transmission and distribution of electrical power - 3 electric transmission line parameters_2 (1)
TRANSCRIPT
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 (4×3)×2+1+1=26 conductors, all of which are
mutually coupled to each other and to earth.
Dr Houssem Rafik El Hana Bouchekara 5
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
𝑅𝐷𝐶 =𝜌𝑙
𝐴 Ω ( 1)
where
ρ 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
Dr Houssem Rafik El Hana Bouchekara 6
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
𝑅𝐴𝐶 = 𝑘𝑅𝐷𝐶 ( 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:
𝑅2 = 𝑅1 𝑇 + 𝑡2
𝑇 + 𝑡1 ( 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 (C°)
Resistivity (𝜌) 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).
Dr Houssem Rafik El Hana Bouchekara 7
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:
𝑅𝑐𝑜𝑛𝑑 =𝜌
𝐴 1 + 𝜋
1
𝑃
2
Ω/𝑚 ( 4)
where
𝑅𝑐𝑜𝑛𝑑 : resistance of wound conductor (Ω)
1 + 𝜋1
𝑃
2: length of wound conductor (m)
𝑃𝑐𝑜𝑛𝑑 =𝑙𝑡𝑢𝑟𝑛
2𝑟𝑙𝑎𝑦𝑒𝑟 relative pitch of wound conductor
𝑙𝑡𝑢𝑟𝑛 : length of one turn of the spiral (m)
2𝑟𝑙𝑎𝑦𝑒𝑟 : diameter of the layer (m)
The parallel combination of n conductors, with same diameter per layer, gives the
resistance per layer as follows:
𝑅𝑙𝑎𝑦𝑒𝑟 =1
1𝑅𝑗
𝑛𝑖=1
Ω/𝑚 ( 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.
Dr Houssem Rafik El Hana Bouchekara 8
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) 20° C
(b) 50° (C)
The resistivity of aluminum at 20° is 2.8 × 10−8Ωm .
1 square centimeter × 197= 1 circular mils.
Solution:
Dr Houssem Rafik El Hana Bouchekara 9
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 20° is 2.8 × 10−8Ωm , determine the
required conductor diameter and the conductor size in circular miles.
Solution:
Dr Houssem Rafik El Hana Bouchekara 10
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
𝐿 =𝜆
𝐼 ( 6)
Where 𝜆 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 𝐻𝑥 , around a circle of radius x, is constant and tangent
to the circle. The Ampere’s law relating 𝐻𝑥 to the current 𝐼𝑥 is given by
𝐻𝑥 .𝑑𝑙2𝜋𝑥
0
= 𝐼𝑥 ( 7)
Or
𝐻𝑥 =𝐼𝑥
2𝜋𝑥 ( 8)
Where 𝐼𝑥 is the current enclosed at radius x. As shown in Figure 4. Equation ( 8) is all
that required for evaluating the flux linkage 𝜆 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.,
Dr Houssem Rafik El Hana Bouchekara 11
𝐼
𝜋𝑟2=
𝐼𝑥𝜋𝑥2
( 9)
Substituting for 𝐼𝑥 in ( 8) yields
𝐻𝑥 =𝐼𝑥
2𝜋𝑟2𝑥 ( 10)
For a nonmagnetic conductor wit constant permeability 𝜇0, the magnetic flux
density is given by 𝐵𝑥 = 𝜇0𝐻𝑥 , or
𝐵𝑥 =𝜇0𝐼
2𝜋𝑟2𝑥 ( 11)
Where 𝜇0 is the permeability of free space (or air) and is equal to4𝜋 × 10−7𝐻/𝑚.
The differential flux 𝑑𝜙 for a small region of thickness 𝑑𝑥 and one meter length of
the conductor is
𝑑𝜙𝑥 = 𝐵𝑥.𝑑𝑥. 𝑙 =𝜇0𝐼
2𝜋𝑟2𝑥𝑑𝑥 ( 12)
The flux 𝑑𝜙𝑥 links only the fraction of the conductor from the center to radius x.
thus, on the assumption of uniform current density, only the fraction 𝜋𝑥2/𝜋𝑟2 of the total
current is linked by the flux, i.e.,
𝑑𝜆𝑥 = 𝑥2
𝑟2 𝑑𝜙𝑥 =𝜇0𝐼
2𝜋𝑟4𝑥3𝑑𝑥 ( 13)
The total flux linkage is found by integrating 𝑑𝜆𝑥 from 0 to r.
𝜆𝑖𝑛𝑡 =𝜇0𝐼
2𝜋𝑟4 𝑥3𝑑𝑥𝑟
0
=𝜇0𝐼
8𝜋 Wb/m ( 14)
From ( 6), the inductance due to the internal flux linkage is
𝐿𝑖𝑛𝑡 =𝜇0 𝐼
8𝜋=
1
2× 10−7 H/m ( 15)
Note that 𝐿𝑖𝑛𝑡 is independent of the conductor radius r.
