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Dr Houssem Rafik El Hana Bouchekara 1
Transmission and Distribution of
Electrical Power
Dr : Houssem Rafik El- Hana BOUCHEKARA
2009/2010 1430/1431
KINGDOM OF SAUDI ARABIA Ministry Of High Education
Umm Al-Qura University College of Engineering & Islamic Architecture
Department Of Electrical Engineering
Dr Houssem Rafik El Hana Bouchekara 2
1 ELECTRIC POWER TRANSMISSION ................................................................................. 3
1.1 BACKGROUND ................................................................................................................ 3
1.2 ELECTRIC TRANSMISSION LINE PARAMETERS ............................................................ 5
1.2.1 Line resistance ....................................................................................................... 5 1.2.1.1 Frequency Effect .......................................................................................................... 5 1.2.1.2 Temperature Effect ...................................................................................................... 6 1.2.1.3 Spiraling and Bundle Conductor Effect ........................................................................ 6 1.2.1.4 Proximity effects .......................................................................................................... 8
1.2.2 Inductance of a single conductor......................................................................... 10 1.2.2.1 Internal Inductance .................................................................................................... 10 1.2.2.2 Inductance Due To External Flux Linkage ................................................................... 11 1.2.2.3 Inductance of a two wire single phase lines............................................................... 12 1.2.2.4 Flux linkage in terms of self and mutual inductance .................................................. 14 1.2.2.5 Inductance of three phase transmission lines -symmetrical spacing- ........................ 15 1.2.2.6 Inductance of three phase transmission lines -asymmetrical spacing- ...................... 16 1.2.2.7 Transpose line ............................................................................................................ 17
1.2.3 Inductance of composite conductors ................................................................... 18 1.2.3.1 GMR of bundled conductors ...................................................................................... 21
1.2.4 Inductance of three phase double circuit line ...................................................... 26
1.2.5 Line capacitance .................................................................................................. 33
1.2.6 Capacitance of single phase lines ........................................................................ 34
1.2.7 Potential difference in a multiconductor configuration ...................................... 35
1.2.8 Capacitance of three phase lines ......................................................................... 36
1.2.9 Effect of bundling ................................................................................................ 38
1.2.10 Capacitance of three phase double circuit lines ................................................ 38
1.2.1 Effect of earth on the capacitance ...................................................................... 39
1.2.2 Magnetic field induction ...................................................................................... 43
1.2.3 Electrostatic induction ......................................................................................... 45
1.2.4 Corona ............................................................................................................... 45
Dr Houssem Rafik El Hana Bouchekara 3
1 ELECTRIC POWER TRANSMISSION
The electric energy produced at generating stations is transported over high-voltage
transmission lines to utilization points. The trend toward higher voltages is motivated by the
increased line capacity while reducing line losses per unit of power transmitted. The
reduction in losses is significant and is an important aspect of energy conservation. Better
use of land is a benefit of the larger capacity.
This chapter develops a fundamental understanding of electric power transmission
systems.
1.1 BACKGROUND
The transmission and distribution of three-phase electrical power on overhead lines
requires the use of at least three-phase conductors. Most low voltage lines use three-phase
conductors forming a single three-phase circuit. Many higher voltage lines consist of a single
three-phase circuit or two three-phase circuits strung or suspended from the same tower
structure and usually called a double-circuit line. The two circuits may be strung in a variety
of configurations such as vertical, horizontal or triangular configurations. Figure 1 illustrates
typical single-circuit lines and double-circuit lines in horizontal, triangular and vertical phase
conductor arrangements. A line may also consist of two circuits running physically in parallel
but on different towers. In addition, a few lines have been built with three, four or even six
three-phase circuits strung on the same tower structure in various horizontal and/or
triangular formations.
In addition to the phase conductors, earth wire conductors may be strung to the
tower top and normally bonded to the top of the earthed tower. Earth wires perform two
important functions; shielding the phase conductors from direct lightning strikes and
providing a low impedance path for the short-circuit fault current in the event of a back
flashover from the phase conductors to the tower structure. The ground itself over which
the line runs is an important additional lossy conductor having a complex and distributed
electrical characteristics. In the case of high resistivity or lossy earths, it is usual to use a
counterpoise, i.e. a wire buried underground beneath the tower base and connected to the
footings of the towers. This serves to reduce the effective tower footing resistance. Where a
metallic pipeline runs in close proximity to an overhead line, a counterpoise may also be
used in parallel with the pipeline in order to reduce the induced voltage on the pipeline from
the power line.
Therefore, a practical overhead transmission line is a complex arrangement of
conductors all of which are mutually coupled not only to each other but also to earth. The
mutual coupling is both electromagnetic (i.e. inductive) and electrostatic (i.e. capacitive).
The asymmetrical positions of the phase conductors with respect to each other, the earth
wire(s) and/or the surface of the earth cause some unbalance in the phase impedances, and
hence currents and voltages. This is undesirable and in order to minimise the effect of line
unbalance, it is possible to interchange the conductor positions at regular intervals along the
line route, a practice known as transposition. The aim of this is to achieve some averaging of
line parameters and hence balance for each phase. However, in practice, and in order to
Dr Houssem Rafik El Hana Bouchekara 4
avoid the inconvenience, costs and delays, most lines are not transposed along their routes
but transposition is carried out where it is physically convenient at the line terminals, i.e. at
substations.
Figure 1: (a) Typical single-circuit and double-circuit overhead lines and (b) double-circuit overhead lines with one earth wire: twin bundle=2 conductors per phase and quad bundle=4 conductors per phase.
Bundled phase conductors are usually used on transmission lines at 220 kV and
above. These are constructed with more than one conductor per phase separated at regular
intervals along the span length between two towers by metal spacers. Conductor bundles of
two, three, four, six and eight are in use in various countries.
The purpose of bundled conductors is to reduce the voltage gradients at the surface
of the conductors because the bundle appears as an equivalent conductor of much larger
diameter than that of the component conductors. This minimizes active losses due to
corona, reduces noise generation, e.g. radio interference, reduces the inductive reactance
and increases the capacitive susceptance or capacitance of the line. The latter two effects
improve the steady state power transfer capability of the line. Figure 1 (a)(ii) shows a typical
400 kV double-circuit line of vertical phase conductor arrangement having four bundled
conductors per phase, one earth wire and one counterpoise wire. The total number of
conductors in such a multi-conductor system is (4Γ3)Γ2+1+1=26 conductors, all of which are
mutually coupled to each other and to earth.
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1.2 ELECTRIC TRANSMISSION LINE PARAMETERS
The power transmission line is one of the major components of an electric power
system. Its major function is to transport electric energy, with minimal losses, from the
power sources to the load centers, usually separated by long distances. The design of a
transmission line depends on four electrical parameters:
1. Series resistance
2. Series inductance
3. Shunt capacitance
4. Shunt conductance
The series resistance relies basically on the physical composition of the conductor at
a given temperature. The series inductance and shunt capacitance are produced by the
presence of magnetic and electric fields around the conductors, and depend on their
geometrical arrangement. The shunt conductance is due to leakage currents flowing across
insulators and air. As leakage current is considerably small compared to nominal current, it is
usually neglected, and therefore, shunt conductance is normally not considered for the
transmission line modeling.
1.2.1 LINE RESISTANCE
The AC resistance of a conductor in a transmission line is based on the calculation of
its DC resistance. If DC current is flowing along a round cylindrical conductor, the current is
uniformly distributed over its cross-section area and its DC resistance is evaluated by
π π·πΆ =ππ
π΄ Ξ© ( 1)
where
Ο is the resistivity of conductor
l is the length
A is the cross-sectional area
If AC current is flowing, rather than DC current, the following factors need to be
considered:
1. Frequency or skin effect
2. Temperature
3. Spiraling of stranded conductors
4. Bundle conductors arrangement
5. Proximity effect
6. Also the resistance of magnetic conductors varies with current magnitude.
1.2.1.1 Frequency Effect
The frequency of the AC voltage produces a second effect on the conductor
resistance due to the nonuniform distribution of the current. This phenomenon is known as
skin effect. As frequency increases, the current tends to go toward the surface of the
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conductor and the current density decreases at the center. Skin effect reduces the effective
cross-section area used by the current, and thus, the effective resistance increases. Also,
although in small amount, a further resistance increase occurs when other current-carrying
conductors are present in the immediate vicinity. A skin correction factor k, obtained by
differential equations and Bessel functions, is considered to reevaluate the AC resistance.
