---
title: "400 kV quadruple bundle line, 20 m spacing"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:quad-bundle-400kv-20m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/quad-bundle-400kv-20m
site: BCT Engineering
---

# 400 kV quadruple bundle line, 20 m spacing

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2073 Bhadra, 10 marks
- 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/quad-bundle-400kv-20m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Figure below shows a quadruple-conductor circuit of a single-circuit, three-phase, 400 kV, 50 Hz line with a horizontal spacing of 20 m. Each sub-conductor of the bundle has a diameter of 40 mm and spacing between the sub-conductors is 0.5 m. Each phase group shares the total current and charge equally and the line is completely transposed. Determine the inductive reactance, capacitive reactance, charging current and charging VAR per phase per km of the line. [Figure: horizontal single-circuit line; each phase A, B, C is a square bundle of four sub-conductors with 0.5 m sides; centres of adjacent phases 20 m apart.]

## Short answer

Using bundle self GMD D_sL = 0.22907 m, D_sC = 0.24384 m and mutual GMD = 25.198 m, per phase per km: inductive reactance X_L = 0.2953 Ω/km, capacitive reactance X_C = 2.654×10⁵ Ω·km, charging current I_c = 0.870 A/km, charging reactive power Q_c = 201.0 kVAR/km.

## Answer

Each phase is a square bundle of four sub-conductors. We find the bundle self-GMD for inductance and capacitance, the mutual GMD, and then the per-km quantities.

```text
   o o         o o         o o
   o o         o o         o o
    A   20 m    B   20 m    C
   d = 0.5 m, diameter 40 mm
```

**Data:** $r = 20$ mm $= 0.02$ m, $r' = 0.7788 \times 0.02 = 0.015576$ m, $d = 0.5$ m, $D = 20, 20, 40$ m, 400 kV, 50 Hz.

#### GMDs

$$
\begin{aligned}
D_m &= \sqrt[3]{20 \times 20 \times 40} = 25.198\ \text{m} \\
D_{sL} &= 1.09\sqrt[4]{r'd^3} = 1.09\sqrt[4]{0.015576 \times 0.5^3} = 0.22907\ \text{m} \\
D_{sC} &= 1.09\sqrt[4]{rd^3} = 1.09\sqrt[4]{0.02 \times 0.5^3} = 0.24384\ \text{m}
\end{aligned}
$$

#### Inductive reactance per phase per km

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{25.198}{0.22907} = 2\times10^{-7}\times4.7005 = 0.9401\ \text{mH/km} \\
X_L &= 2\pi \times 50 \times 0.9401\times10^{-3} = 0.2953\ \Omega/\text{km}
\end{aligned}
$$

#### Capacitive reactance per phase per km

$$
\begin{aligned}
C &= \frac{55.63\times10^{-12}}{\ln(25.198/0.24384)} = \frac{55.63\times10^{-12}}{4.6380} = 11.995\ \text{nF/km} \\
X_C &= \frac{1}{2\pi \times 50 \times 11.995\times10^{-9}} = 2.654\times10^5\ \Omega\text{-km}
\end{aligned}
$$

#### Charging current and VAR

$$
\begin{aligned}
V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\
I_c &= \frac{230\,940}{265\,376} = 0.8702\ \text{A/km} \\
Q_c &= V_p I_c = 230.94 \times 0.8702 = 200.97\ \text{kVAR/km per phase}
\end{aligned}
$$

**Answer:** $X_L = 0.2953\ \Omega$/km, $X_C = 2.654\times10^5\ \Omega$-km, $I_c = 0.870$ A/km, $Q_c = 201.0$ kVAR/km per phase (603 kVAR/km for three phases).

## Check yourself

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

1. For a square four-conductor bundle with side d, D_sC is:
   A. 1.09 x 4th root of (r d^3)
   B. sqrt(r d)
   C. cube root of (r d^2)
   D. 4th root of (r d)
   Answer: A. For a square bundle D_s = 1.09 (r d^3)^(1/4); use r' for inductance and r for capacitance. (Key point: 4-bundle self-GMD)

2. Each sub-conductor has a 40 mm diameter. Which radius goes into D_sC?
   A. 40 mm
   B. 20 mm
   C. 15.576 mm
   D. 10 mm
   Answer: B. r = 20 mm = 0.02 m for capacitance; r' = 0.7788 x 20 = 15.576 mm is used for inductance. (Key point: Radius from diameter)

3. For spacings 20, 20 and 40 m the D_m is:
   A. Cube root of (20 x 20 x 40) = 25.2 m
   B. Mean of 20, 20, 40 = 26.7 m
   C. 20 m
   D. 40 m
   Answer: A. D_m = (D_AB D_BC D_CA)^(1/3) = 25.198 m. (Key point: Mutual GMD)

4. X_C = 2.654e5 ohm-km. How is the charging current found?
   A. I_c = (400 kV/sqrt3) / 2.654e5 = 0.870 A/km
   B. I_c = 400 kV / 2.654e5 = 1.51 A/km
   C. I_c = 2.654e5 x 230.94 kV
   D. I_c = 0.2953 A/km
   Answer: A. I_c = V_p/X_C = 230,940/265,376 = 0.870 A/km, and Q_c = V_p I_c = 201 kVAR/km per phase. (Key point: Capacitive reactance unit and use)

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