Capacitance with and without earth effect

For a transposed line, the capacitance to neutral is $C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r)$ when ground is neglected. The method of images adds a correction term when ground is included.

   R o----8 m----o Y----8 m----o B
     |13 m       |13 m         |13 m
 ======================================= ground
     |13 m       |13 m         |13 m
   R'o           o Y'          o B'   (images)

Data: diameter 2.6 cm, $r = 0.013$ m; $D_{RY} = D_{YB} = 8$ m, $D_{BR} = 16$ m; $h = 13$ m.

$$D_{eq} = \sqrt[3]{8 \times 8 \times 16} = 10.0794\ \text{m}, \qquad \ln\frac{D_{eq}}{r} = \ln\frac{10.0794}{0.013} = 6.6533$$

(i) Neglecting the effect of ground

$$C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{6.6533} = 8.361\times10^{-12}\ \text{F/m} = 8.361\ \text{nF/km}$$

(ii) Considering the effect of earth

Distances to images:

$$\begin{aligned} \sqrt[3]{H_{12}H_{23}H_{31}} &= \sqrt[3]{27.203^2 \times 30.529} = 28.269\ \text{m} \\ \sqrt[3]{H_1H_2H_3} &= 26\ \text{m} \\ \text{Correction} &= \ln\frac{28.269}{26} = 0.08368 \end{aligned}$$ $$\begin{aligned} C_n &= \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}} = \frac{55.63\times10^{-12}}{6.6533 - 0.0837} \\ &= \frac{55.63\times10^{-12}}{6.5696} = 8.468\times10^{-12}\ \text{F/m} = 8.468\ \text{nF/km} \end{aligned}$$

Answer: (i) $C_n = 8.361$ nF/km; (ii) $C_n = 8.468$ nF/km. The earth raises the capacitance by about 1.3%.