For 60 Hz, k is estimated around 1.02
π π΄πΆ = ππ π·πΆ ( 2)
1.2.1.2 Temperature Effect
The resistivity of any conductive material varies linearly over an operating
temperature, and therefore, the resistance of any conductor suffers the same variations. As
temperature rises, the conductor resistance increases linearly, over normal operating
temperatures, according to the following equation:
π 2 = π 1 π + π‘2
π + π‘1 ( 3)
Where
R2 is the resistance at second temperature t2
R1 is the resistance at initial temperature t1
T is the temperature coefficient for the particular material (CΒ°)
Resistivity (π) and temperature coefficient (T) constants depend upon the particular
conductor material. Table 1 lists resistivity and temperature coefficients of some typical
conductor materials
Table 1: Resistivity and Temperature Coefficient of Some Conductors
1.2.1.3 Spiraling and Bundle Conductor Effect
There are two types of transmission line conductors: overhead and underground.
Overhead conductors, made of naked metal and suspended on insulators, are preferred over
underground conductors because of the lower cost and easy maintenance. Also, overhead
transmission lines use aluminum conductors, because of the lower cost and lighter weight
compared to copper conductors, although more cross-section area is needed to conduct the
same amount of current. There are different types of commercially available aluminum
conductors: aluminum-conductor-steel-reinforced (ACSR), aluminum-conductor-alloy-
reinforced (ACAR), all-aluminum-conductor (AAC), and all-aluminumalloy- conductor (AAAC).
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Figure 2: Stranded aluminum conductor with stranded steel core (ACSR).
ACSR is one of the most used conductors in transmission lines. It consists of
alternate layers of stranded conductors, spiraled in opposite directions to hold the strands
together, surrounding a core of steel strands. Figure 13.4 shows an example of aluminum
and steel strands combination. The purpose of introducing a steel core inside the stranded
aluminum conductors is to obtain a high strength-to-weight ratio. A stranded conductor
offers more flexibility and easier to manufacture than a solid large conductor. However, the
total resistance is increased because the outside strands are larger than the inside strands
on account of the spiraling. The resistance of each wound conductor at any layer, per unit
length, is based on its total length as follows:
π ππππ =π
π΄ 1 + π
1
π
2
Ξ©/π ( 4)
where
π ππππ : resistance of wound conductor (Ξ©)
1 + π1
π
2: length of wound conductor (m)
πππππ =ππ‘π’ππ
2ππππ¦ππ relative pitch of wound conductor
ππ‘π’ππ : length of one turn of the spiral (m)
2ππππ¦ππ : diameter of the layer (m)
The parallel combination of n conductors, with same diameter per layer, gives the
resistance per layer as follows:
π πππ¦ππ =1
1π π
ππ=1
Ξ©/π ( 5)
Similarly, the total resistance of the stranded conductor is evaluated by the parallel
combination of resistances per layer.
In high-voltage transmission lines, there may be more than one conductor per phase
(bundle configuration) to increase the current capability and to reduce corona effect
discharge. Corona effect occurs when the surface potential gradient of a conductor exceeds
the dielectric strength of the surrounding air (30 kV/cm during fair weather), producing
ionization in the area close to the conductor, with consequent corona losses, audible noise,
and radio interference.
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As corona effect is a function of conductor diameter, line configuration, and
conductor surface condition, then meteorological conditions play a key role in its evaluation.
Corona losses under rain or snow, for instance, are much higher than in dry weather.
Figure 3: Stranded conductors arranged in bundles per phase of (a) two, (b) three, and (c) four.
Corona, however, can be reduced by increasing the total conductor surface.
Although corona losses rely on meteorological conditions, their evaluation takes into
account the conductance between conductors and between conductors and ground. By
increasing the number of conductors per phase, the total cross-section area increases, the
current capacity increases, and the total AC resistance decreases proportionally to the
number of conductors per bundle. Conductor bundles may be applied to any voltage but are
always used at 345 kV and above to limit corona. To maintain the distance between bundle
conductors along the line, spacers made of steel or aluminum bars are used. Figure 13.5
shows some typical arrangement of stranded bundle configurations.
1.2.1.4 Proximity effects
In a transmission line there is a non-uniformity of current distribution caused by a
higher current density in the elements of adjacent conductors nearest each other than in the
elements farther apart. The phenomenon is known as proximity effect. It is present for
three-phase as well as single-phase circuits. For the usual spacing of overhead lines at 60 Hz,
the proximity effect is neglected.
Example 1:
A solid cylindrical aluminum conductor 25 km long has an area of 336.400 circular
miles. Obtain the conductor resistance at:
(a) 20Β° C
(b) 50Β° (C)
The resistivity of aluminum at 20Β° is 2.8 Γ 10β8Ξ©m .
1 square centimeter Γ 197= 1 circular mils.
Solution:
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Example 2:
A three phase transmission line is designed to deliver 190.5 MVA at 220 kV over a
distance of 63 km. the total transmission loss is not to exceed 2.5 percent of the rated line
MVA. If the resistivity of the conductor material at 20Β° is 2.8 Γ 10β8Ξ©m , determine the
required conductor diameter and the conductor size in circular miles.
Solution:
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1.2.2 INDUCTANCE OF A SINGLE CONDUCTOR
The inductive reactance is by far the most dominating impedance element.
A current-carrying conductor produces a magnetic field around the conductor. The
magnetic flux lines are concentric closed circles with direction given by the right hand rule.
With the thumb pointing in the direction of the current, the fingers of the right hand
encircled the wire point in the direction of the magnetic field. When the current changes,
the flux changes and a voltage is induced in the circuit. By definition, for nonmagnetic
material, the inductance L is the ratio of its total magnetic flux linkage to the current I, given
by
πΏ =π
πΌ ( 6)
Where π is the flux linkage, in Weber turns.
Consider a long round conductor with radius r, carrying a current I as shown in
Figure 4.
Figure 4: Flux linkage of a long conductor.
The magnetic field intensity π»π₯ , around a circle of radius x, is constant and tangent
to the circle. The Ampereβs law relating π»π₯ to the current πΌπ₯ is given by
π»π₯ .ππ2ππ₯
0
= πΌπ₯ ( 7)
Or
π»π₯ =πΌπ₯
2ππ₯ ( 8)
Where πΌπ₯ is the current enclosed at radius x. As shown in Figure 4. Equation ( 8) is all
that required for evaluating the flux linkage π of a conductor. The inductance of the
conductor can be defined as the sum of contributions from flux linkages internal and
external to the conductor.
1.2.2.1 Internal Inductance
A simple can be obtained for the internal flux linkage by neglecting the skin effect
and assuming uniform current density throughout the conductor cross section i.e.,
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πΌ
ππ2=
πΌπ₯ππ₯2
( 9)
Substituting for πΌπ₯ in ( 8) yields
π»π₯ =πΌπ₯
2ππ2π₯ ( 10)
For a nonmagnetic conductor wit constant permeability π0, the magnetic flux
density is given by π΅π₯ = π0π»π₯ , or
π΅π₯ =π0πΌ
2ππ2π₯ ( 11)
Where π0 is the permeability of free space (or air) and is equal to4π Γ 10β7π»/π.
The differential flux ππ for a small region of thickness ππ₯ and one meter length of
the conductor is
πππ₯ = π΅π₯.ππ₯. π =π0πΌ
2ππ2π₯ππ₯ ( 12)
The flux πππ₯ links only the fraction of the conductor from the center to radius x.
thus, on the assumption of uniform current density, only the fraction ππ₯2/ππ2 of the total
current is linked by the flux, i.e.,
πππ₯ = π₯2
π2 πππ₯ =π0πΌ
2ππ4π₯3ππ₯ ( 13)
The total flux linkage is found by integrating πππ₯ from 0 to r.
