Capacitance of line on 5-6-7 m triangle

For a fully transposed three-phase line with unsymmetrical spacing, the capacitance per phase to neutral is

$$C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)}\ \text{F/m}, \qquad D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}$$

Data: diameter 2 cm, so $r = 1$ cm $= 0.01$ m; sides 5 m, 6 m, 7 m; $\varepsilon_0 = 8.854\times10^{-12}$ F/m.

Equivalent spacing

$$D_{eq} = \sqrt[3]{5 \times 6 \times 7} = \sqrt[3]{210} = 5.9439\ \text{m}$$

Capacitance

$$\begin{aligned} C_n &= \frac{2\pi \times 8.854\times10^{-12}}{\ln(5.9439/0.01)} = \frac{55.63\times10^{-12}}{\ln 594.39} \\ &= \frac{55.63\times10^{-12}}{6.3875} = 8.709\times10^{-12}\ \text{F/m} \end{aligned}$$

Per km: $8.709\times10^{-12} \times 1000 = 8.709\times10^{-9}$ F/km.

Answer: $C_n = 8.709$ nF/km $= 0.00871\ \mu$F/km (phase to neutral).