Page 1
Transient Regimes of Power Transmission
Lines with Cyclical Transposition
Luís Manuel Antunes Fiel
M
Abstract— This work describes an analysis method
suitable for the computation of transient waveforms in
networks with overhead multiphase power lines
included. To model the transmission lines we have
used the Fourier Transform, because it allows us to use
non-linear elements on the power network. The main
concern of this work is to analyze the differences of
three types of configuration of the line: line with nontransposition, line with perfect transposition and line
with cyclical transposition. The transient regime is
analyzed with resource of the “The MathWorks, Inc.”,
the MATLAB®, with the objective of calculating and
representing the transient regimes of a multiphase
airline.
much more profitable, therefore allowing the inclusion of
non-linear element in the network, and later, with resource
to the Inverse Fourier Transform, shift these solutions to the
time domain.
For the concretion of the assays above related, we have
elaborated a program with the resource to the tool of the
“The MathWorks, Inc.”, the MATLAB®, with the objective of
calculating the transient regimes of the power line.
II. EQUATIONS OF THE MULTIPHASE LINES
𝒚
Keywords— Theoretical Electrotechnic. Power
Transmission Lines. Transient Analysis. Cyclical
Transposition. Fourier Transform.
I. INTRODUCTION
In this work we intend to compare transient regimes in
networks that include power lines of energy transmission,
namely lines with non-transposition, lines with perfect
transposition and lines with cyclical transposition.
Firstly it is convenient to calculate the characteristic
parameters of the line, such as the series-impedance and
the shunt-admittance, having in consideration the skin effect
in the land, imperfect conductor, and in the aerial
conductors.
For the skin effect corrections in the land we’re going to
use the Dubanton method, which is a sufficiently
reasonable approach of the reality and it is perfectly
suitable for the frequencies that we are going to assay.
For the skin effect correction in the aerial conductors, we
will appeal to the Bessel functions, which are more exact.
Now we will analyze the transitory regimes for the
diverse situations of transposition and load of the line,
namely the assays in the situation of load, open-circuit and
short-circuit.
So we will use the method of the Fourier Transform that
allows us to deal with the solutions in the frequency domain,
2𝑟2
2𝑟𝑛
2𝑟1
𝑕1
𝑥1
𝑕2
…
𝑥2
𝑕𝑛
𝑥𝑛
Figure 1 – Multiphase Line scheme in ground presence
Now we will consider the frequency-domain matrix
equation for a non-uniform multiphase transmission line
[NEVES90]:
𝑑𝑰
= −𝒀𝑽
𝑑𝑧
𝑑𝑽
= −𝒁𝑰
𝑑𝑧
(1)
where 𝒁 is the series-impedance matrix and 𝒀 is the shuntadmittance matrix.
The series-impedance matrix, 𝒁 is given by:
𝒁 𝜔 = 𝑗𝜔𝑳 + ∆𝒁𝒑 𝜔 + ∆𝒁𝑮 𝜔
(2)
Page 2
The external-inductance matrix 𝑳 is a frequencyindependent real symmetric non-singular matrix whose
elements are evaluated according to:
𝐿𝑖𝑖 =
𝐿𝑘𝑖 |𝑘≠𝑖 =
𝜇0
2𝜋
𝑙𝑛
𝜇0 2𝑕𝑖
𝑙𝑛
2𝜋
𝑟𝑖
𝑕 𝑖 +𝑕 𝑘
2+
𝑥 𝑖 −𝑥 𝑘
2
𝑕 𝑖 −𝑕 𝑘
2+
𝑥 𝑖 −𝑥 𝑘
2
(2)
where hi and xi denotes, respectively, the vertical and
horizontal coordinates of conductor 𝑖.
The matrix 𝒁𝒑 is a frequency dependent complex
diagonal matrix that can be determined through the skineffect theory for cylindrical conductors [BRAND06].