ππππ‘ =π0πΌ
2ππ4 π₯3ππ₯π
0
=π0πΌ
8π Wb/m ( 14)
From ( 6), the inductance due to the internal flux linkage is
πΏπππ‘ =π0 πΌ
8π=
1
2Γ 10β7 H/m ( 15)
Note that πΏπππ‘ is independent of the conductor radius r.
1.2.2.2 Inductance Due To External Flux Linkage
Consider π»π₯ external to the conductor at radius π₯ > π as shown in Figure 5. Since
the circle at radius x encloses the entire current πΌπ₯ = πΌ and in ( 8) πΌπ₯ is replaced by I and the
flux density at radius x becomes
π΅π₯ = π0π»π₯ =π0 πΌ
2ππ₯ T ( 16)
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Figure 5: Flux linkage between D1 D2
Since the entire current πΌ is linked by the flux outside the conductor, the flux linkage
πππ₯ is numerically equal to the flux πππ₯ . The differential flux πππ₯ for a small region of
thickness ππ₯ and one meter length of the conductor is then given by
πππ₯ = πππ₯ = π΅π₯ππ₯. 1 =π0πΌ
2ππ₯ππ₯ ( 17)
The external flux linkage between two points π·1 and π·2 is found by integrating
πππ₯ from π·1 to π·2
πππ₯π‘ =π0πΌ
2π
1
π₯
π·2
π·1
ππ₯ = 2 Γ 10β7 πΌ lnπ·2
π·1 Wb/m ( 18)
The inductance between two points external to a conductor is then
πΏππ₯π‘ = 2 Γ 10β7 lnπ·2
π·1 H/m ( 19)
1.2.2.3 Inductance of a two wire single phase lines
Consider one meter length of a single phase line consisting of two solid round
conductors of radius π1 and π2 as shown in Figure 6. The two conductors are separated by a
distance D. conductor 1 carries the phasor current πΌ1referenced into the page and conductor
2 carries return current πΌ2 = βπΌ1. These currents set up magnetic field lines that links
between the conductors as shown.
Figure 6: Single phase two wire line.
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Inductance of conductor 1 due to internal flux is given by ( 15). The flux beyond D
links a net current of zero and does not contribute to the net magnetic flux linkage in the
circuit. Thus, to obtain the inductance of conductor 1 due to the net external flux linkage, it
is necessary to evaluate ( 19) from π·1 = π1 to π·2 = π·.
πΏ1(ππ₯π‘ ) = 2 Γ 10β7 lnπ·
π1 H/m ( 20)
The total inductance of conductor 1 is then
πΏ1 =1
2Γ 10β7 + 2 Γ 10β7 ln
π·
π1 H/m ( 21)
Equation ( 21) is often rearranged as follows:
πΏ1 = 2 Γ 10β7 1
2+ ln
π·
π1
= 2 Γ 10β7 ln π14 + ln
1
π1+ ln
π·
1
= 2 Γ 10β7 ln1
π1πβ1/4
+ lnπ·
1
( 22)
Let π1β² = π1π
β1
4 , the inductance of conductor 1 becomes
πΏ1 = 2 Γ 10β7 ln1
π1β² + 2 Γ 10β7 ln
π·
1 H/m ( 23)
Similarly, the inductance of conductor 2 is
πΏ2 = 2 Γ 10β7 ln1
π2β² + 2 Γ 10β7 ln
π·
1 H/m ( 24)
If the two conductors are identical, π1 = π2 = π and πΏ1 = πΏ2 = πΏ, and the
inductance per phase per meter length of the line is given by
πΏ = 2 Γ 10β7 ln1
πβ²+ 2 Γ 10β7 ln
π·
1 H/m ( 25)
Examination of ( 25) revals that the first term is only a function of the conductor
radius. This term is considered as the inductance due to both the internal flux and that
external to conductor 1 to a radius of 1m. the second term of ( 25) is dependent only upon
conductor spacing. This term is known as the inductance spacing factor.
The term πβ = ππβ1/4 is known mathematically as the self geometric mean distance
of a circle with radius π and is abbreviated by GMR. πβ²can be considered as the radius of a
fictitious conductor assumed to have no internal flux but with the same inductance as the
actual conductor with radius r. GMR is commonly refered to as geometric mean radius and
will be designated by π·π .thus, the inductance per phase in millihenries per kilometer
becomes
πΏ = 0.2 lnπ·
π·π πH/Km ( 26)
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Example 3:
A single-phase transmission line 35 Km long consists of two solid round conductors,
each having a diameter of 0.9 cm. The conductor spacing is 2.5 m. calculate the equivalent
diameter of a fictitious hollow, thin-walled conductor having the same equivalent
inductance as the original line. What is the value of the inductance per conductor?
Solution:
1.2.2.4 Flux linkage in terms of self and mutual inductance
The series inductance per phase for the above sigle phase two wire line can be
expressed in terms of self inductance of each conductor and their mutual inductance.
Consider one meter length of the single phase circuit represented by two coils characterized
by the self inductances πΏ11 and πΏ22 and the mutual inductance πΏ12 . The magnetic polarity is
indicated by dot symbols as shown in Figure 7.
Figure 7: The single phase line viewed as two magnetically coupled coils
The flux linkage π1 and π2 are given by
π1 = πΏ11πΌ1 + πΏ12πΌ2
π2 = πΏ21πΌ1 + πΏ22πΌ2 ( 27)
Since πΌ2 = βπΌ1, we have
π1 = πΏ11 β πΏ12 πΌ1
π2 = βπΏ21 + πΏ22 πΌ2 ( 28)
Comparing ( 28)with ( 18)and ( 19), we conclude the following equivalent
expressions for the self and mutual inductances:
πΏ11 = 2 Γ 10β7 ln1
π1β² ( 29)
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πΏ22 = 2 Γ 10β7 ln1
π2β²
πΏ12 = πΏ21 = 2 Γ 10β7 ln1
π·
The concept of self and mutual inductance can be extended to a group of n
conductors. Consider n conductors carrying phasor currents πΌ1, πΌ2 ,β¦ , πΌπ , such that
πΌ1 + πΌ2 + πΌ3 +β―+ πΌπ +β―+ πΌπ = 0 ( 30)
Generalzing ( 27), the flux linkage of conductor i are
ππ = πΏππ πΌπ + πΏππ πΌπ
π
π=1
π β π ( 31)
Or
ππ = 2 Γ 10β7 πΌπ ln1
ππβ² + πΌπ ln
1
π·ππ
π
π=1
π β π ( 32)
1.2.2.5 Inductance of three phase transmission lines -symmetrical spacing-
Consider one meter length of a three phase line with three conductors, each with
radius r, symmetrically spaced in a triangular configuration as shown in Figure 8.
Figure 8: three phase line with symmetrical spacing.
Assuming balanced three phase currents we have
πΌπ + πΌπ + πΌπ = 0 ( 33)
From ( 32) the total flux linkage of phase a conductor is
ππ = 2 Γ 10β7 πΌπ ln1
πβ²+ πΌπ ln
1
π·+ πΌπ ln
1
π· ( 34)
Substituting β πΌπ = πΌπ + πΌπ
Dr Houssem Rafik El Hana Bouchekara 16
ππ = 2 Γ 10β7 πΌπ ln1
πβ²β πΌπ ln
1
π·
= 2 Γ 10β7πΌπ lnπ·
πβ²
( 35)
Because of symmetry, ππ = ππ = ππ , and the three inductances are identical.
Therefore, the inductance per phase per kilometer length is
πΏ = 0.2 lnπ·
π·π mH/km ( 36)
Where rβ is the geometric mean radius GMR, and is shown by Ds. for a solid round
conductor, π·π = ππβ1
4 for stranded conductor Ds can be evaluated from ( 50).
The comparison of the two inductances expressed by ( 36)and ( 26) shows that
inductance per phase for a three phase circuit with equilateral spacing is the same as for one
conductor of a single phase circuit.
