may be written around the loop passing through conductor a and the earth return. The expression for Va is as follows

Va= (Ra+ jwLa)la + jWLablb + jWLaclc + jWLawlw + jwLavlv + jwLanln +Va' -(Rn+ jWLn )In - jwLanla - jwLbnlb - jwLcnlc - jwLwnlw - jWLvnlv 2.8. l

= (Ra + jwLa - jwLan )la+ (jwLab - jwLbn)lb + (jwLac - jwLcn )Ic +(jWLaw - jWLwn)Iw +(jWLav - jWLvn)lv -(Rn+ jwLn - jwLan)ln +Va'

We can eliminate ground current In from this expression by applying Kirchoffs current

law in the form

Therefore Equation 2.8.1 becomes

Va= (Ra+ Rn+ jwLa - 2jwLan + jwLn )la +(R11 + jWLab - jwLbn + jWLn - jWLan )lb +(Rn+ jwLac - jwLcn + jroLn - jroLan )lc +(Rn+ jroLaw - jroLwn + jroLn - jroLan )lw +(Rn+ jroLav - jroLvn + jroL11 - jroLilll)lv +Va'

2.8.2

2.8.3

Since the effect of the earth return has been included, Equation 2.8.3 may be expressed in the form

2-24

-

Similar equations may be written for conductors b, c, w, and v. Ohin's law may then be expressed for the five-wire circuit as

Va Zaa-g Zab-g Zac-g Zaw-g vb zba-g zbb-g zbc-g Zbw-g Ve = Zca-g zcb-g zcc-g Zcw-g vw Zwa-g Zwb-g Zwc-g Zww-g vv Zva-g Zvb-g Zvc-g Zvw-g

which may be written in abbreviated form as

or

{Vabcwv} [zabcwv-g] {Iabcwv }+ {V'atxwv} {Li V abcwv} = [ Zabcwv-g] {I abcwv}

Zav-g Ia Zbv-g lb zcv-g IC + Zwv-g lw Zvv-g Iv

V' V' v

v

V'

a

b c

w

v

2.8.5

2.8.6

2.8.7

Let us now partition Equation 2.8. 7 to isolate the terms associated with the groundwires. In partitioned form.

2.8.8

Conductors w and v are grounded at both ends, and so

2.8.9

Expansion of Equation 2.8.8 with this identity included yields the pair of matrix equations

Valx: = ZAialx: + Zolwv 0 = Zclabc + Z 0 Iwv

solving 2.8.11 for Iwv , we have

which is substituted into 2.8. l 0 to obtain vabc = ZAiabc + Ze(-Z[)1Zclabc) = (ZA -ZeZiJ'ZcHabc

or Vabc = Z' abc Iatx:

2-25

2.8.10 2.8.11

2.8.12

2.8.13

-

[Z aa-g

where z abc = ZA - ZsZi)1Zc = Z' ba-g Z' ca-g

Z ab-g zbb-s z cb-g

Z' ac-g] Z' bc-g Z' cb-g

2.8.14

The transformation expressed in Equation 2.8.14 allows the five-conductor configuration of Figure 2.8. l to be modeled by the equivalent three-conductor circuit are shown in Figure 2.8.2. This result has considerable practical significance in that it enables us to represent the series parameters of the five-conductor line by a 3 by 3 impedance matrix. This transformation is applicable to lines with any number of groundwires, which will results in a 3 by 3 impedance matrix in each case. The only necessary condition is that each term in the lower portion of the voltage vector must be zero. It will be shown in the following two sections that similar transformations can be developed for bundled conductors and multiple circuits as well.

a'O

c'O Ob'

77777777777777777

Figure 2.8.2 Equivalent three-conductor model of the three-phase line with groundwires

Example 2

A typical three-phase 161 KV line with 636 MCM (2617) ACSR phase conductors and 7 strand No. 8 ACW ground wires is shown in Figure 2.8.3. An earth resistivity of 100 QI m 3 and system frequency of 60 Hz are assumed. We wish to find the series impedance matrix of the three-conductor equivalent of the circuit. The parameters for this circuit can be obtained from tabulated conductor data, and listed in the following table:

Conductor Internal Resistance One Foot Spacing Inductive Diameter Ri (0/mile) Reactance Xa (Wmile) (Inch)

636 MCM(26/7)ACSR 0.1618 0.412 0.990 7 strand No. 8 ACW 2.440 0.749 0.385

2-26

I 7'

48'

Figure 2.8.3 Structure of the transmission line for example 2

Before applying Carson's equations, we first have to convert the one foot spacing inductive reactance to internal inductive reactance. By the equation

Xi= Xa -0.004657f log(rS~)

we have internal inductive reactance 0.025 Q/mile for 636 MCM (2617) ACSR and 0.248 Q/mile for 7 No. 8 ACW. Substitution of the system dimension and cable data into the reduced set of Carson's equations yields the series impedance matrix

Za1x:wv-g =

a b c w v a 0.2537 + jl.3787 0.0919 + j0.6033 0.0919+j05192 0.0914 + j0.6203 0.0913 + j052C b 0.0919 + j0.6033 0.2537 + jl.3787 0.0919 + j0.6033 0.0914 + j05851 0.0914 + j0585 I c 0.0919 + j05192 0.0919 + j0.6033 0.2537 + jl.3787 0.0913 + j052C 0.0914 + j0.6203 Q/mite w 0.0914 + j0.6203 0.0914 + j0.5851 0.0913 + j0.52C 25308 + jl. 7170 0.0908 + j05475 v

0.0913 + j052C 0.0914 + j05851 0.0914 + j0.6203 0.0908 + j05475 25308+ jl.7170

Applying the transformation Zabc-g = ZA -Z8 Z01Zc, the above matrix can be reduced to a 3 by 3 matrix Zabc-g .

where

2-27

[0.2537 + jl.3787 Z A = 0.0919 + j0.6033

0.09 19 + j0.5192

0.0919 + j0.6033 0.2537 + jl.3787 0.0919 + j0.6033

0.0919+ j05192] 0.0919 + j0.6033 0.2537 + jl.3787

[

0.0914 + j0.6203 Z 8 = 0.0914 + j0.5851

0.0913+ j0.5204

0.0913 + j05204] 0.0914 + j0585 l 0.0914 + j0.6203

[0.0914 + j0.6203 0.0914 + j0.5851 0.0913+ j05204]

Zc = 0.0913 + j0.5204 0.0914 + j0585 l 0.0914 + j0.6203

z = [2.5308 + jl.7170 0.0908 + j0.5475] D 0.0908 + j05475 25308 + jl.7170

This five-conductor circuit is then reduced to its equivalent three-conductor model described by series impedance matrix

