DRAFT and INCOMPLETE
Table of Contents
from
A. P. Sakis Meliopoulos
Power System Modeling, Analysis and Control
Chapter 3 _____________________________________________________________ 3
Modeling - Power Transmission Lines ______________________________________ 3
3.1 Introduction____________________________________________________________ 3
3.2 Power Line Designs ______________________________________________________ 3
3.3 Inductance ____________________________________________________________ 10
3.3.1 Magnetic Field Around a Conductor ____________________________________________ 11
3.3.2 Inductive Equations of a Mutilconductor Line _____________________________________ 17
3.3.3 Inductive Equations of an Overhead Multiconductor Line ___________________________ 23
3.3.4 Bundle Conductors __________________________________________________________ 26
3.4 Resistance_____________________________________________________________ 27
Table 3.1: Modulus and Phase of Modified Bessel Functions __________________ 28
3.5 Capacitance ___________________________________________________________ 29
3.5.1 Basic Electric Field Equations Around a Conductor ________________________________ 29
3.5.2. Capacitive Equations of a Multiconductor Line ___________________________________ 32
3.5.3 Capacitive equations of an Overhead Multiconductor Line __________________________ 38
3.6 Power Line Analysis - Single Phase________________________________________ 42
3.6.1 Single-Phase Transmission Line – Time Domain Model _____________________________ 45
3.6.2 Single-Phase Transmission Line – Frequency Domain Model ________________________ 47
3.6.3 Single-Phase Transmission Line – Equivalent Circuit_______________________________ 51
3.7 Power Line Analysis - Three Phase ________________________________________ 54
3.7.1 Three Phase Transmission Line – Time Domain Model______________________________ 55
3.7.2 Three-Phase Transmission Line – Frequency Domain Model_________________________ 60
3.7.3 Three-Phase Transmission Line – Sequence Models _______________________________ 61
3.8 Transmission Line Power Equations_______________________________________ 71
3.9 Transmission Line Power Transfer Limitations _____________________________ 74
3.10 Summary and Discussion _______________________________________________ 76
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
3.11 Problems ____________________________________________________________ 77
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Chapter 3
Modeling - Power Transmission Lines
3.1 Introduction
Power lines are a vital part of the electric power system. They interconnect the various
parts of the system and transfer electric power from generating plants to consumers.
Power lines may be connected in a network configuration or in a radial configuration.
Power lines may be overhead, underground and they may operate at voltages from very
high to very low. The parameters of power lines affect the flow of power, the power
transfer capability and the voltage magnitude along the power line. Therefore power lines
impact the performance of the power system. It is expedient to develop appropriate
mathematical models for power lines that are suitable for a variety of power system
analyses. In this chapter we examine modeling procedures for power lines. Our approach
will be to consider the physical construction of the power lines, consider first principles
of the operation of the line and develop appropriate equivalent circuits.
Before we embark on the modeling issue of power lines, a brief review of the physical
construction of lines will be discussed.
3.2 Power Line Designs
Power lines may be designed to operate under AC or DC voltage. They may be overhead
or underground. AC transmission lines may be three-phase or single-phase. The
components of overhead transmission lines and distribution lines are illustrated in Figure
3.1. A three-phase overhead line consists of three phase conductors HA, HB, and HC,
which are suspended with insulators from towers. Most designs include an overhead
ground wire (OHGW or shield wire) to provide protection against lightning. The OHGW
is typically connected to the neutral of the system and may be grounded at each tower.
The tower grounding system may consist of counterpoise (illustrated in Figure 3.1),
rings, ground rods, etc. A typical overhead transmission line terminates to two
substations. The OHGW is typically connected to the grounding system of the
substations. Figure 3.1 illustrates the termination of the OHGW to the substation ground
mat. A three-phase overhead distribution line is also illustrated in Figure 3.1. It consists
of three phase conductors, indicated as LA, LB, and LC, and a multiply grounded neutral
conductor. The neutral conductor is typically bonded to the substation ground mat and to
the grounds of the distribution poles.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Overhead power lines are suspended on towers or poles. The design of transmission
towers depends on the operating voltage of the line and other mechanical strength
considerations. Three example transmission tower/pole designs are illustrated in Figures
3.2, 3.3, and 3.4 for 230-, 115-, and 12-kV lines, respectively. Note that the 12-kV line,
which is typically used in distribution circuits, does not have an OHGW. Instead, it has a
fourth conductor, the neutral, which is suspended below the phase conductors. The size
of the neutral conductor is comparable to that of the phase conductors and it is intended
to carry the full load current. The reason for this practice is the fact that distribution
circuits may supply single phase loads connected between a phase and the neutral
conductor. This practice generates unbalanced conditions and the neutral conductor may
carry a substantial electrical current.
Sky Wire (OHGW)
HA
LA
HB
LB
LC
HC
Neutral
Counterpoise
Ground Mat
Ground Rod
Ground Rod
Figure 3.1 Overhead Power Lines, A Transmission and A Distribution Line
Connected via a Power Transformer
Recent advances in technology have made DC transmission an economically attractive
alternative over long distances. A typical DC transmission line is illustrated in Figure 3.5.
It consists of two bundle conductors, the positive and negative poles, and an overhead
ground conductor.
Power lines can be also constructed from power cables. Cables may be three phase, or
single phase cables connected in a three phase arrangement. A typical single phase power
cable construction is illustrated in Figure 3.6a and a typical three phase power cable
construction is illustrated in Figure 3.6b.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
A distribution system comprises power lines and voltage-step-down equipment for
electric service at industrial, commercial, and residential sites. A distribution system may
comprise three-phase transmission lines, with typical operating voltages of 12 to 25 kV
line to line, and three-phase, two phase, or single phase tapped lines. The construction of
these lines may be overhead or underground. These possibilities are illustrated in Fig.
3.7.
1'-1"
4'
4'
11'-6"
7'-7"
9'-6"
9'-6"
17'-0"
7'-7"
9'-6"
58'-0"
Figure 3.2 Design of a 230-kV H-frame Transmission Tower
(Courtesy of Georgia Power Co.)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
0.25'
g
9.74'
2
1
8.17'
3
5.50'
5.50'
1.0'
44.21'
40'
Figure 3.3 Design of a 115-kV single-pole Transmission Tower
(Courtesy of Georgia Power Co.)
Figure 3.7 suggests that distribution systems may operate (and in fact they do operate)
under unbalanced conditions. Some of this imbalance may transmit to the transmission
system. This means that distribution systems present some unique analysis problems. In
addition, recent advances in end use equipment technology has resulted in electric loads
that may be interacting with the system dynamically. For example, solid-state motor
controllers, rectifiers, and so on, inject harmonics into the distribution system. Analysis
and understanding of theses phenomena require that the distribution system be modeled
and understood not only for the power frequency (60-Hz in the United States, 50 Hz in
Europe) but also for other frequencies, such as the harmonics of 60 Hz.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
a
b
c
2.16'
3.5'
4.5'
Neutral
31.75'
Figure 3.4 Design of a 12-kV single-pole Distribution Tower
(Courtesy of Georgia Power Co.)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
7.5'
5'
20
37.5'
20'
55
o
o
13.3'
42.5'
Pole conductor::
2 x 1590 kcm
Neutral conductor:
1033.5 kcm
85' - 93''
Figure 3.5 Design of a ± 400-kV HVDC Tower
(Courtesy of the Electric Power Research Institute)
(a)
(b)
Figure 3.6 Typical Power Cables: (a) Single Phase Solid Dielectric, (b)
Three Phase Oil Filled
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
(a)
Distribution
Substation
b'
c'
a'
Loads
Residential/
Commercial
Load
Industrial/Commercial
Loads
Pad-Mounted
Transformer
Residential/
Commercial
Load
Underground Cable
(b)
Figure 3.7 A Power Distribution System
(a) Perspective View
(b) Wire Line Diagram
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
For several technical and safety reasons, electric power installations must be grounded.
Grounding of power systems is achieved by embedding metallic structures (conductors)
into earth and electrically connecting these conductors to the neutral of the power system.
In this way a low impedance is provided between the power system neutral and the vast
conducting soil, which guarantees that the voltage of the neutral, with respect to earth,
will be low under all conditions. Grounding is necessary for several reasons: (a) to assure
correct operation of electrical devices, (b) to provide safety during normal or fault
conditions, (c) to stabilize the voltage during transient conditions, and (d) to dissipate
lightning strokes.
The described physical structures are typically modelled with proper mathematical
models. Depending on the objectives of the analysis the mathematical models may be
different for the same physical structure. As an example for analysis of a power line
under steady state 60 Hz sinusoidal operation, a π-equivalent circuit completely captures
the behavior of the line. However for the same line, this equivalent circuit is inadequate
to describe transients on the line. In general, the following models of transmission lines
and relative applications may be encounter:
1. Three-phase power lines can be modeled in terms of their sequence equivalent
circuits (positive, negative and zero sequence). These models represent an
approximation of the actual behavior of a line. They are extensively used for
power flow studies, short-circuit analysis, and stability studies. These models
will be used in this book.
2. Power line models with explicit representation of transmission tower, neutral
wires or ground wires, grounding systems and substation grounding systems can
be also developed. These models are applicable for ground potential rise
computations and for design of grounding systems. In this book we will not
consider these model. Instead we assume that the power lines are effectively
grounded and the grounding of the power lines is eliminated from the model. For
more information consult reference [???].
3. Distributed parameter models of power lines can be also developed. These models
are applicable for fast electrical transient analysis and the design of overvoltage
protection. These models will not be considered in this book.
The presentation of line modeling will be done in several steps. First, we shall examine
the per unit length parameters of a power line. These parameters are: resistance,
inductance, and capacitance. Next, analysis procedures will be introduced by which the
sequence equivalent circuits of power lines will be developed.
3.3 Inductance
Power circuits generate a magnetic field around the line due to the lectric current that
flows through the conductors. The magnetic field induces voltages on the conductors.
The induced voltages are interrelated to electric currents via the inductance of the line. In
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
this section we develop models for induced voltages and inductances of power lines. The
procedures are applicable to any power line.
3.3.1 Magnetic Field Around a Conductor
Conceptually, the phenomena to be studied can be explained through the simple twoconductor line illustrated in Figure 3.7. Assume that electric current i(t), which is time
dependent, flows through one conductor. The current generates a magnetic field that is
time dependent. Consider an infinitesimal length dx of conductor. Let dλ(t) be the
magnetic flux linking the electric current i(t) flowing in the infinitesimal length dx of the
conductor. By definition, the inductance of the length dx of the conductor is dL, where
dL =
dλ ( t )
i (t )
(3.1)
Since the magnetic flux linkage is time varying, a voltage dv(t) will be induced along
length dx of the conductor:
dv(t ) =
dλ (t )
di(t )
= dL
dt
dt
Now assume that the inductance of the conductor is L henries per meter; then
dL = Ldx
+
i(t)
dv(t)
-
dx
-i(t)
Figure 3.7 A Single Two Conductor Line
Upon substitution in the equations above and subsequent solution for L, we have
dv (t )
L = dx
di (t )
dt
henries/meter
(3.2)
Equation (3.1) or (3.2) defines the inductance of a conductor. Specifically, Equation (3.1)
states that the inductance equals the magnetic flux linkage divided by the electric current.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Alternatively, Equation (3.2) states that the inductance equals the induced voltage per
unit length divided by the time derivative of the electric current.
A transmission line is a complicated structure, comprising two or more conductors above
soil or cables buried in soil. For the purpose of building a generalized model, each
conductor will be characterized with its self inductance and also with its mutual
inductance to any other conductor.
We introduce the basic concepts by considering the magnetic field around an infinitely
long conductor of circular cross section. For simplicity, assume that the conductor
material is nonmagnetic. In other words, the permeability of the conductor material is µ0 .
A cross section of the conductor is shown in Figure 3.8a. The radius of the conductor is a.
Further assume that the conductor carries an electric current i(t), which is uniformly
distributed in the cross section of the conductor (i.e., constant current density). Under
these assumptions, it is relatively easy to compute the magnetic field in and around the
conductor and subsequently the inductance of the line.
Because of the existing cylindrical symmetry, the magnetic field intensity H at a point A,
illustrated in Figure 3.8a, will be perpendicular to the segment OA and the magnitude
will be constant on the circular contour with center O and radius OA. In other words, the
magnitude of the magnetic field intensity, H, is a function of the radius r=OA only, i.e.
H(r). H(r) is computed with a direct application of Ampere's law on the described
configuration. There are two cases.
a
C
o
H
r
(a)
B
(b)
Figure 3.8 Infinitely Long Circular Conductor
(a) Cross Section
(b) Magnetic Flux Density Along a Radial Direction
Case a. Magnetic field outside conductor. In this case, the point A is located outside the
conductor, i.e. r=OA>a.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Application of Ampere's law yields
i (t ) = ∫ H ( r ) ⋅ d l = 2π (OA) H ( r ) = 2πrH ( r )
C
Upon solution for H(r), we obtain:
H (r) =
i (t )
,
2πr
for r > a
(3.3)
The magnetic flux density is given by
B( r ) = µ 0 H ( r ) =
µ 0 i (t )
,
2πr
for r > a
(3.4)
Case b. Magnetic field inside conductor. In this case, the point A is located inside the
conductor, i.e. r=OA<a.
Application of Ampere's law yields
electric current inside contour C = ∫ H ( r ) ⋅ d l = 2π (OA)H ( r ) = 2πrH ( r )
C
Under the assumption that the electric current density is constant inside the conductor, we
have
πr 2
⎛r⎞
i (t ) = ⎜ ⎟ i (t ), r < a
2
πa
⎝a⎠
2
electric current inside contour C =
Substitution and subsequent solution for H(r) yields
H (r) =
ri(t )
, r<a
2πa 2
(3.5)
and
B( r ) = µ 0 H ( r ) =
µ 0 ri(t )
, r<a
2πa 2
(3.6)
The results are summarized in Figure 3.8b, where the magnetic flux density B(r) is
plotted as a function of r along a radial direction.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
From the magnetic flux density B, the magnetic flux Φ crossing any surface S is
computed from the integral
Φ = ∫ B⋅ds
S
If the surface S crosses the conductor and since the electric current is distributed inside
the conductor, the magnetic flux will link variable portions of the electric current. In this
case, the use of the concept of magnetic flux linkage is expedient. The magnetic flux
linkage is defined by
λ = ∫ wB ⋅ d s
S
where w is the portion of electric current linked with the infinitesimal magnetic flux
B ⋅ds .
Given the magnetic flux linkage though a surface S, the induced voltage v(t) along the
perimeter of the surface is computed by
v (t ) =
dλ (t )
dt
l
dr
surface S
D
dr
2a
i(t)
conductor
Figure 3.9 Geometry of surface S
As an example, consider a rectangular surface S, of dimensions l and D, located on a
plane passing through the axis of the conductor. The surface S is defined in Figure 3.9.
Consider the two illustrated infinitesimal strips of the area ldr located on the surface S
and parallel to the axis of the conductor. One infinitesimal strip is located inside the
conductor at a distance r < a from the axis. The magnetic flux through this infinitesimal
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
strip ldr which is located inside the conductor at a distance r links
electric current. Thus the magnetic flux linkage dλint is:
dλint (t ) =
πr 2
portion of the
πa 2
µ 0 r 3i ( t )
πr 2
B
r
l
=
ldr
(
)
πa 2
2πa 4
The magnetic flux linkage of a second infinitesimal strip ldr , which is located outside
the conductor, links the entire electric current through the conductor. The magnetic flux
linkage of this infinitesimal strip dλext is
dλext (t ) =
µ 0 i (t )
ldr
2πr
The total magnetic flux linkage is
D µ i (t )
µ 0 r 3i ( t )
0
l
+
λ (t ) = ∫
dr
∫r =a 2πr ldr
r = 0 2πa 4
a
Evaluation of the integrals provides the following result:
λ (t ) =
µ 0 li ( t ) ⎛ 1
D⎞
⎜ + ln ⎟
2π ⎝ 4
a⎠
(3.7)
Equation (3.7) is usually written in the following compact form:
λ=
µ 0 i (t )l D
ln
2π
d
(3.8)
where
d = ae
−
1
4
µ0 = 4π × 10−7
H /m
(3.9)
The quantity d is known as the geometric mean radius of the conductor. The physical
meaning of the geometric mean radius is that a thin hollow conductor of radius equal to
the geometric mean radius and carrying the same electric current i(t), produces the same
magnetic flux linkage as the conductor under consideration. This interpretation will be
illustrated by the following example.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Example E3.1: An infinitely long hollow conductor of average radius d and
infinitesimal thickness carries as electric current i(t). The conductor is illustrated in
Figure E3.1a. For clarity, it is shown with finite thickness.
d
(a)
B
(b)
Figure E3.1 Magnetic Field Around a Hollow Conductor Carrying Electric
Current
Show that the magnetic flux linking a rectangular surface of dimensions l and D, with
one l -long side located on the axis of the conductor, is
λ=
µ 0 i (t )l D
ln
2π
d
Solution: The magnetic field density around this configuration is illustrated in Fig.
