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400 kV line with 3-conductor bundles: reactances and VAR

Short answer

Compute bundle self GMDs: D_sL = (r' d²)^(1/3) = 0.1356 m, D_sC = (r d²)^(1/3) = 0.1474 m, D_m = 12.599 m. Then X_L = 0.2847 Ω/km, X_C = 2.545×10^5 Ω·km, I_c = 0.907 A/km, Q_c = 209.5 kVAR/km per phase.

Determine the inductive and capacitive reactance, charging current and charging VAR per phase per km of the 400 kV single circuit, 3-φ 50 Hz fully transposed transmission line having three bundle conductors per phase with spacing of sub conductors as d = 40 cm and radius of each sub conductor is 20 mm as shown in figure below. [Figure: horizontal single-circuit line; each phase is a triangular bundle of three sub-conductors with 40 cm sides; centres of adjacent phases 10 m apart.]

Each phase is a 3-conductor bundle at the corners of an equilateral triangle of side dd. We find the bundle self-GMD for inductance and capacitance, the mutual GMD between phases, and then the reactances.

     o          o          o
    o o        o o        o o
     A   10 m   B   10 m   C
   d = 40 cm, r = 20 mm

Data: r=0.02r = 0.02 m, r′=0.7788×0.02=0.015576r' = 0.7788 \times 0.02 = 0.015576 m, d=0.4d = 0.4 m, D=10,10,20D = 10, 10, 20 m, 400 kV, 50 Hz.

GMDs

Dm=10×10×203=12.599 mDsL=r′d23=0.015576×0.423=0.13558 mDsC=rd23=0.02×0.423=0.14736 m\begin{aligned} D_m &= \sqrt[3]{10 \times 10 \times 20} = 12.599\ \text{m} \\ D_{sL} &= \sqrt[3]{r'd^2} = \sqrt[3]{0.015576 \times 0.4^2} = 0.13558\ \text{m} \\ D_{sC} &= \sqrt[3]{rd^2} = \sqrt[3]{0.02 \times 0.4^2} = 0.14736\ \text{m} \end{aligned}

Inductive reactance per km

L=2×10−7ln⁡12.5990.13558=2×10−7×4.5318=0.9064 mH/kmXL=2πfL=2π×50×0.9064×10−3=0.2847 Ω/km\begin{aligned} L &= 2\times10^{-7}\ln\frac{12.599}{0.13558} = 2\times10^{-7}\times4.5318 = 0.9064\ \text{mH/km} \\ X_L &= 2\pi f L = 2\pi \times 50 \times 0.9064\times10^{-3} = 0.2847\ \Omega/\text{km} \end{aligned}

Capacitive reactance per km

C=55.63×10−12ln⁡(12.599/0.14736)=55.63×10−124.4485=12.506 nF/kmXC=12πfC=12π×50×12.506×10−9=2.545×105 Ω-km\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(12.599/0.14736)} = \frac{55.63\times10^{-12}}{4.4485} = 12.506\ \text{nF/km} \\ X_C &= \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 12.506\times10^{-9}} = 2.545\times10^5\ \Omega\text{-km} \end{aligned}

Charging current and VAR per phase per km

Vp=400/3=230.94 kVIc=VpXC=230 940254 533=0.9073 A/kmQc=VpIc=230.94×0.9073=209.53 kVAR/km per phase\begin{aligned} V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\ I_c &= \frac{V_p}{X_C} = \frac{230\,940}{254\,533} = 0.9073\ \text{A/km} \\ Q_c &= V_p I_c = 230.94 \times 0.9073 = 209.53\ \text{kVAR/km per phase} \end{aligned}

(Three phases together: 628.6 kVAR/km.)

Answer: XL=0.2847 ΩX_L = 0.2847\ \Omega/km, XC=2.545×105 ΩX_C = 2.545\times10^5\ \Omega-km, Ic=0.907I_c = 0.907 A/km, Qc=209.5Q_c = 209.5 kVAR/km per phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

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