Transposition; single-phase capacitance with earth

Importance of transposition

Transposition means exchanging the positions of the phase conductors at regular intervals along the line, so that each phase occupies each position for one-third of the length.

• With unsymmetrical spacing, the flux linkages and charges of the three phases differ, so the phases have unequal inductance and capacitance.
• Unequal parameters give unequal voltage drops, so the receiving-end voltages become unbalanced even with balanced load.
• Transposition makes the average inductance and capacitance of each phase equal, so the line behaves as a balanced circuit and can be modelled per phase using $D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}$.
• It reduces electromagnetic and electrostatic interference with nearby telephone and communication lines.

Capacitance of a single-phase line considering ground

Let conductors a and b (radius $r$, spacing $D$) be at height $h$ above ground, carrying charges $+q$ and $-q$ per metre. By the method of images, the earth is replaced by image conductors a' ($-q$) and b' ($+q$) at depth $h$.

   +q a o------- D -------o b -q
        |                 |
        h                 h
 ===============================  ground
        h                 h
        |                 |
   -q a'o                 o b' +q
   a-a' = 2h,  a-b' = sqrt(4h^2 + D^2)

Voltage between a and b due to all four charges, using $V_{ab} = \dfrac{1}{2\pi\varepsilon_0}\sum q_k\ln\dfrac{D_{kb}}{D_{ka}}$:

$$\begin{aligned} V_{ab} &= \frac{1}{2\pi\varepsilon_0}\left[q\ln\frac{D}{r} - q\ln\frac{r}{D} - q\ln\frac{\sqrt{4h^2+D^2}}{2h} + q\ln\frac{2h}{\sqrt{4h^2+D^2}}\right] \\ &= \frac{q}{\pi\varepsilon_0}\left[\ln\frac{D}{r} - \ln\frac{\sqrt{4h^2+D^2}}{2h}\right] \\ &= \frac{q}{\pi\varepsilon_0}\ln\frac{D}{r\sqrt{1+D^2/4h^2}} \end{aligned}$$

Line capacitance (between conductors):

$$C_{ab} = \frac{q}{V_{ab}} = \frac{\pi\varepsilon_0}{\ln\left(\dfrac{D}{r\sqrt{1+D^2/4h^2}}\right)}\ \text{F/m}$$

Phase capacitance (each conductor to neutral, $V_{an} = V_{ab}/2$):

$$C_n = 2C_{ab} = \frac{2\pi\varepsilon_0}{\ln\left(\dfrac{D}{r\sqrt{1+D^2/4h^2}}\right)}\ \text{F/m}$$

Since $\sqrt{1+D^2/4h^2} > 1$, the ground increases the capacitance. When $h \gg D$, the term tends to 1 and the result reduces to the usual $C_{ab} = \pi\varepsilon_0/\ln(D/r)$.