1.2.2.2 Inductance Due To External Flux Linkage
Consider 𝐻𝑥 external to the conductor at radius 𝑥 > 𝑟 as shown in Figure 5. Since
the circle at radius x encloses the entire current 𝐼𝑥 = 𝐼 and in ( 8) 𝐼𝑥 is replaced by I and the
flux density at radius x becomes
𝐵𝑥 = 𝜇0𝐻𝑥 =𝜇0 𝐼
2𝜋𝑥 T ( 16)
Dr Houssem Rafik El Hana Bouchekara 12
Figure 5: Flux linkage between D1 D2
Since the entire current 𝐼 is linked by the flux outside the conductor, the flux linkage
𝑑𝜆𝑥 is numerically equal to the flux 𝑑𝜙𝑥 . The differential flux 𝑑𝜙𝑥 for a small region of
thickness 𝑑𝑥 and one meter length of the conductor is then given by
𝑑𝜆𝑥 = 𝑑𝜙𝑥 = 𝐵𝑥𝑑𝑥. 1 =𝜇0𝐼
2𝜋𝑥𝑑𝑥 ( 17)
The external flux linkage between two points 𝐷1 and 𝐷2 is found by integrating
𝑑𝜆𝑥 from 𝐷1 to 𝐷2
𝜆𝑒𝑥𝑡 =𝜇0𝐼
2𝜋
1
𝑥
𝐷2
𝐷1
𝑑𝑥 = 2 × 10−7 𝐼 ln𝐷2
𝐷1 Wb/m ( 18)
The inductance between two points external to a conductor is then
𝐿𝑒𝑥𝑡 = 2 × 10−7 ln𝐷2
𝐷1 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 𝑟1 and 𝑟2 as shown in Figure 6. The two conductors are separated by a
distance D. conductor 1 carries the phasor current 𝐼1referenced into the page and conductor
2 carries return current 𝐼2 = −𝐼1. These currents set up magnetic field lines that links
between the conductors as shown.
Figure 6: Single phase two wire line.
Dr Houssem Rafik El Hana Bouchekara 13
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 𝐷1 = 𝑟1 to 𝐷2 = 𝐷.
𝐿1(𝑒𝑥𝑡 ) = 2 × 10−7 ln𝐷
𝑟1 H/m ( 20)
The total inductance of conductor 1 is then
𝐿1 =1
2× 10−7 + 2 × 10−7 ln
𝐷
𝑟1 H/m ( 21)
Equation ( 21) is often rearranged as follows:
𝐿1 = 2 × 10−7 1
2+ ln
𝐷
𝑟1
= 2 × 10−7 ln 𝑒14 + ln
1
𝑟1+ ln
𝐷
1
= 2 × 10−7 ln1
𝑟1𝑒−1/4
+ ln𝐷
1
( 22)
Let 𝑟1′ = 𝑟1𝑒
−1
4 , the inductance of conductor 1 becomes
𝐿1 = 2 × 10−7 ln1
𝑟1′ + 2 × 10−7 ln
𝐷
1 H/m ( 23)
Similarly, the inductance of conductor 2 is
𝐿2 = 2 × 10−7 ln1
𝑟2′ + 2 × 10−7 ln
𝐷
1 H/m ( 24)
If the two conductors are identical, 𝑟1 = 𝑟2 = 𝑟 and 𝐿1 = 𝐿2 = 𝐿, and the
inductance per phase per meter length of the line is given by
𝐿 = 2 × 10−7 ln1
𝑟′+ 2 × 10−7 ln
𝐷
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 𝑟’ = 𝑟𝑒−1/4 is known mathematically as the self geometric mean distance
of a circle with radius 𝑟 and is abbreviated by GMR. 𝑟′can 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 𝐷𝑠.thus, the inductance per phase in millihenries per kilometer
becomes
𝐿 = 0.2 ln𝐷
𝐷𝑠 𝑚H/Km ( 26)