E3.1b. Specifically, the magnetic field density is
⎡0
B( r ) = ⎢ µ 0 i (t )
⎢⎣ 2πr
r<d
r >d
The magnetic flux linkage is
µ 0 i (t )
µ i (t )l D
ln
ldr = 0
r = d 2πr
2π
d
λ (t ) = ∫
D
This completes the proof.
The induced voltage across the conductor due to the magnetic flux is readily computed
from
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
v (t ) =
dλ (t ) µ 0 l D di(t )
=
ln
dt
2π
d dt
By definition, the inductance of the conductor is
Lt =
λ (t )
i (t )
=
µ0 l D
ln
2π
d
On a per unit length basis, the inductance is
L=
µ0 D
ln
2π d
(3.10)
One should observe that the inductance of the conductor is dependent on the width D of
the selected surface S. Since the width D can be selected arbitrarily, the result above does
not have any physical meaning. This peculiarity occurs because the path of return of the
electric current i(t) has been neglected. It is apparent that in order to compute the
inductance of the conductor in a unique and meaningful way, it is necessary to consider
the entire circuit, including the path of return of the electric current. In any practical
situation, all conductors or objects carrying electric current will be located in a finite
area. In this case, as we shall see in subsequent sections, the inductance of the conductors
is uniquely defined. Despite the lack of realism of the configuration being considered, the
results obtained are fundamental for the computation of the inductances of realistic
transmission line configurations, as we shall see in subsequent sections.
In summary we have derived expressions for the magnetic field density and magnetic
flux linkage of a current carrying conductor. We will use these results for the analysis of
practical transmission lines.
3.3.2 Inductive Equations of a Mutilconductor Line
A power line configuration comprises multiple conductors. Each conductor carries an
electric current. Such an arrangement is shown in Figure 3.10. The current of each
conductor will establish a magnetic field in and around it which will link all other
conductors. The net result will be an induced voltage on each conductor. Considering
conductor j, the induced voltage will be along the conductor as it is shown in Figure 3.10.
For computing this voltage one must determine the magnetic flux linkage per unit length
of the conductor.
Consider a rectangular frame with one side of the frame located on the axis of conductor
j. The frame extends to a distance x from the axis of the conductor and its length is l .
The flux linkage through this frame with respect to the current through conductor j, i.e.
the flux linkage of conductor j will be
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
λ jx (t ) = λ jjx (t ) + ∑ λ jkx (t )
k
where λ jkx (t ) is the contribution of conductor k current to the flux linkage of conductor j.
I1
Ij
+
Vj
x
Ik
l
In
Figure 3.10 Illustration of Induced Voltage
To compute this term consider Figure 3.11, which illustrates the cross section of the
system of conductors (only conductors j and k are shown) and the frame jx. We would
like to determine the flux linkage through the frame jx defined with the axis of conductor
j and a line parallel to conductor j passing through point x. Note that the contribution to
the magnetic flux linkage from the current of conductor j is:
λ jjx (t ) =
d jx
µ0
i j (t ) ln
2π
dj
Also note that the contribution to the magnetic flux linkage of conductor j from the
electric current of conductor k is the magnetic flux linkage through the surface defined
with the line djx. This magnetic flux equals the flux linkage through the line mx which is
given by
λ jkx =
d
µ0
ik (t ) ln kx
2π
d km
Note that the distance dkm is the same as the distance djk. The total magnetic flux linkage
through the frame jx can be formed from the contribution to the flux from all conductors,
i.e.:
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
j
djk
k
x
m
dkx
M
Fl agn
ux et
ic
Figure 3.11 Illutration of Magnetic Flux Through Plane djx
due to Electric Current ik (t)
λ jx =
d jx
d
µ
µ
+∑
i j (t ) ln
ik (t ) ln kx
2π
d j k ≠ j 2π
d jk
Above equation can be written in compact form as follows.
d
µ
i k (t ) ln kx
d jk
k =1 2π
n
λ jx = ∑
where:
n is the number of conductors
djk is the distance of conductors j, k if j ≠ k
djj is the geometric mean radius of conductor j.
The total magnetic flux linkage of conductor j is computed by stretching the frame jx to
infinity, i.e. x → ∞ . In this case the total magnetic flux linkage is:
µ
1
ik (t ) ln
d jk
k =1 2π
n
λ jx ( x → ∞ ) → λ j = ∑
Proof: In order to prove above equation, it is first noted that the basic physical law of
charge conservation dictates that the sum of all electric currents must be equal to zero,
i.e.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
n
∑i
k
(t ) = 0
k =1
The current of the last conductor n can be written as the negative sum of all other
currents:
n −1
i n ( t ) = − ∑ ik ( t )
k =1
Upon substitution in the expression for the magnetic flux linkage:
n −1
d
d
µ
µ
ik (t ) ln kx − ∑ ik (t ) ln nx
d jk k =1 2π
d jn
k =1 2π
n −1
λ jx = ∑
Above expression can be rewritten in the following form (by simply rearranging):
n −1
n −1
d
µ
1
µ
µ
1
i k (t ) ln
i k (t ) ln
i k (t ) ln kx
−∑
+∑
d jk k =1 2π
d jn k =1 2π
d nx
k =1 2π
n −1
λ jx = ∑
Note that the last sum will vanish as the point x goes to infinity:
⎞
⎛
d kx
(x → ∞ ) → 1.0 therefore ln⎜⎜ d kx ⎟⎟ → 0.0
d nx
⎝ d nx ⎠
The second sum can be expressed in terms of the current in conductor n. Thus:
µ
1
i k (t ) ln
d jk
k =1 2π
n
λj = ∑
This concludes the proof.
The induced voltage along the conductor is computed as the time derivative of the
magnetic flux linkage of the conductor.
v j (t ) =
dλ j
dt
1
µ dik (t )
ln
dt
d jk
k =1 2π
n
=∑
Assuming sinusoidal steady state conditions,
(
~
v j (t ) = Re 2V j e jωt
Page 20
)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
(
~
ik (t ) = Re 2 I j e jωt
)
Upon substitution and subsequent manipulations:
n
n
jωµ ~
1
~
~
Vj = ∑
I k ln
= ∑ x jk I k
d jk k =1
k =1 2π
where
x jk =
jωµ
1
ln
2π
d jk
The previous results can be directly used to determine the induced per unit length of any
line. An example of a three-phase line will be discussed.
Three Phase Symmetric Line: Consider a three-phase line. For simplicity assume the
three phase conductors of the line are placed on the corners of an equilateral triangle as it
is illustrated in Figure 3.12. The distance between any two phase conductors is D. The
neutral is placed in the center of the equilateral triangle. This results in a symmetric three
phase line. Assume that the electric currents at the three phases and the neutral are
~ ~ ~
~
I a , I b , I c , and I n respectively. At any point along the line, the sum of all currents equals
~ ~ ~ ~
zero (i.e., I a + I b + I c + I n = 0 ).
C
Ia
Ib
N
Ic
A
In
Source
l
B
Electric Load
(a)
(b)
Figure 3.12 A Four-Wire, Three-phase Transmission Line
(a) Longitudinal View
(b) Cross Section
The induced voltage on each one of the four conductors per unit length of the line is
obtained by application of Equation (3.23). The induced voltages on phases A, B, C and
the neutral is:
~
~
~
~
~
Va = xaa I a + xab I b + xac I c + xan I n
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 21
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
~
~
~
~
Vb = x ba I a + xbb I b + xbc I c + xbn I n
~
~
~
~
~
Vc = x ca I a + x cb I b + x cc I c + x cn I n
~
~
~
~
~
Vn = x na I a + x nb I b + x nc I c + x nn I n
Note that:
x aa = x bb = x cc = x nn = x s =
jωµ ⎛ 1 ⎞
ln⎜ ⎟
2π ⎝ d ⎠
x ab = x bc = x ca = x m =
jωµ ⎛ 1 ⎞
ln⎜ ⎟
2π ⎝ D ⎠
x an = x bn = x cn = x mn =
jωµ
3
ln
D
2π
Using above notation and substituting the current in the neutral with the negative sum of
the phase conductor currents, the induced voltage along the conductors of the three
phases are:
~
~
~
~
Va = ( x s − x mn ) I a + ( x m − xmn ) I b + ( xm − x mn ) I c
~
~
~
~
Vb = ( xm − x mn ) I a + ( x s − x mn ) I b + ( xm − x mn ) I c
~
~
~
~
Vc = ( x m − xmn ) I a + ( xm − xmn ) I b + ( x s − xmn ) I c
(
~
~ ~ ~
Vn = −( x s − x mn ) I a + I b + I c
)
It is expedient to express the induced phase voltages as the voltage of the phase-neutral
~
~ ~ ~
~ ~
loop, i.e. Van = Va − Vn , Vbn = Vb − Vn , etc. In this case, above equations become:
~
~
~
~
Van = 2( x s − x mn ) I a + ( x s + x m − 2 x mn ) I b + ( x s + x m − 2 x mn ) I c
~
~
~
~
Vbn = ( x s + x m − 2 x mn ) I a + 2( x s − x mn ) I b + ( x s + x m − 2 x mn ) I c
~
~
~
~
Vcn = ( x s + x m − 2 x mn ) I a + ( x s + x m − 2 x mn ) I b + 2( x s − x mn ) I c
Application of the symmetrical transformation on these equations (the details are
omitted), yields:
Page 22
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
~
V1 = z1 I 1 ,
~
~
V2 = z2 I 2 ,
~
~
and V0 = z0 I 0
where:
z1 = z 2 = x s − x m =
jωµ D
ln
d
2π
z 0 = 4 x s + 2 x m − 6 x mn = 4
jωµ ⎛ D ⎞
⎟
ln⎜
2π ⎜⎝ d 4 27 ⎟⎠
The terms z1 , z 2 , and z 0 represent the positive, negative and zero sequence series
reactance of this power line.
3.3.3 Inductive Equations of an Overhead Multiconductor Line
Overhead or underground power lines are characterized by the fact that earth is one of the
paths for the flow of electric current. During normal operating conditions, some electric
current is induced and flows in the conductive earth soil. In general, the magnitude of this
current is comparatively low. During abnormal operating conditions (faults), a substantial
amount of electric current may flow through earth. In any case, this current (earth
current) induces a voltage along the conductors of the power line, thus affecting its
performance. As a matter of fact, most three-phase overhead circuits are designed in such
a way that during ground faults the majority of the fault current may flow through the
earth.
The distribution of the current in the earth follows a complex, nonuniform pattern. As a
result, the computation of the inductive reactance of the earth path and the mutual
inductance between the earth path and overhead conductors is very complex. In this
section a simplified formula will be given which results from the work of Carson [???]
and Rudenberg [???]. This simplified formula is valid only for usual soil resistivities (50
to 500 Ω⋅m), for low frequencies such as the power frequency (50 or 60 Hz), and for
usual overhead line configurations. Consider the simplest configuration of two overhead
conductors, j and k respectively, parallel to the surface of the earth and carrying electric
~
~
currents I j and I k , respectively. The configuration is illustrated in Figure 3.16. Assume
no other conductors exist in the vicinity. Then the current through the soil path, i.e. the
~
~ ~
earth current, is I e = − I j − I k . Carson[???] has given a solution to this problem in terms
of a complex infinite series. For details see Meliopoulos [???]. A simplified version of
~
this results is obtained by retaining the first term of the series. The induced voltage V
along the conductor, using this approximation, is:
~ jωµ ⎛ De ⎞ ~ jωµ ⎛⎜ De
ln ⎜ ⎟ I j +
V ≅
ln
2π ⎝ d ⎠
2π ⎜⎝ d jk
Copyright © A. P. Sakis Meliopoulos – 1990-2006
⎞~
⎟Ik
⎟
⎠
Page 23
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
where
d = geometric mean radius of the overhead conductor j
~
I j = current through the overhead conductor j
~
I k = current through the overhead conductor k
ω = angular frequency of the electric current
De = equivalent depth of return of earth currents which is given by:
De = 2160
ρ
, feet
f
ρ = soil resistivity, ohms⋅meter
f
= frequency of the electric current, hertz (i.e., f =
j
ω
)
2π
d jk
k
(a)
(b)
Figure 3.13 Two Parallel Power Conductors Above Soil
The above result can be used to a general n-conductor configuration above soil by
considering two conductors at a time. Specifically, the inductance per unit length of a
power line is represented by an inductance matrix consisting of the self and mutual
inductances of the line conductors. The inductance matrix of an n-conductor line above
soil is given by
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
⎡ ⎛ De ⎞
⎛ D ⎞⎤
⎛D ⎞
⎟⎟ ln⎜⎜ e ⎟⎟ ... ln⎜⎜ e ⎟⎟⎥
⎢ ln⎜⎜
⎝ d 12 ⎠
⎝ d 1n ⎠⎥
⎢ ⎝ d1 ⎠
µ ⎢ ...
...
...
... ⎥
L=
⎢
2π
...
...
...
... ⎥
⎥
⎢
⎢ln⎛⎜ De ⎞⎟ ln⎛⎜ De ⎞⎟ ... ln⎛⎜ De ⎞⎟ ⎥
⎜d ⎟
⎜ d ⎟⎥
⎢⎣ ⎜⎝ d 1n ⎟⎠
⎝ 2n ⎠
⎝ n ⎠⎦
The series impedance matrix per unit length, Z, of a power line can be obtained from the
inductance matrix as follows:
Z = jωL
Example E3.2: Consider the three-phase electric power line of Figure E3.2. The phase
conductors are ACSR, 556.5 kcm, 26 strands. The line does not have an overhead ground
wire. The soil resistivity is 75 ohm.m. Compute the inductance matrix, and the positive,
negative and zero sequence inductances of the line.
phase
Conductors
ACSR, 556500
cm,26 Strands
b
a
9'
5'
c
14'
45'
Figure E3.2
Solution: The inductance matrix is
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 25
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
⎡ De
⎢ln d
aa
⎢
µ ⎢ De
L=
ln
2π ⎢ d ba
⎢ D
⎢ln e
⎢⎣ d ca
De
d ab
D
ln e
d bb
D
ln e
d cb
ln
De ⎤
d ac ⎥⎥
D
ln e ⎥
d bc ⎥
D ⎥
ln e ⎥
d cc ⎥⎦
ln
where:
d ab = d ba = 14.56 ft
d ac = d ca = 14.87 ft
d bc = d cb = 9.0 ft
d aa = d bb = d cc = 0.0315 ft , (taken from the Tables of ACSR conductors)
75
= 2,415 ft
60
Upon substitution into the inductance matrix:
De = 2160
⎡2.249 1.022 1.018 ⎤
L = 10 −6 ⎢1.022 2.249 1.118 ⎥ Henries / meter
⎥
⎢
⎢⎣1.018 1.118 2.249⎥⎦
To compute the sequence inductances, first the admittance matrix is “symmetrized”, i.e.
all the off-diagonal entries are substituted with their average value of 1.053x10-6
Henries/meter. Then, the symmetrical transformation is applied to the resulting matrix,
yielding:
Lseq = TLT −1
0.0 ⎤
⎡1.196 0.0
= 10 −6 ⎢ 0.0 1.196 0.0 ⎥ Henries / meter
⎥
⎢
0.0 4.355⎦⎥
⎣⎢ 0.0
Thus the positive and negative sequence inductance is 1.196 microhenries per meter and
the zero sequence inductance is 4.355 microhenries per meter.
