---
title: "Power output with 46.85 kV sending, 33 kV receiving"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:power-output-46-85kv-33kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/power-output-46-85kv-33kv
site: BCT Engineering
---

# Power output with 46.85 kV sending, 33 kV receiving

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2076 Baisakh, 8 marks
- Video: [Receiving End Power Circle Diagram of Transmission line| Numerical Solved |Power System| #electrical](https://www.youtube.com/watch?v=AYmQjfNs6Eg) — TechnoSmart
- 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-output-46-85kv-33kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

The sending end voltage of a three phase overhead line with a series impedance of (20.62∠75.96°) ohm is 46.85 kV while the receiving end is 33 kV. Determine the power output at 0.8 pf lagging and the sending end power factor.

## Short answer

Power output is about 20.95 MW at 0.8 pf lagging. Sending end power factor is approximately 0.648 lagging, computed from voltage drop across series impedance (20.62∠75.96° Ω).

## Answer

Short line with $Z = 20.62\angle75.96^\circ = 5.00 + j20.00\ \Omega$ per phase.

#### Method
Take $V_R$ (line, kV) as reference. For three-phase $P + jQ$ in MW/MVAr at the receiving end:
$$
V_S = V_R + \frac{Z\,(P - jQ)}{V_R}
$$
Multiplying by $V_R$ and taking magnitudes:
$$
|V_S|^2|V_R|^2 = (V_R^2 + RP + XQ)^2 + (XP - RQ)^2
$$

#### Substitution
At 0.8 pf lagging, $Q = 0.75P$. With $V_R = 33$, $V_S = 46.85$, $R = 5$, $X = 20$:
$$
\begin{aligned}
(1089 + 5P + 15P)^2 + (20P - 3.75P)^2 &= (46.85\times33)^2 \\
(1089 + 20P)^2 + (16.25P)^2 &= 2\,390\,270
\end{aligned}
$$
Expanding:
$$
664.06P^2 + 43\,560P - 1\,204\,349 = 0
$$
Taking the positive root:
$$
P = 20.95\ \text{MW},\quad Q = 0.75P = 15.71\ \text{MVAr}
$$
Check: $|33 + (5 + j20)(20.95 - j15.71)/33| = 46.85$ kV.

#### Sending-end power factor
$$
\begin{aligned}
I &= \frac{\sqrt{20.95^2 + 15.71^2}}{\sqrt3\times33} = 0.4581\ \text{kA} = 458.1\ \text{A} \\
P_{loss} &= 3I^2R = 3(458.1)^2(5) = 3.148\ \text{MW} \\
Q_{loss} &= 3I^2X = 3(458.1)^2(20) = 12.59\ \text{MVAr} \\
P_S &= 20.95 + 3.15 = 24.10\ \text{MW} \\
Q_S &= 15.71 + 12.59 = 28.30\ \text{MVAr} \\
\text{pf}_S &= \frac{24.10}{\sqrt{24.10^2 + 28.30^2}} = 0.648\ \text{lagging}
\end{aligned}
$$
(Check: $|S_S| = \sqrt3\times46.85\times0.4581 = 37.17$ MVA, and $\sqrt{24.10^2 + 28.30^2} = 37.17$ MVA.)

**Answer: Power output ≈ 20.95 MW at 0.8 pf lag; sending-end pf ≈ 0.648 lagging.**

## Check yourself

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

1. How is P found for a given V_S and V_R with 0.8 pf lagging?
   A. Use |V_S|^2|V_R|^2 = (V_R^2 + RP + XQ)^2 + (XP - RQ)^2 with Q = 0.75P and solve the quadratic
   B. Use P = V_S V_R / X directly
   C. Use P = 3 I^2 R
   D. Use P = V_S - V_R
   Answer: A. Substituting Q = 0.75P gives a quadratic in P. (Key point: Method)

2. At 0.8 pf lagging, Q equals
   A. 0.8 P
   B. 0.75 P
   C. 0.6 P
   D. 1.25 P
   Answer: B. tan(cos^-1 0.8) = 0.75. (Key point: Q for pf)

3. How are sending-end P and Q found?
   A. Subtract line losses
   B. Multiply by 1.05
   C. Add line losses 3I^2R and 3I^2X to P and Q
   D. Use V_S x I only
   Answer: C. P_S = 20.95 + 1.05, Q_S = 15.71 + 4.20. (Key point: Sending power)

4. The sending-end pf is
   A. 0.8 lagging
   B. 0.741 leading
   C. Q_S/P_S
   D. P_S/sqrt(P_S^2 + Q_S^2) = 0.741 lagging
   Answer: D. 22.00/sqrt(22^2 + 19.91^2) = 0.741. (Key point: Sending pf)

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