Short Transmission Lines

Part 2 is finished. Chapter 5 gave the resistance per kilometre and the corrections that make it larger than the handbook's d.c. figure; Chapter 6 gave the inductance \(L=2\times10^{-7}\ln(D_{eq}/D_s)\); Chapter 7 gave the capacitance \(C_n=2\pi\varepsilon/\ln(D_{eq}/r)\); Chapter 8 showed how bundling alters both; Chapter 9 did the same job for a cable. The outcome in every case was three numbers per kilometre — \(r\), \(x=\omega L\) and \(b=\omega C\) — and nothing was done with them.

Part 3 does something with them. The question it asks is the one an operator asks: with a known load drawing a known current at a known power factor at the far end of the line, what voltage must be applied at the near end, how much power is lost on the way, and how far can the load be increased before something gives? Those three questions are line performance, and the quantities that answer them have standard names — the sending-end voltage and current, the voltage regulation, the transmission efficiency, and the power angle.

Every calculation in Part 3 is made per phase. Chapter 3 established the licence: a balanced three-phase network is solved as one single-phase circuit carrying the phase voltage \(V_L/\sqrt3\), the line current \(I_L\) and one third of the three-phase power, with the neutral as the reference. Every \(V\) in this chapter is therefore a phase voltage unless it is explicitly marked \(V_{LL}\), and every impedance is the impedance of one conductor. When a numerical answer is wanted in three-phase terms it is converted back at the end, never in the middle.

The line itself is a distributed object: the resistance and inductance are spread along it, and so is the capacitance to earth, so strictly the voltage and current vary continuously with distance and satisfy a pair of differential equations. Chapter 12 solves them. For most lines that machinery is unnecessary, and lines are classified by how much of it can be dispensed with:

The lengths in that table are conventional, not physical. What actually decides the class is how large the shunt admittance is compared with the load — a quantity that depends on the voltage and the loading as much as on the kilometres, as the next section shows and as Chapter 9's cables demonstrate emphatically.

The short-line model makes exactly one approximation: the shunt admittance \(Y=j\omega C\) is set to zero. That is a strong statement, and it deserves to be tested rather than asserted.

Take a 50 km, 33 kV overhead line with the typical parameters of Chapters 6 and 7: \(C_n = 0.0095\ \mu\text{F/km}\), so the whole line has \(C = 0.475\ \mu\)F. Its charging current, drawn whether or not there is any load, is

A 10 MVA load on the same line draws \(10\times10^{6}/(\sqrt3\times33\,000) = 175\) A. The charging current is 1.6% of it — and worse for the approximation than that figure suggests it is, because the charging current is in quadrature with the load current, so it changes the magnitude of the total by only about \(0.5\times(0.016)^2\), which is one part in eight thousand. Neglecting \(C\) on such a line is not an approximation so much as a rounding.

The second test is independent of the load. Chapter 12 will show that the exact \(A\) constant of a line is \(\cosh(\gamma l)\), which for a lossless line is \(\cos(\beta l)\) with \(\beta=\omega\sqrt{LC}\). For the line above, \(\sqrt{LC} = \sqrt{1.3\times10^{-3}\times9.5\times10^{-9}} = 3.51\times10^{-6}\) s/km and \(\beta = 1.10\times10^{-3}\) rad/km. Even at 80 km, \(\beta l = 0.088\) rad, or \(5.1^\circ\), and \(\cos(\beta l) = 0.9961\). The short-line model takes \(A=1\); the error is four parts in a thousand.

With the shunt branch gone, nothing remains but a series resistance and a series inductance between the two ends. Because there is no path off the line, the current that enters at the sending end is the same current that leaves at the receiving end — the single most useful consequence of the approximation, and the one that makes every calculation in this chapter elementary.

The equation \(\mathbf{V}_s=\mathbf{V}_r+\mathbf{I}\mathbf{Z}\) is a statement about complex numbers, and the only difficulty students meet with it is forgetting that fact. The arithmetic is fixed by choosing \(\mathbf{V}_r\) as the reference phasor, so that \(\mathbf{V}_r = V_r\angle0^\circ = V_r+j0\), and writing the current with the sign its power factor demands:

Multiplying out for the lagging case, which is the usual one, separates the sending-end voltage into its two rectangular components:

Both components have a clear meaning. The in-phase part is what raises the magnitude of the sending-end voltage above the receiving-end value; the quadrature part is what turns \(\mathbf{V}_s\) through the angle \(\delta\) which Section 10-8 identifies as the power angle. Notice already that at a lagging power factor both terms of the in-phase part are positive and both add to \(V_r\), while at a leading power factor \(\sin\phi_r\) changes sign and the reactance term subtracts. That single sign is responsible for most of what the rest of the chapter has to say.

