L and C of triple-bundle line, D=8 m

Each phase is a 3-conductor bundle (equilateral triangle, side $d$). The bundle self-GMD is $\sqrt[3]{r_x d^2}$, with $r_x = r'$ for inductance and $r$ for capacitance.

     o          o          o
    o o        o o        o o
     a   8 m    b   8 m    c
   d = 45 cm, r = 18 mm

Data: $r = 0.018$ m, $r' = 0.7788 \times 0.018 = 0.014018$ m, $d = 0.45$ m, $D = 8, 8, 16$ m, 200 km.

GMDs

$$\begin{aligned} D_m &= \sqrt[3]{8 \times 8 \times 16} = 10.0794\ \text{m} \\ D_{sL} &= \sqrt[3]{0.014018 \times 0.45^2} = 0.14159\ \text{m} \\ D_{sC} &= \sqrt[3]{0.018 \times 0.45^2} = 0.15390\ \text{m} \end{aligned}$$

Inductance per phase

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{10.0794}{0.14159} = 2\times10^{-7}\times4.2653 = 0.8531\ \text{mH/km} \\ L_{200} &= 0.8531 \times 200 = 170.61\ \text{mH} \end{aligned}$$

Capacitance per phase

$$\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(10.0794/0.15390)} = \frac{55.63\times10^{-12}}{4.1820} = 13.30\ \text{nF/km} \\ C_{200} &= 13.30 \times 200 = 2660.5\ \text{nF} = 2.661\ \mu\text{F} \end{aligned}$$

Answer: $L = 170.6$ mH per phase (0.853 mH/km); $C = 2.661\ \mu$F per phase (13.30 nF/km).