[

0.3545 + jl.2128 0.1942 + j0.4343 Zabc-g = 0.1942 + j0.4343 0.3593 + jl.2060

0.1894 + j0.3548 0.1942 + j0.4343

0.1894 + j0.3548] 0.1942 + j0.4343 .Q/mile 0.3545+ jl.2128

2.9 Series Impedance of a Three-Phase Line with Bundled Conductors

A bundled-conductor line with two conductors per phase is shown in Figure 2.9. l(a). Each phase of a bundled-conductor circuit consists of two or more parallel conductors in close proximity to each other and is connected by conducting spacers at regular intervals. Advantages of conductor bundling include reduction of corona loss at higher transmission voltages (as a result of increased surface area) and increased current-carrying capacity (as a result of increased cross-sectional area and decreased skin-effect loss). With reference to Figure 2.9.1 (a), an expression of Ohm's law for the six-conductor circuit with earth return is

2-28

!:::.Va Zaa Zab Zac Zaa Zab' Zac' Ia 6Vb Zba zbb zbc Zba' zbb' Zbc' lb ilVc Zea Zeb Zee Zea' Zeb' Zee' le 2.9.l = 6Ya Zaa Za'b Zac Zaa Za'b' Zac Ia 6Vb' zb'a zb'b zb'c zb'a' zb'b' zb'c' lb' 6V,. Zca z,.b z, cc Zc'a' Zc'b' Zc'c' 1,.

The subscript -g is omitted except where needed for simplicity reasons. We wish to find a transfonnation similar to that of Section 2.8 so that the six-wire circuit of Figure 2.9.l(a) could be reduced to the three-wire equivalent circuit as shown in Figure 2.9. l(b). In another words it is desired to represent the bundled-conductor line by an equation of the fonn

where

and

Z" ab f 6Va"} [Z"aa l6Vb .. = Z"ba Z" bb 6V,.. Z"ca Z" cb ilVa" = 6Va = 6Va' 6Vb" = 6Vb = 6Vb' 6V, .. = 6V, = ilV,. la"=Ia+la lb"= lb +lb' le" =le+ le'

a a' 00

c c' 00

b b' 00

77777777777777

(a) Bundled-conductor line

c" 0

a" 0

b" 0

777777777777

2.9.2

2.9.3

2.9.4

(b) Equivalent three-conductor

Figure 2.9. l Three-phase line with bundled conductors

The transfonnation from Equation 2.9. l ~o Equation 2.9.2 involves three steps. The first step modifies Equation 2. 9. l so that the lower half of the voltage vector becomes zero. This is done by using identity Equations 2.9.3 and subtracting row 1 from row 4, row 2

2-29

from row 5, and row 3 from row 6. Pre-multiplying both sides of Equation 2. 9 .1 by the appropriate row-operation transformation matrix (see Appendix A, Sections A.8 and A.12) after substituting 6.Va for 6Ya, 6.Vb for 6Vb', and 6Vc for 6Vc yields

or

av.

~vb ave

0 0 0

I I I I I I I I l : --~---+---~--~---~--

: 1 : : : : --~---T---r--;---r--

1 I 1 I I I I I I I I

--,---T---r--,---r---1 I I I } I I

I I I I I --,---T---1--1---r--

I -1 I I I 1 I --~---l ___ L __ J ___ L __

: : -1 : : : 1

I I I I I I I I I I -~-- +---~--~- -~--

: 1 : : : : --~---T---r--;---r--

1 I 1 I I I I I I I I

= --1---T---r--,---r---l I I I 1 I I

I I I I I --,---y---,--,---r -

I -1 I I I 1 I --~-- l ___ L_ J ___ L __

: : -1 : : : 1

6.Va 6Vb 6.Vc 6Va 6Vb 6.Vc

z..., z,t> z,. zba zbh Zo.: ze. Zeb Z,"

= z,. -Z"" z . b -z.b z . e - z"' zb .. -zb. zb.b - zbb Zb'e -Z"" z". :--- z,. Zb - Zeb Z,'c -Zee

z..,. zll>. z"". I, zb .. zbb' z"" lb Zea' z,b. Zee. IC 2.9.6

z .. -z ... z . b. -z3b. z.. -Z"'. I .. zb -zb .. Zbb -Zw Zw -Z"". lb' z" . - z, . Zcb -Zeb Zee' - Zee' IC.

In the second step, the upper half of the current vector is modified in accordance with Equations 2. 9 .4 so that its elements represent total phase currents. This cannot be done by row operations without inverting the impedance matrix, since it is the independent vector which must be modified. Instead, Equation 2.9.6 is first rewritten in simpler form as

2-30

-

6Y0 w w.b w.c waa. wab' w.c 6Yb wba wbb wbc wba' ww wbc' 6Yc Wea web wee Wea web' wee'

=

0 w . wa'b w . c w .. w . b. w . c 0 wb'a wb.b wb'c wb ... wb.b' wb. 0 we' a wc'b wc'c wc'a' wc'b' wc'c'

Since we wish to replace the current vector in this equation by

Ia+ la lb+ lb' le +I,

la lb' le

1. lb IC . ..

2.9.7

lb.

le'

we must modify other parts of the equation in order to maintain equality. Consider only the equation for the voltage 6 Ya, which may be expressed in expanded form as

Replacing Ia with Ia+ 13, lb with lb+ Jb. and le with le+ le and adding

corresponding terms to preserve equality, we have

or

6 Ya = W aa (I a + I a + W ab (I b + I b' ) + Wac (I e + I e' ) + W aa' I a' +Wab'lb' +Wac.le -Waala -Wablb' -Wacle'

6 Ya = W aa (I a + I a' ) + W ab (I b + I b' ) + Wac (I e + I e.) +(W aa - W aa)la + (Wab. - Wab )lb +(Wac - W acHe

Similar equations may be written for the other conductors. In matrix form,

6Ya Waa Wab Wac Waa-Waa Wab'-Wab Wac -Wac Ia+ Ia 6Yb wba wbb wbc Wba'-Wba Wbb' -Wbb Wbc'-Wbc lb +lb' 6Ve Wea web wee Wca-Wca Web'-Web Wcc-Wcc le+ le