Dr Houssem Rafik El Hana Bouchekara 14
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 𝐿11 and 𝐿22 and the mutual inductance 𝐿12 . 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 𝜆1 and 𝜆2 are given by
𝜆1 = 𝐿11𝐼1 + 𝐿12𝐼2
𝜆2 = 𝐿21𝐼1 + 𝐿22𝐼2 ( 27)
Since 𝐼2 = −𝐼1, we have
𝜆1 = 𝐿11 − 𝐿12 𝐼1
𝜆2 = −𝐿21 + 𝐿22 𝐼2 ( 28)
Comparing ( 28)with ( 18)and ( 19), we conclude the following equivalent
expressions for the self and mutual inductances:
𝐿11 = 2 × 10−7 ln1
𝑟1′ ( 29)
Dr Houssem Rafik El Hana Bouchekara 15
𝐿22 = 2 × 10−7 ln1
𝑟2′
𝐿12 = 𝐿21 = 2 × 10−7 ln1
𝐷
The concept of self and mutual inductance can be extended to a group of n
conductors. Consider n conductors carrying phasor currents 𝐼1, 𝐼2 ,… , 𝐼𝑛 , such that
𝐼1 + 𝐼2 + 𝐼3 +⋯+ 𝐼𝑖 +⋯+ 𝐼𝑛 = 0 ( 30)
Generalzing ( 27), the flux linkage of conductor i are
𝜆𝑖 = 𝐿𝑖𝑖 𝐼𝑖 + 𝐿𝑖𝑗 𝐼𝑗
𝑛
𝑗=1
𝑗 ≠ 𝑖 ( 31)
Or
𝜆𝑖 = 2 × 10−7 𝐼𝑖 ln1
𝑟𝑖′ + 𝐼𝑗 ln
1
𝐷𝑖𝑗
𝑛
𝑗=1
𝑗 ≠ 𝑖 ( 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
𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 0 ( 33)
From ( 32) the total flux linkage of phase a conductor is
𝜆𝑎 = 2 × 10−7 𝐼𝑎 ln1
𝑟′+ 𝐼𝑏 ln
1
𝐷+ 𝐼𝑐 ln
1
𝐷 ( 34)
Substituting – 𝐼𝑎 = 𝐼𝑏 + 𝐼𝑐
Dr Houssem Rafik El Hana Bouchekara 16
𝜆𝑎 = 2 × 10−7 𝐼𝑎 ln1
𝑟′− 𝐼𝑎 ln
1
𝐷
= 2 × 10−7𝐼𝑎 ln𝐷
𝑟′
( 35)
Because of symmetry, 𝜆𝑏 = 𝜆𝑐 = 𝜆𝑎 , and the three inductances are identical.
Therefore, the inductance per phase per kilometer length is
𝐿 = 0.2 ln𝐷
𝐷𝑠 mH/km ( 36)
Where r’ is the geometric mean radius GMR, and is shown by Ds. for a solid round
conductor, 𝐷𝑠 = 𝑟𝑒−1
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.
𝜆𝑎 = 2 × 10−7 𝐼𝑎 ln1
𝑟′+ 𝐼𝑏 ln
1
𝐷12+ 𝐼𝑐 ln
1
𝐷13
𝜆𝑏 = 2 × 10−7 𝐼𝑎 ln1
𝐷12+ 𝐼𝑏 ln
1
𝑟′+ 𝐼𝑐 ln
1
𝐷23
𝜆𝑐 = 2 × 10−7 𝐼𝑎 ln1
𝐷13+ 𝐼𝑏 ln
1
𝐷23+ 𝐼𝑐 ln
1
𝑟′
( 37)
Or in matrix form
Dr Houssem Rafik El Hana Bouchekara 17
𝜆 = 𝐿𝐼 ( 38)
Where the symmetrical inductance matrix L is given by
𝐿 = 2 × 10−7
ln
1
𝑟′ln
1
𝐷12ln
1
𝐷13
ln1
𝐷12ln
1
𝑟′ln
1
𝐷23
ln1
𝐷13ln
1
𝐷23ln
1
𝑟′
( 39)
For balanced three phase currents with 𝐼𝑎 as reference, we have
𝐼𝑏 = 𝐼𝑎∠240° = 𝑎2𝐼𝑎
𝐼𝑐 = 𝐼𝑎∠120° = 𝑎𝐼𝑎 ( 40)
Where the operator 𝑎 = 1∠120° and 𝑎2 = 1∠240°. Substituting in ( 37) result in
𝐿𝑎 =𝜆𝑎𝐼𝑎
= 2 × 10−7 ln1
𝑟′+ 𝑎2 ln
1
𝐷12+ 𝑎 ln
1
𝐷13
𝐿𝑏 =𝜆𝑏𝐼𝑏
= 2 × 10−7 𝑎 ln1
𝐷12+ ln
1
𝑟′+ 𝑎2 ln
1
𝐷23
𝐿𝑐 =𝜆𝑐𝐼𝑐
= 2 × 10−7 𝑎2 ln1
𝐷13+ 𝑎 ln
1
𝐷23+ ln
1
𝑟′
( 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
“completely transposed.”
