Sixty degrees, and the limit compensation stops raising

The drop a series capacitor can reach measured a series capacitor as a cure for voltage drop and found it reaching only the part of the drop the line’s reactance makes, with the resistance’s part left exactly where it was. On a distribution feeder that is the whole reason to fit one. On a transmission line it is the lesser reason. The greater one is transfer capability: two ends of a long line are each held at their voltage by the machines connected there, the power that flows between them is set by the angle between those voltages, and there is a largest power the line can carry before a larger angle carries less. Cancelling reactance raises that largest power, and the textbook says by how much: the limit is $V^2/X$, so cancelling a fraction $k$ multiplies it by $1/(1-k)$.

That formula has no resistance in it, and the earlier essay found resistance to be the one thing a series capacitor cannot touch. With the resistance kept, the limit does not rise without bound. It rises, peaks at a definite compensation — the one that leaves the line’s impedance angle at sixty degrees — and falls, to nothing at full cancellation. This essay solves two sources and a compensated line directly and finds the peak, its position, what decides it, and a property of the line at its peak that is an old theorem arriving by a new road.

Two sources and a line #

The circuit is the smallest one with the property: two ideal 230 V sources at 50 Hz, the receiving one at zero phase and the sending one leading it by an angle $\delta$, joined by the feeder of the earlier essays — 0.05 Ω of resistance and 0.2 Ω of reactance — with a series capacitor cancelling a fraction $k$ of the reactance. The line’s current is solved, and the real power absorbed by the receiving source is the product of its voltage and the conjugate of that current.

Uncompensated, the received power rises from zero as the angle opens, peaks at 194.37 kW at an angle of 75.96°, and falls after it. With half the reactance cancelled the curve is steeper and higher: its peak is 261.55 kW, at 63.43°. Both peaks sit at an angle equal to the line’s own impedance angle, $\beta = \arctan(X/R)$ — 75.96° for the bare line, 63.43° with half its reactance gone — and both match a closed form to a part in a million.

The closed form comes from the phasors. With the line’s impedance written as $|Z|\angle\beta$ and both sources at magnitude $V$, the power received is

$$P_r = \frac{V^2}{|Z|}\left[\cos(\beta - \delta) - \cos\beta\right],$$

which is largest at $\delta = \beta$, where it is $V^2(1 - \cos\beta)/|Z|$. Since $|Z| = R/\cos\beta$, that is

$$P_{max} = \frac{V^2}{R}\cos\beta\,(1 - \cos\beta).$$

The textbook’s lossless limit is the case $R \to 0$, $\beta \to 90°$, where it becomes $V^2/X$. With any resistance at all it is a function of the impedance angle, and the angle is what a series capacitor changes.

The peak at sixty degrees #

As the capacitor cancels more reactance, $X(1-k)$ falls, the impedance angle falls from its bare value towards zero, and the limit follows $\cos\beta(1 - \cos\beta)$ — a product that is zero at both ends and largest in between.

The product $c(1-c)$ is largest at $c = \tfrac12$, so the limit is largest when $\cos\beta = \tfrac12$ — when the impedance angle is exactly sixty degrees. The compensation that puts it there is

$$k^* = 1 - \sqrt3\,\frac{R}{X},$$

which for this line, with $R/X = 0.25$, is 56.7 per cent, and the limit there is $V^2/4R$ = 264.50 kW, against 194.37 uncompensated: 36 per cent more power through the same line. The search on the solve finds 264.50 at 56.7 per cent to a part in a million, and finds no compensation that beats it.

Past the peak the limit falls. Beyond 78 per cent cancelled it is below its uncompensated value; at 90 per cent it is 36 per cent of it; and at full cancellation it is zero. A purely resistive link between two sources of equal voltage cannot deliver power to the receiving end at any angle, because any current it carries is driven by a difference between the two voltages, and with no reactance to rotate that current the receiving source always receives less than the line dissipates. The lossless formula, dashed in the figure, agrees with the solve for the first few per cent and then diverges without bound. It is not a slightly optimistic estimate of the limit; past a moderate compensation it is the wrong shape.

