Synchronous Motor Performance and V-Curves

Take the terminal voltage as reference and write the current with its sign. A leading current has a positive angle:

The impedance drop:

The real part is negative because the leading current makes the reactive drop point backwards along \(\mathbf V\).

Subtract it — the motor sign convention:

The load angle is the argument of that phasor, and its negative imaginary part says \(\mathbf{E}_f\) lags \(\mathbf{V}\) — the signature of motoring:

Input power from the terminal quantities, and mechanical power by deducting the armature copper loss:

Sanity-check the excitation. The motor is running at 4886 V of internal e.m.f. against a 3300 V supply — 48 % over-excited. That is exactly what a leading power factor demands, and it is the whole basis of the machine's use as a power-factor corrector. A motor drawing lagging current would have shown \(E_f\) below 3300 V.

One quantity is fixed by the load and does not move. With \(R_a = 0\) the power-angle relation gives

Whatever the excitation, \(E_f\) and \(\sin\delta\) adjust so that their product stays at 839.8 V. Raise the field and the load angle falls; weaken it and the angle opens up.

So does the in-phase component of the current. Resolving \(\mathbf{I}_a = (\mathbf{V}-\mathbf{E}_f)/jX_s\) along \(\mathbf V\):

Only the quadrature component varies:

Working one point in full, at \(E_f = 1500\) V per phase:

Repeating for each excitation gives the V-curve:

The current falls, reaches a minimum and rises again — the V shape that names the curve. The load angle, by contrast, decreases monotonically throughout.

Locate the minimum exactly. The quadrature term vanishes when \(E_f\cos\delta = V\), and with \(E_f\sin\delta = 839.8\) already fixed,

Note the row that surprises people. At \(E_f = 1905\) V the excitation e.m.f. exactly equals the terminal voltage, and the power factor is still 0.974 lagging, not unity. Equality of magnitudes is not the unity-power-factor condition; equality of the projections, \(E_f\cos\delta = V\), is. On load the projection always falls short, so unity power factor demands over-excitation.

The power factor at each excitation comes straight from Problem 2, since \(\cos\phi = (P/3V)/I_a\) and the numerator is fixed:

A single-peaked curve, maximum at the same excitation where the V-curve is minimum. Plotted downwards it is the mirror image of the V, which is why it is called the inverted-V or power-factor curve.

The peak is "normal excitation". Below it the machine is under-excited and behaves like an inductive load; above it it is over-excited and behaves like a capacitive one. The terms describe the machine's reactive behaviour and have nothing to do with any rated field current.

Now change the load. The unity-power-factor excitation is the same expression with a new \(P\):

Reading the family. Three things happen together as the load rises:

• the whole V-curve lifts, because its minimum \(P/3V\) is proportional to the load;
• the minimum moves to the right, because more excitation is needed to hold unity power factor;
• the curve becomes flatter near its minimum, so the power factor is less sensitive to small field changes.

The locus of the minima — the line joining the unity-power-factor points — slopes upward and to the right across the family, and it is drawn on every manufacturer's V-curve chart.

A practical consequence. A field setting chosen to give unity power factor at 600 kW leaves the motor over-excited and taking leading current at 300 kW, and under-excited and taking lagging current at 900 kW. A fixed field is a fixed field; if unity power factor is required over a load range, the excitation must be regulated, and that regulator is what an automatic power-factor controller drives.

The armature current is in phase with the terminal voltage, which is what unity power factor means and what makes this the simplest case of all:

Subtract the impedance drop. With \(\mathbf{I}_a\) real, the resistive drop comes off the real part and the reactive drop appears entirely as an imaginary part:

Magnitude and angle:

The excitation e.m.f. is 4.8 % above the terminal voltage. Even at unity power factor the motor must be slightly over-excited, for the reason set out in Problem 2.

Mechanical power developed, after the armature copper loss:

Friction, windage and iron losses would come off this figure to give the shaft output; they are not given here.

What happens if the load rises with the field fixed. The excitation e.m.f. is set by the field current and stays at 242.06 V; the load angle must therefore open up to carry the extra power. As \(\delta\) grows, \(E_f\cos\delta\) falls below \(V\), and by the resolution of Problem 2 the quadrature current turns lagging. A motor set for unity power factor at one load will always be under-excited at any larger load, and its power factor will worsen in the lagging direction.

The power-angle characteristic peaks at 90°, where \(\sin\delta = 1\):

Speed is fixed, so torque follows by division. A 6-pole 50 Hz machine runs at 1000 rev/min:

The pull-out ratio. At 600 kW the running torque is

Because the speed never changes, the torque ratio and the power ratio are the same number — a convenience unique to the synchronous machine.

What pulling out means here. Load the motor beyond 1487.6 kW and \(\delta\) is driven past 90°. Beyond that angle a further advance of the load angle reduces the developed power, so the rotor cannot recover: it falls out of step, the current becomes large and oscillatory, and the machine stalls. There is no slower stable speed to run down to, as there would be in an induction motor.

The minimum excitation. The 600 kW load can be carried only if \(P_{max} \ge 600\) kW, so

This is exactly the constant \(E_f\sin\delta\) found in Problem 2 — and it must be, because at the limit \(\sin\delta = 1\). Weaken the field below 44 % of the per-phase terminal voltage and the machine drops out no matter how carefully it is handled.

Comparing three excitations at the same load:

Excitation buys stability margin as directly as it buys leading kVAR, and the same field rheostat delivers both.

Only the load angle can respond. The speed is fixed by the supply frequency and \(E_f\) is fixed by the untouched field, so the extra power must come from a larger \(\delta\):

The angle has opened from 23.79° to 37.23°. Physically the rotor has slipped back by 13.4 electrical degrees — about 4.5 mechanical degrees on a 6-pole machine — and then re-locked at the new position.

The current components:

Hence the new operating point:

The quadrature term has gone negative: \(E_f\cos\delta\) is now below \(V\), so the machine that was at unity power factor has become under-excited without the field being altered at all.

Restoring unity power factor needs the excitation of Problem 3's table at 900 kW:

At that excitation \(\delta\) settles back to 33.47° and the current falls from 160.5 A to 157.5 A.

Summarising the three states:

The stability cost of leaving the field alone. At \(E_f = 2082\) V the pull-out power is 1487.6 kW, so the margin has fallen from 2.48 at 600 kW to 1487.6/900 = 1.65. Raising the field to 2284 V restores it to 1632/900 = 1.81. Excitation therefore buys margin as well as power factor, which is why a large synchronous motor is normally left over-excited rather than trimmed to unity.