---
title: "UPF power received with 275 kV both ends"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:unity-pf-power-275kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/unity-pf-power-275kv
site: BCT Engineering
---

# UPF power received with 275 kV both ends

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2076 Baisakh, 8 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/unity-pf-power-275kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 275 kV transmission line has the following line constants: A = 0.85∠5°; B = 200∠75°. Determine the power at unity power factor that can be received if the voltage profile at each end is to be maintained at 275 kV.

## Short answer

At unity power factor with both ends at 275 kV, received power is about 117.6 MW. Load angle is approximately 22.0°, found by setting QR=0 in the receiving-end power equations.

## Answer

Use the receiving-end power-circle equations with $|V_S| = |V_R| = 275$ kV, $|A| = 0.85$, $\alpha = 5^\circ$, $|B| = 200\ \Omega$, $\beta = 75^\circ$ (line kV give three-phase MW/MVAr):
$$
\begin{aligned}
P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\
Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)
\end{aligned}
$$

#### Constants
$$
\frac{275\times275}{200} = 378.13,\qquad \frac{0.85\times275^2}{200} = 321.41,\qquad \beta-\alpha = 70^\circ
$$

#### Unity pf condition: $Q_R = 0$
$$
\begin{aligned}
378.13\sin(75^\circ-\delta) &= 321.41\sin70^\circ \\
\sin(75^\circ-\delta) &= 0.85\times0.9397 = 0.7987 \\
75^\circ-\delta &= 53.01^\circ \Rightarrow \delta = 21.99^\circ
\end{aligned}
$$

#### Power received
$$
\begin{aligned}
P_R &= 378.13\cos53.01^\circ - 321.41\cos70^\circ \\
&= 227.51 - 109.93 = 117.58\ \text{MW}
\end{aligned}
$$

**Answer: About 117.6 MW can be received at unity pf, with load angle $\delta$ ≈ 22.0°.**

## Check yourself

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

1. What condition gives unity pf at the receiving end?
   A. Q_R = 0
   B. P_R = 0
   C. delta = 0
   D. Q_R = P_R
   Answer: A. Unity pf means no reactive power at the receiving end. (Key point: Unity pf condition)

2. With Q_R = 0, how is delta found?
   A. cos(75 - delta) = 0.85
   B. 378.13 sin(75 - delta) = 321.41 sin 70, so delta = 21.99 deg
   C. delta = 75 - 70 = 5 deg
   D. delta = alpha
   Answer: B. sin(75 - delta) = 0.7987 gives 75 - delta = 53.01 deg. (Key point: Finding delta)

3. Which expression gives P_R?
   A. 378.13 cos 75 - 321.41 cos 70
   B. 378.13 sin 53.01 - 321.41 sin 70
   C. 378.13 cos 53.01 - 321.41 cos 70 = 117.6 MW
   D. 378.13 + 321.41
   Answer: C. Use the P_R equation with beta - delta = 53.01 and beta - alpha = 70. (Key point: Power received)

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