132 kV horizontal line: susceptance and charging

The conductor radius is not given; it is assumed to be 12 mm (a common value for 132 kV lines in IOE problems). For a transposed line, $C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r)$ and the shunt susceptance is $B = \omega C$.

   a o----4.5 m----o b----4.5 m----o c
     |<----------- 9 m ----------->|

Data: $D_{ab} = D_{bc} = 4.5$ m, $D_{ca} = 9$ m, $r = 0.012$ m, 132 kV, 50 Hz, 120 km.

Equivalent spacing

$$D_{eq} = \sqrt[3]{4.5 \times 4.5 \times 9} = 5.6696\ \text{m}$$

Capacitance per phase

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{\ln(5.6696/0.012)} = \frac{55.63\times10^{-12}}{6.1580} \\ &= 9.034\times10^{-12}\ \text{F/m} = 9.034\ \text{nF/km} \end{aligned}$$

Shunt susceptance per phase

$$\begin{aligned} B &= \omega C = 2\pi \times 50 \times 9.034\times10^{-9} = 2.838\times10^{-6}\ \text{S/km} \\ B_{120} &= 2.838\times10^{-6} \times 120 = 3.406\times10^{-4}\ \text{S} \end{aligned}$$

Charging current per phase per km

$$I_c = V_p B = \frac{132\,000}{\sqrt3} \times 2.838\times10^{-6} = 76\,210 \times 2.838\times10^{-6} = 0.2163\ \text{A/km}$$

Reactive power generated by the line (120 km, 3 phase)

$$Q = V_L^2 B_{120} = (132\times10^3)^2 \times 3.406\times10^{-4} = 5.934\ \text{MVAR}$$

Answer (with $r = 12$ mm): $B = 2.838\ \mu$S/km per phase (340.6 µS for 120 km), $I_c = 0.216$ A/km, $Q = 5.93$ MVAR.