3.3.4 Bundle Conductors
Bundle conductors are conductors that consist of several subconductors placed in a
certain fixed geometric arrangement for the entire length of the conductor. Typically, two
or three subconductors are used. Bundle conductors are used on extra-high-voltage
transmission lines for three reasons: (a) to reduce the inductive reactance of the line, (b)
to increase the power carrying capability of the line, and (c) to reduce corona phenomena
(corona losses, radio and TV interference). Here we are interested in the effects of
Page 26
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
conductor bundling on the inductive reactance of the line. This effect can be quantified in
terms of the equivalent geometric mean radius of the bundled conductor. Two cases of
practical importance are given below:
Case 1: Bundle consisting of two identical conductor. Each conductor has a geometric
mean radius of d and the distance between the two conductors is D. The equivalent
geometric mean radius of the bundle is:
d eq = dD
Case 2: Bundle consisting of three identical conductor. Each conductor has a geometric
mean radius of d, the three conductors are placed at the vertices of an equilateral triangle
and the distance between any two conductors is D. The equivalent geometric mean radius
of the bundle is:
d eq = 3 dD 2
3.4 Resistance
The resistance of power conductors is dependednt upon the frequency of the electric
current. For example the DC resistance per unit length ( rdc , f=0 Hertz) can be directly
computed from the conductor material resistivity:
rdc = ρ
1
A
(3.11)
where ρ is the Resistivity of the conductor material and A is the cross section of the
conductor.
The computation of the AC resistance, rac , of a power conductor can be quite
complicated, depending on the geometry (cross section) of the conductor. For cylindrical
conductors, the AC resistance of the conductor is given in terms of Bessel functions:
rac = rdc
ka M 0 ( ka ) ⎛
π⎞
sin⎜ θ1 ( ka ) − θ 0 ( ka ) − ⎟
2 M 1 ( ka ) ⎝
4⎠
(3.12)
Where:
k = ωµσ , ω = 2πf
a is the radius of the conductor
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 27
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
M 0 (ka ), θ 0 (ka ) : are the magnitude and phase respectively of the modified Bessel
function, order zero, argument ka.
M 1 ( ka ), θ 1 ( ka ) : are the magnitude and phase respectively of the modified Bessel
function, order one, argument ka.
Tabulation of these functions can be found in the references. For easy reference, Table
3.1 provides the numerical values of these functions for the argument range (0 to 10).
Table 3.1: Modulus and Phase of Modified Bessel
Functions
z
M0(z)
θ0(z)
M1(z)
θ1(z)
z
M0(z)
θ0(z)
M1(z)
θ1(z)
0.000
0.025
0.050
0.075
0.100
0.125
0.150
0.175
0.200
0.225
0.250
0.275
0.300
0.325
0.350
0.375
0.400
0.425
0.450
0.475
0.500
0.525
0.550
0.575
0.600
0.625
0.650
0.675
0.700
0.725
0.750
0.775
0.800
0.825
0.850
0.875
0.900
0.925
0.950
0.975
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0001
1.0001
1.0001
1.0002
1.0002
1.0003
1.0004
1.0005
1.0006
1.0008
1.0010
1.0012
1.0014
1.0017
1.0020
1.0024
1.0028
1.0032
1.0037
1.0043
1.0049
1.0056
1.0064
1.0072
1.0081
1.0091
1.0102
1.0114
1.0127
1.0140
0.00
0.01
0.04
0.08
0.14
0.22
0.32
0.44
0.57
0.73
0.90
1.08
1.29
1.51
1.75
2.01
2.29
2.59
2.90
3.23
3.58
3.95
4.33
4.73
5.15
5.59
6.04
6.52
7.01
7.51
8.04
8.58
9.14
9.72
10.31
10.92
11.55
12.19
12.86
13.53
0.0000
0.0125
0.0250
0.0375
0.0500
0.0625
0.0750
0.0875
0.1000
0.1125
0.1250
0.1375
0.1500
0.1625
0.1750
0.1875
0.2000
0.2125
0.2250
0.2375
0.2500
0.2626
0.2751
0.2876
0.3001
0.3126
0.3252
0.3377
0.3502
0.3628
0.3753
0.3879
0.4004
0.4130
0.4256
0.4382
0.4508
0.4634
0.4760
0.4886
135.00
135.00
135.02
135.04
135.07
135.11
135.16
135.22
135.29
135.36
135.45
135.54
135.64
135.76
135.88
136.01
136.15
136.29
136.45
136.62
136.79
136.97
137.17
137.37
137.58
137.80
138.03
138.26
138.51
138.76
139.03
139.30
139.58
139.87
140.17
140.48
140.80
141.12
141.46
141.80
1.300
1.350
1.400
1.450
1.500
1.550
1.600
1.650
1.700
1.750
1.800
1.850
1.900
1.950
2.000
2.050
2.100
2.150
2.200
2.250
2.300
2.350
2.400
2.500
2.600
2.700
2.800
2.900
3.000
3.100
3.200
3.300
3.400
3.500
3.600
3.700
3.800
3.900
4.000
4.500
1.0438
1.0508
1.0586
1.0672
1.0767
1.0871
1.0984
1.1108
1.1242
1.1387
1.1544
1.1712
1.1892
1.2085
1.2290
1.2509
1.2741
1.2986
1.3246
1.3520
1.3808
1.4111
1.4429
1.5111
1.5855
1.6665
1.7541
1.8486
1.9502
2.0593
2.1760
2.3009
2.4342
2.5764
2.7280
2.8894
3.0613
3.2443
3.4391
4.6179
23.75
25.54
27.37
29.26
31.19
33.16
35.17
37.22
39.30
41.41
43.54
45.70
47.88
50.08
52.29
54.51
56.74
58.98
61.22
63.46
65.71
67.95
70.19
74.65
79.09
83.50
87.87
92.21
96.52
100.79
105.03
109.25
113.43
117.60
121.75
125.87
129.99
134.10
138.19
158.59
0.6548
0.6808
0.7070
0.7333
0.7598
0.7866
0.8136
0.8408
0.8684
0.8962
0.9244
0.9530
0.9819
1.0113
1.0412
1.0715
1.1024
1.1339
1.1659
1.1987
1.2321
1.2663
1.3012
1.3736
1.4498
1.5300
1.6148
1.7046
1.7999
1.9011
2.0088
2.1236
2.2458
2.3763
2.5155
2.6640
2.8227
2.9920
3.1729
4.2783
147.07
148.02
148.99
150.00
151.04
152.12
153.23
154.38
155.55
156.76
158.00
159.27
160.57
161.90
163.27
164.66
166.08
167.53
169.00
170.50
172.03
173.58
175.16
178.39
181.70
185.10
188.57
192.11
195.71
199.37
203.08
206.83
210.62
214.44
218.30
222.17
226.07
229.98
233.90
253.67
Page 28
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
1.000
1.025
1.050
1.075
1.100
1.125
1.150
1.175
1.200
1.225
1.250
1.0155
1.0171
1.0188
1.0207
1.0227
1.0248
1.0270
1.0294
1.0320
1.0347
1.0376
14.23
14.94
15.66
16.40
17.16
17.93
18.72
19.52
20.34
21.17
22.02
0.5013
0.5140
0.5267
0.5394
0.5521
0.5648
0.5776
0.5904
0.6032
0.6161
0.6290
142.16
142.52
142.89
143.27
143.66
144.05
144.46
144.87
145.29
145.73
146.17
5.000
5.500
6.000
6.500
7.000
7.500
8.000
8.500
9.000
9.500
10.000
6.2312
8.4473
11.5008
15.7170
21.5479
29.6223
40.8176
56.3586
77.9565
108.0039
149.8476
178.93
199.28
219.62
239.96
260.29
280.61
300.92
321.22
341.52
361.81
382.10
5.8091
7.9253
10.8502
14.8961
20.5003
28.2737
39.0697
54.0807
74.9740
104.0822
144.6705
273.55
293.48
313.45
333.46
353.51
373.59
393.69
413.82
433.96
454.11
474.28
For other conductor cross section geometries, the reader is encourage to consult the
references.
3.5 Capacitance
In this section we discuss methods by which the capacitance of a transmission line can be
computed. For this purpose we employ an approach analogous to the one for computing
the inductive reactance of a transmission line. Recall that for the computation of the
inductive reactance, the magnetic field around the transmission line was examined. For
the computation of the line capacitance, the electric field around the line will be
examined. The source of this electric field is electric charge, which is deposited on the
surface of the line conductors. The analysis of the electric field results in a model relating
the electric charge and the conductor voltage. The time derivative of the total electric
charge on the surface of the conductors is by definition the capacitive current (or the
charging current) of the line. Utilizing this definition, the model can be transformed into
a relationship between the line voltage and the capacitive current. The line capacitance
can be extracted from this model.
This general approach will be utilized to introduce the analysis of capacitive phenomena
in lines in a step-by-step procedure. Specifically, first the simplest case of single circular
conductor will be examined to establish the basic equation. Then the analysis will be
extended to two parallel conductors and the general n-conductor line configuration.
3.5.1 Basic Electric Field Equations Around a Conductor
Consider the simple case of one circular infinitely long conductor. We shall assume that
the conductor is electrically charged and we shall seek the relationship between the
electric charge and the conductor voltage. Specifically, assume that the conductor is
charged with electric charge q (coulombs per meter). Because of symmetry, the electric
charge will be uniformly distributed on the conductor surface. The electric charge
generates an electric field around the conductor. Because of symmetry, the electric field
intensity, E, will be radially directed and the magnitude will depend only on the distance
of the point of observation from the axis of the conductor, as illustrated in Figure 3.19,
r
r
E = E ( r )a r
Copyright © A. P. Sakis Meliopoulos – 1990-2006
(3.13)
Page 29
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
r
Where ar is a unit vector in the radial direction r.
Consider a cylinder of length l and circular bases of radius r. The axis of the cylinder
is taken on the axis of the conductor as it is illustrated Figure 3.14. Let S be the surface of
the cylinder and V its volume. Application of Gauss' s law yields
r r
∫∫∫ ρdv = ∫∫ D.ds
V
(3.14)
S
B
A
r
ds
rB
R
dθ
Surface S
Volume V
(b)
(a)
h
R
(c)
Figure 3.14 An Infinitely Long Circular Conductor
(a) Side View, (b) Cross Section,
(c) Prospective View
where
ρ
r
D
dv
r
ds
Page 30
=
=
=
=
electric charge density, C/m3
electric field density
infinitesimal volume
infinitesimal surface area
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
The volume integral of the electric charge density inside the volume of the cylinder
equals the total electric charge enclosed in the volume. It can be immediately computed
by observing that electric charge exists only on the conductor surface at a density of q
coulombs per meter. Thus
∫∫∫ ρdv =ql
V
The surface integral on the right-hand side of Equation (3.14) is computed as follows:
r r
r r
r r
r r
∫∫ D.ds = ∫∫ D.ds + ∫∫ D.ds + ∫∫ D.ds
S
S1
S2
S3
where S1, S2 are the bases of the cylinder and S3 is the side surface of the cylinder. Note
that because the electric field is radially directed, the contributions of the bases of the
cylinder will vanish, that is,
r r
r r
D
.
d
s
=
D
∫∫
∫∫ .ds = 0.0
S1
S2
r
r
As has been discussed, the magnitude of the electric field intensity E and therefore D is
a function of the radial distance r only. Thus on the surface S3, the magnitude of the
r
electric field density, D(r), is constant. In addition, the vector D is perpendicular to the
r
surface S3 and thus parallel to ds . Thus
r r
D
∫∫ .ds = 2πrlD(r )
S3
Substitution into Equation (4.2) yields
ql = 2πrlD ( r ) = 2πrlεE ( r )
In above equation we used the constitutive relationship: D ( r ) = εE ( r ) . Solution of above
equation for E(r) yields:
E (r) =
q
2πεr
(3.15)
The electric field inside the conductor is zero.
The computed electric field intensity provides the basis for computation of the potential
difference between any two points A and B. This difference is the voltage VAB between
points A and B, defined by:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 31
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
V AB ≡ Φ( A) − Φ( B ) ≡
r
r
∫ E (r).d l
A→ B
The value of above integral depends only on points A and B. (the reader is encouraged to
prove it). Evaluation of the integral yields:
V AB =
r
r
dB
q
∫ E ( r ).d l = 2πε ln d
A→ B
(3.16)
A
where: dA and dB are the distances of points A and B respectively from the axis of the
conductor.
Equation (3.16) relates the electric charge on the conductor to the potential difference
between two points located at a radial distance dA and dB, respectively, from the axis of
the conductor. Equation (3.16) is the basic equation utilized in the analysis of
transmission line capacitance.
3.5.2. Capacitive Equations of a Multiconductor Line
Consider a configuration of n conductors which are parallel and infinitely long. The
conductor cross section is circular. Figure 3.16 shows a cross section of the
configuration. Assume that electric charge qi(t) per unit length has been accumulated on
the surface of conductor i which is uniformly distributed over the surface of the
conductor. As a first step, we consider the potential of conductor i with respect to an
arbitrarily selected point of reference X which is illustrated in Figure 3.16. For this
purpose the principle of superposition and the results of section 3.5.1 are employed to
yield:
q
q
X
1
2
q
q
j
n
Figure 3.15 A General Configuration of n-Parallel Conductors
v ix (t ) = Φ i (t ) − Φ x (t ) =
1
n
∑q
2πε
j =1
j
(t ) ln
d jx
d ij
(3.17)
where
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
dij
djx
Φ i (t )
=
=
=
distance between the axes of conductors i and j
distance between the axis of conductor j and point X
potential of conductor i at time t
Φ x (t )
=
potential of point X at time t
Note that dii = ai, the radius of conductor i.
Equation (3.17) expresses the potential difference between conductor i and an arbitrarily
selected point x. If point x is taken to infinity, the voltage vix will become the absolute
voltage of conductor i, vi . To derive the absolute voltage of conductor i, the general
expression for vi is rewritten as:
vix (t ) =
1
2πε
n
∑ q j (t ) ln
j =1
1
1
−
d ij 2πε
n
∑q
j
(t ) ln
j =1
1
d jx
Now observe that if the n conductors are the only objects with electric charge, the sum of
the electric charges, q 1 ( t ),....., q n ( t ) , must equal zero, that is,
n
∑q
j =1
j
(t ) = 0
(3.18)
In this case it can be shown that (the reader is encouraged to prove it):
lim
x →∞
1
n
∑q
2πε
j
(t ) ln
j =1
1
→ 0.0
d jx
(3.19)
Then the absolute voltage of conductor i is
v i (t ) =
1
n
∑q
2πε
j
(t ) ln
j =1
1
d ij
(3.20)
The proof of the limit of Equation (3.19) follows.
Proof: Equation (3.18) is solved for qn (t ) :
n
q n (t ) = −∑ q j (t )
j =1
Upon elimination of qn(t) in above expression, we have:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
1
2πε
n
∑ q j (t ) ln
j =1
1
1
=
d jx 2πε
n −1
∑q
j =1
j
(t ) ln
d nx
d jx
As x → ∞ , djx , d nx → ∞ and the ratios d jx / d nx → 1.0 . Thus equation (3.19).
It is expedient to repeat the assumptions under which Equation (3.20) has been obtained:
n
∑q
Assumption 1:
j =1
j
= 0.
This assumption is valid for any transmission line
configuration, assuming that all conductors have been accounted for. For overhead lines,
since the conducting soil represents one of the conductors, this means that the earth must
be also accounted for.
Assumption 2: The electric charge is uniformly distributed on the surface of the
conductors. This assumption is equivalent to: d ij , i ≠ j >> ai . This is always valid for
overhead circuits. For cables, it introduces an error depending on cable configuration.
Equation (3.20) can be transformed into an equation relating the conductor capacitive
current to the conductor voltage. For this purpose, Equation (3.20) is differentiated with
respect to time, yielding.
dv i (t ) n 1 dq j (t ) 1
=∑
ln
dt
dt
d ij
j =1 2πε
By definition, the time derivative of the conductor electric charge is the capacitive
current (or charging current):
dq j (t )
dt
≡ i 'j (t )
capacitive current of conductor j
Upon substitution, we have
dv i (t ) n 1 '
1
=∑
i j (t ) ln
dt
d ij
j =1 2πε
(3.21)
Equation (3.21) is the basic equation for modeling the capacitive effects of a
multiconductor power line. For sinusoidal steady-state analysis, Equation (3.21) is
converted into an algebraic equation. For this purpose, recall that under sinusoidal
steady-state conditions, the voltage and currents will have the following general time
variation:
(
~
vi (t ) = Re 2Vi e jωt
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)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
(
~
ii' (t ) = Re 2 I i ' e jωt
)
~
~
where Vi and I j' are complex numbers representing the phasors of the voltage and the
~
capacitive current. Substitution in Equation (3.21) and solution for Vi gives us
n
~
Vi = ∑
j =1
1 ~'
1
I i ln , i = 1, 2,..., n
jω 2πε
d ij
(3.22)
where dii = ai , the radius of the conductor i.
