---
title: "400 km long line, 60 MVA at 220 kV"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:long-line-400km-60mva"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/long-line-400km-60mva
site: BCT Engineering
---

# 400 km long line, 60 MVA at 220 kV

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2078 Chaitra, 10 marks
- 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/long-line-400km-60mva
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 3-Φ line is 400 km long. The line constants are z = 0.105 + j0.3768 Ω per phase per kilometre and y = 0 + j2.822 × 10⁻⁶ ℧ per kilometre. The line delivers 60 MVA at 0.9 power factor lagging at 220 kV. Find the sending end voltage, current, power factor, active and reactive power loss of the line.

## Short answer

Using exact long-line model: total Z=156.46∠74.43° Ω, Y=j1.1288×10⁻³ S, γl=0.4203∠82.21°, Zc=372.3∠−7.79° Ω, A=0.9163∠1.44°, B=152.1∠74.89° Ω. Sending line voltage V_S=231.59 kV, current I_S=152.7 A, pf=0.9244 leading, active loss 2.64 MW, reactive loss −49.53 Mvar.

## Answer

**Given:** $l = 400$ km, $z = 0.105 + j0.3768\ \Omega$/km, $y = j2.822\times10^{-6}$ S/km, load 60 MVA at 0.9 p.f. lagging, 220 kV. A 400 km line is a long line, so the exact (distributed) model is used. Values are per phase.

#### Step 1: Total parameters

$$
\begin{aligned}
Z &= zl = 42 + j150.72 = 156.46\angle 74.43^\circ\ \Omega \\
Y &= yl = j1.1288\times10^{-3}\ \text{S}
\end{aligned}
$$

#### Step 2: Propagation constant and surge impedance

$$
\begin{aligned}
\gamma l &= \sqrt{ZY} = 0.4203\angle 82.21^\circ = 0.05693 + j0.41638 \\
Z_c &= \sqrt{Z/Y} = 372.3\angle -7.79^\circ\ \Omega
\end{aligned}
$$

#### Step 3: ABCD constants

$$
\begin{aligned}
A = D &= \cosh\gamma l = 0.91604 + j0.02304 = 0.9163\angle 1.44^\circ \\
\sinh\gamma l &= 0.05209 + j0.40511 = 0.4084\angle 82.67^\circ \\
B &= Z_c\sinh\gamma l = 152.1\angle 74.89^\circ\ \Omega \\
C &= \frac{\sinh\gamma l}{Z_c} = 1.097\times10^{-3}\angle 90.46^\circ\ \text{S}
\end{aligned}
$$

#### Step 4: Receiving-end quantities

$$
\begin{aligned}
V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\
I_R &= \frac{60\times10^6}{\sqrt3\times220\times10^3} = 157.46\angle -25.84^\circ = 141.71 - j68.64\ \text{A}
\end{aligned}
$$

$P_R = 60 \times 0.9 = 54$ MW, $Q_R = 60 \times 0.4359 = 26.15$ Mvar

#### Step 5: Sending-end voltage and current

$$
\begin{aligned}
V_S &= AV_R + BI_R = (116{,}350 + j2926) + (15{,}695 + j18{,}083) \\
&= 132{,}050 + j21{,}009 = 133.71\angle 9.04^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times133.71 = 231.59\ \text{kV} \\
I_S &= CV_R + DI_R = (-1.11 + j139.34) + (131.40 - j59.61) \\
&= 130.28 + j79.74 = 152.7\angle 31.47^\circ\ \text{A}
\end{aligned}
$$

#### Step 6: Power factor and power
Angle between $V_S$ and $I_S$ $= 9.04^\circ - 31.47^\circ = -22.43^\circ$ (current leads).

Sending-end p.f. $= \cos 22.43^\circ = 0.9244$ **leading**

$$
S_S = 3V_SI_S^* = 56.64 - j23.38\ \text{MVA}
$$

#### Step 7: Losses

$$
\begin{aligned}
P_{loss} &= P_S - P_R = 56.64 - 54 = 2.64\ \text{MW} \\
Q_{loss} &= Q_S - Q_R = -23.38 - 26.15 = -49.53\ \text{Mvar}
\end{aligned}
$$

The negative reactive loss means the line's shunt capacitance generates 49.53 Mvar more than its series inductance absorbs. This is why the sending-end current leads.

**Answer:** $V_S = 231.59$ kV (line), $I_S = 152.7$ A, p.f. $= 0.9244$ leading, active power loss $= 2.64$ MW, reactive power loss $= -49.53$ Mvar (net 49.53 Mvar generated by the line). Efficiency $= 54/56.64 = 95.35\%$.

## Check yourself

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

1. Why is the exact distributed model used for the 400 km line?
   A. Because it is the only model usable for 220 kV
   B. Because it has a lower load
   C. Because the nominal-T cannot be used for 3-phase lines
   D. Because 400 km exceeds 250 km, so it is a long line
   Answer: D. Lines above about 250 km are long and need distributed parameters. (Key point: Method choice)

2. With z = 0.105 + j0.3768 Ω/km and l = 400 km, the total series impedance Z is:
   A. 42 + j150.72 Ω
   B. 0.105 + j0.3768 Ω
   C. 4.2 + j15.07 Ω
   D. 420 + j1507 Ω
   Answer: A. Z = z x l = 400 x (0.105 + j0.3768) = 42 + j150.72 Ω. (Key point: Total parameters)

3. The load current IR for 60 MVA at 220 kV is:
   A. 272.7 A
   B. 157.46 A
   C. 60 A
   D. 472 A
   Answer: B. IR = 60e6/(sqrt(3) x 220e3) = 157.46 A, at an angle of -25.84° for 0.9 lagging. (Key point: Receiving current)

4. The sending-end power factor is 0.9244 leading although the load lags. Why?
   A. Because the load was leading
   B. Because the series resistance is negative
   C. Because the line was assumed lossless
   D. Because the line's charging current exceeds the lagging load current component
   Answer: D. The capacitance generates much reactive power, so IS leads VS. (Key point: Sending-end power factor)

5. Qloss = -49.53 Mvar means:
   A. The line absorbs 49.53 Mvar
   B. The line generates a net 49.53 Mvar
   C. The load absorbs 49.53 Mvar
   D. The line has no reactive effect
   Answer: B. Qloss = QS - QR is negative since shunt capacitance generates more than series inductance absorbs. (Key point: Reactive loss sign)

## Related in Power System Analysis I

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