1.2.2.6 Inductance of three phase transmission lines -asymmetrical
spacing-
Practical transmission lines cannot maintain symmetrical spacing of conductors
because of construction considerations. With asymmetrical spacing, even with balanced
currents, the voltage drop due to the line inductance will be unbalanced.
Consider one meter length of a three phase line with three conductors, each with
radius r. the conductors are asymmetrically spaced with distances shown in Figure 9.
Figure 9: three phase line with asymmetrical spacing.
The application of ( 32) will result in the following flux linkages.
ππ = 2 Γ 10β7 πΌπ ln1
πβ²+ πΌπ ln
1
π·12+ πΌπ ln
1
π·13
ππ = 2 Γ 10β7 πΌπ ln1
π·12+ πΌπ ln
1
πβ²+ πΌπ ln
1
π·23
ππ = 2 Γ 10β7 πΌπ ln1
π·13+ πΌπ ln
1
π·23+ πΌπ ln
1
πβ²
( 37)
Or in matrix form
Dr Houssem Rafik El Hana Bouchekara 17
π = πΏπΌ ( 38)
Where the symmetrical inductance matrix L is given by
πΏ = 2 Γ 10β7
ln
1
πβ²ln
1
π·12ln
1
π·13
ln1
π·12ln
1
πβ²ln
1
π·23
ln1
π·13ln
1
π·23ln
1
πβ²
( 39)
For balanced three phase currents with πΌπ as reference, we have
πΌπ = πΌπβ 240Β° = π2πΌπ
πΌπ = πΌπβ 120Β° = ππΌπ ( 40)
Where the operator π = 1β 120Β° and π2 = 1β 240Β°. Substituting in ( 37) result in
πΏπ =πππΌπ
= 2 Γ 10β7 ln1
πβ²+ π2 ln
1
π·12+ π ln
1
π·13
πΏπ =πππΌπ
= 2 Γ 10β7 π ln1
π·12+ ln
1
πβ²+ π2 ln
1
π·23
πΏπ =πππΌπ
= 2 Γ 10β7 π2 ln1
π·13+ π ln
1
π·23+ ln
1
πβ²
( 41)
Examination of ( 41) shows that the phase inductances are not equal and they
contain an imaginary term due to the mutual inductance.
1.2.2.7 Transpose line
The equilateral triangular spacing configuration is not the only configuration
commonly used in practice. Thus the need exists for equalizing the mutual inductances. One
means for doing this is to construct transpositions or rotations of overhead line wires. A
transposition is a physical rotation of the conductors, arranged so that each conductor is
moved to occupy the next physical position in a regular sequence such as a-b-c, b-c-a, c-a-b,
etc. Such a transposition arrangement is shown in Figure 10. If a section of line is divided
into three segments of equal length separated by rotations, we say that the line is
βcompletely transposed.β
Figure 10: A transposed three phase line
Dr Houssem Rafik El Hana Bouchekara 18
Since a transposed line each takes all three positions, the inductance per phase can
be obtained by finding the average value of ( 41)
πΏ =πΏπ + πΏπ + πΏπ
3 ( 42)
Noting that π + π2 = 1β 120Β° + 1β 240Β° = β1, the average of ( 41) becomes
πΏ =2Γ10β7
3 3 ln
1
π β²β ln
1
π·12β ln
1
π·23β ln
1
π·13 ( 43)
Or
πΏ = 2 Γ 10β7 ln1
πβ²β ln
1
(π·12π·23π·13)13
= 2 Γ 10β7 ln(π·12π·23π·13)
13
πβ²
( 44)
Or the inductance per phase per kilometer length is
πΏ = 0.2 lnπΊππ·
π·π mH/km ( 45)
Where
πΊππ· = (π·12π·23π·13)13 ( 46)
This again is of the same form as the expression for the inductance of one phase a
single-phase line. GMD (geometric mean distance) is the equivalent conductor spacing. For
the above three phase line is the cube root of the product of the tree phase spacings. π·π is
the geometric mean radius, GMR. For stranded conductor π·π is obtained from the
manufactureβs data. For solid conductor π·π = πβ = ππβ1
4.
In modern transmission lines, transposition is not generally used. However, for the
purpose of modeling, it is most practical to treat the circuit as transposed. The error
introduced as a result of this assumption is very small.
1.2.3 INDUCTANCE OF COMPOSITE CONDUCTORS
In the evaluation of inductance, solid round conductors were considered. However,
in practical transmission lines, stranded conductors are used. Also, for reasons of economy,
most EHV lines are constructed with bundled conductors. In this section an expression is
found for the inductance of composite conductors. The result can be used for evaluating the
GMR of stranded or bundled conductors. It is also useful in finding the equivalent GMR and
GMD of parallel circuits.
Dr Houssem Rafik El Hana Bouchekara 19
Figure 11: Single phase line two composite conductors.
Consider a single phase line consisting of two composite conductors π₯ and π¦ as
shown in Figure 11. The current in π₯ is πΌrreferenced into the page, and the return current in
π¦ is β πΌ. Conductor x consists of n identical strands of subconductors, each with radius ππ₯ .
Conductor y consists of m identical strands of subconductors, each with radius ππ¦ . The
current is assumed to be equally divided among the subconductors. The current per strand is
πΌ/π in x and πΌ/π in y. the application of ( 32) will result in the following expression for the
total flux linkage of conductor π.
ππ = 2 Γ 10β7πΌ
π ln
1
ππ₯β² + ln
1
π·ππ+ ln
1
π·ππ+β―+ ln
1
π·ππ
β 2 Γ 10β7πΌ
π ln
1
π·ππ β²+ ln
1
π·ππ β²+ ln
1
π·ππ β²+β―+ ln
1
π·ππ
( 47)
Or
ππ = 2 Γ 10β7πΌ ln π·ππ β²π·ππ β²π·ππ β² βββ π·πππ
ππ₯β²π·πππ·ππ βββ π·ππ
π ( 48)
The inductance of subconductor π is
πΏπ =πππΌ/π
= 2π Γ 10β7 ln π·ππ β²π·ππ β²π·ππ β² βββ π·πππ
ππ₯β²π·πππ·ππ βββ π·ππ
π ( 49)
Using ( 32), the inductance of other subconductors in x are similarly obtained. For
example, the inductance of the subconductor n is
πΏπ =πππΌ/π
= 2π Γ 10β7 ln π·ππ β²π·ππ β²π·ππ β² βββ π·πππ
ππ₯β²π·πππ·ππ βββ π·ππ
π ( 50)
The average inductance of any subconductor in group x is
πΏππ£ =πΏπ + πΏπ + πΏπ +β―+ πΏπ
π ( 51)
Since all subconductors of conductor x are electrically parallel, the inductance of π₯
will be
πΏπ₯ =πΏππ£π
=πΏπ + πΏπ + πΏπ +β―+ πΏπ
π2 ( 52)
Dr Houssem Rafik El Hana Bouchekara 20
Subsitituting the values of πΏπ ,πΏπ ,πΏπ ,β¦ , πΏπ in ( 52) result in
πΏπ₯ = 2 Γ 10β7 lnπΊππ·
πΊππ π₯ H/meter ( 53)
Where
πΊππ· = π·ππ β²π·ππ β² β¦π·ππ β¦ (π·ππ β²π·ππ β² β¦π·ππ )ππ
( 54)
And
πΊππ π₯ = π·πππ·ππ β¦π·ππ β¦ (π·πππ·ππ β¦π·ππ )π2
( 55)
Where π·ππ = π·ππ = π·ππ = ππ₯β
GMD is the mnth root of the product of the mnt distances between n strands of
conductors x and m strands of conductor y.
GMRx is the n2 root of the product of n2 terms consisting of rβ of every strand times
the distance from each strand to all other strands within group x.
The inductance of conductor y can also be similarly obtained. The geometric mean
radius GMRy will be different. The geometric mean distance GMD, however, is the same.
