Chapter 14

Transmission Line Parameters

14.1 Transmission Line Parameters 14-1

Series Resistance • Series Inductance and Series Inductive Reactance • Shunt Capacitance and Capacitive Reactance • Equivalent Circuit of Three-Phase Transmission Lines • Characteristics of Overhead Conductors

References14-36

Manuel Reta-Hernández

Universidad Autónoma de Zacatecas

14.1 Transmission Line Parameters

The power transmission line is one of the main 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, the shunt conductance is normally not considered for the transmission line modeling.

14.1.1 Series Resistance

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, at a given temperature, is evaluated by Equation 14.1:

$R_{DC}=\frac{\rho l}{A}(\Omega )$ (14.1)

where

ρ is the conductor resistivity at a given temperature (Ω m)

l is the conductor length (m)

A is the conductor cross-section area (m2)

If AC current is flowing, rather than DC current, the conductor effective resistance is higher due to frequency or skin effect.

14.1.1.1 Frequency Effect

Frequency of AC voltages produces a second effect on the conductor resistance due to the non-uniform 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 very 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 series resistance as shown in Equation 14.2. For a 60 Hz frequency, it is estimated k = 1.02:

$R_{AC}=R_{DC}k$ (14.2)

Other variations in the series AC resistance are caused by

• Temperature
• Spiraling of stranded conductors
• Bundle conductors’ arrangement

14.1.1.2 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 Equation 14.3:

$R_2=R_1\left(\frac{T+t_2}{T+t_1}\right)\text{(}\Omega \text{)}$ (14.3)

where

R2 is the resistance at second temperature t2

R1 is the resistance at initial temperature t1

T is the temperature coefficient for a specific material (°C)

Resistivity (ρ) and temperature coefficient (T) constants depend upon the particular conductor material. Table 14.1 lists resistivity and temperature coefficients of some typical conductor materials.

14.1.1.3 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 low 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.

Among different types of commercially available aluminum conductors, aluminum-conductor-steel-reinforced (ACSR) conductor is one of the most used conductors in overhead 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.

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 is 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 [1,2]. The resistance of each wound conductor at any layer, per-unit length, is based on its total length as shown in Equation 14.4:

$R_{cond}=\frac{\rho }{A}\sqrt{\text{1}+{\left(\pi \frac{1}{p}\right)}^2}(\Omega \text{/m})$ (14.4)

where

Rcond is the resistance of wound conductor (Ω)

$\sqrt{\text{1}+{(\pi (1\text{/}p))}^2}$ is the length of wound conductor (m)

p = (lturn/2rlayer) is the relative pitch of wound conductor, lturn is the length of one turn of the spiral (m), and 2rlayer is the diameter of the layer (m)

The parallel combination of n conductors, with same diameter per layer, gives the resistance per layer as follows:

$R_{layer}=\frac{1}{{\displaystyle {\sum }_{i=1}^n(1\text{/}{R}_i)}}(\Omega \text{/m})$ (14.5)

Similarly, the total resistance of the stranded conductor is evaluated by the parallel combination of resistances per layer.

In high-voltage overhead 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, meteorological conditions play a key role in its evaluation. Corona losses under rain or snow, for instance, are much higher than in dry weather.

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 series 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 [2,3]. Spacers made of steel or aluminum bars are used to maintain the distance between bundle conductors along the line. Figure 14.1 shows some typical arrangement of stranded bundle configurations.

14.1.1.4 Current-Carrying Capacity (Ampacity)

In overhead transmission lines, the current-carrying capacity is determined mostly by the conductor resistance and the heat dissipated from its surface. The heat generated in a conductor (Joules’s effect) is dissipated from its surface area by convection and radiation given by Equation 14.6:

$I^{2}R=S(w_c+w_r)(\text{W})$ (14.6)

where

R is the conductor resistance (Ω)

I is the conductor current-carrying (A)

S is the conductor surface area (sq. in.)

wc is the convection heat loss (W/sq. in.)

wr is the radiation heat loss (W/sq. in.)

Heat dissipation by convection is defined in Equation 14.7 as

$w_c=\frac{\text{0}\text{.0128}\sqrt{pv}}{T_{air}^{0.123}\sqrt{d_{cond}}}\Delta t(\text{W})$ (14.7)

where

p is the atmospheric pressure (atm)

v is the wind velocity (ft/s)

dcond is the conductor diameter (in.)

Tair is the air temperature (°C)

Δt = Tc − Tair is the temperature rise of the conductor (°C)

Heat dissipation by radiation is obtained from Stefan–Boltzman law and is defined as

$w_r=\text{36}\text{.8}E\left[{\left(\frac{T_c}{1000}\right)}^4-{\left(\frac{T_{air}}{1000}\right)}^4\right](\text{W/sq}\text{.}\text{in}\text{.})$ (14.8)

where

wr is the radiation heat loss (W/sq. in.)

E is the emissivity constant (1 for the absolute black body and 0.5 for oxidized copper)

Tc is the conductor temperature (°C)

Tair is the ambient temperature (°C)

Substituting Equations 14.7 and 14.8 in 14.6 we can obtain the conductor ampacity at given temperatures, as shown in Equations 14.9 and 14.10:

$I=\sqrt{\frac{S(w_c+w_r)}{R}}(\text{A})$ (14.9)

$I=\sqrt{\frac{S}{R}\left(\frac{\Delta t\left(\text{0}\text{.0128}\sqrt{pv}\right)}{T_{air}^{0.123}\sqrt{d_{cond}}}+\text{36}\text{.8}E\left(\frac{T_c^4-T_{air}^4}{{1000}^4}\right)\right)}(\text{A})$ (14.10)

Some approximated current-carrying capacity for overhead ACSR and AAC conductors are presented in Section 14.1.5 [1–3].

14.1.2 Series Inductance and Series Inductive Reactance

A current-carrying conductor produces concentric magnetic flux lines around the conductor. If the current varies with the time, the magnetic flux changes and a voltage is induced. Therefore, an inductance is present, defined as the ratio of the magnetic flux linkage and the current. The magnetic flux produced by the current in transmission line conductors produces a total inductance whose magnitude depends on the line configuration [4–7]. To determine the inductance of the line, it is necessary to calculate, as in any magnetic circuit with permeability μ, the following factors:

1. Magnetic field intensity H
2. Magnetic field density B
3. Flux linkage λ

14.1.2.1 Inductance of a Solid, Round, Infinitely Long Conductor

Consider an infinitely long, solid cylindrical conductor with radius r, carrying-current I as shown in Figure 14.2. If the conductor is made of a nonmagnetic material, and the current is assumed uniformly distributed (no skin effect), then the generated internal and external magnetic field lines are concentric circles around the conductor with direction defined by the right-hand rule.

14.1.2.2 Internal Inductance due to Internal Magnetic Flux

To obtain the internal inductance, a magnetic field with radius x inside the conductor of length l is chosen, as shown in Figure 14.3.

The fraction of the current Ix enclosed in the area of the circle chosen is determined by

$I_x=I\frac{\pi x^2}{\pi r^2}(\text{A})$ (14.11)

Ampere’s law determines the magnetic field intensity Hx, constant at any point along the circle contour as

$H_x=\frac{I_x}{2\pi x}=\frac{I}{2\pi r^2}x(\text{A/m})$ (14.12)

The magnetic flux density Bx is obtained by

$B_x=\mu H_x=\frac{{\mu }_0}{2\pi }\left(\frac{Ix}{r^2}\right)(\text{T})$ (14.13)

where μ = μ0 = 4π × 10−7 H/m is the magnetic permeability for a nonmagnetic material.

The differential flux dϕ enclosed in a ring of thickness dx for a 1 m length of conductor, and the differential flux linkage dλ in the respective area are defined in Equations 14.14 and 14.15:

$d\varphi =B_{x}dx=\frac{{\mu }_0}{2\pi }\left(\frac{Ix}{r^2}\right)dx(\text{Wb/m})$ (14.14)

$d\lambda =\frac{\pi x^2}{\pi r^2}d\varphi =\frac{{\mu }_0}{2\pi }\left(\frac{I{x}^3}{r^4}\right)dx(\text{Wb/m})$ (14.15)

The internal flux linkage is obtained by integrating the differential flux linkage from x = 0 to x = r:

${\lambda }_{int}={\displaystyle \underset{\text{0}}{\overset{r}{\int }}d\lambda }=\frac{{\mu }_{\text{0}}}{\text{8}\pi }I(\text{Wb/m})$ (14.16)

Therefore, the conductor inductance due to internal flux linkage, per-unit length, becomes

$L_{int}=\frac{{\lambda }_{int}}{I}=\frac{{\mu }_{\text{0}}}{\text{8}\pi }(\text{H/m})$ (14.17)

14.1.2.3 External Inductance

The external inductance is evaluated assuming that the total current I is concentrated at the conductor surface (maximum skin effect). At any point, on an external magnetic field circle of radius y (Figure 14.4), the magnetic field intensity Hy and the magnetic field density By, per-unit length, are defined by Equations 14.18 and 14.19:

$H_y=\frac{I}{2\pi y}(\text{A/m})$ (14.18)

$B_y=\mu H_y=\frac{{\mu }_0}{2\pi }\frac{I}{y}(\text{T})$ (14.19)

The differential flux dϕ enclosed in a ring of thickness dy, from point D1 to point D2, for a 1 m length of conductor is

$d\varphi =B_{y}dy=\frac{{\mu }_0}{2\pi }\frac{I}{y}dy(\text{Wb/m})$ (14.20)

As the total current I flows in the surface conductor, then the differential flux linkage dλ has the same magnitude as the differential flux dϕ:

$d\lambda =d\varphi =\frac{{\mu }_0}{2\pi }\frac{I}{y}dy(\text{Wb/m})$ (14.21)

The total external flux linkage enclosed by the ring is obtained by integrating from D1 to D2 becomes

${\lambda }_{1-2}={\displaystyle \underset{D_{\text{1}}}{\overset{D_{\text{2}}}{\int }}d\lambda }=\frac{{\mu }_0}{2\pi }I{\displaystyle \underset{D_{\text{1}}}{\overset{D_{\text{2}}}{\int }}\frac{dy}{y}}=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D_1}{D_2}\right)(\text{Wb/m})$ (14.22)

In general, the total external flux linkage from the surface of the conductor to any point D, per-unit length, is

${\lambda }_{ext}={\displaystyle \underset{r}{\overset{D}{\int }}d\lambda }=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D}{r}\right)(\text{Wb/m})$ (14.23)

The summation of the internal and external flux linkage at any point D permits evaluation of the total inductance of the conductor Ltot, per-unit length, as follows:

${\lambda }_{}+{\lambda }_{ext}=\frac{{\mu }_0}{2\pi }I\left[\frac{1}{4}+\text{ln}\left(\frac{D}{r}\right)\right]=\frac{{\mu }_0}{2\pi }I\left[\text{ln}\left(e^{1/4}\right)+\text{ln}\left(\frac{D}{r}\right)\right]=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D}{e^{-1/4}r}\right)(\text{Wb/m})$ (14.24)

Therefore, the expression of total inductance is given by Equation 14.25:

$L_{tot}=\frac{{\lambda }_{int}+{\lambda }_{ext}}{I}=\frac{{\mu }_0}{2\pi }\text{ln}\left(\frac{D}{GMR}\right)(\text{H/m})$ (14.25)

where GMR (geometric mean radius) = e−1/4 r = 0.7788 r.

GMR 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.

14.1.2.4 Inductance of a Two-Wire, Single-Phase Line

Now, consider a two-wire single-phase line with solid cylindrical conductors A and B with the same radius r, same length l, and separated by a distance D, where D > r, and conducting the same current I, as shown in Figure 14.5. The current flows from the source to the load in conductor A and returns in conductor B (IA = −IB).

The magnetic flux generated by one conductor links the second conductor. The total flux linking conductor A, for instance, has two components: (a) the flux generated by conductor A and (b) the flux generated by conductor B which links conductor A.

As shown in Figure 14.6, the total flux linkage from conductors A and B at point P are

${\lambda }_{AP}={\lambda }_{AAP}+{\lambda }_{ABP}$ (14.26)

${\lambda }_{BP}={\lambda }_{BBP}+{\lambda }_{BAP}$ (14.27)

where

λAAP is the flux linkage from magnetic field of conductor A on conductor A at point P

λABP is the flux linkage from magnetic field of conductor B on conductor A at point P

λBBP is the flux linkage from magnetic field of conductor B on conductor B at point P

λBAP is the flux linkage from magnetic field of conductor A on conductor B at point P

The expressions of the aforementioned flux linkages, per-unit length, are

${\lambda }_{AAP}=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D_{AP}}{GM{R}_A}\right)(\text{Wb/m})$ (14.28)

${\lambda }_{ABP}={\displaystyle \underset{D}{\overset{D_{BP}}{\int }}{B}_{BP}}dP=-\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D_{BP}}{D}\right)(\text{Wb/m})$ (14.29)

${\lambda }_{BAP}={\displaystyle \underset{D}{\overset{D_{AP}}{\int }}{B}_{AP}}dP=-\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D_{AP}}{D}\right)(\text{Wb/m})$ (14.30)

${\lambda }_{BBP}=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D_{BP}}{GM{R}_B}\right)(\text{Wb/m})$ (14.31)

The total flux linkage of the system at point P is the algebraic summation of λAP and λBP:

${\lambda }_P={\lambda }_{AP}+{\lambda }_{BP}=({\lambda }_{AAP}+{\lambda }_{ABP})+({\lambda }_{BAP}+{\lambda }_{BBP})$ (14.32)

${\lambda }_P=\frac{{\mu }_0}{2\pi }I\text{ln}\left[\left(\frac{D_{AP}}{GM{R}_A}\right)\left(\frac{D}{D_{AP}}\right)\left(\frac{D_{BP}}{GM{R}_B}\right)\left(\frac{D}{D_{BP}}\right)\right]=\frac{{\mu }_0}{2\pi }I\text{ln}\left(\frac{D^2}{GM{R}_{A}GM{R}_B}\right)(\text{Wb/m})$ (14.33)

If the conductors have the same radius, rA = rB = r, and the point P is shifted to infinity, then the total flux linkage of the system becomes

$\lambda =\frac{{\mu }_0}{\pi }I\text{ln}\left(\frac{D}{GMR}\right)(\text{Wb/m})$ (14.34)

and the total inductance per-unit length is

$L_{1-phasesystem}=\frac{\lambda }{I}=\frac{{\mu }_0}{\pi }\text{ln}\left(\frac{D}{GMR}\right)(\text{H/m})$ (14.35)

Comparing Equations 14.25 and 14.35, it can be seen that the inductance of the single-phase system is twice the inductance of a single conductor.

For a line with stranded conductors, the inductance is determined using a new GMR value named GMRstranded, evaluated according to the number of conductors. If conductors A and B in the single-phase system are formed by n and m solid cylindrical identical subconductors in parallel, respectively, then the expressions or GMRstranded for conductor A and B are

$GM{R}_{A\_stranded}=\sqrt[n^2]{{\displaystyle \prod _{i=1}^n{\displaystyle \prod _{j=1}^n{D}_{ij}}}}$ (14.36)

$GM{R}_{B\_stranded}=\sqrt[m^2]{{\displaystyle \prod _{i=1}^m{\displaystyle \prod _{j=1}^m{D}_{ij}}}}$ (14.37)

Generally, the GMRstranded for a particular cable can be found in conductors tables given by the manufacturer.

If the line conductor is composed of bundle conductors, the inductance is reevaluated taking into account the number of bundle conductors and the separation among them. The GMRbundle is introduced to determine the final inductance value. Assuming the same separation among bundle conductors, the equation for GMRbundle, up to three conductors per bundle, is defined as

$GM{R}_{nbundleconductors}=\sqrt[n]{d^{n-1}GM{R}_{stranded}}$ (14.38)

where

n is the number of conductors per bundle

GMRstranded is the GMR of the stranded conductor

d is the distance between bundle conductors

In four conductors per bundle with the same separation between consecutive conductors, the GMRbundle, is evaluated, according to their geometry, as

$\begin{array}{l}GM{R}_{4bundleconductors}=\sqrt[\text{16}]{{\left(GM{R}_{stranded}ddd\sqrt{\text{2}}\right)}^4}\\ =\sqrt[\text{16}]{4GM{R}_{stranded}^4{d}^{12}}\\ =4^{1/16}GM{R}_{stranded}^{1/4}{d}^{3/4}\\ =\text{1}\text{.09}\sqrt[\text{4}]{GM{R}_{stranded}{d}^3}\end{array}$ (14.39)

14.1.2.5 Inductance of Three-Phase Transmission Line in Asymmetrical Arrangement

Derivations of inductance in a single-phase system can be extended to obtain the inductance per phase in a three-phase system. Consider a three-phase, three-conductor system with solid cylindrical conductors with identical radius rA, rB, and rC, placed horizontally with separation DAB, DBC, and DCA (where D > r) among them. Corresponding currents IA, IB, and IC flow along each conductor as shown in Figure 14.7.