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)
𝐿 =𝐿𝑎 + 𝐿𝑏 + 𝐿𝑐
3 ( 42)
Noting that 𝑎 + 𝑎2 = 1∠120° + 1∠240° = −1, the average of ( 41) becomes
𝐿 =2×10−7
3 3 ln
1
𝑟 ′− ln
1
𝐷12− ln
1
𝐷23− ln
1
𝐷13 ( 43)
Or
𝐿 = 2 × 10−7 ln1
𝑟′− ln
1
(𝐷12𝐷23𝐷13)13
= 2 × 10−7 ln(𝐷12𝐷23𝐷13)
13
𝑟′
( 44)
Or the inductance per phase per kilometer length is
𝐿 = 0.2 ln𝐺𝑀𝐷
𝐷𝑠 mH/km ( 45)
Where
𝐺𝑀𝐷 = (𝐷12𝐷23𝐷13)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. 𝐷𝑠 is
the geometric mean radius, GMR. For stranded conductor 𝐷𝑠 is obtained from the
manufacture’s data. For solid conductor 𝐷𝑠 = 𝑟’ = 𝑟𝑒−1
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 𝑥 and 𝑦 as
shown in Figure 11. The current in 𝑥 is 𝐼rreferenced into the page, and the return current in
𝑦 is – 𝐼. Conductor x consists of n identical strands of subconductors, each with radius 𝑟𝑥 .
Conductor y consists of m identical strands of subconductors, each with radius 𝑟𝑦 . The
current is assumed to be equally divided among the subconductors. The current per strand is
𝐼/𝑛 in x and 𝐼/𝑚 in y. the application of ( 32) will result in the following expression for the
total flux linkage of conductor 𝑎.
𝜆𝑎 = 2 × 10−7𝐼
𝑛 ln
1
𝑟𝑥′ + ln
1
𝐷𝑎𝑏+ ln
1
𝐷𝑎𝑐+⋯+ ln
1
𝐷𝑎𝑛
− 2 × 10−7𝐼
𝑚 ln
1
𝐷𝑎𝑎 ′+ ln
1
𝐷𝑎𝑏 ′+ ln
1
𝐷𝑎𝑐 ′+⋯+ ln
1
𝐷𝑎𝑚
( 47)
Or
𝜆𝑎 = 2 × 10−7𝐼 ln 𝐷𝑎𝑎 ′𝐷𝑎𝑏 ′𝐷𝑎𝑐 ′ ∙∙∙ 𝐷𝑎𝑚𝑚
𝑟𝑥′𝐷𝑎𝑏𝐷𝑎𝑐 ∙∙∙ 𝐷𝑎𝑛
𝑛 ( 48)
The inductance of subconductor 𝑎 is
𝐿𝑎 =𝜆𝑎𝐼/𝑛
= 2𝑛 × 10−7 ln 𝐷𝑎𝑎 ′𝐷𝑎𝑏 ′𝐷𝑎𝑐 ′ ∙∙∙ 𝐷𝑎𝑚𝑚
𝑟𝑥′𝐷𝑎𝑏𝐷𝑎𝑐 ∙∙∙ 𝐷𝑎𝑛
𝑛 ( 49)
Using ( 32), the inductance of other subconductors in x are similarly obtained. For
example, the inductance of the subconductor n is
𝐿𝑛 =𝜆𝑛𝐼/𝑛
= 2𝑛 × 10−7 ln 𝐷𝑛𝑎 ′𝐷𝑛𝑏 ′𝐷𝑛𝑐 ′ ∙∙∙ 𝐷𝑛𝑚𝑚
𝑟𝑥′𝐷𝑛𝑎𝐷𝑛𝑏 ∙∙∙ 𝐷𝑛𝑐
𝑛 ( 50)
The average inductance of any subconductor in group x is
𝐿𝑎𝑣 =𝐿𝑎 + 𝐿𝑏 + 𝐿𝑐 +⋯+ 𝐿𝑛
𝑛 ( 51)
Since all subconductors of conductor x are electrically parallel, the inductance of 𝑥
will be
𝐿𝑥 =𝐿𝑎𝑣𝑛
=𝐿𝑎 + 𝐿𝑏 + 𝐿𝑐 +⋯+ 𝐿𝑛
𝑛2 ( 52)
Dr Houssem Rafik El Hana Bouchekara 20
Subsitituting the values of 𝐿𝑎 ,𝐿𝑏 ,𝐿𝑐 ,… , 𝐿𝑛 in ( 52) result in
𝐿𝑥 = 2 × 10−7 ln𝐺𝑀𝐷
𝐺𝑀𝑅𝑥 H/meter ( 53)
Where
𝐺𝑀𝐷 = 𝐷𝑎𝑎 ′𝐷𝑎𝑏 ′ …𝐷𝑎𝑚 … (𝐷𝑛𝑎 ′𝐷𝑛𝑏 ′ …𝐷𝑛𝑚 )𝑚𝑛
( 54)
And
𝐺𝑀𝑅𝑥 = 𝐷𝑎𝑎𝐷𝑎𝑏 …𝐷𝑎𝑛 … (𝐷𝑛𝑎𝐷𝑛𝑏 …𝐷𝑛𝑛 )𝑛2
( 55)
Where 𝐷𝑎𝑎 = 𝐷𝑏𝑏 = 𝐷𝑛𝑛 = 𝑟𝑥’
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 r’ 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:
𝐷12 = 𝐷16 = 𝐷17 = 2𝑟
𝐷14 = 4𝑟
𝐷13 = 𝐷15 = 𝐷142 − 𝐷45
2 = 2 3 𝑟
From ( 55) the GMR of the above conductor is
𝐺𝑀𝑅 = 𝑟′ . 2𝑟. 2 3𝑟. 4𝑟. 2 3𝑟. 2𝑟. 2𝑟 6𝑟′ 2𝑟 6
49
= 𝑟 𝑒 −14 2 6 3
67 2
67
7
= 2.1767𝑟
With a large number of strands the calculation of GMR can become very tedious.