(3)
In this work we use the Bessel functions, and
defined as:
𝑃. 𝑟𝑛 . 𝐽0 𝑃. 𝑟
6
2. 𝐽1 𝑃. 𝑟
where 𝑅𝐷𝐶 represents the per-unit-length resistance in dc,
and J0 and J1 denotes, respectively, the Bessel functions of
∆𝒁𝒑 (𝑤) = 𝑅𝑐𝑐
The 𝑟𝑖 denotes the conductor radius, and 𝑕𝑖 and 𝑥𝑖
denotes the vertical and horizontal coordinates of
conductor 𝑖, respectively.
0º and 1º order, and 𝑝 =
The matrix 𝒁𝑮 represents the ground impedance
correction. This is a frequency dependent matrix that can be
determined using the Dubanton’s method [DUBAN69].
1
(7)
where, 𝜸2 is the eigenvectors of the product (𝒁𝒀), which
can be given by:
(4)
𝑗𝜔 𝜇 0 𝜎𝑔
−𝑗𝜔𝜇𝜎.
From (1), we can now write:
𝑑2 𝑰
= 𝜸2 𝑰
𝑑𝑧 2
𝑑2 𝑽
= 𝜸2 𝑽
𝑑𝑧 2
This technique is based on the method of phase images
but where the plan of null potential is located at a complex
depth 𝑃 as:
𝑑=
𝒁𝒑 is
where 𝜎𝑔 ≈ 1 × 10−3 is the ground conductivity.
𝜸2 = 𝑻−𝟏 𝒁𝒀 𝑻
𝑞1
(8)
The modal matrix 𝑻 is a non-singular similarity
transformation, whose columns are the eigenvalues of the
product (𝑍𝑌).
𝑑12
𝑞2
𝑕1
𝑕2
𝑕1
According to [NEVES90]
𝑕2
𝑑
x
𝐷′12
𝜞−1 = 𝑻 𝛾 −1 𝑻−1
= 𝑻 𝑒 (±𝜸𝒛) 𝑻−1
𝑒 (±𝜞𝒛)
(9)
Finally we have the voltage and the current:
−𝑕2
−𝑕1
𝑰 = 𝒀 𝜞−1 𝑒 (−𝜞𝒛) 𝑽1 − 𝒀 𝜞−1 𝑒 (𝜞𝒛) 𝑽2
𝑽 = 𝑒 (−𝜞𝒛) 𝑽1 + 𝑒 (𝜞𝒛) 𝑽2
−𝑞2
(10)
where 𝑧 is a generic point of the power line.
−𝑞1
Figure 2- Representation of Dubanton’s Method.
The matrix of the ground impedance correction 𝒁𝑮 is
achieved with:
𝑗𝜔𝜇0
𝑑
𝒁𝑮 𝑖𝑖 (𝑤) =
ln 1 +
(5)
2𝜋
𝑕𝑖
𝒁𝑮
𝑖𝑘 (𝑤)
=
𝑗𝜔𝜇0
ln
4𝜋
hi + hk + 2P 2 + xi − xk
hi + hk 2 + xi − xk 2
2
(6)
III. THE FOURIER TRANSFORM
The Fourier transform is very useful for the transient
analysis, but only if we have linear time-invariant systems
on the network.