The total magnetic flux enclosing conductor A at a point P away from the conductors, is the sum of the flux produced by conductors A, B, and C as indicated in Equation 14.40:

${\varphi }_{AP}={\varphi }_{AAP}+{\varphi }_{ABP}+{\varphi }_{ACP}$ (14.40)

where

ϕAAP is the flux produced by current IA on conductor A at point P

ϕABP is the flux produced by current IB on conductor A at point P

ϕACP is the flux produced by current IC on conductor A at point P

Considering 1 m length for each conductor, the expressions for the fluxes are

${\varphi }_{AAP}=\frac{{\mu }_{\text{0}}}{\text{2}\pi }{I}_A\ln \left(\frac{D_{AP}}{GM{R}_A}\right)(\text{Wb/m})$ (14.41)

${\varphi }_{ABP}=\frac{{\mu }_{\text{0}}}{\text{2}\pi }{I}_B\ln \left(\frac{D_{BP}}{D_{AB}}\right)(\text{Wb/m})$ (14.42)

${\varphi }_{ACP}=\frac{{\mu }_{\text{0}}}{\text{2}\pi }{I}_C\ln \left(\frac{D_{CP}}{D_{AC}}\right)(\text{Wb/m})$ (14.43)

The corresponding flux linkage of conductor A at point P (Figure 14.8), is evaluated as

${\lambda }_{AP}={\lambda }_{AAP}+{\lambda }_{ABP}+{\lambda }_{ACP}$ (14.44)

having

${\lambda }_{AAP}=\frac{{\mu }_0}{2\pi }{I}_A\text{ln}\left(\frac{D_{AP}}{GM{R}_A}\right)\text{(Wb/m)}$ (14.45)

${\lambda }_{ABP}={\displaystyle \underset{D_{AB}}{\overset{D_{BP}}{\int }}{B}_{BP}}dP=\frac{{\mu }_0}{2\pi }{I}_B\text{ln}\left(\frac{D_{BP}}{D_{AB}}\right)\text{(Wb/m)}$ (14.46)

${\lambda }_{ACP}={\displaystyle \underset{D_{AC}}{\overset{D_{CP}}{\int }}{B}_{CP}}dP=\frac{{\mu }_0}{2\pi }{I}_C\text{ln}\left(\frac{D_{CP}}{D_{AC}}\right)\text{(Wb/m)}$ (14.47)

where

λAP is the total flux linkage of conductor A at point P

λAAP is the flux linkage from magnetic field of conductor A on conductor A at point P

λABP is the flux linkage from magnetic field of conductor B on conductor A at point P

λACP is the flux linkage from magnetic field of conductor C on conductor A at point P

Substituting Equations 14.45 through 14.47 in Equation 14.44 and rearranging, according to natural logarithms law, we have

${\lambda }_{AP}=\frac{{\mu }_{\text{0}}}{\text{2}\pi }\left[I_A\ln \left(\frac{D_{AP}}{GM{R}_A}\right)+I_B\ln \left(\frac{D_{BP}}{D_{AB}}\right)+I_C\ln \left(\frac{D_{CP}}{D_{AC}}\right)\right]\text{(Wb/m)}$ (14.48)

$\begin{array}{c}{\lambda }_{AP}=\frac{{\mu }_{\text{0}}}{\text{2}\pi }\left[I_A\ln \left(\frac{1}{GM{R}_A}\right)+I_B\ln \left(\frac{1}{D_{AB}}\right)+I_C\ln \left(\frac{1}{D_{AC}}\right)\right]\\ \text{+}\frac{{\mu }_{\text{0}}}{\text{2}\pi }[I_A\ln (D_{AP})+I_B\ln (D_{BP})+I_C\ln (D_{CP})]\text{(Wb/m)}\end{array}$ (14.49)

The arrangement of Equation 14.48 into 14.49 is algebraically correct according to natural logarithms law. However, as calculation of any natural logarithm must be dimensionless, the numerator in the expressions ln(1/GMRA), ln(1/DAB), and ln(1/DAC) must have the same dimension as the denominator. The same applies for the denominator in the expressions ln(DAP), ln(DBP), and ln(DCP).

Assuming a balanced three-phase system, where IA + IB + IC = 0, and shifting the point P to infinity in such a way that DAP = DBP = DCP, then the second part of Equation 14.49 is zero, and the flux linkage of conductor A becomes

$\begin{array}{c}{\lambda }_A=\frac{{\mu }_{\text{0}}}{\text{2}\pi }\left[I_A\ln \left(\frac{1}{GM{R}_A}\right)+I_B\ln \left(\frac{1}{D_{AB}}\right)+I_C\ln \left(\frac{1}{D_{AC}}\right)\right]\\ =\frac{{\mu }_0}{2\pi }[I_A{L}_{AA}+I_B{L}_{AB}+I_C{L}_{AC}]\text{(Wb/m)}\end{array}$ (14.50)

Similarly, the flux linkage expressions for conductors B and C are

$\begin{array}{c}{\lambda }_B=\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{1}{D_{BA}}\right)+I_B\text{ln}\left(\frac{\text{1}}{GM{R}_B}\right)+I_C\text{ln}\left(\frac{1}{D_{BC}}\right)\right]\\ =\frac{{\mu }_0}{2\pi }[I_A{L}_{BA}+I_B{L}_{BB}+I_C{L}_{BC}]\text{(Wb/m)}\end{array}$ (14.51)

$\begin{array}{c}{\lambda }_C=\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{1}{D_{CA}}\right)+I_B\text{ln}\left(\frac{1}{D_{CB}}\right)+I_C\text{ln}\left(\frac{\text{1}}{GM{R}_C}\right)\right]\\ =\frac{{\mu }_0}{2\pi }[I_A{L}_{CA}+I_B{L}_{CB}+I_C{L}_{CC}]\text{(Wb/m)}\end{array}$ (14.52)

where

λA, λB, λC are the total flux linkages of conductors A, B, and C

LAA, LBB, LCC are the self-inductances of conductors A, B, and C

LAB, LBC, LCA, LBA, LCB, LAC are the mutual inductance among conductors A, B, and C

The flux linkage of each phase conductor depends on the three currents, and therefore, the inductance per phase is not only one, as in the single-phase system. Instead, there are self and mutual conductor inductances, as shown in Equations 14.50 through 14.52.

14.1.2.6 Inductance of Balanced Three-Phase Transmission Line in Symmetrical Arrangement

If the three-phase transmission line has an asymmetrical arrangement, as in the horizontal arrangement, there are nine different inductances (mutual and self-inductances). However, a single inductance per phase can be obtained if the three conductors are arranged with the same separation among them (symmetrical arrangement), where D = DAB = DBC = DCA. For a balanced three-phase system (IA + IB + IC = 0, or IA = −IB − IC), the flux linkage of each conductor, per-unit length, will be the same. Substituting these values in Equation 14.50, the expression of λA is simplified as follows:

$\begin{array}{c}{\lambda }_A=\frac{{\mu }_{\text{0}}}{\text{2}\pi }\left[(-I_B-I_C)\ln \left(\frac{1}{GM{R}_A}\right)+I_B\ln \left(\frac{1}{D}\right)+I_C\ln \left(\frac{1}{D}\right)\right]\\ =\frac{{\mu }_0}{2\pi }\left[-I_B\text{ln}\left(\frac{D}{GM{R}_A}\right)-I_C\text{ln}\left(\frac{D}{GM{R}_A}\right)\right]\\ =\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{D}{GM{R}_A}\right)\right]\text{(Wb/m)}\end{array}$ (14.53)

If GMR value is the same for all conductors (either single or bundle GMR), the total flux linkage expression is the same for all phases. Therefore, the equivalent inductance per phase is

$L_{phase}=\frac{{\mu }_0}{2\pi }\text{ln}\left(\frac{D}{GM{R}_{phase}}\right)(\text{H/m})$ (14.54)

14.1.2.7 Inductance of Transposed Three-Phase Transmission Lines

In actual transmission lines, the phase conductors cannot maintain symmetrical arrangement along the whole length because of construction considerations, even when bundle conductors spacers are used. With asymmetrical spacing, the inductance will be different for each phase, with a corresponding unbalanced voltage drop on each conductor. Therefore, the single-phase equivalent circuit to represent the power system cannot be used.

However, it is possible to assume symmetrical arrangement in the transmission line by transposing the phase conductors. In a transposed system, each phase conductor occupies the location of the other two phases for one-third of the total line length as shown in Figure 14.9.

