Ferranti Effect, SIL and Compensation

The propagation constant and the two-port constants. With \(z = 0.028+j0.325\) and \(y = j3.6\times10^{-6}\) per kilometre:

Only \(A\) is needed for the open-circuit question; \(B\) and \(C = 1.06115\times10^{-3}\angle90.0872^\circ\) S are recorded here because Problems 4, 6 and 7 want them.

Route one — exact. Open circuit means \(\mathbf{I}_R = 0\), so the first row of the two-port relation is \(\mathbf{V}_S = A\mathbf{V}_R\) with nothing else in it:

No approximation of any kind has been made. This is the number a study would report.

Route two — lossless. Set \(r = 0\). Then \(\gamma = j\beta\) with \(\beta = \sqrt{xb}\), and \(\cosh(j\beta l) = \cos\beta l\):

Eight ten-thousandths of a percentage point from the exact answer. On an EHV line the resistance is so small a fraction of the reactance that ignoring it costs nothing here.

Route three — the lumped estimate. Expanding the secant, or equivalently reading the nominal-\(\pi\) circuit with \(\mathbf{I}_R = 0\), gives \(\mathbf{V}_S = (1+\mathbf{YZ}/2)\mathbf{V}_R\):

A quarter of a percentage point low, because the discarded quartic term \(5(\beta l)^{4}/24 = 0.231\%\) is no longer negligible at 300 km. Problem 2 follows that error out to the length where it becomes fatal.

The ordering, and its cause. Collecting the three:

The lossless figure is the largest of the three. Attenuation reduces the Ferranti rise rather than adding to it — a result that looks wrong until Problem 5 draws it, where it becomes obvious.

May the line be energised? The highest voltage permitted on 400 kV equipment is 420 kV, and 422 kV exceeds it:

Marginally, but the margin is the wrong way round, and the calculation assumed the source holds exactly 400 kV. A real busbar sits at 400–410 kV, which turns a 2 kV exceedance into 12 kV. The line is not energised without a reactor, which Problem 6 sizes.

The three columns. \(\gamma\) and \(\beta\) are per-kilometre constants, so only the length changes:

Where the estimate fails. Take the error of the quadratic against the exact figure:

To 200 km the estimate is right to within a twentieth of a percentage point, which is why the nominal-\(\pi\) model of Set 11 is safe there. At 300 km it is a quarter of a point low; at 400 km, three-quarters. Beyond about 250 km use the secant, and beyond about 500 km use \(1/|\cosh\gamma l|\), because by then the two differ from one another as well.

Why the estimate always under-reads. Every term of the secant series is positive:

Truncating a series of positive terms can only lose, so the lumped model is a systematic under-estimate of the overvoltage — the dangerous direction for an insulation study. There is no length at which it errs on the safe side.

1452 km. That is a quarter wavelength: \(\lambda = 2\pi/\beta = 5808.8\) km, so \(\lambda/4 = 1452.2\) km and \(\beta l = 90^\circ\). There

The lossless model predicts an infinite open-circuit voltage. Physically the line is a series resonance seen from the source: the quarter-wave section inverts impedances about \(Z_s^{2}\), so an open circuit at the far end looks like a short circuit at the near one, and unlimited current flows into an unlimited voltage.

What the exact formula says there, and the difference is instructive:

A rise of 1378%, not infinity. Resistance is what limits a resonance, exactly as in a series \(RLC\) circuit, and here \(\alpha l = 0.0676\) Np is the only thing standing between the line and destruction. That no such line exists at 50 Hz is fortunate; that a 145 km line is a quarter wave at 500 Hz is the reason switching studies look far above power frequency.

Compute \(|A|\) for each. With \(\gamma l = \sqrt{zy}\,l\) and \(A = \cosh\gamma l\):

A spread of fourteen hundredths of a percentage point across a voltage range of three to one. For practical purposes the three lines have the same Ferranti rise.

Why they agree. The rise depends on \(\beta l\) and on nothing else, and \(\beta = \omega\sqrt{LC} = \sqrt{xb}\):

All within 3% of one another, because \(\beta = \omega/v\) and the wave velocity on any overhead line is between \(0.95c\) and \(0.99c\). Geometry that raises \(L\) lowers \(C\) in almost exactly compensating proportion, because both depend logarithmically on the same ratio \(D/r\) — one directly, the other inversely.

The consequences, which do not agree at all. Convert the percentages to kilovolts and compare with the equipment limits:

All three pass at 200 km, but look at what is being spent. The 132 kV line uses 3 kV of a 13 kV allowance; the 400 kV line uses 9.5 kV of a 20 kV allowance. And the allowance is not proportional to the nominal voltage — it is 9.8% at 132 kV, 11.4% at 220 kV and only 5.0% at 400 kV, because EHV insulation is dimensioned much more tightly.

The length at which each runs out of headroom. Solving \(1/|A(l)| = V_{\max}/V_{\text{nom}}\) for each:

The highest-voltage line is the one that runs into the wall first — at 286 km against roughly 400 km for the other two. That inversion is the whole answer to the question.