Example 4:
Find the geometric mean radius of a conductor in terms of the radius r of an
individual strand for
(a) Three equal strands as shown in Figure 12 (a)
(b) Four equal strands as shown in Figure 12 (b)
Figure 12: figure for this example.
Solution:
Dr Houssem Rafik El Hana Bouchekara 21
Example 5:
A stranded conductor consists of seven identical strands each having a radius r as
shown in Figure 13. Determine the GMR of the conductor in terms of r.
Figure 13: Cross section for stranded conductor
Solution:
From Figure 13, the distance from strand 1 to all other strands is:
π·12 = π·16 = π·17 = 2π
π·14 = 4π
π·13 = π·15 = π·142 β π·45
2 = 2 3 π
From ( 55) the GMR of the above conductor is
πΊππ = πβ² . 2π. 2 3π. 4π. 2 3π. 2π. 2π 6πβ² 2π 6
49
= π π β14 2 6 3
67 2
67
7
= 2.1767π
With a large number of strands the calculation of GMR can become very tedious.
Usually these are available in the manufacturerβs data.
1.2.3.1 GMR of bundled conductors
At voltages above 230 kV (extra high voltage) and with circuits with only one
conductor per phase, the corona effect becomes more excessive. Associated with this
phenomenon is a power loss as well as interference with communication links. Corona is the
direct result of high-voltage gradient at the conductor surface. The gradient can be reduced
considerably by using more than one conductor per phase. The conductors are in close
Dr Houssem Rafik El Hana Bouchekara 22
proximity compared with the spacing between phases. A line such as this is called a bundle-
conductor line. The bundle consists of two or more conductors (subconductors)
symmetrically arranged in configuration as shown in Figure 14. Another important
advantage of bundling is the attendant reduction in line reactances, both series and shunt.
The analysis of bundle-conductor lines is a specific case of the general multiconductor
configuration problem.
Figure 14: Examples of bundled arrangements.
The subconductors within a bundle are separated at frequent intervals by spacer
dampers. Spacer-dampers prevent clashing, provide damping and connect the
subconductors in parallel.
The GMR of the equivalent single conductor is obtained by using ( 55). If π·π is the
GMR of each subconductor and d is the bundle spacing we have
For the two subconductor bundle
π·π π = π·π Γ π 24
= π·π Γ π ( 56)
For the three subconudctor bundle
π·π π = π·π Γ π Γ π 39
= π·π Γ π23 ( 57)
For the four subconductor bundle
π·π π = π·π Γ π Γ π Γ π 416
= 1.09 π·π Γ π34 ( 58)
Example 6:
Calculate the inductance per phase for the three-phase, double-circuit line whose
phase conductors have a GMR of 0.06 ft, with the horizontal conductor configuration as
shown in Figure 15.
Figure 15: configuration for this figure.
Dr Houssem Rafik El Hana Bouchekara 23
Solution:
Example 7:
One circuit of a single-phase transmission line is composed of three solid wires, each
0.1 in. in radius. The return circuit is composed of two wires, each 0.2 in. in radius. The
arrangement of conductors is shown in Figure 16. Find the inductance due to the current in
each side of the line and the inductance of the complete line in millihenrys per mile.
Figure 16: Arrangement of conductors for this example
Dr Houssem Rafik El Hana Bouchekara 24
Solution:
Find the GMD between sides X and Y.
Then find the self GMD for side X.
and for side Y
The inductance is,
Example 8:
Evaluate πΏπ₯ and πΏπ¦ then calculate L in H/m for the single phase two conductor line
shown in Figure 17.
Figure 17: Single phase two conductor line for Example 8.
Dr Houssem Rafik El Hana Bouchekara 25
Solution :
Dr Houssem Rafik El Hana Bouchekara 26
1.2.4 INDUCTANCE OF THREE PHASE DOUBLE CIRCUIT LINE
A three double circuit line consists of two identical three phase circuits. The circuits
are opened with π1 β π2 ,π1 β π2 πππ π1 β π2 in parallel. Because of geometrical differences
between conductors, voltage drop due to line inductance will be unbalanced. To achieve
balance, each phase conductor must be transposed within its group and with respect to
parallel three phase line. Consider a three phase double circuit line with relative phase
positions π1π1π1 β π2π2π2, as shown in Figure 18
Figure 18: Transposed double circuit line.
The method of GMD can be used to find inductances per phase. To do this, we group
identical phases together and use ( 54) to find the GMD between each group
π·π΄π΅ = π·π1π1π·π1π2
π·π2π1π·π2π2
4
π·π΅πΆ = π·π1π1π·π1π2
π·π2π1π·π2π2
4
π·π΄πΆ = π·π1π1π·π1π2
π·π2π1π·π2π2
4
( 59)
The equivalent GMD per phase is then
πΊππ· = π·π΄π΅π·π΅πΆπ·π΄πΆ3 ( 60)
Similarly, from ( 55), the GMR of each phase group is
Dr Houssem Rafik El Hana Bouchekara 27
π·ππ΄ = π·π ππ·π1π2
24
= π·π ππ·π1π2
π·ππ΅ = π·π ππ·π1π2
24
= π·π ππ·π1π2
π·ππΆ = π·π ππ·π1π2
24
= π·π ππ·π1π2
( 61)
Where π·π π is the geometric mean radius of the bundled conductors given by ( 56)
( 58). The equivalent geometric mean radius for calculating the per phase inductance to
neutral is
πΊππ πΏ = π·ππ΄π·ππ΅π·ππΆ3 ( 62)
The inductance per phase in millihenries per kilometers is
πΏ = 0.2 lnπΊππ·
πΊππ πΏ mH/km ( 63)
1.2.4.1 Use of tables
It is seldom necessary to calculate GMR or GMD for standard lines. The GMR of
standard conductors is provided by conductor manufactures and can be found in various
handbooks (see Table 2). Also, if the distances between conductors are large compared to
distances between subconductors of each conductor, then the GMD between conductors is
approximately equal to distance between conductor centers.
Inductive reactance rather than inductance is usually desired. The inductive
reactance of one conductor of a single-phase two-conductor line is
ππΏ = 0.2πππΏ = 0.4ππ Γ lnπ·π π·π
m ohms/km ( 64)
Some tables give values of inductive reactance in addition to self GMD. One method
is to expand the logarithmic term of ( 65) as follows:
ππΏ = 0.4ππ Γ ln1
π·π + 0.4ππ Γ lnπ·π m ohms/km
ππΏ = 4.657 Γ 10β3π Γ log1
π·π + 4.657 Γ 10β3π Γ logπ·π ohms/mile
( 65)
If both Ds and Dm are in feet (or in meter), the first term in Equation ( 65) is the
inductive reactance of one conductor of a two-conductor line having a distance of 1 ft (or
one meter) between conductors.
Therefore, the first term of Eq. ( 65) is called the inductive reactance at 1-ft (or one
meter) spacing. It depends upon the self GMD of the conductor and the frequency.
The second term of Equation ( 65) is called the inductive reactance spacing factor.
This second term is independent of the type of conductor and depends on frequency and
spacing only. The spacing factor is equal to zero when Dm is 1 ft (or 1 meter). If Dm is less
than 1 ft (or 1 meter), the spacing factor is negative.
Dr Houssem Rafik El Hana Bouchekara 28
The procedure for computing inductive reactance is to look up the inductive
reactance at 1m or 1ft (or 1 meter) spacing for the conductor under consideration and to
add to this value the inductive reactance spacing factor, both at the desired line frequency.
In the end of this chapter, tables include values of inductive reactance at 1ft (or 1 meter)
spacing.
Example 9:
A single-circuit three-phase line operated at 60 Hz is arranged as shown in Figure 18.
Each conductor is No. 2 single-strand hard-drawn copper wire. Find the inductance and
inductive reactance per phase per mile.
Figure 19: Arrangement of conductors for Example 3.
Solution :
The diameter of No.2 wire is 0.258 in.
Or, from Tables :
Inductive reactance at 1 ft spacing = 0.581
Inductive reactance spacing factor for 5.45 ft = 0.2058
Inductive reactance per phase = 0.7868 ohm/phase/mile
Example 10:
Find the inductive reactance per mile of a two-conductor single-phase line operating
at 60 Hz. The conductors are each No. 1/0 seven-strand hard-drawn copper wire spaced 18
ft bet,veen centers.