The flux linkage of conductor A in each section along the total length is obtained considering the distances related to conductors of phase B and C:

${\lambda }_{Asection1}=\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)+I_B\text{ln}\left(\frac{1}{D_{21}}\right)+I_C\text{ln}\left(\frac{1}{D_{31}}\right)\right]$ (14.55)

${\lambda }_{Asection2}=\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)+I_B\text{ln}\left(\frac{1}{D_{32}}\right)+I_C\text{ln}\left(\frac{1}{D_{12}}\right)\right]$ (14.56)

${\lambda }_{Asection3}=\frac{{\mu }_0}{2\pi }\left[I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)+I_B\text{ln}\left(\frac{1}{D_{13}}\right)+I_C\text{ln}\left(\frac{1}{D_{23}}\right)\right]$ (14.57)

where

λA section 1 is the flux linkage of conductor A in section 1

λA section 2 is the flux linkage of conductor A in section 2

λA section 3 is the flux linkage of conductor A in position 3

D21 is the distance between conductor B and A when conductor A is in section 1

D31 is the distance between conductor C and A when conductor A is in section 1

Then, the average value of flux linkage in phase A is expressed in Equation 14.58:

$\begin{array}{c}{\lambda }_A=\frac{{\lambda }_{Asection1}\text{(}l\text{/}3\text{)}+{\lambda }_{Asection2}(l\text{/}3)+{\lambda }_{Asection3}(l\text{/}3)}{l}\\ =\frac{{\lambda }_{Asection1}+{\lambda }_{Asection2}+{\lambda }_{Asection3}}{3}\\ =\frac{{\mu }_0}{6\pi }\left[3I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)+I_B\text{ln}\left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)+I_C\text{ln}\left(\frac{1}{D_{13}{D}_{21}{D}_{32}}\right)\right]\end{array}$ (14.58)

As D13 = D31, D23 = D32, and D12 = D21, and −IA = IB + IC in a balanced three-phase system, then λA is simplified in Equation 14.59:

$\begin{array}{c}{\lambda }_A=\frac{{\mu }_0}{6\pi }\left[3I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)-I_A\text{ln}\left(\frac{1}{D_{12}{D}_{23}{D}_{31}}\right)\right]\\ =\frac{{\mu }_0}{6\pi }\left[3I_A\text{ln}\left(\frac{1}{GM{R}_A}\right)+I_A\text{ln(}{D}_{12}{D}_{23}{D}_{31}\text{)}\right]\\ =\frac{{\mu }_0}{2\pi }{I}_A\text{ln}\left(\frac{{(D_{12}{D}_{23}{D}_{31})}^{1/3}}{GM{R}_A}\right)\end{array}$ (14.59)

Flux linkages λB and λB are obtained in similar procedure. The expression (D12D23D31)1/3 is defined as the average distance GMD (geometrical mean distance) and substitutes the distance D of the symmetrical arrangement case. As conductor of phase A is assumed the same as conductor of phase B and C, then GMRA = GMRB = GMRC.

Therefore, calculation of phase inductance derived for symmetrical arrangement is valid for transposed transmission lines. The expression of inductance per phase per unit-length becomes

$L_{phase}=\frac{{\mu }_0}{2\pi }\text{ln}\left(\frac{GMD}{GM{R}_{phase}}\right)(\text{H/m})$ (14.60)

Once the inductance per phase is obtained, the inductive reactance per unit-length is expressed as follows:

$X_{L_{phase}}=2\pi f{L}_{phase}={\mu }_{0}f\text{ln}\left(\frac{GMD}{GM{R}_{phase}}\right)(\Omega \text{/m})$ (14.61)

For bundle conductors, the GMRbundle value is determined, as in the single-phase transmission line case, by the number of conductors, and by the number of conductors per bundle and the separation among them. The expression for the total inductive reactance per phase yields

$X_{L_{phase}}={\mu }_{0}f\text{ln}\left(\frac{GMD}{GM{R}_{bundle}}\right)(\Omega \text{/m})$ (14.62)

where

GMRbundle = (dn−1 GMRphase)1/n up to three conductors per bundle

GMRbundle = 1.09(d3 GMRphase)1/4 for four conductors per bundle. GMRphase is the geometric mean radius of phase conductor, either solid or stranded.

14.1.3 Shunt Capacitance and Capacitive Reactance

Capacitance exists among transmission line conductors due to their potential difference. To evaluate the capacitance between conductors in a surrounding medium with permittivity ε, it is necessary to determine the voltage between the conductors, and the electric field strength of the surrounding [4–7].

14.1.3.1 Capacitance of a Single Solid Conductor

Consider a solid, cylindrical, long conductor with radius r, in a free space with permittivity ε0, and with a charge of q+ coulombs per meter, uniformly distributed on the surface. There is constant electric field strength on the surface of cylinder (Figure 14.10). The resistivity of the conductor is assumed to be zero (perfect conductor), which results in zero internal electric field due to the charge on the conductor.

The charge q+ produces an electric field radial to the conductor with equipotential surfaces concentric to the conductor. According to Gauss’s law, the total electric flux leaving a closed surface is equal to the total charge inside the volume enclosed by the surface. Therefore, at an outside point P separated x meters from the center of the conductor, the electric field flux density, and the electric field intensity are

$Densit{y}_P=\frac{q}{A}=\frac{q}{\text{2}\pi x}(\text{C})$ (14.63)

$E_P=\frac{Densit{y}_P}{\epsilon }=\frac{q}{2\pi {\epsilon }_{0}x}(\text{V/m})$ (14.64)

where

DensityP is the electric flux density at point P (C)

EP is the electric field intensity at point P (V/m)

A is the surface of a concentric cylinder with 1 m length and radius x (m2)

ε = ε0 = 10−9/36π is the permittivity of free space assumed for the conductor (F/m)

The potential difference or voltage difference between two outside points P1 and P2 with corresponding distances x1 and x2 from the conductor center is defined by integrating the electric field intensity from x1 to x2:

$V_{P1-P2}={\displaystyle \underset{x_1}{\overset{x_2}{\int }}{E}_P}\frac{dx}{x}={\displaystyle \underset{x_1}{\overset{x_2}{\int }}\frac{q}{2\pi {\epsilon }_0}}\frac{dx}{x}=\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{x_2}{x_1}\right](\text{V})$ (14.65)

Then, the capacitance between points P1 and P2 is evaluated as

$C_{P1-P2}=\frac{q}{V_{P1-P2}}=\frac{2\pi {\epsilon }_0}{\text{ln[}{x}_2\text{/}{x}_1\text{]}}(\text{F/m})$ (14.66)

If point P1 is located at the conductor surface (x1 = r), and point P2 is located at ground surface below the conductor (x2 = h), then the voltage of the conductor, and the capacitance between the conductor and ground are

$V_{cond}=\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{h}{r}\right](\text{V})$ (14.67)

$C_{cond\text{-}ground}=\frac{q}{V_{cond}}=\frac{2\pi {\epsilon }_0}{\text{ln[}h\text{/}r\text{]}}(\text{F/m})$ (14.68)

14.1.3.2 Capacitance of a Single-Phase Line with Two Wires

Consider a two-wire single-phase line with conductors A and B with the same radius r, separated by a distance D > rA and rB. The conductors are energized by a voltage source such that conductor A has a charge q+ and conductor B a charge q− as shown in Figure 14.11.

The charge of each conductor generates independent electric fields. Charge q+ of conductor A generates a voltage VAB−A between both conductors. Similarly, charge q− of conductor B generates a voltage VAB−B between conductors.

VAB−A is calculated by integrating the electric field intensity, due to the charge of conductor A on conductor B from rA to D:

$V_{AB-A}={\displaystyle \underset{r_A}{\overset{D}{\int }}{E}_{A}dx}=\frac{+q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{D}{r_A}\right]$ (14.69)

VAB−B is calculated by integrating the electric field intensity due to the charge of conductor B from D to rB:

$V_{AB-B}={\displaystyle \underset{D}{\overset{r_B}{\int }}{E}_{B}dx}=\frac{-q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{r_B}{D}\right]=\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{D}{r_B}\right]$ (14.70)

The total voltage is the sum of the generated voltages VAB−A and VAB−B:

$V_{AB}=V_{AB-A}+V_{AB-B}=\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{D}{r_A}\right]+\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{D}{r_B}\right]=\frac{q}{2\pi {\epsilon }_0}\text{ln}\left[\frac{D^2}{r_A{r}_B}\right]$ (14.71)

If the conductors have the same radius, rA = rB = r, then the voltage between conductors VAB, and the capacitance between conductors CAB, for a 1 m line length are

$V_{AB}=\frac{q}{\pi {\epsilon }_0}\text{ln}\left[\frac{D}{r}\right](\text{V})$ (14.72)

$C_{AB}=\frac{\pi {\epsilon }_0}{\text{ln[}D\text{/}r\text{]}}(\text{F/m})$ (14.73)

The voltage between each conductor and ground (Figure 14.12) is one-half of the voltage between the two conductors. Therefore, the capacitance from either line to ground is twice the capacitance between lines:

$V_{AG}=V_{BG}=\frac{V_{AB}}{2}(\text{V})$ (14.74)

$C_{AG}=\frac{q}{V_{AG}}=\frac{2\pi {\epsilon }_0}{\text{ln[}D\text{/}r\text{]}}(\text{F/m})$ (14.75)

14.1.3.3 Capacitance of Three-Phase Transmission Line in Asymmetrical Arrangement

Consider a three-phase line with the same voltage magnitude between phases, and assuming a balanced three-phase system with abc (positive) sequence such that qA + qB + qC = 0. The conductors have radii rA, rB, and rC, and the space between conductors are DAB, DBC, and DAC (where DAB, DBC, and DAC > rA, rB, and rC). Also, the effect of earth and neutral conductors is neglected.