It is expedient to define the following quantities:
x ' ij =
1
1
ln , i ≠ j
jω 2πε d ij
ohm⋅meters
(3.23a)
x ' ii =
1
1
ln
jω 2πε ai
ohm⋅meters
(3.23b)
which will be called the separation component of the capacitive reactance and the
conductor component of the capacitive reactance respectively. These quantities depend
on the geometry and material of the components of the capacitive reactance. Using these
variables, Equation (3.22) takes the following simple form:
n
~
~
Vi = ∑ xij' I i '
(3.24)
j =1
It is noted that the components of capacitive reactance for all commercially available
conductors have been tabulated. As in the case of the components of inductive reactance,
note that the mathematically rigorous reader will be offended by the expressions for x'ii
and x'ij since they involve the terms ln(1/ai) and ln(1/dij). It should be observed that if
the quantities ai and dij are expressed in the same unit, the final result will be correct.
For this reason it has been accepted that ai and dij will be expressed in feet under the
understanding that each quantity x'ii, x'ij is meaningless if considered individually.
In summary, the capacitive effects of a power line are represented with Equation (3.21).
Specifically, for each conductor in a power line, one equation can be written relating the
capacitive current of the conductors and the time derivative of the conductor voltage. For
sinusoidal steady-state analysis, these equations are converted into a set of algebraic
equations [Equation (3.24)] relating the phasors of the conductor capacitive currents to
the phasor of the conductor voltage. The computation of line capacitance will be
illustrated for a specific power line example.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Three Phase Symmetric Line: Consider the symmetric three-phase line of Figure 3.12.
The distance between any two phase conductors is D and the neutral is placed in the
center of the equilateral triangle. The distance between any phase conductor and neutral
is: Dn = D / 3 . Assume that the charging currents at the three phases and the neutral are
~ ~ ~
~
I a' , I b' , I c' , and I n' respectively. Figure 3.17 illustrates the geometry and the charging
currents. At any point along the line, the sum of all charging currents equals zero.
Application of equation (3.24) to this system yields.
a
I'a
n
I'n
I'c
c
b
I'b
Figure 3.16 Three-Phase Line with a Symmetrically Placed Neutral
Conductor
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
1 ⎛ ~' 1 ~' 1 ~' 1 ~'
1 ⎞
~
⎟
⎜⎜ I a ln + I b ln + I c ln + I n ln
Vc =
j 2πωε ⎝
D
D
a
Dn ⎟⎠
1 ⎛ ~'
1 ~'
1 ~'
1 ~'
1
~
⎜⎜ I a ln
Vn =
+ I b ln
+ I c ln
+ I n ln
j 2πωε ⎝
Dn
Dn
Dn
an
⎛ ~' 1 ~' 1 ~' 1 ~'
1
⎜⎜ I a ln + I b ln + I c ln + I n ln
j 2πωε ⎝
a
D
D
Dn
1 ⎛ ~' 1 ~' 1 ~' 1 ~'
1
~
⎜⎜ I a ln + I b ln + I c ln + I n ln
Vb =
j 2πωε ⎝
D
a
D
Dn
~
Va =
1
⎞
⎟⎟
⎠
~
~ ~ ~ ~
Note that I a' + I b' + I c' + I n' = 0 . Solving for I n' and substituting in above equations.
⎛ ~ ' Dn ~ ' Dn ~ ' D n ⎞
+ I b ln
+ I c ln
⎜ I a ln
⎟
j 2πωε ⎝
a
D
D⎠
1 ⎛ ~ ' Dn ~ ' Dn ~ ' Dn ⎞
~
Vb =
+ I b ln
+ I c ln
⎜ I a ln
⎟
j 2πωε ⎝
D
a
D⎠
~
Va =
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1
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
⎛ ~ ' D n ~ ' Dn ~ ' Dn ⎞
+ I b ln
+ I c ln
⎜ I a ln
⎟
j 2πωε ⎝
D
D
a ⎠
1 ⎛ ~ ' an ~ ' an ~ ' an
~
⎜ I a ln
Vn = −
+ I b ln
+ I c ln
j 2πωε ⎜⎝
Dn
Dn
Dn
1
~
Vc =
⎞
⎟⎟
⎠
~
~ ~
It is expedient to express the phase voltages with respect to the neutral, i.e. Van = Va − Vn ,
~
~ ~
Vbn = Vb − Vn , etc. In this case, above equations become:
⎛ ~ ' an ~ ' an ~ ' an ⎞
⎜ I a ln + I b ln + I c ln ⎟
j 2πωε ⎝
a
D
D⎠
1 ⎛ ~ ' an ~ ' an ~ ' an ⎞
~
Vbn =
⎜ I a ln + I b ln + I c ln ⎟
j 2πωε ⎝
D
a
D⎠
1 ⎛ ~ ' an ~ ' an ~ ' an ⎞
~
Vcn =
⎜ I a ln + I b ln + I c ln ⎟
j 2πωε ⎝
D
D
a ⎠
1
~
Van =
Above equations can be utilized to derive the capacitive reactance of the line under
various conditions. For example the sequence components of the capacitive reactance of
the line can be computed. For this purpose, the equations are written in the form:
1
~
Vabcn =
j 2πωε
~'
C ' I abc
where
⎡ an
⎢ln a
⎢ a
C ' = ⎢ln n
⎢ D
⎢ln a n
⎢⎣ D
an
D
an
ln
a
an
ln
D
ln
an ⎤
D⎥
a ⎥
ln n ⎥
D⎥
an ⎥
ln
a ⎥⎦
ln
Application of the symmetrical transformation yields
~
V120 =
1
j 2πωε
~'
TC 'T −1 I 120
Note that upon substitution and manipulations
TC 'T −1
⎡ D
⎢ln a
⎢
=⎢ 0
⎢
⎢
⎢⎣ 0
0
ln
D
a
0
⎤
⎥
⎥
0 ⎥
⎥
a n3 ⎥
ln
aD 2 ⎥⎦
0
or
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
D ~'
I1
j 2πωε a
1
D~
~
V2 =
ln I 2'
j 2πωε a
an ~ '
3
~
I0
V0 =
ln
j 2πωε 3 aD 2
~
V1 =
1
ln
(3.25a)
(3.25b)
(3.25c)
Above equations state that the positive, negative and zero sequance capacitive reactance
of the line per unit length is:
x1' = x 2' =
x 0' =
1
j 2πωε
3
j 2πωε
ln
ln
D
a
an
3
aD 2
3.5.3 Capacitive equations of an Overhead Multiconductor Line
Most overhead transmission lines have ground wires to protect them against lightning.
Overhead distribution lines have neutral conductors for unbalanced current return. All
overhead power lines are suspended above earth. Neutral/ground wires and the earth are
conducting media in the vicinity of the line which may be charged with electric charge
due to the electric field of the line. Alternatively, these conducting media alter the
electric field of the line and affect the capacitance of the line. In this section we examine
methods by which the effects of earth and neutral or overhead ground wires on line
capacitance can be quantified.
The effect of neutral/ground wires can be computed in a straightforward way by treating
these wires in the same way as the phase conductors. It should be observed that the
voltage of the neutral/ground wires will be much different from the voltage of the phase
conductors. Actually, the voltage of neutral or ground wires is approximately zero at
normal operating conditions. For usual applications, the approximation is made that the
voltage of neutral or ground wires is exactly zero.
Computation of the effect of earth on the capacitive reactance of a line, in general, is a
difficult problem. To simplify the analysis, it is assumed that the earth is a semi-infinite
perfectly conducting medium. In this case the theory of images is applied directly,
yielding a rather simple analysis procedure. Specifically, the problem of a transmission
line located above earth is replaced with another equivalent problem which does not
include the earth, but includes the images of the conductors with respect to the surface of
the earth.
Consider a multiconductor line above earth. The space around the line consists of two
media: a nonconducting medium (above earth) and a highly conducting medium (earth).
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Assume the interface to be a plane, as illustrated in Figure 3.17. The conductors of the
line are located in the nonconducting medium. The earth is charged in such a way that the
electric field on the surface of the earth is perpendicular to the earth-air interface and the
electric field inside the earth is zero. The charged conductors will establish an electric
field in the nonconducting medium (air). The theory of images [???] guarantees that the
electric field in the space of the nonconducting medium is identical to the electric field
generated by two sets of conductors, the original set of conductors located in the
nonconducting medium, and another set of conductors which are the geometric images of
the actual conductors with respect to the plane interface of the two media. If the electric
charge on an actual conductor is q, the electric charge of its image is -q. This condition
guarantees that the electric field intensity on the interface will be perpendicular to the
plane interface. Thus the boundary conditions of the problem are matched. A
consequence of this condition is that if the voltage of an actual conductor is V, the
voltage of its image will be -V.
q
1
1
2
3
-q
(a)
1
(b)
Figure 3.17 A Multiconductor Line Above Earth
(a) Conductor Arrangement
(b) Conductor and Image Arrangement
Consider the general transmission line suspended above earth, as illustrated in Figure
3.17a. Application of the theory of images results in the equivalent configuration of
Figure 3.17b. Subsequently, the capacitive currents of the conductors are computed as
~ ~
follows: The voltages of the conductors, V1 , V2 ,... , are expressed in terms of the
~ ~
capacitive currents I 1' , I 2' ,... In this analysis the capacitive currents of the images are also
included. The voltage of conductor i will be:
n
~
~ n
~
Vi = ∑ x ' ij I j' − ∑ x ' ij ' I j' , i = 1, 2,....n
j =1
(3.26)
j =1
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
where:
x ' ij =
1
1
ln
jω 2πε d ij
x ' ij ' =
1
1
ln
jω 2πε d ij '
d ij
=
distance between conductors i, j
d ij'
=
distance between conductors i, and the image of conductor j
(which is the same as the distance between conductor j and the
image of conductor i)
Equation (3.26) is rewritten by combining the terms with the same electric current,
yielding the compact form:
n ⎛
d ij ' ⎞~ '
1
~
⎟I j , i = 1, 2,...n
Vi = ∑ ⎜
ln
⎜
⎟
ω
πε
j
2
d
j =1 ⎝
ij ⎠
(3.27)
~
Assuming that the voltage Vi , i = 1, 2,...n , are known, Equation (3.27) is solved to
~
provide the capacitive currents I j' , j = 1, 2,...n . The earth will also carry a capacitive
~
currents, I e' , which is given by the equation
n
~
~
I e' = − ∑ I j'
j =1
The procedure is illustrated with an example involving a single-phase line.
Example E3.3. Consider the three phase line of Figure E3.3. Compute the positive,
negative and zero sequence capacitive reactance of the line per unit length. Consider the
effect of the earth.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
phase
Conductors
ACSR, 556500
cm,26 Strands
b
a
9'
5'
c
14'
45'
Figure E3.3
Solution: Recall that the distances among the phases of this line are:
d ab = d ba = 14.56 ft
d ac = d ca = 14.87 ft
d bc = d cb = 9.0 ft
The radii opf the phase conductors are (from the Tables of ACSR conductors):
a aa = a bb = a cc = 0.04 feet
The relationship between voltages and charging currents is:
~
⎡Va ⎤
1
⎢~ ⎥
⎢Va ⎥ = jω 2πε
⎢V~a ⎥
⎣ ⎦
⎡ ⎛ 100.0 ⎞
⎛ 104.94 ⎞
⎛ 96.02 ⎞⎤
⎢ ln⎜ 0.04 ⎟ ln⎜ 14.56 ⎟ ln⎜ 14.87 ⎟⎥ ~ '
⎠
⎝
⎠
⎝
⎠⎥ ⎡ I a ⎤
⎢ ⎝
104
.
94
109
.
0
99
.
0
⎞
⎛
⎞
⎛
⎞ ⎥⎢ ~ ' ⎥
⎢ln⎛⎜
⎟ ln⎜
⎟ ln⎜
⎟ ⎢Ib ⎥
⎢ ⎝ 14.56 ⎠
⎝ 0.04 ⎠
⎝ 9 .0 ⎠ ⎥ ⎢ ~ ' ⎥
I
⎢ ⎛ 96.02 ⎞
⎛ 99.0 ⎞
⎛ 90.0 ⎞ ⎥ ⎣ c ⎦
⎢ ln⎜
ln⎜
ln⎜
⎟
⎟
⎟⎥
⎢⎣ ⎝ 14.87 ⎠
⎝ 9 .0 ⎠
⎝ 0.04 ⎠ ⎥⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 41
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
or
~
⎡Va ⎤
1
⎢~ ⎥
⎢Vb ⎥ = jω 2πε
⎢V~c ⎥
⎣ ⎦
~'
⎡7.8241 1.9751 1.8652 ⎤ ⎡ I a ⎤
⎢1.9751 7.9102 2.3979⎥ ⎢ I~ ' ⎥
⎢
⎥ ⎢ ~b ⎥
⎢⎣1.8652 2.3979 7.7187⎥⎦ ⎢⎣ I c' ⎥⎦
To compute the sequence capacitances, first the impedance matrix is “symmetrized”, i.e.
all the diagonal entries are substituted with their average value of 7.8177 and the offdiagonal entries are substituted with their average value of 2.0794. Then, the symmetrical
transformation to the voltages and currents yielding:
~
⎡V1 ⎤
1
⎢~ ⎥
⎢V2 ⎥ = jω 2πε
⎢V~0 ⎥
⎣ ⎦
~
0.0
0.0 ⎤ ⎡ I 1' ⎤
⎡5.7383
⎢~ ⎥
⎢ 0.0
5.7383
0.0 ⎥ ⎢ I 2' ⎥
⎢
⎥ ~
0.0
11.9765⎦⎥ ⎢⎣ I 0' ⎥⎦
⎣⎢ 0.0
or by expanding the matrix equation:
1 ~'
~ 5.7383 ~ '
V1 =
I1 =
I1
jω 2πε
jωC1
1 ~'
~ 5.7383 ~ '
V2 =
I2 =
I2
jω 2πε
jωC 2
1 ~'
~ 11.9765 ~ '
V0 =
I0 =
I0
jω 2πε
jωC 0
Upon computation of the capacitance values above, the sequence values of the
capacitances are:
C1 = C 2 = 9.6947 pF / m
C0 = 4.645 pF / m
3.6 Power Line Analysis - Single Phase
In previous sections we have examined the computation of power line parameters on a
per unit length basis, i.e. resistance, inductance and capacitance per unit length. In typical
applications we are interested in the behavior of the entire line. It is therefore important
to develop appropriate power line models in terms of the terminal voltages and currents.
This section examines the development of these models from the per unit length
parameters of the line. These models are applicable to a variety of power system analysis
Page 42
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
problems, such as (a) Power flow analysis, (b) Short-circuit analysis, (c) Stability
analysis, (d) Harmonic analysis, (e) Electrical transients analysis, etc.
The procedure of developing power line models from the per nuit length parameters of
the line is dependent upon the intended application, i.e. the physical phenomena that must
be captured. The conditions involved in these analysis problems range from the steadystate low-frequency (50 or 60 Hz) nearly balanced operation to very fast transients
(switching surges or lightning-induced transients). In addition, certain phenomena (such
as asymmetric operation, fault conditions, or transients) involve the earth as one of many
paths of electric current flow. In these cases, accurate modeling of the earth path is
required. Transmission line modeling procedures are drastically simplified by developing
specific transmission line models for specific applications. These models result from
specific simplifying approximations which are reasonable for the application under
consideration. For example, the sequence models of a transmission line, computed at the
power frequency, are utilized for power flow, short-circuit analysis or stability analysis.
On the other hand, for fast electromagnetic transients the distributed nature of the
transmission line parameters must be modeled explicitly.
In this book we focus on near steady state operation of power systems. Thus, the
objective of this section is to develop transmission line equivalent models under steady
state low frequency operation and to spell out the specific assumptions and
simplifications leading to these models. For this purpose we shall consider the basic
equations of a transmission line. These equations are in the form of first-order partial
differential equations. From these equations we shall develop transmission line models
suitable for representing the line during steady-state operation. The traditional approach
is to neglect asymmetries among the phases of a power line. This is a reasonable
approximation for the applications covered in this book. For coverage of nonsymmetric
lines see reference [???].
The procedure for computing the symmetric line models is llustrated in Figure 3.18.