And one further asymmetry. A 132 kV line that is 400 km long is a curiosity; a 400 kV line that is 400 km long is ordinary. EHV lines exist precisely to cover long distances, so they meet the Ferranti problem both because their permitted rise is smaller and because they are longer. The two effects multiply, which is why every 400 kV and 765 kV circuit has reactors and no 132 kV circuit does.

The exact current. The second row of the two-port with \(\mathbf{I}_R = 0\) gives \(\mathbf{I}_S = C\,\mathbf{V}_R\), and \(V_R\) is the risen voltage of Problem 1, not the nominal one:

The lumped rule, which treats the line as a lumped capacitor of susceptance \(B = bl\) at nominal voltage:

3.7% low. Both of the neglected effects push the same way: the line actually sits above 400 kV over most of its length, and the distributed capacitance nearest the open end is charged at the highest voltage of all.

The reactive power, exactly. The sending-end current leads the sending-end voltage by very nearly \(90^\circ\), so the complex power entering the line is almost purely negative reactive:

The real part, 0.57 MW, is the no-load loss: \(3I^{2}r l\) with a current that is nowhere near uniform along the line. The line returns 179.1 MVAr to the source.

By the lumped rule:

Again 3.7% low, and for the same reason — the rule evaluates \(V^{2}\omega C\) at 400 kV when the line averages about 411 kV. Note that the error in \(Q\) is not twice the error in \(V\): the current error and the voltage error do not compound, because the exact \(Q\) is \(3|V_S||I_S|\) with \(V_S\) at exactly 400 kV.

A relation worth keeping. Divide the charging by the natural loading of Problem 9:

Exact, not approximate: a line's charging in MVAr is its SIL in MW multiplied by its electrical length in radians. A 300 km EHV line has \(\beta l \approx 0.32\), so its charging is about a third of its natural loading, and that ratio is the same at every voltage class.

What it does to the generator. A 500 MVA machine asked to absorb 179 MVAr is being run at 0.36 pu leading:

The underexcited limit of a large turbogenerator is typically 0.25–0.35 pu, set by end-region heating of the stator core and by the steady-state stability limit. This one line, empty, takes the machine to or past its underexcited capability. The field is weak, the internal EMF is low, and the synchronising torque is at its smallest — precisely the state in which one does not want a fault.

Which is the second reason for the reactor. Problem 1 gave the insulation reason; this is the machine reason, and it applies even where the voltage would have been acceptable. Reactors are sized against the larger of the two demands, and on long circuits it is usually this one.

The two waves. Set 12 wrote the voltage at a distance \(x\) measured back from the receiving end as the sum of a wave travelling towards the load and one returning from it:

Put \(\mathbf{I}_R = 0\). Both coefficients collapse to \(\mathbf{V}_R/2\): at an open end the reflected wave equals the incident wave exactly, which is what total reflection from an open circuit means.

Carry each to the sending end, where \(x = l\) and \(\gamma l = 0.0139655+j0.324800\). Take \(\mathbf{V}_R\) as the reference phasor:

The incident half has grown by \(e^{\alpha l} = 1.014063\) and turned forward; the reflected half has shrunk by the same factor and turned back. They have opened out by \(2\beta l = 37.22^\circ\).

Add them by components. The real parts reinforce; the imaginary parts nearly cancel:

Which is the identity \(\cosh(\alpha l+j\beta l) = \cosh\alpha l\cos\beta l + j\sinh\alpha l\sin\beta l\) drawn as a parallelogram. The sum of the two lengths is \(\cosh\alpha l\); the difference is \(\sinh\alpha l\).

Now the question about attenuation. Hold \(\beta l\) fixed at \(18.610^\circ\) and let \(\alpha l\) go to zero. The two halves become equal at \(0.5\) each, the imaginary parts cancel exactly, and

Attenuation reduces the rise. Not by much — 0.0115 of a percentage point here — but the sign is unambiguous, and it is the opposite of what most people guess.

Why, read off the figure. With \(\alpha = 0\) the two phasors are equal, so their resultant lies exactly on the axis with length \(\cos\beta l\) — the pure scissors effect and nothing else. Switching on attenuation lengthens one and shortens the other by the same factor, and since

the magnitude can only grow. The added term \(\sinh^{2}\alpha l = 1.95\times10^{-4}\) is tiny, but it is positive, always, for every line ever built. A lossy line therefore has a smaller Ferranti rise than the lossless line of the same electrical length.

The identity also settles the angle. \(\angle A = \arctan(\tanh\alpha l\tan\beta l)\), which is zero when \(\alpha\) is zero:

So the small angle of \(A\) is entirely a loss effect, while its magnitude is almost entirely a phase effect. The two attributes of one complex number have completely different origins.

Derive the rule on the lossless line. A reactor of susceptance \(B_L\) across the receiving end draws \(\mathbf{I}_R = -jB_L\mathbf{V}_R\). Substituting into \(\mathbf{V}_S = \mathbf{V}_R\cos\beta l + jZ_s\mathbf{I}_R\sin\beta l\):

Both terms are real, so \(\mathbf{V}_S\) is in phase with \(\mathbf{V}_R\) — a reactor at the far end removes the angle as well as the magnitude difference.

Demand equality of magnitudes and solve:

Using the half-angle identity \((1-\cos\theta)/\sin\theta = \tan(\theta/2)\). The rule needs only the two numbers of the Method Recap.

