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L and C of double circuit, 0.75 cm radius

Short answer

Compute equivalent mutual GMD and self GMD for each phase group using r' for inductance and r for capacitance. L = 0.622 mH/km per phase, C = 18.64 nF/km per phase to neutral.

A 3-Φ double circuit line is arranged as shown in figure below. The conductors are transposed. The radius of each conductor is 0.75 cm. Phase sequence is abc. Find the inductance and capacitance per phase per km. [Figure: vertical double-circuit arrangement; top row a and c' 4 m apart; middle row b and b' 5.5 m apart; bottom row c and a' 4 m apart; vertical spacing 3 m between rows.]

For a transposed double-circuit line, the two conductors of each phase act as a two-conductor group. We find the equivalent mutual GMD DmD_m and the self-GMD DsD_s of each phase group.

   a o------ 4 m ------o c'    --
                                3 m
  b o------- 5.5 m -----o b'   --
                                3 m
   c o------ 4 m ------o a'    --

Data: r=0.75r = 0.75 cm =0.0075= 0.0075 m, r′=0.7788×0.0075=0.005841r' = 0.7788 \times 0.0075 = 0.005841 m. Horizontal offset of b from a is (5.5−4)/2=0.75(5.5-4)/2 = 0.75 m.

Distances

PairDistance (m)
ab, bc, a'b', b'c'32+0.752=3.0923\sqrt{3^2+0.75^2} = 3.0923
ab', a'b, bc', b'c32+4.752=5.6181\sqrt{3^2+4.75^2} = 5.6181
ac, a'c'6
ac', a'c4
aa', cc'62+42=7.2111\sqrt{6^2+4^2} = 7.2111
bb'5.5

Mutual GMD

DAB=(3.0923×5.6181×5.6181×3.0923)1/4=4.1681 mDBC=4.1681 mDCA=(6×4×4×6)1/4=4.8990 mDm=4.1681×4.1681×4.89903=4.3987 m\begin{aligned} D_{AB} &= (3.0923 \times 5.6181 \times 5.6181 \times 3.0923)^{1/4} = 4.1681\ \text{m} \\ D_{BC} &= 4.1681\ \text{m} \\ D_{CA} &= (6 \times 4 \times 4 \times 6)^{1/4} = 4.8990\ \text{m} \\ D_m &= \sqrt[3]{4.1681 \times 4.1681 \times 4.8990} = 4.3987\ \text{m} \end{aligned}

Inductance

Self GMD of each phase using r′r':

DsA=r′Daa′=0.005841×7.2111=0.20523 mDsB=0.005841×5.5=0.17924 m,DsC=0.20523 mDsL=0.20523×0.17924×0.205233=0.19617 mL=2×10−7ln⁡4.39870.19617=2×10−7×3.1101=6.220×10−7 H/m=0.622 mH/km\begin{aligned} D_{sA} &= \sqrt{r'D_{aa'}} = \sqrt{0.005841 \times 7.2111} = 0.20523\ \text{m} \\ D_{sB} &= \sqrt{0.005841 \times 5.5} = 0.17924\ \text{m}, \quad D_{sC} = 0.20523\ \text{m} \\ D_{sL} &= \sqrt[3]{0.20523 \times 0.17924 \times 0.20523} = 0.19617\ \text{m} \\ L &= 2\times10^{-7}\ln\frac{4.3987}{0.19617} = 2\times10^{-7}\times3.1101 \\ &= 6.220\times10^{-7}\ \text{H/m} = 0.622\ \text{mH/km} \end{aligned}

Capacitance

Self GMD using actual radius rr:

DsA=0.0075×7.2111=0.23256 mDsB=0.0075×5.5=0.20310 m,DsC=0.23256 mDsC,eq=0.23256×0.20310×0.232563=0.22229 mC=2πε0ln⁡(4.3987/0.22229)=55.63×10−122.9851=18.64×10−12 F/m=18.64 nF/km\begin{aligned} D_{sA} &= \sqrt{0.0075 \times 7.2111} = 0.23256\ \text{m} \\ D_{sB} &= \sqrt{0.0075 \times 5.5} = 0.20310\ \text{m}, \quad D_{sC} = 0.23256\ \text{m} \\ D_{sC,eq} &= \sqrt[3]{0.23256 \times 0.20310 \times 0.23256} = 0.22229\ \text{m} \\ C &= \frac{2\pi\varepsilon_0}{\ln(4.3987/0.22229)} = \frac{55.63\times10^{-12}}{2.9851} \\ &= 18.64\times10^{-12}\ \text{F/m} = 18.64\ \text{nF/km} \end{aligned}

Answer: L=0.622L = 0.622 mH/km per phase; C=18.64C = 18.64 nF/km (0.01864 µF/km) per phase to neutral.

✓ Answer checked by Claude Opus 5.5 (high effort)

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