Ferranti Effect, Surge Impedance Loading and Tuned Lines

In 1890 the Deptford station of the London Electric Supply Corporation was sending power into the city over a 10 kV cable, and Sebastian Ziani de Ferranti found the voltage at the far end of a lightly loaded circuit standing above the voltage at the generator. That is not what a circuit made of resistance and inductance can do. Every intuition built on Ohm's law says a load pulls the voltage down, and the further away it is, the further down.

The resolution is that a transmission line is not only resistance and inductance. Chapter 7 gave every metre of it a capacitance to earth, and Chapter 12 spread that capacitance uniformly along the length rather than parking it in a lump. The current that flows into that distributed capacitance does not vanish when the load is switched off — it is drawn by the line itself. And because it is a capacitive current, it leads the voltage by very nearly \(90^\circ\), so when it passes through the line's series inductive reactance it produces a drop that is very nearly \(180^\circ\) out of phase with the voltage. A negative drop is a rise.

Chapter 13 already met the result in a different guise. Regulation was written \(\big(|V_s|/|A| - |V_r|\big)/|V_r|\), with the no-load voltage \(|V_s|/|A|\); and \(A = \cosh\gamma l\), whose magnitude is less than one for every real line. Every line therefore has a no-load voltage above its sending-end voltage. This chapter asks how large, why, and what to do about it — and then follows the same reactive-power argument to its two other consequences: the loading at which the effect exactly disappears, and the length at which it would disappear on its own.

Open the far end of the line, so \(\mathbf{I}_r = 0\), and read the first row of the two-port relation of Chapter 12:

That is the whole effect, stated exactly and with no approximation whatever. To see why the ratio exceeds unity, drop the resistance — legitimate on an EHV line, where Chapter 8's bundling has already made \(r\) small — so that \(\gamma = j\beta\) with \(\beta = \omega\sqrt{LC}\). The hyperbolic cosine of an imaginary argument is a circular cosine:

Expanding the secant for a moderate line recovers the estimate used in Chapter 11, where the shunt admittance was still a lump:

The same result falls out of the nominal-π circuit directly. With \(\mathbf{I}_r=0\), that circuit gives \(\mathbf{V}_s = (1+\mathbf{YZ}/2)\mathbf{V}_r\); putting \(r=0\) so that \(\mathbf{Z}=j\omega Ll\) and \(\mathbf{Y}=j\omega Cl\),

Three routes, one answer. The nominal circuit is the two-term expansion, the secant is the exact lossless result, and \(1/|\cosh\gamma l|\) is exact for a real line. How much the routes differ is a fair measure of when a lumped model may be trusted, and the table says: not far.

Up to 200 km the estimate is good to a twentieth of a percentage point. At 400 km it is three-quarters of a point low, which is already enough to matter when an insulation margin is being checked. Beyond 600 km it collapses entirely, because the quartic term it discarded has become comparable with the quadratic one it kept.

The engineering consequence is severe and immediate. A 400 km, 400 kV line energised from one end with its far end open settles at 440 kV — a 40 kV overvoltage on equipment specified for a maximum of 420 kV. No 400 kV line is ever switched in without shunt reactors already connected, and the switching sequence at commissioning is written around exactly this number.

The formula \(\sec\beta l\) is correct but not yet an explanation. Chapter 12 gave the physical picture: the voltage anywhere on a line is the sum of an incident wave travelling toward the load and a reflected wave travelling back from it,

Set \(\mathbf{I}_r = 0\). Both coefficients collapse to \(\mathbf{V}_r/2\): at the open end the reflected wave equals the incident wave exactly, which is what "total reflection from an open circuit" means. Writing \(\gamma = \alpha+j\beta\),

Follow the two terms as \(x\) grows from the open end toward the source. The incident term grows in magnitude as \(e^{\alpha x}\) and rotates forward through \(+\beta x\). The reflected term shrinks as \(e^{-\alpha x}\) and rotates backward through \(-\beta x\). At \(x=0\) they are two equal, co-linear phasors that add to \(\mathbf{V}_r\). At any \(x>0\) they have separated by an angle \(2\beta x\), and two phasors of nearly equal length that have opened out like scissors have a resultant shorter than their arithmetic sum. The sending-end voltage is therefore smaller than \(V_r\), which is the Ferranti effect drawn rather than calculated.