Through the Fourier transform one signal could be
represented as a linear combination of complex
exponentials as [ARIEH05]:
Page 3
𝑓 𝑡 =
1 𝑐+𝑗 ∞
2𝜋 𝑐−𝑗 ∞
𝑒 𝑠𝑡 𝐹(𝑠) 𝑑𝑠
According to (15), (16) and (10) the transfer functions for
a generic point of the power line with non-transposition are
given as:
(11)
The 𝐹 𝑠 is the Inverse Fourier transform, which could be
expressed as:
𝐹 𝑠 =
+∞
−∞
𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡
𝑰𝑧 = 𝒀𝜞−1 𝑒
𝑼𝑧 = 𝑒
(12)
For the transient analysis we have to assure that the
transient takes place in 𝑡 = 0 or 𝑡 = 𝑡0, so we will use an
auxiliary function:
𝑢 𝑡 =
0,
1,
𝑡 < 0 (𝑡 < 𝑡0 )
𝑡 > 0 (𝑡 > 𝑡0 )
+∞
−∞
𝑒 −𝑠𝑡 𝑢(𝑡)𝑓(𝑡) 𝑑𝑡 =
+∞
0−
−𝜞𝑧
𝑩𝑨−1 𝑽𝐺 − 𝒀𝜞−1 𝑒 𝜞𝑧 𝑨−1 𝑽𝐺
𝑩𝑨−1 𝑽𝐺 + 𝑒 𝜞𝑧 𝑨−1 𝑽𝐺
(17)
where
𝑨 = 𝒁𝐺 𝒀𝜞−1 + 𝑬 𝑒 (𝜞𝒍) 𝒁𝐶 𝒀𝜞−1 − 𝑬 −1 𝒁𝐶 𝒀𝜞−1 + 𝑬 𝑒 (𝜞𝒍)
+ 𝑬 − 𝒁𝐺 𝒀𝜞−1
𝜞𝒍
and 𝑩 = 𝑒
𝒁𝐶 𝒀𝜞−1 − 𝑬 −1 𝒁𝐶 𝒀𝜞−1 + 𝑬 𝑒 𝜞𝒍 .
𝑬 is the identity matrix.
(13)
Now we can represent the generator voltage 𝑽𝐺 in
frequency-domain with the Fourier transform, as showed in
the (Figure 5):
This turns the equation (12) into:
𝐹 𝑠 =
−𝜞𝑧
𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡
(14)
1000
With this the generator voltages are represented as
(Figure 3):
0
-1000
-500
-400
-300
-200
-100
0
100
Frequency [Hz]
200
300
400
500
-400
-300
-200
-100
0
100
Frequency [Hz]
200
300
400
500
-400
-300
-200
-100
0
100
Frequency [Hz]
200
300
400
500
1
1000
0.8
0
0.6
-1000
Voltage [pu]
0.4
-500
0.2
0
1000
-0.2
0
-0.4
-1000
-500
-0.6
-0.8
Figure 5 - Fourier Transform of the generator voltages.
-1
-0.01
-0.005
0
0.005
0.01
Time[s]
0.015
0.02
0.025
Figure 3 - Generator voltages.
IV. TRANSIENT ANALYSIS IN A NON-TRANSPOSED LINE
Now we will analyze the transfer functions for the power
line with non-transposition with the help of (Figure 4) and
Kirchoff laws, and so we can write:
𝑽𝐺
~
𝐺1
𝑰𝐶
𝑰𝑆
𝒁𝐺
𝒚
𝐺2
𝑧=𝑙
𝑧=0
𝑽𝑠
𝒁𝐶
𝑽𝐶
𝐶1
~
𝐶3
𝐶2
Yg1
Yg2
Figure 4 - Power line equivalent scheme with generator and load.
Yc2
Yc1
𝑽𝑠 = 𝑽𝐺 − 𝒁𝐺 𝑰𝑠
(15)
𝑽𝐶 = 𝒁𝐶 𝑰𝐶
(16)
Xc1
Xg1
Xc2
Yc3
Xg2
𝒛
Figure 6 - Scheme of studied power line structure.
Xc3
𝒙
Page 4
-3
Current [pu]
x 10
2
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
-3
x 10
Current [pu]
We will now represent the transient regimens of the nontransposition line, according to (Table 1), which gives us the
standard structure of the power line that we will assay along
this work. We will also consider the length of the line as
300𝐾𝑚, operating in a frequency of 50𝐻𝑧, being the
generator impedance 𝑍𝐺 = 25Ω and the load impedance
𝑍𝑐 = 400Ω.
The blue line represents the real part and the red line
represents the imaginary part.
2
0
-2
-0.005
-3
Conductor
Radius
Coordinate
Coordinate
Conductivity
[𝒎𝒎]
𝑿 [𝒎]
𝒀 [𝒎]
[𝑺/𝒎]
C1
15.9
−12
26
4.5 × 106
C2
15.9
0
26
4.5 × 106
6
C3
15.9
12
26
4.5 × 10
G1
7.3
−8
36
4.5 × 106
G2
7.3
8
36
4.5 × 106
Table 1 - Configuration of the assayed power line
Current [pu]
x 10
2
0
-2
-0.005
Figure 8- Current in z=300km of the non-transposed power line
with load.
