Inductance of 3-conductor bundled line, 6.1 m spacing

For a bundled, transposed line the inductance per phase is

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

Here $D_s^b$ is the self-GMD (GMR) of the bundle and $D_m$ is the mutual GMD between phases. The bundle spacing (0.305 m) is much smaller than the phase spacing, so distances between phases are taken centre to centre. The line is assumed to be fully transposed.

Data

• Sub-conductor radius: $r = 2.54/2 = 1.27$ cm $= 0.0127$ m
• GMR of a solid sub-conductor: $r' = 0.7788r = 0.7788\times0.0127 = 0.009891$ m
• Sub-conductor spacing: $d = 0.305$ m (equilateral triangle)

Self-GMD of the three-conductor bundle

$$\begin{aligned} D_s^b &= \sqrt[9]{(r'\,d\,d)^3} = \sqrt[3]{r'd^2}\\ &= \sqrt[3]{0.009891\times0.305^2}\\ &= 0.09726\ \text{m} \end{aligned}$$

Mutual GMD

Horizontal arrangement: $D_{ab} = D_{bc} = 6.1$ m and $D_{ca} = 12.2$ m.

$$D_m = \sqrt[3]{6.1\times6.1\times12.2} = 6.1\times\sqrt[3]{2} = 7.686\ \text{m}$$

Inductance

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{7.686}{0.09726}\ \text{H/m}\\ &= 2\times10^{-7}\times\ln(79.02) = 2\times10^{-7}\times4.3697\\ &= 0.8739\times10^{-6}\ \text{H/m} = 0.8739\ \text{mH/km} \end{aligned}$$

If the frequency is 50 Hz, the reactance is $X_L = 2\pi\times50\times0.8739\times10^{-3} = 0.2746\ \Omega$/km per phase.

Answer: $L \approx 0.874$ mH per phase per km ($X_L \approx 0.275\ \Omega$/km at 50 Hz).