---
title: "Equivalent-T of 300 km long line"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:long-line-equivalent-t-300km"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/long-line-equivalent-t-300km
site: BCT Engineering
---

# Equivalent-T of 300 km long line

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2075 Baisakh, 6 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-equivalent-t-300km
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 220 kV, three phase transmission line is 300 km long. The line has resistance of 0.12 ohm per phase per km, line inductance of 1.5 mH per phase per km and shunt capacitance of 2 μF per phase per km. Calculate ABCD parameters of the line with long line model in equivalent T-model. If the line is excited by 220 kV from the sending end, calculate the receiving voltage.

## Short answer

Per-phase total Z=145.9∠75.71° Ω, Y=j0.1885 S, γl=5.244∠82.86°, Zc=27.82∠-7.14° Ω. ABCD: A=D=0.8433∠-47.0°, B=31.31∠-80.12° Ω, C=0.04046∠-65.83° S. Equivalent-T: Z'/2=18.51∠-58.73° Ω, Y'=0.04046∠-65.83° S. With VS=220 kV, open-circuit receiving voltage VR(L)=260.88 kV.

## Answer

**Given (per phase, per km):** $r = 0.12\ \Omega$, $L = 1.5$ mH, $C = 2\ \mu$F, $l = 300$ km, $f = 50$ Hz (assumed), $V_S = 220$ kV.

#### Step 1: Line constants

$$
\begin{aligned}
z &= 0.12 + j(2\pi\times50\times1.5\times10^{-3}) = 0.12 + j0.4712\ \Omega/\text{km} \\
y &= j2\pi\times50\times2\times10^{-6} = j6.283\times10^{-4}\ \text{S/km} \\
Z &= zl = 36 + j141.37 = 145.9\angle 75.71^\circ\ \Omega \\
Y &= yl = j0.1885\ \text{S}
\end{aligned}
$$

#### Step 2: $\gamma l$ and $Z_c$

$$
\begin{aligned}
\gamma l &= \sqrt{ZY} = 5.244\angle 82.86^\circ = 0.6521 + j5.2032 \\
Z_c &= \sqrt{Z/Y} = 27.82\angle -7.14^\circ\ \Omega
\end{aligned}
$$

#### Step 3: ABCD constants (long line)

$$
\begin{aligned}
A = D &= \cosh\gamma l = \cosh\alpha l\cos\beta l + j\sinh\alpha l\sin\beta l \\
&= 0.5751 - j0.6167 = 0.8433\angle -47.0^\circ \\
\sinh\gamma l &= 0.3296 - j1.0762 = 1.126\angle -72.97^\circ \\
B &= Z_c\sinh\gamma l = 31.31\angle -80.12^\circ\ \Omega \\
C &= \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S}
\end{aligned}
$$

#### Step 4: Equivalent-T circuit
In the equivalent T, each series arm is $Z'/2$ and the shunt branch is $Y'$, with $A = 1 + Z'Y'/2$ and $C = Y'$:

$$
\begin{aligned}
Y' &= C = \frac{\sinh\gamma l}{Z_c} = 0.04046\angle -65.83^\circ\ \text{S} \\
\frac{Z'}{2} &= Z_c\tanh\frac{\gamma l}{2} = \frac{A - 1}{C} = 18.51\angle -58.73^\circ\ \Omega
\end{aligned}
$$

```text
 IS     Z'/2          Z'/2     IR
 o-->--[18.51<-58.7]-+-[18.51<-58.7]-->--o
                     |
 VS            Y' = 0.04046<-65.8 S        VR
                     |
 o-------------------+-------------------o
```

#### Step 5: Receiving-end voltage (open circuit, $I_R = 0$)

$$
\begin{aligned}
V_R &= \frac{V_S}{A} = \frac{127.02\angle 0^\circ}{0.8433\angle -47.0^\circ} = 150.62\angle 47.0^\circ\ \text{kV/phase} \\
V_{R(L)} &= \sqrt3\times150.62 = 260.88\ \text{kV}
\end{aligned}
$$

The receiving voltage is higher than 220 kV (Ferranti effect).

**Answer:** $A = D = 0.8433\angle -47.0^\circ$, $B = 31.31\angle -80.12^\circ\ \Omega$, $C = 0.04046\angle -65.83^\circ$ S; equivalent T: $Z'/2 = 18.51\angle -58.73^\circ\ \Omega$ per arm, $Y' = 0.04046\angle -65.83^\circ$ S; no-load $V_R = 260.88$ kV.

Note: 2 μF/km is about 200 times the capacitance of a real overhead line (typically about 0.01 μF/km), which is why $\gamma l$ and the angles are unusual. With $C = 0.01\ \mu$F/km the same steps give $A = 0.9342\angle 1.02^\circ$, $B = 142.7\angle 76.04^\circ\ \Omega$, $C = 9.217\times10^{-4}\angle 90.33^\circ$ S and $V_R = 235.49$ kV.

## Check yourself

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

1. The shunt admittance per km is y = jωC. Which expression is used for it?
   A. j2π x 50 x 2e-6 = j6.283e-4 S/km
   B. 1/(2π x 50 x 2e-6)
   C. 2π x 50 x 1.5e-3
   D. 0.12 + j0.4712
   Answer: A. y = j2πfC = j6.283e-4 S/km. (Key point: Line constants)

2. A, B and C of the long line are:
   A. cosh(γl), sinh(γl), Zc
   B. Zc, cosh(γl), sinh(γl)
   C. cosh(γl), Zc sinh(γl), sinh(γl)/Zc
   D. sinh(γl), cosh(γl), 1/Zc
   Answer: C. A = D = cosh γl, B = Zc sinh γl, C = sinh γl / Zc. (Key point: ABCD of long line)

3. In the equivalent T circuit, the shunt branch Y' and series arm Z'/2 are:
   A. Y' = B, Z'/2 = Zc
   B. Y' = A, Z'/2 = B
   C. Y' = C, Z'/2 = (A - 1)/C
   D. Y' = 1/C, Z'/2 = A + 1
   Answer: C. Matching C = Y' and A = 1 + Z'Y'/2 gives Z'/2 = (A - 1)/C = Zc tanh(γl/2). (Key point: Equivalent-T parameters)

4. With IR = 0, VR is found from:
   A. VR = VS x A
   B. VR = VS / B
   C. VR = VS / A
   D. VR = VS x C
   Answer: C. VS = A VR at no load, giving VR = 150.62 kV per phase (260.88 kV line), higher than VS (Ferranti). (Key point: No-load receiving voltage)

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