Inductance of 420 kV line, transposed and untransposed

For EHV lines, the inductance matrix is $[L] = \dfrac{\mu_0}{2\pi}[M] = 0.2\,[M]$ mH/km. Here $[M]$ is Maxwell's coefficient matrix (image method, equivalent bundle radius; internal flux neglected). Transposition replaces it by averaged self and mutual terms.

Data

• 420 kV, flat horizontal, $S = 11$ m, $H = 15$ m
• Bundle $2 \times 3.18$ cm diameter, so $r = 1.59$ cm; $B = 45.72$ cm
• $\mu_0/2\pi = 2 \times 10^{-7}$ H/m $= 0.2$ mH/km

Step 1: Equivalent radius

$$r_{eq} = \sqrt{rB} = \sqrt{1.59 \times 45.72} = 8.526\ \text{cm}$$

Step 2: Maxwell coefficients

$$\begin{aligned} M_{ii} &= \ln\frac{2H}{r_{eq}} = \ln\frac{30}{0.08526} = 5.8632 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11^2 + 30^2}}{11} = 1.0664 \\ M_{13} &= \ln\frac{\sqrt{22^2 + 30^2}}{22} = 0.5253 \end{aligned}$$

Untransposed line

$$[L] = 0.2\begin{bmatrix} 5.8632 & 1.0664 & 0.5253 \\ 1.0664 & 5.8632 & 1.0664 \\ 0.5253 & 1.0664 & 5.8632 \end{bmatrix} = \begin{bmatrix} 1.1726 & 0.2133 & 0.1051 \\ 0.2133 & 1.1726 & 0.2133 \\ 0.1051 & 0.2133 & 1.1726 \end{bmatrix}\ \text{mH/km}$$

The mutual terms are not all equal ($L_{12} = L_{23} \ne L_{13}$). So with balanced currents the voltage drops of the three phases are unequal, and the line is unbalanced.

Transposed line

Average the self and mutual terms:

$$\begin{aligned} L_s &= 0.2 \times 5.8632 = 1.1726\ \text{mH/km} \\ M_m &= \frac{1.0664 + 1.0664 + 0.5253}{3} = 0.8860 \\ L_m &= 0.2 \times 0.8860 = 0.1772\ \text{mH/km} \end{aligned}$$ $$[L_t] = \begin{bmatrix} 1.1726 & 0.1772 & 0.1772 \\ 0.1772 & 1.1726 & 0.1772 \\ 0.1772 & 0.1772 & 1.1726 \end{bmatrix}\ \text{mH/km}$$

Sequence inductances:

$$\begin{aligned} L_1 = L_2 &= L_s - L_m = 1.1726 - 0.1772 = 0.9954\ \text{mH/km} \\ L_0 &= L_s + 2L_m = 1.1726 + 0.3544 = 1.5271\ \text{mH/km} \end{aligned}$$

Check with the GMD formula

$D_{eq} = \sqrt[3]{11 \times 11 \times 22} = 13.86$ m, so

$$L_1 = 0.2\ln\frac{13.86}{0.08526} = 1.018\ \text{mH/km}$$

This is close to 0.995 mH/km; the small difference is due to the effect of the ground images.

Answer: untransposed $L_{ii} = 1.1726$, $L_{12} = 0.2133$, $L_{13} = 0.1051$ mH/km; transposed $L_s = 1.1726$, $L_m = 0.1772$ mH/km, positive-sequence $L_1 = 0.995$ mH/km.