Evaluate. With \(\beta l = 18.5925^\circ\) and \(Z_s = 300.463\ \Omega\):

Rating and current, both quoted for a star-connected three-phase bank at 400 kV:

A modest device: 87 MVAr and 126 A, oil-filled, gapped-core, sitting in the line bay. Compare it with the 179 MVAr the source was absorbing in Problem 4.

As a fraction of the line's charging. The line's own susceptance is \(bl = \beta l/Z_s\), so the ratio is a function of electrical length alone:

A reactor that exactly cancels the Ferranti rise absorbs just over half the line's charging. The ratio is 50.1% at 150 km, 50.4% here, 50.8% at 400 km and 52% at 600 km — it creeps up with length but never far from a half. Real EHV lines carry 50–80% compensation, the excess above a half being there to hold the voltage down at light load rather than at no load.

Check against the exact lossy line. With the reactor connected, \(\mathbf{I}_R = \mathbf{V}_R/(jX_L)\) and the first row of the two-port gives

Solving \(|A + B/(jX_L)| = 1\) numerically with \(A = 0.947817\angle0.2694^\circ\) and \(B = 96.153\angle85.163^\circ\):

The lossless rule is right to four significant figures. That is not luck: the rule depends on \(\beta\) and \(Z_s\), and Problem 5 showed that resistance perturbs both only in the fourth decimal on an EHV line.

Two practical points. First, 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. Second, a line reactor connected inside the circuit breaker follows the line automatically when it is de-energised, whereas a bus reactor must be switched separately — and a bus reactor left connected to a dead busbar is a way to lose a transformer.

The relation to evaluate. A reactor of rating \(Q\) at 400 kV has \(X_L = V_{LL}^{2}/Q\), and the exact receiving voltage with it connected is

Using the true \(A\) and \(B\) rather than the lossless model, because the answer is now being compared with a hard equipment limit.

The five candidates, with \(V_S = 400\) kV:

The 87.17 MVAr of Problem 6 sits between the 80 and 100 MVAr units, which is why neither of them lands on zero.

Apply the real criterion. The busbar is not held at 400 kV but somewhere up to 410 kV, and the rise is a ratio, so scale each residual:

Select 63 MVAr. The 50 MVAr unit passes arithmetically and fails as engineering: 0.67 kV is inside the measurement uncertainty of the busbar voltage, let alone the tolerance on \(b\), which is itself a few per cent depending on sag and on how far the conductors are from the ground on the day.

What the two largest do. At 100 and 125 MVAr the residual rise has gone negative — the far end is now below the sending end at no load:

The reactor is now over-compensating: it draws more lagging current than the line's capacitance supplies, and the surplus is drawn through the line's series reactance, producing an ordinary voltage drop. That is not a fault — a 125 MVAr unit is a perfectly sensible choice if the line must also be held down at light load, which is a longer-lasting condition than no load — but it must be switched out as the load rises, or it will cost regulation at full load.

The usual industrial answer is a single switchable 80 MVAr line reactor. It covers the energisation duty with 8 kV of margin, covers most of the light-load duty, and is one standard unit rather than two. The 63 MVAr choice above is the answer to the question as asked — the smallest that passes — and the two answers differ because the question and the specification differ.

Distributed compensation changes the line's own constants. Absorbing a fraction \(k\) of the shunt susceptance everywhere leaves an effective capacitance \(C(1-k)\), so

The line becomes electrically shorter, which helps, but its surge impedance rises and its natural loading falls, which does not. That penalty is the subject of Set 14's Problem 6 and is not re-derived here.

The same MVAr, spread out. 87.17 MVAr out of the line's 172.80 MVAr charging is \(k = 0.5044\):

Identical hardware, less than half the benefit. Position matters more than quantity for the no-load duty.

Why the end position wins. The two act by different mechanisms. Distributed reactors reduce the cause: less capacitance means less charging current means a smaller rise, but a shorter line is still a line and still has \(\sec\beta'l > 1\). The end reactor cancels the symptom: it draws a lagging current through the whole of \(B\), producing a drop that is subtracted from the rise:

One term is made to annihilate the other, and annihilation is exact. Reduction is not.

The degrees required. Solving \(\sec(\beta l\sqrt{1-k}) - 1 = \text{target}\):

Read the last row. Zero rise needs \(k = 1\) — every scrap of the line's capacitance cancelled — and at that point \(Z_s' = \infty\) and the natural loading is zero. The line has been turned into a pure inductance, which indeed has no Ferranti effect and also carries no power at any sensible angle.

The formal statement. For any \(k < 1\) the compensated electrical length is strictly positive:

So a uniformly compensated line always has some Ferranti rise. The rise can be made as small as one likes, at a cost in natural loading that grows without bound.

Where the highest voltage sits. With the end reactor connected and both terminals at 400 kV, the profile is \(V(x) = V_R(\cos\beta x + Z_sB_L\sin\beta x)\) measured from the receiving end, which at the mid-point \(x = l/2\) becomes

A clean result: an end reactor sized to null the terminal rise leaves a mid-line bulge of exactly \(\sec(\beta l/2)\). Both ends are at 400 kV and the middle is 5.3 kV higher, because the reactors cannot reach it.