= 0 Waa Wa'b Wac Waa-Waa Wab-Wa'b Wac-Wa'c Ia 0 wb'a wb'b wb'e Wb'a'-Wb'a Wb'b'-Wb'b Ww-Wb'e lb' 0 Wca we'b we'c Wca-Wc'a Wc'b'-Wc'b Wce-Wc'e le

or

2-31

2.9.9

2.9.lO

2.9.11

L\Va L\Vb L\Vc 0 0 0

=

Ia -Ia lb+ lb' le+ le

13 2.9.12 lb' le

The third step is the actual reduction of the 6 by 6 impedance matrix of Equation 2.9.12 to a 3 by 3 matrix. This reduction can be performed in exactly the same manner as in Section 2.8, since the lower part of the voltage vector contains zeros only and the upper parts of both the voltage and current vectors are in exactly the form in which we wish them to appear in the reduced equation. If Equation 2.9.12 is expressed in partitioned form as

{.Y~~c-} = [~!'-f-~~] {~>~~} 0 Zc 1 Zo I abc then we can find Z" abc as

where the Ohm's law expression for the three-wire equivalent circuit is

Circuits composed of bundled phase conductor normally have two or more ground wires as well. A reduction for eliminating the variables associated with these ground wires may be carried out at the same time as the reduction which accounts for conductor bundling, or it may be performed as a separate step. In any case, a 3 by 3 equivalent impedance matrix results.

The techniques developed in this section are applicable to three-phase lines having any number of bundled conductor. When forming the current and voltage vectors and the impedance matrix, it is customary to consider the conductors in groups of three, such that

{a b a' b' c' a" b" c" } ' ' c, ' ' ' ' ' '

2-32

It is worthwhile to note that if the original impedance matrix is symmetrical, as it is usually the case. the transformations developed in this section preserve this symmetry throughout each step of the process.

2.10 General Philosophy of the Calculation of Line Constants

An impedance matrix [Z] may be formed for any three-phase line, regardless of the number of conductors present in the line. The dimension of [Z] is N by N, where N is the total number of conductors (including the ground wires but excluding the earth return). A bundle of conductors associated with a particular phase is considered to be composed of a "main conductor" and one or more "bundled conductors." When forming the series impedance matrix of a line, it is customary to consider the conductor in the following order:

1. Main conductors 2. Bundled conductors 3. Ground wires

The impedance matrix is then of the form

Main Conductors

[z] = Bundled Conductors

Ground Wires

Main Bundled Ground Conductors Conductors Wires

I I I I I I I I

---------L---------L--------1 I I I I I I I I I I I I I ---------~---------~--------! I

I I I I I I I I I I ! !

2.10.1

As an example, the line shown in Figure 2.10. l has two conductors per phase and one ground wire, or a total of seven conductors. With the conductors numbered according to the scheme of Equation 2. l 0.1, the 7 by 7 impedance matrix of the line is of the form

2-33

[z] =

Main Conductors

Bundled Conductors

{ {

GroundW ires {

2 5 00

1, 1 : 1, 2 : 1, 3 : 1, 4 1, 5 ~ 1, 6 : 1, 7 .... : ......... + ................... : ......... ~

2, l 2, 2 l 2, 3 : 2, 4 2, 5 . 2, 6 : 2, 7 ....... ! ........ 't......... .... 1 .. .

3, l : 3, 2 ~ 3, 3 : 3, 4 3, 5 3, 6 : 3, 7 --- ---~----,---- -- --,----4, 1 4, 2 : 4, 3 I 4, 4 4, 5 4, 6 I 4, 7

I I . . .. : ......... , . . . . . . . . . . . . ....... , ....... . 5, 1 . 5, 2 : 5, 3 I 5, 4 5, 5 5, 6 I 5, 7 . . . . . . ... : .......... l...... .. . . ........ : ......... J

6, 1 6, 2 ~ 6, 3 : 6, 4 6, 5 ~ 6, 6 : 6, 7 ---~--- ~---- ----~---~----

7, 1 . 7, 2 : 7, 3 : 7, 4 7, 5 ~ 7, 6: 7, 7

Main Conductors

I 4 00

3 6 00

7 0

...._.,.....,

Bundled Conductors G.W.

Figure 2.10. l Example of line with bundled conductors and ground wire

After a complete impedance matrix is formed, it is operated on to reduce its size. The number of main conductors on a transmission line detennined the final dimensions of the modified impedance matrix.

2.11 The Symmetrical-Component Transformation

It is clear from the discussions in previous sections of this chapter that practically every three-phase transmission or distribution line can be modeled by an equivalent circuit consisting of three phase-conductors and an earth return, and can therefore be mathematically modeled by a series impedance matrix of the form

2.11.1

The subscript -g may be dropped, since an earth return is present in all practical three-phase circuits. The fact that we can model each of our lines by a 3 by 3 impedance matrix

2-34

regardless of their individual construction is extremely important, since it greatly facilitates the analysis of an interconnected power system consisting of different types of lines, generators, transformers, and loads.

We shall show in this section that a further simplification of the series impedance model of the three-phase line can be achieved through the use of a similarity transformation (see Appendix A, Section A.14 ). lf a tine is perfectly symmetrical, then the impedance matrix Zatx:-g is perfectly symmetrical, all off-diagonal terms of Zatx:-g are the same, and all diagonal terms of Zatx:-g are the same, i.e.

[z M MJ

Zabc-g = M Z M M M Z

2.11.2

Equation 2.1 l .2 represents the series impedance of a completely-transposed line. Since most transmission lines may be considered to be completely-transposed for all practical purposes, all of the three-phase lines discussed in this book will be modeled by the perfectly-symmetrical impedance matrix of Equation 2.1 l.2 unless otherwise stated.

In order to develop a similarity transformation for matrix Zatx:-g, it is first necessary to find the eigenvalues and eigenvectors of the matrix. The eigenvalues of Zabc-g (see Appendix A, Section A. l 4) may be found by setting

2.11.3

The cubic characteristic equation itself can be written as

2.11.4

and can be solved for A.; however, we can solve instead by using row operations to reduce the determinant of Equation 2.11.3 to triangular form as follows:

2-35

Z-A.+M Z-A.

