Question 1 of 7: Line Transposition and the Long-Line Model

Question 1 (Problem 1): Line Transposition and the Long-Line Model (5 + 10 + 10 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — what transposition is and why it is needed. A three-phase overhead line is almost never built with its three conductors at the vertices of an equilateral triangle. On a flat or vertical-configuration tower the spacings between the three pairs of phases are unequal, so the flux linking each phase differs and the three phases end up with different series inductances and different shunt capacitances to earth. Transposition is the deliberate rotation of the physical positions of the three conductors: the route is divided into three equal sections and each conductor occupies each of the three tower positions for exactly one third of the length, usually by means of a transposition structure at the section boundaries.

The purpose is to make the average inductance and capacitance identical for the three phases, so that the line can legitimately be represented by a single per-phase impedance and a single shunt admittance — which is exactly what the calculation in parts (b) and (c) assumes. Without transposition a balanced set of source voltages produces unbalanced currents; the resulting negative-sequence current heats generator and motor rotors, the residual zero-sequence current inductively couples into parallel communication and pipeline circuits, and protective relays that assume balance mis-measure. On short lines the imbalance is small enough that utilities often accept it rather than pay for transposition towers; on long extra-high-voltage circuits, or where a line parallels a telephone route, transposition is standard Canadian practice.

Given. A 150 km, 138 kV three-phase line has distributed constants z = 0.17 + j0.79 Ω/km and y = j5.4 × 10⁻⁶ S/km, and delivers 15 MW at 132 kV line to line at unity power factor.

Find. The characteristic impedance, propagation constant and its two components; then the sending-end voltage, current, power factor and the transmission efficiency using the exact long-line model.

Approach. Compute Z_c and γ from the distributed constants, form the ABCD constants from the hyperbolic functions of γl, and apply them to the receiving-end phasors.

1. Part (b) — characteristic (surge) impedance. The characteristic impedance is the ratio of voltage to current in a travelling wave on the line, and for a uniform line it depends only on the two distributed constants: $$\begin{aligned}Z_{c}&=\sqrt{\dfrac{z}{y}}=\sqrt{\dfrac{0.17+j0.79}{j5.4\times10^{-6}}} \\ &=\boxed{386.84\angle -6.07^\circ\ \Omega}\end{aligned}$$ The small negative angle is the signature of a practical line: a lossless line would give a purely resistive 386.8 Ω, and the series resistance rotates it a few degrees into the fourth quadrant.
2. Propagation constant, attenuation constant and phase constant. The propagation constant follows from the same two quantities: $$\begin{aligned}\gamma&=\sqrt{zy}=\sqrt{(0.17+j0.79)(j5.4\times10^{-6})} \\ &=0.002089\angle 83.928^\circ\ \text{km}^{-1}\end{aligned}$$ Resolving it into rectangular form separates the two physical effects, because $\gamma=\alpha+j\beta$: $$\begin{gathered}\alpha=0.0002210\ \text{Np/km}, \\ \beta=0.0020772\ \text{rad/km}\end{gathered}$$
3. Reading the two constants physically. The attenuation constant says the wave amplitude falls by a factor $e^{-\alpha l}$ over the route, and the phase constant fixes the wavelength $\lambda = 2\pi/\beta = 3024.8$ km. Over the whole line $$\begin{aligned}\gamma l &= 0.313340\angle 83.928^\circ \\ &= 0.033145 + j0.311582\end{aligned}$$ so the electrical length is $\beta l = 17.852°$ — well under the quarter-wave point, which is why the exact hyperbolic model and a nominal-π model will not differ dramatically here, though only the former is asked for.
4. Part (c) — assemble the ABCD constants of the long-line model. The exact solution of the transmission-line equations gives the two-port constants in terms of the hyperbolic functions of $\gamma l$: $$\begin{gathered}A=D=\cosh\gamma l = 0.952427\angle 0.6114^\circ, \\ B=Z_{c}\sinh\gamma l = 119.2831\angle 78.0542^\circ\ \Omega, \\ C=\dfrac{\sinh\gamma l}{Z_{c}} = 0.00079711\angle 90.1985^\circ\ \text{S}\end{gathered}$$ As a check on the arithmetic, $\sinh\gamma l = 0.308353\angle 84.1264^\circ$, and the reciprocity condition $AD-BC=1$ is satisfied to the last figure carried.
5. Express the receiving-end conditions as phasors. Working per phase and taking the receiving-end voltage as the reference, $$\begin{gathered}V_{R}=\dfrac{132\times 10^{3}}{\sqrt{3}}=76210.24\ \text{V}, \\ I_{R}=\dfrac{15\times 10^{6}}{\sqrt{3}\,(132\times 10^{3})(1.0)}=65.6080\ \text{A}\end{gathered}$$ Unity power factor puts $I_{R}$ exactly in phase with $V_{R}$, so both are real.
6. Sending-end voltage. Substituting into $V_{S}=AV_{R}+BI_{R}$ term by term, $$\begin{gathered}AV_{R}=72580.54 + j774.53\ \text{V}, \\ BI_{R}=1619.85 + j7656.45\ \text{V}\end{gathered}$$ $$\begin{aligned}&V_{S} \\ &=74200.39 + j8430.97\ \text{V (per phase)}\; \\ &\Rightarrow\; \boxed{|V_{S}|_{LL}=129.346\ \text{kV at }6.482°}\end{aligned}$$ Notice that the sending-end voltage comes out below the receiving-end value. That is not an error: at 15 MW the line is loaded well below its surge-impedance loading of about 45.0 MW, so the line charging current dominates and the line behaves capacitively — the Ferranti effect in its mild, partial form.
7. Sending-end current and power factor. The second two-port equation gives $$\begin{aligned}I_{S}&=CV_{R}+DI_{R} \\ &=(-0.2105 + j60.7473)+(62.4832 + j0.6668) \\ &=87.4619\angle 44.6022^\circ\ \text{A}\end{aligned}$$ The angle between $V_{S}$ and $I_{S}$ is $6.482°-44.602°=-38.120°$, so the current leads and $$\boxed{\text{p.f.}_{S}=\cos(38.120°)=0.7867\ \text{leading}}$$
8. Transmission efficiency. The three-phase sending-end power and the efficiency follow directly: $$\begin{gathered}P_{S}=3|V_{S}||I_{S}|\cos\phi_{S}=15.4153\ \text{MW}, \\ \boxed{\eta=\dfrac{P_{R}}{P_{S}}=\dfrac{15}{15.4153}=97.31\%}\end{gathered}$$ The 0.415 MW difference is the copper loss in the 150 km of conductor, which is what the real part of $z$ was carrying all along.