Which is where distributed compensation earns its place. The bulge grows fast with length:

At 300 km a 1.3% bulge is irrelevant and end reactors are the whole answer. At 700 km the middle of the line is 7.6% above both ends even with the ends perfectly held, and a reactor at an intermediate switching station becomes necessary — not because it is more efficient per MVAr, but because there is no other way to reach the point that is highest. Distributed compensation on the 300 km line, meanwhile, leaves the far end at 410.7 kV and the mid-point at 408.0 kV: worse everywhere.

The reactive balance. Per unit length the shunt capacitance generates \(V^{2}\omega C\) and the series inductance consumes \(I^{2}\omega L\). Ask for the loading at which they cancel:

\(\omega\) cancels, so the balance point is a property of the line's geometry alone and is the same at 50 Hz and 60 Hz. And \(Z_s\) comes out real, though it is built from two purely reactive elements — which is what makes it a terminating resistance rather than a reactance.

From \(Z_s\) to SIL. A line terminated in \(Z_s\) carries \(|I| = |V_{LL}|/(\sqrt3Z_s)\) per phase at unity power factor, so

With \(V_{LL}\) in kilovolts the answer is in megawatts directly, which is the form worth memorising.

The 400 kV line. At \(\omega = 314.159\) rad/s:

The 220 kV line, same arithmetic:

The ratio, and where the surplus comes from.

The extra factor of 1.215 is the ratio of surge impedances, and it comes entirely from bundling. Two sub-conductors 450 mm apart have a much larger equivalent radius than one, which raises \(C\) and lowers \(L\) at the same time — and \(Z_s = \sqrt{L/C}\) feels both movements in the same direction. Set 29 works the bundle geometry that produces the numbers used here.

Why \(Z_s\) hardly ever leaves the range 250–400 Ω. Both parameters depend on the same logarithm of the same geometric ratio:

So \(Z_s\) is proportional to the logarithm itself, not to its square, and a logarithm is a slow function. Quadrupling the spacing changes \(Z_s\) by about a fifth. Every overhead line ever built, at every voltage, lands between about 250 and 400 Ω, and the only way to leave that band is to leave the geometry — which is exactly what a cable does, at 40 Ω.

And why the velocity is always near \(c\). The same cancellation, the other way round:

0.9688\(c\) and 0.9566\(c\) here, and no overhead line falls outside 0.95–0.99\(c\). This is why every 50 Hz line has a wavelength near 5800 km and why the tuned length of Problem 19 is the same order for all of them.

The table. Reactances from the data of Problem 3, converted through \(L = x/\omega\) and \(C = b/\omega\):

The velocities are \(0.984c\), \(0.957c\), \(0.969c\) and \(0.979c\) for the four overhead lines, and \(0.334c\) for the cable.

Read the overhead rows first. The voltage rises by a factor of 5.8 from top to bottom and the SIL by a factor of 52:

The extra factor of 1.56 is \(394.41/252.40\), the ratio of surge impedances, and it is bundling again: single conductor, single conductor, twin, quad. Higher voltage buys the square; bundling buys a further half on top of it.

The identity. Both charging and SIL scale as \(V^{2}\), so their ratio is a pure function of the line:

Exact in every row, as it must be — the identity is algebra, not a fit. A line's charging in MVAr is its natural loading in MW times its electrical length in radians. Since \(\beta l \approx 0.32\) for a 300 km line at any voltage, the charging of a 300 km circuit is always about a third of its SIL.

Now the cable row, which breaks every pattern. Its capacitance is 22 times that of the 400 kV overhead line, because the conductor sits millimetres from an earthed sheath rather than metres from the ground:

A surge impedance a tenth of the overhead value and a wave velocity a third of it. The nominal SIL of 435.6 MW at 132 kV is a fiction: it implies 1906 A, which is beyond the thermal rating of any 132 kV cable. A cable is never operated near its surge impedance loading, and its natural loading is not a useful design figure.

The charging current limit. Per kilometre, at 132 kV:

At 83.5 km the charging current alone equals the conductor's 500 A rating and the cable can deliver no useful load at all. Practically the limit is reached at half that, since the charging current must be a small fraction of the rating for the cable to be worth laying. This is the reason long submarine links are d.c.: not the loss, not the insulation, but the charging current. Set 30 works the cable capacitance that produces it.

Substitute the terminating condition. On a lossless line, with \(x\) measured back from the receiving end,

SIL means \(\mathbf{I}_R = \mathbf{V}_R/Z_s\) with \(Z_s\) real. Put that in:

Both quantities are the terminal value multiplied by a unit-magnitude rotation. Every claim follows from those two lines.

Claim by claim.

The reflected wave is absent as well: its coefficient \((\mathbf{V}_R - \mathbf{I}_RZ_c)/2\) vanishes identically when \(\mathbf{I}_R = \mathbf{V}_R/Z_c\). A line terminated in its surge impedance is indistinguishable from an infinite line, and an infinite line has nowhere to send a reflection.

The numerical check. With \(V_R = 400/\sqrt3 = 230.94\) kV per phase and \(P = 532.51\) MW at unity power factor:

Flat to the fourth figure, and the angle is the electrical length to three decimal places. The line has become a pure phase-shifter: it moves 532.5 MW and rotates it by 18.6°, and does nothing else at all.

