The reactance the capacitor cannot reach

Sixty degrees, and the limit compensation stops raising put a series capacitor in a line joining two sources and found the transfer limit — the most power the receiving end can take before a larger angle between the sources carries less — rising with compensation only until the line’s impedance angle reached sixty degrees, and falling after. For a 230 V tie of 0.05 + j0.2 Ω the peak was at 56.7 per cent of the reactance cancelled, where the limit is $V^2/4R$ = 264.5 kW, 36 per cent above the uncompensated 194.4. The lossless textbook formula, which rises without bound, was not a slightly optimistic estimate but the wrong shape.

Its two sources were ideal: each held its voltage at its own terminals whatever current flowed. A real tie joins machines or grids, and a machine holds its EMF behind a reactance of its own — the synchronous or transient reactance of a generator, the short-circuit impedance of a grid. The capacitor sits in the line and cancels the line’s reactance; it cannot reach the machines’. That was the first of the sixty-degree essay’s open questions, and it changes the answer more than the question suggests.

The same tie, with machines at its ends #

The circuit is the earlier one with a reactance $X_m$ added behind each source: the sending EMF leads the receiving one by an angle δ, both at 230 V, and the power counted is what the receiving EMF absorbs. The line is still 0.05 + j0.2 Ω, the capacitor still cancels a fraction k of the line’s 0.2 Ω, and nothing else changes. The earlier closed form still holds with the totals in it: the loop’s resistance is the line’s, its reactance is $X + 2X_m - kX$, and the limit is $V^2\cos\beta\,(1-\cos\beta)/R$ for the loop’s impedance angle β, peaking where β is sixty degrees. A search on the solved netlist, angle by angle, agrees with it to a part in a hundred thousand at every compensation.

With 0.05 Ω behind each EMF the uncompensated limit is lower — 145.3 kW, against 194.4 for the bare tie, because the loop is longer — and compensation now raises it at every step: to 261.6 kW at full cancellation, a gain of 80 per cent, with no peak along the way. The sixty-degree angle would need the loop’s reactance down to $\sqrt3 R$ = 0.087 Ω, which is 0.2 + 0.1 − 0.087 = 0.213 Ω of cancellation, 106.7 per cent of the line’s reactance. The capacitor would have to cancel more than all of the line to get there, and it can only reach the line.

The slider shows the boundary. With no machine reactance the peak is the earlier essay’s, 56.7 per cent; with 0.02 Ω at each end it moves to 76.7 per cent and is still $V^2/4R$ when reached; with 0.1 Ω the sixty degrees would need 157 per cent and the limit rises from 115.0 to 194.4 kW; with 0.2 Ω, 257 per cent, and from 80.6 to 115.0.

Where the loop’s angle starts #

The reason is visible in the angle the loop starts at. The bare tie’s impedance angle is 75.96°, with 0.2 Ω of reactance against 0.05 of resistance; a capacitor that cancels 56.7 per cent of that reactance brings it to sixty. With 0.05 Ω behind each EMF the loop has 0.3 Ω of reactance and starts at 80.5°; with 0.1 Ω it has 0.4 and starts at 82.9°. Every machine’s reactance tilts the loop further towards ninety degrees, where the lossless formula is exact, and every one of them is reactance the capacitor sees only as a fixed part of the loop it cannot shorten.

Sixty degrees means the loop’s reactance is $\sqrt3$ times its resistance, 0.087 Ω for this line, and nothing about the machines changes that target: they change only how much reactance there is to remove before reaching it. A capacitor in the line can remove at most the line’s own 0.2 Ω, so the target is reachable only while the machines contribute less than 0.113 Ω between them. The whole of the earlier essay’s peak, and its fall to nothing at full cancellation, lives inside that narrow allowance.

When the peak leaves reach #

The optimum compensation is

$$k^* = \frac{X + 2X_m - \sqrt3\,R}{X},$$

the bare tie’s $1 - \sqrt3R/X$ plus $2X_m/X$. It reaches full cancellation exactly where $2X_m = \sqrt3 R$ — each machine’s reactance equal to $\sqrt3/2$ of the line’s resistance.

For a line with R/X of 0.25 the boundary is at $X_m/X$ = 0.217; for 0.1 it is 0.087; for 0.5, 0.433. Those are small numbers. A generator’s reactance is usually many times a short tie’s, and even the short-circuit impedance of a stiff grid is rarely a small fraction of a line long enough to need compensation. On most ties the sixty-degree peak is out of reach, and the earlier essay’s warning — that compensation past a moderate fraction lowers the limit — applies only to a line between sources stiffer than its own resistance, which is to say a short, lossy line between two strong buses.