The expression for voltage between two conductors in a single-phase system can be extended to obtain the voltages between conductors in a three-phase system. The expressions for VAB and VAC are

$V_{AB}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{BA}}{r_A}\right]+q_B\text{ln}\left[\frac{r_B}{D_{AB}}\right]+q_C\text{ln}\left[\frac{D_{BC}}{D_{AC}}\right]\right]\text{(V)}$ (14.76)

$V_{AC}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{CA}}{r_A}\right]+q_B\text{ln}\left[\frac{D_{CB}}{D_{AB}}\right]+q_C\text{ln}\left[\frac{r_C}{D_{AC}}\right]\right]\text{(V)}$ (14.77)

With all conductors with the same radii rA = rB = rC = r, and assuming a balanced system line-to-line voltages with sequence abc, where $V_{AB}=\sqrt{\text{3}}{V}_{AN}\angle {30}^{\circ }$ and $V_{AC}=-V_{CA}=\sqrt{\text{3}}{V}_{AN}\angle -{30}^{\circ }$ , the line-to-neutral voltage VAN can be expressed in terms of VAB and VAC as

$V_{AN}=\frac{V_{AB}+V_{AC}}{3}$ (14.78)

$\begin{array}{c}{V}_{AN}=\frac{1}{6\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{BA}}{r_A}\right]+q_B\text{ln}\left[\frac{r_B}{D_{AB}}\right]+q_C\text{ln}\left[\frac{D_{BC}}{D_{AC}}\right]+q_A\text{ln}\left[\frac{D_{CA}}{r_A}\right]+q_B\text{ln}\left[\frac{D_{CB}}{D_{AB}}\right]+q_C\text{ln}\left[\frac{r_C}{D_{AC}}\right]\right]\\ =\frac{1}{6\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{BA}{D}_{CA}}{r^2}\right]+q_B\text{ln}\left[\frac{r{D}_{BC}}{D_{AB}^2}\right]+q_C\text{ln}\left[\frac{r{D}_{BC}}{D_{AC}^2}\right]\right]\end{array}$ (14.79)

Similar procedure can be followed to obtain VBN and VCN. However, it can be observed from Equation 14.79 that, for asymmetrical arrangement, as in the inductance evaluation case, there is no a single-phase capacitance value.

14.1.3.4 Capacitance of Three-Phase Transmission Line in Symmetrical Arrangement

If the three-phase system has triangular arrangement with equidistant conductors (symmetrical arrangement) such that DAB = DBC = DAC = D, with the same radii in all conductors such that rA = rB = rC = r (where D > r), then the expressions for VAB and VAC from Equations 14.76 and 14.77, are

$\begin{array}{c}{V}_{AB}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_B\text{ln}\left[\frac{r}{D}\right]+q_C\text{ln}\left[\frac{D}{D}\right]\right]\\ =\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_B\text{ln}\left[\frac{r}{D}\right]\right](\text{V})\end{array}$ (14.80)

$\begin{array}{c}{V}_{AC}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_B\text{ln}\left[\frac{D}{D}\right]+q_C\text{ln}\left[\frac{r}{D}\right]\right]\\ =\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_C\text{ln}\left[\frac{r}{D}\right]\right](\text{V})\end{array}$ (14.81)

Assuming again balanced line-to-line voltages with sequence abc, expressed in terms of the line-to-neutral voltage, where VAN = (VAB + VAC)/3, and substituting VAB and VAC from Equations 14.80 and 14.81, we have

$\begin{array}{c}{V}_{AN}=\frac{1}{6\pi {\epsilon }_0}\left[\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_B\text{ln}\left[\frac{r}{D}\right]\right]+\left[q_A\text{ln}\left[\frac{D}{r}\right]+q_C\text{ln}\left[\frac{r}{D}\right]\right]\right]\\ =\frac{1}{6\pi {\epsilon }_0}\left[2q_A\text{ln}\left[\frac{D}{r}\right]+(q_B+q_C)\text{ln}\left[\frac{r}{D}\right]\right]\text{(V)}\end{array}$ (14.82)

Under balanced conditions qA + qB + qC = 0, or −qA = (qB + qC) then, the final expression for the line-to-neutral voltage is

$V_{AN}=\frac{1}{2\pi {\epsilon }_0}{q}_A\text{ln}\left[\frac{D}{r}\right](\text{V})$ (14.83)

The positive sequence capacitance per unit-length between phase A and neutral can now be obtained. The same result is obtained for capacitance between phases B and C to neutral:

$C_{AN}=\frac{q_A}{V_{AN}}=\frac{2\pi {\epsilon }_0}{\text{ln[}D\text{/}r\text{]}}(\text{F/m})$ (14.84)

14.1.3.5 Capacitance of Stranded Bundle Conductors

The calculation of the capacitance in Equation 14.84 is based on

1. Solid conductors with zero resistivity (zero internal electric field)
2. Charge uniformly distributed
3. Equilateral spacing of phase conductors

In actual transmission lines, the resistivity of the conductors produces a small internal electric field, and therefore, the electric field at the conductor surface is smaller than the estimated. However, the difference is negligible for practical purposes.

Because of the presence of other charged conductors, the charge distribution is nonuniform, and therefore the estimated capacitance is different. However, this effect is negligible for most practical calculation. In a line with stranded conductors, the capacitance is evaluated assuming a solid conductor with the same radius as the outside radius of the stranded conductor. This produces a negligible difference.

Most transmission lines do not have equilateral spacing of phase conductors. This causes differences between the line-to-neutral capacitances of the three phases. However, transposing the phase conductors balances the system resulting in equal line-to-neutral capacitance for each phase. Consider a transposed three-phase line with conductors having the same radius r, and with space between conductors DAB, DBC, and DAC, where DAB, DBC, and DAC > r.

Assuming abc positive sequence, the expressions for VAB on the first, second, and third sections of the transposed line (refer to Figure 14.9) are

$V_{ABsection\text{1}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AB}}{r}\right]+q_B\text{ln}\left[\frac{r}{D_{AB}}\right]+q_C\text{ln}\left[\frac{D_{BC}}{D_{AC}}\right]\right]\text{(V)}$ (14.85)

$V_{ABsection\text{2}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{BC}}{r}\right]+q_B\text{ln}\left[\frac{r}{D_{BC}}\right]+q_C\text{ln}\left[\frac{D_{AC}}{D_{AB}}\right]\right]\text{(V)}$ (14.86)

$V_{ABsection\text{3}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AC}}{r}\right]+q_B\text{ln}\left[\frac{r}{D_{AC}}\right]+q_C\text{ln}\left[\frac{D_{AB}}{D_{BC}}\right]\right]\text{(V)}$ (14.87)

Similarly, the expressions for VAC on the first, second, and third sections of the transposed line are

$V_{ACsection\text{1}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AC}}{r}\right]+q_B\text{ln}\left[\frac{D_{BC}}{D_{AB}}\right]+q_C\text{ln}\left[\frac{r}{D_{AC}}\right]\right]\text{(V)}$ (14.88)

$V_{ACsection\text{2}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AB}}{r}\right]+q_B\text{ln}\left[\frac{D_{AC}}{D_{BC}}\right]+q_C\text{ln}\left[\frac{r}{D_{AB}}\right]\right]\text{(V)}$ (14.89)

$V_{ACsection\text{3}}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{BC}}{r}\right]+q_B\text{ln}\left[\frac{D_{AB}}{D_{AC}}\right]+q_C\text{ln}\left[\frac{r}{D_{BC}}\right]\right]\text{(V)}$ (14.90)

Taking the average value of the three sections, we have the final expressions of VAB and VAC of the transposed line in Equations 14.91 and 14.92:

$\begin{array}{c}{V}_{AB}{}_{transp}=\frac{V_{AB}+V_{AB}+V_{AB}}{\text{3}}\\ =\frac{1}{6\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AB}{D}_{BC}{D}_{AC}}{r^3}\right]+q_B\text{ln}\left[\frac{r^3}{D_{AB}{D}_{BC}{D}_{AC}}\right]+q_C\text{ln}\left[\frac{D_{BC}{D}_{AC}{D}_{AB}}{D_{AC}{D}_{AB}{D}_{BC}}\right]\right]\text{(V)}\end{array}$ (14.91)

$\begin{array}{c}{V}_{AC}{}_{transp}=\frac{V_{AC}+V_{AC}{}_{section\text{2}}+V_{AC}}{\text{3}}\\ =\frac{1}{6\pi {\epsilon }_0}\left[q_A\text{ln}\left[\frac{D_{AC}{D}_{AB}{D}_{BC}}{r^3}\right]+q_B\text{ln}\left[\frac{D_{BC}{D}_{AC}{D}_{AB}}{D_{AB}{D}_{BC}{D}_{AC}}\right]+q_C\text{ln}\left[\frac{r^3}{D_{AC}{D}_{AB}{D}_{BC}}\right]\right]\text{(V)}\end{array}$ (14.92)

For a balanced system where −qA = (qB + qC), the phase-to-neutral voltage VAN (phase voltage) becomes

$\begin{array}{c}{V}_{AN}{}_{transp}=\frac{V_{AB}{}_{transp}+V_{AC}{}_{transp}}{\text{3}}\\ =\frac{1}{18\pi {\epsilon }_0}\left[2q_A\text{ln}\left[\frac{D_{AB}{D}_{BC}{D}_{AC}}{r^3}\right]+(q_B+q_C)\text{ln}\left[\frac{r^3}{D_{AC}{D}_{AB}{D}_{BC}}\right]\right]\\ =\frac{1}{18\pi {\epsilon }_0}\left[2q_A\text{ln}\left[\frac{D_{AB}{D}_{BC}{D}_{AC}}{r^3}\right]-q_A\text{ln}\left[\frac{r^3}{D_{AC}{D}_{AB}{D}_{BC}}\right]\right]\\ =\frac{1}{6\pi {\epsilon }_0}{q}_A\text{ln}\left[\frac{D_{AB}{D}_{BC}{D}_{AC}}{r^3}\right]\\ =\frac{1}{2\pi {\epsilon }_0}{q}_A\text{ln}\left[\frac{GMD}{r}\right](\text{V})\end{array}$ (14.93)

where $GMD=\sqrt[\text{3}]{D_{AB}{D}_{BC}{D}_{CA}}$ is the geometrical mean distance for a three-phase line.

