Power Angle and Power-Factor Correction

Per-phase voltage and synchronous speed first:

The amplitude of the power-angle curve:

This is the whole characteristic in one number: \(P = 3353.2\sin\delta\) kilowatts.

Invert it for the working angle:

On an 8-pole machine that is \(20.97/4 = 5.24\) mechanical degrees — the rotor sits about five degrees behind where it would be on no load, and a strobe light would show exactly that shift.

Torque, from the fixed speed:

The armature current, resolved about \(\mathbf V\):

The positive quadrature term marks the machine as over-excited, and it is already supplying \(3\times3810.5\times19.9 = 227\) kVAR of leading reactive power to the system while doing its mechanical work.

The stability margin follows without further work:

Comfortable. Below about 1.5 a drive is considered fragile, because a supply voltage dip reduces \(P_{max}\) in proportion to \(V\) and can wipe the margin out in a cycle.

The real power does not change, and that is the pivot of every correction calculation. Only the reactive component moves:

The difference is the condenser's job:

Rating the machine. A synchronous condenser carries no mechanical load, but it still has friction, windage, iron loss and field loss to feed — taken here as 25 kW. Its own kVA is therefore

The losses raise the rating by 0.05 %, so an 830 kVA machine is specified. Their real cost is the 25 kW of energy they consume continuously, which Problem 5 prices.

The current before and after:

What the reduction buys. Losses in the supply cables, transformer windings and switchgear go as \(I^2\), so they fall by

Nearly half the distribution copper loss, removed without touching a single load. The same 26 % margin also releases transformer and cable capacity for future load, which is often worth more than the energy saving.

A caution on the target. Correcting all the way to unity looks tempting but is rarely done. From 0.95 to 1.00 costs a further 394 kVAR — almost half as much again as the whole job so far — and saves only the last 10 % of current. Tariffs are normally written to reward 0.95, and that is where the economics stop.

The motor's electrical input, not its shaft output, is what the supply sees:

Forgetting the efficiency here is the single commonest slip in this type of question, and it puts every later figure out by 8 %.

Existing load, in the two currencies:

The new total real power, which the motor increases whether we like it or not:

The reactive power the works is allowed to draw at the target factor:

The motor must make up the difference, and the sign says which way:

Negative, meaning the motor must supply 265.6 kVAR of leading reactive power rather than draw lagging reactive power. It must therefore run over-excited.

Combine the motor's own two components:

Checking the whole works, which is worth doing because the sign convention is easy to lose:

What has been avoided. Had the new load been driven by an induction motor at, say, 0.85 lagging, it would have added 135 kVAR instead of removing 266, and a separate 800 kVAR correction plant would have been needed to reach 0.95. The synchronous motor does the mechanical work and the correction with one machine and one set of losses.

Only one quantity in the power-angle relation changes, so the comparison is a single division repeated:

The per-unit base, for context:

Tabulating the three designs:

The limiting reactance, found by setting \(P_{max} = P\):

Read the middle rows carefully. Between 15 Ω and 40 Ω the reactance has not quite tripled, but the load angle has more than tripled and the margin has collapsed from 2.79 to 1.05. The reason is the sine: near 20° the characteristic is almost linear, so a small change in \(P_{max}\) produces a proportional change in \(\delta\); near 70° the curve is flattening and the angle runs away for very little further loading. A machine at 72.6° would be thrown out of step by the first voltage dip.

Where the reactance comes from. Problem 5 of Set 35 showed that most of \(X_s\) is armature reaction, which is inversely proportional to the air-gap reluctance. A machine built with a long air gap has a small \(X_s\), a large short-circuit ratio, a large stability margin and an expensive field winding; a machine built with a short gap is cheap and weak. The three rows above are the same trade-off priced in stability rather than in copper.

The remedy when \(X_s\) is fixed is more excitation, since \(P_{max} \propto E_f\). To restore the 40 Ω machine to a margin of 2.0 would need

That is 2.2 times the terminal voltage, far beyond what the field winding and the iron will stand. Stability lost to a large \(X_s\) cannot usually be bought back with excitation.

Each phase of a delta bank carries a third of the total kVAR at the full line voltage:

In star each phase carries the same kVAR but only \(V_L/\sqrt3\), so three times the capacitance is needed:

Capacitance is the expensive quantity, so medium-voltage banks are connected in delta almost without exception.

The currents:

Running energy, over a year of continuous service:

A factor of 60. On energy alone the capacitor bank wins outright, and that is why it wins in almost every industrial installation.

The full comparison, since energy is not the only consideration:

The row that decides most real cases is the voltage one. A capacitor's output falls as the square of the voltage, so exactly when the system is sagging and needs reactive support most, the bank delivers least. A synchronous condenser behaves the opposite way: its field can be forced, and its output can be temporarily raised above rating. That is why the machine survives in transmission service long after it disappeared from factories.

The margin fixes the amplitude of the power-angle curve:

The load angle comes free with a margin of exactly 2:

A margin of 2 always means a load angle of 30°, whatever the machine. It is worth carrying that pairing in mind as a landmark.

The current, resolved as before:

Notice what has happened. The excitation e.m.f. is 3936 V against a terminal voltage of 3810 V — the machine is over-excited — and yet the power factor is lagging. The reason is the projection: at 30°, \(E_f\cos\delta = 3409\) V has fallen below \(V\), and it is that projection, not \(E_f\) itself, that decides the sign of the reactive current.

What unity power factor would have demanded instead:

So unity power factor here is not in conflict with the margin at all — it exceeds it. The two demands pull in the same direction, both calling for more field.

The comparison, laid out:

The lesson for the works. At 1200 kW this machine supplied 227 kVAR of leading correction (Problem 1). Re-rated to 1500 kW at the same field, the leading current falls to 8.3 A and the correction to 95 kVAR; and if the field were trimmed to the bare minimum for a margin of 2.0 it would absorb 26.8 A, or 306 kVAR lagging, instead. A synchronous motor's value as a corrector therefore shrinks as its mechanical load grows, and a plant that has come to depend on it must either raise the excitation or find its kilovars elsewhere.