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

# Maximum power with 138.5∠8° kV and 132 kV

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2070 Bhadra, 4 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-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 maximum amount of active power that can be delivered by the line. What will be the reactive power delivered at that condition?

## Short answer

Maximum active power is 346.7 MW, occurring when load angle equals 71.5°. At this condition, the receiving end must supply about 443.2 MVAr (leading reactive power).

## Answer

For fixed voltage magnitudes, the receiving-end power
$$
P_R = \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)
$$
is maximum when $\delta = \beta$. (Line kV give three-phase MW/MVAr.)

Data: $|V_S| = 138.5$ kV, $|V_R| = 132$ kV, $|A| = 0.983$, $\alpha = 0.68^\circ$, $|B| = 36.5\ \Omega$, $\beta = 71.5^\circ$, so $\beta-\alpha = 70.82^\circ$.

#### Maximum active power
$$
\begin{aligned}
\frac{|V_S||V_R|}{|B|} &= \frac{138.5\times132}{36.5} = 500.88 \\
\frac{|A||V_R|^2}{|B|} &= \frac{0.983\times132^2}{36.5} = 469.25 \\
P_{R,max} &= 500.88 - 469.25\cos70.82^\circ \\
&= 500.88 - 154.17 = 346.71\ \text{MW}
\end{aligned}
$$

#### Reactive power at this condition
At $\delta = \beta$, $\sin(\beta-\delta) = 0$:
$$
\begin{aligned}
Q_R &= -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \\
&= -469.25\sin70.82^\circ = -443.21\ \text{MVAr}
\end{aligned}
$$
The negative sign means the receiving end must **supply** about 443 MVAr to the line (the load must be strongly leading, pf ≈ 0.62 lead) for maximum power transfer. The load angle at this condition is $\delta = 71.5^\circ$.

**Answer: $P_{R,max}$ ≈ 346.7 MW; reactive power at the receiving end ≈ −443.2 MVAr (443.2 MVAr must be supplied at the receiving end).**

## Check yourself

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

1. For fixed voltage magnitudes, at what load angle is P_R maximum?
   A. delta = alpha
   B. delta = 90 degrees always
   C. delta = beta
   D. delta = beta + alpha
   Answer: C. P_R contains cos(beta - delta), which is largest (=1) when delta = beta. (Key point: Condition for maximum power)

2. What is P_R,max in terms of the given constants?
   A. |Vs||Vr|/|B| + |A||Vr|^2 cos(beta-alpha)/|B|
   B. |Vs||Vr|/|B| - |A||Vr|^2 cos(beta-alpha)/|B|
   C. |Vs||Vr|/|B| x cos(beta-alpha)
   D. |A||Vr|^2/|B| - |Vs||Vr|/|B|
   Answer: B. Setting cos(beta-delta)=1 leaves 500.88 - 469.25 cos70.82 = 346.7 MW. (Key point: Maximum power value)

3. Why does Q_R depend only on the A term at maximum power?
   A. The sending voltage is zero
   B. sin(beta - delta) = 0 so the first term vanishes
   C. B becomes zero
   D. cos(beta - alpha) = 0
   Answer: B. At delta = beta, sin(beta-delta) = 0, so Q_R = -|A||Vr|^2 sin(beta-alpha)/|B|. (Key point: Reactive power at maximum power)

4. Q_R = -443 MVAr at maximum power. What does this mean?
   A. The receiving end absorbs 443 MVAr (lagging load)
   B. The receiving end must supply about 443 MVAr (strongly leading load)
   C. The line has no reactive flow
   D. The sending end absorbs 443 MVAr
   Answer: B. Negative Q_R means the receiving end must deliver reactive power to the line, i.e. a leading pf of about 0.62. (Key point: Meaning of negative Q_R)

## Related in Power System Analysis I

- [Line power loss, 138.5∠8° kV and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-loss-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
- [Maximum power with 132 kV at both ends](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amax-power-132kv-both-ends) — Power System Analysis I, chapter 6
- [Receiving end power, 140.2∠18.2° kV and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-140-2kv) — Power System Analysis I, chapter 6
- [Power delivered from sending end data, nominal-π](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-pi-power-delivered) — Power System Analysis I, chapter 5
- [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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