Transposition and inductance of transposed line

Why transposition is needed

When the phase spacings are unequal ($D_{12} \ne D_{23} \ne D_{31}$), each phase has a different flux linkage. Its inductance therefore differs from the other phases and even contains an imaginary part. This causes unequal voltage drops, an unbalanced receiving-end voltage, and interference with nearby communication lines. Transposition means each phase takes each position for one-third of the length. This makes the average inductance of all phases equal, so the line becomes balanced and can be analysed per phase.

          Sec I      Sec II     Sec III
 pos 1     a           c           b
 pos 2     b           a           c
 pos 3     c           b           a
        D12 between pos1-2, D23 pos2-3, D31 pos3-1

Derivation

Flux linkage of conductor a, of radius r, in a group of conductors (per metre):

$$\lambda_a = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{ab}} + I_c\ln\frac{1}{D_{ac}}\right], \quad r' = 0.7788r$$

Section I (a at 1, b at 2, c at 3):

$$\lambda_{a1} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{12}} + I_c\ln\frac{1}{D_{31}}\right]$$

Section II (a at 2, b at 3, c at 1):

$$\lambda_{a2} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{23}} + I_c\ln\frac{1}{D_{12}}\right]$$

Section III (a at 3, b at 1, c at 2):

$$\lambda_{a3} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{31}} + I_c\ln\frac{1}{D_{23}}\right]$$

Average over the cycle:

$$\begin{aligned} \lambda_a &= \frac{\lambda_{a1} + \lambda_{a2} + \lambda_{a3}}{3}\\ &= \frac{2\times10^{-7}}{3}\left[3I_a\ln\frac{1}{r'} + (I_b + I_c)\ln\frac{1}{D_{12}D_{23}D_{31}}\right] \end{aligned}$$

For balanced currents, $I_b + I_c = -I_a$:

$$\begin{aligned} \lambda_a &= 2\times10^{-7}\left[I_a\ln\frac{1}{r'} - I_a\ln\frac{1}{\sqrt[3]{D_{12}D_{23}D_{31}}}\right]\\ &= 2\times10^{-7}I_a\ln\frac{\sqrt[3]{D_{12}D_{23}D_{31}}}{r'} \end{aligned}$$

Inductance per phase:

$$L_a = \frac{\lambda_a}{I_a} = 2\times10^{-7}\ln\frac{D_{eq}}{r'}\ \text{H/m}, \qquad D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}$$

By symmetry, $L_b = L_c = L_a$. The transposed line behaves like an equilaterally spaced line with spacing $D_{eq}$ (the geometric mean distance), and the inductance is real and equal in all phases.

In mH/km: $L = 0.2\ln\frac{D_{eq}}{r'}$ mH/km per phase. For stranded or bundled conductors, $r'$ is replaced by the GMR $D_s$.

Example: horizontal spacing 4 m, 4 m, 8 m gives $D_{eq} = \sqrt[3]{4\times4\times8} = 5.04$ m.