The drop a series capacitor can reach

The reactance cancelled, and the resonance it buys put a capacitor in series with a feeder and measured the trade: a quarter of the line’s reactance cancelled took the regulation of a 20 kVA load at 0.9 lagging from 5.17 to 4.26 per cent, and put a resonance at exactly half the line frequency into a loop that had none. It noted in passing that even full compensation could not remove all the regulation at that power factor, because the capacitor cancels the reactive part of the drop and leaves the resistive part alone.

That aside is the whole of the next question, because the split between the two parts is not a property of the feeder. It is a property of the load. A load drawing mostly reactive current makes most of its drop in the line’s reactance, and a series capacitor can take nearly all of it away; a load at unity power factor makes nearly all of its drop in the line’s resistance, and the capacitor can take almost nothing. And a load that leads — a capacitive load, an over-corrected installation — was using the line’s reactance the other way, to lift its far end, and the capacitor takes that away too. This essay measures the same capacitor against loads of every power factor, and then asks what happens past full cancellation and whether it matters where on the line the capacitor goes.

The same capacitor, every load #

The feeder is the earlier essay’s: 230 V at the source, 0.05 Ω of resistance and 0.2 Ω of reactance in the line, and 20 kVA of load at the far end, solved at 50 Hz. The load’s power factor is swept from 0.71 leading, through unity, to 0.3 lagging, and the regulation — the far end’s rise from full load to no load, over the loaded voltage — is solved at each. The solve is checked against the voltage divider written in closed form, and agrees to a billionth of a per cent.

Uncompensated, the curve runs from −3.78 per cent at 0.71 leading, through 2.17 at unity, to 7.78 at 0.3 lagging. With half the line’s reactance cancelled it runs from −1.26 through 1.96 to 4.18. The capacitor has taken 3.61 points off the heavily lagging load, 1.79 off the 0.9 load of the earlier essay, 0.21 off the unity load — and has added 2.52 points to the leading one, bringing its far end down from 3.8 per cent above the source to 1.3 above it.

Move the slider to full cancellation and the curve flattens almost completely: 1.35 per cent at 0.71 leading, 1.89 at unity, 1.70 at 0.9 lagging and 0.58 at 0.3. What is left is the drop the line’s resistance makes, which depends on the load’s real current alone and so falls as the power factor falls. The capacitor does not lower voltage drops in general. It removes the line’s reactance from the arithmetic, and every load is pulled towards the regulation the resistance alone would give it.

Why the load decides #

The arithmetic behind the curve is the phasor drop, and it is worth writing down because it shows which of the feeder’s numbers enter. The load’s current $I$ at a power-factor angle $\varphi$ flows through the line’s $R + jX$, and the far end falls short of the source by, to first order,

$$\Delta V \approx I\,(R\cos\varphi + X\sin\varphi),$$

with $\varphi$ positive for a lagging load and negative for a leading one. The resistance works on the load’s in-phase current and the reactance on its quadrature current. A series capacitor multiplies $X$ by $1 - k$ and leaves $R$ alone, so it scales the second term and cannot touch the first. The share of the drop within its reach is

$$\frac{X\sin\varphi}{R\cos\varphi + X\sin\varphi},$$

which is the load’s power factor and the line’s ratio of reactance to resistance, together, and nothing else. For this feeder, whose reactance is four times its resistance, that share is two-thirds at 0.9 lagging, 93 per cent at 0.3 lagging and zero at unity — and for a leading load it is negative, because the reactive term is then a rise.

The same expression says why the device suits some lines and not others. An overhead line whose reactance is ten times its resistance has 83 per cent of a 0.9 load’s drop within reach of a series capacitor; an underground cable, whose resistance and reactance are comparable, has about a third; a low-voltage cable run, dominated by resistance, has almost nothing to cancel at any power factor a building draws. Series compensation is a transmission and overhead-distribution technique because of that ratio, not because of the voltage. On a cable the drop is copper, and the only cure for copper is more of it — the same conclusion the reactance that is not free reached about cancelling a source’s reactance with a capacitor that has its own losses: cancellation removes reactance and nothing else.

Straight lines whose slope is the reactive power #

Turned the other way — regulation against the fraction cancelled, at fixed power factors — the same measurement makes the rule quantitative.