Usually these are available in the manufacturer’s 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 𝐷𝑠 is the
GMR of each subconductor and d is the bundle spacing we have
For the two subconductor bundle
𝐷𝑠𝑏 = 𝐷𝑠 × 𝑑 24
= 𝐷𝑠 × 𝑑 ( 56)
For the three subconudctor bundle
𝐷𝑠𝑏 = 𝐷𝑠 × 𝑑 × 𝑑 39
= 𝐷𝑠 × 𝑑23 ( 57)
For the four subconductor bundle
𝐷𝑠𝑏 = 𝐷𝑠 × 𝑑 × 𝑑 × 𝑑 416
= 1.09 𝐷𝑠 × 𝑑34 ( 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 𝐿𝑥 and 𝐿𝑦 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 𝑎1 − 𝑎2 ,𝑏1 − 𝑏2 𝑎𝑛𝑑 𝑐1 − 𝑐2 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 𝑎1𝑏1𝑐1 − 𝑐2𝑏2𝑎2, 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
𝐷𝐴𝐵 = 𝐷𝑎1𝑏1𝐷𝑎1𝑏2
𝐷𝑎2𝑏1𝐷𝑎2𝑏2
4
𝐷𝐵𝐶 = 𝐷𝑏1𝑐1𝐷𝑏1𝑐2
𝐷𝑏2𝑐1𝐷𝑏2𝑐2
4
𝐷𝐴𝐶 = 𝐷𝑎1𝑐1𝐷𝑎1𝑐2
𝐷𝑐2𝑐1𝐷𝑎2𝑐2
4
( 59)
The equivalent GMD per phase is then
𝐺𝑀𝐷 = 𝐷𝐴𝐵𝐷𝐵𝐶𝐷𝐴𝐶3 ( 60)
Similarly, from ( 55), the GMR of each phase group is
Dr Houssem Rafik El Hana Bouchekara 27
𝐷𝑆𝐴 = 𝐷𝑠𝑏𝐷𝑎1𝑎2
24
= 𝐷𝑠𝑏𝐷𝑎1𝑎2
𝐷𝑆𝐵 = 𝐷𝑠𝑏𝐷𝑏1𝑏2
24
= 𝐷𝑠𝑏𝐷𝑏1𝑏2
𝐷𝑆𝐶 = 𝐷𝑠𝑏𝐷𝑐1𝑐2
24
= 𝐷𝑠𝑏𝐷𝑐1𝑐2
( 61)
Where 𝐷𝑠𝑏 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
𝐺𝑀𝑅𝐿 = 𝐷𝑆𝐴𝐷𝑆𝐵𝐷𝑆𝐶3 ( 62)
The inductance per phase in millihenries per kilometers is
𝐿 = 0.2 ln𝐺𝑀𝐷
𝐺𝑀𝑅𝐿 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
𝑋𝐿 = 0.2𝜋𝑓𝐿 = 0.4𝜋𝑓 × ln𝐷𝑚 𝐷𝑠
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:
𝑋𝐿 = 0.4𝜋𝑓 × ln1
𝐷𝑠+ 0.4𝜋𝑓 × ln𝐷𝑚 m ohms/km
𝑋𝐿 = 4.657 × 10−3𝑓 × log1
𝐷𝑠+ 4.657 × 10−3𝑓 × log𝐷𝑚 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)
𝐷𝑠 = 0.4114 𝐴 =0.4114 105500
12× 10−3ft = 0.01113 ft
Which is the value listed in Tables for 𝐷𝑠 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
𝑋𝐿 = 4.657 × 10−3 × 60 log18
0.01113= 0.897 ohms/mile
If only 𝐷𝑠 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
𝑋𝐿 = 2 × 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 𝐷𝑠 = 0.0159 m
Then 𝐷𝑒𝑞 = 12.6 m
Then, 𝐿𝑎 = 0.267 H
Thus, 𝑋𝑎 = 101 Ω
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 𝐷𝑠 = 0.0159 m
Then 𝐷𝑒𝑞 = 12.6 m
Then, 𝐿𝑎 = 0.267 𝐻
Thus, 𝑋𝑎 = 101 𝛺
2)
From tables 𝐷𝑠 = 0.0114 m
The two conductor bundle GMR is 𝐷𝑆𝐿 = 0.0676 m
Since 𝐷𝑒𝑞 = 12.6 m from the first question:
Then, 𝐿𝑎 = 0.209 𝐻
Thus, 𝑋𝑎 = 78.8 𝛺
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
𝐶 =𝑞
𝑉 ( 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 Gauss’s law for one meter length of
the conductor, the electric flux density at a cylinder of a radius x is given by
𝐷 =𝑞
𝐴=
𝑞
2𝜋𝑥(𝑙) ( 67)
The electric field intensity E may be found from the relation
𝐸 =𝐷
𝜀0 ( 68)
Where 𝜀0 is the permittivity of free space and is equal to 8.85 × 10−16𝐹/𝑚.