Why a resistance cannot carry it #

The zero at full cancellation deserves a picture, because it is where the lossless intuition fails most completely. With two sources of equal magnitude, the voltage across the line is the chord between two points on a circle, and the current is that chord divided by the line’s impedance. A reactive line turns the current ninety degrees from the chord, which puts it nearly in phase with both sources’ voltages when the angle between them is small — so both ends see a current in phase with their own voltage, one sending and one receiving, and very little is lost between them. A resistive line leaves the current along the chord, which is perpendicular to the bisector of the two voltages; at the receiving end the current then has a component against the receiving voltage, and the receiving source is pushed to send rather than receive. Three voltages that close on one drew Kirchhoff’s law as a closed polygon of phasors; here the polygon is a triangle of two equal voltages and a line drop, and whether the receiving end gains power is a question of which way the drop’s current points.

Between the two extremes the line’s impedance angle sets how far the current is rotated, and sixty degrees is where the gain from rotation and the loss in the resistance balance in the receiving end’s favour most. It is a property of the angle alone — the same for a 230 V feeder and a 400 kV line — which is why the optimum is a formula in $R/X$ and nothing else.

The line decides how much there is to gain #

The optimum depends on the line through one number, its ratio of resistance to reactance, and so does the gain.

A line whose resistance is a twentieth of its reactance — a large overhead transmission line — can have its limit raised 5.27 times, at 91.3 per cent compensation. At a tenth, 2.79 times at 82.7 per cent. The distribution feeder of these essays, at a quarter, gains 1.36 times at 56.7. At a half, the gain is 1.011 — one per cent — at 13.4 per cent compensation. And a line whose resistance exceeds $1/\sqrt3$, 0.577, of its reactance has an impedance angle already below sixty degrees before any capacitor is fitted: it is past its own best angle, and every series capacitor lowers its limit.

This is the quantitative form of the earlier essay’s observation that series compensation belongs to overhead lines and not to cables. There the argument was about voltage drop and the share of it the reactance makes; here it is about transfer, and the ratio $R/X$ appears again, now as a threshold with an exact value. The 1/(1 − k) of the textbook describes the first line in the figure to within six per cent up to half compensation and overstates it by nearly a third at eighty per cent; it describes the last line not at all.

A worked high-voltage case makes the scale concrete. An overhead line whose resistance is three per cent of its reactance has its optimum at 94.8 per cent compensation, where its limit is 8.59 times the uncompensated one. Compensated to seventy per cent, which is a common figure for such lines, the solve’s closed form gives a gain of 3.08 against the textbook’s 3.33; at half compensation, 1.94 against 2.00. For such a line the lossless formula is a fair tool over the range compensation is actually used in, and it is useful to know that it is fair because the line sits a long way from its sixty degrees, not because resistance does not matter. For a distribution feeder the same formula is wrong at every compensation worth installing.

The angle it buys, and the headroom it does not #

A transmission line is not run at its limit. It is run at an angle well below the limit’s, and the difference is the margin against a disturbance pushing the angle over the top. So the other reading of compensation is the angle a fixed transfer needs.

A 100 kW transfer needs 25.2° uncompensated, 18.5° at 30 per cent and 14.6° at half: the capacitor lets the same power pass at a smaller angle, which is what it is sold for. But the limit’s angle is the line’s impedance angle, and the capacitor lowers that too, from 76.0° to 63.4° at half compensation and to 50.2° at 70 per cent. The headroom between operating angle and limit angle is 50.8° uncompensated, 48.8° at half compensation and 38.0° at 70 per cent: nearly constant at first, and then shrinking, because past the peak the operating angle stops falling — at 80 per cent 100 kW needs 13.1°, more than the 12.2° it needed at 70 — while the limit’s angle keeps falling.

So the margin argument and the power argument part company. By the lossless reasoning, compensation always helps both: smaller operating angle, same 90° limit angle, more headroom. With resistance, the limit’s angle moves down with the compensation, and the headroom in degrees is roughly conserved until the peak and lost after it. A designer reading the operating angle alone would conclude that more compensation is always safer; the solve says the extra safety is spent as it is bought.

The shape is familiar from a different limit in these essays. The load that has two voltages or none found a fixed-power load on a line delivered at two voltages below a limit and at none above it, the limit sitting at the load the maximum-power theorem picks. The angle between two sources is the same kind of variable: every power below the limit is carried at two angles, one on each side of the impedance angle, and only the smaller is a stable operating point, because on the far side a larger angle carries less power and a disturbance that opens the angle is not opposed. Compensation moves the top of that curve up and to the left at once, and the headroom between the operating point and the top is what a stability margin measures.