We will now consider the situations of open-circuit and
short-circuit for the non-transposed line.
In the open-circuit case we will use the transfer functions
expressed in (17), where now, 𝑨 = 𝒁𝐺 𝒀𝜞−1 + 𝑬 𝑒 2𝜞𝒍 +
𝑬−𝒁𝐺 𝒀𝜞−1 and 𝑩 = 𝑒 2𝜞𝒍 .
In case of the short-circuit assay, the 𝑨 = − 𝒁𝐺 𝒀𝜞−1 +
𝑬𝑒2𝜞𝒍+𝑬−𝒁𝐺𝒀𝜞−1 and 𝑩=−𝑒2𝜞𝒍, so we have:
Voltage [pu]
1
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Voltage [pu]
1
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
1
Voltage [pu]
Voltage [pu]
Voltage [pu]
Voltage [pu]
For the non-transposed line with load we have:
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Figure 7 - Voltage in z=300km of the non-transposed power line
with load.
2
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
2
0
-2
-0.005
2
0
-2
-0.005
Figure 9 - Voltage in z=300km of the non-transposed power line
with open-circuit.
Current [pu]
Current [pu]
Current [pu]
Page 5
0.01
0
-0.01
-0.005
𝒁 𝑧 =
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
𝒁𝐴
, 0 < 𝑧 < 𝐿𝐴
𝒁𝐵 = 𝑃𝒁𝐴 𝑃 −1
, 𝐿𝐴 < 𝑧 < 𝐿𝐴 + 𝐿𝐵
𝒁𝐶 = 𝑃 −1 𝒁𝐴 𝑃 , 𝐿𝐴 + 𝐿𝐵 < 𝑧 < 𝐿𝐴 + 𝐿𝐵 + 𝐿𝐶
(18)
0.01
And the shunt-admittance is stepwise function:
0
-0.01
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
𝒀 𝑧 =
0.01
0
-0.01
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
𝒀𝐴
, 0 < 𝑧 < 𝐿𝐴
𝒀𝐵 = 𝑃𝒀𝐴 𝑃−1
, 𝐿𝐴 < 𝑧 < 𝐿𝐴 + 𝐿𝐵
𝒀𝐶 = 𝑃 −1 𝒀𝐴 𝑃 , 𝐿𝐴 + 𝐿𝐵 < 𝑧 < 𝐿𝐴 + 𝐿𝐵 + 𝐿𝐶
The standard approach to deal with the line structure
represented in (Figure 11) uses a homogenization
technique leading to a uniform line model described by
balanced per-unit-length series-impedance 𝒁 and shuntadmittance 𝒀 matrices:
Figure 10 - Current in z=300km of the non-transposed power line
with short-circuit.
V. TRANSIENT ANALYSIS IN A PERFECT TRANSPOSED LINE
𝒁=
For balancing reasons some airlines have their
conductors transposed along the line length, a complete
transposition cycle consisting of three cascaded line
sections differing on the conductor sequence order (Figure
11).
1
3
2
2
1
3
3
𝐿𝐴
2
𝐿𝐵
1
𝐿
Figure 11 - Transposition for a three-phase line structure.
The line structure in (Figure 11) includes three line
sections A, B and C, of length 𝐿𝐴 , 𝐿𝐵 and 𝐿𝐶 ,
respectively.
1
3
𝒀=
𝐿𝑐
0
(19)
𝑧
𝒁𝐴 + 𝒁𝐵 + 𝒁𝐶
1
3
𝒀𝐴 + 𝒀𝐵 + 𝒀𝐶
𝑍𝑝 𝑍𝑚 𝑍𝑚
= 𝑍𝑚 𝑍𝑝 𝑍𝑚
𝑍𝑚 𝑍𝑚 𝑍𝑝
(20)
𝑌𝑝 𝑌𝑚 𝑌𝑚
= 𝑌𝑚 𝑌𝑝 𝑌𝑚
𝑌𝑚 𝑌𝑚 𝑌𝑝
(21)
Now we can evaluate the propagation parameters
characterizing the normal modes of the structure, one
ground mode and two aerial modes.
