---
title: "Maximum power with 132 kV at both ends"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:max-power-132kv-both-ends"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/max-power-132kv-both-ends
site: BCT Engineering
---

# Maximum power with 132 kV at both ends

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2070 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/max-power-132kv-both-ends
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 132 kV, 3-phase transmission line has constants A = 0.98∠3° and B = 110∠75° ohms per phase. The line is to be operated with both sending end and receiving end voltages at 132 kV. Determine the maximum power which the line can deliver. What will be the power factor at the receiving end and efficiency of the line under this condition?

## Short answer

Maximum deliverable power is 110.4 MW, with receiving-end power factor 0.599 leading and efficiency 59.6%.

## Answer

Data: $|A| = 0.98$, $\alpha = 3^\circ$, $|B| = 110\ \Omega$, $\beta = 75^\circ$, $|V_S| = |V_R| = 132$ kV. Take $D = A$. Line kV give three-phase MW/MVAr.

#### Maximum power
Maximum $P_R$ occurs at $\delta = \beta = 75^\circ$:
$$
\begin{aligned}
P_{R,max} &= \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\
&= \frac{132^2}{110} - \frac{0.98\times132^2}{110}\cos72^\circ \\
&= 158.40 - 155.23\times0.3090 = 158.40 - 47.97 = 110.43\ \text{MW}
\end{aligned}
$$

#### Receiving-end power factor
$$
Q_R = -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) = -155.23\sin72^\circ = -147.63\ \text{MVAr}
$$
$$
\text{pf}_R = \frac{110.43}{\sqrt{110.43^2 + 147.63^2}} = 0.599\ \text{leading}
$$

#### Efficiency
Sending-end power at $\delta = 75^\circ$ ($\beta+\delta = 150^\circ$):
$$
\begin{aligned}
P_S &= \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta) \\
&= 155.23\cos72^\circ - 158.40\cos150^\circ \\
&= 47.97 + 137.18 = 185.15\ \text{MW}
\end{aligned}
$$
$$
\eta = \frac{110.43}{185.15}\times100 = 59.64\%
$$
The low efficiency shows that operating at the theoretical maximum (steady-state stability limit) is not practical.

**Answer: $P_{max}$ ≈ 110.4 MW, receiving-end pf ≈ 0.599 leading, efficiency ≈ 59.6 %.**

## Check yourself

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

1. At which load angle does the line deliver maximum power with fixed |Vs| and |Vr|?
   A. delta = 3 degrees
   B. delta = beta = 75 degrees
   C. delta = 90 degrees
   D. delta = 0
   Answer: B. P_R is maximum when delta = beta, here 75 degrees. (Key point: Condition for maximum power)

2. How is the receiving-end pf found at maximum power?
   A. pf = cos(beta)
   B. pf = P_R / sqrt(P_R^2 + Q_R^2), with Q_R = -|A||Vr|^2 sin(beta-alpha)/|B|
   C. pf = |A|
   D. pf = 1 since P is maximum
   Answer: B. Q_R = -147.63 MVAr and P_R = 110.43 MW give pf = 0.599 leading. (Key point: Receiving-end power factor)

3. Which expression gives P_S at delta = 75 degrees?
   A. |A||Vs|^2 cos(beta-alpha)/|B| - |Vs||Vr| cos(beta+delta)/|B|
   B. |Vs||Vr| cos(beta-delta)/|B| only
   C. |A||Vr|^2 cos(beta+delta)/|B|
   D. |Vs||Vr| sin(delta)/|B|
   Answer: A. The sending-end circle equation gives 47.97 + 137.18 = 185.15 MW, since cos150 is negative. (Key point: Sending-end power)

4. What does the efficiency of about 59.6% at maximum power show?
   A. Operating at maximum power is the most efficient condition
   B. Efficiency is independent of power
   C. Operating at the theoretical maximum is impractical because losses are very large
   D. The line has no losses
   Answer: C. Efficiency = 110.43/185.15 = 59.6%, so the stability limit is not a practical operating point. (Key point: Efficiency at maximum power)

## Related in Power System Analysis I

- [Sending end from A = 0.98∠0.3°, B = 110∠75°](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-132kv-50mva-a098-0-3) — Power System Analysis I, chapter 6, asked 2×
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- [110 km, 110 kV line: sending end, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A110km-line-performance) — Power System Analysis I, chapter 6, asked 2×
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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

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