Power, and the part that does no work

The headroom that is the line's own charge

A line of 50 + j100 ohms delivers at most 0.6180 of a matched resistive line's power to a unity-power-factor load, with 0.5878 of the source voltage left. Give the line its own shunt capacitance — a π, half at each end, B|Z| = 0.4 in all — and the nose moves to 0.7025 at 0.6895: 13.7 per cent more power and 17.3 per cent more voltage. It is headroom, but not the headroom the far end advertises, which with no load rises 21.1 per cent. The capacitance turns the source and line into a Thevenin equivalent with more voltage behind more impedance, and it moves a voltage threshold's meaning in opposite directions depending on whether it is referred to the source or to the unloaded far end.

15 min read 6 figures Every model has an edge Two routes to a number The parasitic is the component

Assumes: The far end that rises · The current that does no work

The load that has two voltages or none found the limit every line carries. A load that takes a fixed power draws more current as its voltage falls, and a line with impedance in it delivers every power below a maximum at two voltages and every power above it at none. On fifty ohms and a hundred of reactance a unity-power-factor load reaches that maximum — the nose — at 0.6180 of the power a matched resistive line would deliver, with 0.5878 of the source’s voltage left at the far end. A load leading by sixty degrees reaches its nose with the far end above the source, which is why a far end that reads high is not a far end with margin.

That line was two elements, a resistance and an inductance in series. A real line has a third: the capacitance between its conductors, distributed along its length, which charges and discharges with every cycle and draws a current of its own whether or not anything is connected at the far end. The same essay named it as the first thing left out. The far end that rises found a far end rising above its source because a leading load and a line’s reactance resonate partially; a line’s own capacitance is a leading load that is always there.

This essay puts the capacitance in and asks what it does to the nose. It raises it — that much is expected. The measurement is how much, for which loads, and what it does to the voltage a margin is usually judged by.

A line with its own charge

The line here is the one the nose was drawn on: fifty ohms of resistance and a hundred of reactance at the frequency of interest. Its capacitance is lumped as a π — half at the sending end, half at the receiving end — which is the standard short-line model, and its size is stated as $B|Z|$ , the total shunt susceptance times the magnitude of the series impedance. That product is the line’s charging current as a fraction of the current its own impedance would pass with the full voltage across it, and it has no dimensions; a long overhead line or a cable run has a value of a few tenths.

The sending end’s half sits directly across the source. A source that holds its voltage whatever it supplies does not notice it, so it changes nothing the load sees. The receiving end’s half sits across the load, and it matters. Seen from the load, the source, the series impedance and that capacitor are a Thevenin equivalent:

V_{th} = \frac{V}{1 + jBZ/2}, \qquad Z_{th} = \frac{Z}{1 + jBZ/2}.

For an inductive line the denominator’s magnitude is below one, so both the equivalent’s voltage and its impedance rise. A load connected to the far end is connected to a stiffer-looking voltage behind a larger impedance, and whether that helps depends on which of the two wins at the nose.

The noses, with and without the charge

Fig. 1 Four load angles on a line of 50 Ω + j100 Ω with no capacitance. A load lagging by 30° reaches 0.4222 of a matched resistive line’s power with 0.5221 of the voltage left; at unity power factor, 0.6180 at 0.5878; leading by 30°, 0.8240 at 0.7293; leading by 60°, 0.9960 at 1.0553.

The figure without capacitance is the starting point, and every nose on it was found by searching the solved network and held to the closed form. The unity-power-factor nose is at 0.6180 of a matched line’s power and 0.5878 of the source voltage.

Fig. 2 A unity-power-factor load on the same line carrying its own capacitance as a π, at B|Z| of 0, 0.1, 0.2 and 0.4. The nose moves from 0.6180 of a matched resistive line’s power at 0.5878 of the source voltage to 0.6376 at 0.6108, 0.6582 at 0.6353 and 0.7025 at 0.6895. With no load the far end reads 1.0465, 1.0969 and 1.2107 of the source. Each nose is searched on the solved network and agrees with the nose of the Thevenin equivalent.