For bundle conductors, an equivalent radius re replaces the radius r of a single conductor and is determined by the number of conductors per bundle and the spacing of conductors. The expression of re is similar to GMRbundle used in the calculation of the inductance per phase, except that the actual outside radius of the conductor is used instead of the GMRphase. Therefore, the expression for VAN is

$V_{AN}{}_{transp}=\frac{1}{2\pi {\epsilon }_0}{q}_A\text{ln}\left[\frac{GMD}{r_e}\right](\text{V})$ (14.94)

where

$r_e=\sqrt[n]{d^{n-1}r}$ is the equivalent radius for up to three conductors per bundle (m)

$r_e=1.09\sqrt[4]{d^{3}r}$ is the equivalent radius for four conductors per bundle (m), d is the distance between bundle conductors (m), and n is the number of conductor per bundle

Finally, the capacitance and capacitive reactance, per-unit length, from phase to neutral can be evaluated as

$C_{AN}{}_{transp}=\frac{q_A}{V_{AN}{}_{transp}}=\frac{2\pi {\epsilon }_0}{\text{ln[}GMD\text{/}{r}_e\text{]}}(\text{F/m})$ (14.95)

$X_{AN}{}_{transp}=\frac{1}{2\pi f{C}_{AN}}=\frac{1}{4\pi f{\epsilon }_0}\text{ln}\left[\frac{GMD}{r_e}\right](\Omega \text{/m})$ (14.96)

14.1.3.6 Capacitance due to Earth’s Surface

Considering a single overhead conductor with a return path through the earth, separated a distance H from earth’s surface, the charge of the earth would be equal in magnitude to that on the conductor but of opposite sign. If the earth is assumed as a perfectly conductive horizontal plane with infinite length, then the electric field lines will go from the conductor to the earth, perpendicular to the earth’s surface (Figure 14.13).

To calculate the capacitance, the negative charge of the earth can be replaced by an equivalent charge of an image conductor with the same radius as the overhead conductor, lying just below the overhead conductor (Figure 14.14).

The same principle can be extended to calculate the capacitance per phase of a three-phase system. Figure 14.15 shows an equilateral arrangement of identical single conductors for phase A, B, and C carrying the charges qA, qB, and qC and their respective image conductors A′, B′, and C′.

DA, DB, and DC are perpendicular distances from phases A, B, and C to earth’s surface. DAA′, DBB′, and DCC′ are perpendicular distances from phases A, B, and C to image conductors A′, B′, and C′. Voltage VAB can be obtained as

$V_{AB}=\frac{1}{2\pi {\epsilon }_0}\left[\begin{array}{l}{q}_A\text{ln [}{D}_{AB}\text{/}{r}_A\text{]}+q_B\text{ln [}{r}_B\text{/}{D}_{AB}\text{]}+q_C\text{ln [}{D}_{BC}\text{/}{D}_{AC}\text{]}-\\ -q_A\text{ln [}{D}_{AB\prime }\text{/}{D}_{A{A}^{\prime }}\text{]}-q_B\text{ln [}{D}_{B{B}^{\prime }}\text{/}{D}_{A{B}^{\prime }}\text{]}-q_C\text{ln [}{D}_{B{C}^{\prime }}\text{/}{D}_{A{C}^{\prime }}\text{]}\end{array}\right]\text{(V)}$ (14.97)

With identical overhead conductors (r = rA = rB = rC) in equilateral arrangement, D = DAB = DBC = DCA, the expression for VAB becomes

$V_{AB}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\left(\text{ln}\left[\frac{D}{r}\right]-\text{ln}\left[\frac{D_{A{B}^{\prime }}}{D_{A{A}^{\prime }}}\right]\right)+q_B\left(\text{ln}\left[\frac{r}{D}\right]-\text{ln}\left[\frac{D_{B{B}^{\prime }}}{D_{A{B}^{\prime }}}\right]\right)-q_C\text{ln}\left[\frac{D_{B{C}^{\prime }}}{D_{A{C}^{\prime }}}\right]\right]\text{(V)}$ (14.98)

Similarly, the expressions for VBC and VAC are

$V_{BC}=\frac{1}{2\pi {\epsilon }_0}\left[-q_A\text{ln}\left[\frac{D_{C{A}^{\prime }}}{D_{B{A}^{\prime }}}\right]+q_B\left(\text{ln}\left[\frac{D}{r}\right]-\text{ln}\left[\frac{D_{C{B}^{\prime }}}{D_{B{B}^{\prime }}}\right]\right)+q_C\left(\text{ln}\left[\frac{r}{D}\right]-\text{ln}\left[\frac{D_{C{C}^{\prime }}}{D_{B{C}^{\prime }}}\right]\right)\right]\text{(V)}$ (14.99)

$V_{AC}=\frac{1}{2\pi {\epsilon }_0}\left[q_A\left(\text{ln}\left[\frac{D}{r}\right]-\text{ln}\left[\frac{D_{C{A}^{\prime }}}{D_{A{A}^{\prime }}}\right]\right)-q_B\text{ln}\left[\frac{D_{C{B}^{\prime }}}{D_{A{B}^{\prime }}}\right]+q_C\left(\text{ln}\left[\frac{r}{D}\right]-\text{ln}\left[\frac{D_{C{C}^{\prime }}}{D_{A{C}^{\prime }}}\right]\right)\right]\text{(V)}$ (14.100)

The phase voltage VAN becomes, through algebraic reduction,

$\begin{array}{c}{V}_{AN}=\frac{V_{AB}+V_{AC}}{3}\\ =\frac{1}{2\pi e_0}{q}_A\left(\text{ln}\left[\frac{D}{r}\right]-\text{ln}\left[\frac{\sqrt[3]{D_{A{B}^{\prime }}{D}_{B{C}^{\prime }}{D}_{C{A}^{\prime }}}}{\sqrt[3]{D_{A{A}^{\prime }}{D}_{B{B}^{\prime }}{D}_{C{C}^{\prime }}}}\right]\right)(\text{V})\end{array}$ (14.101)

Therefore, the phase capacitance CAN, per-unit length, is

$C_{AN}=\frac{q_A}{V_{AN}}=\frac{2\pi {\epsilon }_0}{\text{ln}\left[\frac{D}{r}\right]-\text{ln}\left[\frac{\sqrt[3]{D_{A{B}^{\prime }}{D}_{B{C}^{\prime }}{D}_{C{A}^{\prime }}}}{\sqrt[3]{D_{A{A}^{\prime }}{D}_{B{B}^{\prime }}{D}_{C{C}^{\prime }}}}\right]}(\text{F/m})$ (14.102)

Equations 14.84 and 14.102 have similar expressions, except for the term ln[(DAB′, DBC′, DCA′)1/3/(DAA′, DBB′, DCC′)1/3] included in Equation 14.102. That term represents the effect of the earth on phase capacitance, increasing its total value. However, the capacitance increment is really small, and is usually neglected, because distances from overhead conductors to ground are always greater than distances among conductors.

14.1.4 Equivalent Circuit of Three-Phase Transmission Lines

Once evaluated, the line parameters are used to model the transmission line and to perform design calculations. The arrangement of the parameters (equivalent circuit model) representing the line depends upon the length of the line [4–9].

A transmission line is defined as a short-length line if its length is less than 80 km (50 miles). In this case, the shut capacitance effect is negligible and only the series resistance and series inductive reactance are considered. Assuming balanced conditions, the line can be represented by the equivalent circuit of a single phase with resistance R, and inductive reactance XL in series, as shown in Figure 14.16.

Sending voltage and current VS and IS can be expressed in terms of receiving voltage and current VR and IR at the load side as

$V_S=V_R+Z{I}_R$ (14.103)

$I_S=I_R$ (14.104)

where Z = R + jXL is the total series impedance of the transmission line (Ω).

If the transmission line has a length between 80 km (50 miles) and 240 km (150 miles), the line is considered a medium-length line and its single-phase equivalent circuit can be represented by a nominal π circuit configuration. The shunt capacitance of the line is divided into two equal parts, each placed at the sending and receiving ends of the line. Figure 14.17 shows the equivalent circuit for a medium-length line.

