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Inductance of unsymmetrical transposed line

Short answer

Deq = 5.7044 m, r' = 0.7788×5 = 3.894 mm. Per-phase inductance L = 2×10⁻⁷ ln(Deq/r') = 1.458 mH/km.

A 3-phase, 50 Hz, overhead line has three phase conductors with the following configuration. Compute the inductance per phase per km if each of the phase conductors is round solid of 10 mm diameter. Assume completely transposed line. [Figure: conductors a, b, c at the corners of a triangle with Dab = 4.5 m, Dbc = 5.5 m, Dca = 7.5 m.]

For a completely transposed line with unsymmetrical spacing, the inductance per phase is L=2×10−7ln⁡(Deq/r′)L = 2\times10^{-7}\ln(D_{eq}/r') H/m, where Deq=DabDbcDca3D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}.

          a
         / \
   4.5 m/   \7.5 m
       /     \
      b-------c
        5.5 m

Data: diameter 10 mm, r=5r = 5 mm =0.005= 0.005 m, r′=0.7788×0.005=0.003894r' = 0.7788 \times 0.005 = 0.003894 m.

Equivalent spacing

Deq=4.5×5.5×7.53=185.6253=5.7044 mD_{eq} = \sqrt[3]{4.5 \times 5.5 \times 7.5} = \sqrt[3]{185.625} = 5.7044\ \text{m}

Inductance per phase

L=2×10−7ln⁡5.70440.003894=2×10−7×7.2896=1.4579×10−6 H/m=1.458 mH/km\begin{aligned} L &= 2\times10^{-7}\ln\frac{5.7044}{0.003894} = 2\times10^{-7}\times7.2896 \\ &= 1.4579\times10^{-6}\ \text{H/m} = 1.458\ \text{mH/km} \end{aligned}

Answer: L=1.458L = 1.458 mH per phase per km.

✓ Answer checked by Claude Opus 5.5 (high effort)

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