Capacitance depends on conductor size and spacing

The capacitance to neutral of a transposed three-phase line is $C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)}$. It contains only the conductor radius r and the spacings (through $D_{eq}$), apart from the constant $\varepsilon_0$. The derivation below shows this.

Potential difference between two conductors

For a long conductor with charge q C/m, the potential difference between two points at distances $D_1$ and $D_2$ is $V_{12} = \frac{q}{2\pi\varepsilon_0}\ln\frac{D_2}{D_1}$. For a group of conductors, using superposition:

$$V_{ab} = \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D_{ab}}{r} + q_b\ln\frac{r}{D_{ab}} + q_c\ln\frac{D_{bc}}{D_{ca}}\right]$$

Equilateral spacing D

$$\begin{aligned} V_{ab} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_b\ln\frac{r}{D}\right]\\ V_{ac} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_c\ln\frac{r}{D}\right] \end{aligned}$$

Adding and using $q_b + q_c = -q_a$ (balanced charges):

$$V_{ab} + V_{ac} = \frac{3q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}$$

For a balanced system, $V_{ab} + V_{ac} = 3V_{an}$. So

$$V_{an} = \frac{q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}, \qquad C_n = \frac{q_a}{V_{an}} = \frac{2\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}$$

Unsymmetrical spacing (transposed)

Averaging $V_{ab}$ over the three transposition sections gives the same form, with D replaced by the geometric mean spacing:

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r}}, \qquad D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$$

For bundled conductors, r is replaced by the bundle's self-GMD for capacitance, for example $\sqrt{rd}$.

Conclusion

• $\varepsilon_0$ is a constant (air). So C depends only on the ratio $D_{eq}/r$, that is, on conductor size and spacing.
• Larger r (or bundling) gives higher C. Larger spacing gives lower C. The dependence is logarithmic, so a big change in geometry causes only a modest change in C.
• Current, voltage and frequency do not appear in C. The earth's effect (images) is small for normal conductor heights and is neglected. That is why size and spacing are the primary factors.

Example: r = 1 cm and D = 4 m give $C_n = \frac{5.563\times10^{-11}}{\ln 400} = 9.29$ nF/km. Doubling D to 8 m gives 8.32 nF/km.