Now the regulation trap. Two different quantities are both called regulation, and only one of them is zero here. The voltage drop along the line is zero:

But regulation as Set 13 defined it is the rise when the load is thrown off with \(|V_S|\) held:

Which is exactly the Ferranti rise of Problem 1, and no accident: throwing the load off a line at SIL returns it precisely to the open-circuit condition. A flat profile at the operating point and a 5.5% excursion when the load disappears are entirely compatible.

Which resolves an apparent paradox. "SIL is the ideal loading" and "a line at SIL still needs reactors" are both true. The first is about the steady state; the second is about what happens when the breaker at the far end opens. A 400 kV line running happily at its natural loading with a perfectly flat 400 kV profile will go to 422 kV within a cycle of losing its load, and the reactor of Problem 6 exists for that cycle.

The relation. With \(\mathbf{V}_R\) and \(\mathbf{I}_R\) both real (unity power factor at the receiving end),

The real part is fixed by the receiving voltage; the imaginary part grows in proportion to the load. The two add in quadrature, so \(|V_S|\) starts below \(|V_R|\) and climbs past it.

The table, with \(V_R\cos\beta l = 218.89\) kV/phase fixed throughout:

The sending-end voltage spans 45.5 kV across the loading range while the receiving end never moves. That span is the daily work of voltage control, and SIL is the point about which it is symmetric.

Read the tilt. Below SIL the profile rises towards the far end, because the capacitance is generating more reactive power than the inductance consumes; above SIL it falls, for the opposite reason. At SIL it is flat, and the flatness is exact rather than approximate. That is the reactive balance of Problem 9 drawn as a picture.

The two limits. Thermally, a twin-bundle ACSR Moose line carries about 1.25 kA per phase:

By the angle limit, with both ends held at 400 kV and \(\delta\) capped at 30° for stability margin:

The angle limit binds, but only just — 835 MW against 866 MW, a difference of 3.6%. The 300 km line sits almost exactly on the crossover.

Find the crossover exactly. The two limits are equal when

Below 289 km the conductor temperature is the constraint and the line will never be angle-limited; above it the angle is the constraint and the conductors will never get hot. This is the St Clair loadability curve, obtained from three lines of algebra rather than from the empirical plots of the 1950s.

Set up the lossless two-port with equal terminal magnitudes. Let \(\mathbf{V}_S = V\angle\delta\) and \(\mathbf{V}_R = V\angle0\), with \(B = jZ_s\sin\beta l\) and \(A = D = \cos\beta l\). The standard receiving-end complex power is

With \(\theta_B = 90^\circ\) and \(\theta_A = 0\), and writing \(V\) in kilovolts so that the answers come out in MW and MVAr:

The first is the loadability relation of the Method Recap. The second is the one this problem is about, and it is remarkable for containing no reference to the load's power factor — with both terminal voltages fixed, the reactive flow is determined.

By symmetry the sending end is the mirror image. Repeating for \(\mathbf{S}_S\) with \(|A| = |D|\):

\(Q_S\) is the reactive power flowing into the line at the sending end. When \(Q_R\) is positive the line is delivering reactive power to the load and returning an equal amount to the source: it is a VAr generator feeding both ends at once. When \(Q_R\) is negative it is a VAr sink drawing from both.

The table. Solve \(\sin\delta = (P/\mathrm{SIL})\sin\beta l\) for \(\delta\), then evaluate \(Q_R\):

The no-load value. At \(\delta = 0\),

The same 87.17 MVAr as the reactor of Problem 6, and necessarily so. The reactor's job is to absorb exactly the reactive power the line insists on delivering. Problem 6 got the number by demanding a voltage; this problem gets it by counting megavars; the two are the same statement.

The crossover is at SIL, exactly. \(Q_R = 0\) requires \(\cos\delta = \cos\beta l\), so \(\delta = \beta l\), and then

Which is a third independent characterisation of SIL, after the reactive balance of Problem 9 and the flat profile of Problem 11: it is the loading at which a line with equal terminal voltages exchanges no reactive power with the network.

The asymmetry either side. The curve is not straight, and it is much steeper above SIL than below:

Because the generation term goes as \(V^{2}\), which is pinned by the terminal voltages, while the absorption term goes as \(I^{2}\), which is free to grow. Doubling the loading quadruples the consumption and leaves the production alone. This is why heavily loaded lines are such voracious consumers of reactive power and why shunt capacitors appear at the receiving ends of them.

What it means for the two busbars. Below SIL both must absorb; above SIL both must supply:

And the equipment for the two duties is different, which is the practical sting. Reactors and capacitors are separate assets, both must be paid for, and both spend most of the year switched out.

The two integrals. Each element \(dx\) of line contributes shunt generation and series absorption:

The factors of three convert per-phase quantities to three-phase totals. Conservation of reactive power then demands that the surplus leave through the two ends.

The profiles to integrate. From the lossless solution with \(x\) measured back from the receiving end,

with \(\mathbf{V}_R = 230.94\angle0^\circ\) kV per phase and \(\mathbf{I}_R = \left[(P+jQ_R)/(3V_R)\right]^{*}\), the \(Q_R\) being Problem 13's.

