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Inductance of 4-conductor bundle line, 6 m spacing

Short answer

Using Deq = 7.56 m and square bundle self-GMD = 0.1863 m, inductance = 0.741 mH/km, yielding 148 mH per phase for 200 km.

Compute the inductance per phase for a 200 km line. Each sub-conductor is of 12 mm radius and configured as given below. [Figure: horizontal single-circuit line; each phase is a square bundle of four sub-conductors with 45 cm sides; centres of adjacent phases 6 m apart.]

Each phase is a square 4-conductor bundle. For inductance, its self-GMD is Ds=(r′⋅d⋅d⋅2d)1/4=1.09r′d34D_s = (r' \cdot d \cdot d \cdot \sqrt2 d)^{1/4} = 1.09\sqrt[4]{r'd^3}. The line is taken as transposed.

   o o        o o        o o
   o o        o o        o o
    A    6 m   B    6 m   C
   d = 45 cm, r = 12 mm

Data: r=0.012r = 0.012 m, r′=0.7788×0.012=0.009346r' = 0.7788 \times 0.012 = 0.009346 m, d=0.45d = 0.45 m, D=6,6,12D = 6, 6, 12 m, 200 km.

Mutual GMD

Dm=6×6×123=7.5595 mD_m = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}

Self GMD of bundle

Ds=1.090.009346×0.4534=0.18629 mD_s = 1.09\sqrt[4]{0.009346 \times 0.45^3} = 0.18629\ \text{m}

Inductance

L=2×10−7ln⁡7.55950.18629=2×10−7×3.7033=7.4065×10−7 H/m=0.7407 mH/kmL200=0.7407×200=148.13 mH\begin{aligned} L &= 2\times10^{-7}\ln\frac{7.5595}{0.18629} = 2\times10^{-7}\times3.7033 \\ &= 7.4065\times10^{-7}\ \text{H/m} = 0.7407\ \text{mH/km} \\ L_{200} &= 0.7407 \times 200 = 148.13\ \text{mH} \end{aligned}

Answer: L=148.1L = 148.1 mH per phase for 200 km (0.7407 mH/km).

✓ Answer checked by Claude Opus 5.5 (high effort)

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