The capacitor two-thirds of the way out

The current that does no work computed reactive power on a solved network and found it doing exactly what its name says: flowing out to a reactance and back twice a cycle, delivering nothing on average, and still carried by every wire between the source and the load. The energy a unity power factor doubles then corrected a load to a power factor of one with a capacitor across it and found that the correction does not remove the stored energy, it relocates its exchange: the capacitor and the inductor now trade it locally, and the source sees none of it.

Both essays put the capacitor at the load, where there was one load. On a distribution feeder there is not one load. A rural or suburban feeder runs for kilometres with transformers tapped off along its whole length, each drawing current at a lagging power factor, and the reactive current in any section of the feeder is the sum of the reactive currents of every load beyond it. That current is largest at the substation and falls to nothing at the far end. A utility that wants to cut the loss the reactive current causes has one capacitor — or two, or five — to place somewhere on that run, and the question is where.

The answer has a classical form: two-thirds of the reactive load, two-thirds of the way out. This essay solves the feeder as a network, finds how close the classical form is, and measures what it leaves out.

Moving one capacitor along the line #

The feeder is one phase of an 11 kV line, ten kilometres long, with 0.3 ohms of resistance and 0.35 of reactance per kilometre. Its load is 2 MW at 0.8 lagging — so 1.5 Mvar of reactive power — divided equally among thirty taps, one at the end of each third of a kilometre. Every number is read from the phasor solve of the whole ladder network.

Uncorrected, the feeder loses 16.8 kilowatts a phase. If every load were corrected to unity at its own terminals the loss would be 11.1 kilowatts, so the reactive current’s share — the part any capacitor can hope to remove — is 5.66 kilowatts. Now place a capacitor of two-thirds of the load’s reactive power, a megavar in three phases, at each tap in turn.

At the substation end it removes nothing. The capacitor there supplies reactive power that the substation would otherwise have supplied, but it supplies it at the substation, so the current in every section of the feeder is unchanged and so is the loss. As it moves out the loss falls: the capacitor now supplies the reactive current of the loads beyond it locally, and the sections between it and the substation carry less. The loss is least, 11.8 kilowatts, two-thirds of the way out, where 87.6 per cent of the reactive share has gone. Beyond that it rises again: at the far end the capacitor supplies more reactive power than the last few taps need, and the excess flows back towards the substation through sections that were carrying little before.

Change the capacitor’s size and the best place moves. A third of the reactive load does best at 83 per cent of the way out and removes 68.3 per cent; the whole reactive load does best halfway and removes 72.2 per cent. The best place for a capacitor of fraction $c$ of the reactive load is, near enough, $1 - c/2$ of the way out: far enough that the loads beyond it absorb half of what it supplies and the loads before it the other half.

The best size, and how close the rule comes #

Do that search at every size, and each size has a best place and a best share removed.

The best single capacitor is 65 per cent of the reactive load, placed 67 per cent of the way out, and it removes 87.6 per cent of the reactive share of the loss. The textbook rule for this problem says two-thirds of the reactive load at two-thirds of the length, removing eight-ninths, 88.9 per cent. It is derived for a feeder with a continuous, uniform load that draws a constant current whatever its voltage, on a line with resistance and no reactance — none of which the solve assumes. The loads here are impedances, so the current they draw falls with the voltage along the feeder; the line has reactance, so the capacitor’s current changes the voltage profile it is itself drawing against; and the load is thirty taps rather than a continuum. Together those move the answer by a point and a third. The rule is right to the precision a utility planning a capacitor bank ever works to, and wrong in the third figure for reasons that can be named.

The curve is also flat at the top. Anything from about half to four-fifths of the reactive load, placed at its own best position, removes more than eighty per cent of the reactive share, and a utility choosing among standard bank sizes loses little by rounding. Position matters more than size: the hero figure shows a two-thirds capacitor at the substation removing nothing and the same capacitor two-thirds of the way out removing nearly nine-tenths.

Two capacitors, and five #

One capacitor can only make the reactive current zero at one point. Several capacitors can approximate the distributed load more closely, and the textbook rule extends: $n$ equal capacitors, each $2/(2n + 1)$ of the reactive load, placed at $2/(2n + 1)$, $4/(2n + 1)$, … of the length, remove $1 - 1/(2n + 1)^2$ of the reactive share.

