Skip to main content

Charging current of 132 kV line, 6 m equilateral

Short answer

Capacitance per phase is 9.577 nF/km. Charging current per phase is 0.229 A/km.

Determine the charging capacitance and charging current per phase in a 50 Hz, 132 kV, 3-phase overhead line spaced at 6 m from each other and using 18 mm radius conductors per phase.

For equilateral spacing, the capacitance per phase to neutral is Cn=2πε0/ln⁡(D/r)C_n = 2\pi\varepsilon_0/\ln(D/r).

Data: D=6D = 6 m, r=18r = 18 mm =0.018= 0.018 m, 132 kV, 50 Hz.

Capacitance per phase

Cn=2π×8.854×10−12ln⁡(6/0.018)=55.63×10−125.8091=9.577×10−12 F/m=9.577 nF/km\begin{aligned} C_n &= \frac{2\pi \times 8.854\times10^{-12}}{\ln(6/0.018)} = \frac{55.63\times10^{-12}}{5.8091} \\ &= 9.577\times10^{-12}\ \text{F/m} = 9.577\ \text{nF/km} \end{aligned}

Charging current per phase

XC=12πfCn=12π×50×9.577×10−9=3.324×105 Ω-kmVp=132 0003=76 210 VIc=VpXC=76 210332 386=0.2293 A/km\begin{aligned} X_C &= \frac{1}{2\pi f C_n} = \frac{1}{2\pi \times 50 \times 9.577\times10^{-9}} = 3.324\times10^5\ \Omega\text{-km} \\ V_p &= \frac{132\,000}{\sqrt3} = 76\,210\ \text{V} \\ I_c &= \frac{V_p}{X_C} = \frac{76\,210}{332\,386} = 0.2293\ \text{A/km} \end{aligned}

(The charging reactive power for all 3 phases is VL2/XC=52.4V_L^2/X_C = 52.4 kVAR/km.)

Answer: Cn=9.577C_n = 9.577 nF/km (0.00958 µF/km) per phase; Ic=0.229I_c = 0.229 A/km per phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 4 questions →