The construction also shows why attenuation is not the cause. If \(\alpha\) were zero, the two halves would keep their length exactly and the resultant would be \(V_r\cos\beta l\) — still less than \(V_r\), and by more than the lossy case. Attenuation slightly reduces the Ferranti rise, because it makes the incident half longer than the reflected half and so brings the parallelogram back toward the axis. The rise is a phase phenomenon, not an amplitude one.

The current that produces all this deserves its own accounting. With the far end open, the second row of the two-port gives \(\mathbf{I}_s = C\,\mathbf{V}_{r,\text{NL}}\), and since \(C=\sinh\gamma l/Z_c\), which for a lossless line is \(j\sin\beta l/Z_s\), the sending-end current is purely capacitive. For a short enough line the whole calculation collapses to something one can do in the head: the line looks like a capacitor of total susceptance \(B = \omega C l\), and

For the 400 km, 400 kV line of Chapter 12, with \(b = \omega C = 3.6\times10^{-6}\) S/km, that is \(Q_{\text{ch}} = (400\times10^{3})^{2}\times1.44\times10^{-3} = 230\) MVAr — nearly half the line's own natural loading, produced by the line whether anyone wants it or not. That reactive power has to go somewhere. At no load it flows back into the source, which must be able to absorb it; at light load the generators are pushed toward the underexcited region of their capability curve, where Chapter 26 will show their stability margin is smallest.

The remedy is a shunt reactor, and its size follows from a short calculation that is worth doing properly rather than guessing. Connect a reactor of susceptance \(B_L\) across the open receiving end, so \(\mathbf{I}_r = -jB_L\mathbf{V}_r\), and demand that the receiving voltage come back to the sending value. On a lossless line:

Two practical points follow. A reactor sized for no load makes the full-load regulation slightly worse, because at full load the line needs all the reactive help it can get; EHV reactors are therefore switchable, and are taken out as the load rises. And a reactor connected permanently at the line end — a bus reactor — must be disconnected before the line is de-energised on that side, whereas a line reactor connected inside the circuit breaker follows the line automatically. Chapter 34 develops the switching philosophy; here it is enough to know that the size comes from \(\tan(\beta l/2)/Z_s\).

Sections 14-1 to 14-4 have been about the case where the line's capacitance dominates. Turn the question round: is there a load at which the two reactive effects cancel, so that the line neither generates nor absorbs?

Write the balance directly. Per unit length, the shunt capacitance produces \(V^{2}\omega C\) and the series inductance consumes \(I^{2}\omega L\). Setting them equal:

The ratio of voltage to current at which the balance holds is exactly the surge impedance that Chapter 12 obtained by an entirely different route — solving the wave equation and taking the ratio of the two solution constants. That the reactive-balance argument and the wave argument give the same number is not a coincidence: both are statements that a line terminated in \(Z_s\) is indistinguishable from an infinitely long line, and an infinite line has nowhere to store a reactive imbalance.

Two things about \(Z_s\) deserve emphasis. First, it is resistive, even though it is built entirely from two reactive elements. That is what makes it a natural terminating load: a resistor of \(Z_s\) ohms absorbs the incident wave completely and reflects nothing, which is why Chapter 12 called such a line "infinite" and why surge impedance is the impedance that matters for the lightning and switching surges of Chapter 38.

Second, it is remarkably insensitive to everything. Both \(L\) and \(C\) depend logarithmically on the ratio \(D/r\) of Chapters 6 and 7 — \(L\) proportional to \(\ln(D/r')\) and \(C\) inversely proportional to \(\ln(D/r)\) — so their ratio depends on the square of that logarithm and moves very little as the geometry changes. Every overhead line, at every voltage, ends up with a surge impedance between about 250 and 400 Ω. Bundling reduces it, because it raises \(C\) and lowers \(L\) together, and that is the reason a bundled 400 kV line has a natural loading four times that of a single-conductor 220 kV line rather than the \((400/220)^{2}=3.3\) that voltage alone would give.

