Inductance of transposed double circuit, 0.9 cm radius

For a transposed double-circuit line the inductance per phase is

$$L = 2\times10^{-7}\ln\frac{D_m}{D_s}\ \text{H/m}$$

Here $D_m$ is the equivalent mutual GMD between phase groups and $D_s$ is the self-GMD of each phase (its two parallel conductors). Both are worked out for the section shown. By transposition, the other sections give the same result.

Geometry (section shown)

The tower is symmetrical, so the middle conductors b and b' each stand $(7-6)/2 = 0.5$ m further out than the top and bottom conductors. Coordinates in metres, with c at the origin:

   a (0,6) ------- 6 m ------- c' (6,6)
    |3 m
 b (-0.5,3) ------ 7 m ------- b' (6.5,3)
    |3 m
   c (0,0) ------- 6 m ------- a' (6,0)

Self-GMD

$r' = 0.7788\times0.009 = 0.007009$ m (solid copper).

$$\begin{aligned} D_{sa} &= \sqrt{r'\,D_{aa'}} = \sqrt{0.007009\times8.485} = 0.2439\ \text{m}\\ D_{sb} &= \sqrt{r'\,D_{bb'}} = \sqrt{0.007009\times7} = 0.2215\ \text{m}\\ D_{sc} &= \sqrt{r'\,D_{cc'}} = 0.2439\ \text{m}\\ D_s &= \sqrt[3]{0.2439\times0.2215\times0.2439} = 0.2362\ \text{m} \end{aligned}$$

Mutual GMD

$$\begin{aligned} D_{AB} &= \sqrt[4]{D_{ab}D_{ab'}D_{a'b}D_{a'b'}} = \sqrt[4]{3.041\times7.159\times7.159\times3.041} = 4.666\ \text{m}\\ D_{BC} &= \sqrt[4]{D_{bc}D_{bc'}D_{b'c}D_{b'c'}} = \sqrt[4]{3.041\times7.159\times7.159\times3.041} = 4.666\ \text{m}\\ D_{CA} &= \sqrt[4]{6\times6\times6\times6} = 6\ \text{m}\\ D_m &= \sqrt[3]{4.666\times4.666\times6} = 5.074\ \text{m} \end{aligned}$$

Inductance

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{5.074}{0.2362} = 2\times10^{-7}\times3.0673\\ &= 0.6135\times10^{-6}\ \text{H/m} = 0.6135\ \text{mH/km per phase} \end{aligned}$$

At 50 Hz this gives $X_L = 314.16\times0.6135\times10^{-3} = 0.193\ \Omega$/km per phase.

Answer: $D_s = 0.236$ m, $D_m = 5.074$ m, and the inductance is 0.614 mH per phase per km.