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Inductance of twin-bundle line, 16 mm radius

Short answer

Inductance per phase is 0.968 mH/km, giving 145.1 mH for 150 km.

Compute the inductance per phase for a 150 km line using phase conductors of 16 mm radius and configured as given below. Assume complete transposition of the phases. [Figure: horizontal single-circuit line; each phase a, b, c is a two-conductor bundle with sub-conductor spacing d = 45 cm; centres of adjacent phases D = 7.5 m apart.]

Each phase is a two-conductor bundle, so its self-GMD for inductance is Ds=r′dD_s = \sqrt{r'd}. The line is transposed, so Dm=DabDbcDca3D_m = \sqrt[3]{D_{ab}D_{bc}D_{ca}}.

   o--o         o--o         o--o
    a    7.5 m   b    7.5 m   c
   d = 45 cm, r = 16 mm

Data: r=0.016r = 0.016 m, r′=0.7788×0.016=0.012461r' = 0.7788 \times 0.016 = 0.012461 m, d=0.45d = 0.45 m, D=7.5,7.5,15D = 7.5, 7.5, 15 m, 150 km.

Mutual GMD

Dm=7.5×7.5×153=9.4494 mD_m = \sqrt[3]{7.5 \times 7.5 \times 15} = 9.4494\ \text{m}

Self GMD of bundle

Ds=0.012461×0.45=0.074882 mD_s = \sqrt{0.012461 \times 0.45} = 0.074882\ \text{m}

Inductance

L=2×10−7ln⁡9.44940.074882=2×10−7×4.8378=9.676×10−7 H/m=0.9676 mH/kmL150=0.9676×150=145.13 mH\begin{aligned} L &= 2\times10^{-7}\ln\frac{9.4494}{0.074882} = 2\times10^{-7}\times4.8378 \\ &= 9.676\times10^{-7}\ \text{H/m} = 0.9676\ \text{mH/km} \\ L_{150} &= 0.9676 \times 150 = 145.13\ \text{mH} \end{aligned}

Answer: L=145.1L = 145.1 mH per phase for 150 km (0.968 mH/km).

✓ Answer checked by Claude Opus 5.5 (high effort)

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