With the capacitance in, the same search on the network — the load swept in size from open circuit to short circuit, the most power found by golden section on its magnitude — finds the nose moving up and to the right. At $B|Z| = 0.1$ it is 0.6376 of a matched line’s power at 0.6108 of the source voltage; at 0.2, 0.6582 at 0.6353; at 0.4, 0.7025 at 0.6895. The figure checks every one against the nose of the Thevenin equivalent to a part in a hundred million, and checks that a far end with almost no load reads the equivalent’s own voltage.

That unloaded reading is the third number worth having beside each nose. With no load the far end is at 1.0465 of the source at $B|Z| = 0.1$ , 1.0969 at 0.2, and 1.2107 at 0.4. The line’s own charging current, flowing through its own inductance, lifts the far end above the source before anything is connected — the rise is named after Ferranti, who met it on a cable, and it is the unloaded end of the same resonance the far end that rises found for a leading load.

The headroom the far end advertises

Set the three rises side by side at $B|Z| = 0.4$ . The unloaded far end is 21.1 per cent above the line without capacitance. The nose voltage is 17.3 per cent higher. The nose power is 13.7 per cent higher.

The line's own capacitance raises the nose power by 1.137 at B|Z| = 0.4 — and its unloaded far end to 1.211 of the source. Closed form, held to the solved network at three capacitances. For a line of 50 Ω + j100 Ω carrying its own capacitance as a π, against B|Z|: the most power a unity-power-factor load can draw and the voltage left at it, each as a ratio to the same line without capacitance; the same for a load 30° lagging; and the unloaded far end's voltage. At B|Z| = 0.2: nose power ×1.065, nose voltage ×1.081, unloaded far end 1.0969. At 0.4: ×1.137, ×1.173, 1.2107. At 0.8: ×1.299, ×1.396, 1.5000. For the lagging load the nose power ratio at 0.4 is ×1.176. Every ratio is the Thevenin equivalent's |1 + jBZ/2| entering the power and the voltage. — Fig. 3 The nose’s gains against the line’s own capacitance, each as a ratio to the same line without it. At B|Z| = 0.2: nose power ×1.065, nose voltage ×1.081, unloaded far end 1.0969. At 0.4: ×1.137, ×1.173, 1.2107. At 0.8: ×1.299, ×1.396, 1.5000. A load lagging by 30° gains ×1.176 in nose power at 0.4.

The three are not in proportion, and the gap between them grows with the capacitance. At $B|Z| = 0.8$ — a long line — the unloaded far end is at 1.5000 of the source, the nose voltage 39.6 per cent higher than without capacitance, and the nose power only 29.9 per cent higher.

The reason is in the equivalent. Its voltage rises as $1/|1 + jBZ/2|$ , which is exactly the unloaded far end’s rise. Its impedance rises by the same factor, and the most power a load can draw from an equivalent goes as the square of its voltage over its impedance — so the magnitudes alone would give the power the same factor as the voltage. What they leave out is the angle. The capacitance also rotates the equivalent’s impedance: dividing by $1 + jBZ/2$ , whose angle is positive, makes $Z_{th}$ less inductive than $Z$ , and a less inductive source impedance gives a unity-power-factor load a smaller nose relative to its magnitude. The rotation takes back part of what the voltage gave.

So the capacitance is real headroom, and it is less headroom than the far end appears to have. A designer who reads a twenty-one-per-cent rise at the unloaded far end and credits the line with twenty-one per cent more capacity has credited it with half as much again as it has.

The margin, measured against the far end

The more consequential error is about the voltage rather than the power.

The load that has two voltages or none found that a far-end voltage says little about how near a load is to its nose, because at nine tenths of the nose power the far end reads anywhere from 0.68 to 1.17 of the source depending on the load’s angle. A common response is to judge the margin not against the source but against the far end’s own voltage with no load: a far end that has fallen to some fraction of its unloaded value is near the nose.