Sending voltage and current VS and IS can be expressed in terms of receiving voltage and current VR and IR at the load side as

$V_S=V_R+V_Z=V_R+Z[I_R+V_R(Y_R)]=V_R\left[1+Z\left(\frac{Y}{2}\right)\right]+I_{R}Z$ (14.105)

$I_S=V_{Ys}+I_Z=V_S(Y_S)+[I_R+V_R(Y_R)]=V_R\left[Y\left(1+Z\left(\frac{Y}{4}\right)\right)\right]+I_R\left[1+Z\left(\frac{Y}{2}\right)\right]$ (14.106)

In matrix form, we have

$\begin{array}{c}\left[\begin{array}{c}{V}_S\\ I_S\end{array}\right]=\left[\begin{array}{cc}1+Z(Y\text{/}2)& Z\\ Y[1+Z(Y\text{/}4)]& 1+Z(Y\text{/}2)\end{array}\right]\left[\begin{array}{c}{V}_R\\ I_R\end{array}\right]\\ =\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{c}{V}_R\\ I_R\end{array}\right]\end{array}$ (14.107)

where

A = 1 + Z (Y/2) pu

B = Z Ω

C = Y [1 + Z(Y/4)] S

D = 1 + Z(Y/2) pu

A, B, C, and D are defined as approximated lumped parameters models, and are used in both short-length and medium-length transmission lines. However, if the line is larger than 240 km, the model must consider parameters uniformly distributed along the line. The appropriate series impedance and shunt capacitance are found by solving the corresponding differential equations, where voltages and currents are described as a function of distance and time. Figure 14.18 shows the distributed parameters along the line, neglecting the shunt conductance.

In Figure 14.18, the total length l of the transmission line is taken from the receiving end (x = 0) to the sending end (x = l). It is shown a small length Δx in which there are a series impedance zΔx and a shunt admittance yΔx, where z = R + jωL in Ω/m and y = jωC in S/m. The expression of V(x + Δx) is obtained applying voltage Kirchhoff’s law:

$V(x+\Delta x)=V(x)+(z\Delta x)I(x)$ (14.108)

Rearranging terms,

$\frac{V(x+\Delta x)-V(x)}{\Delta x}=zI(x)\text{or}\frac{dV(x)}{dx}=zI(x)\text{if}\text{lim}\Delta x\to 0$ (14.109)

Similarly, the expression for current I(x + Δx) is obtained applying current Kirchhoff’s law:

$I(x+\Delta x)=I(x)+(y\Delta x)V(x+\Delta x)$ (14.110)

Rearranging terms,

$\frac{I(x+\Delta x)-I(x)}{\Delta x}=yV(x+\Delta x)\text{or}\frac{dI(x)}{dx}=yV(x)\text{if}\text{lim}\Delta x\to 0$ (14.111)

Taking the derivative with respect to x in Equation 14.109 and substituting Equation 14.111 gives Equation 14.112. Similarly, taking the derivative with respect to x in Equation 14.110 and substituting Equation 14.109 gives Equation 14.113:

$\frac{d^{2}V(x)}{d{x}^2}=zyV(x)={\gamma }^{2}V(x)\text{or}\frac{d^{2}V(x)}{d{x}^2}-{\gamma }^{2}V(x)=0$ (14.112)

$\frac{d^{2}I(x)}{d{x}^2}=zyI(x)={\gamma }^{2}I(x)\text{or}\frac{d^{2}I(x)}{d{x}^2}-{\gamma }^{2}I(x)=0$ (14.113)

where $\gamma =\sqrt{zy}$ is defined as the propagation constant.

Equations 14.112 and 14.113 are second-order linear homogenous differential equations. Solving Equation 14.112 for V(x) we have

$V(x)=k_1{e}^{\gamma x}+k_2{e}^{-\gamma x}$ (14.114)

The solution for I(x) is obtained by taking the derivative of Equation 14.114 with respect to x and substituting the value in Equation 14.109:

$I(x)=\frac{k_1{e}^{\gamma x}-k_2{e}^{-\gamma x}}{(z\text{/}y)}=\frac{k_1{e}^{\gamma x}-k_2{e}^{-\gamma x}}{\sqrt{z\text{/}y}}=\frac{k_1{e}^{\gamma x}-k_2{e}^{-\gamma x}}{Z_c}$ (14.115)

where $Z_c=\sqrt{z\text{/}y}$ is defined as the characteristic impedance of the transmission line.

The values of constants k1 and k2 are defined by initial conditions. At the receiving end (x = 0), V(0) = VR and I(0) = IR. Therefore, substituting these values in Equations 14.114 and 14.115 we have

$k_1=\frac{V_R+Z_c{I}_R}{2}\text{,}{k}_2=\frac{V_R-Z_c{I}_R}{2}$ (14.116)

The reduced expressions of V(x) and I(x) along any point of the transmission lines become

$\begin{array}{c}V(x)=\left(\frac{V_R+Z_c{I}_R}{2}\right)e^{\gamma x}+\left(\frac{V_R-Z_c{I}_R}{2}\right)e^{-\gamma x}\\ =\left(\frac{e^{\gamma x}+e^{-\gamma x}}{2}\right)V_R+Z_c\left(\frac{e^{\gamma x}-e^{-\gamma x}}{2}\right)I_R\\ =\cosh (\gamma x)V_R+Z_c\sinh (\gamma x)I_R\end{array}$ (14.117)

$\begin{array}{c}I(x)=\left(\frac{V_R+Z_c{I}_R}{2Z_c}\right)e^{\gamma x}+\left(\frac{V_R-Z_c{I}_R}{2Z_c}\right)e^{-\gamma x}\\ =\frac{1}{Z_c}\left(\frac{e^{\gamma x}-e^{-\gamma x}}{2}\right)V_R+\left(\frac{e^{\gamma x}+e^{-\gamma x}}{2}\right)I_R\\ =\frac{1}{Z_c}\sinh (\gamma x)V_R+\cosh (\gamma x)I_R\end{array}$ (14.118)

Describing Equations 14.117 and 14.118 in matrix form,

$\left[\begin{array}{c}V(x)\\ I(x)\end{array}\right]=\left[\begin{array}{cc}A(x)& B(x)\\ C(x)& D(x)\end{array}\right]\left[\begin{array}{c}{V}_R\\ I_R\end{array}\right]$ (14.119)

where

A(x) = cosh(γx) pu

B(x) = Zc sinh(γx) Ω

C(x) = (1/Zc) sinh(γx) S

D(x) = cosh(γx) pu

A(x), B(x), C(x), and D(x) are known as exact ABCD parameters for a transmission line in steady state.

V(x) and I(x) can give the exact value of voltage and current at any distance x from the receiving end, no matter what is the total length of the transmission line. However, in most of the cases, the only points of interest are at the receiving and sending ends. If VR and IR are known, then the matrix form of equations for VS and IS, where x = l, are

$\left[\begin{array}{c}{V}_S\\ I_S\end{array}\right]=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{c}{V}_R\\ I_R\end{array}\right]$ (14.120)

where

A = cosh(γl) pu

B = Zc sinh(γl) Ω

C = (1/Zc) sinh(γl) S

D = cosh(γl) pu

In this case, a long-length transmission line can be represented with an equivalent π circuit model, similar to the nominal π circuit representing the medium-length transmission line, as shown in Figure 14.19, where Z′ is the modified value of total series impedance (Ω), and Y′ is the modified value of total shunt admittance (S).

From the equivalent π circuit model, the corresponding matrix array for VS and IS, following the same procedure as in the medium-length transmission line, are

$\begin{array}{c}\left[\begin{array}{c}\begin{array}{l}{V}_S\\ \end{array}\\ I_S\end{array}\right]=\left[\begin{array}{cc}1+Z^{\prime }\left(\frac{Y^{\prime }}{2}\right)& Z^{\prime }\\ Y^{\prime }\left[1+Z^{\prime }\left(\frac{Y^{\prime }}{4}\right)\right]& 1+Z^{\prime }\left(\frac{Y^{\prime }}{2}\right)\end{array}\right]\left[\begin{array}{c}\begin{array}{l}{V}_R\\ \end{array}\\ I_R\end{array}\right]\\ =\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{c}{V}_R\\ I_R\end{array}\right]\end{array}$ (14.121)

As parameters A, B, C, and D from Equations 14.120 and 14.121 are the same, we have

$A=\cosh (\gamma \text{}l)=1+Z^{\prime }\left(\frac{Y^{\prime }}{2}\right)(\text{pu)}$ (14.122)

$B=Z_c\sinh (\gamma \text{}l)=Z^{\prime }(\Omega )$ (14.123)

$C=\frac{1}{Z_c}\sinh (\gamma \text{}l)=Y^{\prime }\left[1+Z^{\prime }\left(\frac{Y^{\prime }}{4}\right)\right](\text{S)}$ (14.124)

$D=\cosh (\gamma \text{}l)=1+Z^{\prime }\left(\frac{Y^{\prime }}{2}\right)(\text{pu)}$ (14.125)

The equivalent expression of Y′/2 is obtained from Equations 14.122 through 14.124 as

$\frac{Y^{\prime }}{2}=\frac{\tanh (\gamma \text{}l)}{Z_c}=\frac{Y}{2}\frac{\tanh (\gamma \text{}l)}{\gamma \text{}l\text{/}2}$ (14.126)

Example 14.1

A 60 Hz, 350 km, three-phase, transposed transmission line, is designed with three bobolink conductors per phase, in horizontal arrangement. Distance between consecutive phases is D = 9.5 m and distance among bundle conductors is d = 0.45 m. The transmission line is connected to a 250 MVA load at 0.8 lagging power factor and at a line-to-line voltage of 400 kV. Determine (a) the series impedance and shunt admittance, (b) the ABCD parameters of the π equivalent circuit, and (c) the voltage and current at the sending end.

