Skip to main content

Inductive reactance of vertical 132 kV line

Short answer

Inductive reactance per phase is 49.1 Ω for 120 km (0.409 Ω/km).

A 120 km long 50 Hz, 132 kV, 3-phase overhead line has flat vertical configuration and uses conductors with radii 12 mm. The spacing between adjacent phases is 5 m. Assuming complete transposition, compute the inductive reactance per phase for the line.

For a transposed line, L=2×10−7ln⁡(Deq/r′)L = 2\times10^{-7}\ln(D_{eq}/r') per phase and XL=2πfLX_L = 2\pi fL.

   a o   --
         5 m
   b o   --      D_ab = D_bc = 5 m
         5 m     D_ca = 10 m
   c o   --

Data: r=0.012r = 0.012 m, r′=0.7788×0.012=0.009346r' = 0.7788 \times 0.012 = 0.009346 m, 50 Hz, 120 km.

Equivalent spacing

Deq=5×5×103=6.2996 mD_{eq} = \sqrt[3]{5 \times 5 \times 10} = 6.2996\ \text{m}

Inductance

L=2×10−7ln⁡6.29960.009346=2×10−7×6.5133=1.3027×10−6 H/m=1.3027 mH/kmL120=1.3027×120=156.32 mH\begin{aligned} L &= 2\times10^{-7}\ln\frac{6.2996}{0.009346} = 2\times10^{-7}\times6.5133 \\ &= 1.3027\times10^{-6}\ \text{H/m} = 1.3027\ \text{mH/km} \\ L_{120} &= 1.3027 \times 120 = 156.32\ \text{mH} \end{aligned}

Inductive reactance

XL=2π×50×1.3027×10−3=0.4092 Ω/kmXL,120=0.4092×120=49.11 Ω\begin{aligned} X_L &= 2\pi \times 50 \times 1.3027\times10^{-3} = 0.4092\ \Omega/\text{km} \\ X_{L,120} &= 0.4092 \times 120 = 49.11\ \Omega \end{aligned}

Answer: XL=49.1 ΩX_L = 49.1\ \Omega per phase for 120 km (0.409 Ω/km).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 4 questions →