--- title: "330 kV quad-bundle line: inductance and charging" subject: "Power System Analysis I" chapter: 4 key: "semester_4/power_system_analysis_i:quad-bundle-330kv-6m" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/quad-bundle-330kv-6m site: BCT Engineering --- # 330 kV quad-bundle line: inductance and charging Power System Analysis I · Chapter 4: Line Parameter Calculations - Asked in: 2072 Magh, 10 marks - Video: [PS15 Bundled Conductors](https://www.youtube.com/watch?v=l982pc8h0OU) — Lectures in Electrical Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/quad-bundle-330kv-6m - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4 ## Question A 3 phase overhead line has 4 sub-conductors per phase separated from each other by 45 cm and placed at the vertices of a square. The phases have flat horizontal configuration with centre to centre distance between adjacent phases equalling 6 m and between far end phases equalling 12 m. Assuming complete transposition determine the inductance per phase, charging current per phase per km and total charging VAR of the line. The line is operated at 330 kV and 50 Hz. ## Short answer Assuming sub-conductor radius 12 mm, with D_sL = 0.18629 m, D_sC = 0.19830 m, mutual GMD = 7.5595 m, per phase per km: inductance 0.7407 mH/km, charging current 0.915 A/km, total three-phase charging reactive power 0.523 MVAR/km. ## Answer Each phase is a square bundle of four sub-conductors with side $d = 0.45$ m. The sub-conductor radius is not given; it is **assumed to be 12 mm** (the usual value in this IOE problem). ```text o o o o o o o o o o o o A 6 m B 6 m C |<------- 12 m ------->| ``` **Data:** $r = 0.012$ m, $r' = 0.7788 \times 0.012 = 0.009346$ m, $d = 0.45$ m, 330 kV, 50 Hz. #### GMDs $$ \begin{aligned} D_m &= \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m} \\ D_{sL} &= 1.09\sqrt[4]{r'd^3} = 1.09\sqrt[4]{0.009346 \times 0.45^3} = 0.18629\ \text{m} \\ D_{sC} &= 1.09\sqrt[4]{rd^3} = 1.09\sqrt[4]{0.012 \times 0.45^3} = 0.19830\ \text{m} \end{aligned} $$ #### Inductance per phase $$ L = 2\times10^{-7}\ln\frac{7.5595}{0.18629} = 2\times10^{-7}\times3.7033 = 0.7407\ \text{mH/km} $$ #### Charging current per phase per km $$ \begin{aligned} C &= \frac{55.63\times10^{-12}}{\ln(7.5595/0.19830)} = \frac{55.63\times10^{-12}}{3.6408} = 15.28\ \text{nF/km} \\ X_C &= \frac{1}{2\pi \times 50 \times 15.28\times10^{-9}} = 2.083\times10^5\ \Omega\text{-km} \\ I_c &= \frac{330\,000/\sqrt3}{208\,316} = \frac{190\,526}{208\,316} = 0.9146\ \text{A/km} \end{aligned} $$ #### Total charging VAR $$ Q_c = \frac{V_L^2}{X_C} = \frac{(330\times10^3)^2}{208\,316} = 0.5228\ \text{MVAR per km} $$ The line length is not given, so the total is per km; for a line of length $\ell$ km, $Q = 0.5228\,\ell$ MVAR (e.g. 104.6 MVAR for 200 km). **Answer:** $L = 0.7407$ mH/km per phase, $I_c = 0.915$ A/km per phase, $Q_c = 0.523$ MVAR/km (all three phases), with $r = 12$ mm assumed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a flat configuration with 6 m spacing and 12 m between far phases, D_m is: A. Cube root of (6 x 6 x 12) = 7.56 m B. 8 m (arithmetic mean) C. 6 m D. 12 m Answer: A. D_m = (6 x 6 x 12)^(1/3) = 7.5595 m. (Key point: Mutual GMD) 2. Which expression gives L per phase? A. 2e-7 ln(7.5595/0.18629), using D_sL = 1.09 (r' d^3)^(1/4) B. 2e-7 ln(7.5595/0.19830) C. 2e-7 ln(7.5595/0.012) D. 4e-7 ln(7.5595/0.18629) Answer: A. Inductance uses r' in D_sL, giving L = 0.7407 mH/km. (Key point: Inductance per phase) 3. Line length is not given. How is total charging VAR expressed? A. 0.5228 MVAR per km (all three phases) using V_L^2/X_C, times length for a given line B. 0.5228 MVAR in total C. 0.174 MVAR/km per phase only D. 9.146 MVAR Answer: A. Q = V_L^2/X_C = 0.5228 MVAR/km; multiply by the length, e.g. 104.6 MVAR for 200 km. 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