Solution:
The area of stranded conductor is A=105500 circular mils (from Tables)
π·π = 0.4114 π΄ =0.4114 105500
12Γ 10β3ft = 0.01113 ft
Which is the value listed in Tables for π·π at 60 Hz. Arrangement of calculated and
tabulated values indicates that skin effect is negligible for this case.
Dr Houssem Rafik El Hana Bouchekara 29
For one conductor
ππΏ = 4.657 Γ 10β3 Γ 60 log18
0.01113= 0.897 ohms/mile
If only π·π is given in the tables, the above method is used. An alternative method
follows:
The latter method is preferred if tables are available giving inductive reactances a 1ft
(or 1meter) spacing and the inductive reactances spacing factor, for then it is necessary only
to add these two values found in tables.
Since the conductors composing the two sides of the line are identical, the inductive
reactance of the line is
ππΏ = 2 Γ 0.897 = 1.794 ohms/mile
Example 11:
One circuit of a single-phase transmission line is composed of three solid 0.5cm
radius wires. The return circuit is composed of two solid 2.5-cm radius wires.
The arrangement of conductors is as shown in Figure 35. Applying the concept of the
GMD and GMR, find the inductance of the complete line in millihenry per kilometer.
Figure 20: Conductor layout for this example.
Solution:
Dr Houssem Rafik El Hana Bouchekara 30
Example 12:
A three-phase transposed line is composed of one ACSR 159,000 cmil, 54/19
Lapwing conductor per phase with flat horizontal spacing of 8 meters as shown in Figure 13.
The GMR of each conductor is 1.515 cm.
(a) Determine the inductance per phase per kilometer of the line.
(b) This line is to be replaced by a two-conductor bundle with 8-m spacing measured
from the center of the bundles as shown in Figure 14. The spacing between the conductors
in the bundle is 40 cm. If the line inductance per phase is to be 77 percent of the inductance
in part (a), what would be the GMR of each new conductor in the bundle?
Example 13: conductor layout for question (a)
Example 14: conductor layout for this example for question (b).
Solution:
Dr Houssem Rafik El Hana Bouchekara 31
Example 15:
A completely transposed 60H-z three phase line has flat horizontal phase spacing
with 10 m between adjacent conductors. The conductors are 1,590,000 cmil ACSR with 54/3
stranding. Line length is 200 Km. Determine the inductance in H and inductive reactance in
ohms.
Solution:
Form tables π·π = 0.0159 m
Then π·ππ = 12.6 m
Then, πΏπ = 0.267 H
Thus, ππ = 101 Ξ©
Example 16:
A completely transposed 60H-z three phase line has flat horizontal phase spacing
with 10 m between adjacent conductors. The conductors are 1,590,000 cmil ACSR with 54/3
stranding. Line length is 200 Km. Determine the inductance in H and inductive reactance in
ohms.
Now, each of the 1,590,000 cmil conductors is replaced by two 795,000 cmil ACSR
26/2 conductors as shown in Figure 21. Bundle spacing is 0.40m. flat horizontal spacing is
retained, with 10 m between adjacent bundle centers. Calculate the inductive reactance of
the line and compare it with the previous question.
Figure 21: For this example.
Dr Houssem Rafik El Hana Bouchekara 32
Solution:
1)
Form tables π·π = 0.0159 m
Then π·ππ = 12.6 m
Then, πΏπ = 0.267 π»
Thus, ππ = 101 πΊ
2)
From tables π·π = 0.0114 m
The two conductor bundle GMR is π·ππΏ = 0.0676 m
Since π·ππ = 12.6 m from the first question:
Then, πΏπ = 0.209 π»
Thus, ππ = 78.8 πΊ
The reactance of the bundled line 78.7 ohms, is 22% less than of the first question,
even though the two conductor buddle has the same amount of conductor material (that is,
the same cmil per phase). One advantage of reduced series reactance is smaller line voltage
drops. Also, the loadability of medium and long EHV lines is increased.
Dr Houssem Rafik El Hana Bouchekara 33
1.2.5 LINE CAPACITANCE
Transmission line conductors exhibit capacitance with respect to each other due to
the potential difference between them. The amount of capacitance between conductors is a
function of conductor size, spacing, and height above ground. By definition, the capacitance
C is the ratio of charge q to the voltage v, given by
πΆ =π
π ( 66)
Consider a long round conductor with radius r, carrying a charge of q coulombs per
meter length as shown in Figure 22.
Figure 22: Electric field around a long round conductor.
The charge on the conductor gives rise to an electric field with radial flux lines. The
total electric flux is numerically equal to the value of charge on the conductor. The intensity
of the field at any point defined as the force per unit charge and is termed electric field
intensity designated as E. Concentric cylinders surrounding the conductor are equipentential
surfaces and have the same electric flux density. From Gaussβs law for one meter length of
the conductor, the electric flux density at a cylinder of a radius x is given by
π· =π
π΄=
π
2ππ₯(π) ( 67)
The electric field intensity E may be found from the relation
πΈ =π·
π0 ( 68)
Where π0 is the permittivity of free space and is equal to 8.85 Γ 10β16πΉ/π.
Substituting ( 67)in ( 68) result in
πΈ =π
2ππ0π₯ ( 69)
The potential difference between cylinders from position π·1to π·2 is defined as the
work done in moving a unit charge of one coulomb from π·2to π·1 through the electric field
produced by the charge on the conductor. This is given by
Dr Houssem Rafik El Hana Bouchekara 34
π12 = πΈππ₯π·2
π·1
= π
2ππ0π₯ππ₯
π·2
π·1
=π
2ππ0lnπ·2
π·1 ( 70)
The notation π12 implies the voltage drop from 1 relative to 2, that is, 1 is
understood to be positive relative to 2. The charge q carries its own sign.
1.2.6 CAPACITANCE OF SINGLE PHASE LINES
Consider one meter length of a single phase line consisting of two long solid round
conductors each having a radius r as shown in Figure 23.
Figure 23: Single phase two wire line.
The two conductors are separated by a distance D. Conductor 1 carries a charge of
π1coulombs/meter and conductor 2 carries a charge of π2coulombs/meter. The presence of
the second conductor and ground disturbs the field of the first conductor. The distance of
separation of the wires D is great with respect to r and the height of conductors is much
larger compared with D. Therefore, the distortion effect is small and the charge is assumed
to be uniformly distributed on the surface of the conductors.
Assuming conductor 1 alone to have a charge of π1, the voltage between conductor
1 and 2 is
π12(π1) =π1
2ππ0lnπ·
π ( 71)
Now assuming only conductor 2, having a charge of π2, the voltage between
conductor 2 and 1 is
π21(π2) =π2
2ππ0lnπ·
π ( 72)
Since π12(π2) = βπ21(π2), we have
π12(π2) =π2
2ππ0lnπ
π· ( 73)
From the principal of superposition, the potential difference due to presence of both
charges is
π12 = π12 π1 + π12(π2) =π1
2ππ0lnπ·
π+
π2
2ππ0lnπ
π· ( 74)
For a single phase line π2 = βπ1 = βπ, and ( 74) reduces to
π12 =π
ππ0lnπ·
π ( 75)
Dr Houssem Rafik El Hana Bouchekara 35
From ( 66), the capacitance between conductors is
πΆ12 =ππ0
lnπ·π
F/m ( 76)
Equation ( 76) gives the line to line capacitance between conductors. For the
purpose of transmission line modeling, we find convenient to define a capacitance C
between each conductor and a neural as illustrated in Figure 24.
Figure 24: Illustration of capacitance to neutral.
Since the voltage to neutral is half of π12, the capacitance to neutral πΆ = 2πΆ12, or
πΆ =2ππ0
lnπ·π
F/m ( 77)
Recalling π0 = 8.85 Γ 10β12πΉ/π and converting to ππΉ per kilometer, we have
πΆ =0.0556
lnπ·π
πF/km ( 78)
The capacitance per phase contains terms analogous to those derived for
inductance per phase. However, unlike inductance where the conductor geometric mean
radius (GMR) is used, in capacitance formula the actual conductor radius r is used.