On the solved feeder the rule’s placements remove 87.6, 95.6, 97.8, 98.7 and 99.2 per cent for one to five capacitors, against the rule’s 88.9, 96.0, 98.0, 98.8 and 99.2. The departure shrinks as the number grows, because more capacitors make each one’s local effect smaller and the idealisations matter less. The second capacitor is worth eight percentage points; the fifth is worth less than half of one. A utility has the same diminishing return in money: each bank costs about the same, and after the second or third the loss it saves is worth less than the bank.

The rule’s structure is worth seeing once. With $n$ capacitors the reactive current along the feeder becomes a sawtooth: each capacitor resets it to minus half a step, the loads between capacitors carry it back up through zero to plus half a step, and the loss in each stretch is the integral of the square of a current that runs symmetrically about zero. The sawtooth’s amplitude is $1/(2n + 1)$ of the uncorrected maximum, its mean square is a third of the amplitude squared, and the uncorrected triangle’s mean square is a third of the maximum squared — whence $1/(2n + 1)^2$.

The voltage the capacitor lifts #

Loss is not the only reason a utility places capacitors on a feeder, and on many feeders it is not the main one. The reactive current also drops voltage across the line’s reactance, and the far end of a long feeder is where customers complain.

Uncorrected, the far end of this feeder sits 4.62 per cent below the substation’s voltage. With the two-thirds capacitor at two-thirds of the way out it sits 2.84 per cent below, and with every load corrected at its own terminals, 2.54. One capacitor recovers most of the voltage that perfect correction would, as it recovers most of the loss. The profile kinks at the capacitor’s node: between the substation and the capacitor the feeder carries a smaller net reactive current, so the voltage falls more slowly there; beyond the capacitor it still carries its loads’ reactive current and falls at the uncorrected rate.

The far end that rises and the reactance cancelled, and the resonance it buys measured the two other ways a utility lifts a far end: by the line’s own charging and by a capacitor in series with the line. A shunt capacitor on a distributed load is the third, and the cheapest; the series capacitor cancels the line’s reactance for all current and the shunt capacitor cancels the load’s reactive current for part of the line.

Sized for full load, left in at light load #

A capacitor’s reactive power is fixed by its size and the voltage; a feeder’s load is not. Overnight, the load on a residential feeder can fall to a fifth of its peak, and a capacitor sized for the peak then supplies more reactive power than the loads draw.

At full load the capacitor cuts the loss from 16.8 to 11.8 kilowatts. As the load falls, the loads’ reactive current falls with it and the capacitor’s does not, so the feeder carries the capacitor’s surplus back towards the substation. Below half of full load the feeder loses more with the capacitor than without it, and at a fifth of full load it loses 4.07 kilowatts against 0.71 — nearly six times as much — while its far end sits 0.96 per cent above nominal. The capacitor that was right once found the same structure for a single load: a correction computed exactly for one condition is wrong at every other, and can be worse than no correction.

That is why distribution capacitors are switched. A bank that is connected when the load is high and disconnected when it is low — by time of day, by voltage, or by measured reactive power — keeps the benefit of the full-load figure and avoids the light-load penalty. The half-load crossover is the arithmetic a switching controller needs: a fixed bank sized at two-thirds of the peak reactive load must come off before the load falls below about half its peak.

How many taps make a continuum #

The classical rule is for a load spread continuously along the line. Real feeders have a handful of transformers, and a capacitor can only sit at one of them.

With three taps the best tap is the second, at two-thirds of the length, and the capacitor removes 86.3 per cent; with five taps the nearest tap to two-thirds is at 60 or 80 per cent, and the best is 80, removing 87.4; with eight, 75 per cent; and from twelve taps on the best position sits at two-thirds to within the spacing and the share removed has settled at 87.6. The share removed is remarkably insensitive to the coarseness — three taps lose little more than a point against fifty — because the loss curve is flat near its minimum, and a capacitor a tenth of the feeder away from the ideal spot costs only a second-order amount.

The triangle under every distributed load #

The two-thirds rule is one instance of an arithmetic that appears wherever a current is drawn off a conductor along its length rather than at its end, and it is worth stating once in general.

On a feeder whose load is spread evenly, the current in each section is proportional to the load beyond it, so it falls linearly from its full value at the source to zero at the far end. The voltage drop is the integral of that current times the resistance per unit length, and it is half what the same total current would drop if the whole load sat at the far end. The loss is the integral of the current’s square times the resistance, and it is a third of what the lumped load would cost. Both numbers come from integrating a triangle; the capacitor rules come from integrating a triangle with a step cut out of it, and the eight-ninths is what is left of a third when the step is placed best.