The result depends only on ratios, so it reads the same at any voltage. In per-unit terms, a tie whose line has a reactance of 0.3 and a resistance of 0.03, joining two machines each with a reactance two-thirds of the line’s, would need its sixty-degree compensation at $(0.3 + 0.4 - 0.052)/0.3$ = 216 per cent of the line — more than twice everything the capacitor could cancel. For that tie the earlier essay’s curve is irrelevant: its limit rises with every per cent of compensation, and the question is not where the peak is but how much of the line the system can stand to have cancelled.

That reverses the practical lesson without contradicting the physics. The limit is still $V^2\cos\beta(1-\cos\beta)/R$ and still peaks at sixty degrees; what the machines change is which part of that curve a capacitor in the line can reach. A loop whose angle starts far above sixty degrees — long because of its machines rather than its line — can only be brought down part of the way, and along that part the limit rises.

The lossless estimate comes back #

The earlier essay found the textbook gain, $1/(1-k)$, diverging from the solve after the first few per cent. With machines in the loop the textbook’s version of full cancellation’s gain is $(X + 2X_m)/2X_m$: the ratio of the loop’s reactance before and after the line’s share is removed.

With 0.01 Ω behind each EMF the lossless estimate says full cancellation multiplies the limit by 11; the solve says it multiplies it by 0.385 — full cancellation of a line that dominates a resistive loop is ruinous — and the best compensation, at 67 per cent, gains 1.45. With 0.1 Ω the estimate says 2.02 and the line gives 1.70, at full cancellation; with 0.2 Ω, 1.50 against 1.43; with 0.5 Ω, 1.20 against 1.19. The more of the loop’s reactance belongs to the machines, the smaller the gain compensation can buy, and the more accurately the lossless formula prices it, because the loop’s angle stays near ninety degrees where the resistance hardly matters.

There is a largest gain in between. The best compensation’s gain peaks at about 1.8 with 0.05 Ω machines, where the machines are just long enough to push the peak out of reach and not yet long enough to dilute the line’s share. A tie with machine reactance of that order — a short line between moderately strong sources — is the one compensation helps most.

The location of that best case can be found by sweeping the machine reactance, and it sits a little past the boundary where the sixty-degree peak leaves reach: the gain is largest, 1.81, with 0.058 Ω behind each EMF. Below that the line’s resistance is a large enough share of the loop that full cancellation would overshoot sixty degrees, so the best compensation is partial and its gain is capped by the resistance; above it the machines’ reactance is a large enough share that cancelling all of the line removes a shrinking fraction of the loop. At the best case the two limits meet: the line can be fully cancelled without overshooting, and the line is still most of the loop.

A higher limit at a lower terminal #

The limit is the most power the loop can pass, and it is reached at the angle where the power stops rising. What the receiving machine’s terminals see there is a separate question, and it is the one an operator watches.

With 0.1 Ω machines the receiving terminal sits at 180 V at the uncompensated limit, 168 V with half the line cancelled and 148 V with all of it; with 0.05 Ω machines 195, 181 and 149; with 0.2 Ω, 169, 162 and 154. The compensated tie reaches its larger limit with a larger current, and that current crosses the receiving machine’s own reactance, which the capacitor does not touch: the terminal is pulled further from its EMF.

So the extra power a capacitor buys is delivered with the receiving bus sagging further at the edge. A tie run near its limit is a tie whose receiving voltage is already low, and compensation lowers it more for the same margin. That is the same trade the drop a series capacitor can reach found from the other side: the capacitor props up the voltage at a given load and makes a larger load possible, and at the larger load the voltage is lower than it was at the old limit. Whether the receiving bus can tolerate 148 V is a question the transfer limit does not ask.

The angle at which the limit is reached moves the same way. The limit always falls at an angle between the two EMFs equal to the loop’s impedance angle, so with 0.1 Ω machines the uncompensated tie reaches its limit at 82.9° between the EMFs and the fully compensated one at 76.0°. Compensation raises the limit and lowers the angle at which it is reached, which means the tie arrives at its edge with less angular travel to spare: a disturbance that swings the machines apart by a fixed number of degrees uses a larger share of what is left. That, rather than the steady-state limit, is usually what a transmission engineer is protecting when a compensated line’s loading is restricted, and it is the same physics as the terminal sag — the capacitor shortens the loop, and a shorter loop carries more power at a smaller angle and a lower terminal voltage. None of the three is a defect of the capacitor; they are what a shorter loop is.