1.2.7 POTENTIAL DIFFERENCE IN A MULTICONDUCTOR CONFIGURATION
Consider n parallel long conductors with charges π1 ,π2 ,β¦ , ππ coulombs/meter as
shown in
Figure 25: Multiconductor configuration.
Assume that the distortion effect is negligible and charge is uniformly distributed
around the conductor, with the following constraint
Dr Houssem Rafik El Hana Bouchekara 36
π1 + π2 +β―+ ππ = 0 ( 79)
Using superposition and ( 70) potential difference between conductor I and j due to
the presence of all charges is
πππ =1
2ππ0 ππ ln
π·ππ
π·ππ
π
π=1
( 80)
When k=I, π·ππ is the distance between the surface of the conductor and its center,
namely its radius r.
1.2.8 CAPACITANCE OF THREE PHASE LINES
Consider one meter length of a three phase line with three long conductors, each
with radius r, with conductor spacing as shown in
Figure 26: three phase transmission line.
Since we have a balanced three phase system
ππ + ππ + ππ = 0 ( 81)
We shall neglect the effect of ground and the shield wires. Assume that the lines is
transposed. We proceed with the calculation of the potential difference between a and b for
each section of transposition. Applying ( 80) to the first section of the transposition, π_ππ is
πππ (πΌ) =1
2ππ0 ππ ln
π·12
π+ ππ ln
π
π·12+ ππ ln
π·23
π·13 ( 82)
Similarly, for the second section of the transposition, we have
πππ (πΌπΌ) =1
2ππ0 ππ ln
π·23
π+ ππ ln
π
π·13+ ππ ln
π·13
π·12 ( 83)
And for the last section
πππ (πΌπΌπΌ) =1
2ππ0 ππ ln
π·13
π+ ππ ln
π
π·13+ ππ ln
π·12
π·23 ( 84)
The average value of πππ is
πππ =1
(3)2ππ0 ππ ln
π·12π·23π·13
π3+ ππ ln
π3
π·12π·23π·13+ ππ ln
π·12π·23π·13
π·12π·23π·13 ( 85)
Dr Houssem Rafik El Hana Bouchekara 37
Or
πππ =1
2ππ0 ππ ln
π·12π·23π·13 13
π+ ππ ln
π
π·12π·23π·13 13
( 86)
Note that the GMD of the conductor appears in the logarithm arguments and is
given by
πΊππ· = π·12π·23π·133 ( 87)
Therefore, πππ is
πππ =1
2ππ0 ππ ln
πΊππ·
π+ ππ ln
π
πΊππ· ( 88)
Similarly, we find the average voltage πππ as
πππ =1
2ππ0 ππ ln
πΊππ·
π+ ππ ln
π
πΊππ· ( 89)
Adding ( 88) and ( 89) and substituting for ππ + ππ = βππ , we have
πππ + πππ =1
2ππ0 2ππ ln
πΊππ·
πβ ππ ln
π
πΊππ· =
3ππ2ππ0
lnπΊππ·
π ( 90)
For balanced three phase voltages,
πππ = πππβ 0Β°β πππβ β 120Β°
πππ = πππβ 0Β°β πππβ β 240Β° ( 91)
Therefore,
πππ + πππ = 3πππ ( 92)
Subsisting in ( 90) the capacitance per phase to neutral is
πΆ =πππππ
=2ππ0
lnπΊππ·π
πΉ/π ( 93)
Or capacitance to neutral in ππΉ per kilometers is
πΆ =0.0556
lnπΊππ·π
ππΉ/ππ ( 94)
This is the same form as the expression the capacitance of one phase of a single
phase line. GMD (geometric mean distance) is the equivalent conductor spacing. For the
above three phase line is the cube root of the product of three phase spacings.
Dr Houssem Rafik El Hana Bouchekara 38
1.2.9 EFFECT OF BUNDLING
The procedure for finding the capacitance per phase for a three phase transposed
line with bundle conductors follows the same steps as the procedure in the precedent
section. The capacitance per phase is found to be
πΆ =2ππ0
lnπΊππ·ππ
πΉ/π ( 95)
The effect of the bundling is to introduce an equivalent radius ππ . The equivalent
radius ππ is similar to the GMR (geometric mean radius) calculated earlier for the inductance
with the exception that radius r of each subconductor is used instead of π·π . If d is the bundle
spacing, we obtain for the two-subconductor bundle
ππ = π Γ π ( 96)
For the three-subconductor bundle
ππ = π Γ π23 ( 97)
For the four-subconductor bundle
ππ = 1.09 π Γ π24 ( 98)
1.2.10 CAPACITANCE OF THREE PHASE DOUBLE CIRCUIT LINES
Consider a three-phase double-circuit line with relative phase positions
π1π1π1π2π2π2 , as shown in Figure 18β1.2.8. Each phase conductor is transposed within its
group and with respect to the parallel three-phase line. The effect of shield wires and the
ground are considered to be negligible for this balanced condition. Following the procedure
of section 1.2.8, the average voltages ππππππ and πππ are calculated and the per-phase
equivalent capacitance to neutral is obtained to be
πΆ = 2ππ0
lnπΊππ·πΊππ π
F/m
( 99)
Or capacitance to neutral in ππΉ per kilometer is
πΆ =0.0556
lnπΊππ·πΊππ π
πF/km ( 100)
The expression for πΊππ· is the same as was found for inductance calculation and is
given by (4.55). The πΊππ π of each phase group is similar to theπΊππ π , with the exception
that in (4.56) ππ is used instead ofπ·π π . This will result in the following equations
Dr Houssem Rafik El Hana Bouchekara 39
ππ΄ = πππ·π1π2
ππ΅ = πππ·π1π2
ππΆ = πππ·π1π2
( 101)
Where ππ is the geometric mean radius of the bundled conductors given by ( 96),
( 97) and ( 98). The equivalent geometric mean radius for calculating the per-phase
capacitance to neutral is
πΊππ π = ππππππ3 ( 102)
1.2.1 EFFECT OF EARTH ON THE CAPACITANCE
For an isolated charged conductor the electric flux lines are radial and are
orthogonal to the cylindrical equipotential surfaces. The presence of earth will alter the
distribution of electric flux lines and equipotential surfaces, which will change the effective
capacitance of the line.
The earth level is an equipotential surface, therefore the flux lines are forced to cut
the surface of the earth orthogonally. The effect of the presence of earth can be accounted
for by the method of image charges introduced by Kelvin. To illustrate this method, consider
a conductor with a charge q coulombs/meter at a high H above ground. Also, imagine a
charge βq at a depth βH below the surface of earth. This configuration without the presence
of the earth surface will produce the same field distribution as a single charge and the earth
surface. Thus, the earth can be replaced for the calculation of electric field potential by a
fictitious charged conductor with charge equal and opposite to the charge on the actual
conductor and at a depth below the surface of the earth the same as the height of the actual
conductor above earth. This, imaginary conductor is called the image of the actual
conductor. The procedure of section β1.2.8 can now be used for the computation of the
capacitance.
Figure 27: Distribution of electric field lines from an overhead conductor to earthβs surface.
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Figure 28: Equivalent image conductor representing the charge of the earth.
The effect of the earth is to increase the capacitance. But normally the height of the
conductor is large as compared to the distance between the conductors, and the earth
effect is negligible. Therefore, for all line models used for balanced steady state analysis, the
effect of earth on the capacitance can be neglected. However, for unbalanced analysis such
as unbalanced faults, the earthβs effect as well as the shield wires should be considered.
Example 17:
If a single phase line haz as parameters D=5ft, r=0.023 ft, and a flat horizontal
spacing H=18ft average line height, determine the effect of the earth on capacitance.
Assume a perfectly conducting earth plane.
Figure 29: for this example.