With the line’s own capacitance in, both references move, and they move in opposite directions. Without capacitance the unity-power-factor nose is at 0.5878 of the source, and the unloaded far end is the source, so the two readings agree. With capacitance the nose voltage rises against the source — 0.6353 at $B|Z| = 0.2$ , 0.6895 at 0.4 — while against the unloaded far end it falls: $0.6353/1.0969 = 0.5792$ at 0.2, $0.6895/1.2107 = 0.5695$ at 0.4. The nose voltage rises, and the unloaded voltage rises faster.

Put a threshold at 59 per cent, set on a line without charge, where it sits a hair above the unity-power-factor nose by either reference, and move it to the line with $B|Z| = 0.4$ . Referred to the source, the threshold is now below the nose voltage of 0.6895: the far end passes the nose, where the load collapses, while it still reads above 59 per cent, and a relay set that way never acts in time. Referred to the unloaded far end, the threshold is now above the nose’s 57 per cent: the relay acts well before the nose, at a power further below the limit than it was set for, and on a line whose capacity has in fact increased it trips earlier than it did on the line without charge.

One reference fails dangerously and the other fails conservatively, and neither is the margin. The setting did not change; the line’s own charge moved both of the things it could have been set against, in opposite directions. The quantity that stays attached to the nose is the power, which is why the load that has two voltages or none measured margins as fractions of the nose power rather than of any voltage.

The same measurement at other load angles says the direction holds and the size varies, which is the next thing to draw.

A lagging load gains most

Fig. 4 A load lagging by 30° on the same line at B|Z| of 0, 0.1, 0.2 and 0.4. The nose moves from 0.4222 of a matched resistive line’s power at 0.5221 of the source voltage to 0.4389 at 0.5445, 0.4567 at 0.5686 and 0.4964 at 0.6228.

A load lagging by thirty degrees, which is what an induction motor or a transformer’s magnetising current looks like, starts from a lower nose — 0.4222 of a matched line’s power at 0.5221 of the voltage — and gains more from the line’s capacitance: at $B|Z| = 0.4$ it reaches 0.4964 at 0.6228, a power 17.6 per cent higher. That is more than the unity-power-factor load’s 13.7.

Fig. 5 A load leading by 30° on the same line. The nose moves from 0.8240 of a matched resistive line’s power at 0.7293 of the source voltage to 0.8414 at 0.7539, 0.8590 at 0.7799 and 0.8944 at 0.8360.

A load leading by thirty degrees starts high — 0.8240 at 0.7293 — and gains least: 0.8944 at 0.8360 at $B|Z| = 0.4$ , 8.5 per cent. Its own leading current already supplies much of what the line’s inductance wants, and a charging current that supplies more has less left to cancel.

The ordering is the one reactive compensation would predict, and it is worth saying why it appears here without anyone having chosen to compensate anything. The capacitance at the far end supplies a reactive current that flows back through the line’s inductance, and that current’s effect on a load depends on whether the load was already drawing reactive current in the same sense. The current that does no work described reactive power as the part of a load’s demand that is carried by the line and does nothing; a line’s own capacitance delivers some of it at the far end, for free, to lagging loads, and adds to the surplus of leading ones. The line is partially compensating itself, and the compensation is fixed by its geometry rather than by the load.

A leading load gains voltage and no power

The ordering has an end point, and the load that reaches it is the one the earlier essay used as its warning: a load leading by sixty degrees, whose nose on this line already sat above the source’s voltage.

Fig. 6 A load leading by 60° on the same line. At B|Z| of 0, 0.1, 0.2 and 0.4 the nose is at 0.9960, 0.9986, 0.9999 and 0.9978 of a matched resistive line’s power, with 1.0553, 1.0809, 1.1074 and 1.1622 of the source voltage left.

Its nose power hardly moves. Without capacitance it is 0.9960 of a matched line’s power; at $B|Z| = 0.1$ , 0.9986; at 0.2, 0.9999; at 0.4, 0.9978 — up by four parts in a thousand and then back down. Its nose voltage moves a great deal: 1.0553 of the source without capacitance, 1.1622 at $B|Z| = 0.4$ . The line’s charge gives this load no more power at all, and sixteen per cent more voltage at the point where it collapses.