The capacitor’s own resistance #

The sixty-degree essay’s third question was whether the capacitor’s losses abolish the peak. A capacitor of quality factor Q has a series resistance equal to its reactance over Q, and since its reactance is $kX$, its resistance grows with the compensation — raising the loop’s resistance exactly as the capacitor is lowering the loop’s reactance.

The effect is real and, for a power capacitor, negligible. At a Q of 10 the best limit is 0.848 of $V^2/4R$, at 40.8 per cent cancellation; at 30, 0.938 at 51.5 per cent; at 100, 0.978 at 55.0; and a Q of 222 keeps the peak within one per cent of the lossless capacitor’s. The reactance that is not free found a cancelling capacitor’s resistance eating into what cancellation buys on a feeder; on a transmission tie the capacitor’s resistance is compared with the line’s, and power capacitors, whose dissipation factors put their Q in the thousands at 50 Hz, are thousands of times better than they need to be. The peak survives intact, which leaves the machines’ reactance as the thing that moves it.

What the three numbers say together #

Read together, the figures give a designer three questions to ask of a tie before fitting a capacitor. First, how does each end’s source reactance compare with $\sqrt3/2$ of the line’s resistance: if it is larger, as it nearly always is, the sixty-degree optimum is out of reach and compensation raises the limit monotonically. Second, how large is the line’s reactance against the whole loop’s: that fraction bounds the gain, and the lossless estimate of it is good once the machines are long. Third, how low may the receiving terminal go: the capacitor raises the limit by letting the loop carry more current, and the receiving machine’s reactance turns that current into a sag the capacitor cannot correct.

None of the three is the resistance of the line, which was the earlier essay’s subject, and none is the capacitor’s own loss, which is negligible. The earlier result was right for its circuit and the circuit was unusual — a lossy line between two infinitely stiff sources. Add the machines that every real tie has, and the classic picture of series compensation as a straightforward gain in transfer capability, limited by the terminal voltage, returns.

Applied to the tie of the figures, the three questions take a minute. With 0.1 Ω behind each EMF, each machine’s reactance is more than twice $\sqrt3/2$ of the line’s 0.05 Ω, so the peak is out of reach and every per cent of compensation helps. The line is half of the loop’s reactance, so full cancellation can at most double the limit, and the solve gives 1.70. And the receiving terminal at the new limit is 148 V — sixty-four per cent of nominal — which is the number that decides whether the 194 kW the compensated tie can carry is a power anyone would let it carry.

How the numbers were obtained #

The tie is a netlist at 50 Hz: a sending EMF at 230 V and angle δ, a reactance $X_m$, the line’s resistance and inductance, the capacitor, a second reactance $X_m$ and the receiving EMF at 230 V and zero angle. The power absorbed by the receiving EMF is the real part of its voltage times the conjugate of the line current. The limit is found by golden-section search over δ on the solved netlist and compared with the closed form at every compensation; the capacitor’s resistance, when present, is in series with it in the netlist. The terminal voltage is the magnitude of the node between the line and the receiving machine’s reactance, at the angle of the limit.

What it leaves out #

Unequal ends. Both EMFs are 230 V. A receiving end held lower than the sending one changes the closed form: the limit becomes a function of the two voltages as well as the angle, and the sixty degrees is a statement about equal voltages that need not survive.

The machines’ dynamics. A machine’s EMF is not held constant through a swing; its reactance changes from subtransient to synchronous, and a limit computed with one reactance is the limit for disturbances of one duration. The steady-state limit here is the slowest of them.

Still open: unequal ends, a load in the middle, and the capacitor that resonates #

Unequal voltages at the two ends. With the receiving voltage below the sending, the received power at the best angle is the receiving voltage times the sending voltage less its own projection, over the impedance, and the impedance angle that maximises it is no longer sixty degrees. Whether the peak’s height — $V^2/4R$ for equal ends — depends on the receiving voltage at all is the question that decides whether a sagging receiving bus costs the tie its best compensation. The peak a sagging receiver keeps measures it.

A load tapped along the line. Power taken off between the ends changes the current in each section, and the capacitor’s position along the line then matters, as it did for the feeder in the drop essay. The limit becomes a surface over position and compensation.

The resonance the capacitor makes with the machines. The reactance cancelled, and the resonance it buys found a compensated feeder acquiring a series resonance that was not there before. With machines in the loop, a series capacitor and the machines’ inductance resonate below the system frequency, and a generator’s shaft has torsional modes in the same range. Whether the compensation that maximises the transfer limit puts that resonance near a mode is the question that, in practice, sets how much of a line is ever cancelled.