Capacitance of Three-Phase Transmission Lines

Lecture Video

SECTION 01

Capacitance of a three-phase line

Overview of the topic:

equilateral spacing
unsymmetrical spacing & Transposition
Effect of earth on capacitance

\[\begin{aligned} V_{ab}&=\dfrac{1}{2\pi\varepsilon}\left(q_{a}~ln\dfrac{D}{r}+q_{b}~ln\dfrac{r}{D}+\underset{zero}{\underbrace{q_{c}~ln\dfrac{D}{D}}}\right)~V \\ V_{ac}&=\dfrac{1}{2\pi\varepsilon}\left(q_{a}~ln\dfrac{D}{r}+\underset{zero}{\underbrace{q_{b}~ln\dfrac{D}{D}}}+q_{c}~ln\dfrac{r}{D}\right)~V \end{aligned}\]

\[V_{ab}+V_{ac}=\dfrac{1}{2\pi\varepsilon}\left[2q_{a}~ln\dfrac{D}{r}+\left(q_{b}+q_{c}\right)~ln\dfrac{r}{D}\right]~V\]

\[V_{ab}+V_{ac}=\dfrac{3q_{a}}{2\pi\varepsilon}ln\dfrac{D}{r}~V\]

Since,

Phasor diagram of the balanced voltages of a \(3\phi\) line

Adding, \(V_{ab}+V_{ac}=3V_{an}\).
\[\begin{aligned} V_{an}&=\dfrac{q_a}{2\pi\epsilon}~ln \dfrac{D}{r} \\ C_{n}&=\dfrac{q_a}{V_{an}}=\dfrac{2\pi\epsilon}{ln(D/r)}~F/m \end{aligned}\]
On substitution,
\[I_{chg} = j \omega C_{ab}V_{ab}\]
line, the charging currentFor
\[I_{chg} = j \omega C_{n}V_{an}\]
line, the charging current per phaseFor

SECTION 02

Capacitance of a three-phase line with unsymmetrical spacing

\[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{12}}{r}+q_{b}ln\dfrac{r}{D_{12}}+q_{c}ln\dfrac{D_{23}}{D_{31}}\right)~V\]
in position 3, in position 2, and in position 1, With phase
\[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{23}}{r}+q_{b}ln\dfrac{r}{D_{23}}+q_{c}ln\dfrac{D_{31}}{D_{12}}\right)~V\]
in position 1, in position 3, and in position 2, With phase

With phase \(a\) in position 3, \(b\) in position 1, and \(c\) in position 2,

\[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{31}}{r}+q_{b}ln\dfrac{r}{D_{31}}+q_{c}ln\dfrac{D_{12}}{D_{23}}\right)V\]
\[\begin{aligned} V_{ab} & =\dfrac{1}{6\pi\varepsilon}\left(q_{a}ln\dfrac{D_{12}D_{23}D_{31}}{r^{3}}+q_{b}ln\dfrac{r^{3}}{D_{12}D_{23}D_{31}}+q_{c}ln\dfrac{D_{12}D_{23}D_{31}}{D_{12}D_{23}D_{31}}\right)\\ & =\dfrac{1}{2\pi\varepsilon}\left(qln\dfrac{D_{eq}}{r}+qln\dfrac{r}{D_{eq}}\right) \end{aligned}\]
The average voltage drop
where \(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\)

\[\begin{aligned} V_{ac} & =\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{eq}}{r}+q_{c}ln\dfrac{r}{D_{eq}}\right)V \end{aligned}\]
The average voltage drop
\[3V_{an}=V_{ab}+V_{ac}=\dfrac{1}{2\pi\varepsilon}\left(2q_{a}ln\dfrac{D_{eq}}{r}+q_{b}ln\dfrac{r}{D_{eq}}+q_{c}ln\dfrac{r}{D_{eq}}\right)V\]
The voltage to neutral
\[V_{an}=\dfrac{1}{2\pi\varepsilon}q_{a}ln\dfrac{D_{eq}}{r}V\]
Since,

equation* C_n==F/m