M

Z-A+2MJ [I M =(Z-A+2M) M

Z-A M Z-A

M

RrMR1:;)R2 R J-MR1::;) R i

[

I I (Z-A+2M) 0 Z A-M

0 0 z ~ J=(Z-A+2M)(Z-A-M)(Z-A-M) A-M ,

Then, the characteristic function of Equation 2.11.3 can be written in the following fonn

(Z-A. +2M)(Z-A.-M)(Z-A.-M) = 0

where the roots can be writteri as follows

A. 1 = Z.+2M A. 2 =Z-M A. 3 = Z-M

2.11.5

The set of eigenvectors corresponding to each of these eigenvalues may also be found with the help of row operations. The eigenvector {X}1 corresponding to .A. 1 may be found from the equation

We have

or

M Z-(Z+ 2M)

M

2. l l.6

M ]{x1} {o} M X 2 = 0 Z-(Z+2M) X3 1 0

2.11.7

By using row operations, the coefficient matrix of Equation 2.11. 7 is reduced to upper-triangular form in the following manner:

[-2~ -2~ ~J M M -2M 2-36

lR ~R J 2 2

n _:] -2 I I 2 I r-2 I l] _l l -R 1 ~ R3 RJ+R~~R 1 0 3 - -I ) 0 -I 3 3 2.11.8 l l 0 0 0 3 3

Equation 2.11.6 may now be written as

[-2 1 1]{x1

} -{o} 0 -1 1 X 2 - 0 0 0 0 X3 1 0

2.11.9

which in expanded fom1 becomes

-2X1 +X2 +X3 =0 -X2 +X3 =0

0=0

This set of equations possesses an infinite number of solutions. One of the Xi must be arbitrarily selected before the other two can be calculated. The solution is of the form

X 1 =X2 =X3 =K where K is an arbitrary constant. Thus the eigenvector corresponding to A. 1 = Z+ 2M is

{X}, =rn The eigenvector corresponding to /.. 2 = z- M is found in a similar manner. For this eigenvalue,

[Z-(~-M) M M Jr1 n Z-(Z-M) M X 2 = 0 M Z-(Z-M) X3 2 0 [~ M ~]~:t =m or M 2.11.10 M

Using row operations,

2-37

[~ ~] R1 {: :J [~ ~] M M:::)R1 R1-R1~R2 l R2 R; M M ~R2. M:::)R3 R,-R 1 ~R, 0 M 0 Therefore

[~ l ~1rn1 =rn 0 2.ll.ll 0 from which a single significant equation can be obtained as

2.l l.12

Two of the unknowns in this equation can be arbitrary selected. Any choice of X 1 , X2 and X 3 which satisfies this equation is a solution vector {X}i. Some possible solution vectors are

ttl Since A. 3 = A. 2 , the eigenvector {Xh must satisfy the san1e equation as that satisfied by {X}i; thus it is not necessary to show the derivation of {Xh. A set of the eigenvectors of the perfectly-symmetrical coefficient matrix such as

may be used as the basis of a transformation for diagonalizing this matrix. The transformation matrix required for a similarity transformation must be a square matrix of the form

2-38

-

where the {Xt are the eigenvectors of the original matrix Zabc-g, each distinct eigenvector corresponding to a neigenvalue of Zabc-g. An infinite number of such similarly transfonnation matrices exist for the perfectly symmetrical matrix Zabc-g, since Zabc-g has an infinite number of distinct eigenvector sets. Note that none of these

eigenvectors is a function of Z or M, therefore the eigenvector selected is applicable for any transmission line. For the purpose of calculating the symmetrical components for both lines and rotating machines, we shall consider a particular similarity transformation matrix T5 called the symmetrical-component transfonnatio11 matrix which is defined as

T, =[: a2 .~] a where a= 1Ll20 and a 2 = lL'.240

Note that the elements of T5 are complex numbers and that each column is in fact an eigenvector of the perfectly-symmeuical matrix Zabc-g. It is important to observe that distinct (rather than identical) columns were chosen for the vectors {X}i and {Xh which correspond to the repeated roots A2 = A. 3 , since the matrix T5 would be singular if two identical columns were present. T5 can be shown to be non-singular; its inverse may

be found by elementary techniques as

[l 1 l ]

Ts-l = 1 i a a2 1 a2 a

Ohm's law for the completely-transposed line may be expressed as

2.11.13

or 2.11.14

Using the transformation matrix T5 , a new voltage vector tJ,. V012 and a new current vector 1012 are defined in terms of the vector of phase voltages and the vector of phase

currents, respectively, by the relations

2-39

2. 1 l.15

2.11.16

Substitution of these equations into Equation 2.11.14 yields

Multiplying both sides of this equation by the inverse of Ts, we have

or 2.11.17

Since the matrix product T;1Zabc-g T5 relates the new current vector I012 to the new

voltage vector ~ V012 , it may be defined as the impedance matrix Z012 of the hypothetical system represented by the vectors ~ V012 and 1012 . Thus

2.11.18

where 2.11.19

Equations 2. l l. l 8 and 2.11.19 actually define the similarity transformation associated with the matrix Zatxg. The matrix Z01 :: is diagonal if the transformation matrix Ts is a nonsingular matrix whose columns are the eigenvectors of Zatic-g . Furthermore, the

elements lying along the principal diagonal of this diagonal matrix are the eigenvalues of the matrix Zabc-g . This can be easily verified by expanding Equation 2.11.19 to get

Zo12 = 1s1Zatic.g Ts

=~ l a a2 M [1 l 1 Iz l a2 a M

l l lZ+2M a = 0 a2 0

0 Z-M

0

2.11.20

The results of this similarity transformation have an important physical significance. Through the use of this transformation, the mutually-coupled network described by the equation

2-40

-

2.11.21

may be reduced to a set of uncoupled networks described by the equations

{tl.V

0} [Z+2M 0 0 ]{10 } ti V1 = 0 Z- M 0 11

tl.V2 0 0 Z-M 12 2. 11.22

Physically, this means that we can represent lhe mulually-couplcd three-phase nelwork of Figure 2.1 l.l(a) by the three single-phase, uncoupled networks of Figure 2.11.l(b). Expanding Equation 2. l l.22, we have

ti V0 = (Z+ 2M)I0 ti V1 = (Z- M)l1 ti V2 = (Z- M)l2

(a) Coupled 3-phase network

or

fl.Vo= Zolo ti V1 = Z111 tl.V2 = Z2I2

2.11.23

Z2 ~~

v2{ __ }v2 (b) Uncoupled equivalent of the 3-cp network

Figure 2.11. l Representation of a coupled three-phase network by three uncoupled single-phase networks

The quantities ti V 0, ti V1 and ti V2 are called the zero-sequence, positive-sequence, and negative-sequence voltages, respectively. Similarly, the currents 10 , 11 and 12 are, respectively, designated zero-, positive- and negative-sequence currents and the terms Z0 , Z1 and Z2 are called the zero-, positive-, and negative-sequence impedance. Collectively, these groups of terms are often called the sequence voltages, sequence

2-41

currents, and sequence impedance, respectively. The transformation defined by Equations 2. l L l 5 and 2.1 L 16 is known as the symmetrical-component transformation; hence 6V012 , 1012 and Z012 are symmetrical-component quantities. Note that 6V0 is a function of only Z0 and 10 , 6 V1 is a function of only Z 1 and 11 and 6 V2 is a function of only Z 2 and 12 ; i.e., the currents and voltages in one sequence network are independent

of the currents and voltages in the other sequences.

