---
title: "Nominal-T 100 km line, 50 MVA at 132 kV"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:nominal-t-100km-50mva"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-t-100km-50mva
site: BCT Engineering
---

# Nominal-T 100 km line, 50 MVA at 132 kV

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2074 Bhadra, 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/nominal-t-100km-50mva
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 100 km long 3 phase 50 Hz overhead line delivers 50 MVA at 0.8 pf lagging and at 132 kV. The resistance and inductive reactance are 0.1 Ω per phase per km and j0.2 per phase per km respectively. The shunt admittance is j4 × 10⁻⁶ S/phase per km. Calculate sending end current, sending end voltage, and transmission efficiency using nominal T-model.

## Short answer

I_S = 201.07∠−29.78° A, V_S = 80.30∠1.67° kV per phase (139.09 kV line-to-line), efficiency ≈96.80%.

## Answer

#### Line constants (total, per phase)
$$
\begin{aligned}
Z &= (0.1 + j0.2)\times100 = 10 + j20\ \Omega,\quad Z/2 = 5 + j10\ \Omega \\
Y &= j4\times10^{-6}\times100 = j4\times10^{-4}\ \text{S}
\end{aligned}
$$

#### Receiving end
$$
\begin{aligned}
V_R &= 132\,000/\sqrt3 = 76\,210.2\angle0^\circ\ \text{V} \\
I_R &= \frac{50\times10^6}{\sqrt3\times132\,000} = 218.69\angle-36.87^\circ\ \text{A}
\end{aligned}
$$

#### Nominal-T calculation
```text
 Vs   Z/2        V1        Z/2   Vr
 o---/\/\--+------+------/\/\---o
   Is      |      |  Ir
           |    Y (shunt at middle)
           |      |
 o---------+------+-------------o
```
$$
\begin{aligned}
V_1 &= V_R + I_R\frac{Z}{2} = 78\,404.8\angle0.80^\circ\ \text{V} \\
I_C &= YV_1 = 31.36\angle90.80^\circ\ \text{A} \\
I_S &= I_R + I_C = 201.07\angle-29.78^\circ\ \text{A} \\
V_S &= V_1 + I_S\frac{Z}{2} = 80\,302.4\angle1.67^\circ\ \text{V}
\end{aligned}
$$
Sending-end line voltage $= \sqrt3\times80.302 = 139.09$ kV.

#### Efficiency
$$
\begin{aligned}
S_S &= 3V_SI_S^* = 41.32 + j25.27\ \text{MVA} \\
P_R &= 50\times0.8 = 40\ \text{MW} \\
\eta &= \frac{40}{41.32}\times100 = 96.80\%
\end{aligned}
$$
(Check: loss $= 3(218.69^2 + 201.07^2)(5) = 1.32$ MW.)

**Answer: $I_S$ = 201.07∠−29.78° A, $V_S$ = 80.30∠1.67° kV per phase (139.09 kV line), efficiency ≈ 96.80 %.**

## Check yourself

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

1. Where is the shunt admittance Y placed in the nominal-T model?
   A. Half at each end
   B. At the sending end only
   C. At the receiving end only
   D. In the middle, between two Z/2 halves
   Answer: D. Nominal-T splits series impedance into Z/2 on each side with the full Y at the midpoint. (Key point: Nominal-T layout)

2. How is the midpoint voltage V1 found?
   A. V1 = V_R + I_R Z
   B. V1 = V_S - I_S Z
   C. V1 = V_R / 2
   D. V1 = V_R + I_R Z/2
   Answer: D. V1 is the receiving voltage plus the drop in the receiving-side half impedance. (Key point: Midpoint voltage)

3. How is I_S found?
   A. I_S = I_R - Y V1
   B. I_S = I_R + Y V1 (capacitor current at V1)
   C. I_S = I_R + Y V_R
   D. I_S = I_R
   Answer: B. The shunt branch draws I_C = Y V1, which adds to I_R: I_S = 201.07 angle -29.78 deg A. (Key point: Sending current)

4. How is efficiency obtained?
   A. S_R/S_S using 50 and 48.4
   B. P_R / P_S = 40/41.32 = 96.80 %
   C. Q_R / Q_S
   D. 50/41.32
   Answer: B. P_R = 50 x 0.8 = 40 MW and P_S = 3 Re(V_S I_S*) = 41.32 MW. (Key point: Efficiency)

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