Capacitance of Three-Phase Transmission Lines
Demonstrative Video
Capacitance of a three-phase line
Overview of the topic:
equilateral spacing
unsymmetrical spacing & Transposition
Effect of earth on capacitance
\[\begin{aligned}
V_{ab}&=\dfrac{1}{2\pi\varepsilon}\left(q_{a}~ln\dfrac{D}{r}+q_{b}~ln\dfrac{r}{D}+\underset{zero}{\underbrace{q_{c}~ln\dfrac{D}{D}}}\right)~V
\\
V_{ac}&=\dfrac{1}{2\pi\varepsilon}\left(q_{a}~ln\dfrac{D}{r}+\underset{zero}{\underbrace{q_{b}~ln\dfrac{D}{D}}}+q_{c}~ln\dfrac{r}{D}\right)~V
\end{aligned}\]
\[V_{ab}+V_{ac}=\dfrac{1}{2\pi\varepsilon}\left[2q_{a}~ln\dfrac{D}{r}+\left(q_{b}+q_{c}\right)~ln\dfrac{r}{D}\right]~V\]
Since, \((q_{b}+q_{c})=-q_{a}\) \[V_{ab}+V_{ac}=\dfrac{3q_{a}}{2\pi\varepsilon}ln\dfrac{D}{r}~V\]
\[\begin{aligned} V_{ab} & =\sqrt{3}V_{an}\angle30^{0}\\ &=\sqrt{3}V_{an}\left(0.866+j0.5\right)\\ V_{ac}&=-V_{ca} =\sqrt{3}V_{an}\angle-30^{0}\\ &=\sqrt{3}V_{an}\left(0.866-j0.5\right) \end{aligned}\]
Adding, \(V_{ab}+V_{ac}=3V_{an}\).
On substitution, \[\begin{aligned} V_{an}&=\dfrac{q_a}{2\pi\epsilon}~ln \dfrac{D}{r} \\ C_{n}&=\dfrac{q_a}{V_{an}}=\dfrac{2\pi\epsilon}{ln(D/r)}~F/m \end{aligned}\]
For \(1-\phi\) line, the charging current
\[I_{chg} = j \omega C_{ab}V_{ab}\]For \(3-\phi\) line, the charging current per phase
\[I_{chg} = j \omega C_{n}V_{an}\]
Capacitance of a three-phase line with unsymmetrical spacing
With phase \(a\) in position 1, \(b\) in position 2, and \(c\) in position 3, \[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{12}}{r}+q_{b}ln\dfrac{r}{D_{12}}+q_{c}ln\dfrac{D_{23}}{D_{31}}\right)~V\]
With phase \(a\) in position 2, \(b\) in position 3, and \(c\) in position 1, \[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{23}}{r}+q_{b}ln\dfrac{r}{D_{23}}+q_{c}ln\dfrac{D_{31}}{D_{12}}\right)~V\]
With phase \(a\) in position 3, \(b\) in position 1, and \(c\) in position 2,
\[V_{ab}=\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{31}}{r}+q_{b}ln\dfrac{r}{D_{31}}+q_{c}ln\dfrac{D_{12}}{D_{23}}\right)V\]
The average voltage drop \(V_{ab}\) \[\begin{aligned} V_{ab} & =\dfrac{1}{6\pi\varepsilon}\left(q_{a}ln\dfrac{D_{12}D_{23}D_{31}}{r^{3}}+q_{b}ln\dfrac{r^{3}}{D_{12}D_{23}D_{31}}+q_{c}ln\dfrac{D_{12}D_{23}D_{31}}{D_{12}D_{23}D_{31}}\right)\\ & =\dfrac{1}{2\pi\varepsilon}\left(qln\dfrac{D_{eq}}{r}+qln\dfrac{r}{D_{eq}}\right) \end{aligned}\]
where \(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\)
The average voltage drop \(V_{ac}\) \[\begin{aligned} V_{ac} & =\dfrac{1}{2\pi\varepsilon}\left(q_{a}ln\dfrac{D_{eq}}{r}+q_{c}ln\dfrac{r}{D_{eq}}\right)V \end{aligned}\]
The voltage to neutral \[3V_{an}=V_{ab}+V_{ac}=\dfrac{1}{2\pi\varepsilon}\left(2q_{a}ln\dfrac{D_{eq}}{r}+q_{b}ln\dfrac{r}{D_{eq}}+q_{c}ln\dfrac{r}{D_{eq}}\right)V\]
Since, \(q_{a}+q_{b}+q_{c}=0\) \[V_{an}=\dfrac{1}{2\pi\varepsilon}q_{a}ln\dfrac{D_{eq}}{r}V\]
equation* C_n==F/m