The modal attenuation constants 𝛼 and modal phase
velocities 𝑣 obtained with this approach make the frequency
functions differ smoothly [FARIA02].
The ground (𝑔) mode is given by:
𝛾𝑔 = 𝛼𝑔 𝜔 + 𝑗
𝜔
𝜗𝑔 𝜔
=
𝑍𝑝 + 2𝑍𝑚 (𝑌𝑝 + 2𝑌𝑚 )
(22)
The Aerial (𝑎) modes are given by:
The Permutation matrix operator 𝑃 is defined as:
010
𝑃 = 001
100
𝛾𝑎 = 𝛼𝑎 𝜔 + 𝑗
(18)
=
𝑍𝑝 − 𝑍𝑚 (𝑌𝑝 − 𝑌𝑚 )
(23)
So the transient regimens for the line described in (Table
1) in the conditions of the preceding assays for the line with
perfect transposition with load are:
obeying the following proprieties[FARIA02]:
𝑃 2 = 𝑃 −1 = 𝑃 𝑡
𝑃3 = 𝑬
𝜔
𝜗𝑎 𝜔
(19)
𝑬 is the identity matrix and superscript 𝑡 denotes
transposition.
Assuming that each individual line section is
uniform, the series-impedance is stepwise function:
Current [pu]
1
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Current [pu]
1
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
1
Current [pu]
Voltage [pu]
Voltage [pu]
Voltage [pu]
Page 6
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Figure 12- Voltage in z=300km of the perfect transposed power
line with load.
0.01
0
-0.01
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0.01
0
-0.01
-0.005
0.01
0
-0.01
-0.005
Figure 15- Current in z=300km of the perfect transposed power
line with short-circuit.
-3
Current [pu]
x 10
2
VI. TRANSIENT ANALYSIS IN A CYCLE TRANSPOSED LINE
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
-3
Current [pu]
x 10
2
0
-2
-0.005
Considering the line structure in (Figure 11), which
includes three line sections A, B and C, of length 𝐿𝐴 , 𝐿𝐵
and 𝐿𝐶 , respectively, we will calculate the new transfer
functions for the cycle transposed power line, so we will
define 𝑻 as the transmission matrix.
-3
Current [pu]
x 10
2
𝑻 = 𝑻𝐴 𝑻𝐵 𝑻𝐶 =
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Figure 13- Current in z=300km of the perfect transposed power
line with load.
Voltage [pu]
Voltage [pu]
Voltage [pu]
Now considering the situations of open-circuit and shortcircuit for the perfect transposed line, we have:
0
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
2
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
2
0
-2
-0.005
𝑩
𝑫
(24)
The transmission matrix for the section 𝐴, 𝑇𝐴 , provides
the relationship between line currents and voltages
measured at ports 𝑍 = 0 and 𝑍 = 𝐿𝐴
𝑽(𝐿𝐴 )
𝑽0
= 𝑻𝐴
𝑰0
𝑰(𝐿𝐴 )
with:
𝑻𝐴 =
𝑨𝐴
𝑪𝐴
𝑩𝐴
𝑫𝐴
(25)
(26)
The square complex matrices 𝑨𝐴 , 𝑩𝐴 , 𝑪𝐴 and 𝑫𝐴 can be
obtained from 𝒁𝐴 and 𝒀𝐴 expressed in (18) and (19):
2
-2
-0.005
𝑨
𝑪
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Figure 14- Voltage in z=300km of the perfect transposed power
line with open-circuit.
𝑨𝐴 = 𝑐𝑜𝑠𝑕 𝜞𝐴 𝐿𝐴
𝑩𝐴 = 𝜞𝐴 𝒀𝑨 −1 𝑠𝑖𝑛𝑕 𝜞𝐴 𝐿𝐴
𝑪𝐴 = 𝜞𝐴 𝒁𝑨 −1 𝑠𝑖𝑛𝑕 𝜞𝐴 𝐿𝐴
𝑫𝐴 = 𝒀𝑨 𝜞𝐴 −1 𝑐𝑜𝑠𝑕 𝜞𝐴 𝐿𝐴 𝜞𝐴 𝒀𝑨 −1
(26)
where 𝜞𝐴 represents the square root of the product (𝒁𝑨 𝒀𝑨 ).
