Charging current of 60 Hz horizontal 132 kV line

For a transposed horizontal line, $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$ and $C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r)$. Since no length is given, the results are per km.

   a o-----5 m-----o b-----5 m-----o c
     |<----------- 10 m ---------->|

Data: $r = 0.012$ m, $D = 5, 5, 10$ m, 132 kV, $f = 60$ Hz.

Capacitance

$$\begin{aligned} D_{eq} &= \sqrt[3]{5 \times 5 \times 10} = 6.2996\ \text{m} \\ C_n &= \frac{55.63\times10^{-12}}{\ln(6.2996/0.012)} = \frac{55.63\times10^{-12}}{6.2633} \\ &= 8.882\times10^{-12}\ \text{F/m} = 8.882\ \text{nF/km} \end{aligned}$$

Capacitive reactance at 60 Hz

$$X_C = \frac{1}{2\pi \times 60 \times 8.882\times10^{-9}} = 2.986\times10^5\ \Omega\text{-km}$$

Charging current per phase

$$I_c = \frac{V_p}{X_C} = \frac{132\,000/\sqrt3}{298\,645} = \frac{76\,210}{298\,645} = 0.2552\ \text{A/km}$$

Charging VAR

$$\begin{aligned} Q_{3\phi} &= \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{298\,645} = 58.34\ \text{kVAR/km} \\ Q_{ph} &= 58.34/3 = 19.45\ \text{kVAR/km} \end{aligned}$$

Answer: $I_c = 0.255$ A/km per phase; charging $= 58.3$ kVAR/km for the three-phase line (19.45 kVAR/km per phase).