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Solution:
The earth plane is replaced by a separate image for each overhead conductor, and
the conductors are charged as shown in Figure 29. The voltages between conductors x and y
is:
The line to line capacitance is
πΆπ₯π¦ =π
ππ₯π¦=
ππ
lnπ·π β ln
π»π₯π¦π»π₯π₯
πΉ/m
Compared with 5.169 Γ 10β12 F/m (without the earth effect). The earth effect
plane is to slightly increase the capacitance. Note that as the line height H increases, the
ratio π»π₯π¦
π»π₯π₯β 0, and the effect of the earth becomes negligible.
Example 18:
A 500kV three phase transposed line is composed of one ACSR 1,272,000 cmil, 45/7
Bittern conductor per phase with horizontal conductor configuration as shown in Figure 30.
The conductors have a diameter of 1.345 in and a GMR of 0.5328 in. find the inductance and
capacitance per phase per kilometer of the line.
Figure 30: conductor layout for this example.
Solution:
Conductor radius is
π =1.345
2 Γ 12= 0.056 ft
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And
πΊππ πΏ =0.5328
12= 0.0444 ft
And
πΊππ· = 35 Γ 35 Γ 703
= 44.097 ft
The inductance per phase is
πΏ = 0.2 ln44.097
0.0444= 1.38 mH/km
And the capacitance per phase is
πΆ =0.0556
ln44.0970.056
= 0.0083 ΞΌF/km
Example 19:
Figure 31: Conductor layout for this example.
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1.2.2 MAGNETIC FIELD INDUCTION
Transmission line magnetic fields affect objects in the proximity of the line. The
magnetic fields related to the currents in the line induces voltage in objects that have a
considerable length parallel to the line, such as fences ,pipelines, and telephone wires.
The magnetic field is affected by the presence of earth return currents. Carson
presents an equation for computation of mutual resistance and inductance which are
functions of the earth's resistivity. For balanced three-phase systems the total earth return
current is zero. Under normal operating conditions, the magnetic field in proximity to
balanced three-phase lines may be calculated considering the currents in the conductors and
neglecting earth currents.
Magnetic fields have been reported to affect blood composition, growth, behavior,
immune systems, and neural functions. There are general concerns regarding the biological
effects of electromagnetic and electrostatic fields on people. Long-term effects are the
subject of several worldwide research efforts.
Example 20:
A three-phase untransposed transmission line and a telephone line are supported on
the same towers as shown in Figure 32. The power line carries a 60-Hz balanced current of
200 A per phase. The telephone line is located directly below phase b. Assuming balanced
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three-phase currents in the power line, find the voltage per kilometer induced in the
telephone line.
Solution:
The flux linkage between conductors 1 and 2 due to current πΌπ is
π12 πΌπ
= 0.2πΌπ lnπ·π2
π·π1 m Wb/km
Since π·π1 = π·π2,π12
due to πΌπ is zero. The flux linkage between conductors 1 and 2
due to current πΌπ is
π12 πΌπ
= 0.2πΌπ lnπ·π2
π·π1 m
Wb
km
Figure 32: Conductor layout for this Example.
Total flux linkage between conductor 1 and 2 due to all currents is
π12
= 0.2πΌπ lnπ·π2
π·π1+ 0.21πΌπ ln
π·π2
π·π1m Wb/Km
For positive phase sequence, with πΌπ as reference, πΌπ = πΌπβ β 240Β° and we have
π12 = 0.2πΌπ + lnπ·π2
π·π1 + 1β β 240Β° ln
π·π2
π·π1 mH/Km
With πΌπ as reference, the instantaneous flux linkage is
πΌ12 π‘ = 2 π12 cos ππ‘ + πΌ
Thus, the induced voltage in the telephone line per kilometer length is
π =ππ12 π‘
ππ‘= 2π π12 cos ππ‘ + πΌ + 90Β°
The rms voltage induced in the telephone line per kilometer is
π = π π12 β πΌ + 90Β° = πππ12
From the circuits geometry
π·π1 = π·π2 = 32 + 42 12 = 5 m
π·π2 = π·π1 = 4.22 + 42 12 = 5.8 m
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The total flux linkage is
π12
= 0.2 Γ 200β 0Β° ln5.8
5+ 0.2 Γ 200β β 240Β° ln
5
5.8
= 10.283β β 30Β° m Wb/Km
The voltage induced in the telephone line per kilometer is
π = πππ12 = π2π60 10.283β β 30Β° 10β3 = 3.88β 60Β° V/km
1.2.3 ELECTROSTATIC INDUCTION
Transmission line electric fields affect objects in the proximity of the line. The
electric field produced by high voltage lines induces current in objects which are in the area
of the electric fields. The effects of electric fields becomes of increasing concern at higher
voltage. Electric fields, related to the voltage of the line, are the primary cause of induction
to vehicles, buildings, and objects of comparable size. The human body is affected with
exposure to electric discharges from charged objects in the field of the line. These may be
steady current or spark discharges. The current densities in humans induced by electric
fields of transmission lines are known to be much higher than those induced by magnetic
fields.
The resultant electric field in proximity to a transmission line can be obtained by
representing the earth effect by image charges located below the conductors at a depth
equal to the conductor height.
1.2.4 CORONA
When the surface potential gradient of a conductor exceeds the dielectric strength
of the surrounding air, ionization occurs in the area close to the conductor surface.
This partial ionization is known as corona. The dielectric strength of air during fair
weather and at NTP (25Β°C and 76 cm of Hg) is about 30 kV/cm.
Corona produces power loss, audible hissing sound in the vicinity of the line, ozone
and radio and television interference. The audible noise is an environmental concern and
occurs in foul weather. Radio interference occurs in the AM band. Rain and snow may
produce moderate TVI in a low signal area. Corona is a function of conductors diameter, line
configuration, type of conductor, and condition of its surface. Atmospheric conditions such
as air density, humidity, and wind influence the generation of corona. Corona losses in rain
or snow are many times the losses during fair weather. On a conductor surface, an
irregularity such as a contaminating particle causes a voltage gradient that may become the
point source of a discharge. Also, insulators are contaminated by dust or chemical deposits
which will lower the disruptive voltage and increase the corona loss. The insulators are
cleaned periodically to reduce the extent of the problem. Corona can be reduced by
increasing the conductor size and the use of conductor bundling.
The power loss associated with corona can be represented by shunt conductance.
However, under normal operating conditionsπ, which represents resistive leakage between
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a phase and ground, has negligible effect on performance and is customarily neglected.
(I,e,π = 0 ) .
Example 21:
A three-phase, 60-Hz transposed transmission line has a flat horizontal configuration
as shown in Figure 33. The line reactance is 0.486 ohms per kilometer. The conductor
geometric mean radius is 2.0 cm. Determine the phase spacing D in meter.
Figure 33: for this example.
Solution:
Example 22:
A three-phase transposed line is composed of one ACSR 1,431,000 cmil, 47/7
Bobolink conductor per phase with flat horizontal spacing of 11 meters as shown in Figure
34. The conductors have a diameter of 3.625 cm and a GMR of 1.439 cm. The line is to be
replaced by a three-conductor bundle of ACSR 477,000 cmil, 26/7 Hawk conductors having
the same cross-sectional area of aluminum as the single-conductor line. The conductors
have a diameter of 2.1793 cm and a GMR of 0.8839 cm. The new line will also have a flat
horizontal configuration, but it is to be operated at a higher voltage and therefore the phase
spacing is increased to 14 m as measured from the center of the bundles as shown in Figure
34 . The spacing between the conductors in the bundle is 45 cm. Determine
(a) The percentage change in the inductance.
(b) The percentage change in the capacitance.
Dr Houssem Rafik El Hana Bouchekara 47
(a)
(b)
Figure 34: for this example.
Solution:
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Table 2: Characteristics of copper conductors, hard-drawn, 97.3 % conductivity.
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Table 3: Characteristics of ACSR
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Table 4: Characteristics of ACSR
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Table 5: Characteristics of AAC
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Table 6: Characteristics of AAC
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Table 7: inductive reactance spacing factor at 60hz