132 kV triangular line: L, C, charging MVAR

For a transposed line with unsymmetrical spacing, use the equivalent spacing $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

Data: diameter 18 mm, $r = 0.009$ m, $r' = 0.7788 \times 0.009 = 0.0070092$ m; spacings 4.0, 5.0, 4.5 m; 132 kV, 50 Hz.

$$D_{eq} = \sqrt[3]{4 \times 5 \times 4.5} = \sqrt[3]{90} = 4.4814\ \text{m}$$

Inductance per phase

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{D_{eq}}{r'} = 2\times10^{-7}\ln\frac{4.4814}{0.0070092} \\ &= 2\times10^{-7}\times6.4605 = 1.2921\times10^{-6}\ \text{H/m} = 1.292\ \text{mH/km} \end{aligned}$$

Capacitance per phase

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{\ln(4.4814/0.009)} = \frac{55.63\times10^{-12}}{6.2105} \\ &= 8.958\times10^{-12}\ \text{F/m} = 8.958\ \text{nF/km} \end{aligned}$$

Charging current per km

$$\begin{aligned} X_C &= \frac{1}{2\pi \times 50 \times 8.958\times10^{-9}} = 3.553\times10^5\ \Omega\text{-km} \\ V_p &= 132/\sqrt3 = 76.21\ \text{kV} \\ I_c &= \frac{76\,210}{355\,349} = 0.2145\ \text{A/km} \end{aligned}$$

Charging MVAR per km (three phase)

$$Q_c = \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{355\,349} = 49\,033\ \text{VAR} = 0.0490\ \text{MVAR/km}$$

Answer: $L = 1.292$ mH/km, $C = 8.958$ nF/km, $I_c = 0.2145$ A/km, $Q_c = 0.0490$ MVAR/km (49.0 kVAR/km).