The reactance cancelled, and the resonance it buys

A feeder’s far end sits below its near end because the load’s current works against the line’s impedance. Most of that impedance is reactance, and most of the drop is therefore the product of a current and a reactance rather than of a current and a resistance — which means the drop can be cancelled by a component that dissipates nothing at all.

Put a capacitor in series with the line. Its reactance is negative where the line’s is positive, and whatever fraction of the line’s reactance the capacitor cancels is a fraction of the regulation that goes away. It is cheap, it is lossless, and it has been standard practice on long transmission lines for a century.

It also puts a resonance into a loop that did not have one, and the frequency of that resonance is chosen by the compensation and by nothing else.

What it buys #

The feeder drawn is a small one: fifty milliohms of resistance and two hundred milliohms of reactance, feeding twenty kilovolt-amperes at 0.9 lagging on 230 V. Its regulation — the rise from full load to no load, as a fraction of the loaded voltage — is 5.168% uncompensated.

With the capacitor in, solved at the fundamental:

The fall is close to linear in the fraction, which is what the phasor argument predicts: the drop is $I(R\cos\phi + X\sin\phi)$ to a good approximation, and the compensation removes a fraction of the second term while leaving the first alone. At 0.9 power factor the reactive term is the smaller of the two here, so even full compensation cannot remove all of the regulation — which is worth knowing before the capacitor is bought.

The other half of the ledger is that the capacitor dissipates nothing. Every alternative repair — a larger conductor, a tap changer, a shunt bank — either costs copper or costs a control loop. This one costs a component and a resonance.

The resonance, and the one number in it #

A series capacitor in a line whose other elements are inductive makes a series resonant loop, and the figure finds its frequency by sweeping the current the source delivers rather than by rooting a polynomial.

The clean case is the far end shorted, which is what a fault produces. Then the loop is the line’s inductance and the capacitor, and the capacitor’s value was chosen so that $X_C = k X_L$ at the line frequency $\omega_0$. That fixes $C = 1/(k\omega_0^2 L)$, so

$$\omega_{\text{res}} = \frac{1}{\sqrt{LC}} = \omega_0\sqrt k .$$

The line’s inductance has cancelled out. So has the voltage, and so has everything about the conductor. The figure asserts this against the swept current at every setting of the slider, and separately sweeps the line reactance over eight times its range and requires the answer not to move — because a law evaluated once is an arithmetic coincidence until it is asked somewhere else.

The numbers are unnerving in how round they are. Compensation of $1/n^2$ puts the resonance at exactly $f_0/n$:

• 25% → 25.00 Hz, exactly half the line frequency
• 11.1% → 16.67 Hz, exactly a third
• 4% → 10.00 Hz, exactly a fifth

A resonance at an integer submultiple of the fundamental is the worst place it could be, and the compensation fraction chooses it.

Where it sits when there is a load on the end #

The shorted case is the clean one and it is not the usual one, so the figure measures both and the general law contains the difference.

With a load connected, the loop contains the load’s own reactance as well as the line’s, and the capacitor was sized against the line’s alone. The resonance is then

$$f_{\text{res}} = f_0\sqrt{\frac{k\,X_{\text{line}}}{X_{\text{total}}}}$$

with $X_{\text{total}}$ the whole loop’s inductive reactance. At a quarter compensation on this feeder, where the load’s reactance is 1.15 Ω against the line’s 0.2, that gives 9.59 Hz rather than 25 — the extra inductance has pulled the resonance down by the root of the ratio.

Two things follow, and they are the reason the shorted case is the one to design against.

The resonant frequency moves with the load. Solved at five, twenty and sixty kilovolt-amperes the same compensated feeder resonates at 5.1, 9.6 and 14.6 hertz, because a heavier load is a smaller inductance as well as a smaller resistance. A resonance that walks with the load cannot be avoided by choosing a compensation that keeps it away from a known frequency, because there is no single frequency to keep it away from.

It walks upward towards the shorted value as the load grows, and the shorted value is the largest it can reach. So $f_0\sqrt k$ is not one case among several; it is the upper bound over every load the feeder will ever see, and it is the number a study has to clear.

Always below the fundamental, which is the awkward part #

The resonance is $f_0\sqrt k$ and $k < 1$, so it is always below the line frequency. That is awkward for two reasons that have nothing to do with each other.