Data

Conductor

Bobolink (Table 14.2): Diam = 36.25 mm, RAC,75°C = 0.0503 Ω/km, GMRc = 14.39 mm, l = 350 km

Arrangement

Horizontal, Cond. per phase = 3, D = 9.5 m, d = 0.45 m

Load

S3-phase = 250 MVA, VLL = 400 kV, PF = 0.8 lagging, f = 60 Hz

Solution

a. Series impedance and shunt admittance

$GMD=\sqrt[\text{3}]{D\times D\times 2D}=\sqrt[\text{3}]{9.5\times 9.5\times 9.5}=11.969\text{m}$

$GM{R}_{bundle}=\sqrt[\text{3}]{d^{2}GM{R}_c}=\sqrt[\text{3}]{{0.45}^2\times 0.01439}=0.143\text{m}$

$r_{ebundle}=\sqrt[\text{3}]{d^2{r}_c}=\sqrt[\text{3}]{{0.45}^2\times 0.018}=0.154\text{m}$

$R_{phase}=\frac{R_c}{\text{Cond}\text{per}\text{phase}}=\frac{0.0503\Omega \text{/km}}{\text{3}}=1.677\times {10}^{-5}\Omega \text{/m}$

$X_L{}_{phase}=j\omega L_{phase}=j2\pi f\left(\frac{{\mu }_{\text{0}}}{\text{2}\pi }\ln \left(\frac{GMD}{GM{R}_{bundle}}\right)\right)=j377\times \frac{4\pi \times {10}^{-7}}{2\pi }\times \ln \left(\frac{11.969}{0.143}\right)=j3.33\times {10}^{-4}\Omega \text{/m}$

$\begin{array}{l}{Y}_C=j\omega C_{phase}=j2\pi f\left(\frac{\text{2}\pi {\epsilon }_0}{\ln \left(\frac{GMD}{r_{ebundle}}\right)}\right)=j377\times \frac{8.85\times {10}^{-7}}{\ln (11.69\text{/}0.154)}=j4.817\times {10}^{-9}\text{S/m}\\ z_{phase}=R_{}+j{X}_{Lphase}=1.677\times {10}^{-5}+j3.33\times {10}^{-4}\Omega \text{/m}\end{array}$

$y_{phase}=G_{phase}+j{Y}_{Cphase}=0+j4.817\times {10}^{-9}\text{S/m}$

b. ABCD parameters of the π equivalent circuit

$Z_C=\sqrt{\frac{z}{y}}=\sqrt{\frac{1.677\times {10}^{-5}+j3.33\times {10}^{-4}}{0+j4.817\times {10}^{-9}}}=263.364+j6.61\Omega$

$\gamma =\sqrt{z\text{y}}=\sqrt{(1.677\times {10}^{-5}+j3.33\times {10}^{-4})(0+j4.817\times {10}^{-9})}=3.184\times {10}^{-8}+j1.269\times {10}^{-6}1\text{/m}$

$A=\cosh (\gamma \text{}l)=\cosh ((3.184\times {10}^{-8}+j1.269\times {10}^{-6})(3,50,000))=0.903+j4.787\times {10}^{-3}\text{pu}$

$\begin{array}{c}B=Z_C\sinh (\gamma \text{}l)=(263.364+j6.61)\times \sinh ((3.184\times {10}^{-8}+j1.269\times {10}^{-6})(350,000))\\ =5.49+j113.074\Omega \end{array}$

$\begin{array}{c}C=\frac{1}{Z_C}\sinh (\gamma \text{}l)=\frac{1}{263.364+j6.61}\sinh ((3.184\times {10}^{-8}+j1.269\times {10}^{-6})(350,000))\\ =-2.726\times {10}^{-6}+j1.631\times {10}^{-3}\text{S}\end{array}$

$D=\cosh (\gamma \text{}l)=\cosh ((3.184\times {10}^{-8}+j1.269\times {10}^{-6})(3,50,000))=0.903+j4.787\times {10}^{-3}\text{pu}$

c. Voltage and current at the sending end

$\begin{array}{c}{V}_{RLL}=400\angle {30}^{\circ }\text{kV}\\ V_{Rphase}=\frac{V_{R\text{LL}}}{\sqrt{3}}\angle -{30}^{\circ }=\text{230}\text{.94}\angle {\text{0}}^{\circ }\text{kV}\end{array}$

$\begin{array}{c}{S}_{Load3-phase}=250\angle -\text{acos(PF)}=250\angle \text{acos(0}\text{.8)}=\text{250}\angle {36.87}^{\circ }\text{MVA}\\ S_{Loadphase}=\frac{S_{Load3-phase}}{3}=\frac{\text{250}\angle {25.842}^{\circ }}{3}=\text{83}\text{.333}\angle {36.87}^{\circ }\text{MVA}\end{array}$

$I_{Rphase}=I_{Load}=\frac{\overline{S_{Loadphase}}}{V_{Rphase}}=\frac{\overline{\text{83}\text{.333}\angle {36.87}^{\circ }}}{\text{230}\text{.94}\angle 0^{\circ }}=\frac{\text{83}\text{.333}\angle -{36.87}^{\circ }}{\text{230}\text{.94}\angle 0^{\circ }}=\text{360}\text{.844}\angle -{36.87}^{\circ }\text{A}$

$\begin{array}{c}\left[\begin{array}{l}{V}_S\\ I_S\end{array}\right]=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{l}{V}_R\\ I_R\end{array}\right]=\left[\begin{array}{cc}0.903+j4.787\times {10}^{-3}& 5.49+j113.074\\ -2.726\times {10}^{-6}+j1.631\times {10}^{-3}& 0.903+j4.787\times {10}^{-3}\end{array}\right]\left[\begin{array}{l}\text{230}\text{.94}\angle 0^{\circ }\\ 360.844\angle -{36.87}^{\circ }\end{array}\right]\\ =\left[\begin{array}{l}236.874\angle {7.9}^{\circ }\\ 318.592\angle {34.959}^{\circ }\end{array}\right]\end{array}$

$\begin{array}{c}{V}_{Sphase}=236.874\angle {7.9}^{\circ }\text{kV}\\ I_{Sphase}=318.592\angle {34.959}^{\circ }\text{A}\end{array}$

$V_{SLL}=\sqrt{3}{V}_{Sphase}\angle {30}^{\circ }=\sqrt{3}\times 236.874\angle {7.9}^{\circ }\text{= 410}\text{.277}\angle {\text{30}}^{\circ }\text{kV}$

14.1.5 Characteristics of Overhead Conductors

Table 14.2 presents typical values of resistance, inductive reactance, and capacitance reactance, per-unit length, of ACSR conductors. The size of the conductors (cross-section area) is specified in square millimeters and kcmil, where a cmil is the cross-section area of a circular conductor with a diameter of 1/1000 in. The tables include also the approximate current-carrying capacity of the conductors assuming 60 Hz, wind speed of 1.4 miles/h, and conductor and air temperatures of 75°C and 25°C, respectively. Table 14.3 presents the corresponding characteristics of AAC conductors.

References

1. Electric Power Research Institute, Transmission Line Reference Book 345 kV and Above, 2nd edn., EPRI, Palo Alto, CA, 1987.

2. Barnes, C. C., Power Cables. Their Design and Installation, 2nd edn., Chapman and Hall Ltd., London, U.K., 1966.

3. Glover, J. D. and Sarma, M. S., Power System Analysis and Design, 3rd edn., Brooks/Cole, Pacific Grove, CA, 2002.

4. Stevenson W. D. Jr., Elements of Power System Analysis, 4th. edn., McGraw-Hill, New York, 1982.

5. Saadat, H., Power System Analysis, McGrawHill, New York, 1999.

6. Gross, C. A., Power System Analysis, John Wiley and Sons, Inc., New York, 1979.

7. Yamayee, Z. A. and Bala J. L. Jr., Electromechanical Energy Devices and Power Systems, John Wiley and Sons, Inc., New York, 1994.

8. Gungor, B. R., Power Systems, Harcourt Brace Jovanovich, Orlando, FL, 1988.

9. Zaborszky, J. and Rittenhouse, J. W., Electric Power Transmission. The Power System in the Steady State, The Ronald Press Company, New York, 1954.