Substituting ( 67)in ( 68) result in
𝐸 =𝑞
2𝜋𝜖0𝑥 ( 69)
The potential difference between cylinders from position 𝐷1to 𝐷2 is defined as the
work done in moving a unit charge of one coulomb from 𝐷2to 𝐷1 through the electric field
produced by the charge on the conductor. This is given by
Dr Houssem Rafik El Hana Bouchekara 34
𝑉12 = 𝐸𝑑𝑥𝐷2
𝐷1
= 𝑞
2𝜋𝜖0𝑥𝑑𝑥
𝐷2
𝐷1
=𝑞
2𝜋𝜖0ln𝐷2
𝐷1 ( 70)
The notation 𝑉12 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
𝑞1coulombs/meter and conductor 2 carries a charge of 𝑞2coulombs/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 𝑞1, the voltage between conductor
1 and 2 is
𝑉12(𝑞1) =𝑞1
2𝜋𝜖0ln𝐷
𝑟 ( 71)
Now assuming only conductor 2, having a charge of 𝑞2, the voltage between
conductor 2 and 1 is
𝑉21(𝑞2) =𝑞2
2𝜋𝜖0ln𝐷
𝑟 ( 72)
Since 𝑉12(𝑞2) = −𝑉21(𝑞2), we have
𝑉12(𝑞2) =𝑞2
2𝜋𝜖0ln𝑟
𝐷 ( 73)
From the principal of superposition, the potential difference due to presence of both
charges is
𝑉12 = 𝑉12 𝑞1 + 𝑉12(𝑞2) =𝑞1
2𝜋𝜖0ln𝐷
𝑟+
𝑞2
2𝜋𝜖0ln𝑟
𝐷 ( 74)
For a single phase line 𝑞2 = −𝑞1 = −𝑞, and ( 74) reduces to
𝑉12 =𝑞
𝜋𝜖0ln𝐷
𝑟 ( 75)
Dr Houssem Rafik El Hana Bouchekara 35
From ( 66), the capacitance between conductors is
𝐶12 =𝜋𝜖0
ln𝐷𝑟
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 𝑉12, the capacitance to neutral 𝐶 = 2𝐶12, or
𝐶 =2𝜋𝜖0
ln𝐷𝑟
F/m ( 77)
Recalling 𝜖0 = 8.85 × 10−12𝐹/𝑚 and converting to 𝜇𝐹 per kilometer, we have
𝐶 =0.0556
ln𝐷𝑟
𝜇F/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 𝑞1 ,𝑞2 ,… , 𝑞𝑛 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
𝑞1 + 𝑞2 +⋯+ 𝑞𝑛 = 0 ( 79)
Using superposition and ( 70) potential difference between conductor I and j due to
the presence of all charges is
𝑉𝑖𝑗 =1
2𝜋𝜀0 𝑞𝑘 ln
𝐷𝑘𝑗
𝐷𝑘𝑖
𝑛
𝑘=1
( 80)
When k=I, 𝐷𝑖𝑖 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
𝑞𝑎 + 𝑞𝑏 + 𝑎𝑐 = 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, 𝑉_𝑎𝑏 is
𝑉𝑎𝑏 (𝐼) =1
2𝜋𝜀0 𝑞𝑎 ln
𝐷12
𝑟+ 𝑞𝑏 ln
𝑟
𝐷12+ 𝑞𝑐 ln
𝐷23
𝐷13 ( 82)
Similarly, for the second section of the transposition, we have
𝑉𝑎𝑏 (𝐼𝐼) =1
2𝜋𝜀0 𝑞𝑎 ln
𝐷23
𝑟+ 𝑞𝑏 ln
𝑟
𝐷13+ 𝑞𝑐 ln
𝐷13
𝐷12 ( 83)
And for the last section
𝑉𝑎𝑏 (𝐼𝐼𝐼) =1
2𝜋𝜀0 𝑞𝑎 ln
𝐷13
𝑟+ 𝑞𝑏 ln
𝑟
𝐷13+ 𝑞𝑐 ln
𝐷12
𝐷23 ( 84)
The average value of 𝑉𝑎𝑏 is
𝑉𝑎𝑏 =1
(3)2𝜋𝜀0 𝑞𝑎 ln
𝐷12𝐷23𝐷13
𝑟3+ 𝑞𝑏 ln
𝑟3
𝐷12𝐷23𝐷13+ 𝑞𝑐 ln
𝐷12𝐷23𝐷13
𝐷12𝐷23𝐷13 ( 85)
Dr Houssem Rafik El Hana Bouchekara 37
