--- title: "Inductance of 420 kV line, transposed and untransposed" subject: "High Voltage Engineering" chapter: 5 key: "semester_8/high_voltage_engineering:inductance-420kv-transposed" source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/inductance-420kv-transposed site: BCT Engineering --- # Inductance of 420 kV line, transposed and untransposed High Voltage Engineering · Chapter 5: High Stress Electric Fields - Asked in: 2077 Chaitra, 8 marks - Video: [Inductance of 3phase line with unsymmetrical spacing but transposed](https://www.youtube.com/watch?v=9xgcTCH8Weo) — Yashwin Creations - 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-420kv-transposed - Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-5 ## Question For the 3-phase 420 kV horizontal line having conductor diameter of 2 × 3.18 cm and bundle spacing 45.72 cm. The ground clearance is 15 m and phase spacing is 11 m. Calculate the inductance in both un-transposed and transposed condition. ## Short answer Equivalent radius is 8.526 cm. Untransposed: self-inductance 1.1726 mH/km, mutuals 0.2133 and 0.1051 mH/km. Transposed: L_s=1.1726, L_m=0.1772 mH/km, giving positive-sequence inductance L1 = 0.995 mH/km. ## Answer For EHV lines, the inductance matrix is $[L] = \dfrac{\mu_0}{2\pi}[M] = 0.2\,[M]$ mH/km. Here $[M]$ is Maxwell's coefficient matrix (image method, equivalent bundle radius; internal flux neglected). Transposition replaces it by averaged self and mutual terms. #### Data - 420 kV, flat horizontal, $S = 11$ m, $H = 15$ m - Bundle $2 \times 3.18$ cm diameter, so $r = 1.59$ cm; $B = 45.72$ cm - $\mu_0/2\pi = 2 \times 10^{-7}$ H/m $= 0.2$ mH/km #### Step 1: Equivalent radius $$ r_{eq} = \sqrt{rB} = \sqrt{1.59 \times 45.72} = 8.526\ \text{cm} $$ #### Step 2: Maxwell coefficients $$ \begin{aligned} M_{ii} &= \ln\frac{2H}{r_{eq}} = \ln\frac{30}{0.08526} = 5.8632 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11^2 + 30^2}}{11} = 1.0664 \\ M_{13} &= \ln\frac{\sqrt{22^2 + 30^2}}{22} = 0.5253 \end{aligned} $$ #### Untransposed line $$ [L] = 0.2\begin{bmatrix} 5.8632 & 1.0664 & 0.5253 \\ 1.0664 & 5.8632 & 1.0664 \\ 0.5253 & 1.0664 & 5.8632 \end{bmatrix} = \begin{bmatrix} 1.1726 & 0.2133 & 0.1051 \\ 0.2133 & 1.1726 & 0.2133 \\ 0.1051 & 0.2133 & 1.1726 \end{bmatrix}\ \text{mH/km} $$ The mutual terms are not all equal ($L_{12} = L_{23} \ne L_{13}$). So with balanced currents the voltage drops of the three phases are unequal, and the line is unbalanced. #### Transposed line Average the self and mutual terms: $$ \begin{aligned} L_s &= 0.2 \times 5.8632 = 1.1726\ \text{mH/km} \\ M_m &= \frac{1.0664 + 1.0664 + 0.5253}{3} = 0.8860 \\ L_m &= 0.2 \times 0.8860 = 0.1772\ \text{mH/km} \end{aligned} $$ $$ [L_t] = \begin{bmatrix} 1.1726 & 0.1772 & 0.1772 \\ 0.1772 & 1.1726 & 0.1772 \\ 0.1772 & 0.1772 & 1.1726 \end{bmatrix}\ \text{mH/km} $$ Sequence inductances: $$ \begin{aligned} L_1 = L_2 &= L_s - L_m = 1.1726 - 0.1772 = 0.9954\ \text{mH/km} \\ L_0 &= L_s + 2L_m = 1.1726 + 0.3544 = 1.5271\ \text{mH/km} \end{aligned} $$ #### Check with the GMD formula $D_{eq} = \sqrt[3]{11 \times 11 \times 22} = 13.86$ m, so $$ L_1 = 0.2\ln\frac{13.86}{0.08526} = 1.018\ \text{mH/km} $$ This is close to 0.995 mH/km; the small difference is due to the effect of the ground images. **Answer: untransposed $L_{ii} = 1.1726$, $L_{12} = 0.2133$, $L_{13} = 0.1051$ mH/km; transposed $L_s = 1.1726$, $L_m = 0.1772$ mH/km, positive-sequence $L_1 = 0.995$ mH/km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the inductance matrix of the EHV line obtained? A. [L] = 0.2 [M]^-1 mH/km B. [L] = 0.2 [M] mH/km, with [M] from the image method C. [L] = 2πε0 [M] mH/km D. [L] = [M] / 0.2 mH/km Answer: B. [L] = (μ0/2π)[M] = 0.2[M] mH/km. (Key point: Inductance matrix) 2. How is the mutual inductance Lm of the transposed line found? A. Use only M12 B. Use only M13 C. Average the mutual coefficients (1.0664, 1.0664, 0.5253) and multiply by 0.2 D. Add the three mutual coefficients without dividing Answer: C. Transposition replaces the mutuals by their average, giving Lm = 0.1772 mH/km. (Key point: Transposed mutual term) 3. What is the positive-sequence inductance of the transposed line? A. L1 = Ls - Lm = 0.995 mH/km B. L1 = Ls + 2Lm = 1.527 mH/km C. L1 = Ls + Lm = 1.35 mH/km D. L1 = Lm = 0.177 mH/km Answer: A. L1 = 1.1726 - 0.1772 = 0.9954 mH/km; Ls + 2Lm is the zero-sequence value. 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