If the original impedance matrix Zabc-g is not perfectly symmetrical, the use of the

symmetrical-component transformation

will result in nonzero off-diagonal terms in the impedance matrix Z012 , in this case the sequence networks will no longer be uncoupled but will instead be represented by an equation of the form

The magnitudes of the mutual terms depend upon the dissymmetry of the matrix Zabc-g. If Zabc-g is almost perfectly symmetrical, Z 012 will be diagonally dominant. Since the

purpose of the symmetrical-component method is to diagonalize the system coefficient matrix and thereby simplify the analysis of the system, no advantage is gained by applying this method to systems having significant dissymmetries.

The value of the symmetrical-component transformation is most clearly illustrated by considering the application of a balanced set of three-phase abc-sequence source voltages, Vabc , to the completely-transposed line. Suppose that such a line is shorted to ground at

one end as shown in Figure 2. l l.2(a) and is energized at the other end by the balanced set of line-to-ground voltages shown in Figure 2. l l .2(b ).

Then, from Equation 2. l l.15, V012 = T;1 V abc, we have

2-42

a Ia-- a'

b ,b ___ h'

Ya le--- c'

vb

(a) Three-phase line shorted at one end

(b) Balanced set of 3-phase voltages, abc sequence

( c) Positive-sequence equivalent circuit

Figure 2.11.2 Transformation of a balanced 3-phase system

In addition, since the source voltages and effective load impedance are balanced, the line currents must be balanced, so 1012 = T5- 1Iabc

{1

0

} [l l ]{I} [ I+a2

+a] {O} {O} 11 =* l a2

a2

a2I =*I l+a:+a: =jI 3 = I

I 2 1 a a al 1 + a + a - 0 0 2.11.25

Here only the positive-sequence network of Figure 2.11. l(b) is active. The negative- and zero-sequence networks have no source voltages and carry no currents, and therefore need not be considered. Thus the completely-transposed three-phase line described by Equation 2.11.21 may be completely represented by the positive-sequence network alone when the line is energized by a balanced set of source voltages of abc sequence. As indicated by Equations 2.11.24 and 2.11.25, the positive-sequence voltage is equal to Va in this case, and the positive-sequence current is equal to I.. The positive-sequence network corresponding to the three-phase network of Figure 2. l l .2(a) is shown in Figure 2. l 1.2(c). It is obvious that this positive-sequence network does in fact completely represent the three-phase line, since any phase voltage or line current may be calculated from the positive-sequence model by means of Equations 2.11.24 and 2.11.25. It is easy to see from these equations why the abc phase sequence is called the positive sequence. If we consider as an alternative the application of the set of balanced voltages of acb sequence (the only other possibility) shown in Figure 2.11.3 for the network of Figure 2. l l .2(a), then Equation 2.11.15 becomes

2-43

Only the negative-sequence network is active in this case. Thus the acb phase sequence is called the negative sequence.

Figure 2. l l .3 Balanced acb-sequence voltage set

The representation of a balanced network by the positive-sequence network alone is not limited to transmission lines but is applicable also to static loads, rotating machines, transformers, and other elements of the power system (the shorted transmission line of Figure 2. l l .2(a) is indistinguishable from a static load when viewed from the source). When the power system is operating under balanced conditions in this manner, the performance of the entire network may be analyzed by considering only the positive-sequence network. Note that if unbalances in source voltages or line currents are present, the negative-sequence and zero-sequence networks are in general active and cannot be ignored.

Example 3

A system consisting ideal voltage sources Va, Vb and Ve, transmission line and ideal load Ra.,Rb and Re is shown in Figure 2.11.4. The transmission line is assumed to be completed-transposed and its equivalent three-conductor series impedance matrix is

[

0.3560 + jl.2103 Z aoc-g = 0.1925 + j0.4076

0.1925 + j0.4076

0.1925 + j0.4076 0.1925 + j0.4076] 0.3560+j1.2103 O.l 925 + j0.4076 0.1925+ j0.4076 0.3560+ jl.2103

2-44

-

Transmission

Line

Figure 2.11.4 System configuration for example 3

Case l

Assume that the ideal voltage sources and load are balanced, 1.e., Va = 120L0,

vb= 120L240 and Ve = 120Ll20' and Ra= Rb = RC= 25 Q

The symmetrical components of voltage are:

{Vo} l l20L0 } [1 Vi =T~- 1 ll20L240 =~ l a V2 l 20LI 20 l a2

I l{ l 20L0 } { 0 } a

2 a 2 l20L0 = l20L0

a a 120L0 o volts

Since Zabc-g is perfectly synunetrical, the symmetrical components of the series impedance can be obtained directly by applying Z 0 = Z + 2M and Z 1 = Z 2 = Z- M with Z = 0.3560 + jl.2103 and M = 0.1925 + j0.4076, or, if one wishes, by equation Z 012 = Ts-lzabc-g T5 The results would be the same.

Zline.012 = T;'zabc-g Ts l [I I I 10.3560 + jl.2103 0.1925 + j0.4076 0.1925 + j0.407611 I

=- l a a2 0.1925+ j0.4076 0.3560+ jl.2103 0.1925+ j0.4076 I a2 3 I a2 a 0.1925+ j0.4076 0.1925+ j0.4076 0.3560+ jl.2103 l a

[

0.74!0 + j2.0255 0 0 ] = 0 0.1635 + j0.8027 0

0 0 0.1635 + j0.8027

Similarly, the symmetrical components of the balanced load are:

2-45

.~] Q

~] n 25

Examining V012 , we will find that only the positive voltage source exists and is equal to Va, and, zero and negative sources are zero. The off-diagonal terms of Zline.ol 2 and Z 1oad,Ol 2 are all zero, which means no coupling between any two sequence networks.

Thus the sequence networks of the above system are as follows:

Zero sequence Positive sequence Negative sequence

2.5 120LO' 2.5 2.5

0.74 IO+j2.0255 O. l 635+j0.8027 O. l 635+j0.8027

Figure 2.11.5 Transformed sequence networks of the system in Figure 2.11.4 -- balanced voltage sources and load

Case 2

Assume that the ideal voltage sources are unbalanced, for instance, Va = l20L0.

