Inductance matrix of 500 kV bundled line

For an EHV line, the inductance matrix is $[L] = \dfrac{\mu_0}{2\pi}[M] = 0.2\,[M]$ mH/km. $[M]$ is Maxwell's coefficient matrix from the image method, using the equivalent radius of each bundle (internal flux linkage neglected).

Data

• 500 kV, flat horizontal, $S = 15$ m, $H = 20$ m
• $N = 2$, sub-conductor radius $r = 5$ cm, bundle spacing $B = 30$ cm
• $\mu_0/2\pi = 2 \times 10^{-7}$ H/m $= 0.2$ mH/km

    1          2          3
    o--- 15 m--o--- 15 m--o
    |                       H = 20 m
 ////////////////////////////////

Step 1: Equivalent radius

$$r_{eq} = \sqrt{rB} = \sqrt{5 \times 30} = 12.247\ \text{cm} = 0.12247\ \text{m}$$

Step 2: Maxwell coefficients ($2H = 40$ m)

$$\begin{aligned} M_{ii} &= \ln\frac{40}{0.12247} = \ln 326.6 = 5.7887 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{15^2 + 40^2}}{15} = \ln\frac{42.72}{15} = 1.0466 \\ M_{13} &= \ln\frac{\sqrt{30^2 + 40^2}}{30} = \ln\frac{50}{30} = 0.5108 \end{aligned}$$

Step 3: Inductance matrix (untransposed)

$$[L] = 0.2\begin{bmatrix} 5.7887 & 1.0466 & 0.5108 \\ 1.0466 & 5.7887 & 1.0466 \\ 0.5108 & 1.0466 & 5.7887 \end{bmatrix} = \begin{bmatrix} 1.1577 & 0.2093 & 0.1022 \\ 0.2093 & 1.1577 & 0.2093 \\ 0.1022 & 0.2093 & 1.1577 \end{bmatrix}\ \text{mH/km}$$

Step 4: If transposed (for comparison)

$$\begin{aligned} L_s &= 1.1577\ \text{mH/km} \\ L_m &= 0.2 \times \frac{1.0466 + 1.0466 + 0.5108}{3} = 0.1736\ \text{mH/km} \\ L_1 &= L_s - L_m = 0.9841\ \text{mH/km} \\ L_0 &= L_s + 2L_m = 1.5049\ \text{mH/km} \end{aligned}$$

Positive-sequence reactance at 50 Hz: $X_1 = 2\pi \times 50 \times 0.9841 \times 10^{-3} = 0.309\ \Omega$/km.

Answer: $[L] = \begin{bmatrix} 1.158 & 0.209 & 0.102 \\ 0.209 & 1.158 & 0.209 \\ 0.102 & 0.209 & 1.158 \end{bmatrix}$ mH/km.