Or
𝑉𝑎𝑏 =1
2𝜋𝜀0 𝑞𝑎 ln
𝐷12𝐷23𝐷13 13
𝑟+ 𝑞𝑏 ln
𝑟
𝐷12𝐷23𝐷13 13
( 86)
Note that the GMD of the conductor appears in the logarithm arguments and is
given by
𝐺𝑀𝐷 = 𝐷12𝐷23𝐷133 ( 87)
Therefore, 𝑉𝑎𝑏 is
𝑉𝑎𝑏 =1
2𝜋𝜀0 𝑞𝑎 ln
𝐺𝑀𝐷
𝑟+ 𝑞𝑏 ln
𝑟
𝐺𝑀𝐷 ( 88)
Similarly, we find the average voltage 𝑉𝑎𝑐 as
𝑉𝑎𝑐 =1
2𝜋𝜀0 𝑞𝑎 ln
𝐺𝑀𝐷
𝑟+ 𝑞𝑐 ln
𝑟
𝐺𝑀𝐷 ( 89)
Adding ( 88) and ( 89) and substituting for 𝑞𝑏 + 𝑞𝑐 = −𝑞𝑎 , we have
𝑉𝑎𝑏 + 𝑉𝑎𝑐 =1
2𝜋𝜀0 2𝑞𝑎 ln
𝐺𝑀𝐷
𝑟− 𝑞𝑎 ln
𝑟
𝐺𝑀𝐷 =
3𝑞𝑎2𝜋𝜀0
ln𝐺𝑀𝐷
𝑟 ( 90)
For balanced three phase voltages,
𝑉𝑎𝑏 = 𝑉𝑎𝑛∠0°− 𝑉𝑎𝑛∠ − 120°
𝑉𝑎𝑐 = 𝑉𝑎𝑛∠0°− 𝑉𝑎𝑛∠ − 240° ( 91)
Therefore,
𝑉𝑎𝑏 + 𝑉𝑎𝑐 = 3𝑉𝑎𝑛 ( 92)
Subsisting in ( 90) the capacitance per phase to neutral is
𝐶 =𝑞𝑛𝑉𝑎𝑛
=2𝜋𝜀0
ln𝐺𝑀𝐷𝑟
𝐹/𝑚 ( 93)
Or capacitance to neutral in 𝜇𝐹 per kilometers is
𝐶 =0.0556
ln𝐺𝑀𝐷𝑟
𝜇𝐹/𝑘𝑚 ( 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
𝐶 =2𝜋𝜀0
ln𝐺𝑀𝐷𝑟𝑏
𝐹/𝑚 ( 95)
The effect of the bundling is to introduce an equivalent radius 𝑟𝑏 . The equivalent
radius 𝑟𝑏 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 𝐷𝑠. If d is the bundle
spacing, we obtain for the two-subconductor bundle
𝑟𝑏 = 𝑟 × 𝑑 ( 96)
For the three-subconductor bundle
𝑟𝑏 = 𝑟 × 𝑑23 ( 97)
For the four-subconductor bundle
𝑟𝑏 = 1.09 𝑟 × 𝑑24 ( 98)
1.2.10 CAPACITANCE OF THREE PHASE DOUBLE CIRCUIT LINES
Consider a three-phase double-circuit line with relative phase positions
𝑎1𝑏1𝑐1𝑐2𝑏2𝑎2 , as shown in Figure 181.2.8. 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 𝑉𝑎𝑏𝑉𝑎𝑐 and 𝑉𝑎𝑛 are calculated and the per-phase
equivalent capacitance to neutral is obtained to be
𝐶 = 2𝜋𝜀0
ln𝐺𝑀𝐷𝐺𝑀𝑅𝑐
F/m
( 99)
Or capacitance to neutral in 𝜇𝐹 per kilometer is
𝐶 =0.0556
ln𝐺𝑀𝐷𝐺𝑀𝑅𝑐
𝜇F/km ( 100)
The expression for 𝐺𝑀𝐷 is the same as was found for inductance calculation and is
given by (4.55). The 𝐺𝑀𝑅𝑐 of each phase group is similar to the𝐺𝑀𝑅𝑙 , with the exception
that in (4.56) 𝑟𝑏 is used instead of𝐷𝑠𝑏 . This will result in the following equations
Dr Houssem Rafik El Hana Bouchekara 39
𝑟𝐴 = 𝑟𝑏𝐷𝑎1𝑎2
𝑟𝐵 = 𝑟𝑏𝐷𝑏1𝑏2
𝑟𝐶 = 𝑟𝑏𝐷𝑐1𝑐2
( 101)
Where 𝑟𝑏 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
𝐺𝑀𝑅𝑐 = 𝑟𝑎𝑟𝑏𝑟𝑐3 ( 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 –q at a depth –H 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 1.2.8 can now be used for the computation of the
capacitance.