The algebra of Section 10-3 is complete, but it hides the geometry, and the geometry is what makes the results memorable. Draw \(\mathbf{V}_r\) horizontally from an origin \(O\) to a point \(F\). The current lags it by \(\phi_r\), so draw \(\mathbf{I}\) from \(O\) at that angle below the horizontal. From \(F\) add \(\mathbf{I}R\), which is parallel to the current because a resistance produces a drop in phase with the current it carries; call its tip \(B\). From \(B\) add \(\mathbf{I}X\), which leads the current by \(90^\circ\); call its tip \(D\). Then \(OD\) is \(\mathbf{V}_s\).

Drop a perpendicular from \(D\) onto the horizontal, meeting it at \(G\). The triangle \(OGD\) is right-angled, so

The second line is the same statement made with the current, rather than the voltage, taken as the reference direction — the form in which it is usually quoted, and the form in which the two right-hand terms are recognisably "the component of everything along the current" and "the component of everything across it". Either version expands to the same thing as the rectangular resolution of Section 10-3, and the power factor at the sending end follows from the same triangle:

Exact as these expressions are, they are more machinery than the problem needs. On any real line \(FG\) and \(GD\) are small compared with \(OF\) — the drop is a few per cent of the voltage — and \(D\) lies very nearly on the horizontal. Then \(OD \approx OG\), and the sending-end magnitude is obtained by simply adding the in-phase part of the drop to \(V_r\):

A transmission line, seen from its two ends, is a four-terminal network: two terminals at the sending end, two at the receiving end. It is linear, it is passive, and it contains no independent sources. For any network of that kind the sending-end quantities must be linear combinations of the receiving-end quantities, and the four coefficients of the combination are the generalised circuit constants, universally written \(A\), \(B\), \(C\) and \(D\):

The short line's constants are read straight off Section 10-3. Comparing \(\mathbf{V}_s=\mathbf{V}_r+\mathbf{Z}\mathbf{I}_r\) with the first relation gives \(A=1\) and \(B=\mathbf{Z}\); comparing \(\mathbf{I}_s=\mathbf{I}_r\) with the second gives \(C=0\) and \(D=1\).

Three things make the ABCD form worth adopting even for a network as simple as this one. First, it is the form in which Chapters 11 and 12 will state their results, so the short line becomes the special case \(A=1,\ C=0\) of a single framework rather than a separate piece of theory. Second, the constants are exactly what is needed to invert the problem: given the sending-end conditions, the receiving-end quantities follow by inverting the matrix, and because the determinant is unity the inverse is written down at sight:

Third, networks in cascade multiply. A transformer of series impedance \(\mathbf{Z}_T\) feeding a line of impedance \(\mathbf{Z}_L\) is the product of two matrices, and because both are of the same form the product is trivial — but the same rule handles a line feeding a transformer feeding another line, and it is how a whole transmission path is reduced to four numbers. Chapter 4's per-unit system makes that cascade painless by removing the turns ratios from the matrices altogether.

Consumers' equipment is designed for a nominal voltage and tolerates a few per cent of departure from it. The load on a line, however, varies through the day by a factor of two or more (Chapter 30), and every change in load changes the drop along the line. The measure of that sensitivity is the voltage regulation: the rise in receiving-end voltage, expressed as a percentage of the full-load value, when the full load at a specified power factor is thrown off while the sending-end voltage is held constant.

The definition is easy to apply to a short line because throwing off the load makes \(I=0\), and with no current there is no drop: the no-load receiving-end voltage is simply \(V_s\).

Substituting the approximate drop of Section 10-4 gives the working formula and, with it, the whole story of how regulation depends on the load:

At a lagging power factor both terms are positive and regulation is large. As the power factor improves the reactive term shrinks. At unity it vanishes and only \(IR\cos\phi_r\) remains. At a leading power factor the reactive term reverses and begins to cancel the resistive one — and at one particular leading power factor it cancels it exactly. Setting the approximate drop to zero gives that condition immediately as \(\tan\phi_r = R/X\); the exact condition follows from demanding \(|V_s|=|V_r|\) in the full expression:

Beyond the zero-regulation point the regulation is negative: the receiving-end voltage is higher than the sending-end voltage under load, and throwing the load off makes it fall. A leading load draws current that flows backwards through the reactance in the phasor sense and produces a rise instead of a drop. It is the same mechanism as the Ferranti effect of Chapter 14, where the leading current is drawn not by a capacitive load but by the line's own capacitance — which is why a short line, having no capacitance in its model, cannot show the Ferranti effect at all.

