220 kV line on 6-6-12 m triangle

Sides of 6, 6 and 12 m mean the conductors are actually in a straight line (flat configuration with 6 m between adjacent conductors). For a transposed line use $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

Data: $r = 1.81$ cm $= 0.0181$ m, $r' = 0.7788 \times 0.0181 = 0.014096$ m, 220 kV, 50 Hz, 200 km.

$$D_{eq} = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}$$

Inductance per phase per km

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{7.5595}{0.014096} = 2\times10^{-7}\times6.2847 \\ &= 1.2569\times10^{-6}\ \text{H/m} = 1.257\ \text{mH/km} \end{aligned}$$

Capacitance per phase per km

$$\begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(7.5595/0.0181)} = \frac{55.63\times10^{-12}}{6.0347} \\ &= 9.219\times10^{-12}\ \text{F/m} = 9.219\ \text{nF/km} \end{aligned}$$

Capacitive reactance per phase (200 km)

$$\begin{aligned} C_{200} &= 9.219 \times 200 = 1843.7\ \text{nF} = 1.8437\ \mu\text{F} \\ X_C &= \frac{1}{2\pi \times 50 \times 1.8437\times10^{-6}} = 1726.4\ \Omega \end{aligned}$$

Charging current

$$I_c = \frac{V_p}{X_C} = \frac{220\,000/\sqrt3}{1726.4} = \frac{127\,017}{1726.4} = 73.57\ \text{A}$$

Total charging MVAR

$$Q = 3V_pI_c = \frac{V_L^2}{X_C} = \frac{(220\times10^3)^2}{1726.4} = 28.03\ \text{MVAR}$$

Answer: $L = 1.257$ mH/km, $C = 9.219$ nF/km, $X_C = 1726.4\ \Omega$, $I_c = 73.6$ A, $Q = 28.0$ MVAR.