The cable row is the exception that proves the rule, and it is the reason a.c. cables cannot be run over long distances. Its capacitance is thirty times that of an overhead line, because the conductor sits millimetres from an earthed sheath rather than metres from the ground. Its surge impedance is a tenth as large, its charging current per kilometre is enormous, and beyond a few tens of kilometres the charging current alone reaches the conductor's thermal rating with no room left for load. Long submarine links are therefore d.c., which is where Chapter 38 begins.

Chapter 12 substituted \(\mathbf{I}_r = \mathbf{V}_r/Z_s\) into the lossless solution and obtained \(\mathbf{V}(x) = \mathbf{V}_re^{\,j\beta x}\) and \(\mathbf{I}(x) = \mathbf{I}_re^{\,j\beta x}\). Both magnitudes are constant along the entire line; only the phase advances. Everything else follows from those two lines of algebra.

A caution about vocabulary is needed here, because two different quantities both get called "regulation". At SIL there is no voltage drop: \(|V_s| = |V_r|\), and the profile between them is flat. But the regulation as Chapter 13 defined it — the rise when the load is thrown off with \(|V_s|\) held — is not zero at SIL. Throwing off the load returns the line to the open-circuit condition of Section 14-2, so

which for the 400 km line is 9.98% — precisely the Ferranti rise. A flat profile at the operating point and a large excursion when the load disappears are entirely compatible, and confusing the two is a reliable way to lose marks.

Away from SIL the profile tilts, and the direction of the tilt is the reactive balance again. Below SIL the capacitance wins and the voltage rises toward the far end; above SIL the inductance wins and it falls. The figure shows all four cases on the same 400 km line, with the receiving end held at 400 kV.

Read the sending-end values off the left edge: 364 kV at no load, 373 kV at half SIL, 400 kV at SIL, 441 kV at one and a half SIL. The same physical line requires a source voltage spanning 77 kV, purely because of what is hanging on its far end. Managing that span is the daily work of voltage control, and the SIL is the point about which the whole problem is symmetric.

Because SIL is a natural unit, line loading is quoted in multiples of it throughout the industry, and the practice is not merely convenient — the loading limit itself turns out to have a clean expression in those units. Take the lossless line, whose \(B = jZ_s\sin\beta l\) has magnitude \(Z_s\sin\beta l\) and angle \(90^\circ\), and put it into the power-transfer formula of Chapter 11:

A practical operating limit needs an angle limit, since \(\delta\) cannot be pushed to \(90^\circ\) without losing all stability margin. Chapter 28 will justify \(\delta \le 30^\circ\) for a line that must survive a fault; adopting that here gives a loadability curve.

This is the St Clair curve of line loadability, obtained here from three lines of algebra rather than from the empirical plots of the 1950s. A twin-bundle 400 kV line can carry about 1.25 kA per phase thermally, which is \(\sqrt3\times400\times1.25 = 866\) MW, or 1.61 SIL. Comparing that with the table: the two limits are equal where \(\sin\beta l = 0.5/1.61 = 0.311\), that is at \(\beta l = 18.1^\circ\) or 294 km. Below that length the conductor temperature binds; above it the angle does. That crossover is the answer to the question Chapter 10 first raised — when does a line stop being limited by how hot the conductor gets? — and for a 400 kV circuit the answer is: at about 290 km, sooner if the permitted angle is smaller.

Take the lossless ABCD matrix and ask what would make the line perfect — no voltage drop, no rise, no regulation at all, for any load whatever. With \(\gamma = j\beta\), Chapter 12's constants are

The off-diagonal terms are what couple the voltage to the current and therefore what create the drop. Kill them: set \(\sin\beta l = 0\), which happens whenever \(\beta l = n\pi\). Then \(\cos\beta l = \pm1\) and the matrix becomes \(\pm\) the identity.

How long is such a line? The wavelength of Chapter 12 is \(\lambda = 2\pi/\beta = v/f\), where \(v = 1/\sqrt{LC}\) is between \(0.95c\) and \(0.98c\) for any overhead line. So

Three thousand kilometres. No transmission line of that length has ever been built as a single a.c. circuit, and there is no economic reason to build one: the loss over 3000 km at any useful loading would be a quarter of the power sent, and the right-of-way cost alone would pay for an HVDC link several times over. The tuned line is a limiting case, not a design.

