---
title: "300 km line: short line vs nominal-π"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:300km-short-vs-nominal-pi"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/300km-short-vs-nominal-pi
site: BCT Engineering
---

# 300 km line: short line vs nominal-π

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2076 Baisakh, 10 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/300km-short-vs-nominal-pi
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 50 Hz transmission line 300 km long has a total series impedance of 40 + j125 ohms and a total shunt admittance of 10⁻³ mho. The receiving-end load is 50 MW at 220 kV with 0.8 lagging power factor. Find the sending-end voltage, current, power and power factor using: (i) short line approximation (ii) nominal-π method.

## Short answer

Short line: V_S=251.33 kV, I_S=164.0 A, pf=0.7455 lag, P_S=53.23 MW. Nominal-π: V_S=238.07 kV, I_S=128.1 A, pf=0.9881 lead, P_S=52.21 MW.

## Answer

**Given:** $Z = 40 + j125 = 131.24\angle 72.26^\circ\ \Omega$, $Y = j10^{-3}$ S (totals), load 50 MW at 220 kV, 0.8 p.f. lagging. Per-phase working.

$$
\begin{aligned}
V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\
I_R &= \frac{50\times10^6}{\sqrt3\times220\times10^3\times0.8} = 164.02\angle -36.87^\circ = 131.22 - j98.41\ \text{A}
\end{aligned}
$$

$P_R = 50$ MW, $Q_R = 37.5$ Mvar.

#### (i) Short-line approximation ($A = D = 1$, $B = Z$, $C = 0$)

$$
\begin{aligned}
I_S &= I_R = 164.0\angle -36.87^\circ\ \text{A} \\
ZI_R &= (40 + j125)(131.22 - j98.41) = 17{,}550 + j12{,}466\ \text{V} \\
V_S &= V_R + ZI_R = 144{,}567 + j12{,}466 = 145.10\angle 4.93^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times145.10 = 251.33\ \text{kV}
\end{aligned}
$$

p.f. $= \cos(4.93^\circ + 36.87^\circ) = \cos 41.80^\circ = 0.7455$ lagging

$$
P_S = 3V_SI_S\cos\phi_S = 3\times145.10\times10^3\times164.02\times0.7455 = 53.23\ \text{MW}
$$

#### (ii) Nominal-Π method

$$
\begin{aligned}
ZY &= (40 + j125)(j10^{-3}) = -0.125 + j0.04 \\
A = D &= 1 + \frac{ZY}{2} = 0.9375 + j0.02 = 0.9377\angle 1.22^\circ \\
B &= Z = 131.24\angle 72.26^\circ\ \Omega \\
C &= Y\left(1 + \frac{ZY}{4}\right) = -1.0\times10^{-5} + j9.6875\times10^{-4} = 9.688\times10^{-4}\angle 90.59^\circ\ \text{S}
\end{aligned}
$$

$$
\begin{aligned}
V_S &= AV_R + BI_R = (119{,}080 + j2540) + (17{,}550 + j12{,}466) \\
&= 136{,}630 + j15{,}006 = 137.45\angle 6.27^\circ\ \text{kV/phase} \\
V_{S(L)} &= 238.07\ \text{kV} \\
I_S &= CV_R + DI_R = (-1.27 + j123.05) + (124.98 - j89.64) \\
&= 123.71 + j33.41 = 128.1\angle 15.11^\circ\ \text{A}
\end{aligned}
$$

p.f. $= \cos(6.27^\circ - 15.11^\circ) = \cos 8.85^\circ = 0.9881$ **leading** (current leads voltage)

$$
P_S = 3\times137.45\times10^3\times128.1\times0.9881 = 52.21\ \text{MW}
$$

#### Comparison

| Quantity | Short line | Nominal-Π |
|---|---|---|
| $V_S$ (line) | 251.33 kV | 238.07 kV |
| $I_S$ | 164.0 A | 128.1 A |
| Sending p.f. | 0.7455 lag | 0.9881 lead |
| $P_S$ | 53.23 MW | 52.21 MW |
| Efficiency | 93.94 % | 95.76 % |

The short-line model ignores about 123 A of charging current, so for a 300 km line it badly overestimates $V_S$ and $I_S$. The nominal-Π (or better, the exact long-line) model must be used.

**Answer:** (i) $V_S = 251.33$ kV, $I_S = 164.0$ A, p.f. 0.7455 lag, $P_S = 53.23$ MW. (ii) $V_S = 238.07$ kV, $I_S = 128.1$ A, p.f. 0.9881 lead, $P_S = 52.21$ MW.

## Check yourself

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

1. In the short-line approximation, the sending current IS is:
   A. Equal to IR, since capacitance is ignored
   B. IR plus charging current
   C. Zero
   D. VS/Z
   Answer: A. With C = 0 there is no shunt path, so IS = IR. (Key point: Short-line sending current)

2. In the short-line method VS is found from:
   A. VS = VR / Z
   B. VS = VR + Z IR
   C. VS = (1 + ZY/2) VR
   D. VS = VR - Z IR
   Answer: B. A = 1 and B = Z give VS = VR + Z IR (251.33 kV line here). (Key point: Short-line VS)

3. In the nominal-Π result, the sending p.f. is 0.9881 leading because:
   A. The load is leading
   B. Series resistance is negligible
   C. The charging current (about 123 A) exceeds the reactive part of the load current
   D. VS is smaller than VR
   Answer: C. IS = 128.1 A at +15.11° leads VS at +6.27°. (Key point: Nominal-Π power factor)

4. Compared with nominal-Π, the short-line model for this 300 km line:
   A. Underestimates VS and IS
   B. Gives identical values
   C. Gives identical efficiency
   D. Overestimates VS and IS, because charging current is ignored
   Answer: D. Short: 251.33 kV and 164 A; Π: 238.07 kV and 128.1 A. (Key point: Comparison of results)

## Close variants set in other exams

- [280 km, 60 Hz line: nominal-π vs long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A280km-60hz-nominal-pi-long) — Power System Analysis I, chapter 5, 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
- [70 km, 40 MW line: regulation and zero-regulation pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-40mw-regulation) — Power System Analysis I, chapter 6
- [Equivalent-T of 300 km long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-equivalent-t-300km) — Power System Analysis I, chapter 5
- [150 km line with mid-point shunt capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150km-mid-shunt-capacitor) — 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

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