Vb= 125L240 and Ve= I 20Ll30, and load is still balanced, Ra= Rb= Re = 25 n

Since there is imbalance of the three-phase voltages, three sequence voltage sources exist

V1 =T;1 l25L240 =.!. 1 a a2 -6250- jl08.25 = 12l.26L3.28 volts {

Vol { 120L0) [I l 11{ 120.00+ j0.00 l {851L-140.25) V2 120L- l 30

3 l a2 a -77.13+ j91.93 5.68L-15.31

Line impedance and load resistance are same as Case 1. The sequence networks are shown in Figure 2.22.6.

Case 3

Assume that the ideal voltage sources are balanced, i.e., Ya= 120L0, Vb= 120L240 and Ve= 120Ll20, but the load is unbalanced, Ra= 3.0 Q, Rb= 2.7 Q and Re= 2.s n.

2-46

Zero sequence Positive sequence

8.5 IL-140.25 2.5 121.26L3.28 2.5

0.74IO+j2.0255 0.1635+j0.8027

Negative sequence

5.68L-15.31 2.5

O. I 635+j0.8027

Figure 2. l 1.6 Sequence networks of the system in Figure 2.11.4 -- unbalanced voltage sources and balaQced load

Symmetrical components of voltage source and line series impedance are the same as in case l, i.e.

and [o. 7410 + j2.0255

Zline,012 = 0 0

0 0.1635 + j0.8027

0 0.1635 ~ j0.8027} However, symmetrical components of the load become

Zioad,012 = ~[: :2 a 12 ][3~ 2~5 ~ ][~ a12 a~] I a a 0 0 2.50 I a [2.7500 + j0.0000 0.1250- j0.0722 01250 + j0.0722]

= o.i250 + jo.0122 2.1500 + jo.oooo o.i25o- jo.0122 n 0.1250- j0.0722 0.1250 + j0.0722 2.7500 + j0.0000

There are mutual couplings among zero, positive and negative components. Therefore, we cannot have three separated sequence networks similar to those in case l or 2.

2-47

2.12 The Clarke~Component Transformation

The symmetrical component transformation matrix

[

l 1 T5 = 1 a2

l a

introduced in Section 2.11 is the most widely used similarity-transfom1ation matrix for diagonalizing a perfectly-symmetrical coefficient matrix, but it is not the only possibility. A different set of eigenvalues is used in defining the Clarke-component transformation matrix

T, ~[: f] 2. 12. l _J_ 2 _J_ J3

2 --,,

which is useful in analyzing the performance of transient systems. The Clarke-component voltage and current vectors, 8 Yoa:p and l0a:p respectively, are defined by the relations

8 Yabc = Tc.:1 Y0a:p Iabc = Tcloop

2.12.2 2.12.3

Since 8 Yabc = Zabc-g labc , we may derive the Clarke-component similarity-transformation

equations

8 V OoP = Zoafl loall

Zoap = T.;-1zabc-g Tc

2.12.4

2.12.5

in a manner exactly analogous to that of the symmetrical-component equations of Section 2.11. For perfectly symmetrical systems, Equation 2.12.5, can be expressed as

~ = ~1Zaoc-g Tc

[l l ~ r M Mr 0 ]lZ+2M 0 0 3 3 3 1 = 0 - 2. .=l .=l M Z M l .=l Z-M 0 ]2.12.6 - 3 3 3 2

0 -1 ..=l. M M Z 1 =.l -.f3 0 0 Z-M .fj .fj 2 2

Thus the Clarke-component expression can be presented as follows

2-48

2.12.7

and the corresponding sequence network are shown in Figure 2.12. l

a a'

Vo{ }v, b b'

c c' v.{ }v. v~{ }vi

(a) Three phase system (b) Clarke-component networks

Figure 2.12.1 Clarke transformation of a three-phase system

where V0 , Vu and VP called the 0-sequence, a-sequence and ~-sequence voltage. Similar expressions can be used for the current and for the impedances.

However, if the original phase matrix is not perfectly symmetrical, use of the Clarke component transformation will result in mutual terms, depending upon the non-symmetry of the original matrix. In this case

2.12.8

Many transformations can be used for the transforming of a perfectly symmetrical three phase network to uncoupled, single-phase networks. However since lines are not the only components in a power system, the choice of components is dependent upon other components like generators, transformers, loads, etc. Both symmetrical components and Clarke components are used to decouple other elements in power systems, beside decoupling the transmission lines.

2-49

2.13 Groups of Three-Phase Circuits Lying Along the Same Right-of-Way

The placing of several three-phase transmission circuits on a single-right-of-way is common practice. Steel towers accommodating two circuits are commonplace, and it is not unusual to encounter four or more three-phase circuits terminating at generation facilities or near the perimeters of major cities. A voltage-drop equation for an N-circuit right-of-way can be written as

6Yp = Zplp 2.13. l

{6 y(I)} abc z(ll( I) abc

zm(2) abc

z(l)(Nl abc

{1 (I) } abc

{6 y

where

A similar approach may be used fo{the Clarke-component transformation.

Example 4

For the case of multi-circuit transmission lines shown in Figure 2.13. l. Assume that earth resistivity is 100 0./ M 3 , system frequency is 60 Hz, conductors are 795 MCM (45n) ACSR and ground wires arc 7 strand No. 8 ACW.

In the previous examples, we did not consider the effects of the sags of phase conductors and ground wires. Sags should be taken into account when we calculate line constants of a transmission line, especially for the long-span tower lines. In this exan1ple we assume that the sag of phase conductors is 25 feet and that of ground wires is 20 feet. For the purpose of comparison, a series impedance matrix for the case of without considering sags will be given at the end of this example.

Conductor j Internal Resistance One Foot Spacing Inductive Diameter I Ri (Q/mile) Reactance X 3 (Q/mile) (Inch)

795 MCM(45n)ACSR J 0.1428 0.406 1.063 7 strand No. 8 ACW l 2...\40 ! 0.749 0.385

15' ! ----------!

1 GW2 G\Vl

11.s r I I

-1---r 1.5' j ... 1 ,,..