Specifically, two approaches are represented in Figure 3.18. In the first approach, given
the geometry of the transmission line, the positive, negative and zero sequence (per unit
length) resistance, inductance and capacitance of the line is computed. Having these
parameters, each set of parameters (positive, negative or zero sequence parameters) is
treated as a sinlge phase transmission line. Transmission line analysis will provide the
model in terms of the terminal voltages and currents and the equivalent circuit. Since this
procedure is applied separately on each set of parameters (positive, negative and zero
sequence) the end result will be three models and three equivalent circuits (positive,
negative and zero sequence equivalent circuit). In the second approach, the per unit
length resistance, inductance and capacitance matrices of the line are computed first.
Next transmission line analysis is applied to develop the three phase line model. The
resulting model is a set of coupled equations. The symmetrical transformation is applied
to this model. By imposing symmetry on the model matrices, this transformation results
in three sets of decoupled equations which represent the positive, negative and zero
sequence models of the three-phase line. Each one of these models represents an
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 43
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
equivalent circuit. In this way the positive, negative and zero sequence equivalent circuits
are obtained.
Note that the end result of both approaches is the three equivalent circuits, i.e. positive,
negative and zero sequence equivalent circuit. Both procedures provide identical results.
It should be emphasized that the assumption of symmetry results in an approximate
model of the line.
Symmetry
Assumption
3 Phase
Line
R 1, L 1, C 1
+ seq equivalent
R 2=R 1, L 2=L 1, C 2=C 1
- seq equivalent
R 0, L 0, C 0
0 seq equivalent
(a)
Yes
Symmetry
Assumption
No
3 Phase
Line
Matrices
R, L, C
+ Seq Model
- Seq Model
0 Seq Model
Coupled
Differential
Equations
(b)
Figure 3.18. Two Alternate Procedures for Deriving Sequence Models of
Three Phase Transmission Lines
(a) Computation of Per Unit Sequence Parameters and Then Equivalents
(b) Transformation of Three Phase Line Model into Sequence Models
Page 44
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
3.6.1 Single-Phase Transmission Line – Time Domain Model
A single-phase electric power transmission line is characterized by the following
constants:
r:
L:
C:
g:
series resistance (in ohms/meter)
inductance (in henries/meter)
capacitance (in farads/meters)
shunt conductance (in siemens/meter)
The last parameter is typically zero for power lines, but it was added for completeness
and generality. It should be obvious that above parameters may represent the parameters
of a physically single-phase line or they may be the positive, negative or zero sequence
parameters of a three-phase line.
Consider a single-phase transmission line of length l and parameters r, L, C, and g. The
transmission line is illustrated in Figure 3.19a. We will call one end of the line the
‘sending’ end and the other the ‘receiving’ end. This is simply for notation purposes and
does not mean that the receiving end always ‘receives’ power. Consider an infinitesimal
section of the single-phase line, of length dy, at a distance y from the receiving end. The
equivalent-circuit representation of this section is illustrated in Figure 3.19b. Application
of Kirchhoff's current and voltage laws on the circuit of Figure 3.19b yields:
y=0
y
i(0,t)
y=
dy
i( ,t)
v(0,t)
v( ,t)
rdy
i(y,t)
+
v(y,t)
-
Ldy
i(y+dy,t)
+
gdy
Cdy
v(y+dy,t)
-
Figure 3.19 General Representation of a Single Phase Line
(a) Schematic Representation and Notation
(b) Equivalent Circuits of an Infinitesimal Length
Kirchhoff's voltage law:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
d⎞
⎛
− v ( y, t ) − ⎜ rdy + Ldy ⎟i ( y + dy, t ) + v( y + dy, t ) = 0
dt ⎠
⎝
Kirchhoff's current law:
i ( y, t ) + gdy v( y, t ) + Cdy
dv( y, t )
− i ( y + dy, t ) = 0
dt
Above equations are rewritten as follows (by dividing by “dy” and rearranging):
v ( y + dy, t ) − v ( y , t )
d
= ri( y + dy, t ) + L i ( y + dy, t )
dy
dt
i ( y + dy, t ) − i ( y, t )
dv( y, t )
= gv( y, t ) + C
dy
dt
Taking the limit as dy → 0 yields:
∂v ( y, t )
dy
∂i ( y, t )
dy
= ri ( y , t ) + L
di ( y, t )
dt
(3.28a)
= gv( y, t ) + C
dv( y, t )
dt
(3.28b)
The partial differential equations (3.28) define the model of a single-phase transmission
line in terms of the parameters r, L, g, and C.
Equations (3.28) are coupled first-order partial differential equations. It is possible to
obtain an equivalent set of equations which are not coupled. For this purpose Equation
(3.28b) is differentiated with respect to y:
∂ 2 i( y, t )
dy
2
=g
dv ( y , t )
d dv ( y, t )
+C
dy
dt dy
Then the term dv(y,t)/dy is substituted from Equation (3.28a) to yield
∂ 2 i( y, t )
dy
2
= CL
∂ 2 i( y, t )
dt
2
+ ( gL + Cr )
di ( y, t )
+ gri( y, t )
dt
(3.29a)
Note that Equation (3.29a) is decoupled (i.e., it is an equation in terms of the current
function only). A similar procedure yields a differential equation in terms of the voltage
function only:
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∂ 2 v( y, t )
dy
2
= LC
∂ 2 v( y, t )
dt
2
+ ( Lg + rC )
dv ( y, t )
+ rgv( y, t )
dt
(3.29b)
The partial differential equations (3.28) or (3.29) described the general model of a singlephase line.
3.6.2 Single-Phase Transmission Line – Frequency Domain
Model
Most of the time, power systems operate under sinusoidal steady-state conditions. In this
case the imposed voltages and currents on the transmission line vary sinusoidally with
frequency f. Since the transmission line is a linear system, the currents and voltages at
any point, y, in the transmission line will vary sinusoidally with time. Thus, in general,
(
(
)
)
~
i ( y, t ) = Re 2 I ( y )e jωt
~
v( y, t ) = Re 2V ( y )e jωt
(3.32a)
(3.32b)
~
~
where I ( y ) , V ( y ) are complex numbers (phasors) and ω=2πf. The models of a singlephase line under these conditions is developed next.
Upon substitution of Equations (3.32) into (3.28), we obtain
~
⎛
d 2V ( y ) ⎞
~
~
~
⎟⎟ = 2 Re [ −ω 2 LCV ( y ) + jω ( Lg + rC )V ( y ) + rgV ( y )]e jωt
2 Re⎜⎜ e jωt
2
dy
⎠
⎝
(
~
⎛
dV ( y ) ⎞
~
⎟⎟ = 2 Re e jωt ( r + jωL) I ( y )
2 Re⎜⎜ e jωt
dy ⎠
⎝
(
)
)
The equation above must be satisfied for any time t. Thus the coefficients of the time
functions on the two sides of the equation must be identical, yielding.
~
d 2V ( y )
~
= [ −ω 2 LC + jω ( Lg + rC ) + rg ]V ( y )
2
dy
~
dV ( y )
~
= ( r + jωL) I ( y )
dy
Upon factorization of the right-hand-side expression, we have
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
d 2V ( y )
~
= ( r + jωL)( g + jωC )V ( y )
2
dy
(3.33a)
~
dV ( y )
~
= ( r + jωL) I ( y )
dy
(3.33b)
Now let's define
z ≡ r + jωL = series impedance per unit length of the line at frequency ω
y ' ≡ g + jωC = shunt admittance per unit length of the line at frequency ω
With the new notation, Equation (3.33) become
~
d 2V ( y )
~
= zy 'V ( y )
2
dy
(3.34a)
~
dV ( y )
~
= zI ( y )
dy
(3.34b)
Equations (3.34) represent the single-phase line model at sinusoidal steady state. The
general solution of Equation (3.34a) is
~
V ( y ) = ae py + be − py
(3.35)
where a, b are constants dependent on the boundary conditions of the line, and
p = zy ' = − ω 2 LC + jω ( Lg + rC ) + rg
(3.36)
Note that p is dependent on the angular frequency ω = 2πf . The dimension of the
constant p is inverse of length. The constant p characterizes the propagation of voltage
through the transmission line. For this reason it is called the propagation constant. The
real and imaginary parts of the propagation constant is referred to as the attenuation and
phase constant, respectively. That is, p = κ + jη , where κ is the attenuation constant
andη is the phase constant.
~
The general solution for the electric current phasor I ( y ) is obtained by substituting
Equation (3.35) into Equation (3.34b). The result is
p
~
I ( y ) = (ae py − be − py )
z
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Observe that
p
=
z
y'
z
Z0 =
z
1
=
'
y
Y0
Define
(3.37)
Note that the quantity Z0 has dimensions of impedance and it is a characteristic of the
transmission line under consideration. We refer to it as the characteristic impedance of
the line. The quantity Y0 will the the characteristic admittance of the line. In terms of the
characteristic impedance or admittance, the equation for the line current becomes
a py b − py
~
I ( y) =
e − e = aY0 e py − bY0 e − py
Z0
Z0
(3.38)
while:
~
V ( y ) = ae py + be − py
Above equation provides the general solution for the voltage and current phasors at a
location y of a single-phase line. The solution is expressed in terms of the propagation
constant p, the characteristic impedance Z0, and two constants a and b. The quantities p
and Z0 depend on the parameters of the line, while the constants a, b are dependent on the
boundary conditions. If enough boundary conditions are given, for example the terminal
voltage and current at one end of the line, the constants a and b can be expressed as a
function of the boundary data.
As an example, we will assume that the voltage and current at the receiving end of the
~
~
line of Figure 3.19 are known to be VR and I R . Note that the receiving end of this line is
characterized with y=0. Therefore:
~
~
V ( y = 0) = VR = a + b
a
b
~
~
I ( y = 0) = I R =
−
Z0 Z0
Upon solution of two equations above for the constants a and b we obtain
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
~
VR + Z 0 I R
a=
2
~
~
VR − Z 0 I R
b=
2
Substitution into Eqs. (3.8) and (3.9) gives us
~
~ e py + e − py
~ e py − e − py ~
~
V ( y ) = VR
+ Z0I R
= VR cosh( py ) + Z 0 I R sinh( py )
2
2
(3.39a)
~
VR e py − e − py ~ e py + e − py
~
~
~
+ IR
= Y0VR sinh( py ) + I R cosh( py )
I ( y) =
2
2
Z0
(3.39b)
Equations (3.39) provide the voltage and current phasors at any location y along the line
in terms of the voltage and current at the receiving end of the line (y = 0). Of special
interest are the voltage and current at the other end of the line (y = l ):
~ ~
~
~
VS = V ( y = l) = VR cosh( pl) + Z 0 I R sinh( pl)
~ ~
~
~
I S = I ( y = l) = Y0VR sinh( pl) + I R cosh( pl)
In compact matrix notation:
~
~
⎡VS ⎤ ⎡ cosh( pl) Z 0 sinh( pl)⎤ ⎡VR ⎤
⎢~ ⎥ = ⎢
⎥⎢ ~ ⎥
⎣ I S ⎦ ⎣Y0 sinh( pl) cosh( pl) ⎦ ⎣ I R ⎦
This equation states that the sending-end voltage and current are a linear combination of
the receiving-end voltage and current, and vice versa. Three parameters describe this
model completely: (a) the characteristic impedance of the line Z0; (b) the propagation
constant of the line, p; and (c) the length of the line, l . Note that the model depends only
on the product p l and the characteristic impedance Z0. Alternatively, the following
parameters completely describe the single-phase transmission line: (a) A = cosh p l , (b)
B = Z0sinh p l , and (c) C = Y0sinh p l . In terms of the parameters A, B, C, the line
equations (3.39) become:
~
~
~
VS = AVR + BI R
(3.40a)
~
~
~
I S = CVR + AI R
(3.40b)
These parameters are known as the A, B, C constants of the line. Note that
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A2 − BC = cosh2 ( py ) − sinh 2 ( py ) = 1.0
Thus the parameters A, B, and C are not independent. Knowledge of the two is enough to
determine the third.
3.6.3 Single-Phase Transmission Line – Equivalent Circuit
In previous sections we have developed the model of a single-phase line under steadystate conditions. The model is in terms of the A, B, C parameters or alternatively, in
terms of the characteristic impedance, propagation constant, and line length. An
alternative representation of the transmission line under steady-state conditions is by
means of an equivalent circuit. This approach is more attractive because of the familiarity
of engineers with circuits. This section presents the computation of equivalent circuits
from the transmission line parameters.
Consider Equations (3.40) of a single-phase line in terms of the terminal currents and
voltages. From realization theory it is known that a two-port circuit can be found which is
described with the same equations. Furthermore, this two-port circuit is not unique. From
the multiplicity of equivalent circuits, two particular circuits have been popular among
power engineers: (a) the π equivalent, and (b) the T equivalent. In this textbook we will
develop and use the π equivalent circuit. This circuit is introduced next.
To a transmission line with constants A, B, C, corresponds a π-equivalent circuit with
elements Yπ, Y'π as in Figure 3.20. The elements Yπ, Y'π of the π-equivalent circuit are
computed by the following simple procedure. First, the line model equations (3.40) are
rewritten in trems of the terminal currents and voltages as defined in Figure 3.20. Note
that:
~
~
I1 = −I R ,
~ ~
I2 = IS
~ ~
V1 = VR ,
~ ~
V2 = Vs
Upon substitution, and expressing the line terminal currents as a function of line terminal
voltages and the parameters A, B and C:
~ A~ 1 ~
I 1 = V1 − V2
B
B
(3.51a)
1 ~ A~
~
I 2 = − V1 + V2
B
B
(3.51b)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
I1
~
I2
Yπ
+
+
~
V1
~
V2
Y'π
Y'π
-
-
Figure 3.20 π -equivalent Circuit
Second, the equations expressing the terminal currents of the circuit 3.20 as a function of
the terminal voltages and the parameters of the pi-equivalent circuit are:
~
~
~
I 1 = (Yπ + Yπ' )V1 − Yπ V2
~
~
~
I 1 = −Yπ V1 + (Yπ' + Yπ )V2
For equivalence, the two sets of equations should be identical. Therefore:
Yπ =
1
B
(3.52a)
Yπ' =
A −1
B
(3.52b)
Equations (3.52) define the parameters of the π-equivalent circuit of a line. These
parameters are the series admittance of the equivalent circuit, Yπ, and the shunt
admittance of the equivalent circuit, Y'π. The impedance parameters of the π equivalent
circuit will be:
Zπ =
1
= B = Z 0 sinh( pl)
Yπ
Z π' =
Z sinh( pl)
1
B
=
= 0
'
Yπ A − 1 cosh( pl) − 1
Nominal π-Equivalent Circuit. The nominal π-equivalent circuit of a transmission line
is an approximation of the exact equivalent. In general, this approximation is valid for
short lines; thus the name short-line equivalent is alternatively used. Consider the πequivalent circuit described by the parameters Zπ, and Z'π, as derived earlier. The
nominal π-equivalent circuit is obtained by approximating the hyperbolic sine and cosine
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
functions. Specifically, assuming that p l <<1, (this assumption is equivalent to the
assumption of short lines, i.e. l is small), the functions are expanded into a series and
then only the major terms are retained:
sinh pl ≅ pl
( pl) 2
cosh pl ≅ 1 +
2
Substitution of the approximations above in the equations for the parameters Zπ, and Z'π,
yields:
Z π ≅ Z 0 pl = zl = ( r + jωL)l
Z 'π ≅
Z 0 pl
2Z
2
2
= 0 = ' =
2
pl
( pl)
y l ( g + jωC )l
1+
−1
2
(3.53a)
(3.53b)
The computation of equivalent circuits for a transmission line will be illustrated with an
example.
Problem E3.4. A three phase transmission line has the following per unit length positive
sequence parameters
Resistance: R = 0.08 ohms/mile
Inductance: L = 1.1×10-6 Henries/meter
Capacitance: C = 10.8×10-12 Farads/meter
The line is 200 miles long.
a)
b)
Compute the positive sequence π-equivalent circuit of the line.
Compute the positive sequence nominal π equivalent circuit of the line.
