---
title: "Inductive reactance of vertical 132 kV line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:vertical-132kv-12mm-5m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/vertical-132kv-12mm-5m
site: BCT Engineering
---

# Inductive reactance of vertical 132 kV line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2069 Bhadra, 4 marks
- Video: [T&D 4_1.Inductance Claculation for 3 phase line](https://www.youtube.com/watch?v=wAezzXHXfAw) — FOR STUDENTS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/vertical-132kv-12mm-5m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

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.

## Short answer

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

## Answer

For a transposed line, $L = 2\times10^{-7}\ln(D_{eq}/r')$ per phase and $X_L = 2\pi fL$.

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

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

#### Equivalent spacing

$$
D_{eq} = \sqrt[3]{5 \times 5 \times 10} = 6.2996\ \text{m}
$$

#### Inductance

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

#### Inductive reactance

$$
\begin{aligned}
X_L &= 2\pi \times 50 \times 1.3027\times10^{-3} = 0.4092\ \Omega/\text{km} \\
X_{L,120} &= 0.4092 \times 120 = 49.11\ \Omega
\end{aligned}
$$

**Answer:** $X_L = 49.1\ \Omega$ per phase for 120 km (0.409 Ω/km).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With conductors vertically 5 m apart in a flat vertical arrangement, the spacings are:
   A. 5 m, 5 m and 10 m
   B. 5 m, 5 m, 5 m
   C. 10 m, 10 m, 5 m
   D. 5 m, 10 m, 15 m
   Answer: A. Adjacent spacings are 5 m and the outer two are 10 m apart. (Key point: Vertical spacing)

2. The inductance per phase uses:
   A. L = 2e-7 ln(D_eq / r)
   B. L = 2e-7 ln(D_eq / r')
   C. L = 2e-7 ln(r' / D_eq)
   D. L = 4e-7 ln(D_eq / r')
   Answer: B. Inductance includes internal flux via r' = 0.7788r. (Key point: L formula)

3. The inductive reactance per km is found from:
   A. X_L = 1/(2*pi*f*L)
   B. X_L = L/(2*pi*f)
   C. X_L = 2*pi*f*L
   D. X_L = 2*pi*L
   Answer: C. Inductive reactance is omega times L; multiply by length for the whole line. (Key point: Reactance)

## Related in Power System Analysis I

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- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
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