The results, integrated numerically over the 300 km:

All in MVAr. The last two columns agree in every row, as conservation requires, and the factor of two in \(2Q_R\) is because the surplus leaves through both terminals in equal shares.

The SIL row is exact and can be checked by hand. At SIL the profiles are flat, so both integrands are constants:

Equal to the last digit — which is the reactive balance of Problem 9 done as an accounting exercise rather than as a per-unit-length equation. Note also that both equal the line's charging \(V_{LL}^{2}bl\): at SIL the line consumes precisely its own charging.

Why the generation barely moves. It is governed by \(|V|^{2}\), and the voltage is pinned at 400 kV at both ends. Between them it can only bulge or sag by a few per cent:

At no load the middle of the line bulges to 405.3 kV (Problem 8), which lifts the integral slightly; at 1.4 SIL it sags below 400 kV, which lowers it slightly. Neither movement is large, because the terminals will not let it be.

Why the consumption moves so violently. It is governed by \(|I|^{2}\), and nothing pins the current at all:

At no load the only current is the charging current, small and varying along the length from zero at the open end to its maximum at the source — which is why \(Q_{\text{abs}}\) is 1.55 MVAr rather than zero. Load it, and the current is the load current, which is free to be anything. The whole variability of a line's reactive behaviour lives in one of the two integrals.

And the shape of the answer follows immediately. Write the net as one nearly-constant term minus one quadratic term:

Right to within 6%, and it explains the curve of Problem 13 without any trigonometry: the net reactive output of a line falls as one minus the square of the loading in SIL units, and it is zero at SIL because that is where the square reaches one. The residual 6% error is the profile bulge that this simple version ignores.

The capacitor. The line's total series reactance is

Built as series-parallel strings of individual units in a fenced yard on an insulated platform at line potential — the platform is the expensive part, not the capacitance.

Its rating. A series capacitor's duty is set by the current through it, not the voltage across the line:

Note the size: 183 MVAr of capacitors to compensate a line whose charging is 173 MVAr. Series compensation is not a small purchase, and its rating rises as the square of the current, so it must be specified against the line's emergency rating rather than its normal one.

And its protection, which dominates the design. During a through fault the same 39 Ω carries the fault current:

Sixteen times the normal duty, in a device whose insulation is rated for 49 kV. Every series bank therefore carries a metal-oxide varistor across it that conducts above about 2.3 pu of protective level, plus a triggered spark gap and a bypass breaker behind that. The capacitor is out of circuit within a quarter cycle of a nearby fault and is reinserted a few hundred milliseconds later.

The effect on the line constants. The capacitor cancels part of \(L\) and leaves \(C\) untouched, so with \(m = \sqrt{1-k} = \sqrt{0.6} = 0.774597\):

Both movements help. The line is electrically shorter and its natural loading is 29% larger, and the second is the effect that shunt compensation cannot reproduce.

The two headline results.

A 65.5% increase in transfer capability and a 41% reduction in the no-load overvoltage, from one bank of capacitors in one substation. Nothing else available to a transmission planner does this much for the money.

What it does not change. The charging is untouched, because \(C\) is untouched:

And the identity of Problem 10 still holds, with both factors having moved:

SIL rose by \(1/m\) and the electrical length fell by \(m\), so their product is invariant under series compensation. A neat check on any answer.

The reactor of Problem 6, recomputed. The rule \(B_L = \tan(\beta l/2)/Z_s\) uses both changed quantities:

Essentially unchanged — 86.9 against 87.17 MVAr. The two factors move in opposite directions and very nearly cancel, so series compensation reduces the Ferranti rise without reducing the reactor you need. That surprises people, and the reason is that the reactor's job is to absorb the charging, and the charging did not change.

The table, every column following from \(m = \sqrt{1-k}\):

The returns accelerate. Take the increment in transfer capability per ten points of compensation:

Each extra point of compensation is worth more than the last, because \(P \propto 1/(m\sin(\beta l\,m))\) and both factors shrink together. That is the opposite of the usual engineering pattern and it is exactly what makes high degrees tempting — and dangerous, since the design is being pushed towards a region where small changes have large effects.

Reason one: subsynchronous resonance. The electrical resonant frequency \(f_0\sqrt{X_C/X_L}\) rises with \(k\), and its complement \(f_0 - f_{er}\) sweeps downward through the band 15–35 Hz where turbine-generator torsional modes live. High compensation puts the complement into the middle of that band and keeps it there. Problem 17 computes the frequencies.

Reason two: fault current and protection. The compensated series reactance is \(X_L(1-k)\):

Fault levels along the line rise by up to a factor of three, and — worse — a distance relay looking through a capacitor can see a negative apparent reactance for a fault just beyond the bank. Voltage inversion and current inversion both become possible, the relay's directional decision reverses, and the protection has to be redesigned around the bank's location.

Reason three: it is the wrong tool past a point. At \(k = 0.7\) the line's transfer capability is 2750 MW while its conductors carry 866 MW:

Beyond the point where the thermal limit takes over, further compensation buys nothing whatever. On the reference line that happens at about \(k = 0.04\) for a 30° angle; the useful range for compensation here is set by the desire for a larger stability margin, not by capability. On a 600 km line, whose uncompensated 30° transfer is only 441 MW, the crossover is at \(k \approx 0.51\), and that is why the long lines get the heavy compensation.