There is nothing to filter it with. A harmonic problem is above the fundamental and can be attacked by something that passes the fundamental and rejects what is above it. A subharmonic cannot be filtered without also attacking the fundamental, because it is on the wrong side.

The machines have modes down there. A turbine-generator shaft is a torsional spring-mass system with natural frequencies in the tens of hertz. An electrical resonance at $f_0\sqrt k$ appears in the machine’s torque at the difference from the fundamental, $f_0(1-\sqrt k)$, and if that lands on a shaft mode the electrical circuit and the mechanical one exchange energy. At 25% compensation the difference frequency is 25 Hz; at 11%, 33.4 Hz; at 4%, 40 Hz. A shaft with a mode anywhere between twenty and forty hertz — which is where they are — has a compensation fraction it cannot tolerate.

This collection has no mechanical model in it and does not compute what the shaft does. What it can compute is the electrical half exactly, and the electrical half is where the choice is made.

The damping points the wrong way round #

A resonance is only dangerous if it is lightly damped, and the damping here is the loop’s resistance, which produces a result the wrong way round from intuition.

In a series loop, a heavy load is a small resistance in the loop, so heavy loading raises the quality factor rather than lowering it. Measured on the faulted feeder — the extreme case of a heavy load — the quality factor of the resonance is

so the resonance is more sharply defined the more of the reactance has been cancelled. Both of the things that make a compensated line dangerous — the resonance climbing towards the fundamental, and the quality factor rising — get worse with the same knob that makes the regulation better.

The mechanism is worth being explicit about, because the shunt case trains the opposite instinct. A quality factor is $\omega L/R$ with $R$ the resistance in the loop, and in a series arrangement the load’s resistance is in the loop. So the lightly loaded feeder is the well-damped one and the heavily loaded feeder is the sharp one, which is exactly backwards from a shunt resonance, where the load resistance appears across the tank and damps it. Anybody carrying an intuition from harmonic filters into this problem will get the sign wrong.

What a fault does, which is the hard limit #

The regulation argument is about the steady state. The limit on how much compensation can be fitted is about the other case, and it is severe.

With the far end shorted, the loop’s impedance at the fundamental is $R + j(X_L - X_C) = R + jX_L(1-k)$. The fault current is the voltage divided by that magnitude, and as $k \to 1$ the reactance vanishes and the current is limited by resistance alone. Solved:

Ninety per cent compensation nearly quadruples the fault current on this feeder, and full compensation would put the loop exactly on resonance at the fundamental, where the only thing between the source and the short circuit is a hundred milliohms.

So the boundary is two-sided and both ends are hard. Too little compensation and the regulation is not worth the capacitor; too much and the fault current outgrows the switchgear that was sized for the uncompensated line. In practice the fraction lives between about a quarter and a half, and this is the arithmetic that puts it there — not a convention.

What the capacitor itself has to survive #

There is a component-level consequence of the fault table that is worth stating, because it is the reason series compensation is expensive in a way shunt compensation is not.

The capacitor carries the whole line current, all of it, at all times. At the fault currents above, the voltage across it is $I \cdot X_C = I \cdot kX_L$ — at 50% compensation and 2036 A of fault current on this feeder, that is 204 volts across a component whose steady-state duty is 96. On a real transmission line the same arithmetic gives tens of kilovolts against a rating of a few, and the capacitor cannot be built for it.

So every series capacitor is fitted with a bypass: a gap, or a varistor, or a thyristor switch, that short-circuits it when the voltage across it exceeds its rating. The compensation therefore removes itself during a fault, which changes the line’s impedance in the middle of the event the protection is trying to measure, and re-inserts afterwards.

That is a second boundary of a kind this collection does not usually meet: not a frequency or an amplitude at which a model stops being true, but a circuit that reconfigures itself at a stated amplitude, so that the network being analysed is not the network that exists during the disturbance. The bypass level is the boundary, it is a voltage, and it is the number that decides how much of the fault-current increase in the table above is actually seen.

Two capacitors that are not the same repair #

The distinction above is worth drawing out, because both are called “power factor” work and only one of them is.

A shunt capacitor across the load supplies the load’s reactive current locally, so the line no longer carries it. The line current falls and the drop falls with it. Its failure mode is a parallel resonance with the line’s inductance, at a frequency above the fundamental — typically a few hundred hertz — where harmonics live, and the standard defence is a detuning reactor that moves it below the lowest harmonic present.

