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Charge matrix of 400 kV line

Short answer

Compute Maxwell's coefficient matrix using average height (9.858 m) and equivalent bundle radius (8.513 cm). The capacitance matrix has diagonal values 10.39 and 10.53 pF/m. The RMS charge matrix gives outer phases 2.60 μC/m, centre 2.73 μC/m (peak 3.67, 3.86, 3.67 μC/m).

A 400 kV three phase transmission line has flat horizontal configuration with 2 × 3.17 cm diameter bundle in each phase and a phase spacing of 11.3 m. The heights of the phase conductors are 11.875 m and 8.85 m at tower and at midspan respectively. The separation between the sub-conductors is 45.72 cm. Calculate the charge matrix of this line.

The charge matrix is [Q]=[C][V]=2πε0[M]−1[V][Q] = [C][V] = 2\pi\varepsilon_0 [M]^{-1}[V]. Here [M][M] is Maxwell's coefficient matrix, built from the average height of the sagging conductor and the equivalent bundle radius.

Data

  • 400 kV, flat horizontal, S=11.3S = 11.3 m
  • Bundle 2×3.172 \times 3.17 cm diameter, so r=1.585r = 1.585 cm; B=45.72B = 45.72 cm
  • Heights 11.875 m (tower) and 8.85 m (midspan)
  • 2πε0=55.562\pi\varepsilon_0 = 55.56 pF/m

Step 1: Average height

Hav=8.85+11.875−8.853=9.858 m,2H=19.717 mH_{av} = 8.85 + \frac{11.875 - 8.85}{3} = 9.858\ \text{m}, \quad 2H = 19.717\ \text{m}

Step 2: Equivalent radius

req=rB=1.585×45.72=8.513 cmr_{eq} = \sqrt{rB} = \sqrt{1.585 \times 45.72} = 8.513\ \text{cm}

Step 3: Maxwell coefficient matrix

Mii=ln⁡19.7170.08513=5.4451M12=M23=ln⁡11.32+19.717211.3=0.6987M13=ln⁡22.62+19.717222.6=0.2830\begin{aligned} M_{ii} &= \ln\frac{19.717}{0.08513} = 5.4451 \\ M_{12} = M_{23} &= \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} [M]=[5.44510.69870.28300.69875.44510.69870.28300.69875.4451],det⁡=155.96[M] = \begin{bmatrix} 5.4451 & 0.6987 & 0.2830 \\ 0.6987 & 5.4451 & 0.6987 \\ 0.2830 & 0.6987 & 5.4451 \end{bmatrix}, \quad \det = 155.96

Step 4: Capacitance (charge-coefficient) matrix

[M]−1=[0.18697−0.02312−0.00675−0.023120.18959−0.02312−0.00675−0.023120.18697][M]^{-1} = \begin{bmatrix} 0.18697 & -0.02312 & -0.00675 \\ -0.02312 & 0.18959 & -0.02312 \\ -0.00675 & -0.02312 & 0.18697 \end{bmatrix} [C]=55.56 [M]−1=[10.387−1.285−0.375−1.28510.533−1.285−0.375−1.28510.387] pF/m[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 10.387 & -1.285 & -0.375 \\ -1.285 & 10.533 & -1.285 \\ -0.375 & -1.285 & 10.387 \end{bmatrix}\ \text{pF/m}

Step 5: Charge matrix

Balanced phase voltages: V=400/3=230.94V = 400/\sqrt{3} = 230.94 kV rms at 0°,−120°,+120°0°, -120°, +120°.

q1=230.94 (10.387−1.285∠−120°−0.375∠120°)×10−3=2.597∠4.02° μC/m\begin{aligned} q_1 &= 230.94\,(10.387 - 1.285\angle{-120°} - 0.375\angle{120°}) \times 10^{-3} \\ &= 2.597\angle 4.02°\ \mu\text{C/m} \end{aligned}

Similarly:

[Q]=[2.597∠4.02°2.729∠−120°2.597∠115.98°] μC/m (rms)[Q] = \begin{bmatrix} 2.597\angle 4.02° \\ 2.729\angle -120° \\ 2.597\angle 115.98° \end{bmatrix}\ \mu\text{C/m (rms)}

Peak values: 3.67, 3.86 and 3.67 μC/m.

PhaseCharge (rms)Charge (peak)
Outer 12.597 μC/m3.67 μC/m
Centre2.729 μC/m3.86 μC/m
Outer 32.597 μC/m3.67 μC/m

The centre phase carries about 5% more charge than the outer phases, so its conductor surface gradient is the highest.

Answer: [C][C] as above (diagonal 10.39/10.53 pF/m); charges ≈ 2.60, 2.73, 2.60 μC/m rms (3.67, 3.86, 3.67 μC/m peak).

✓ Answer checked by Claude Opus 5.5 (high effort)

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