Practice. Real fixed banks are 25–50%; the 70% figure appears only on very long radial lines with dedicated SSR countermeasures, and even there it is usually split between a fixed bank and a thyristor-controlled one. Anything above 50% is designed jointly with the machines at each end, not by the line engineer alone.

The resonant loop. A series capacitor forms an \(LC\) circuit with everything inductive in the path: the generator subtransient reactance, the transformer, and the line. Its natural frequency is

The reactances are all quoted at 50 Hz and the ratio is frequency-independent, so no scaling is needed inside the root.

Assemble the inductive path in ohms. On a 500 MVA, 400 kV base:

The line is less than half of it. That single fact is the source of most of the wrong answers to this question.

The frequencies. With \(X_C = k\times97.5\):

Why the complement is the frequency that matters. A subsynchronous current at \(f_{er}\) in the stator produces, in the rotor's own reference frame, a torque at the difference frequency:

The rotor is turning at \(f_0\), so a stator field rotating at \(f_{er}\) slips backward past it at \(f_0-f_{er}\) and pulses the shaft at that rate. If the pulse lands on a torsional natural frequency the shaft is driven at resonance, the oscillation grows because the electrical circuit presents negative damping at that frequency, and the coupling between two turbine sections fatigues. Two shafts were destroyed this way at Mohave in 1970 and 1971, which is how the phenomenon came to be understood.

The verdict. 40% is the natural choice: it clears every mode by more than 2 Hz, it delivers the 1382 MW of Problem 15, and it sits in the middle of a safe band rather than at its edge. 50% must be rejected outright — a 0.2 Hz margin is no margin, since the torsional frequencies themselves are known only to a few tenths and drift with temperature and with blade erosion. 70% is likewise rejected.

The common error. Computing \(f_{er}\) from the line alone gives

An error of 9.5 Hz — and it points the wrong way about safety, because 18.38 Hz is close to the 20.2 Hz mode while the true 27.92 Hz is not close to anything. The \(50\sqrt{k}\) formula would have condemned a safe design and, at another degree, cleared a dangerous one. Never use it: the machine and its transformer are part of the resonant circuit whether or not the question mentions them.

What is done when the required degree is unsafe. Four measures, in ascending order of cost:

A thyristor-controlled series capacitor is the modern answer. In capacitive vernier mode it gives the required compensation at 50 Hz while presenting an apparent inductance at subsynchronous frequencies, so the resonance simply does not form. Splitting the bank into two smaller banks at different locations, which is often proposed, does not help — the loop sees the total \(X_C\) regardless of where it sits.

The two transformations. With \(m = \sqrt{1-k} = 0.774597\) in both cases:

The first row is identical, which is why the Ferranti reduction is identical. The second row has opposite signs in the exponent, and that single difference generates everything else.

The comparison in numbers.

The two results that decide everything. Look at the SIL row and the transfer row:

Shunt compensation makes the line very slightly worse at carrying power. The electrical shortening helps and the fall in SIL hurts, and the two nearly cancel — the net ratio is \(m\sin\beta l/\sin(\beta l\,m) = 0.9930\). A shunt reactor is therefore not a capability measure in any sense; it is a voltage-control device that costs a little capability.

The opposite duty cycles. The devices are needed at opposite ends of the loading range:

A series capacitor is self-regulating in a way the shunt devices are not: its output \(3I^{2}X_C\) rises as the square of the load, which is exactly when the reactive support is wanted. A shunt capacitor's output \(V^{2}B_C\) falls when the voltage sags, which is exactly when it is wanted — a perverse characteristic and the reason shunt capacitors are poor at supporting a collapsing bus.

The costs are not comparable either. The series bank here is 183 MVAr against the reactor's 69 MVAr, but the megavar is the wrong unit for a comparison:

A shunt reactor is a transformer without a secondary and is priced like one. A series capacitor is a small substation. Per megavar the series bank is several times dearer, and its megavars buy something the reactor's megavars cannot buy at any price.

The specification, in one sentence each. Buy a series capacitor when the problem is how much power the corridor can carry; buy shunt reactors when the problem is how high the voltage goes when the corridor is empty. The reference line has both problems, so it gets both devices, and Problem 20 puts them together. Set 14 works the same pair of devices as a regulation and efficiency question, and its Problems 8 to 12 give the load-flow consequences that this comparison leaves out.

The condition. The lossless two-port is

The off-diagonal terms are what couple the voltage to the current, so kill them:

Zero regulation, zero Ferranti rise, no drop. The sign alternates: odd \(n\) reverses the phase by 180°, even \(n\) reproduces it.

The length. From Problem 9, \(\lambda = 5808.8\) km, so

The loss over the shortest, carrying its natural loading of 532.5 MW at 768.6 A per phase:

Perfect regulation at the cost of a quarter of the power. And that is the loss at SIL; at 1.5 SIL it would be 324 MW, or 41% of what was sent. No tuned a.c. line has ever been built and there is no reason to build one — the right-of-way alone would pay for an HVDC link several times over.