A series capacitor cancels part of the line’s own reactance. The line current is unchanged and the drop falls because the impedance did. Its failure mode is a series resonance at a frequency below the fundamental, where nothing can filter it.

Same component, opposite topology, opposite side of the fundamental. It is one of the cleanest cases in this subject of a fact that cannot be carried between two circuits that look alike.

What the cancellation is measured against #

Series compensation trades a regulation figure for a resonance, and both ends of that trade are measured elsewhere in the field. The far end that rises is the regulation, and it shows that the figure being improved does not even carry the load angle. The current that does no work is the reactive current the compensation is cancelling. The capacitor that was right once is the shunt version of the same idea, exact at one load and leading at every other. Resonance, and the bandwidth it sets exactly is what the resonance this buys actually is, and The first cycle, which no steady state contains is the transient a fault puts through it.

What is checked #

Four assertions, and the second and third are the essay.

That the regulation falls when the compensation is fitted, which is the thing the capacitor is for and the frame the rest is measured against.

That the resonance is at the line frequency times the root of the compensation fraction, scaled by the share of the loop’s reactance the line holds — $f_0\sqrt{kX_{\text{line}}/X_{\text{total}}}$, asserted against the peak found by sweeping the source current, at every setting of the slider. With the far end shorted the scaling factor is one and the answer is exactly $f_0\sqrt k$.

That it does not depend on the line’s own inductance, swept over eight times its range with the answer moving by less than five parts in a thousand. That is what turns the expression from a fitted formula into a statement, and it is the assertion that would have caught a coincidence.

And that with nothing cancelled the regulation is the uncompensated one and there is no resonance to find — the zero setting of the slider being a real frame rather than a special case that is quietly skipped.

A resonance that was not there before #

The shape of this result — a repair that removes what it was fitted for and introduces a resonance nobody asked about — is one the collection meets three times, and the three together say what to look for when a reactance is added anywhere.

The capacitor that was right once is the same component in the shunt position rather than the series one, and its failure is complementary: cancelling a load’s reactive power needs one division and no iteration, the answer is exact for the load it was computed from at the frequency it was computed at, and everywhere else it can make the installation worse than it was before anything was fitted.

The pair that is worse than either is the same arithmetic at six decades higher: a bulk capacitor and a ceramic each good where the other is not, with a frequency between them at which the pair presents six times the impedance either does alone, and 5.87 amps exchanged between the two for every amp the load draws.

And the inductance that limits, and lifts is the series case in the applied field, where adding leakage inductance to a rectifier divides the first peak by seven and the energy by five while dissipating nothing — and buys a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

The rule the three suggest is short. Adding a reactance to cancel another one creates a series or parallel resonance with everything else reactive in the loop, its frequency is set by the fraction cancelled rather than by the components’ absolute values, and the quantity to compute afterwards is not the improved one but the impedance at that frequency.

Why the resonance lands where the harmonics are #

A quarter compensation resonating at exactly half the line frequency sounds harmless, and it is the fractions above a half that make this result a design constraint rather than an arithmetic curiosity. Compensating four ninths puts the resonance at two thirds of the line frequency; compensating nine-sixteenths puts it at three quarters. The resonance is always below the line frequency, which is a region a fundamental-only analysis regards as empty.

It is not empty in an installation with rectifier loads on it, and it is not empty for the reason the neutral that carries more than a line gives: a harmonic of order three is shifted by 360° between phases, so third harmonics add rather than cancelling, and the currents circulating between phase and neutral are at the fundamental’s odd multiples rather than at fractions of it. What lands below the fundamental instead is inter-harmonic — the beat between a variable-speed drive’s switching and the line, the flicker of a cyclic load — and those are the currents a sub-synchronous resonance amplifies.

Which is the honest statement of what this essay adds to a compensation decision. The fraction cancelled is chosen for regulation and it sets a resonant frequency as a side effect; the frequency is computable from the fraction alone, with the line’s inductance and voltage absent from the answer; and whether it matters depends entirely on what else is connected to the feeder.

That the answer contains neither the inductance nor the voltage is what makes it worth stating as a rule rather than as a result for one line. A compensation of one quarter resonates at half the line frequency on every feeder there is, whatever its length, its conductor size or its voltage — so the decision about how much to cancel is a decision about a resonant frequency, made in a quantity that looks like a percentage of regulation.