Synchronous Motors Questions and Answers

Answer: A

Correct answer: cheaper, simpler and self-starting

The cage machine has no field winding, no exciter, no slip rings and no starting sequence, so it costs less to buy and less to maintain, and it starts by being switched on. That combination is why the great majority of industrial motors are induction machines despite the synchronous machine’s advantages.

Answer: B

Correct answer: the efficiency gain and the reactive support both pay at that size

At several megawatts a point or two of efficiency is a large sum each year, and a machine that supplies rather than absorbs reactive power can improve the whole plant’s power factor. Both benefits scale with rating while the extra cost of the excitation system does not, so the balance tips at large sizes.

Answer: C

Correct answer: each start is a lengthy and thermally severe operation

Starting means accelerating on the damper winding, which is not designed for continuous duty, and then applying the field at the right moment. The sequence takes time and heats the damper heavily, so a machine started many times an hour would overheat. Loads with frequent starts are given induction machines.

Answer: D

Correct answer: 1.25 per unit

\[ P_{max} = \frac{EV}{X_s} = \frac{1.5 \times 1.0}{1.2} = 1.25\ \mathrm{per\ unit}, \] reached at a load angle of ninety degrees. In practice the machine is never loaded near that point, since the synchronising stiffness falls to zero there and any disturbance would pull it out of step.

Answer: A

Correct answer: 53 degrees

\[ 1.0 = 1.25\sin\delta \quad\Rightarrow\quad \sin\delta = 0.8 \quad\Rightarrow\quad \delta = 53^\circ. \] Eighty per cent of the pull-out power is being delivered at 53 degrees rather than at 72, because the relation is sinusoidal - and the synchronising stiffness has already fallen to \( \cos 53^\circ \), or 60 per cent of its maximum.

Answer: B

Correct answer: 585 kVA

Two steps: the input power exceeds the output by the efficiency, and the apparent power exceeds the input power by the power factor. \[ \frac{500}{0.95} = 526\ \mathrm{kW}, \qquad \frac{526}{0.9} = 585\ \mathrm{kVA}. \] Dividing by only one of the two gives 526 or 555, both plausible-looking wrong answers.

Answer: C

Correct answer: 255 kVAr

At 0.9 power factor the angle is 25.8 degrees, so \[ Q = S\sin\phi = 585 \times 0.436 = 255\ \mathrm{kVAr}. \] Because the power factor is leading, this reactive power is supplied to the network rather than drawn from it - the machine is acting as a capacitor of that rating while also doing its mechanical work.

Answer: D

Correct answer: 105 kW

\[ \omega = \frac{2\pi \times 500}{60} = 52.4\ \mathrm{rad/s}, \qquad P = T\omega = 2000 \times 52.4 = 105\ \mathrm{kW}. \] Multiplying the torque by the speed in revolutions per minute without converting gives a figure ten times too large, which is the usual slip in this calculation.

Answer: A

Correct answer: 6.7 degrees

Electrical and mechanical angles differ by the number of pole pairs: \[ \theta_{mech} = \frac{\theta_{elec}}{P/2} = \frac{20}{3} = 6.7\ \mathrm{degrees}. \] The load angle is always quoted in electrical degrees, and the physical displacement of the rotor is correspondingly smaller in a machine with many poles.

Answer: B

The angle and the current both take up the load, but the speed is rigidly locked to the supply frequency and does not move at all. Confusing the angle with a change of speed is the standard error in comparing this machine with an induction one.