Could a shorter line be made tuned? Since \(\beta = \omega\sqrt{LC}\), tuning a 400 km line at 50 Hz needs \(\beta = \pi/400 = 7.854\times10^{-3}\) rad/km against the natural \(1.073\times10^{-3}\) — a factor of 7.3, which means the product \(LC\) must be raised 53-fold by adding series inductance or shunt capacitance along the route. That is absurd for a power line, and it also destroys the thing one wanted: raising \(LC\) at fixed \(L/C\) leaves \(Z_s\) alone but multiplies the loss and the cost. The technique is used at high frequency in telephony, where the required \(\lambda/2\) is metres rather than megametres, and Heaviside's loading coils are exactly this idea applied to a telephone cable. At power frequency it is a curiosity.

Two special lengths are worth carrying away even so, because they occur in problems and in surge analysis. At \(\beta l = \pi\), a half wavelength, \(A=D=-1\) and \(B=C=0\): the line is a perfect repeater with a phase reversal. At \(\beta l = \pi/2\), a quarter wavelength, \(A=D=0\), \(B = jZ_s\), \(C = j/Z_s\), and the line inverts impedances:

An open circuit at the far end appears as a short at the near end and vice versa, which is why \(\sec\beta l\) blew up at \(90^\circ\) in Section 14-2: a quarter-wave line on open circuit is a series resonance seen from the source. At 50 Hz that length is about 1460 km, comfortably beyond anything built, but a switching surge contains frequencies of several hundred hertz, and at 500 Hz the quarter wavelength is only 146 km. Resonant overvoltages on a line of ordinary length are therefore entirely possible during switching — which is why Chapter 37's insulation coordination study looks at frequencies far above 50 Hz.

Tuning is out of reach, but its two ingredients — the electrical length \(\beta l = \omega l\sqrt{LC}\) and the surge impedance \(Z_s = \sqrt{L/C}\) — can each be moved, and moving them is what compensation does.

A series capacitor cancels part of the line's inductive reactance. With a compensation degree \(k = X_C/X_L\), the effective inductance becomes \(L(1-k)\) while \(C\) is untouched, so

Both changes help. The line becomes electrically shorter, so \(\sin\beta l\) falls and loadability rises; and the surge impedance falls, so the natural loading itself rises. The two multiply: at 40% compensation the 400 km line's \(\beta l\) falls from 24.6° to 19.1°, its \(Z_s\) from 298 to 231 Ω, its SIL from 537 to 693 MW, and its transfer at \(\delta=30^\circ\) from 645 MW to 1061 MW — a 65% increase for a device that sits in one substation. The Ferranti rise falls too, from 9.98% to 5.80%. This is the most cost-effective single measure available for a long line, and its price is the subsynchronous resonance risk and the protection complexity that Chapter 38 examines.

A shunt reactor cancels part of the line's capacitance. With a degree \(k_{sh}\) of the charging susceptance absorbed, the effective capacitance is \(C(1-k_{sh})\), so

Here the two effects pull against each other. The line is shortened electrically, which helps, but its surge impedance rises and its natural loading falls, which does not. Shunt reactors are therefore a light-load measure: they exist to hold the voltage down when the line is nearly empty, and they are switched out as soon as it fills. That is the opposite duty from a series capacitor, and a well-designed EHV corridor has both.

Which brings Part 3 to a close. Chapter 10 modelled a line as a series impedance and got the voltage drop. Chapter 11 lumped its shunt admittance and met the charging current. Chapter 12 distributed that admittance and found waves, a characteristic impedance and a wavelength. Chapter 13 turned the resulting two-port into performance figures. This chapter has read the same two-port at its two extremes — nothing on the end, and the one load that makes the line invisible — and found the Ferranti effect and the surge impedance loading waiting there. Chapter 15 leaves the circuit model entirely and asks what happens when the electric field at the conductor surface becomes strong enough to ionise the air around it.

Take 50 Hz unless told otherwise, treat lines as lossless where the problem says so, and quote voltages line-to-line. Problems 5 to 8 are of GATE standard.