The transmission matrix for the section 𝐵 and 𝐶 is
achieved with the permutation matrix operator.
𝑃 0
𝑃 0 −1
𝑇𝐵 =
𝑇𝐴
(27)
0 𝑃
0 𝑃
𝑇𝐶 =
𝑃
0
0
𝑃
thus generating[FARIA02]:
−1
𝑇𝐴
𝑃
0
0
𝑃
(28)
𝑇=
𝐴
𝐶
𝐴 𝑃
𝐵
= 𝐴
𝐶𝐴 𝑃
𝐷
𝐵𝐴 𝑃
𝐷𝐴 𝑃
3
Voltage [pu]
Page 7
(29)
Voltage [pu]
Voltage [pu]
1
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
-1
-0.005
0
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
2
0
-2
-0.005
2
0
-2
-0.005
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
-1
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Current [pu]
1
-3
x 10
2
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
-3
x 10
2
0
-2
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
Current [pu]
Current [pu]
-2
-0.005
Figure 18- Voltage in z=300km of the transposed power line with
open-circuit.
0
Figure 16- Voltage in z=300km of the transposed power line with
load.
Current [pu]
0
1
Current [pu]
Voltage [pu]
Voltage [pu]
Voltage [pu]
So the transient regimens for the line described in (Table
1) in the conditions of the preceding assays for the line with
transposition with load are:
2
0.01
0
-0.01
-0.005
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045
Time [s]
0.01
0
-0.01
-0.005
0.01
0
-0.01
-0.005
Figure 19- Current in z=300km of the transposed power line with
short-circuit.
-3
Current [pu]
x 10
2
0
-2
-0.005
VII. RESULTS
Figure 17- Current in z=300km of the transposed power line with
load.
Considering the situations of open-circuit and shortcircuit for the transposed line, we have to do two
approaches: in case of open-circuit we assume 𝒁𝐶 = 1𝑀Ω
and for short-circuit we assume 𝒁𝐶 = 0Ω.
As we analyzed in the previous chapters, the transient
regimens of a non-transposed line and a perfect transposed
line are practically equals, so we will focus on the eventual
differences between the perfect transposition and the
transposition.
For a better analysis we have chosen the half period of
the wave, so that we can understand if there is any
difference between the two transposition methods.
Considering the situation of load, we have:
Page 8
1
2
0.8
1.5
0.6
1
0.5
0.2
Voltage [pu]
Voltage [pu]
0.4
0
-0.2
0
-0.5
-0.4
-1
-0.6
-1.5
-0.8
-1
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
-2
0.01
Figure 20- Voltage in z=300km of the perfect transposed power
line with load.
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
0.01
Figure 23- Voltage in z=300km of the transposed power line with
open-circuit.
Considering the situation of short-circuit, we have:
1
0.8
-3
0.6
8
x 10
0.4
4
0
-0.2
2
Current [pu]
Voltage [pu]
6
0.2
-0.4
-0.6
0
-2
-0.8
-1
-4
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
0.01
-6
Figure 21- Voltage in z=300km of the transposed power line with
load.
-8
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
0.01
Figure 24 - Current in z=300km of the perfect transposed power
line with short-circuit.
Considering the situation of open-circuit, we have:
2
-3
8
x 10
1.5
6
1
4
2
Current [pu]
Voltage [pu]
0.5
0
-0.5
0
-2
-1
-4
-1.5
-6
-2
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
0.01
Figure 22- Voltage in z=300km of the perfect transposed power
line with open-circuit.
-8
0
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009
Time [s]
0.01
Figure 25- Current in z=300km of the transposed power line with
short-circuit.