Figure 27: Distribution of electric field lines from an overhead conductor to earth’s surface.
Dr Houssem Rafik El Hana Bouchekara 40
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 earth’s 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.
Dr Houssem Rafik El Hana Bouchekara 41
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
𝐶𝑥𝑦 =𝑞
𝑉𝑥𝑦=
𝜋𝜀
ln𝐷𝑟 − ln
𝐻𝑥𝑦𝐻𝑥𝑥
𝐹/m
Compared with 5.169 × 10−12 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 𝐻𝑥𝑦
𝐻𝑥𝑥→ 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
𝑟 =1.345
2 × 12= 0.056 ft
Dr Houssem Rafik El Hana Bouchekara 42
And
𝐺𝑀𝑅𝐿 =0.5328
12= 0.0444 ft
And
𝐺𝑀𝐷 = 35 × 35 × 703
= 44.097 ft
The inductance per phase is
𝐿 = 0.2 ln44.097
0.0444= 1.38 mH/km
And the capacitance per phase is
𝐶 =0.0556
ln44.0970.056
= 0.0083 μF/km
Example 19:
Figure 31: Conductor layout for this example.
Dr Houssem Rafik El Hana Bouchekara 43
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
Dr Houssem Rafik El Hana Bouchekara 44
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 𝐼𝑎 is
𝜆12 𝐼𝑎
= 0.2𝐼𝑎 ln𝐷𝑎2
𝐷𝑎1 m Wb/km
Since 𝐷𝑏1 = 𝐷𝑏2,𝜆12
due to 𝐼𝑏 is zero. The flux linkage between conductors 1 and 2
due to current 𝐼𝑐 is
𝜆12 𝐼𝑎
= 0.2𝐼𝑎 ln𝐷𝑐2
𝐷𝑐1 m
Wb
km
Figure 32: Conductor layout for this Example.
Total flux linkage between conductor 1 and 2 due to all currents is
𝜆12
= 0.2𝐼𝑎 ln𝐷𝑎2
𝐷𝑎1+ 0.21𝐼𝑐 ln
𝐷𝑐2
𝐷𝑐1m Wb/Km
For positive phase sequence, with 𝐼𝑎 as reference, 𝐼𝑐 = 𝐼𝑎∠− 240° and we have
𝜆12 = 0.2𝐼𝑎 + ln𝐷𝑎2
𝐷𝑎1 + 1∠ − 240° ln
𝐷𝑐2
𝐷𝑐1 mH/Km
With 𝐼𝑎 as reference, the instantaneous flux linkage is
𝐼12 𝑡 = 2 𝜆12 cos 𝜔𝑡 + 𝛼
Thus, the induced voltage in the telephone line per kilometer length is
𝜐 =𝑑𝜆12 𝑡
𝑑𝑡= 2𝜔 𝜆12 cos 𝜔𝑡 + 𝛼 + 90°
The rms voltage induced in the telephone line per kilometer is
𝑉 = 𝜔 𝜆12 ∠𝛼 + 90° = 𝑗𝜔𝜆12
From the circuits geometry
𝐷𝑎1 = 𝐷𝑐2 = 32 + 42 12 = 5 m
𝐷𝑎2 = 𝐷𝑐1 = 4.22 + 42 12 = 5.8 m
Dr Houssem Rafik El Hana Bouchekara 45
The total flux linkage is
𝜆12
= 0.2 × 200∠0° ln5.8
5+ 0.2 × 200∠ − 240° ln
5
5.8
= 10.283∠ − 30° m Wb/Km
The voltage induced in the telephone line per kilometer is
𝑉 = 𝑗𝜔𝜆12 = 𝑗2𝜋60 10.283∠ − 30° 10−3 = 3.88∠60° 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 (25°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 conditions𝑔, which represents resistive leakage between
Dr Houssem Rafik El Hana Bouchekara 46
a phase and ground, has negligible effect on performance and is customarily neglected.
(I,e,𝑔 = 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:
Dr Houssem Rafik El Hana Bouchekara 48
Dr Houssem Rafik El Hana Bouchekara 49
Table 2: Characteristics of copper conductors, hard-drawn, 97.3 % conductivity.
Dr Houssem Rafik El Hana Bouchekara 50
Dr Houssem Rafik El Hana Bouchekara 51
Table 3: Characteristics of ACSR
Dr Houssem Rafik El Hana Bouchekara 52
Table 4: Characteristics of ACSR
Dr Houssem Rafik El Hana Bouchekara 53
Table 5: Characteristics of AAC
Dr Houssem Rafik El Hana Bouchekara 54
Table 6: Characteristics of AAC
Dr Houssem Rafik El Hana Bouchekara 55
Table 7: inductive reactance spacing factor at 60hz