For small drops the regulation is $(R P + X(1-k)Q)/V^2$, with $P$ and $Q$ the load’s real and reactive power and $k$ the fraction cancelled. That is a straight line in $k$ with a slope of $-XQ/V^2$: the load’s reactive power in the line’s reactance, and nothing else. The solved lines are straight to the eye and their total falls follow that slope within the second-order terms: 7.20 points at 0.3 lagging, where $Q$ is 19.1 kvar, 6.56 at 0.5, 5.46 at 0.7, 3.46 at 0.9, 2.57 at 0.95, and 0.28 at unity, where $Q$ is zero and all that moves is the second-order term — the drop’s quadrature component, which also passes through the reactance.

Where the lines end is the other half of the rule. At full cancellation each sits at its $RP/V^2$, so the fan at the right of the figure is ordered by real power — 1.89 per cent at unity, where the whole 20 kVA is real, down to 0.58 at 0.3, where six kilowatts are. The heavily lagging load, the worst uncompensated, ends up the best compensated. A designer choosing a compensation for a feeder whose loads vary in power factor is choosing a slope for each of them, and the loads with the most reactive current gain the most.

It follows that the two ways of fighting a reactive drop compete for the same term. A shunt bank at the load removes the load’s reactive current, which flattens its line in this figure to the unity one; a series capacitor removes the line’s reactance, which tilts every line down to its end point. A load already corrected to unity has 0.28 points left for a series capacitor to take, and a line already fully compensated has nothing left for a shunt bank to do except move reactive power it no longer needs to move. The energy a unity power factor doubles is the accounting of what the shunt route stores and exchanges to get there; the series route exchanges energy between the capacitor and the line’s inductance instead, and it is that exchange which rings at $f_0\sqrt k$.

A leading load and its lagging twin #

A leading load is where the rule gives the least intuitive answer, and it is not rare: a power-factor correction bank left connected when the motors it was sized for switch off, a lightly loaded cable, a rectifier with a large reservoir on a weak supply. The far end that rises showed that past a computable angle of leading load the far end goes above the source, because the leading current through the line’s reactance adds to the voltage rather than subtracting from it.

Uncompensated, 20 kVA at 0.95 leading lifts the far end 0.26 per cent above the source, at 0.7 leading 3.85 per cent, and at 0.3 leading 6.56 per cent. A series capacitor cancels the reactance doing the lifting, so it brings each far end down: through zero at 14 per cent cancellation for the 0.95 load, 75 per cent for the 0.7 load and 92 per cent for the 0.3 load, and on to a positive regulation. At full cancellation each leading load’s regulation is exactly its lagging twin’s — 1.80, 1.33 and 0.58 per cent — to a billionth of a per cent, because a line that is only resistance sees a load of $R_L + jX_L$ and one of $R_L - jX_L$ as the same magnitude of impedance and divides the voltage identically.

That is the plainest way to state what the capacitor does. It does not add voltage or subtract it. It removes the line’s reactance, and with it the only element that could tell the sign of the load’s reactive current. An installation whose far end sits too high because its load leads can be brought down by a series capacitor — a use the device is rarely credited with — and one whose far end sits too low because its load lags can be brought up, by the same component and the same amount per kilovar.

Past full cancellation #

Nothing stops a capacitor from cancelling more reactance than the line has, and the curves say what happens if it does.

Beyond full cancellation the line’s net reactance is capacitive, and a lagging load now behaves as a leading one did: its current through the net negative reactance lifts the far end. The regulation of the 0.9 load keeps falling past 1.70 per cent at 100, through zero at 154.8 per cent, and below it — the far end above the source and rising further as the load grows, which is the opposite of the behaviour every piece of downstream equipment is designed around.

The dashed curve is the reason nobody goes there on purpose. The current into a short at the far end is the source voltage over the line’s net impedance, 1,116 A uncompensated. At full cancellation the net impedance is the line’s resistance alone, and the current is 4,600 A — 4.1 times as much. The earlier essay found the resonance of a compensated feeder at $f_0\sqrt k$, below the fundamental for any compensation less than one. At $k = 1$ it is the fundamental: the capacitor and the line’s inductance resonate at 50 Hz, and a fault at the far end is a fault into a series-resonant circuit driven at its own frequency. Past one the resonance moves above the fundamental, the fault current falls again, and the line is simply capacitive. The boundary between compensation and over-compensation is not a regulation boundary. It is a resonance crossing the frequency the feeder carries.

