--- title: "Inductance matrix of 500 kV bundled line" subject: "High Voltage Engineering" chapter: 5 key: "semester_8/high_voltage_engineering:inductance-matrix-500kv" source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/inductance-matrix-500kv site: BCT Engineering --- # Inductance matrix of 500 kV bundled line High Voltage Engineering · Chapter 5: High Stress Electric Fields - Asked in: 2073 Magh, 8 marks - Video: [Inductance Calculation of Bundled Conductors](https://www.youtube.com/watch?v=x9VKEP5CMLM) — Moving Electrons - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/inductance-matrix-500kv - Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-5 ## Question A 500 kV, three phase transmission line has flat horizontal configuration with two sub-conductors per bundle. The inter-phase spacing of the line is 15 m. The radius of each sub-conductor in a bundle is 5 cm and the separation between sub-conductors in a bundle is 30 cm. Calculate the inductance matrix if the average ground clearance of the line is 20 m. ## Short answer Self-inductance 1.158 mH/km, adjacent mutual 0.209 mH/km, outer-outer mutual 0.102 mH/km. Matrix: [[1.158, 0.209, 0.102], [0.209, 1.158, 0.209], [0.102, 0.209, 1.158]] mH/km. ## Answer For an EHV line, the inductance matrix is $[L] = \dfrac{\mu_0}{2\pi}[M] = 0.2\,[M]$ mH/km. $[M]$ is Maxwell's coefficient matrix from the image method, using the equivalent radius of each bundle (internal flux linkage neglected). #### Data - 500 kV, flat horizontal, $S = 15$ m, $H = 20$ m - $N = 2$, sub-conductor radius $r = 5$ cm, bundle spacing $B = 30$ cm - $\mu_0/2\pi = 2 \times 10^{-7}$ H/m $= 0.2$ mH/km ```text 1 2 3 o--- 15 m--o--- 15 m--o | H = 20 m //////////////////////////////// ``` #### Step 1: Equivalent radius $$ r_{eq} = \sqrt{rB} = \sqrt{5 \times 30} = 12.247\ \text{cm} = 0.12247\ \text{m} $$ #### Step 2: Maxwell coefficients ($2H = 40$ m) $$ \begin{aligned} M_{ii} &= \ln\frac{40}{0.12247} = \ln 326.6 = 5.7887 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{15^2 + 40^2}}{15} = \ln\frac{42.72}{15} = 1.0466 \\ M_{13} &= \ln\frac{\sqrt{30^2 + 40^2}}{30} = \ln\frac{50}{30} = 0.5108 \end{aligned} $$ #### Step 3: Inductance matrix (untransposed) $$ [L] = 0.2\begin{bmatrix} 5.7887 & 1.0466 & 0.5108 \\ 1.0466 & 5.7887 & 1.0466 \\ 0.5108 & 1.0466 & 5.7887 \end{bmatrix} = \begin{bmatrix} 1.1577 & 0.2093 & 0.1022 \\ 0.2093 & 1.1577 & 0.2093 \\ 0.1022 & 0.2093 & 1.1577 \end{bmatrix}\ \text{mH/km} $$ #### Step 4: If transposed (for comparison) $$ \begin{aligned} L_s &= 1.1577\ \text{mH/km} \\ L_m &= 0.2 \times \frac{1.0466 + 1.0466 + 0.5108}{3} = 0.1736\ \text{mH/km} \\ L_1 &= L_s - L_m = 0.9841\ \text{mH/km} \\ L_0 &= L_s + 2L_m = 1.5049\ \text{mH/km} \end{aligned} $$ Positive-sequence reactance at 50 Hz: $X_1 = 2\pi \times 50 \times 0.9841 \times 10^{-3} = 0.309\ \Omega$/km. | Element | Value (mH/km) | |---|---| | Self, $L_{11} = L_{22} = L_{33}$ | 1.1577 | | Mutual, adjacent ($L_{12}$, $L_{23}$) | 0.2093 | | Mutual, outer–outer ($L_{13}$) | 0.1022 | **Answer: $[L] = \begin{bmatrix} 1.158 & 0.209 & 0.102 \\ 0.209 & 1.158 & 0.209 \\ 0.102 & 0.209 & 1.158 \end{bmatrix}$ mH/km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is r_eq for r = 5 cm and B = 30 cm? A. 35 cm B. 15 cm C. sqrt(5 + 30) = 5.9 cm D. sqrt(5 x 30) = 12.247 cm Answer: D. r_eq = sqrt(rB) = 12.247 cm. (Key point: Equivalent radius) 2. How are self and mutual coefficients found with images? A. M_ii = ln(2H/r_eq), M_ij = ln(D_ij'/D_ij) where D' is the distance to the image B. M_ii = ln(r_eq/2H), M_ij = ln(D_ij/D_ij') C. M_ii = ln(H/r_eq), M_ij = ln(D_ij) D. M_ii = 2H/r_eq, M_ij = D'/D Answer: A. Image-method coefficients use the logarithm of image distance over actual distance. (Key point: Maxwell coefficient) 3. Why is L13 (0.1022) smaller than L12 (0.2093)? A. They carry a smaller current B. They are at different heights C. Conductors 1 and 3 are farther apart, so coupling is weaker D. The bundle spacing differs Answer: C. Mutual inductance falls as spacing rises (30 m vs 15 m). 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