Synchronous Motors Questions and Answers

Answer: A

Correct answer: over-excited for that load

A leading power factor means the machine is supplying reactive power, which happens when its excitation exceeds the value giving unity power factor at that load. Under-excited it would draw a lagging current instead, and at normal excitation the power factor would be unity.

Answer: B

Neither is required by a cage induction machine, and together they are what confine the synchronous motor to large or special installations. Its speed is exactly constant and its power factor is adjustable, both of which are advantages.

Answer: C

Correct answer: its efficiency and power factor are better at that rating

A large fan runs continuously at a fixed speed, which is exactly the duty a synchronous machine suits: the efficiency advantage is earned every hour, and the leading power factor supports the rest of the plant. Its poor starting behaviour hardly matters for a machine started once and left running.

Answer: B

For a three phase motor the input power is \(P = \sqrt{3}\,V_L I_L \cos\phi\). With a line voltage of 3300 V, 800 kW and a power factor of 0.707:

\[ \begin{aligned} I_L \\ &= \frac{800 \times 10^3}{\sqrt{3} \times 3300 \times 0.707} \\ &= \frac{800000}{4041} = 198\ \text{A} \end{aligned} \]

The synchronous reactance is not needed for this part, since the power and power factor fix the current. The value 330 A would be for a power factor of about 0.42, and the larger values do not follow from the data. The answer is 198 A, option B.

Answer: C

The efficiency excluding field and stator losses is 96 per cent, so the developed (air gap) power is the output divided by the efficiency. The output is 100 horse power, which is 74.6 kW.

\[ P_{dev} = \frac{74.6}{0.96} = 77.7\ \text{kW} \]

The synchronous speed of a 4 pole, 60 Hz motor is \(N_s = 120 \times 60/4 = 1800\) rpm, so \(\omega = 188.5\) rad/s.

\[ \begin{aligned} T = \frac{P_{dev}}{\omega} = \frac{77700}{188.5} \\ &= 412\ \text{N m} = 412 \times 0.7376 \\ &= 304\ \text{lb-ft} \end{aligned} \]

The answer is 304 lb-ft, option C. Without the efficiency correction the torque would be about 292 lb-ft, which is not offered.

Answer: A

A synchronous motor can be used for increasing the power factor. When it is over-excited it draws a leading current, so it can be run on no load as a synchronous condenser to improve the power factor of a plant with lagging loads.

A DC shunt motor, a single phase induction motor and a stepper motor do not draw leading current and cannot improve the power factor of a system. So the first option is correct.

Answer: B

In the V-curves of a synchronous motor, the field current is plotted on the x-axis and the armature current on the y-axis, with a separate curve for each constant load.

The armature current is the quantity on the vertical axis, and the power factor is on the vertical axis in the inverted V-curves. The load is held constant for each curve. So the second option is the answer.

Answer: A

The correct statement is that a synchronous motor can be operated on a leading as well as a lagging power factor. By changing the DC field excitation, an under-excited motor draws lagging current and an over-excited one draws leading current.

The other statements are false: it needs DC excitation for its field, it is not self-starting (it needs an auxiliary means such as damper windings), and its power factor depends on the excitation and, to some extent, the load. So the first option is correct.

Answer: C

The motor has 4 poles. The synchronous speed of a motor depends on the supply frequency and the number of poles.

\[ N_s = \frac{120 f}{P} \]

\[ P = \frac{120 \times 60}{1800} = 4 \]

Eight poles would give 900 rpm, six poles 1200 rpm, and two poles 3600 rpm at 60 Hz, so only four poles gives 1800 rpm. The third option is the answer. As a check, 4 poles on a 60 Hz supply give 120 x 60 / 4 = 1800 rpm, which is the speed in the question.

Answer: D

The minimum armature current occurs at unity power factor with no load. In the V-curve the armature current is lowest when the power factor is unity, because then the current is wholly the active component. For a lighter load the minimum is lower, and on no load it is the smallest of all.

At leading or lagging power factors the armature current is larger because of the added reactive component, and a full load needs more active current than a half load. So the fourth option is correct.