---
title: "Line power loss, 138.5∠8° kV and 132 kV"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:power-loss-138-5kv-132kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/power-loss-138-5kv-132kv
site: BCT Engineering
---

# Line power loss, 138.5∠8° kV and 132 kV

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2071 Magh, 6 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/power-loss-138-5kv-132kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

For a 3-phase overhead line, the following data is available: VS,L = 138.5∠8° kV, VR,L = 132∠0° kV, A = 0.983∠0.68°, B = 36.5∠71.5° Ω. Determine the transmission line active and reactive power loss.

## Short answer

Active power loss is 9.12 MW and reactive loss is −9.60 MVAr, meaning the line is a net source of 9.60 MVAr. This occurs because the line's shunt capacitive generation exceeds its series inductive absorption.

## Answer

Use the power-circle equations (with $D = A$). Line voltages in kV give three-phase MW and MVAr.

Data: $|V_S| = 138.5$ kV, $|V_R| = 132$ kV, $\delta = 8^\circ$, $|A| = 0.983$, $\alpha = 0.68^\circ$, $|B| = 36.5\ \Omega$, $\beta = 71.5^\circ$.

#### Constants
$$
\begin{aligned}
\frac{|V_S||V_R|}{|B|} &= \frac{138.5\times132}{36.5} = 500.88 \\
\frac{|A||V_R|^2}{|B|} &= \frac{0.983\times132^2}{36.5} = 469.25 \\
\frac{|A||V_S|^2}{|B|} &= \frac{0.983\times138.5^2}{36.5} = 516.61
\end{aligned}
$$
Angles: $\beta-\delta = 63.5^\circ$, $\beta-\alpha = 70.82^\circ$, $\beta+\delta = 79.5^\circ$.

#### Receiving end
$$
\begin{aligned}
P_R &= 500.88\cos63.5^\circ - 469.25\cos70.82^\circ = 223.49 - 154.17 = 69.32\ \text{MW} \\
Q_R &= 500.88\sin63.5^\circ - 469.25\sin70.82^\circ = 448.25 - 443.21 = 5.05\ \text{MVAr}
\end{aligned}
$$

#### Sending end
$$
\begin{aligned}
P_S &= 516.61\cos70.82^\circ - 500.88\cos79.5^\circ = 169.72 - 91.28 = 78.45\ \text{MW} \\
Q_S &= 516.61\sin70.82^\circ - 500.88\sin79.5^\circ = 487.93 - 492.49 = -4.56\ \text{MVAr}
\end{aligned}
$$

#### Losses
$$
\begin{aligned}
P_{loss} &= P_S - P_R = 78.45 - 69.32 = 9.12\ \text{MW} \\
Q_{loss} &= Q_S - Q_R = -4.56 - 5.05 = -9.60\ \text{MVAr}
\end{aligned}
$$
The negative reactive "loss" means the line generates about 9.6 MVAr net (its shunt charging exceeds the $I^2X$ absorbed in its series reactance).

**Answer: Active power loss ≈ 9.12 MW; reactive power loss ≈ −9.60 MVAr (line is a net source of 9.60 MVAr).**

## Check yourself

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

1. How are active and reactive losses found?
   A. P_R - P_S
   B. P_S x P_R
   C. Q_S + Q_R
   D. P_loss = P_S - P_R, Q_loss = Q_S - Q_R
   Answer: D. Compute both ends from power-circle equations and subtract. (Key point: Loss definition)

2. Which formula gives P_R with delta = 8 degrees?
   A. |V_S|^2 cos(delta)
   B. |V_S||V_R|cos(beta-delta)/|B| - |A||V_R|^2 cos(beta-alpha)/|B|
   C. |V_S||V_R|cos(beta+delta)/|B|
   D. |A||V_S|^2/|B|
   Answer: B. That gives 223.49 - 154.17 = 69.32 MW. (Key point: Receiving-end equation)

3. Which formula gives P_S?
   A. |V_S||V_R|cos(beta-delta)
   B. |V_R|^2 cos(alpha)
   C. |A||V_S|^2 cos(beta-alpha)/|B| - |V_S||V_R| cos(beta+delta)/|B|
   D. |A||V_R|^2 cos(beta-delta)
   Answer: C. Gives 169.72 - 91.28 = 78.45 MW. (Key point: Sending-end equation)

4. What does Q_loss = -9.60 MVAr mean?
   A. Line absorbs 9.60 MVAr
   B. Line is a net source of 9.60 MVAr since charging exceeds I^2X
   C. Load is leading
   D. Error in computation
   Answer: B. Negative loss means shunt capacitance generates more than series reactance absorbs. (Key point: Negative Q loss)

## Related in Power System Analysis I

- [Maximum power with 138.5∠8° kV and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amax-power-138-5kv-132kv) — Power System Analysis I, chapter 6
- [Sending end active power from ABCD, 132/130.5 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asending-power-132kv-130-5kv) — Power System Analysis I, chapter 6
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- [Sending and receiving power, 140∠10° and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-140kv-132kv-a099) — Power System Analysis I, chapter 6
- [Significance of transmission voltage level, 132 vs 400 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avoltage-level-132kv-vs-400kv) — Power System Analysis I, chapter 1
- [132 kV horizontal line: susceptance and charging](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-132kv-4-5m-susceptance) — Power System Analysis I, chapter 4

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