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400 kV quadruple bundle line, 20 m spacing

Short answer

Using bundle self GMD D_sL = 0.22907 m, D_sC = 0.24384 m and mutual GMD = 25.198 m, per phase per km: inductive reactance X_L = 0.2953 Ω/km, capacitive reactance X_C = 2.654×10⁵ Ω·km, charging current I_c = 0.870 A/km, charging reactive power Q_c = 201.0 kVAR/km.

Figure below shows a quadruple-conductor circuit of a single-circuit, three-phase, 400 kV, 50 Hz line with a horizontal spacing of 20 m. Each sub-conductor of the bundle has a diameter of 40 mm and spacing between the sub-conductors is 0.5 m. Each phase group shares the total current and charge equally and the line is completely transposed. Determine the inductive reactance, capacitive reactance, charging current and charging VAR per phase per km of the line. [Figure: horizontal single-circuit line; each phase A, B, C is a square bundle of four sub-conductors with 0.5 m sides; centres of adjacent phases 20 m apart.]

Each phase is a square bundle of four sub-conductors. We find the bundle self-GMD for inductance and capacitance, the mutual GMD, and then the per-km quantities.

   o o         o o         o o
   o o         o o         o o
    A   20 m    B   20 m    C
   d = 0.5 m, diameter 40 mm

Data: r=20r = 20 mm =0.02= 0.02 m, r′=0.7788×0.02=0.015576r' = 0.7788 \times 0.02 = 0.015576 m, d=0.5d = 0.5 m, D=20,20,40D = 20, 20, 40 m, 400 kV, 50 Hz.

GMDs

Dm=20×20×403=25.198 mDsL=1.09r′d34=1.090.015576×0.534=0.22907 mDsC=1.09rd34=1.090.02×0.534=0.24384 m\begin{aligned} D_m &= \sqrt[3]{20 \times 20 \times 40} = 25.198\ \text{m} \\ D_{sL} &= 1.09\sqrt[4]{r'd^3} = 1.09\sqrt[4]{0.015576 \times 0.5^3} = 0.22907\ \text{m} \\ D_{sC} &= 1.09\sqrt[4]{rd^3} = 1.09\sqrt[4]{0.02 \times 0.5^3} = 0.24384\ \text{m} \end{aligned}

Inductive reactance per phase per km

L=2×10−7ln⁡25.1980.22907=2×10−7×4.7005=0.9401 mH/kmXL=2π×50×0.9401×10−3=0.2953 Ω/km\begin{aligned} L &= 2\times10^{-7}\ln\frac{25.198}{0.22907} = 2\times10^{-7}\times4.7005 = 0.9401\ \text{mH/km} \\ X_L &= 2\pi \times 50 \times 0.9401\times10^{-3} = 0.2953\ \Omega/\text{km} \end{aligned}

Capacitive reactance per phase per km

C=55.63×10−12ln⁡(25.198/0.24384)=55.63×10−124.6380=11.995 nF/kmXC=12π×50×11.995×10−9=2.654×105 Ω-km\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(25.198/0.24384)} = \frac{55.63\times10^{-12}}{4.6380} = 11.995\ \text{nF/km} \\ X_C &= \frac{1}{2\pi \times 50 \times 11.995\times10^{-9}} = 2.654\times10^5\ \Omega\text{-km} \end{aligned}

Charging current and VAR

Vp=400/3=230.94 kVIc=230 940265 376=0.8702 A/kmQc=VpIc=230.94×0.8702=200.97 kVAR/km per phase\begin{aligned} V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\ I_c &= \frac{230\,940}{265\,376} = 0.8702\ \text{A/km} \\ Q_c &= V_p I_c = 230.94 \times 0.8702 = 200.97\ \text{kVAR/km per phase} \end{aligned}

Answer: XL=0.2953 ΩX_L = 0.2953\ \Omega/km, XC=2.654×105 ΩX_C = 2.654\times10^5\ \Omega-km, Ic=0.870I_c = 0.870 A/km, Qc=201.0Q_c = 201.0 kVAR/km per phase (603 kVAR/km for three phases).

✓ Answer checked by Claude Opus 5.5 (high effort)

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