Page 9
Attenuation
𝜸2 = 𝑻−𝟏 𝒁𝒀 𝑻 = 𝑑𝑖𝑎𝑔 𝛾𝑡 , 𝛾𝑎′ , 𝛾𝑎′′
(30)
4.5
4
3.5
3
Attenuation[dB]
To get a clear understanding of the propagation
proprieties of the two transposition structures, we will
analyze the modal attenuation constants 𝛼 and modal
phase velocities 𝑣, according to (22) and (23).
For the perfect transposition line the propagation
constant 𝜸 is given by [NEVES90]:
2.5
2
1.5
1
Attenuation
4.5
0.5
4
0
400
3.5
Attenuation[dB]
3
600
800
1000
1200
1400
Frequency [Hz]
1600
1800
2000
Figure 28 - Attenuation of a transposition line.
2.5
Phase Velocity
1.05
2
1.5
1
Normalized Phase Velocity
1
0.5
0
400
600
800
1000
1200
1400
Frequency [Hz]
1600
1800
2000
Figure 26 - Attenuation of a perfect transposition line.
0.95
0.9
Phase Velocity
1.05
0.85
400
Normalized Phase Velocity
1
600
800
1000
1200
1400
Frequency [Hz]
1600
1800
2000
Figure 29- Phase velocity of a transposition line.
0.95
VIII. CONCLUSION
0.9
0.85
400
600
800
1000
1200
1400
Frequency [Hz]
1600
1800
2000
Figure 27- Phase velocity of a perfect transposition line.
And for a cycle transposition line is given by[FARIA02]:
𝑒 +𝛾𝑡 𝐿 , 𝑒 −𝛾𝑡 𝐿 , 𝑒 +𝛾 𝑎 ′𝐿 , 𝑒 −𝛾 𝑎 ′𝐿 , 𝑒 +𝛾 𝑎 ′′𝐿 , 𝑒 −𝛾 𝑎 ′′𝐿
(31)
In cycle transposition the resonant frequencies can be
found in entire multiples of half wavelength 𝐿 ≈ 𝑚𝜆 2.
𝑓𝑟𝑒𝑠 ≈
𝑚𝑐
2𝐿
(32)
where 𝐿 is the length of the line and 𝑐 is the velocity of light
in vacuum.
For balancing purposes many overhead power lines have
their phase conductors periodically transposed along the
line length.
With this work we have compared the transient analyses
of two kinds of transpositions, the perfect transposition and
cycle transposition, which revealed that there are no such
major differences between the two methods. As the perfect
transposition is less difficult to implement, it is fair to say
that we can use it for transient analyses.
Concerning the attenuation and the phase velocity, there
are major differences. In cycle transposition we have
resonant frequencies, which does not interfere in the
transmission system, as this one operates at 50𝐻𝑧, and the
first resonant frequency appears only at 500𝐻𝑧, but we
should take resonances into account if we want to transmit
information through the power line.
Page 10
IX. REFERENCES
[NEVES90] M. V. Guerreiro Neves, “Cálculo de
Transitórios em Linhas de Transmissão de energia
baseado no Emprego de um Esquema Equivalente por
Troços – Comparação com o Método da Transformada de
Laplace”, IST, UTL, Lisboa, Abril 1990.
[FARIA02] J. A. Brandão Faria and M. V. Guerreiro das
Neves, “Resonance Effects due to Conductor Transposition
in Three-Phase Power Lines”, 14th PSCC, Sevilla, 24-28,
June 2002.
[DUBAN69] C. Dubanton, “Calcul Approché dês
Paramètres Primaires et Secundaires d’une Ligne de
Transport”, EDF Bulletin de la Direction dês Études et
Recherches, no.1, pp. 53-62, 1969.
[BRAND06] J. A. Brandão Faria and M. E. Almeida,
“Accurate Calculation of Magnetic-Field Intensity due to
Overhead Power Lines with or without Mitigation Loops with
or without Capacitor Compensation”, IEEE, Jan. 2006.
[ARIEH05]
ARIEH L. SHENKMAN, “Transient Analysis
of Electric Power Circuits Handbook”, Holon Academic
Institute of Technology, Holon, Israel, 2005.