The second performance figure is the fraction of the power fed in at one end that emerges at the other. Since the short-line model has no shunt branch, there is no leakage and no charging current, and the only loss is the \(I^2R\) of the three conductors:

Substituting \(I = P_r/(\sqrt3\,V_{r(LL)}\cos\phi_r)\) turns the loss into a statement about the quantities the designer controls:

Chapter 2 arrived at the proportionality \(1/(V^2\cos^2\phi)\) by an argument about conductor volume; here it reappears as a statement about loss at fixed conductor size. The two are the same fact seen from opposite sides: for a given loss, raising the voltage lets the conductor shrink; for a given conductor, raising the voltage cuts the loss. Both scale as the inverse square, and both are equally sensitive to the power factor.

One structural point separates a line from the transformer of an earlier course. A transformer has iron loss that is present whether or not it is loaded, so its efficiency rises to a maximum where copper loss equals iron loss and falls again. The short line has no no-load loss whatever, so its efficiency simply falls as the load rises — it is highest when the line is doing nothing. Quoting a line's efficiency therefore means nothing unless the load at which it was computed is quoted too, and the convention is to state it at full load.

So far the load has been given and the line asked to carry it. Turn the question round: with both ends held at fixed voltages — as they are in a real network, where generators and tap-changers regulate the busbars — how much power flows, and what determines it?

Let \(\mathbf{V}_s = V_s\angle\delta\), \(\mathbf{V}_r=V_r\angle0^\circ\) and \(\mathbf{Z}=|Z|\angle\theta\). The current is fixed by the two voltages, and the complex power delivered at the receiving end follows from its definition:

Separating real and imaginary parts gives the two quantities that the receiving-end busbar actually sees:

The maximum receivable power is found by inspection: \(P_r\) is largest when \(\cos(\theta-\delta)=1\), that is when \(\delta=\theta\).

For an overhead line the resistance is small enough that setting \(R=0\) costs little accuracy, and then \(|Z|=X\) and \(\theta=90^\circ\). The two expressions collapse into the pair that the rest of this book uses constantly:

The theoretical maximum at \(\delta=\theta\) is never reached on a short line, and the reason is worth stating plainly. Example 5 evaluates \(P_{r,\max}\) for a 20 km, 132 kV line and finds it to be over 700 MW, while the thermal rating of the same conductors is perhaps 100 MW. The line melts long before it becomes unstable. A short line is therefore thermally limited. As length grows, \(|Z|\) grows with it and \(P_{r,\max}\) falls in inverse proportion, until on a long line the stability limit falls below the thermal one and the line becomes stability limited — the transition that Chapters 14 and 28 examine, and the reason series capacitors are installed on long lines to reduce the effective \(X\).

Sections 10-6 and 10-7 both reduce to the same complaint: the line is carrying more current than the real power requires, because the load draws reactive power as well. The reactive component contributes nothing to \(P_r\) but is fully charged for in \(I^2R\) and, through \(IX\sin\phi_r\), dominates the voltage drop.

The remedy is to supply the reactive power locally. A capacitor bank connected at the receiving-end busbar generates \(Q_c\) on the spot, so the line carries only the difference. Requiring the load's real power \(P\) to be delivered at an improved power factor \(\cos\phi_2\) instead of \(\cos\phi_1\) fixes the rating:

The effect on the line is immediate and quadratic in the current: raising the power factor from 0.8 to 0.95 cuts the current by a factor \(0.8/0.95=0.842\) and the loss by \(0.842^2=0.709\), a 29% reduction, while the drop falls by rather more because the \(X\sin\phi_r\) term shrinks faster than the current does. Example 6 works both figures through.

Two other forms of compensation appear later in the book and are mentioned here so that their place in the argument is clear. A series capacitor inserted in the line subtracts from \(X\), which reduces the drop and raises \(P_{r,\max}\) directly — the natural remedy for a line that is long rather than heavily loaded (Chapter 38). A shunt reactor does the opposite of a capacitor bank and is used on lightly loaded lines whose own capacitance is producing too much reactive power, which is a problem the short-line model cannot even express (Chapter 14).

That last point is the honest summary of this chapter's limits. Everything derived here follows from setting \(Y=0\), so every phenomenon that lives in the shunt branch is invisible: no charging current, no Ferranti rise on open circuit, no surge impedance, no reactive generation by the line itself, no travelling waves. All of them return in Chapters 11 to 14. What survives unchanged is the framework — per-phase working, the phasor equation, the ABCD matrix, the definitions of regulation and efficiency, and the power-angle relation — and that framework is why the short line is studied first even though few interesting lines are short.

Work per phase with the receiving-end voltage as reference, and state at the end whether the exact or the approximate drop was used. Difficulty rises down the list.