14' a

18' ! a I . .....:;:__-~--+---~, 15' :.. b

'

--f--- b 18'

Circuit #I Circuit #2 51'

Figure 2.13.1 Configuration of transmission line for example 4

2-51

The symmetrical components of series impedance will be 6 by 6:

Symm: component self - : Symm. component impedance impedance of Circuit #1 : coupled from Circuit #2 to

z I '= (012) (012)" (3x3) : Circuit#l (3x3)

syil1rli.- co111poncnl"il11redancc -:-- -sy111nL-can1ponen1 se1T-=-- -coupled from Circuit #1 to : impedance of Circuit #2

Circuit#2 (3x3) : (3x3)

which is transformed from three-phase series impedance

Three - phase self -impedance of Circuit #1

(3 x 3) -tilrec-- rl1:i"scl111pc

Imaginary part of z(abd(abcJ2glg2 =

1.374 .6164 .5316 .5573 5345 .4973 :-:9188 .6165 5318 5636 .5389 .4998: 5928 .5303 .6164 1.373 .6151 .5345 .5480 .5273: .6154 .9174 .6152 5389 .5539 5317 : 5200 .4920

I I 5316 .6151 1.371 .4973 5273 .5392 I 5312 .6\41 .9161 .4998 5317 5448 I .4744 .4586

I I 5513 .5345 .4973 1.374 .6164 5316 I 5636 .5389 .4998 .9188 .6165 .5318 I 5303 5928

I I 5345 5480 5273 .6164 1.373 .6151 I 5389 5539 5317 .6154 .9174 .6152 I .4920 52()()

I I :.:i2v_...;~2n._ 2~91 _J l~--2 !..5l _ J .. n1_L'!.9J~ _ :?~ u __ 5_41~ _ 2~.1 _.:.6J1L _2L61 J.:.4.2~6_ ...;'!.7~ .9188 .6154 5312 5636 5389 .4998: 1.374 .6164 5316 5703 .5435 5023: .5924 .5343 .6165 .9174 .6141 .5389 5539 .5317: .6164 1.373 .6151 5435 5602 5361 : 5200 .4942 5318 .6152 .9161 .4998 5317 5448: 5316 .6151 1371 5023 5361 5506: .4744 .4599 5636 5389 .4998 .9188 .6154 .5312: 5703 5435 .5023 1.374 .6164 5316: 5343 5924 .5389 5539 5317 .6165 .9174 .6141 : 5435 .5602 536'1 .6164 1.373 .6151 : .4942 5200

I I .4998 5317 .5448 5318 .6152 .9161 I 5023 .5361 5506 .5316 .6151 1.371 I .4599 .4744 --------------------------r-------------------------T--------5928 5200 .4744 .5303 .4920 .4586 I 5924 .5200 .4744 .5343 .4942 .4599 I 1.719 5569

I I 5303 .4920 .4586 5928 .52!X) .4744 : _''i.343 .4942 .4599 5924 52(X) .4744 : .5569 1.719

The impedance shown above is in Q/mile. Since the voltages of two conductors of the same phase are same, we apply some row operations on Z abc , .... 2 1 , as described in ( ) (a.,.;) g g

--Equations 2.9.6 through 2.9.12. This arrangement does not change the dimension of the impedance matrix, but makes the dimension ready for reduction. The new impedance matrix Z'(abc)1(abc)2g1g2 is again separated into two parts.

Real part of Z' 1 , , = !abc.1 (abc)" "'g """

.2333 .09!0 .0916 .()CX)4 .0910 .0916 : -.1428 .0000 .0000 .0000 .0000 .0000 l .0898 .0897

.0910 .2344 .0922 .0910 .0916 .0922 : .0000 -.1428 .0000 .0000 .0000 .0000 : .0903 .0903

.0916 .0922 .2356 .0916 .0922 .0928 : .0000 .0000 -.1428 .0000 .0000 .0000 : ,OCX)9 ,OCX)9 I I .~ .0910 .0916 .2333 .09!0 .0916 I .0000 .0000 .0000 -.1428 .0000 .0000 I .0897 .0898

I I .0910 .0916 .0922 .0910 .2344 .0922 I .0000 .0000 .0000 .0000 -.1428 .0000 I .0903 .0903

I I ..:~l._..:Q9l~ __ .Q'B. __ f!!_lj __ .:.O'A2 ___ }]~~l..:

Imaginary part ofZ'(abd(abc>"g'g" = 1.374 .6164 5316 5573 .5345 4973 : -.4552 .0001 .0002 .0063 .0045 .0025 : .5928 5303 .6164 1.373 .6151 5345 .5480 .5273 : -.0010 -.4552 .0001 .0045 .0059 .0043 : .5200 .4920

I I .5316 .6151 1.371 .4973 .5273 .5392 I -.0004 -.0010 -.4552 .0025 .0043 .0056 I .4744 .4586

I I .5573 5345 4973 1.374 .6164 5316 I .0063 .0045 .0025 -.4552 .0001 .0002 I .5303 5928

I I .5345 5480 .5273 .6164 IJ73 .6151 : .0045 .0059 .0043 -.OOIO -.4552 .0001 : .4920 .5200

...;-!_91 ~ __ :E.7} __ J~'!_2 __ ..:~}.! L _ !.?L_ .!XU_ L ..c-9.L _ Q" = .3295 .1751 .1671 .1854 .1745 .1667 : -.1425 .0002 .O

Further reduction is to include the effects of bundled conductors and generate a 6 by 6 impedance matrix:

z I '= (abcgJ (abcg) 0.2585 + j0.9855 0.17 54 + j0.4694 0.1673 + j0.3948 l OJ 858 + j0.4035 0.1748 + j0.3925 0.1670 + j0.3632

I 0.1754+j0.4694 0.2367+jl.Olll 0.1583+j0.4901 :o.t748+j0.3925 0.1650+j0.4203 0.1581+j0.4070

I 0.1673 + j0.3948 0.1583 + j0.4901 0.2234 + jl.0274 : 0.1670 + j0.3632 0.1581 + j0.4070 0.1519 + j0.4286 -----------------------------------------.-----------------------------------------0.1858 + j0.4035 0.1748+ j0.3925 0.1670+ j0.3632: 0.2585+ j0.9855 0.1754+ j0.4694 0.1673+ j0.3948 0.1748 + j0.3925 0.1650 + j0.4203 0.1581 + j0.4070 i 0.1754 + j0.4694 0.2367 + jl.0111 0.1583 + j0.4901 0.1670 + j0.3632 0.1581 + j0.4070 0.1519 + j0.4286 : 0.1673 + j0.3948 0.1583 + j0.4901 0.2234 + jl.0274

The last step is to transform three-phase components to symmetrical components by applying

The symmetrical components of series impedance are

z I '= (012) (012)" 0 2 0 2

0 05735 + jl.9108 0.0307 - j0.0367 --0.0030 - j0.0244 : 0.5008 + jl.1926 0.0194 - j0.0227 0.0073- j0.0107 I

