Capacitance matrix of 400 kV bundled line

The capacitance matrix of the line is found from Maxwell's potential coefficients, using the image method and replacing each bundle by its equivalent radius: $[C] = 2\pi\varepsilon_0 [M]^{-1}$.

Data

• 3 phases in flat horizontal configuration, phase spacing $S = 11$ m
• Height $H = 15$ m
• Sub-conductor radius $r = 3.18$ cm; bundle spacing $B = 45.72$ cm; $N = 2$
• $2\pi\varepsilon_0 = \dfrac{1}{18 \times 10^9}$ F/m $= 55.56$ pF/m

    1          2          3
    o--- 11 m--o--- 11 m--o
    |                       H = 15 m
 ////////////////////////////////

Step 1: Equivalent radius of the bundle

For $N = 2$:

$$\begin{aligned} r_{eq} &= \sqrt{r\,B} = \sqrt{3.18 \times 45.72} \\ &= 12.058\ \text{cm} = 0.12058\ \text{m} \end{aligned}$$

Step 2: Maxwell coefficient matrix $[M]$ (dimensionless)

Self terms:

$$M_{11} = M_{22} = M_{33} = \ln\frac{2H}{r_{eq}} = \ln\frac{30}{0.12058} = 5.5167$$

Mutual terms ($A_{ij}$ = distance to conductor, $I_{ij}$ = distance to image):

$$\begin{aligned} M_{12} = M_{23} &= \ln\frac{\sqrt{11^2 + 30^2}}{11} = \ln\frac{31.953}{11} = 1.0664 \\ M_{13} &= \ln\frac{\sqrt{22^2 + 30^2}}{22} = \ln\frac{37.202}{22} = 0.5253 \end{aligned}$$ $$[M] = \begin{bmatrix} 5.5167 & 1.0664 & 0.5253 \\ 1.0664 & 5.5167 & 1.0664 \\ 0.5253 & 1.0664 & 5.5167 \end{bmatrix}$$

(The potential coefficient matrix is $[P] = 18 \times 10^9 [M]$ m/F, i.e. diagonal $99.30 \times 10^9$, $P_{12} = 19.20 \times 10^9$, $P_{13} = 9.46 \times 10^9$.)

Step 3: Inverse of $[M]$

Determinant of $[M]$ = 155.02.

$$[M]^{-1} = \begin{bmatrix} 0.18899 & -0.03434 & -0.01136 \\ -0.03434 & 0.19454 & -0.03434 \\ -0.01136 & -0.03434 & 0.18899 \end{bmatrix}$$

Step 4: Capacitance matrix

$$\begin{aligned} [C] &= 2\pi\varepsilon_0 [M]^{-1} = 55.56 \times [M]^{-1}\ \text{pF/m} \\[4pt] [C] &= \begin{bmatrix} 10.499 & -1.908 & -0.631 \\ -1.908 & 10.808 & -1.908 \\ -0.631 & -1.908 & 10.499 \end{bmatrix}\ \text{pF/m (= nF/km)} \end{aligned}$$

Interpretation

• Diagonal terms are the self (Maxwell) capacitances. The centre phase has the largest value, 10.81 nF/km, because it is closely coupled to both outer phases.
• Off-diagonal terms are negative, as expected in a Maxwell capacitance matrix.
• If the line were transposed, $M_s = 5.5167$ and $M_m = 0.8860$ (averages), giving a positive-sequence capacitance $C_1 = 55.56/(5.5167 - 0.8860) = 12.00$ nF/km.

Answer: $[C] = \begin{bmatrix} 10.50 & -1.91 & -0.63 \\ -1.91 & 10.81 & -1.91 \\ -0.63 & -1.91 & 10.50 \end{bmatrix}$ nF/km (pF/m).