Capacitance matrices, untransposed and transposed (15 m)

The capacitance matrix is $[C] = 2\pi\varepsilon_0[M]^{-1}$ with Maxwell's coefficients from the image method. For the transposed line, the self and mutual coefficients are averaged before inversion.

Data

• 400 kV, flat horizontal, $S = 11$ m, $H = 15$ m
• $r = 3.18$ cm, $B = 45.72$ cm, $N = 2$
• $2\pi\varepsilon_0 = 1/(18 \times 10^9) = 55.56$ pF/m

Step 1: Equivalent radius

$$r_{eq} = \sqrt{rB} = \sqrt{3.18 \times 45.72} = 12.058\ \text{cm}$$

Step 2: Maxwell coefficients

$$\begin{aligned} M_{ii} &= \ln\frac{30}{0.12058} = 5.5167 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11^2 + 30^2}}{11} = 1.0664 \\ M_{13} &= \ln\frac{\sqrt{22^2 + 30^2}}{22} = 0.5253 \end{aligned}$$

Untransposed condition

$$[M] = \begin{bmatrix} 5.5167 & 1.0664 & 0.5253 \\ 1.0664 & 5.5167 & 1.0664 \\ 0.5253 & 1.0664 & 5.5167 \end{bmatrix}, \quad \det = 155.02$$ $$[M]^{-1} = \begin{bmatrix} 0.18899 & -0.03434 & -0.01136 \\ -0.03434 & 0.19454 & -0.03434 \\ -0.01136 & -0.03434 & 0.18899 \end{bmatrix}$$ $$[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 10.499 & -1.908 & -0.631 \\ -1.908 & 10.808 & -1.908 \\ -0.631 & -1.908 & 10.499 \end{bmatrix}\ \text{pF/m}$$

Transposed condition

Average the coefficients:

$$\begin{aligned} M_s &= 5.5167 \\ M_m &= \frac{1.0664 + 1.0664 + 0.5253}{3} = 0.8860 \end{aligned}$$ $$[M_t] = \begin{bmatrix} 5.5167 & 0.8860 & 0.8860 \\ 0.8860 & 5.5167 & 0.8860 \\ 0.8860 & 0.8860 & 5.5167 \end{bmatrix}$$

For a matrix with equal off-diagonal terms, the inverse has

$$C_s = 2\pi\varepsilon_0\frac{M_s + M_m}{(M_s - M_m)(M_s + 2M_m)}, \quad C_m = -2\pi\varepsilon_0\frac{M_m}{(M_s - M_m)(M_s + 2M_m)}$$ $$[C_t] = \begin{bmatrix} 10.539 & -1.458 & -1.458 \\ -1.458 & 10.539 & -1.458 \\ -1.458 & -1.458 & 10.539 \end{bmatrix}\ \text{pF/m}$$

Sequence capacitances:

$$\begin{aligned} C_1 &= C_s - C_m = \frac{55.56}{M_s - M_m} = \frac{55.56}{4.6307} = 12.00\ \text{nF/km} \\ C_0 &= C_s + 2C_m = \frac{55.56}{M_s + 2M_m} = \frac{55.56}{7.2887} = 7.62\ \text{nF/km} \end{aligned}$$

Transposition makes all phases identical and gives a balanced positive-sequence capacitance of 12.0 nF/km.

Answer: untransposed $[C]$ diagonal 10.50/10.81, off-diagonal −1.91/−0.63 pF/m; transposed $C_s = 10.54$, $C_m = -1.46$ pF/m, $C_1 = 12.0$ nF/km.