---
title: "110 km, 110 kV line: sending end, regulation, efficiency"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:110km-line-performance"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/110km-line-performance
site: BCT Engineering
---

# 110 km, 110 kV line: sending end, regulation, efficiency

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2076 Bhadra, 8 marks
- Video: [PS25 Numerical on Medium Transmission Lines](https://www.youtube.com/watch?v=Dt4Y5BlylMo) — Lectures in Electrical Engineering
- 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/110km-line-performance
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 3-phase 50 Hz, 110 km long overhead line has the following line constants: resistance per phase per km = 0.153 ohm, inductance per phase per km = 1.21 mH, capacitance per phase per km = 0.00958 μF. The line supplies a load of 20 MW at 0.9 power factor lagging at a line voltage of 110 kV at the receiving end. Calculate the sending end voltage, current, power factor, regulation and efficiency.

## Short answer

Sending end voltage is 116.16 kV, current 108.38 A, power factor 0.947 lagging. Voltage regulation is 6.33% and efficiency 96.91%, using the nominal-π model for the 110 km line.

## Answer

A 110 km line is a **medium line**; it is solved here with the **nominal-π model** (half the shunt admittance at each end).

#### Line constants
$$
\begin{aligned}
R &= 0.153\times110 = 16.83\ \Omega \\
X &= 2\pi(50)(1.21\times10^{-3})(110) = 41.81\ \Omega \\
Y &= 2\pi(50)(0.00958\times10^{-6})(110) = 3.311\times10^{-4}\ \text{S} \\
Y/2 &= 1.655\times10^{-4}\ \text{S}
\end{aligned}
$$

#### Receiving end
$$
\begin{aligned}
V_R &= 110\,000/\sqrt3 = 63\,508.5\angle0^\circ\ \text{V} \\
I_R &= \frac{20\times10^6}{3\times63\,508.5\times0.9} = 116.64\angle-25.84^\circ\ \text{A}
\end{aligned}
$$

#### Nominal-π steps
$$
\begin{aligned}
I_{C1} &= j\frac{Y}{2}V_R = j10.51\ \text{A} \\
I_L &= I_R + I_{C1} = 112.45\angle-21.02^\circ\ \text{A} \\
V_S &= V_R + I_L Z = 63\,508.5 + 5068.8\angle47.06^\circ \\
&= 67\,064.3\angle3.17^\circ\ \text{V} \\
I_{C2} &= j\frac{Y}{2}V_S = 11.10\angle93.17^\circ\ \text{A} \\
I_S &= I_L + I_{C2} = 108.38\angle-15.65^\circ\ \text{A}
\end{aligned}
$$

#### Results
- Sending-end line voltage $= \sqrt3\times67.064 = 116.16$ kV
- Sending-end current $= 108.38$ A
- Sending-end pf $= \cos(3.17^\circ + 15.65^\circ) = 0.947$ lagging

Regulation (no-load $V_R = V_S/|A|$, $A = 1 + YZ/2 = 0.9931\angle0.16^\circ$):
$$
\%VR = \frac{67\,064.3/0.9931 - 63\,508.5}{63\,508.5}\times100 = \frac{67\,531.4 - 63\,508.5}{63\,508.5}\times100 = 6.33\%
$$

Efficiency:
$$
\begin{aligned}
P_{loss} &= 3|I_L|^2R = 3(112.45)^2(16.83) = 0.638\ \text{MW} \\
\eta &= \frac{20}{20.638}\times100 = 96.91\%
\end{aligned}
$$

**Answer: $V_S$ = 116.16 kV, $I_S$ = 108.38 A, pf 0.947 lag, regulation 6.33 %, efficiency 96.91 %.** (A nominal-T solution gives almost the same: 116.13 kV, 6.30 %, 96.90 %.)

## Check yourself

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

1. Why is the nominal-π model used for this 110 km, 50 Hz line?
   A. It is a short line, so shunt capacitance is neglected
   B. It is a medium line, so shunt admittance is included, half at each end
   C. It is a long line, so distributed constants are needed
   D. The load is above SIL, so only a lumped series reactor is needed
   Answer: B. A 110 km line is a medium line; the nominal-π model places Y/2 at each end to account for the shunt capacitance. (Key point: Choice of model)

2. In the nominal-π solution, how is the current in the series impedance Z found?
   A. I_L = I_R - I_C1, with I_C1 = j(Y/2)V_S
   B. I_L = I_R, since shunt current is ignored
   C. I_L = I_R + I_C1, with I_C1 = j(Y/2)V_R
   D. I_L = I_S + I_C2, with I_C2 = j(Y/2)V_R
   Answer: C. The receiving-end half admittance draws I_C1 = j(Y/2)V_R, which adds to I_R to give the series current I_L; then V_S = V_R + I_L Z. (Key point: Order of nominal-π steps)

3. How is the regulation computed here?
   A. (V_S - V_R)/V_R using the loaded V_R
   B. With no-load V_R = V_S/|A|, regulation = (V_S/|A| - V_R)/V_R
   C. (V_S - V_R)/V_S
   D. Using V_S/|D| with D = 1 + YZ/4
   Answer: B. No-load receiving voltage is V_S/|A| with A = 1 + YZ/2; regulation is the rise from loaded V_R to this value, giving 6.33 %. (Key point: Voltage regulation)

4. Which current is used for the copper loss in the efficiency calculation, and what is the efficiency?
   A. I_R, giving about 95 %
   B. I_S, giving about 98 %
   C. I_C1, giving 0.6 %
   D. I_L (through R), giving loss 0.638 MW and eta = 96.91 %
   Answer: D. Only the current through the series resistance, I_L, causes I^2R loss: 3(112.45)^2(16.83) = 0.638 MW, so eta = 20/20.638 = 96.91 %. (Key point: Efficiency)

## Close variants set in other exams

- [Regulation and efficiency of 100 km, 110 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aregulation-efficiency-100km-110kv) — Power System Analysis I, chapter 6, asked 2×

## Related in Power System Analysis I

- [12 km line at 32.5 kV: regulation and capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-32-5kv-regulation-capacitor) — Power System Analysis I, chapter 6
- [Short line 40 km, 381 MVA: regulation and efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ashort-line-40km-381mva) — Power System Analysis I, chapter 6
- [450 km 400 kV line, 420 MW, 0.9 pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A450km-400kv-420mw-090) — Power System Analysis I, chapter 5
- [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6
- [150 mile, 60 Hz line: ABCD and sending end](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150-mile-60hz-line-abcd) — Power System Analysis I, chapter 5
- [Capacitor for zero regulation, 70 km 220 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-220kv-zero-regulation-capacitor) — Power System Analysis I, chapter 6

## Related in other subjects

- [220 kV double circuit efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Adouble-circuit-220kv-efficiency-regulation) — Transmission and Distribution System Design, chapter 4
- [Peterson coil for 33 kV 60 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apeterson-coil-33kv-60km) — Switchgear and Protection, chapter 5, asked 6×
- [Arc suppression coil for 66 kV 90 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aarc-suppression-coil-66kv-90km) — Switchgear and Protection, chapter 5, asked 6×