Capacitive reactance and charging of 2-conductor bundle line

The voltage and frequency are not printed. They are assumed as 400 kV (line), 50 Hz, the usual data for this two-conductor-bundle configuration. Reactance per km does not depend on the voltage; only the charging current and VAR scale with it.

Data

• Sub-conductor radius $r = 9$ mm $= 0.009$ m. Bundle spacing $d = 0.40$ m.
• Phase spacing: 10 m, 10 m, 20 m (horizontal), transposed.
• $V_{ph} = 400/\sqrt3 = 230.94$ kV; $\omega = 314.16$ rad/s.

Self-GMD for capacitance

For capacitance the actual radius r is used, not $r'$:

$$D_{sc} = \sqrt{r\,d} = \sqrt{0.009\times0.40} = 0.06\ \text{m}$$

Mutual GMD

$$D_{eq} = \sqrt[3]{10\times10\times20} = 12.599\ \text{m}$$

Capacitance and capacitive reactance per phase per km

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/D_{sc})} = \frac{5.563\times10^{-11}}{\ln(12.599/0.06)} = \frac{5.563\times10^{-11}}{5.3470}\\ &= 1.0404\times10^{-11}\ \text{F/m} = 0.010404\ \mu\text{F/km}\\ X_C &= \frac{1}{\omega C} = \frac{1}{314.16\times1.0404\times10^{-8}} = 3.059\times10^5\ \Omega\cdot\text{km} \end{aligned}$$

Charging current and charging VAR (per km)

$$\begin{aligned} I_c &= \frac{V_{ph}}{X_C} = \frac{230940}{3.059\times10^5} = 0.7548\ \text{A/km}\\ Q_{c,ph} &= V_{ph}I_c = 230940\times0.7548 = 174.32\ \text{kVAR/km per phase}\\ Q_{c,3\phi} &= 3\times174.32 = 522.97\ \text{kVAR/km} \end{aligned}$$

Answer: $X_C \approx 3.06\times10^5\ \Omega$·km per phase. At 400 kV, $I_c \approx 0.755$ A/km and the charging VAR is about 174.3 kVAR per phase per km. (At another line voltage V, scale $I_c$ by $V/400$ and the VAR by $(V/400)^2$.)