--- title: "Capacitance of double circuit line, 6 mm radius" subject: "Power System Analysis I" chapter: 4 key: "semester_4/power_system_analysis_i:double-circuit-capacitance-6mm" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/double-circuit-capacitance-6mm site: BCT Engineering --- # Capacitance of double circuit line, 6 mm radius Power System Analysis I · Chapter 4: Line Parameter Calculations - Asked in: 2077 Chaitra, 10 marks - Video: [44. Capacitance of Double Circuit Line in Hexagonal Configuration](https://www.youtube.com/watch?v=NuFnQ1AL3YE) — Bright Engineers - 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/double-circuit-capacitance-6mm - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4 ## Question Find the capacitance per phase per km of the double circuit three phase line shown in figure below. The line is completely transposed and operates at a frequency of 50 Hz. Radius of each conductor is 6 mm. Also compute the capacitive reactance per phase and total VAR generated by the line of 100 km and operating at 132 kV. [Figure: vertical double-circuit arrangement; left column top to bottom a, b, c; right column top to bottom c', b', a'; vertical spacing 3 m between rows; horizontal distances a–c' = 5 m, b–b' = 6 m, c–a' = 5 m.] ## Short answer Treat each phase as a two-conductor bundle. Equivalent mutual GMD =4.7082 m, self GMR =0.20717 m. Capacitance C =2*pi*epsilon0/ln(4.7082/0.20717)=17.81 nF/km. For 100 km, C_total=1.781 µF, Xc=1/(2π*50*1.781e-6)=1787 Ω. At 132 kV, total charging VAR = (132e3)^2/1787 =9.75 MVAR. ## Answer For a transposed double-circuit line, the two conductors of a phase are treated as a two-conductor bundle. The capacitance per phase is $C = 2\pi\varepsilon_0/\ln(GMD/GMR)$, where GMD is the equivalent mutual distance between phase groups and GMR is the self-GMD of each phase group (using actual radius $r$). ```text a o------- 5 m -------o c' -- 3 m b o-------- 6 m --------o b' -- 3 m c o------- 5 m -------o a' -- ``` **Data:** $r = 6$ mm $= 0.006$ m, vertical spacing 3 m, $f = 50$ Hz, 132 kV, 100 km. #### Distances (from geometry) Horizontal offset between a and b is $(6-5)/2 = 0.5$ m. | Pair | Distance (m) | |---|---| | ab, bc, a'b', b'c' | $\sqrt{3^2+0.5^2} = 3.0414$ | | ab', a'b, bc', b'c | $\sqrt{3^2+5.5^2} = 6.2650$ | | ac, a'c' | 6 | | ac', a'c | 5 | | aa', cc' | $\sqrt{6^2+5^2} = 7.8102$ | | bb' | 6 | #### Mutual GMD $$ \begin{aligned} D_{AB} &= (D_{ab}D_{ab'}D_{a'b}D_{a'b'})^{1/4} = (3.0414^2 \times 6.2650^2)^{1/4} = 4.3651\ \text{m} \\ D_{BC} &= 4.3651\ \text{m} \quad (\text{by symmetry}) \\ D_{CA} &= (6 \times 5 \times 5 \times 6)^{1/4} = 5.4772\ \text{m} \\ D_m &= \sqrt[3]{4.3651 \times 4.3651 \times 5.4772} = 4.7082\ \text{m} \end{aligned} $$ #### Self GMD of each phase (for capacitance) $$ \begin{aligned} D_{sA} &= \sqrt{r \cdot D_{aa'}} = \sqrt{0.006 \times 7.8102} = 0.21648\ \text{m} \\ D_{sB} &= \sqrt{0.006 \times 6} = 0.18974\ \text{m} \\ D_{sC} &= D_{sA} = 0.21648\ \text{m} \\ D_s &= \sqrt[3]{0.21648 \times 0.18974 \times 0.21648} = 0.20717\ \text{m} \end{aligned} $$ #### Capacitance per phase per km $$ \begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_s)} = \frac{55.63\times10^{-12}}{\ln(4.7082/0.20717)} = \frac{55.63\times10^{-12}}{3.1235} \\ &= 17.81\times10^{-12}\ \text{F/m} = 17.81\ \text{nF/km} \end{aligned} $$ #### Capacitive reactance for 100 km $$ \begin{aligned} C_{100} &= 17.81 \times 100 = 1781\ \text{nF} = 1.781\ \mu\text{F} \\ X_C &= \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 1.781\times10^{-6}} = 1787.2\ \Omega \end{aligned} $$ #### Total VAR generated Phase voltage $V_p = 132/\sqrt{3} = 76.21$ kV. $$ \begin{aligned} I_c &= \frac{V_p}{X_C} = \frac{76\,210}{1787.2} = 42.64\ \text{A} \\ Q &= 3V_p I_c = \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{1787.2} = 9.749\ \text{MVAR} \end{aligned} $$ **Answer:** $C = 17.81$ nF/km per phase; $X_C = 1787.2\ \Omega$ per phase (100 km); total charging = 9.75 MVAR. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a transposed double-circuit line treated for capacitance per phase? A. Each phase's two conductors are treated as a two-conductor group with a mutual GMD and a self-GMD B. Only one circuit is used and the other is ignored C. The two circuits are added in series D. Each conductor is treated as an independent phase Answer: A. The two conductors of a phase act as a bundle-like group, so C = 2*pi*eps0/ln(GMD/GMR) with group GMDs. (Key point: Double circuit as bundle) 2. For phase A with conductors a and a' separated by 7.8102 m, which expression gives the self-GMD used for capacitance? A. sqrt(0.7788 r x 7.8102) B. sqrt(r x 7.8102) C. r x 7.8102 D. (r x 7.8102)^(1/4) Answer: B. For capacitance the actual radius r is used (no internal flux), so D_sA = sqrt(r x D_aa'). (Key point: Self GMD for capacitance) 3. How is D_AB (between phase groups A and B) obtained? A. Arithmetic mean of the four distances B. Fourth root of the product of the four distances ab, ab', a'b, a'b' C. Smallest of the four distances D. Square root of the product of ab and a'b' Answer: B. Each of the 2 conductors of A to each of the 2 of B gives 4 distances, and D_AB is their geometric mean (4th root of product). (Key point: Mutual GMD between phases) 4. For the 100 km line at 132 kV with X_C = 1787.2 ohm, the total three-phase VAR generated is found from: A. V_L^2 / X_C giving about 9.75 MVAR B. V_p / X_C giving 42.6 VAR C. 3 x V_L^2 / X_C giving about 29 MVAR D. V_p x X_C giving about 136 MVAR Answer: A. Q = 3 V_p I_c = V_L^2/X_C = (132e3)^2/1787.2 = 9.749 MVAR. 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