The peak a sagging receiver keeps

Sixty degrees, and the limit compensation stops raising found that a series capacitor raises the most power a lossy tie can pass between two sources only until the tie’s impedance angle is sixty degrees, and that the limit there is $V^2/4R$. The reactance the capacitor cannot reach then put machines at both ends and found the sixty-degree point moving out of reach, the limit rising at every step of compensation, and the receiving terminal sagging as it rose — 180 volts falling to 148 at the edge of a tie with 0.1 Ω machines.

Both essays held the two ends at the same voltage. Real ties rarely do: a long line’s receiving end sits below its sending end under load, a grid’s two halves run at different voltages, and the machine essay’s sagging terminal is itself a receiving end held low. Its open question was whether sixty degrees survives unequal ends, and whether the peak’s height — $V^2/4R$ when both ends are equal — depends on the receiving voltage at all. The answer to the first is no and to the second is also no, and the two together say something about what a compensated tie is.

A parabola in the cosine #

Keep the bare tie of the sixty-degree essay — 0.05 + j0.2 Ω at 50 Hz, a capacitor cancelling a fraction k of the reactance — and hold the receiving source at a voltage $V_r$ different from the sending $V_s$. With the line’s impedance written as $|Z|\angle\beta$ and the sending source leading by δ, the power the receiving source absorbs is

$$P_r = \frac{V_sV_r}{|Z|}\cos(\beta - \delta) - \frac{V_r^2}{|Z|}\cos\beta.$$

The second term is the receiving source’s own current through the line’s resistance, and it no longer has the sending voltage in it. The first is largest at $\delta = \beta$, as before, and with $|Z| = R/\cos\beta$ the limit is

$$P_{\max} = \frac{V_r\cos\beta}{R}\left(V_s - V_r\cos\beta\right).$$

As a function of $c = \cos\beta$ that is a downward parabola, zero at $c = 0$ and at $c = V_s/V_r$, with its top halfway between, at $c = Vₛ/2Vᵣ$. For equal ends the top is at $c = \tfrac12$, which is sixty degrees. For anything else it is not.

With the receiving source at 0.9 of the sending voltage, 207 V against 230, the limit’s top is at $\cos\beta = 0.5556$, an impedance angle of 56.25°, reached at 62.58 per cent cancellation rather than 56.7. The search on the solved netlist agrees with the closed form to a part in a hundred thousand at every compensation. And the top’s height is 264.50 kW — exactly what it was for equal ends.

The slider shows both properties together. A receiver at 0.8 of the sender peaks at 51.3° and 68.8 per cent cancellation; at 0.6, at 33.6° and 83.4 per cent; at 1.1, above the sender, at 63.0° and 51.0 per cent. Every peak is 264.5 kW.

Why the height does not move #

The parabola’s top is at $c = Vₛ/2Vᵣ$, and substituting it back gives

$$P_{\text{peak}} = \frac{V_r}{R}\cdot\frac{V_s}{2V_r}\cdot\frac{V_s}{2} = \frac{V_s^2}{4R},$$

with the receiving voltage cancelled out. That number is not new. It is the most power a source of voltage $V_s$ can deliver through a resistance R into any load whatever — the maximum power theorem, which the load that takes the most drew for a source and a resistor. At its best compensation the tie delivers what the sending source would deliver into a load matched to the line’s resistance, and the receiving source, whatever its voltage, is behaving as that load.

The reason the receiving voltage drops out is that the receiving source enters the limit twice, with opposite signs. It multiplies the useful term, because the power it absorbs is its own voltage times the current, and it sets the wasted term, because its own voltage drives a current back through the line’s resistance. At the top of the parabola the two exactly trade: a lower receiving voltage absorbs less per ampere and wastes less per ampere, and the compensation that reaches the top adjusts the current until the sending source sees the line’s resistance as its whole internal resistance and the receiver as a matched load. The sending source cannot tell, at that point, what voltage the receiving end is held at.

From a receiving voltage of half the sending upwards, the best compensated limit is $V_s^2/4R$ exactly. Below half it falls — to 253.9 kW at 0.4 and 222.2 at 0.3 — because the parabola’s top would need $\cos\beta > 1$, an angle below zero, which no amount of cancellation reaches: at full cancellation the line is pure resistance and the angle is zero, and that is as far as the capacitor can go. The uncompensated limit, by contrast, varies everywhere: 207.0 kW with the receiver at 1.1 of the sender, 194.4 at equal ends, 165.5 at 0.8, 131.6 at 0.6, 92.7 at 0.4.