Solution:
a) The series impedance and shunt admittance of the line are:
z = 0.08 + j0.667 Ω/mi
y’ = j6.551×10-6 S/mi
Thus:
Z0 =
0
z
= 320.2e − j 3.42 = 319.63 − j19.1 Ω
'
y
pl = l zy ' = 0.4195e j 86.58 = 0.025 + j 0.4187
0
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Z π = Z 0 sinh pl = 130.417e j 83.36 Ω = 15.08 + j129.54 Ω
0
Yπ' =
1
⎛ pl ⎞
tanh⎜ ⎟ = (0.0115 + j 6.646)10 −4 S
Z0
⎝ 2 ⎠
Z π' =
1
= 2.626 − j1504.66 Ω
Yπ'
b) The nominal pi-equivalent circuit parameters are:
Z πn = zl = 16 + j133.4 Ω = 134.35e j 83.16 Ω
0
l
= j 6.551 × 10 −4 S
2
1
= ' = − j1526.48 Ω
Yπn
Yπ'n = y '
Z π' n
The results for this example are summarized in Figure E3.4.
15.08
j129.54
8.489
16.00
j133.4
8.489
-j1504.66
-j1504.66
-j1526.48
-j1526.48
(a)
(b)
Figure E3.4 The Positive Sequence Equivalent Circuits (a) pi-equivalent
Circuit, (b) Nominal pi-equivalent Circuit
(All indicated values are in ohms)
3.7 Power Line Analysis - Three Phase
Analusis of single phase power lines has been presented in section 3.6. In this section we
examine the analysis of three phase lines. As in the case of single phase lines, the
analysis will be focused on developing equivalent circuits that describe the behavior of
the power line.
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3.7.1 Three Phase Transmission Line – Time Domain Model
In this section we extend the analysis of a single-phase transmission line to three-phase or
multiphase transmission lines. Figure 3.21 illustrates a three-phase transmission line. The
transmission line of Figure 3.21 is the simplest three-phase overhead line. It comprises
four paths of electric current flow: the three phase conductors and the earth. Shield or
neutral conductors, which are present in most power lines, are omitted for the purpose of
minimizing the complexity of the presentation. Now consider the three phase conductors.
Each conductor is above earth. By utilizing the models developed in earlier sections, each
conductor i is characterized with a series self-resistance rii ; series mutual resistance to
any other conductor j, rij ; series self-inductance Liie ; series mutual inductance to any
Lije ; self-capacitance Cii; and mutual capacitance to any other
conductor j, Cij. The series self- and mutual resistance and inductance can be computed
using Carson's method or other alternative methods. For simplicity, we shall use the
approximate method referred to as the method of equivalent depth of return. According to
this method, the series parameters will be given by the following equations:
other conductor j,
Self-resistance:
rii = ri + re
Mutual resistance:
rij = re
Self-inductance:
Liie =
µ De
ln
d
2π
Mutual inductance:
Lije =
µ De
ln
2π d ij
where
re
ri
di
= ωµ /8
= conductor i resistance per unit length, ohms per meter
= conductor i geometric mean radius
De = 2,160
ρ
f
d ij
ρ
feet, equivalent depth of return
f
= soil resistivity
= frequency of currents
= distance between conductors i and j
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
y=0
y
dy
y=
i a(0,t)
i a( ,t)
i b(0,t)
i b( ,t)
i c(0,t)
i c( ,t)
(a)
(b)
Figure 3.21 General Representation of a Three Phase Line
(a) Schematic Representation and Notation
(b) Equivalent Circuits of an Infinitesimal Length
The self- and mutual capacitances are computed using the methods developed in earlier
sections. First the matrix C' is computed:
'
⎡C aa
⎢ '
C ' = ⎢C ba
⎢C ca'
⎣
'
C ab
'
C bb
C cb'
C ac' ⎤
⎥
C bc' ⎥
C cc' ⎥⎦
where:
Cij' =
1
2πε
ln
d ij
d ij '
d ij = distance between conductors i and j (dii equals the radius of conductor i)
d ij' = distance between conductors i and the image of conductor j with
respect to the earth interface
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Then the matrix C' is inverted to yield the matrix C:
C = (C
)
' −1
⎡C aa
= ⎢C ba
⎢
⎢⎣C ca
C ab
C bb
C cb
C ac ⎤
C bc ⎥
⎥
C cc ⎥⎦
The matrix above defines the self- and mutual capacitances. The self and mutual
parameters of the three phase conductors are represented schematically in Figure 3.21b.
For completeness, we also assume that there is self- and mutual conductance between the
phase conductors, represented by the conductance matrix G.
Now application of Kirchhoff's voltage and current law on the circuit of Figure 3.21b
yields
d⎞
⎛
− v ( y , t ) − ⎜ Rdy + Ldy ⎟i ( y + dy, t ) + v ( y + dy, t ) = 0
dt ⎠
⎝
i ( y, t ) + Gdy v( y, t ) + Cdy
dv( y, t )
− i ( y + dy, t ) = 0
dt
where
⎡ia ( y, t )⎤
i ( y , t ) = ⎢i b ( y , t ) ⎥
⎥
⎢
⎢⎣ic ( y, t ) ⎥⎦
⎡v a ( y , t )⎤
v( y, t ) = ⎢ v b ( y, t )⎥
⎥
⎢
⎢⎣ v c ( y , t ) ⎥⎦
⎡ re + ra
R = ⎢ re
⎢
⎢⎣ re
⎡ Laae
L = ⎢ Lbce
⎢
⎢⎣ Lcae
re
re + rb
re
Labe
Lbbe
Lcbe
re ⎤
re ⎥
⎥
re + rc ⎥⎦
Lace ⎤
Lbce ⎥
⎥
Lcce ⎥⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
⎡ g aae
G = ⎢ g bae
⎢
⎢⎣ g cae
g abe
g bbe
g cbe
⎡C aa
C = ⎢Cba
⎢
⎢⎣C ca
C ab
Cbb
C cb
g ace ⎤
g bce ⎥
⎥
g cce ⎥⎦
C ac ⎤
Cbc ⎥
⎥
C cc ⎥⎦
Upon division of the matrix equations above by “dy” and taking the limit as dy → 0, the
following matrix differential equations are obtained:
∂v ( y, t )
dy
∂i( y, t )
dy
= Ri( y, t ) + L
di( y , t )
dt
(3.30a)
= Gv( y, t ) + C
dv( y , t )
dt
(3.30b)
Again, as in the case of the single-phase line, the foregoing first-order matrix coupled
partial differential equations can be transformed into uncoupled second-order matrix
differential equations.
The final result is
∂ 2i( y, t )
dy
2
∂ 2 v( y, t )
dy
2
= CL
∂ 2 i( y, t )
= LC
dt
2
+ (GL + CR )
∂ 2 v( y, t )
dt
2
di ( y, t )
+ GRi ( y , t )
dt
+ ( LG + RC )
dv( y, t )
+ RGv( y, t )
dt
(3.31a)
(3.31b)
It should be observed that the differential equations describing a multiphase transmission
line are similar to the equations of a single-phase transmission line. The parameters of the
single-phase line have been substituted by appropriate matrices. In above derivation we
have considered a three phase line with three conductors and no neutral or ground
conductor. It should be apparent that the derivation will be similar to a transmission line
with any number of conductors. In this case the size of the matrices will be equal to the
number of conductors.
The computation of the matrices L, R, C, and G of the equations above are now
demonstrated with an example.
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Example E3.5. Consider a three-phase transmission line that is suspended on the
transmission tower illustrated in Figure 3.3. The position of the phase conductors is as
follows: phase A, position 1; phase B, position 2; phase C, position 3. the phase
conductors are ACSR, 336.4 kcm, 30 strand. The shield conductor is steel, 5/16 inches in
diameter. The soil resistivity is 150 ohm.meters. Compute the resistance matrix, the
inductance matrix, and the capacitance matrix of the line. The units should be ohms per
meter, henries per meter, and microfarads per meter, respectively. Use the equivalent
depth of return method. Neglect the shield conductor.
Solution: The resistance matrix of the line is computed with
Conductors:
rc = 0.278 ohms / mi = 0.000173 ohms / m
Earth:
re = 0.0954 ohms / mi = 0.0000593 ohms / m
⎡0.2323
R = ⎢0.0593
⎢
⎢⎣0.0593
0.0593
0.2323
0.0593
0.0593⎤
0.0593⎥ × 10 −3 Ω / m
⎥
0.2323⎥⎦
Inductance:
Liie =
µ 0 De
ln
2π d i
d i = 0.0255 ft
Lije =
µ 0 De
ln
2π Dij
De = 2,160
ρ
ft
f
Dab = 12.636 ft
Dac = 12.717 ft
Dbc = 8.170 ft
Upon substitution of above values, we get
⎡ 2.361
L = ⎢1.1199
⎢
⎢⎣1.2071
1.1199
2.361
1.1186
1.2071⎤
1.1186⎥ µH/m
⎥
2.361 ⎥⎦
Capacitance: The matrix C' is computed:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
⎡8.0457
1 ⎢
C' =
1.9977
2πε ⎢
⎢⎣2.3788
1.9977
7.9591
1.9004
2.3788⎤
1.9004⎥
⎥
7.859 ⎥⎦
Upon inversion, the capacitance matrix is
⎡ 7.8844
C = ⎢ − 1.4954
⎢
⎢⎣− 2.0248
− 1.4954
7.7016
− 1.4097
− 2.0248⎤
− 1.4097 ⎥ pF/m
⎥
8.0324 ⎥⎦
3.7.2 Three-Phase Transmission Line – Frequency Domain
Model
As in the case of a single phase transmission line, we consider the operation of the line
under sinusoidal steady state conditions. Since the three-phase transmission line is a
linear system, the currents and voltages at any point, y, in the transmission line will vary
sinusoidally with time. Thus, in general,
(
(
)
)
~
i ( y, t ) = Re 2 I ( y )e jωt
~
v( y, t ) = Re 2V ( y )e jωt
(3.40a)
(3.40b)
~
~
where I ( y ), V ( y ) , are vectors of complex numbers (phasors) and ω=2πf.
Above expressions are substituted into Equations (3.31b) and (3.30a) and after
eliminating the common factors, we obtain:
~
d 2V ( y )
~
= ( R + jωL)(G + jωC )V ( y )
2
dy
(3.41a)
~
dV ( y )
~
= ( R + jωL) I ( y )
dy
(3.41b)
Define the following matrices:
Z ≡ R + jωL
Y ' ≡ G + jωC
Then
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
d 2V ( y )
~
= ZY 'V ( y )
2
dy
(3.42a)
~
dV ( y )
~
= ZI ( y )
dy
(3.42b)
The foregoing matrix differential equations in complex variables fully describe the
performance of a general three-phase transmission line. Solution of these equations for
specified boundary conditions will yield the electric currents and voltages of any phase at
any location of the line. However, solution of the equations above is rather difficult. In
the following section we discuss transformations that decompose the matrix equations
(3.42) into scalar equations. In this way, the solution of the matrix equations (3.42)
reduces to the solution of a set of scalar equations.
3.7.3 Three-Phase Transmission Line – Sequence Models
The model of a three-phase transmission line under sinusoidal steady state condition is
defined by Equations (3.42). Solution of these equations is in general complex because
the matrices Z, Y' are full matrices resulting in a set of three coupled differential
equations. To simplify the solution, observe that it is possible to find a transformation T
~
~
of the voltage and current vector V ( y ) and I ( y ) as follows:
~
~
TV ( y ) = V m ( y )
or
~
~
V ( y ) = T −1V m ( y )
(3.43a)
~
~
TI ( y ) = I m ( y )
or
~
~
I ( y ) = T −1 I m ( y )
(3.43b)
~
where T is a 3×3 matrix, V m ( y ) are the transformed voltages at location y of the line,
~
and I m ( y ) are the transformed currents at location y of the line. Substitution of the
transformation above into Equations (3.42) and subsequent premuliplication of the
resulting equation by T results in
~
d 2V ( m ) ( y )
~
= TZY ' T −1V ( m ) ( y )
2
dy
(3.44a)
~
dV ( m ) ( y )
~
= TZT −1 I ( m ) ( y )
dy
(3.44b)
Now assume that T has been selected in such a way that the matrices TZY ' T −1 and
TZT −1 are diagonal matrices. In this case Equations (3.44) represent six uncoupled
~
differential equations. The voltages V m ( y ) are called the modal voltages of the line and
the transformation T is called a modal transformation matrix. Similarly, the currents
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~
I m ( y ) are called the modal currents. The procedures is called the modal decomposition.
The advantages of modal decomposition are obvious. Solution of the decoupled
equations (3.44) is identical to solution methods for single phase lines.
A special case of modal decomposition is the symmetrical component transformation,
introduced in Chapter 2. For convenience, we repeat the symmetrical component
transformation matrix, T, here:
⎡1
1⎢
T = ⎢1
3
⎢1
⎣
a2⎤
⎥
a ⎥ , where: a = e j120 ,
1 ⎥⎦
a
a
2
0
1
and
T −1
⎡1
= ⎢a 2
⎢
⎢⎣ a
1 1⎤
a 1⎥
⎥
a 2 1⎥⎦
In general however, the symmetrical component transformation, T, will not make the
matrices, TZY ' T −1 and TZT −1 , diagonal. For this purpose, we introduce an
approximation that will make the transformed matrices diagonal. Specifically, many
transmission lines are transposed or their construction is such that the mutual parameters
(inductance, capacitance) are approximately the same for any pair of phases and the
phase self-parameters are also approximately the same for all three phases. For some
applications it is justifiable to approximate a three-phase power line with a symmetric
line. Mathematically, this is equivalent to assuming that the matrices Z and Y' have the
following symmetric structure:
⎡ zs
Z = ⎢ zm
⎢
⎣⎢ z m
zm
zs
zm
⎡ y s'
⎢
Y ' = ⎢ y m'
⎢ y m'
⎣
y m'
y s'
y m'
zm ⎤
zm ⎥
⎥
z s ⎦⎥
y m' ⎤
⎥
y m' ⎥
y s' ⎥⎦
Note that if the matrices Z and Y' do not have this form, which is usually the case, they
are put in this form using the following equations:
1
z s = ( z aa + zbb + z cc )
3
1
z m = ( z ab + zbc + z ca )
3
1
y ' s = ( y ' aa + y ' bb + y ' cc )
3
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
1
y ' m = ( y ' ab + y ' bc + y ' ca )
3
The product ZY' of the two matrices is computed to be
⎡α 1
ZY ' = ⎢α 2
⎢
⎢⎣α 2
α2
α1
α2
α2 ⎤
α2 ⎥
⎥
α1 ⎥⎦
where
α1 = z s y s' + 2 z m y m'
α 2 = z m y m' + z s y m' + z m y s'
Now under the discussed assumption of symmetry, the symmetrical component
transformation will make the transformed matrices diagonal and the resulting three sets
of uncoupled equations will be referred to as the positive, negative and zero sequence
models of the three-phase transmission line.
~
The modal voltages V m ( y ) in this case will be the symmetrical compenents (positive,
negative and zero sequence) and they will be denoted by:
~
⎡V1 ( y ) ⎤
~
⎢~
⎥
V120 ( y ) = ⎢V2 ( y )⎥
⎢V~0 ( y )⎥
⎣
⎦
~
Similarly, the modal current I m ( y ) in this case will be the symmetrical compenents
(positive, negative and zero sequence) and they will be denoted by:
~
⎡ I1 ( y) ⎤
~
⎢~
⎥
I 120 ( y ) = ⎢ I 2 ( y )⎥
⎢ I~0 ( y )⎥
⎣
⎦
Upon substitution into Equations (3.44), we obtain
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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~
d 2V120 ( y )
~
= M seqV120 ( y )
2
dy
(3.45a)
~
dV120 ( y )
~
= Z seq I 120 ( y )
dy
(3.45b)
where
M seq
Z seq
⎡m1
=⎢0
⎢
⎢⎣ 0
⎡ z1
= ⎢0
⎢
⎢⎣ 0
0
m1
0
0
z1
0
0⎤
0⎥
⎥
m0 ⎥⎦
0⎤
0⎥
⎥
z 0 ⎥⎦
m1 = α1 − α 2 = p12
m0 = α1 + 2α 2 = p02
z1 = z s − z m
z0 = z s + 2 zm
The matrix equations (3.45) represent six scalar equations which can be grouped into
three sets of uncoupled equations. It is expedient to write the six scalar equations
explicitly as three sets (positive, negative and zero sequence) of uncoupled equations
with two equations per set:
Positive sequence set:
~
d 2V1 ( y )
~
= p12V1 ( y )
2
dy
~
dV1 ( y )
~
= z1 I 1 ( y )
dy
(3.46a)
(3.46b)
Negative sequence set:
~
d 2V2 ( y )
~
= p12V2 ( y )
2
dy
~
dV2 ( y )
~
= z1 I 2 ( y )
dy
Page 64
(3.47a)
(3.47b)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Zero sequence set:
~
d 2V0 ( y )
~
= p 02V0 ( y )
2
dy
~
dV 0 ( y )
~
= z 0 I 0 ( y)
dy
(3.48a)
(3.48b)
It is now apparent that Equations (3.46), (3.47) and (3.48) represent three single-phase
lines. Note that Equations (3.46) are the positive sequence model of the line, Equations
(3.47) are the negative sequence model, and Equations (3.48) are the zero sequence
model of the line. Note that the parameters ( p1 , z1 ) of the negative sequence model are
identical to those of the positive sequence model. Collectively, we shall refer to
Equations (3.45) or equivalently Equations (3.46), (3.47), and (3.48) as the sequence
model of a three-phase line. The modal voltages and currents will be referred to as the
symmetrical components of the currents and voltages. In addition, the parameters of the
sequence models are defined as follows:
p1 , p0 will be called the positive (or negative) and zero sequence propagation constants
of the line.
z1 , z0 will be called the per unit length positive (or negative) and zero sequence series
impedance of the line.
