Synchronous Motor Drives Questions and Answers

Answer: A

Correct answer: 60 per cent of its apparent power

With a power factor of 0.8, \( \cos\phi = 0.8 \) and so

The reactive power is \( S\sin\phi \), that is 60 per cent of the apparent power, and because the current leads, that reactive power is delivered to the supply. An over-excited synchronous motor is therefore a rotating capacitor as well as a mechanical drive.

Answer: B

Correct answer: 484 kVAr

The apparent power is

and with \( \sin\phi = \sqrt{1 - 0.81} = 0.436 \),

Because the power factor is leading, that 484 kVAr is supplied to the network, offsetting the lagging demand of induction motors elsewhere in the plant.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: the speed is rigidly fixed by the supply frequency and pole number whatever the load, and over-excitation makes the machine supply reactive power so that its current leads. Statement 3 is false and is the machine’s chief practical drawback - at standstill the stator field sweeps past the rotor too quickly for any net torque, so a damper winding or a variable-frequency start is required.

Answer: D

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: the torque angle grows with load, and for a cylindrical rotor machine the pull-out point is at 90 degrees where \( \sin\delta \) is greatest. Statement 3 confuses the synchronous machine with the induction machine. The speed does not fall at all - it cannot, since the rotor is locked to the stator field; what changes is the angle at which it is locked.

Answer: A

Correct answer: 1-R, 2-Q, 3-P

A wound field machine carries a direct current winding on the rotor, fed through slip rings or a brushless exciter. A permanent magnet machine has magnets instead, so no excitation power and no field losses, but no controllable power factor either. A synchronous reluctance machine has neither: its rotor is simply shaped so that the reluctance differs along two axes, and torque arises from the rotor’s tendency to align with the field.

Answer: B

Correct answer: 1-R, 2-Q, 3-P

Load commutated inverters suit large high-speed machines - compressors, pumps, wind tunnel drives - where the back emf is ample for natural commutation. Cycloconverters are limited to about a third of the supply frequency and therefore to very low speeds, which is exactly what a large mill or mine hoist wants, at very high torque. Brushless dc drives dominate the small end: servos, fans, appliances and tools.

Answer: C

Correct answer: 1, 3, 2

The V-curve has its minimum at unity power factor, so the armature current is least there. Moving away from that point in either direction adds a reactive component and increases the current, and the further the excursion the larger the current. Slightly over-excited therefore draws more than unity power factor, and heavily over-excited more again. The same order would apply on the under-excited side.

Answer: D

The reactive power it supplies offsets what the induction machines and transformers around it draw, raising the overall power factor. That is exactly the mechanism, and it is why large synchronous drives are often run over-excited.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is the mechanism. In a self-controlled drive the inverter is switched according to the measured position of the rotor, so the stator field is always placed at the correct angle relative to where the rotor actually is. There is no independent frequency for the rotor to fall out of step with - the machine defines the frequency itself, exactly as a dc machine’s commutator does mechanically. That is why such a drive is often called an electronically commutated machine.

Answer: B

The angle is indeed smaller at light load, but the speed is exactly synchronous whatever the loading. Confusing the angle with a change of speed is the standard trap in comparing the synchronous machine with the induction one.