The tuning frequency of the actual line. Solve \(\omega l\sqrt{LC} = \pi\) with \(\sqrt{LC} = \beta/\omega = 3.4431\times10^{-6}\) s/km:

Equivalently, to tune it at 50 Hz the product \(LC\) would have to rise by

Ninety-four times, by adding series inductance or shunt capacitance along the route. Absurd for a power line, and self-defeating: raising \(LC\) at fixed \(L/C\) leaves \(Z_s\) alone while multiplying the loss and the cost. Heaviside's loading coils are exactly this idea applied to a telephone cable, where the required half-wavelength is metres rather than megametres.

The real half-wave line. Restore \(r = 0.028\ \Omega\)/km over 2904.4 km, so \(\gamma l = 0.135205+j3.144501\):

\(B\) is not zero. The perfect repeater exists only on the lossless line, and the resistance leaves a residual series impedance of 41 Ω — small compared with the 947 Ω of raw series impedance \(|zl|\), but not nothing. Note also \(|A| > 1\), so the far end of a real half-wave line on open circuit is below the sending voltage: no Ferranti effect at all, and a small anti-Ferranti one instead.

Its regulation at SIL, the figure the whole idea was supposed to make zero:

Worse than the 300 km line's 5.5%. The tuned line's zero regulation does not survive contact with resistance, because at 2904 km the current is being pushed through 81 Ω of accumulated resistance and the drop is real even though the reactive terms have cancelled. The idea is exact and useless in the same breath.

The quarter-wave section, \(\beta l = \pi/2\) at 1452.2 km. There \(A = D = 0\), \(B = jZ_s\), \(C = j/Z_s\), so

The section inverts impedances about \(Z_s^{2}\): a load heavier than \(Z_s\) looks lighter from the far end and vice versa, and the two answers above are very nearly each other's partners because 600 and 150 straddle 300.5. An open circuit inverts to a short, which is why \(\sec\beta l\) blew up at 90° in Problem 2.

Which exposes the deeper objection. A half-wave line's terminals are perfect, but its middle is the quarter-wave point, where \(\mathbf{V} = jZ_s\mathbf{I}_R\):

The mid-point voltage is proportional to the loading. Empty, the middle of the line sits at zero volts; at twice SIL it sits at 800 kV on a 400 kV line. No insulation system and no protection scheme can accommodate that, which is a stronger argument against the tuned line than the 27% loss.

Step 1 — the uncompensated line. Same per-kilometre constants, so \(Z_s = 300.46\ \Omega\) and SIL \(= 532.5\) MW as before, but the length has changed everything else:

Both requirements fail, and not marginally: 517 MW against 800 required, and 466 kV on open circuit against a 420 kV ceiling. The line as built is unusable in both directions.

Step 2 — which requirement sets the series compensation. Only a series capacitor can raise the transfer, so size it on requirement 1. With \(m = \sqrt{1-k}\),

36.5% is the bare minimum. Round up to a standard degree with margin, and take \(k = 0.40\) — a number that must now survive the subsynchronous check before anything else is decided.

Step 3 — the subsynchronous check, done before the design is committed. On a 1000 MVA, 400 kV base, \(Z_{\text{base}} = 160\ \Omega\):

Safe, and comfortably so. For contrast, \(k = 0.50\) would give a complement of 19.17 Hz, within 0.9 Hz of the 20.1 Hz mode — so the rounding-up must stop at 40%, and a design that had needed 50% would have required a thyristor-controlled bank instead.

Step 4 — specify the capacitor.

One bank at the mid-point, on an insulated platform, with metal-oxide varistors, a triggered gap and a bypass breaker. The compensated constants are then

Step 5 — verify requirement 1.

And the conductors can carry it:

The actual angle at 800 MW is \(\arcsin(800\sin24.003^\circ/687.5) = 28.2^\circ\), so there is 1.8° of margin below the limit as well.

Step 6 — the reactors, sized on requirement 2. With the capacitor in, the exact constants of the compensated line are \(A = 0.913582\) and \(B = 95.64\ \Omega\), giving an open-circuit rise of 9.46%. The full nulling reactor would be

But the requirement is 420 kV with the busbar at 410 kV, that is a permitted rise of \(420/410 - 1 = 2.44\%\). Testing standard ratings:

Specify two 63 MVAr line reactors, one at each end, totalling 126 MVAr — 43.8% of the line's 288 MVAr charging. Splitting them is not cosmetic: it halves the mid-line bulge and it leaves the far end protected when only one end is energised, which is exactly the condition the requirement was written for.

Step 7 — check the mid-line voltage, since Problem 8 showed the terminals are not where the peak is. With both ends held at 400 kV on the compensated line,

And with the busbar at 410 kV and the reactors only part-compensating, the highest point on the line stays under 419 kV. No intermediate reactor is needed at this length; the calculation of Problem 8 says one would become necessary beyond about 700 km.

The specification, complete.

What was decided in what order, because the order is the method. The series degree came from the capability requirement; it was then checked against the machine before anything was bought, and that check — not the capability calculation — is what fixed it at 40% rather than 50%. Only then were the reactors sized, against the residual rise that the capacitor had already reduced from 16.65% to 9.46%. Sizing the reactors first would have produced a 190 MVAr answer that the capacitor then made unnecessary.