Capacitance of double circuit line, 9 mm radius

The line is transposed, so each phase pair (a, a'), (b, b'), (c, c') is treated as a two-conductor group. $C_n = 2\pi\varepsilon_0/\ln(D_m/D_s)$ with $D_s$ using the actual radius.

   a o------ 5.25 m ------o a'   --
                                  3.05 m
  b o------- 6.40 m -------o b'  --
                                  3.05 m
   c o------ 5.25 m ------o c'   --

Data: $r = 0.009$ m. Horizontal offset of b from a $= (6.40-5.25)/2 = 0.575$ m.

Distances

Mutual GMD

$$\begin{aligned} D_{AB} &= (3.1037^2 \times 6.5752^2)^{1/4} = 4.5175\ \text{m} \\ D_{BC} &= 4.5175\ \text{m} \\ D_{CA} &= (6.10^2 \times 8.0481^2)^{1/4} = 7.0067\ \text{m} \\ D_m &= \sqrt[3]{4.5175^2 \times 7.0067} = 5.2292\ \text{m} \end{aligned}$$

Self GMD (capacitance)

$$\begin{aligned} D_{sA} &= \sqrt{0.009 \times 5.25} = 0.21737\ \text{m} \\ D_{sB} &= \sqrt{0.009 \times 6.40} = 0.24000\ \text{m} \\ D_{sC} &= 0.21737\ \text{m} \\ D_s &= \sqrt[3]{0.21737^2 \times 0.24} = 0.22467\ \text{m} \end{aligned}$$

Capacitance per phase

$$\begin{aligned} C_n &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_s)} = \frac{55.63\times10^{-12}}{\ln(5.2292/0.22467)} = \frac{55.63\times10^{-12}}{3.1474} \\ &= 17.675\times10^{-12}\ \text{F/m} \end{aligned}$$

Answer: $C_n = 17.68$ pF/m $= 17.68$ nF/km per phase to neutral.