For the purpose of completing the discussion of the sequence model, recall that
M seq = TZY ' T −1
Consider the following:
'
M seq = TZY ' T −1 = TZT −1TY ' T −1 = Z seqTY ' T −1 = Z seqYseq
where:
Y ' seq = TY ' T −1
Upon evaluation of Y'seq , we have
⎡ y1'
⎢
Y ' seq = ⎢ 0
⎢0
⎣
0
y1'
0
0⎤
⎥
0⎥
y 0' ⎥⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
where
y1' = y s' − y m'
y 0' = y s' + 2 y m'
Note that y1' , y 0' are the per unit length positive (or negative) and zero sequence shunt
admittance of the line.
In terms of the parameters y'1, y'0, the propagation constants p1, p2 and p0 are
p1 = p 2 = z1 y1'
(3.49a)
p 0 = z 0 y 0'
(3.49b)
and the characteristic impedances (positive, negative and zero sequence):
Z 01 = Z 02 =
Z 00 =
z1
y1'
z0
y 0'
(3.50a)
(3.50b)
In summary, application of the symmetrical component transformation on the three-phase
line equations results in the sequence models of the line (i.e., the positive, negative, and
zero sequence models). Each model is identical to a single-phase line model. The
parameters of the positive sequence model are equal to the parameters of the negative
sequence model.
A physical interpretation of the sequence models of a three-phase transmission line is
expedient. For this purpose, assume that only one symmetrical component of the voltage
or current is present. As an example, assume that only the positive sequence current is
~
~
~
~
present [i.e., I 1 ( y ) ≠ 0 , I 2 ( y ) = 0 , and I 0 ( y ) = 0 ]. The actual phase currents I a ( y ) ,
~
~
I ( y ) , I ( y ) are obtained from the inverse transformation T-1:
b
c
~
~
⎡ I 1 ( y )⎤ ⎡ I 1 ( y ) ⎤
o ⎥
~
⎢
⎥ ⎢~
I abc ( y ) = T −1 ⎢ 0 ⎥ = ⎢ I 1 ( y )e − j120 ⎥
o
⎢ 0 ⎥ ⎢ I~1 ( y )e − j 240 ⎥
⎣
⎦ ⎣
⎦
As is evident from the equation above, the three phase currents are balanced and of the
positive sequence. The case is depicted in Figure 3.22a, which illustrates the three phase
currents. Note that the electric current in the earth is zero.
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Similarly, if we assume that only the negative sequence component is present [i.e.,
~
~
~
I 1 ( y ) = 0 , I 2 ( y ) ≠ 0 , and I 0 ( y ) = 0 ], the actual phase currents are
~
⎡ 0 ⎤ ⎡ I 2 ( y) ⎤
o ⎥
~
~
⎢~
I abc ( y ) = T −1 ⎢ I 2 ( y )⎥ = ⎢ I 2 ( y )e j120 ⎥
⎢
⎥
o
~
⎢⎣ 0 ⎥⎦ ⎢⎣ I 2 ( y )e j 240 ⎥⎦
Again, as is evident from equation above, the three phase currents are balanced but of the
negative sequence. The case is depicted in Figure 3.22b, which illustrates the three phase
currents. Note that the electric current in the earth is zero.
~
Finally, if we assume that only the zero sequence component is present, [ i.e., I 1 ( y ) = 0 ,
~
~
I 2 ( y ) = 0 , and I 0 ( y ) ≠ 0 ], the actual phase currents are
~
⎡ 0 ⎤ ⎡ I 0 ( y )⎤
~
⎢~
⎥
I abc ( y ) = T −1 ⎢ 0 ⎥ = ⎢ I 0 ( y )⎥
⎢~
⎥
~
⎢⎣ I 0 ( y )⎥⎦ ⎢⎣ I 0 ( y )⎥⎦
As is evident from the equation above, all three phase currents are identical. Sequence
cannot be defined for these currents-thus the name "zero sequence". The earth current
~
will be the negative sum of the phase currents [i.e., − 3I 0 ( y ) ]. The case is depicted in
Figure 3.22c.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
y
I 1(y)
o
I 1(y) e
-j120
I 1(y) e
-j240
o
(a)
y
I 2(y)
o
I 2(y) e
-j120
I 2(y) e
-j240
o
(b)
y
I 0(y)
I 0(y)
I 0(y)
-3 I 0(y)
(c)
Figure 3.22 Illustration of the Symmetrical Components on a Transmission
Line, (a) Positive Sequence Components, (b) Negative Sequence
Components,(c) Zero Sequence Components
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Example E3.6. Consider the transmission line of Example E3.5. Compute the sequence
parameters of the line. Neglect the shield wire.
Solution: The Z matrix of the line, computed at 60 Hz, and then symmetrized is:
⎡ 0.2323 + j 0.8901
Z = ⎢0.0593 + j 0.4330
⎢
⎢⎣0.0593 + j 0.4330
0.0593 + j 0.4330
0.2323 + j 0.8901
0.0593 + j 0.4330
0.0593 + j 0.4330⎤
0.0593 + j 0.4330⎥ × 10 −3 ohms / meter
⎥
0.2323 + j 0.8901⎥⎦
The Y' matrix of the line, computed at 60 Hz, and then symmetrized is:
⎡ j 3.0331
Y ' = ⎢ − j 0.5556
⎢
⎢⎣ − j 0.5556
− j 0.5556
j 3.0331
− j 0.5556
− j 0.5556 ⎤
− j 0.5556 ⎥ × 10 −9 S / m
⎥
j 3.0331⎥⎦
The product ZY' is
⎡- 2.2186 + jo.6387
ZY ' = ⎢- 0.5782 + j0.0178
⎢
⎢⎣- 0.5782 + j0.0178
- 0.5782 + j0.0178
- 2.2186 + jo.6387
- 0.5782 + j0.0178
- 0.5782 + j0.0178⎤
- 0.5782 + j0.0178⎥ × 10 −12 m -2
⎥
- 2.2186 + jo.6387⎥⎦
Upon application of the transformation T, we have
z1 = (0.173 + j0.4571)(10 −3 ) Ω/m
z 0 = (0.3509 + j1.7561)(10 −3 ) Ω/m
y1' = ( j3.5887)(10 −9 ) S/m
y 0' = ( j1.9219)(10 −9 ) S/m
m1 = (− 1.6404 + j0.6209)(10 −12 ) m -2
m0 = (− 3.375 + j 0.6743)(10 −12 ) m -2
The characteristic impedance and propagation constants of the sequence components are:
Positive (or negative) sequence components:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Z 01 =
o
z1
= 369.0e − j10.365
y '1
ohms
p1 = p 2 = z1 y1' = (1.3244)(10 −6 )e j 79.63 m -1
o
Zero sequence components:
Z 00 =
o
z0
= 965.0e − j 5.65
'
y0
ohms
o
p 0 = z 0 y 0' = 1.8552 × 10 −6 e j 84.35 m -1
In summary of this section, the symmetrical component transformation provides a tool
for the simplified solution of the equations of a multiphase line. It also yields the
sequence models of a three-phase line. In this case the analysis of three-phase lines is
reduced to the analysis of three single-phase transmission lines, the positive sequence,
negative sequence and zero sequence line models. The analysis of each one of the
sequence models is identical to that of a single-phase transmission line. This topic has
been covered in section 3.6. An example will illustrate the procedure.
Example E3.7. Consider the positive sequence model of the three-phase transmission
line of example E3.6. The line is 85 miles long. Compute:
(a) The π-equivalent circuit.
(b) The nominal π-equivalent circuit.
(c) Compare the two models
Solution: The positive sequence parameters of the line are:
o
Z 01 = 369.0e − j10.365
ohms
p1 = (1.3244)(10 −6 )e j 79.63 m -1
o
The A, B, C parameters of the line are computed as follows:
o
pl = 0.18113e j 79.63
cosh pl = 0.984659 + j 0.005809 = 0.984676e j 0.34
o
sinh pl = 0.032092 + j 0.177323 = 0.180204e j 79.74
o
o
A = 0.984676e j 0.34
o
B = 66.4953e j 69.38
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
o
C = 0.000488e j 90.1
(a)
Zπ = B = (23.417 + j62.235) Ω
B
= (8.489 − j 4053.59) Ω
A −1
The π-equivalent circuit is illustrated in Figure E3.7a.
Z π' =
(b)
o
Z πn = zl = Z 0 pl = 66.837e j 69.27 ohms = ( 23.658 + j 62.509) ohms
o
2 z' 2Z 0
=
= 4079.42e − j 90 ohms = − j 4074.4 ohms
l
pl
The nominal π-equivalent circuit is illustrated in Figure E3.7b.
Z π' n =
(c)
The two equivalent circuits are very close.
15.08
j129.54
8.489
16.00
j133.4
8.489
-j1504.66
-j1504.66
-j1526.48
(a)
-j1526.48
(b)
Figure E3.7 Positive Sequence Equivalent Circuits (a) pi-Equivalent Circuit,
(b) Nominal pi-Equivalent Circuit
(All indicated values are in ohms)
3.8 Transmission Line Power Equations
There are a number of applications, such as the power flow problem, stability analysis,
and so on, in which the power transmitted through a transmission line is required in the
formulation of the problem. In these applications it is assumed that the line operates
under sinusoidal or nearly sinusoidal steady-state conditions. In this section we derive
equations that describe the power flow through a transmission line. For this purpose the
transmission line is typically represented with the sequence models. Depending on the
application, the line may operate under balanced conditions (e.g., the power flow
problem) or unbalanced conditions (e.g., short circuit analysis, stability analysis of
unbalanced faults, etc.). In general, then, the power line may operate under unbalanced
~ ~ ~
~ ~ ~
conditions. Let VS 1 ,VS 2 ,VS 0 and VR1 ,VR 2 ,VR 0 be the sending and receiving bus positive,
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
~ ~ ~
~ ~ ~
negative, and zero sequence voltages, respectively, and I S1 , I S 2 , I S 0 and I R1 , I R 2 , I R 0 be
the sending and receiving bus positive, negative, and zero sequence electric currents,
respectively. The power injected at the sending end is SS:
~
⎡ I S∗1 ⎤
−1 ⎢ ~ ⎥
~T ~∗
~
~
~
S S = Vabc
I abc = VS 1 VS 2 VS 0 T −T (T * ) ⎢ I S∗2 ⎥
⎢ I~S∗0 ⎥
⎣ ⎦
Note that (by direct computation):
[
T
−T
(T )
* −1
]
⎡ 3 0 0⎤
= ⎢ 0 3 0⎥
⎥
⎢
⎢⎣0 0 3⎥⎦
~ ~
~ ~
~ ~
Thus: S S = 3VS1 I S∗1 + 3VS 2 I S∗2 + 3VS 0 I S∗0
(3.54a)
Similarly, the power delivered at the receiving end is SR:
~ ~
~ ~
~ ~
S R = 3VR1 I R∗1 + 3VR 2 I R∗2 + 3VR 0 I R∗0
(3.54b)
Equations (3.44) state that the power transmitted through a transmission line is computed
from the symmetrical components of the voltages and currents. In addition, the
contribution to the power flow from a specific symmetrical component is independent
from the other two. As example, for the power flow problem it is assumed that the line
operation is balanced (i.e., only the positive sequence component is present). In this case
the power flow equations become
~ ~
S S = 3VS1 I S∗1
and
~ ~
S R = 3VR1 I R∗1
It is expedient to express the power equations in terms of the line terminal voltages only.
For this purpose the line terminal currents (each symmetrical component separately) are
expressed in terms of the line terminal voltages (symmetrical components) and the line
parameters. Subsequently, they are substituted into Equations (3.44) to yield the power
equations in terms of the line terminal voltages only. Subsequently, the π − equivalent
circuit of the power line will be used to demonstrate the procedure.
The power equations in terms of the terminal voltages and the parameters of the
π − equivalent circuit, Zπ and Z'π, are obtained as follows: First, the line terminal
currents are expressed in terms of the parameters Yπ, Y'π and the line terminal voltages.
A rather popular form of the power equations is based on representing the parameters Yπ,
Y'π with their cartesian coordinates and the line terminal voltages with their polar
coordinates. Consider for example a line between buses k and m as it is illustrated in
Figure 3.34. In this case
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Pkm + jQkm
Pmk + jQmk
~
Ik
Yπ = g + jb
+
~
Vk
~
Im
+
Y'π = g' + jb'
_
~
Vm
_
Figure 3.23
Yπ = g + jb
Yπ' = g s + jbs
~
Vk = Vk e jδ k
~
Vm = Vm e jδ m
Upon substitution and separation of the real and imaginary parts of the two equations, the
following power flow equations are obtained:
Pkm = 3( g + g s )Vk2 − 3gVkVm cos(δ k − δ m ) − 3bVkVm sin(δ k − δ m )
(3.55a)
Qkm = −3(b + bs )Vk2 − 3gVkVm sin(δ k − δ m ) + 3bVkVm cos(δ k − δ m )
(3.55b)
Pmk = 3( g + g s )Vm2 − 3gVkVm cos(δ m − δ k ) − 3bVkVm sin(δ m − δ k )
(3.55c)
Qmk = −3(b + bs )Vm2 − 3gVkVm sin(δ m − δ k ) + 3bVkVm cos(δ m − δ k )
(3.55d)
It is expedient to rewrite above equations in the following compact form:
Pkm = 3( g + g s )Vk2 − 3α kmVkVm
Qkm = −3(b + bs )Vk2 − 3β kmVkVm
Pmk = 3( g + g s )Vm2 − 3α mkVmVk
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Qmk = −3(b + bs )Vm2 − 3β mkVmVk
Where:
α km = g cos(δ k − δ m ) + b sin(δ k − δ m )
bkm = g sin(δ k − δ m ) − b cos(δ k − δ m )
The transmission line power equations (3.55) are the most popular form for load flow
applications.
In most applications, the power flow equations are written in per unit. The per unit
system is defined in Chapter 4. Whenthe above equations are transfiormed into the per
unit system, the final equations are in the same form as above, except that the constant
term 3 is eliminated. Specifically, the per unitized power flow equations are:
Pkmu = ( g u + g su )Vku2 − α kmuVkuVmu
Qkmu = −(bu + bsu )Vku2 − β kmuVkuVmu
2
Pmku = ( g u + g su )Vmu
− α mkuVmuVku
2
Qmku = −(bu + bsu )Vmu
− β mkuVmuVku
Where:
α kmu = g u cos(δ k − δ m ) + bu sin(δ k − δ m )
bkmu = g u sin(δ k − δ m ) − bu cos(δ k − δ m )
and the subscript u denotes a perunitized quantity.
For more details on the per unitization procedure, see Chapter 4. Many times in this book
we will use the per unitized equations without the subscript u.
3.9 Transmission Line Power Transfer Limitations
The ability of transmission lines to transfer electric power is limited by a variety of
factors. These factors can be classified into two categories: (a) line design factors and (b)
system factors.
The line design factors are typically the thermal limitation of the line conductors.
