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330 kV quad-bundle line: inductance and charging

Short answer

Assuming sub-conductor radius 12 mm, with D_sL = 0.18629 m, D_sC = 0.19830 m, mutual GMD = 7.5595 m, per phase per km: inductance 0.7407 mH/km, charging current 0.915 A/km, total three-phase charging reactive power 0.523 MVAR/km.

A 3 phase overhead line has 4 sub-conductors per phase separated from each other by 45 cm and placed at the vertices of a square. The phases have flat horizontal configuration with centre to centre distance between adjacent phases equalling 6 m and between far end phases equalling 12 m. Assuming complete transposition determine the inductance per phase, charging current per phase per km and total charging VAR of the line. The line is operated at 330 kV and 50 Hz.

Each phase is a square bundle of four sub-conductors with side d=0.45d = 0.45 m. The sub-conductor radius is not given; it is assumed to be 12 mm (the usual value in this IOE problem).

   o o        o o        o o
   o o        o o        o o
    A    6 m   B    6 m   C
   |<------- 12 m ------->|

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, 330 kV, 50 Hz.

GMDs

Dm=6×6×123=7.5595 mDsL=1.09r′d34=1.090.009346×0.4534=0.18629 mDsC=1.09rd34=1.090.012×0.4534=0.19830 m\begin{aligned} D_m &= \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m} \\ D_{sL} &= 1.09\sqrt[4]{r'd^3} = 1.09\sqrt[4]{0.009346 \times 0.45^3} = 0.18629\ \text{m} \\ D_{sC} &= 1.09\sqrt[4]{rd^3} = 1.09\sqrt[4]{0.012 \times 0.45^3} = 0.19830\ \text{m} \end{aligned}

Inductance per phase

L=2×10−7ln⁡7.55950.18629=2×10−7×3.7033=0.7407 mH/kmL = 2\times10^{-7}\ln\frac{7.5595}{0.18629} = 2\times10^{-7}\times3.7033 = 0.7407\ \text{mH/km}

Charging current per phase per km

C=55.63×10−12ln⁡(7.5595/0.19830)=55.63×10−123.6408=15.28 nF/kmXC=12π×50×15.28×10−9=2.083×105 Ω-kmIc=330 000/3208 316=190 526208 316=0.9146 A/km\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(7.5595/0.19830)} = \frac{55.63\times10^{-12}}{3.6408} = 15.28\ \text{nF/km} \\ X_C &= \frac{1}{2\pi \times 50 \times 15.28\times10^{-9}} = 2.083\times10^5\ \Omega\text{-km} \\ I_c &= \frac{330\,000/\sqrt3}{208\,316} = \frac{190\,526}{208\,316} = 0.9146\ \text{A/km} \end{aligned}

Total charging VAR

Qc=VL2XC=(330×103)2208 316=0.5228 MVAR per kmQ_c = \frac{V_L^2}{X_C} = \frac{(330\times10^3)^2}{208\,316} = 0.5228\ \text{MVAR per km}

The line length is not given, so the total is per km; for a line of length ℓ\ell km, Q=0.5228 ℓQ = 0.5228\,\ell MVAR (e.g. 104.6 MVAR for 200 km).

Answer: L=0.7407L = 0.7407 mH/km per phase, Ic=0.915I_c = 0.915 A/km per phase, Qc=0.523Q_c = 0.523 MVAR/km (all three phases), with r=12r = 12 mm assumed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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