Capacitance of Overhead Transmission Lines

Demonstrative Video

Overview

Lecture-12: Capacitance of Single-Phase TL
Lecture-13: Capacitance of a Three-Phase TL
Lecture-14: Effect of Earth on Capacitance of the TL
Lecture-15: Capacitance Calculations for Bundle Conductors

Introduction to Capacitance of a TL

Due to potential difference between the conductor

$C = \dfrac{q}{V}$
Gauss’s law for electric field:
- total $q$ within a closed surface equal total electric flux emerging from the surface
Between parallel conductors
- $C$ is a constant
- depends on the size and spacing of the conductors

Power lines less than 80 Km $\Longrightarrow$ $C$ is slight and neglected
For longer HV lines $C$ becomes important
AC voltage $\Longrightarrow$ charge $\Longrightarrow$ Charging current $(I_c)$
$I_c$ flows even when the TL is open-circuited
1. affects the voltage drop along the lines
2. Efficiency and power factor
3. Stability of the system

Electric field of a long straight conductor

If the conductor lies in uniform medium such as air and isolated from other charges then $q$ is uniformly distributed around its periphery and the flux is radial

Lines of electric flux originating from positive charge

The electric flux density at $x$ m from the conductor: flux leaving the conductor per m of length divided by the area of the surface in an axial length of 1m $D_f = \dfrac{q}{2 \pi x}~C/m^2$

The electric field intensity $E = \dfrac{q}{2 \pi x \epsilon}~V/m$ where $q =$ charge per m of length
$\epsilon =$ permittivity of the medium

Potential difference between two points due to a charge

$V = W/q$ and $E = F/q$
$+ve$ charge on the conductor exerts a repelling force on a $+ve$ charge placed in the field

path of integration between two points external to the conductor

$P_1$ is at higher potential than $P_2$
work must be done to move from $P_2$ to $P_1$

Instantaneous voltage drop between $P_1$ and $P_2$ $v_{12}=\intop_{D_{1}}^{D_{2}}E\cdot dx=\intop_{D_{1}}^{D_{2}}\dfrac{q}{2\pi\varepsilon x}\cdot dx=\dfrac{q}{2\pi\varepsilon}ln\dfrac{D_{2}}{D_{1}}~V$
Voltage drop may be +ve or -ve, depends on
- $q$ is +ve or -ve
- voltage drop is computed near to far or vice versa

Capacitance of a two-wire line

cross-section of a parallel wire line

By principle of superposition the voltage drop from $a$ to $b$ is the sum of each charge on the conductor alone $v_{ab}=\begin{array}{c} \underbrace{\dfrac{q_{a}}{2\pi\varepsilon}ln\dfrac{D}{r_{a}}}\\ \mbox{due to }q_{a} \end{array}+\begin{array}{c} \underbrace{\dfrac{q_{b}}{2\pi\varepsilon}ln\dfrac{r_{b}}{D}}\\ \mbox{due to }q_{b} \end{array}~V$

Since, $q_{a} = -q_{b}$ for a two-wire line $v_{ab} =\dfrac{q_{a}}{2\pi\varepsilon}\left(ln\dfrac{D}{r_{a}}-ln\dfrac{r_{b}}{D}\right)=\dfrac{q_{a}}{2\pi\varepsilon}ln\dfrac{D^{2}}{r_{a}r_{b}}~V$
The capacitance between conductors
$C_{ab}=\dfrac{q_{a}}{V_{ab}}=\dfrac{2\pi\varepsilon}{ln\left(D^{2}/r_{a}r_{b}\right)}~F/m$

If $r_{a}=r_{b}=r$

equation* C_ab= F/m

equation* C_n=C_an=C_bn== F/m

NOTE: $r$ used in $C$ calculation is actual radius of the conductor, unlike use of GMR in $L$

The capacitive reactance between one conductor and neutral $X_{c}=\dfrac{1}{2\pi fC_{n}}=\dfrac{2.862}{f}\times10^{9}ln\dfrac{D}{r}~\Omega m$