400 kV twin-bundle line: reactances and charging VAR

Each phase is a two-conductor bundle. The self-GMD of a 2-bundle is $\sqrt{r'd}$ for inductance and $\sqrt{rd}$ for capacitance.

   o--o        o--o        o--o
   a  a'       b  b'       c  c'
    A    10 m   B    10 m   C
   d = 40 cm, r = 20 mm

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

GMDs

$$\begin{aligned} D_m &= \sqrt[3]{10 \times 10 \times 20} = 12.599\ \text{m} \\ D_{sL} &= \sqrt{0.015576 \times 0.4} = 0.07893\ \text{m} \\ D_{sC} &= \sqrt{0.02 \times 0.4} = 0.08944\ \text{m} \end{aligned}$$

Inductive reactance per phase per km

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{12.599}{0.07893} = 2\times10^{-7}\times5.0728 = 1.0146\ \text{mH/km} \\ X_L &= 2\pi \times 50 \times 1.0146\times10^{-3} = 0.3187\ \Omega/\text{km} \end{aligned}$$

Capacitive reactance per phase per km

$$\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(12.599/0.08944)} = \frac{55.63\times10^{-12}}{4.9478} = 11.244\ \text{nF/km} \\ X_C &= \frac{1}{2\pi \times 50 \times 11.244\times10^{-9}} = 2.831\times10^5\ \Omega\text{-km} \end{aligned}$$

Charging VAR per phase per km

$$\begin{aligned} V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\ I_c &= V_p/X_C = 230\,940/283\,101 = 0.8158\ \text{A/km} \\ Q_c &= \frac{V_p^2}{X_C} = \frac{(230\,940)^2}{283\,101} = 188.39\ \text{kVAR/km} \end{aligned}$$

Answer: $X_L = 0.3187\ \Omega$/km, $X_C = 2.831\times10^5\ \Omega$-km, charging $Q_c = 188.4$ kVAR per phase per km (565.2 kVAR/km for all three phases).