That is the whole difficulty of judging a heavily compensated line by its voltage, arriving from the side of the line rather than the load. A load that already leads has had its line’s reactance cancelled by its own current; the line’s capacitance cancels more than is there to cancel, and the surplus appears as voltage at the far end without adding any capacity to deliver power through the resistance that remains. A far end at 1.16 of the source, on the point of collapse, is a far end that every instrument on it will describe as healthy.

The power curve’s small peak is worth a sentence because it is real rather than rounding. The nose power of this load is greatest at a charge near $B|Z| = 0.2$ , where the line’s charging current and the load’s leading current together have cancelled the line’s reactance as nearly as they can; more charge overshoots. The figure checks each of the four values against the Thevenin equivalent’s closed form to a part in a hundred million, so the rise and fall of four parts in a thousand is well inside what the arithmetic resolves.

The charging current, in amperes

The dimensionless numbers hide how large the charging current is beside the load’s, and the comparison is short. At $B|Z| = 0.4$ the far-end half of the capacitance is $0.2/|Z|$ of susceptance, and at the unity-power-factor nose it has 0.6895 of the source voltage across it, so it draws 0.138 of the source voltage over $|Z|$ . The load at the same nose takes 0.7025 of $V^2/4R$ at 0.6895 of the voltage, which for this line’s 50 ohms and 111.8 of impedance magnitude is a current of 0.570 of $V/|Z|$ . The charging current is a quarter of the load current at the nose, flowing at right angles to it.

A quarter of the current, at right angles, is not small, and it is the right size to explain everything above. It is large enough to lift the unloaded far end by a fifth, because with no load it is the only current and it flows through all of the line’s reactance. It is small enough that at the nose the load’s own current dominates the line’s drop, which is why the nose power moves by less than the unloaded voltage. And its direction — leading, at right angles — is why it helps a lagging load’s reactive current most and a leading load’s not at all.

What the measurement assumes

A lumped π. A real line’s capacitance is distributed, and a π with half at each end is exact only while the line is short against a wavelength. For the numbers here that restriction is loose — $B|Z|$ of a few tenths is what a line of a few hundred kilometres of overhead conductor carries at mains frequency, still a small fraction of a wavelength — but the distributed line’s Thevenin equivalent has hyperbolic functions where the π has a first-order expansion, and at $B|Z|$ near one the two differ.

A stiff source. The sending end’s capacitance was dropped because the source holds its voltage. A source with impedance of its own — a generator’s reactance, a transformer’s leakage — puts that capacitance back into the equivalent, and the headroom depends on both.

A load that takes a fixed power at a fixed angle. That is the load the nose is defined for, and it is the load that makes a nose exist at all. A real load is part impedance, part current, part power, and its own dynamics decide whether the nose is ever reached — which is what a regulator between the line and the load does to the question, and the next essay on this subject takes up.

Nothing marched. Every number is a steady state at the operating frequency. The charging current also means a line energised with no load at the far end sits above its source from the first cycle, and what that does during switching is a transient this solve does not contain.

Still open: the regulator that pushes, the distributed line, and the capacitance a designer adds

A regulator between the line and the load. A tap changer or an automatic voltage regulator raises the load’s voltage when it sags, and a load that restores its power when its voltage is restored turns that correction against the line. Below the nose raising the voltage helps; past it, the same action lowers the voltage. The regulator that pushes past the nose marches the two together and finds how long the collapse takes.

The distributed line. Replacing the π with the exact hyperbolic equivalent of a line of the same total impedance and admittance would say at what $B|Z|$ the π’s headroom is wrong by a per cent, and in which direction — which is the same question a ladder is not a line asks of a chain of lumped inductors and capacitors, asked of a single one.

Capacitance put there on purpose. A shunt capacitor bank at the far end is the same element with a value chosen rather than inherited. The measurement here says a unity-power-factor load’s nose rises by less than the far end’s no-load voltage; whether a bank sized for a stated nose power leaves the unloaded far end at a voltage the equipment on it can stand is the design question the two numbers together answer, and it is not measured here.

Part 3 on voltage regulation

One argument about Voltage regulation, and one of 4 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it: