Download pdf - Transmission and Distribution of Electrical Power

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


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


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


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.


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


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).


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.


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:


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:


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.,


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)


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.


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)


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)


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


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


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


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.


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)


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:


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


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.


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


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.


Solution :


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


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.


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.


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:


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:


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.


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.


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


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)


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


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


𝐷23

𝑟+ 𝑞𝑏 ln

𝑟

𝐷13+ 𝑞𝑐 ln

𝐷13

𝐷12 ( 83)

And for the last section

𝑉𝑎𝑏 (𝐼𝐼𝐼) =1


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)


Or

𝑉𝑎𝑏 =1


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


𝐺𝑀𝐷

𝑟+ 𝑞𝑏 ln

𝑟

𝐺𝑀𝐷 ( 88)

Similarly, we find the average voltage 𝑉𝑎𝑐 as

𝑉𝑎𝑐 =1


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.


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


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.


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.


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


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.


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


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


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


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.


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.


(a)

(b)


Solution:



Table 2: Characteristics of copper conductors, hard-drawn, 97.3 % conductivity.



Table 3: Characteristics of ACSR


Table 4: Characteristics of ACSR


Table 5: Characteristics of AAC


Table 6: Characteristics of AAC


Table 7: inductive reactance spacing factor at 60hz