Chapter 4, Problem 4.48P

The capacitance of a single-circuit, three-phase transposed line with the configuration shown in Figure 4.38, including ground effect, and with conductors not equilaterally spaced is given by $C_{\alpha \eta }\frac{2\pi {\epsilon }_0}{\ln \frac{D_{eq}}{r}-\ln \frac{H_m}{H_8}}$

F/m line-to-neutral

where $D_{eq}=\sqrt[3]{D_{12}{D}_{23}{D}_{13}}=GMD$

$r=$ conductor’s outside radius $\begin{array}{l}{H}_m={\left(H_{12}{H}_{23}{H}_{13}\right)}^{1/3}\\ H_S={\left(H_1{H}_2{H}_3\right)}^{1/3}\end{array}$

1. Now consider Figure 4.39 in which the configuration of a three-phase, single circuit, 345-kV line with conductors having an outside diameter of 1.065 in. is shown. Determine the capacitance to neutral in F/m, including the ground effect.
2. Next, neglecting the effect of ground, see how the value changes.