A control that responds to a shortfall by opening the angle further — a governor asked for more power — is stable only on the near side of the peak, for the reason the regulator that pushes past the nose found for a tap changer that raises its ratio whenever the voltage is low: past the peak, the action that used to raise the quantity lowers it, and the loop that was correcting now runs away. Compensation does not change that structure. It moves the peak, and past the optimum it moves it the wrong way, to a lower power at a smaller angle, which is the least comfortable place for a peak to be when the controller does not know where it is.

Half of what is sent #

The last reading is the one that turns a transmission-line result into an old friend.

At the angle of its largest delivery, a line delivers a fraction of what it is sent that depends only on its impedance angle: the sent power at $\delta = \beta$ is $V^2\cos\beta(\cos\beta - \cos2\beta)/R$, and the ratio of received to sent is $1/(1 + 2\cos\beta)$. Uncompensated, with $\cos\beta = 0.2425$, the line at its limit delivers 67.3 per cent of what goes in. At the compensation that makes the limit largest, $\cos\beta = \tfrac12$, it delivers exactly 50 per cent — the solve gives 50.0000 — and as the line is cancelled towards a pure resistance the fraction falls towards a third.

Half is not a coincidence. The load that takes the most found a load resistance equal to its source’s taking more power than any other, at exactly fifty per cent efficiency, with the source burning as much as the load receives. Here there is no load resistance to choose: the receiving end is a source, and what is chosen is the angle and the compensation. The two together play the same role, and the largest power that can be delivered is delivered, as always, with as much again lost on the way. The load that may be complex reached the conjugate match by cancelling reactance with an opposite one; series compensation is that cancellation made in the line instead of at the load, and it stops short of full cancellation for the reason the maximum-power theorem gives — the best transfer is at a sixty-degree line, not a zero-degree one, because the receiving source’s voltage is fixed and cannot be matched the way a load resistance can.

How the numbers were obtained #

Every figure solves one netlist at 50 Hz: a 230 V source at an angle $\delta$, the line’s resistance and inductance, a series capacitor sized to cancel the stated fraction of the reactance, and a 230 V source at zero phase at the far end. The line current is read from the voltage across the resistance, and the received and sent powers are the real parts of each source’s voltage times the conjugate of that current. The largest received power is found by golden-section search over $\delta$ from zero to 180 degrees on the solve, and is compared with $V^2\cos\beta(1 - \cos\beta)/R$, never replaced by it; the angle for 100 kW is bisected on the solve. The ratio figure evaluates the closed form and checks it against the search at 30 per cent compensation on every line.

What it leaves out #

The machines at the two ends. Both ends are ideal sources. A real transfer limit is set by generators whose internal reactance adds to the line’s, which raises the impedance angle and moves the optimum compensation higher; and the dynamic limit — whether the angle recovers after a fault — depends on their inertia, which this solve does not have.

Unequal voltages. Both sources are at 230 V. With the receiving end lower, the closed form gains a term in the product of the two magnitudes and the peak moves; the sixty-degree result is for equal magnitudes, which is the condition a transmission system is operated near.

The resonance. A compensated line has the sub-synchronous resonance the reactance cancelled, and the resonance it buys found at $f_0\sqrt k$ — 37.6 Hz at the optimum here — and a generator’s shaft has torsional modes at such frequencies. That interaction, not the steady-state limit, is what usually caps compensation on a real line well below the peak this essay finds.

Still open: the machines’ reactance, unequal ends, and the capacitor’s own losses #

The generator’s reactance in the loop. Adding each machine’s internal reactance to the line’s raises the total impedance angle, which moves $k^*$ up — the capacitor then cancels a fraction of a larger reactance of which it can reach only the line’s part. Whether the sixty-degree optimum survives when part of the reactance is out of reach is a solve with two more inductors in it.

A capacitor with losses. The reactance that is not free found a cancelling capacitor’s own resistance, its reactance over its Q, eating into what cancellation buys. Here that resistance adds to the line’s and so raises $R/X$ as $k$ rises — pushing the peak to a lower compensation, and for a poor enough capacitor abolishing it. How good a capacitor has to be for the peak to stay near $1 - \sqrt3 R/X$ is a small closed form this essay did not write down.

Unequal ends and a load in the middle. With the receiving voltage below the sending, and with load tapped along the tie, the transfer limit becomes a surface over two voltages and a position, and the compensation that maximises it is no longer a single number.