Capacitance and charge matrices, transposed and untransposed

The capacitance matrix is found from Maxwell's coefficients, using the average height over the span and the equivalent bundle radius. The charge matrix then follows from $[Q] = [C][V]$.

Data

• 400 kV, flat horizontal, $S = 11.3$ m, $N = 2$
• Sub-conductor radius $r = 3.17$ cm (as given); bundle spacing $B = 45.72$ cm
• Height at tower 11.875 m, at midspan 8.85 m
• $2\pi\varepsilon_0 = 1/(18 \times 10^9) = 55.56$ pF/m

Step 1: Average height (parabolic sag)

$$\begin{aligned} H_{av} &= H_{min} + \frac{1}{3}(H_{tower} - H_{min}) \\ &= 8.85 + \frac{11.875 - 8.85}{3} = 9.858\ \text{m} \end{aligned}$$

Step 2: Equivalent radius

$$r_{eq} = \sqrt{r B} = \sqrt{3.17 \times 45.72} = 12.04\ \text{cm}$$

Step 3: Maxwell coefficients

$$\begin{aligned} M_s &= \ln\frac{2H}{r_{eq}} = \ln\frac{19.717}{0.1204} = 5.0985 \\ M_{12} &= \ln\frac{\sqrt{11.3^2 + 19.717^2}}{11.3} = 0.6987 \\ M_{13} &= \ln\frac{\sqrt{22.6^2 + 19.717^2}}{22.6} = 0.2830 \end{aligned}$$

Untransposed line

$$[M] = \begin{bmatrix} 5.0985 & 0.6987 & 0.2830 \\ 0.6987 & 5.0985 & 0.6987 \\ 0.2830 & 0.6987 & 5.0985 \end{bmatrix}, \quad \det = 127.42$$ $$[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 11.121 & -1.467 & -0.416 \\ -1.467 & 11.298 & -1.467 \\ -0.416 & -1.467 & 11.121 \end{bmatrix}\ \text{pF/m}$$

Charge matrix. Take balanced phase voltages $V = 400/\sqrt{3} = 230.94$ kV rms ($\angle 0°, \angle -120°, \angle 120°$). Then $[Q] = [C][V]$:

$$[Q] = \begin{bmatrix} 2.794\angle 4.31° \\ 2.948\angle -120° \\ 2.794\angle 115.69° \end{bmatrix}\ \mu\text{C/m (rms)}$$

Peak values are 3.95, 4.17 and 3.95 μC/m. The centre phase carries the most charge, so it has the highest surface gradient.

Transposed line

Average the coefficients:

$$\begin{aligned} M_s &= 5.0985 \\ M_m &= \frac{0.6987 + 0.6987 + 0.2830}{3} = 0.5601 \end{aligned}$$ $$[C_t] = \begin{bmatrix} 11.139 & -1.103 & -1.103 \\ -1.103 & 11.139 & -1.103 \\ -1.103 & -1.103 & 11.139 \end{bmatrix}\ \text{pF/m}$$

The positive-sequence capacitance is

$$C_1 = \frac{55.56}{M_s - M_m} = \frac{55.56}{4.5384} = 12.24\ \text{pF/m (nF/km)}$$

and each phase charge is equal in magnitude:

$$q = C_1 V = 12.24 \times 10^{-12} \times 230.94 \times 10^3 = 2.827\ \mu\text{C/m (rms)} \ (4.00\ \mu\text{C/m peak})$$ $$[Q_t] = \begin{bmatrix} 2.827\angle 0° \\ 2.827\angle -120° \\ 2.827\angle 120° \end{bmatrix}\ \mu\text{C/m}$$

Answer: untransposed $C_{11} = 11.12$, $C_{22} = 11.30$, $C_{12} = -1.47$, $C_{13} = -0.42$ pF/m, with charges 2.79/2.95/2.79 μC/m rms; transposed $C_s = 11.14$, $C_m = -1.10$ pF/m, $C_1 = 12.24$ nF/km, with $q = 2.83$ μC/m rms per phase.