So compensation’s relative gain depends heavily on the receiving voltage — 1.28 times with the receiver above the sender, 1.36 at equal ends, 1.60 at 0.8, 2.01 at 0.6 — while what it achieves at best does not. A sagging receiving end has a lower uncompensated limit and the same best one, and the capacitor makes up the difference.

The capacitor a sagging receiver needs #

The compensation that reaches the top is the one that makes $\cos\beta = Vₛ/2Vᵣ$,

$$k^* = 1 - \frac{R}{X}\tan\beta^*, \qquad \cos\beta^* = \frac{V_s}{2V_r},$$

and the sixty-degree essay’s $1 - \sqrt3R/X$ is the equal-ended case.

For the line of R/X 0.25 the best compensation is 45.5 per cent with the receiver at 1.2 of the sender, 56.7 at equal ends, 62.6 at 0.9, 68.8 at 0.8, 75.5 at 0.7 and 83.4 at 0.6, reaching full cancellation at exactly half the sending voltage on every line. A lower receiving voltage wants a smaller impedance angle: the receiving source’s own current through the resistance, the term with $V_r^2$ in it, is smaller relative to the useful term, so the line can afford to be more resistive before the loss overtakes the gain.

That has a practical edge the equal-ended analysis hides. The receiving voltage of a loaded tie is not a constant: it sags as the load rises, and it sags most near the limit, as the machine essay’s terminal voltages showed. A capacitor sized for the sixty-degree optimum at nominal voltage is undersized for the same tie at the edge of its capability, where the receiving end has dropped; the optimum it was sized for has moved to more compensation as the voltage fell. Because the peak is flat at its top, the cost of being off is small — the parabola is quadratic about its maximum — but the direction is consistent: a sagging receiver wants more capacitor, never less.

How flat the top is can be said exactly. Writing the limit about its peak,

$$P_{\max} = \frac{V_s^2}{4R}\left[1 - \left(\frac{\cos\beta - \cos\beta^*}{\cos\beta^*}\right)^2\right],$$

so a ten per cent error in the cosine of the impedance angle costs one per cent of the limit, a twenty per cent error four. For the equal-ended tie a cosine ten per cent high is an angle of 56.6° rather than 60, reached at 62.0 per cent compensation rather than 56.7; for a receiver at 0.8 of the sender it is the difference between 68.8 and 73.6 per cent. A capacitor sized for nominal voltage and run with a sagging receiver is therefore off by a few per cent of compensation and loses a fraction of a per cent of the limit, which is why the equal-ended optimum has served as a rule of thumb without anyone noticing that it moves.

Half of what is sent, at any receiving voltage #

The matched-load reading has a second consequence that the solve confirms to four decimal places: at the best compensation the receiving end takes exactly half of what the sending end puts in, for every receiving voltage tried — 0.5000 at 0.6, 0.8, 1 and 1.2 of the sending voltage. A matched load always takes half of what its source delivers, the other half heating the source’s resistance, and here the line’s resistance plays the source’s part. The sixty-degree essay found the efficiency at its peak to be one half for equal ends and called it an old theorem arriving by a new road; the road turns out not to depend on where the receiving end is held.

Away from the peak the efficiency at the limit depends on the receiving voltage. On the equal-ended tie it falls through one half as the compensation passes the peak, from 0.673 uncompensated to 0.485 at 60 per cent and 0.35 at 90; with the receiver at 0.6 of the sender it touches one half at the peak and rises on either side, to 0.664 uncompensated and 0.538 at 90 per cent. A tie run at its best compensation is, by construction, one whose line dissipates as much as it delivers at the limit, and that is a statement about operating at the limit rather than at the compensation: a tie loaded to a fraction of its limit is far more efficient than either number.

What a compensated tie is #

Put together, the three essays on the tie describe it more simply than any one of them did. A lossy line between two sources, compensated as well as it can be, is a source driving a matched load through the line’s resistance: its limit is $V_s^2/4R$, its efficiency there is one half, and neither depends on the receiving voltage. What depends on everything else is how much capacitor it takes to get there. Equal voltages need the line’s impedance angle at sixty degrees; a sagging receiver needs it lower; machines behind either end push the target out of reach; and below half the sending voltage the target is out of reach for any line.

This also explains why the lossless formula, $V^2/X$, fails as badly as it does. It is the limit of a line with no resistance, and with no resistance $V_s^2/4R$ is infinite: a lossless tie has no matched-load ceiling, so compensation appears to raise its limit without bound. Every real tie has a ceiling at $V_s^2/4R$, set by its resistance and its sending voltage alone. Compensation can bring the limit up to that ceiling and no further, and the receiving voltage, the line’s reactance and the machines’ reactances only decide whether the capacitor can bring it all the way.

