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Capacitance and charge matrices (9.81 m clearance)

Short answer

Equivalent bundle radius is 12.04 cm. Maxwell's matrix inversion gives capacitance matrix (diagonals 11.13 and 11.31 pF/m). With 230.94 kV rms phase voltages, the RMS charge matrix is 2.795∠4.3°, 2.949∠-120°, 2.795∠115.7° μC/m (peak 3.95, 4.17, 3.95 μC/m).

A 400 kV, three phase transmission has flat horizontal configuration with two sub-conductors per bundle. The average ground clearance is 9.81 m and the inter phase spacing is 11.3 m. The radius of each sub-conductor in the bundle is 3.17 cm and the separation between the sub-conductors is 45.72 cm. Calculate the capacitance and charge matrices of this line.

The capacitance matrix is [C]=2πε0[M]−1[C] = 2\pi\varepsilon_0[M]^{-1}, where [M][M] is Maxwell's coefficient matrix built with the image method and the equivalent bundle radius. The charge matrix is [Q]=[C][V][Q] = [C][V].

Data

  • 400 kV, flat horizontal, S=11.3S = 11.3 m, H=9.81H = 9.81 m (average)
  • Sub-conductor radius r=3.17r = 3.17 cm; B=45.72B = 45.72 cm; N=2N = 2
  • 2πε0=1/(18×109)=55.562\pi\varepsilon_0 = 1/(18 \times 10^9) = 55.56 pF/m

Step 1: Equivalent radius

req=rB=3.17×45.72=12.04 cm=0.1204 mr_{eq} = \sqrt{rB} = \sqrt{3.17 \times 45.72} = 12.04\ \text{cm} = 0.1204\ \text{m}

Step 2: Maxwell coefficients (2H=19.622H = 19.62 m)

Mii=ln⁡19.620.1204=5.0936M12=M23=ln⁡11.32+19.62211.3=ln⁡22.64111.3=0.6950M13=ln⁡22.62+19.62222.6=ln⁡29.92822.6=0.2809\begin{aligned} M_{ii} &= \ln\frac{19.62}{0.1204} = 5.0936 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11.3^2 + 19.62^2}}{11.3} = \ln\frac{22.641}{11.3} = 0.6950 \\ M_{13} &= \ln\frac{\sqrt{22.6^2 + 19.62^2}}{22.6} = \ln\frac{29.928}{22.6} = 0.2809 \end{aligned} [M]=[5.09360.69500.28090.69505.09360.69500.28090.69505.0936],det⁡=127.10[M] = \begin{bmatrix} 5.0936 & 0.6950 & 0.2809 \\ 0.6950 & 5.0936 & 0.6950 \\ 0.2809 & 0.6950 & 5.0936 \end{bmatrix}, \quad \det = 127.10

Step 3: Inverse and capacitance matrix

[M]−1=[0.20033−0.02632−0.00746−0.026320.20351−0.02632−0.00746−0.026320.20033][M]^{-1} = \begin{bmatrix} 0.20033 & -0.02632 & -0.00746 \\ -0.02632 & 0.20351 & -0.02632 \\ -0.00746 & -0.02632 & 0.20033 \end{bmatrix} [C]=55.56 [M]−1=[11.129−1.462−0.414−1.46211.306−1.462−0.414−1.46211.129] pF/m (nF/km)[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 11.129 & -1.462 & -0.414 \\ -1.462 & 11.306 & -1.462 \\ -0.414 & -1.462 & 11.129 \end{bmatrix}\ \text{pF/m (nF/km)}

Step 4: Charge matrix

Balanced phase voltages V=400/3=230.94V = 400/\sqrt{3} = 230.94 kV rms at 0°,−120°,120°0°, -120°, 120°; [Q]=[C][V][Q] = [C][V]:

q1=230.94 (11.129−1.462∠−120°−0.414∠120°)×10−3=2.795∠4.30° μC/m\begin{aligned} q_1 &= 230.94\,\big(11.129 - 1.462\angle{-120°} - 0.414\angle{120°}\big) \times 10^{-3} \\ &= 2.795\angle 4.30°\ \mu\text{C/m} \end{aligned} [Q]=[2.795∠4.30°2.949∠−120°2.795∠115.70°] μC/m (rms)[Q] = \begin{bmatrix} 2.795\angle 4.30° \\ 2.949\angle -120° \\ 2.795\angle 115.70° \end{bmatrix}\ \mu\text{C/m (rms)}

Peak values: 3.95, 4.17 and 3.95 μC/m.

If the line is transposed (for reference)

Ms=5.0936,Mm=2(0.6950)+0.28093=0.5569M_s = 5.0936,\quad M_m = \frac{2(0.6950) + 0.2809}{3} = 0.5569 C1=55.565.0936−0.5569=12.25 nF/km,q=12.25×230.94×10−3=2.83 μC/m (rms)C_1 = \frac{55.56}{5.0936 - 0.5569} = 12.25\ \text{nF/km}, \quad q = 12.25 \times 230.94 \times 10^{-3} = 2.83\ \mu\text{C/m (rms)}
ItemValue
reqr_{eq}12.04 cm
C11=C33C_{11} = C_{33}11.13 pF/m
C22C_{22}11.31 pF/m
C12C_{12}, C13C_{13}−1.46, −0.41 pF/m
qq outer / centre2.79 / 2.95 μC/m rms

Answer: [C][C] as above; charges ≈ 2.79∠4.3°, 2.95∠−120°, 2.79∠115.7° μC/m rms.

✓ Answer checked by Claude Opus 5.5 (high effort)

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