I --0.0030 - j0.0244 0.0725 + j05566 --0.0470 + j0.0282 : 0.0073 - j0.0107 0.0009 + j0.0299 --0.0187 + j0.0132 2 _o.:.ol~~-J~~~6_7 __ ~~~6-~!o.:.1~~~--~~1_2~_:_J~~~~6-~-~~1~~--';:i:.?~~1 __ _?:.?~~~!i~~1!!__~-~-+_-!_?.:.0!~ 0 0.5008 + jl.1926 0.0194 - j0.0227 0.0073 - jO.O I 07 I 0.5735 + jl.9108 0.0307 - j0.0367 --0.0030 - j0.0244

I I 0.0073 - j0.0107 0.0009 + j0.0299 --0.0187 + j0.0132 i --0.0030- j0.0244 0.0725 + j05566 --0.0470 + j0.0282 2 0.0194- j0.0227 0.0199 + j0.0117 0.0009 + j0.0299 : 0.0307 - j0.0367 0.0486 + j0.0266 0.0725+ j05566

Several observations about the above impedance matrix are listed below:

(1) The self-impedance submatrices of circuit #1 and circuit #2 are identical and the mutual-impedance submatrices between circuits #l and #2 are identical. This is because the arrangement of two circuits is symmetrical with respect to the center line of the tower, including phase sequence, both circuits have the sequence of phase a, b then c from the top to the bottom.

(2) Off-diagonal terms of each submatrix are not zero because we did not assume a complete-transposition. For a completely-transposed line, the matrix becomes

2-55

05735+ jl.9108 0

0 0 i 05008 + j 1.1926 I

0.0725 + j05566 0 : 0 I

0 0.0009 + j0.0299

0 0

0 0 0.0725 + j05566 : 0 0 0.0009 + j0.0299 -----------------------------------------T-----------------------------------------05008 + jl.!926 0 0 :05735+jl.9108 0 0

0 0.0009 + j0.0299 0 : 0 0.0725 + j0-5566 0 I

0 0 0.0009 + j0.0299 : 0 0 0.0725 + j05566

(3) Zero-sequence impedance of mutual impedance between 2 circuits, 0.5008+j l.1926 in the example, is much greater than other impedances in the same submatrix, which means that there exists stronger coupling between zero sequences of two circuits.

If we

Conductor Internal Resistance One Foot Spacing Inductive Diameter Ri (Q/mile) Reactance X3 (Q/mile) (Inch) .

7 strand No. 6 ACW 1.536 0.721 0.486

7 strand No. 8 ACW 2.440 0.749 0.385

The smaller internal resistance of 7 slrand No. 6 ACW rcsulls in a smaller zero-sequence impedance, in both self- and mutual-impedances between circuits. Its effects on reactance and coupling among zero-, positive- and negative- sequences are trivial.

Similar effects of earth resistance can be found on symmetrical components of series impedance. Suppose that earth resistivety is increased to 200 QI M 3 , from the following impedance matrix, we can find this change significantly affects zero-sequence impedance. In this case, we use 7 strand No. 8 ACW as ground wires.

Z I ' = (012) (012) 0.6105+jl.9765 0.0309-j0.0386 -0.0015-J0.0254l 05379+jl.2583 0.0196-j0.0246 0.0089-j0.0117

-0.0015-j0.0254 0.0725+ j05566 -00471 + j0.0282 \ 0.0089-jO.Ol 17 0.0009 + j0.0299 -0.0188+ j0.0131 I

0.0309- J0.0386 0.0485 + j0.0266 0.0725 + j05566 l 0.0196- j0.0246 0.0199 + j0.0117 0.0009 + j0.0299 -05-379 ~ -11.2583--o.o I%-: jo~o2:i6- -o~&i89~jo.o111- : -0:61-05 ~-Jl.9765- -o.o3w--=-JQ.o386-:0:00 l5-:jo~o254

I 0.0089 - j0.01 17 0.0009 + j0.0299 -0.0188 + jO.O 131 : -0.0015 - j0.0254 0.0725 + j05566 -0.0471 + j0.0282 0.0196 - j0.0246 0.0199 + jO.O 117 0.0009 + j0.0299 : 0.0309 - j0.0386 0.0485 + j0.0266 0.0725 + j05566

2.14 Lines Containing Fewer than Three Phase.Conductors

The major emphasis of this chapter has been placed upon the development of series-pararneter models for the balanced three-phase line. Such modeling is directly applicable to all three-phase transmission lines and most primary distribution circuits which, if not actually transposed, are at least capable of representation by the perfectly-symmetrical 3 by 3 impedance matrix.

In many distribution circuits, however, only one or two phase-conductors are carried beyond a certain distance from the substation. Such distribution lines often extend for inordinate lengths, especially in rural areas.

The impedance associated with such lines has traditionally been obtained by extrapolation of tabulated data prepared specifically for distribution circuits. Since most data of this type is empirically obtained, the accuracy attainable through its application to widely-differing line types is questionable. The widespread use of line-impedance tables is partially attributable to the previous absence of available computing facilities in the

2-57

'-,',~

smaller, rural utilities, where the accurate modeling of long lines of fewer than three -phase-conductors is most essential. One of the more obvious flaws in the impedance-table modeling of distribution lines is the omission of the mutual-coupling terms associated with the circuit carrying two phase-conductors, since it is obvious that these terms are not actually zero. The accuracy obtained through the use of these approximations is, of course, highly dependent upon line length.

Fortunately, the impedance of these lines may be obtained in a straightforward (but sometimes computationally tedious) manner through the use of Carson's equations and associated matrix reduction techniques. Recall from Section 2.6 that the self impedance term Zii-g associated with any metallic conductor above ground and the mutual impedance term Zij-g associated with any pair of conductors above ground may be

calculated from the geometric configuration of the circuit, the operating frequency, and the conductivity of the earth. These equations are applicable to circuits containing any number of conductors above ground, including the circuit containing only one phase-conductor ,md a neutral wire. The impedance associated with any distribution line may be accurately calculated by this method if even a small digital computer is available. The widespread availability of both the mini-computer and the time-shared terminal now makes it unneceE,sary to restrict the accuracy of one's models to that obtainable through the handbook and desk-calculator approaches of the past.

Example i

The technique of dealing with a multi-conductor circuit can be applied to a power line with conductors of non-regular shape, such as electric railroad. In this example, a 4/0 (7 strand) hard drawn copper wire (97.3 percent conductivity) is the power line and two solid steel rails serve as return.

We cannot find the electrical characteristics of the steel rails from any conductor characteristics' tables. By mean of bundled conductors, we can simulate each rail by filling the cros.:; secti