Synchronous Motor Drives Questions and Answers

Answer: A

Correct answer: exactly the synchronous speed of the supply

The rotor field is locked to the rotating stator field, so the machine turns at

and at no other speed, whatever the load. Increasing the load does not slow it; it merely increases the torque angle by which the rotor lags the stator field. This constancy is the machine’s defining property and the reason it is used where speed must be exact. Running a few per cent below synchronous speed describes the induction motor.

Answer: B

Load changes alter the torque angle rather than the speed, and excitation alters the power factor. Only frequency and pole number appear in the synchronous speed relation, which is why variable-speed operation needs a converter.

Answer: C

Correct answer: the rotor field and the stator field

The torque angle \( \delta \) is the electrical angle by which the rotor field lags the rotating stator field. It is the machine’s load indicator: at no load the two fields are almost aligned and \( \delta \) is near zero, and it grows as load is applied, since power is transferred in proportion to \( \sin\delta \). It is not an angle between voltages or currents, though the excitation emf phasor does lag the terminal voltage by the same angle in the motoring case.

Answer: D

Correct answer: \( \sin\delta \)

For a cylindrical rotor machine with negligible armature resistance,

where \( E \) is the excitation emf, \( V \) the terminal voltage and \( X_s \) the synchronous reactance. The power therefore rises with \( \delta \) to a maximum at 90 degrees and falls thereafter. A salient pole machine adds a reluctance term in \( \sin 2\delta \), which shifts its maximum below 90 degrees, but the cylindrical rotor machine has only the fundamental term.

Answer: A

Correct answer: 90 degrees

Since \( P = (EV/X_s)\sin\delta \), the power and hence the torque is greatest where \( \sin\delta = 1 \), that is at \( \delta = 90 \) electrical degrees, giving

Beyond that angle the developed torque falls while the load torque does not, so the rotor cannot recover and slips poles. Steady operation is therefore confined to angles well below 90 degrees, typically 20 to 30 at full load, to leave a margin against transient disturbances.

Answer: B

Difference between the direct and quadrature reactances gives a reluctance torque that peaks at forty-five degrees. It shifts the pull-out point below ninety degrees and lets the machine develop some torque even without excitation.

Answer: C

Once the torque angle exceeds the pull-out value the rotor slips poles, and the torque then alternates in sign and the machine draws heavy pulsating current. Protection trips it, and the load must be reduced before it can be resynchronised.

Answer: D

Correct answer: at standstill the stator field sweeps past the rotor too quickly

Switched on to a fixed supply at rest, the stator field rotates at synchronous speed past a stationary rotor. In the first half-cycle the torque acts one way and in the next the other, and the rotor’s inertia is far too great for it to follow, so the average torque over a cycle is zero and the machine simply hums. Starting requires either a damper winding to bring it up as an induction motor, or a converter whose frequency is raised gradually from zero.

Answer: A

Bars in the pole faces act as a cage rotor and bring the machine up to near synchronous speed, at which point the field is applied and the rotor pulls into step. The field must be short-circuited or shorted through a resistance during that run-up.

Answer: B

Correct answer: raising the frequency slowly from zero

With a variable-frequency converter the synchronous speed can be started at zero and increased gradually, so the rotor field never has to catch a field that is already rotating quickly - it accelerates in step from rest. The voltage is raised with the frequency to hold the flux constant. This is the standard start for a converter-fed drive and removes the need for a damper winding to act as a starting cage.