The over-compensated feeder has one more property worth naming, because it inverts a stability argument. A load whose voltage rises as it draws more current is the mirror image of the fixed-power load the load that has two voltages or none studied, whose voltage falls as it draws more and which therefore has a limit beyond which no voltage will do. Over-compensation removes that limit’s mechanism from the reactance and puts a positive feedback in its place: more load, more voltage, more power drawn by any load whose power rises with its voltage. On an ordinary feeder a lamp dims when a neighbour’s load comes on; on an over-compensated one it brightens.

Where the capacitor goes #

The last question is one a designer faces physically: the capacitor has to be put somewhere, at the substation, halfway along the feeder or next to the load.

For the load it does not matter at all. With the line built as ten equal sections and the capacitor inserted at the source end, in the middle or at the load end, the far-end voltage is 222.4770 V in all three to a part in a billion, against 218.70 V uncompensated. Series elements with nothing connected across them commute: the current through every one of them is the load’s current, and the far-end voltage is the source’s less the sum of their drops, in whatever order they come.

For the line it matters. The capacitor’s drop is a rise — a leading voltage added by the load’s lagging current — and it happens where the capacitor sits. At the source end the conductor just past the capacitor is at 233.97 V, 1.7 per cent above the source, and the voltage falls from there along the whole line. In the middle the line falls to the midpoint, steps up, and falls again, never exceeding the source. At the load end the line falls all the way and the capacitor lifts the last node to the far-end value. The same far end, three different insulation stresses, and three different sets of voltages at any customer tapped off the line between the two ends — which is where the position actually matters, on a real feeder with loads along it and not only at its end.

How the numbers were obtained #

Every figure solves the feeder at 50 Hz as a netlist: a 230 V source, the line’s resistance and inductance, the series capacitor sized to cancel the stated fraction of the line’s reactance, and a load of the stated kilovolt-amperes as a resistance in series with an inductance (lagging) or a capacitance (leading). Regulation is taken against the source voltage, since with no shunt element the unloaded far end is the source. The closed-form route is the voltage divider between the load’s impedance and the line’s net impedance, computed without the solver. The fault current is the source voltage over the line’s net impedance at the fundamental. The profile builds the line as ten equal series sections and records the voltage magnitude at each junction and on both sides of the capacitor.

What it leaves out #

Loads along the line. Every load here is at the far end. With load tapped at points along the feeder, the commutation argument fails — current differs from section to section — and the capacitor’s position changes every customer’s voltage, not only the stress on the line.

The line’s own capacitance. A long line or a cable has shunt capacitance distributed along it, which breaks the commutation: current differs along the line, and the capacitor’s position then matters to the far end as well. The headroom that is the line’s own charge measured what that shunt capacitance does to how much power a line can deliver, and a series capacitor on such a line changes both at once.

Harmonics. The regulation is solved at the fundamental. A lagging load with a rectifier in it draws harmonic currents that see the capacitor’s reactance fall with frequency, which is the concern the earlier essay’s resonance raises, not this one’s.

Still open: the limit a capacitor raises, loads along the line, and the shunt it replaces #

What the capacitor does to how much power the line can carry. Regulation is one reason to cancel reactance and the lesser one on a transmission line, where the reason is the power that can be transferred between two sources before they fall out of step. That limit rises as the reactance falls — but the resistance, which this essay found left untouched by the capacitor, is in the limit too, and whether it caps the gain is a solve on two sources joined by the compensated line.

Distributed load. A feeder with customers spread along it has a best capacitor position that depends on how the load is spread, and the commutation that made position irrelevant here is exactly what distributed load removes. Where the capacitor should go for a uniformly loaded feeder is a small optimisation on the ten-section model.

Series against shunt, for the same kilovars. The current that does no work is what a shunt bank removes at the load, and the capacitor that was right once is what that bank does when the load changes. A shunt bank cancels the load’s reactive current; a series capacitor cancels the line’s reactance. For a given number of kilovars installed, which lowers the regulation more depends on the load’s power factor in a way this essay’s slopes suggest and do not yet measure.