---
title: "Regulation and efficiency of 100 km, 110 kV line"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:regulation-efficiency-100km-110kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/regulation-efficiency-100km-110kv
site: BCT Engineering
---

# Regulation and efficiency of 100 km, 110 kV line

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2082 Kartik (new course), 6 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/regulation-efficiency-100km-110kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 3-phase, 50 Hz, 100 km long overhead line has the following line constants: resistance per phase per km = 0.153 Ω, 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 regulation and efficiency.

## Short answer

Using nominal-T model for this 100 km line, total series impedance Z = 15.3 + j38.01 Ω, shunt admittance Y = 3.01×10⁻⁴ S. Sending voltage found 115.62 kV, giving voltage regulation 5.72 %. Total losses 0.585 MW, efficiency 97.16 %.

## Answer

A 100 km line is a **medium line**, so it is solved here with the **nominal-T model**: total series impedance split into two halves, with the total shunt admittance at the middle.

#### Line constants (per phase, total)
$$
\begin{aligned}
R &= 0.153 \times 100 = 15.3\ \Omega \\
X_L &= 2\pi (50)(1.21\times10^{-3})(100) = 38.01\ \Omega \\
Y &= 2\pi (50)(0.00958\times10^{-6})(100) = 3.010\times10^{-4}\ \text{S} \\
Z/2 &= 7.65 + j19.01\ \Omega
\end{aligned}
$$

#### Receiving-end quantities
$$
\begin{aligned}
V_R &= \frac{110\,000}{\sqrt3} = 63\,508.5\angle 0^\circ\ \text{V} \\
I_R &= \frac{20\times10^6}{3 \times 63\,508.5 \times 0.9} = 116.64\angle -25.84^\circ\ \text{A}
\end{aligned}
$$

#### Nominal-T solution
$$
\begin{aligned}
V_1 &= V_R + I_R \frac{Z}{2} = 65\,297.6\angle 1.41^\circ\ \text{V} \\
I_C &= jY V_1 = 19.65\angle 91.41^\circ\ \text{A} \\
I_S &= I_R + I_C = 109.05\angle -16.62^\circ\ \text{A} \\
V_S &= V_1 + I_S \frac{Z}{2} = 66\,754.4\angle 2.88^\circ\ \text{V}
\end{aligned}
$$

Sending-end line voltage $= \sqrt3 \times 66.754 = 115.62$ kV.

#### Regulation
At no load the receiving voltage is $V_S/|A|$, with $A = 1 + \frac{YZ}{2} = 0.9943\angle 0.13^\circ$:
$$
\begin{aligned}
V_{R0} &= \frac{66\,754.4}{0.9943} = 67\,138.3\ \text{V} \\
\%VR &= \frac{67\,138.3 - 63\,508.5}{63\,508.5}\times 100 = 5.72\%
\end{aligned}
$$
(If the no-load Ferranti rise is ignored, $(V_S - V_R)/V_R = 5.11\%$.)

#### Efficiency
$$
\begin{aligned}
P_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2} \\
&= 3\,(116.64^2 + 109.05^2)(7.65) = 0.585\ \text{MW} \\
\eta &= \frac{20}{20 + 0.585}\times 100 = 97.16\%
\end{aligned}
$$

**Answer: Voltage regulation ≈ 5.72 % (sending end 115.62 kV), efficiency ≈ 97.16 %.**

## Check yourself

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

1. Why is the nominal-T model used for the 100 km line?
   A. 100 km is a medium line, so shunt capacitance is included, lumped at the middle with Z split in halves
   B. It is a short line so capacitance is ignored
   C. It is a long line needing hyperbolic functions
   D. T is used for cables only
   Answer: A. Medium lines (about 80-250 km) use nominal T or pi. (Key point: Model choice)

2. How is regulation computed?
   A. (V_S - V_R)/V_S x 100
   B. Using no-load receiving voltage V_S/|A|, then (V_R0 - V_R)/V_R x 100 = 5.72 %
   C. (V_R - V_S)/V_R x 100
   D. Using I_R only
   Answer: B. At no load V_R0 = V_S/|A|, with A = 1 + YZ/2 for nominal T. (Key point: Regulation basis)

3. How are total line losses found in nominal T?
   A. 3 |I_R|^2 R only
   B. 3 |I_C|^2 R
   C. 3(|I_R|^2 + |I_S|^2)(R/2), as each half of resistance carries I_R or I_S
   D. P_R times regulation
   Answer: C. The two R/2 halves carry I_R and I_S respectively; result 0.585 MW. (Key point: Loss computation)

4. How is efficiency computed?
   A. P_loss/P_R
   B. P_R/P_S with P_S = P_R + regulation
   C. 1 - regulation
   D. P_R/(P_R + P_loss) = 20/20.585 = 97.16 %
   Answer: D. Efficiency = output / (output + losses). (Key point: Efficiency)

## Close variants set in other exams

- [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×

## Related in Power System Analysis I

- [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [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
- [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
- [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

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