ACSR 132 kV line: parameters and charging MVA

The line length is not given, so all results are per km. The ACSR conductor is treated as a solid round conductor of the given effective diameter, with GMR $r' = 0.7788r$ (no strand data is given).

Data

• $V_{LL} = 132$ kV, so $V_{ph} = 132/\sqrt3 = 76.21$ kV; f = 50 Hz, $\omega = 314.16$ rad/s
• $r = 39.8/2 = 19.9$ mm $= 1.99$ cm; D = 8 m = 800 cm (equilateral)
• Copper-equivalent area $A = 1.5$ cm²; $\rho = 1.73\times10^{-6}\ \Omega$-cm

(i) Line parameters

Resistance (l = 1 km = $10^5$ cm):

$$R = \frac{\rho l}{A} = \frac{1.73\times10^{-6}\times10^5}{1.5} = 0.1153\ \Omega/\text{km}$$

Inductance: $r' = 0.7788\times1.99 = 1.5498$ cm.

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{D}{r'} = 2\times10^{-7}\ln\frac{800}{1.5498}\\ &= 2\times10^{-7}\times6.2465 = 1.2493\ \text{mH/km}\\ X_L &= 314.16\times1.2493\times10^{-3} = 0.3925\ \Omega/\text{km} \end{aligned}$$

Capacitance to neutral:

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D/r)} = \frac{5.563\times10^{-11}}{\ln(800/1.99)} = \frac{5.563\times10^{-11}}{5.9965}\\ &= 9.277\times10^{-12}\ \text{F/m} = 9.277\times10^{-3}\ \mu\text{F/km}\\ X_C &= \frac{1}{\omega C} = \frac{1}{314.16\times9.277\times10^{-9}} = 3.431\times10^5\ \Omega\cdot\text{km} \end{aligned}$$

(ii) Charging current and charging MVA

$$\begin{aligned} I_c &= \frac{V_{ph}}{X_C} = 76210\times314.16\times9.277\times10^{-9} = 0.2221\ \text{A/km}\\ Q_c &= 3V_{ph}I_c = \sqrt3\times132\times10^3\times0.2221\\ &= 50.78\ \text{kVAR/km} = 0.0508\ \text{MVAR/km} \end{aligned}$$

Answer: R = 0.115 Ω/km, L = 1.249 mH/km, C = 0.00928 μF/km; charging current 0.222 A/km and charging MVA 0.0508 MVAR/km. For example, a 100 km line would draw 22.2 A and 5.08 MVAR.