The same triangle sits under a printed-circuit supply rail feeding a row of chips, a ground return collecting the currents of a row of loads, and a heating element tapped along its length. The millivolts in the wire measured a shared return as a lumped impedance between one load and one sensor; spread the loads along the return and the voltage a sensor at the far end sees is half the lumped figure, and a sensor halfway along sees three-eighths of it. The rail the load moves gave an amplifier’s supply an impedance; a rail feeding a row of amplifiers drops the triangle’s voltage, and decoupling capacitors placed along it play exactly the role of the feeder’s capacitors — supplying locally a current that would otherwise travel the length of the rail.

The analogy is closer than it looks. A decoupling capacitor supplies the transient current of the chips beyond it; a feeder capacitor supplies the reactive current of the loads beyond it; in both cases the capacitor at the source end does nothing for the conductor between source and load, and in both cases one capacitor does best somewhere along the run and several do better in diminishing proportion. The feeder makes the arithmetic clean because its loads are steady and its frequency single.

Reactive power that flows backwards #

The light-load figure has a second lesson, about the direction of reactive power. With the capacitor oversized for the load, the sections between the capacitor and the substation carry reactive power towards the substation: the capacitor is generating vars that the loads do not want, and the substation absorbs them. The loss that results is identical in kind to the loss the uncorrected feeder suffered — reactive current in a resistive conductor — and only its direction has changed.

That symmetry is the reason power engineers speak of reactive power being “generated” by capacitors and “absorbed” by inductors, a convention that the energy a unity power factor doubles showed is a bookkeeping of where stored energy is exchanged rather than a statement that anything is made. On a feeder the bookkeeping has a cost: every var that travels costs loss in the sections it travels through, whichever way it goes, and the best capacitor is the one whose vars travel least. The load that has two voltages or none is the other side of the same coin at heavy load, where a feeder’s voltage depends on reactive power so strongly that the capacitor becomes the difference between a load that can be supplied and one that cannot.

Where the numbers come from, and what the feeder omits #

The feeder is a ladder network of thirty sections of series resistance and inductance with a load at the end of each, driven from an ideal source at the substation and solved as a phasor network at 50 Hz. Each load is a series resistance and inductance chosen to draw its share of the power at the nominal voltage; the capacitor is chosen to supply its stated reactive power at nominal voltage. The loss is the sum of the squared current magnitude times the resistance over every line section, and the reference losses — uncorrected, and every load at unity — are solves of the same network. Every position, size and count is a separate solve; nothing is taken from the rule except the placements the many-capacitor figure tests.

The loads are constant impedances, which is right for heating and roughly right for lighting and wrong for motors, which draw nearly constant power; a constant-power load draws more current as its voltage falls and makes the far end’s sag, and the capacitor’s benefit there, a little larger. The feeder is uniform and its load evenly spread, which few real feeders are. There is one phase, balanced with the other two, and no harmonics; the capacitor that was right once is the warning that a capacitor on a feeder with rectifier loads can resonate with the line’s inductance at a harmonic, which no loss figure here would show.

Still open: loads that are not uniform, a switched bank’s timing, and harmonics on the feeder #

A load concentrated at the far end. Many feeders run several kilometres before reaching their customers. The reactive current is then constant along the empty stretch and falls only over the populated one, and the best capacitor sits nearer the populated end by an amount that depends on the ratio of the two lengths. The rule for a uniform load gives the wrong answer for that feeder, and the size of the error is the same network solved with the taps moved.

When a switched bank should come on. A bank switched by the load’s reactive power has a threshold, and the half-load crossover found above is the break-even for a single fixed bank. With two banks the thresholds for each, and the hysteresis that stops them hunting, are a question about the loss curve at every combination of banks and load — the light-load figure generalised to a switching schedule.

Harmonic currents on the corrected feeder. A capacitor two-thirds of the way out and the inductance of the line between it and the substation form a resonance, and its frequency falls as the capacitor grows. Where that resonance sits relative to the fifth and seventh harmonics of the feeder’s rectifier loads decides whether the capacitor that cuts the fundamental’s loss amplifies a harmonic’s, and the loss figure for a feeder with real loads is the sum over harmonics that this essay, at one frequency, has not done.