---
title: "150 km line by T-model, 60 MW"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:150km-t-model-60mw"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/150km-t-model-60mw
site: BCT Engineering
---

# 150 km line by T-model, 60 MW

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2075 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/150km-t-model-60mw
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 150 km long three phase overhead line has a resistance of 45 ohms per phase, inductive reactance of 85 ohms per phase and capacitance (line to neutral) 9.00 nF per km. It supplies a load of 60 MW at a voltage of 132 kV and pf 0.9 lagging. Find (i) sending end active power (ii) efficiency (iii) line losses and (iv) voltage regulation. Use T-model.

## Short answer

Per-phase VR=76.21 kV, IR=291.6 A lagging 0.9. T-model with Z/2=22.5+j42.5 Ω, Y=j4.241e-4 S gives VS=98.69 kV (170.94 kV line). Results: sending power 70.81 MW, line loss 10.81 MW, efficiency 84.73%, regulation 31.87%.

## Answer

**Given:** $R = 45\ \Omega$, $X = 85\ \Omega$ per phase (totals); $C = 9$ nF/km × 150 km $= 1.35\ \mu$F; load 60 MW, 132 kV, 0.9 p.f. lagging. Frequency assumed 50 Hz.

$$
\begin{aligned}
Z &= 45 + j85\ \Omega, \quad \frac{Z}{2} = 22.5 + j42.5\ \Omega \\
Y &= j2\pi\times50\times1.35\times10^{-6} = j4.241\times10^{-4}\ \text{S} \\
V_R &= \frac{132}{\sqrt3} = 76.21\angle 0^\circ\ \text{kV} \\
I_R &= \frac{60\times10^6}{\sqrt3\times132\times10^3\times0.9} = 291.6\angle -25.84^\circ = 262.43 - j127.10\ \text{A}
\end{aligned}
$$

#### Nominal-T working

```text
 IS     Z/2            Z/2     IR
 o-->--[R/2 jX/2]--+--[R/2 jX/2]-->--o
                   |
 VS              [ Y ] Ic   (VC)     VR
                   |
 o-----------------+-----------------o
```

$$
\begin{aligned}
V_C &= V_R + I_R\frac{Z}{2} = 87{,}517 + j8294 = 87.91\angle 5.41^\circ\ \text{kV} \\
I_C &= YV_C = -3.52 + j37.12 = 37.28\angle 95.41^\circ\ \text{A} \\
I_S &= I_R + I_C = 258.91 - j89.98 = 274.1\angle -19.17^\circ\ \text{A} \\
V_S &= V_C + I_S\frac{Z}{2} = 97{,}167 + j17{,}273 = 98.69\angle 10.08^\circ\ \text{kV} \\
V_{S(L)} &= \sqrt3\times98.69 = 170.94\ \text{kV}
\end{aligned}
$$

Sending p.f. $= \cos(10.08^\circ + 19.17^\circ) = 0.8725$ lagging.

#### (i) Sending-end active power

$$
P_S = 3|V_S||I_S|\cos\phi_S = 3\times98.69\times10^3\times274.1\times0.8725 = 70.81\ \text{MW}
$$

#### (iii) Line losses (series resistance, $R/2$ in each arm)

$$
\begin{aligned}
P_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2} = 3(291.6^2 + 274.1^2)(22.5) \\
&= 10.81\ \text{MW}
\end{aligned}
$$

Check: $P_S - P_R = 70.81 - 60 = 10.81$ MW.

#### (ii) Efficiency

$$
\eta = \frac{P_R}{P_S}\times100 = \frac{60}{70.81}\times100 = 84.73\ \%
$$

#### (iv) Voltage regulation
$A = 1 + ZY/2 = 0.982\angle 0.56^\circ$. No-load receiving voltage:

$$
\begin{aligned}
V_{R(NL)} &= \frac{|V_S|}{|A|} = \frac{98.69}{0.982} = 100.50\ \text{kV/phase} = 174.07\ \text{kV (line)} \\
\%\text{VR} &= \frac{174.07 - 132}{132}\times100 = 31.87\ \%
\end{aligned}
$$

**Answer:** (i) $P_S = 70.81$ MW, (ii) $\eta = 84.73\%$, (iii) line loss $= 10.81$ MW, (iv) regulation $= 31.87\%$ (with $V_S = 170.94$ kV, $I_S = 274.1$ A).

The large regulation and loss arise because $R = 45\ \Omega$ and $X = 85\ \Omega$ are high for a 132 kV line carrying 60 MW.

## Check yourself

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

1. In the nominal-T model, the first quantity found is:
   A. The shunt voltage VC = VR + IR (Z/2)
   B. VS directly
   C. The loss
   D. IC = Y VS
   Answer: A. VC across the shunt is VR plus the drop in the receiving-side half impedance. (Key point: T-model structure)

2. How is IS found in the T model?
   A. IS = IR
   B. IS = Y VR
   C. IS = IR + Y VC
   D. IS = VC / Z
   Answer: C. IC = Y VC flows in the shunt branch, so IS = IR + IC. (Key point: Sending current)

3. In the T model, line loss is:
   A. 3|IR|^2 R only
   B. 3(|IR|^2 + |IS|^2)(R/2)
   C. 3|IC|^2 R
   D. 3|VC|^2 / R
   Answer: B. Each arm carries a different current (IR and IS) and has resistance R/2; result is 10.81 MW. (Key point: Power loss)

4. Efficiency here is PR / PS = 60/70.81, which equals:
   A. 94.7 percent
   B. 70.8 percent
   C. 100 - 31.87 percent
   D. 84.73 percent
   Answer: D. η = 60/70.81 x 100 = 84.73 percent. (Key point: Efficiency)

5. Voltage regulation uses the no-load receiving voltage |VS|/|A| because:
   A. With IR = 0, VS = A VR
   B. A is the load p.f.
   C. VS is zero at no load
   D. The T model has no A
   Answer: A. VR(NL) = 98.69/0.982 = 100.5 kV per phase, giving 31.87 percent regulation. (Key point: Voltage regulation)

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