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Charging VAR of 132 kV equilateral line

Short answer

With symmetrical spacing 6 m and r=1.1 cm, per-phase capacitance = 8.828 nF/km. Three-phase charging VAR = V_L² ω C_n = 48.3 kVAR/km (16.1 kVAR/km per phase).

A 50 Hz, 132 kV, 3-phase overhead transmission line has symmetrical phase spacing of 6 m. The effective radius of the phase conductor is 1.1 cm. Compute the charging VAR per unit length generated by the line.

With symmetrical spacing, Cn=2πε0/ln⁡(D/r)C_n = 2\pi\varepsilon_0/\ln(D/r) per phase, and the three-phase charging VAR is Q=VL2 ωCnQ = V_L^2\,\omega C_n.

Data: D=6D = 6 m, r=1.1r = 1.1 cm =0.011= 0.011 m, 132 kV, 50 Hz.

Capacitance per phase

Cn=55.63×10−12ln⁡(6/0.011)=55.63×10−126.3016=8.828×10−12 F/m=8.828 nF/km\begin{aligned} C_n &= \frac{55.63\times10^{-12}}{\ln(6/0.011)} = \frac{55.63\times10^{-12}}{6.3016} \\ &= 8.828\times10^{-12}\ \text{F/m} = 8.828\ \text{nF/km} \end{aligned}

Capacitive reactance

XC=12π×50×8.828×10−9=3.606×105 Ω-kmX_C = \frac{1}{2\pi \times 50 \times 8.828\times10^{-9}} = 3.606\times10^5\ \Omega\text{-km}

Charging current and VAR

Ic=132 000/3360 564=0.2114 A/kmQ3ϕ=VL2XC=(132×103)2360 564=48 324 VAR/km\begin{aligned} I_c &= \frac{132\,000/\sqrt3}{360\,564} = 0.2114\ \text{A/km} \\ Q_{3\phi} &= \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{360\,564} = 48\,324\ \text{VAR/km} \end{aligned}

Per phase: 48.32/3=16.1148.32/3 = 16.11 kVAR/km.

Answer: charging VAR =48.3= 48.3 kVAR per km for the three-phase line (16.1 kVAR/km per phase).

✓ Answer checked by Claude Opus 5.5 (high effort)

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