Inductance and capacitance of 4-conductor bundle line

Each phase is a 4-conductor square bundle, so we find the bundle GMR (self-GMD) and the mutual GMD between phase centres, then use the standard formulas for a transposed line.

   o o        o o        o o
   o o        o o        o o
    A    6 m   B    6 m   C
   |<------- 12 m ------->|
 square side d = 0.45 m, r = 12 mm

Data: $r = 0.012$ m, $d = 0.45$ m, $D_{AB} = D_{BC} = 6$ m, $D_{CA} = 12$ m, length 200 km.

Mutual GMD

$$D_m = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}$$

Self GMD of a 4-conductor bundle

For a square bundle, $D_s = (r_x \cdot d \cdot d \cdot \sqrt{2}d)^{1/4} = 1.09\,(r_x d^3)^{1/4}$.

• Inductance: $r' = 0.7788 \times 0.012 = 0.009346$ m

$$D_{sL} = 1.09\,(0.009346 \times 0.45^3)^{1/4} = 0.18629\ \text{m}$$

• Capacitance (uses actual radius $r$):

$$D_{sC} = 1.09\,(0.012 \times 0.45^3)^{1/4} = 0.19830\ \text{m}$$

Inductance per phase

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{D_m}{D_{sL}} = 2\times10^{-7}\ln\frac{7.5595}{0.18629} \\ &= 2\times10^{-7}\times 3.7033 = 7.4065\times10^{-7}\ \text{H/m} \\ &= 0.7407\ \text{mH/km} \end{aligned}$$

For 200 km: $L = 0.7407 \times 200 = 148.13$ mH.

Capacitance per phase (to neutral)

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_{sC})} = \frac{2\pi \times 8.854\times10^{-12}}{\ln(7.5595/0.19830)} \\ &= \frac{55.63\times10^{-12}}{3.6408} = 15.28\times10^{-12}\ \text{F/m} \\ &= 15.28\ \text{nF/km} \end{aligned}$$

For 200 km: $C = 15.28 \times 200 = 3056$ nF $= 3.056\ \mu$F.

Answer: $L = 0.7407$ mH/km per phase (148.13 mH for 200 km); $C = 15.28$ nF/km per phase to neutral (3.056 µF for 200 km).