Two roads to the same ceiling #

The parabola has a second reading that the capacitor hides. It is a function of $V_r\cos\beta$, not of either separately, so the top can be reached by changing the receiving voltage at a fixed angle as well as by changing the angle at a fixed voltage. The uncompensated tie, with its impedance angle fixed at 75.96°, reaches $V_s^2/4R$ when the receiving end is held at $V_s/2\cos\beta$ — 2.06 times the sending voltage for this line. A receiving end held that high would draw the same matched-load power through the bare line that a capacitor delivers at equal voltages.

No real receiving end is held at twice its sender, and the reading is not offered as a design. It says what the capacitor is doing. Cancelling reactance makes the cosine larger; raising the receiving voltage makes $V_r$ larger; the limit depends only on their product, and the ceiling is where that product is half the sending voltage. A receiving end above the sender is partway along the second road already, which is why it needs less capacitor — 51.0 per cent at 1.1 against 56.7 — and why its uncompensated limit, 207.0 kW, is already closer to the ceiling. A receiving end below the sender has gone backwards along it, and the capacitor has further to take it.

How far the ceiling is from a real line’s rating #

The ceiling is worth placing against what a line can actually carry, because for many lines it is academic. In per-unit terms, a transmission line with a resistance of 0.02 and a reactance of 0.4 between two stiff unit-voltage buses has an uncompensated limit of 2.37 and a matched-load ceiling of $1/4R$ = 12.5 — five times higher — reached only at 91 per cent compensation. Long before the limit approaches the ceiling, the conductors’ thermal rating, the terminal voltages and the system’s angular stability will have stopped the transfer. For such a line the lossless formula is not merely good enough; the matched-load ceiling never enters the design.

The ceiling matters where the line’s resistance is a substantial fraction of its reactance: distribution feeders, cables, short ties between nearby substations, and low-voltage interconnections like the 230 V tie of these essays, whose R/X of 0.25 puts the ceiling only 36 per cent above the uncompensated limit. There the capacitor can reach the ceiling with ordinary compensation, and every result above applies in full: the peak is $V_s^2/4R$, it moves with the receiving voltage, and at it half of what is sent is lost.

The sixty-degree essay’s figure, above, is the equal-ended member of the family the hero figure slides through. Its fall to zero at full cancellation is also a statement about equal ends: a purely resistive link between two sources of equal voltage cannot deliver power at any angle. With the receiver lower, a resistive link delivers power at zero angle — current flows from the higher voltage to the lower through the resistance — and at half the sending voltage the limit at full cancellation is the peak itself.

How the numbers were obtained #

The tie is the sixty-degree essay’s netlist with the receiving source’s magnitude set separately: a sending source at 230 V and angle δ, the line’s resistance and inductance, the capacitor, and a receiving source at $V_r$ and zero angle, solved at 50 Hz. The power absorbed by each source 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 solve and compared with the closed form at every compensation; the efficiency is the ratio of the two sources’ powers at the angle of the limit.

What it leaves out #

Receiving voltage that depends on the power. The receiving source here holds its voltage whatever it absorbs. A real receiving bus sags as it takes more, so its voltage is a function of the power drawn, and the limit is then the point where that function and the parabola meet — the nose of a power–voltage curve, which the load that has two voltages or none drew for a load rather than a source.

Reactive power at the receiving end. Holding a bus at a voltage takes reactive power, which the receiving source supplies freely here. A receiving end that must import its reactive power through the same line changes what the line’s current is spent on, and the matched-load ceiling then applies to the real part of a current that is partly reactive.

Still open: the load in the middle, the nose of a voltage curve, and the resonance #

A load tapped along the line. Power taken off between the ends splits the line into two sections carrying different currents, and the capacitor’s position along the line then decides which section it compensates. The matched-load ceiling should apply section by section, and whether the best position for the capacitor is where the larger current flows or where the voltage is lowest is a question the tie, with its single section, cannot ask.

The receiving voltage as a curve rather than a number. A receiving bus that sags with its load has a voltage that moves along the parabola as the power rises, and the optimum compensation moves with it. Whether there is a compensation that keeps the tie at its matched-load ceiling as the load changes — a capacitor that follows the voltage — is the dynamic version of this essay’s static answer.

The resonance a large capacitor makes. A sagging receiver wants more cancellation, and more cancellation puts the capacitor’s resonance with the line’s and the machines’ inductance closer to the system frequency. Whether the compensation a low receiving voltage calls for is one a system with turbine-generators can tolerate is, in practice, the question that decides how far towards the matched-load ceiling a real tie is ever taken.