---
title: "150 mile, 60 Hz line: ABCD and sending end"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:150-mile-60hz-line-abcd"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/150-mile-60hz-line-abcd
site: BCT Engineering
---

# 150 mile, 60 Hz line: ABCD and sending end

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2072 Asoj, 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/150-mile-60hz-line-abcd
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A single circuit, 60 Hz, three phase transmission line is 150 miles long. The line is connected to a load of 50 MVA at a lagging power factor of 0.85 at 138 kV. The line constants are given as R = 0.1858 Ω/mile, L = 2.6 mH/mile, and C = 0.012 μF/mile. Determine the ABCD constants, sending end voltage, current and power.

## Short answer

Use long-line model with Z = 149.6∠79.27° Ω, Y = j6.786×10⁻⁴ S; γl = 0.3187∠84.63°, Zc = 469.6∠−5.37° Ω. A = D = 0.9506∠0.56°, B = 147.2∠79.45° Ω, C = 6.674×10⁻⁴∠90.18° S. Sending-end: V_S = 171.97 kV (line), I_S = 177.0 A, P_S = 45.64 MW at 0.8655 p.f. lagging.

## Answer

**Given:** $l = 150$ miles, 60 Hz; $R = 0.1858\ \Omega$/mi, $L = 2.6$ mH/mi, $C = 0.012\ \mu$F/mi; load 50 MVA at 0.85 p.f. lagging, 138 kV. 150 miles (241 km) is treated with the exact long-line model.

#### Step 1: Per-mile and total constants

$$
\begin{aligned}
z &= 0.1858 + j(2\pi\times60\times2.6\times10^{-3}) = 0.1858 + j0.9802\ \Omega/\text{mi} \\
y &= j2\pi\times60\times0.012\times10^{-6} = j4.524\times10^{-6}\ \text{S/mi} \\
Z &= zl = 27.87 + j147.03 = 149.6\angle 79.27^\circ\ \Omega \\
Y &= yl = j6.786\times10^{-4}\ \text{S}
\end{aligned}
$$

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

$$
\begin{aligned}
\gamma l &= \sqrt{ZY} = 0.3187\angle 84.63^\circ = 0.02981 + j0.31727 \\
Z_c &= \sqrt{Z/Y} = 469.6\angle -5.37^\circ\ \Omega
\end{aligned}
$$

#### Step 3: ABCD constants

$$
\begin{aligned}
A = D &= \cosh\gamma l = 0.95051 + j0.00930 = 0.9506\angle 0.56^\circ \\
\sinh\gamma l &= 0.02832 + j0.31211 = 0.3134\angle 84.82^\circ \\
B &= Z_c\sinh\gamma l = 147.2\angle 79.45^\circ\ \Omega \\
C &= \frac{\sinh\gamma l}{Z_c} = 6.674\times10^{-4}\angle 90.18^\circ\ \text{S}
\end{aligned}
$$

#### Step 4: Receiving-end values

$$
\begin{aligned}
V_R &= \frac{138}{\sqrt3} = 79.67\angle 0^\circ\ \text{kV} \\
I_R &= \frac{50\times10^6}{\sqrt3\times138\times10^3} = 209.2\angle -31.79^\circ = 177.81 - j110.19\ \text{A}
\end{aligned}
$$

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

$$
\begin{aligned}
V_S &= AV_R + BI_R = (75{,}732 + j741) + (20{,}735 + j22{,}755) \\
&= 96{,}466 + j23{,}496 = 99.29\angle 13.69^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times99.29 = 171.97\ \text{kV} \\
I_S &= CV_R + DI_R = (-0.17 + j53.17) + (170.03 - j103.09) \\
&= 169.86 - j49.92 = 177.0\angle -16.38^\circ\ \text{A}
\end{aligned}
$$

#### Step 6: Sending-end power
p.f. $= \cos(13.69^\circ + 16.38^\circ) = \cos 30.07^\circ = 0.8655$ lagging

$$
\begin{aligned}
P_S &= 3|V_S||I_S|\cos\phi_S = 3\times99.29\times10^3\times177.0\times0.8655 = 45.64\ \text{MW} \\
Q_S &= 26.42\ \text{Mvar}
\end{aligned}
$$

($P_R = 42.5$ MW, so the line loss is 3.14 MW and efficiency 93.12 %.)

**Answer:** $A = D = 0.9506\angle 0.56^\circ$, $B = 147.2\angle 79.45^\circ\ \Omega$, $C = 6.674\times10^{-4}\angle 90.18^\circ$ S; $V_S = 171.97$ kV (line), $I_S = 177.0$ A, $P_S = 45.64$ MW at 0.8655 p.f. lagging.

## Check yourself

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

1. How are the per-mile series impedance and shunt admittance obtained from R, L, C at 60 Hz?
   A. z = R + j(1/(2πfL)), y = j(1/(2πfC))
   B. z = R + j2πfL, y = j2πfC
   C. z = R + jL, y = jC
   D. z = R + j2πfL, y = 1/(j2πfC)
   Answer: B. The series reactance is ωL = 2πfL and the shunt susceptance is ωC = 2πfC; here z = 0.1858 + j0.9802 Ω/mi and y = j4.524e-6 S/mi. (Key point: Per-unit-length parameters)

2. After finding Z = zl and Y = yl, how are the ABCD constants found for this 150-mile line?
   A. γl = sqrt(ZY), Zc = sqrt(Z/Y); A = D = cosh(γl), B = Zc sinh(γl), C = sinh(γl)/Zc
   B. A = D = 1, B = Z, C = 0
   C. A = D = 1 + ZY/2, B = Z, C = Y
   D. A = D = sinh(γl), B = Z/Zc, C = Zc
   Answer: A. The answer uses the exact long-line model with hyperbolic functions of γl. (Key point: Method for ABCD of long line)

3. How are VR and IR found from the load data (50 MVA, 0.85 lagging, 138 kV)?
   A. VR = 138 kV and IR = 50 MVA/138 kV
   B. VR = 138/sqrt3 kV as reference and IR = S/(sqrt3 VL) at an angle of -cos^-1(0.85)
   C. VR = 138*sqrt3 kV and IR = S/(3 VL)
   D. VR = 138/sqrt3 kV and IR leads by 31.79°
   Answer: B. The phasor model is per phase: VR = VL/sqrt3, IR = S/(sqrt3 VL) = 209.2 A lagging by 31.79° for 0.85 p.f. (Key point: Receiving-end phase quantities)

4. Having VS = A VR + B IR = 99.29 kV per phase, what is the sending-end line voltage?
   A. 99.29 kV
   B. About 57.3 kV
   C. About 172 kV
   D. About 298 kV
   Answer: C. Line voltage = sqrt3 x phase voltage = 1.732 x 99.29 = 171.97 kV. (Key point: Sending-end voltage and line value)

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