Specifically, a transmission line is designed to operate at a nominal voltage and the
temperature of the conductors should not be higher than a permissible value. The
permissible value is determined by two factors: (1) the conductor should not loose its
mechanical strength properties and (2) the conductor sag should be limited by the
recommended clearances at mid spans of the line. A heat transfer analysis from the
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conductor to its surroundings under worst conditions (typically, the worst conditions are
assumed to be 40 degrees Celsius and no wind) determines the maximum amount of
current through the conductor that will bring the temperature of the conductor to the
permissible value. This current value is referred to as the conductor ampacity. Since the
operation of the line is typically characterized as near nominal voltage operation, the
ampacity of the conductor can be translated into power flow through the line, i.e.
Pmax = 3VN I ampacity
The quantity above is the power carrying capability of the line.
It should be apparent that when the line is operating not at nominal voltage, then its
power carrying capability will be different. It should also be apparent that the assumption
for determining the power carrying capability of the line may be too pessimistic. This is
indeed the case. Most of the times the ambient temperature is much lower than 40
degrees and there is almost always a breeze. For this reason, technology is being
developed to monitor the conductor temperature in real time. In this way one knows
exactly when the power carrying limit of the line has been reached. Direct monitoring of
the line temperature results in dynamically determining the power carrying capability.
There are three different technologies for this purpose: (a) use of temperature sensors on
the conductor, (b) use of a conductor with impedded fiber optic/temperature sensor in the
conductor, and (c) monitoring of the conductor sag and indirectly deriving the conductor
temperature from the sag. These technologies provide the temperature of the conductor in
real time. There are also technologies that focus on development of new conductors that
are capable of operating in higher permissible temperatures. The idea here is to use
materials with a very small heat expansion factor. This means that as the conductor
temperature increases due to increased current, the conductor will not expand as much
thus limiting the conductor sag. Therefore these conductors can be operated to much
larger temperatures with practically very small sag. Since the lines are operated at higher
electric current, the power transefred via the line is increased. This technology is
presently under evaluation.
It is also important to note that many times the power carrying capability of a line may be
limited by the ratings of other equipment that operate in series with the line. For example
consider a transmission line that has a series capacitor. If the current carrying capability
of the capacitor is lower than the ampacity of the line conductors, then the series
capacitor limits the current carrying capability of the line.
Another factor that limits the power carrying capability is the desired abililty to maintain
the voltage magnitude at the remote end at permissible levels.
System factors also affect the ability of a power line to transfer power. These
considerations may be voltage stability issues and transient stability issues. Typically
long transmission lines are limited by system issues. The level of limitation depends on
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system parameters and operating conditions. Some of these issues are discussed in
subsequent sections.
3.10 Summary and Discussion
In this chapter we examined models of transmission lines. Specifically, we derived
equations for the parameters of a line and the general model of single-phase as well as
multiple-phase transmission lines was developed in terms of a set of differential
equations. The single-phase transmission-line model, under sinusoidal steady-state
conditions, can be represented as a two-port network with three constants A, B, C, or
with an equivalent circuit. The π − equivalent circuit of the line was developed. For short
transmission lines, the approximate nominal π − equivalent circuit has been also
introduced.
Three-phase or multiple-phase transmission lines are modeled by a set of coupled
differential equations. Using the symmetrical components transformation, and the
approximation of a symmetric three-phase transmission line, a three-phase transmission
line is represented with three equivalent single phase models: the positive, negative, and
zero sequence models. Each sequence model can be represented with a two-port network
with constants, A, B, and C or with the π − equivalent circuit.
For many applications, such as power flow, transfer capability, etc., it is assumed that the
system is symmetric and it operates under balanced conditions. In this case, only the
positive sequence model of the power lines is used.
The representation of a three-phase line with the sequence models is an approximation.
Specifically, it neglects line asymmetries, and the explicit representation of the
shield/neutral wires and grounding structures is lost. In several applications it may be
desirable to explicitly include the shield/neutral wires and the grounding of the line. For
example, design of grounding systems, lightning performance analysis of the line, etc.
These applications are not discussed in this book. The interested reader may consult
reference [???].
Finally, equations for the power flow through a line have been developed under
sinusoidal conditions. The power equations are useful for several applications, such as
power flow studies and stability analysis.
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3.11 Problems
Problem P3.1. Consider an infinitely long conduit with a single solid copper conductor
of 2 cm diameter placed on the center of the conduit. The inner diameter of the conduit is
5 cm and the outer diameter is 5.3 cm. Assume that the conductor carries an electric
current of 1000 A which returns via the conduit. Assume the current is uniformly
distributed over the cross section of the conductor and the conduit.
(a) Compute the resistance of the system per unit length. Use copper and aluminium
conductivity from Tables.
(b) Plot the variation of the magnetic flux density B (in webers per square meter) along a
radial line perpendicular to the axis of the conduit.
(c) Calculate the inductance of the system in henries per meter.
5.3 cm
5 cm
2 cm
Figure P3.1
Problem P3.2. Compute the geometric mean radius (GMR) of the conduit of Problem
3.1. Assume constant current density inside the conduit.
Problem P3.3. Compute the inductive reactance of a 15 kV cable with concentric
neutral consisting of eight #14 copper wires symmetrically arranged around the
conductor as it is illustrated in Figure P3.3 assuming that the electric current returns via
the eight wires of the neutral, each wire carries the same current and the electric current
distribution over the cross-section of each conductor is uniform. The dimensions are
shown in the Figure.
b
a
a = 1 cm
b = 0.5 cm
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Figure P3.3. Approximate Model of a Concentric Neutral Cable
Problem P3.4:
A dc transmission line consists of two solid copper conductors a, b
of diameter 3 cm. The conductors are 50 ft above ground and at a distance of 30 ft
between centers. A third (ground return) conductor, g, is located 60 ft above ground and
symmetrically arranged with respect to conductors a and b. The diameter of conductor g
is 1.0 cm. The voltages of conductors a and b are Ea = 250 kV and Eb = -250 kV,
respectively. The voltage of conductor g is Eg = 0.0 .
(a) Calculate the electric charge of conductors a, b, and g, neglecting the effect of
ground.
(b) Calculate the electric charge of conductors a, b, and g, without neglecting the
effect of ground.
Problem P3.5: A single-phase 7.2 kV distribution line is illustrated in Figure P3.5. The
phase conductor is a solid copper conductor of 1.5 cm diameter and the neutral conductor
is solid aluminum conductor of 1 cm diameter. The phase conductor is the top conductor
in Figure P3.5 and the neutral conductor is the conductor under the phase conductor.
3'
30'
Figure P3.5
(a) Compute the capacitive currents of the line. Do not neglect the effects of earth.
(b) Compute the maximum electric charge accumulated on the surface of the phase
and neutral conductors in coulombs per meter.
(c) Compute the maximum electric field in volts per meter.
Hint: The neutral conductor is under zero potential.
Problem P3.6: Consider the three-phase transmission line of the Figure P3.6. Each phase
conductor has the following parameters:
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Radius:
Geometric Mean Radius:
Resistance:
0.691 inches
0.0456 feet
0.085 ohms per mile
The transmission line is symmetrically transposed, 125 miles long and the soil resistivity
along the line is 185 ohm⋅meters.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Compute the series resistance matrix of the line (per unit length).
Compute the series inductance matrix of the line (per unit length).
Compute the capacitance matrix of the line (per unit length).
Compute the r1, L1, and C1 parameters of the line per unit length. r1, L1, and C1 are the
positive sequence resistance, inductance and capacitance respectively.
Compute r0, L0, and C0 parameters of the line per unit length. r0, L0, and C0 are the
zero sequence resistance, inductance and capacitance respectively.
Compute and draw the positive sequence π-equivalent circuit of the line.
Compute and draw the zero sequence π-equivalent circuit of the line.
Compute the positive sequence characteristic impedance and propagation constant
of the line.
Compute the zero sequence characteristic impedance and propagation constant of
the line.
If the line carries a set of three phase balanced currents of 850 Amperes rms,
compute the magnetic flux density at a point directly below the outer phase on the
right and 5 feet above ground.
12'
12'
70'
Figure P3.6
Problem P3.7: Consider the three-phase overhead distribution line illustrated in the
Figure P3.7. The line is 12.3 miles long and it is symmetrically transposed. The soil
resistivity is 225 ohm.meters. Each phase conductor is an ACSR conductor with the
following parameters:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Resistance:
Radius:
Geometric Mean Radius:
a)
b)
0.08 ohms/mile
0.375 inches
0.0243 feet
Compute the positive sequence pi-equivalent circuit of the line.
Compute the zero sequence pi-equivalent circuit of the line.
3'
2'
40'
Figure P3.7
Problem 3.8: Consider a three-phase transmission line with tower design as illustrated in
Figure P3.8. The line is effectively grounded. Compute the zero sequence series
impedance of the line assuming that all the zero sequence current returns through the
earth. Assume a 175-Ω⋅m soil resistivity. The phase conductors are bundle conductors
consisting of two subconductors spaced 12 in. apart. Each subconductor is ACSR, 795
kcm, 54 strands. The subconductor has the following parameters:
Resistance:
Geometric Mean Radius:
Diameter:
0.1376 ohms per mile
0.0368 feet
1.093 inches
(a) Compute the positive sequence series impedance in ohms per meter.
(b) Compute the zero sequence series impedance in ohms per mile.
(c) Compute the positive sequence capacitance of the line in Farads per meter.
(d) Compute the zero sequence capacitance of the line in Farads per meter.
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24'
24'
80'
Figure P 3.8
Problem 3.9: Consider the three phase, symmetrically transposed, transmission line of
Figure P3.9. Each phase conductor consists of two ACAR subconductors. Each
subconductor has the following parameters :
Radius :
Geometric Mean Radius :
Distance between Subconductors:
0.7 inches
0.046 feet
12 inches
a) Compute the positive sequence series inductance of the line in Henries per meter.
b) Compute the positive sequence capacitance of the line in Farads per meter
Given:
eo = 8.854 . 10-12
F/m and µo = 4 p 10-7
20'
a
H/m
20'
b
c
80'
Figure P 3.9
Problem 3.10: What is the propagation constant of a one inch diameter aluminum
conductor above earth at 60 Hertz?
Given :
Conductivity of Aluminum σ Al = (0.35) 108 S
( )
Soil conductivity
Copyright © A. P. Sakis Meliopoulos – 1990-2006
σ soil = 0.005 S
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
Problem 3.11: Consider a single phase transmission line 185 miles long with the
following parameters :
R = 0.00173 ohm/meter
L = 1.212 mH/meter
C = 9.519 pF/meter
a) Compute the pi equivalent circuit of the line at the fifth harmonic (f=300 Hz)
b) Compute the nominal pi equivalent circuit of the line at the fifth harmonic (f=300 Hz)
c) Compare the two equivalent circuits and state your conclusions.
Problem 3.12: Consider the 230 kV transmission line of Figure P3.12. The line carries a
balanced three phase electric current of 1200 Amperes, i. e.
ia(t) =
2 1200 A cos( ω t + ϕ )
ib(t) =
2 1200 A cos( ω t + ϕ - 120o)
ic(t) =
2 1200 A cos( ω t + ϕ + 120o)
Compute the magnitude of the magnetic field density in Teslas at point A of the Figure.
Neglect the presence of earth.
a
20'
b
20'
c
50'
6'
Figure P3.12
Problem 3.13: Consider a power system comprising an electric load, a three phase
transmission line and a generating unit. The positive sequence impedance of the line is 1
+ j8 ohms. The shunt admittance is negligible. The total electric load (three phases) is
108 MW + j35 MVARS. The voltage at the electric load is 165 kV line to line.
a) Compute the power factor of the electric load.
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b) Compute the electric current of the load (line current).
c) Compute the real and reactive power generated by the generating unit. (total 3 phase)
d) Compute the real power loss of the transmission line (total 3 phase)
Problem 3.14: A three phase overhead transmission line has a horizontal arrangement of
the phases. The distance between two adjacent phases in D. The geometric mean radius
of each phase conductor is 0.04 feet. The line is symmetrically transposed.
Compute the positive sequence inductance of the line per unit length (in Henries per
meter) for the following D values: D = 10 feet, 20 feet and 50 feet.
Problem 3.15: Consider a three-phase transmission line of the Figure P3.15. The line is
symmetrically transposed. Each phase conductor consists of ACSR subconductors. Each
subconductor has the following parameters:
Radius:
Geometric Mean Radius:
Distance between Subconductor:
a)
b)
c)
0.6 inches
0.042 feet
12 inches
Compute the positive sequence series inductance in Henries per meter.
Compute the positive sequence capacitance of the line in Farads per meter.
Compute the speed of propagation of the positive sequence voltage and current
along the line in meters per second.
12'
12'
1 2 '‘
Figure P3.15
Problem 3.16: A 200 mile long transmission line has the following zero sequence
parameters:
R=0
L = 0.3125 µH/m
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
C = 0.125 nF/m
Compute the zero sequence π-equivalent parameters of the line.
12'
12'
1 2 '‘
Figure P3.16
Problem 3.17: A telephone line wire runs parallel to a three phase line as is indicated in
the Figure P3.22. The power transmission line carries a balanced set of three phase
currents. The magnitude of the phase A current is 560 Amperes. Compute the induced
voltage on the telephone line wire in Volts per mile.
a
18'
b
18'
c
8'
20'
50'
Figure P3.17
Problem 3.18: A 15 kV power cable consists of an aluminum conductor and a concentric
copper neutral as is illustrated in the Figure P3.18. The insulation is cross linked
polyethylene of thickness 175 miles (=0.175 inches). Compute the inductance of this
cable in Henries per meter assuming uniform current distribution inside conductor and
neutral.
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0.425"
0.25"
0.475"
Al
XLP
Cu
Figure P3.18
Problem 3.19: A three phase, symmetrically transposed, 500 kV transmission line
consists of bundled phase conductors as is illustrated in the Figure P3.24. Each
subconductor is
ACSR, 954000 cm, 54 aluminum strands, 7 steel strands,
Separation between any two subconductor is = 18".
a) Compute the positive sequence resistance of the line in ohms per meter
b) Compute the positive sequence capacitance in Farads per meter.
c) Compute the positive sequence inductance in Henries per meter.
30'
30'
100'
Figure P3.19
Problem 3.20: A telephone line parallels a three phase power line for a distance of one
mile (1609 meters). The geometric arrangement is illustrated in the Figure P3.20. The
three phase power line carries a balanced electric current of 35 Amperes (rms), positive
sequence, and of frequency 420 Hz, i.e. the seventh harmonic of 60 Hz.
(a)
(b)
Write the time functions expressing the instantaneous current in each of the three
phases of the power line, ia(t), ib(t), and ic(t).
Compute the voltage induced on the telephone line.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
a
12'
b
12'
c
8'
16'
1'
40'
Figure P3.20
Problem 3.21: Consider the three-phase transmission line of Figure P3.21. The line is
symmetrically transposed. Each phase conductor consists of two ACSR subconductors.
Each subconductor has the following parameters:
Resistance:
Radius:
Geometric Mean Radius:
Distance between Subconductors:
a)
b)
c)
d)
e)
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0.085 ohms per mile
0.691 inches
0.0465 feet
18 inches
Compute the positive sequence series resistance of the line in ohms per meter.
Compute the positive sequence series inductance of the line in Henries per meter.
Compute the positive sequence capacitance of the line in Farads per meter.
Compute the characteristic impedance and propagation constant of the line for f =
60 Hz.
Compute and draw the π-equivalent circuit of the line if the line is 87 miles long. (
f = 60 Hz )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
20'
20'
18'‘
Figure P3.21
Problem 3.22: Consider a three-phase, 60 Hz, transmission line of the Figure P3.22. The
phase conductors are ACSR with the following parameters:
Radius:
Geometric Mean Radius:
0.5 inches
0.0324 feet
The ground wire is neglected. The soil resistivity is 235 ohm.meters. Compute the
positive sequence sequence series inductance in Henries per meter and the positive
sequence capacitance in µF per meter.
20'
20'
50'
Figure P3.22
Problem P3.23: Consider the 230 kV transmission line of Figure P3.23. The phase
conductors are ACSR, 1272 (PHAESANT), with the following parameters: resistance at
60 Hz=0.0751 ohms per mile, Geometric Mean Radius= 0.04672 feet, and Diameter=
1.382 inches. The soil resistivity is 185 ohm.meters. For simplicity ignore the shield wire.
The line is 89 miles long.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 3, Meliopoulos
a) Compute the positive sequence pi-equivalent circuit of the line at 60 Hz.
b) What is the percent error of the impedances due to “symmetrization” of the line
model? Hint: Compute the asymmetry factor.
a
20'
b
20'
c
50'
6'
Figure P3.23
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