Synchronous Generators Questions and Answers

Answer: A

Correct answer: suppress a chosen harmonic in the output waveform

Choosing the coil span sets the pitch factor for every harmonic independently, and a span of five-sixths of a pole pitch reduces the fifth and seventh very substantially while costing the fundamental only about four per cent. Trading that small loss for a much cleaner waveform is almost always worth it.

Answer: B

Correct answer: The speed of its prime mover and the system frequency

\[ P = \frac{120f}{N}. \] The frequency is fixed by the system and the speed by the turbine, so the pole number is not a free choice at all - it follows from the two. This is why steam-driven machines have two or four poles and water-driven machines have dozens.

Answer: C

Correct answer: about 98 to 99 per cent

A large turbo-alternator is among the most efficient machines ever built, losing only one or two per cent in copper, iron, windage and excitation. At several hundred megawatts even that fraction is several megawatts of heat, which is precisely why hydrogen and direct water cooling are needed to carry it away.

Answer: D

The rotor field is supplied rather than induced, so there is no slip-dependent rotor loss of the kind an induction machine suffers. That absence is a large part of why the synchronous machine is the more efficient of the two.

Answer: D

Running alternators in parallel instead of one large unit gives reduced operating cost, because units can be switched on and off to follow the load, better reliability, because the loss of one unit does not stop the supply, and easier maintenance, because one machine can be taken out for service while the others carry the load. A set of machines can also be run near full load, where each is most efficient.

So poor efficiency is not an effect of parallel operation, and it is the incorrect statement. The answer is option D.

Answer: D

The distribution factor of a winding is \(k_d = \dfrac{\sin(m\beta/2)}{m \sin(\beta/2)}\), where \(m\) is the number of slots per pole per phase and \(\beta\) is the slot angle in electrical degrees.

\[ \begin{aligned} m = \frac{36}{6 \times 3} = 2,\qquad \beta \\ &= \frac{180^\circ \times P}{S} \\ &= \frac{180^\circ \times 6}{36} = 30^\circ \end{aligned} \]

\[ \begin{aligned} k_d \\ &= \frac{\sin(2 \times 30^\circ / 2)}{2 \sin(30^\circ / 2)} \\ &= \frac{\sin 30^\circ}{2 \times \sin 15^\circ} \approx 0.966 \end{aligned} \]

That is the fourth option. The other fractions have the wrong denominators for \(m = 2\): the third option would be right for \(m = 3\) with a 30 degree angle, and the first two use angles of 20 degrees that do not occur here. The answer is option D.

Answer: C

The turbine torque equals the input power divided by the angular speed, because losses are neglected. A two pole machine on 60 Hz runs at 3600 rpm.

\[ \begin{aligned} \omega = \frac{2\pi \times 3600}{60} \\ &= 377\ \text{rad/s} \end{aligned} \]

\[ \begin{aligned} T = \frac{112 \times 10^3}{377} = 297\ \text{N m} \\ &= 297 \times 0.7376 = 219\ \text{lb-ft} \end{aligned} \]

The synchronous reactance and the power angle fix the terminal conditions but are not needed once the power is known. The answer is 219 lb-ft, option C.

Answer: A

The frequency is 40 Hz. The frequency of the voltage generated by an alternator depends on the number of poles and the speed.

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

\[ = \frac{4 \times 1200}{120} = 40\ \text{Hz} \]

A frequency of 50 Hz would need 1500 rpm for 4 poles, 60 Hz would need 1800 rpm, and 100 Hz would need 3000 rpm. At 1200 rpm none of these frequencies results, so the first option is the answer.

Answer: A

For connecting an alternator to the grid, the phase angle between the generated voltage and the grid voltage must be 0°, that is the two voltages must be in phase. Along with equal voltage, equal frequency and the same phase sequence, this lets the breaker be closed without a surge of current.

A phase angle of 45°, 60° or 90° would mean a voltage difference across the breaker at the moment of closing, which would cause a large circulating current and a mechanical shock to the machine. So the first option is correct.

Answer: C

A synchroscope is used for synchronizing an alternator with the bus bars or the grid. It shows whether the incoming machine is faster or slower than the system and when the two are in phase, so the operator knows the right instant to close the breaker.

It does not start, brake or reverse a machine. With the lamp method the lamps go dark together when the machine is in step, and the synchroscope shows the same thing on a dial: the pointer turns one way if the incoming machine is fast and the other way if it is slow, and stands still at the top when the two are in phase. So the third option is the answer.