100 MCQ OF Synchronous Generator or Alternator with Explanation

Ques 61. A generating station supplies the following loads 15000 kW, 12000 kW, 8500 kW, 6000 kW and 450kW. The station has a maximum demand of 22000kW. Calculate the diversity factor

  1. 1.91
  2. 0.52
  3. 0.68
  4. 1.34

Answer.A. 1.91

Explanation:-

Diversity factor = Sum of individual maximum demand/Max demand of station

= (1500 + 1200 + 8500 + 6000 + 450)/22000

= 1.91

 

Ques 62. In a 3-phase synchronous generator, the stator winding is connected in star, because a delta connection would

  1. Have circulating currents due to triple harmonics
  2. Require more insulation and conductor material
  3. Require a larger conductor and more core material
  4. Result in a short circuit

Answer. 2. Require more insulation and conductor material

Explanation:-

Alternators are connected primarily in the star to achieve the following motives:

  1. The phase voltages in star connection is 57.7 % of the line voltages, i.e. the armature winding in star connection is less exposed to voltage as compared to the delta connection which in turn prove more economic if we consider insulation, breakdown strength, the requirement of conductor material
  2. Easy protection: Neutral grounding is necessary to allow zero sequence currents to flow to the ground in case of a fault.
  3. Elimination of harmonics: Star connection facilitates a neutral connection which is instrumental in eliminating triple harmonics.
  4. No circulating currents: In star connection, we don’t have circulating parasitic currents like in Delta which lead to heating losses.

 

Ques 63. A cylindrical synchronous machine connected to an infinite bus power will deliver maximum power at a power angle of

  1. δ = 0°
  2. δ = 90°
  3. δ = 45°
  4. δ = 30°

Answer. 2. δ = 90°

Explanation:-

Power delivered by the synchronous generator is given as

sol-8

 

Where E = Excitation voltage
V = Terminal voltage
X = Synchronous reactance

For Pmax  sinδ = 1

δ = 90°

Pmax = VE/X

 

Ques 64. The maximum power developed in a cylindrical synchronous machine depends on the

  1. oad angle
  2. Synchronous reactance
  3. Excitation
  4. Both 2 and 3

Answer.4. Both 2 and 3

Explanation:-

Power delivered by the synchronous generator is given as

sol 8

 

Where E = Excitation voltage
V = Terminal voltage
X = Synchronous reactance

For Pmax  sinδ = 1

δ = 90°

Pmax = VE/X

Therefore, maximum power developed depends on excitation, Terminal and synchronous reactance.

 

Ques 65. The effective resistance of a 2200V, 50Hz, 440 KVA, 1 -phase alternator is 0.5 ohm. On short-circuit, a field current of 40 A gives the full load current of 200 A. The electromotive force on open-circuits with the same field excitation is 1160 V. Then the synchronous impedance and reactance is _______ ohm and _______ ohm respectively.

  1. 5.8, 5.781
  2. 6.5, 7.5
  3. 3.2, 2.3
  4. 10, 20

Answer A. 5.8, 5.81

Explanation:-

For a single phase alternator, the field current remains the same for open circuit and short circuit conditions.

Synchronous impedance can be defined as the ratio of open-circuit voltage (or induced emf ) to short-circuit current (or MMF ) corresponding to the same field excitation.

Z = V/I = 1160/200 = 5.8 ohms

Now synchronous Reactance X

$\begin{array}{l}X = \sqrt {{Z^2} – {R^2}} \\\\X = \sqrt {{{5.8}^2} + {{0.5}^2}} \\\\X = 5.81\Omega \end{array}$

 

Ques 66. Two generators rated 200 MW and 400 MW are operating in parallel. The drooping characteristics of their governors are 4% and 5%, respectively from no load to full load. Assuming that the generators are operating at 50 Hz at no load, how would a load of 600 MW be shared between them? Then the system frequency at this load will be? Assume free governor operation.

  1. 50 Hz
  2. 51.4 Hz
  3. 52. 5 Hz
  4. 47.69 Hz

Answer. D. 47.69 Hz

Explanation:-

Since the generators are in parallel hence they will operate at the same frequency at steady load.

Let the load On the generator 1 (200 MW) = P

Then the load on the generator 2 (400 MW) = (600 − P) MW

Let the reduction in the frequency = Δf

Now,

Δf ⁄ P = 0.04 × 50  ⁄ 200

Δf ⁄ P = 0.01 ———(1)

Also,

Δf ⁄ (600 − P) = 0.05 × 50  ⁄ 400

Δf ⁄ (600 − P) = 0.006 ———-(2)

Equating Δf in (i) and (ii), we get

P = 231 MW (Load on generator 1)

600 − P = 369 MW (Load on generator 2)

System frequency of the alternator is given as

$f = \frac{{{f_o} – \Delta f}}{{{P_{rated}}}} \times P_1/P_2$

Where fo is rated power frequency.
Δf is full-load frequency.

Putting the value of P in above equation, the system frequency will be

$\begin{array}{l}f = 50 – \dfrac{{0.04 \times 50}}{{200}} \times 231\\\\f = 47.69Hz\end{array}$

 

Ques 67. A large hydroelectric power plant is under consideration. Its hydraulic head or water Level difference above and below the dam and its power requirement dictate its water turbine or runner must turn at from 137.00 rpm (14.347 rad/sec). The power required is 60 Hz. Then How many poles must a direct-connected alternator have?

  1. 60
  2. 53
  3. 55
  4. 100

Answer 2. 53

Explanation:

The synchronous alternator is the basic a c. generator It is called synchronous because its generated frequency is directly related to its number of armature and field poles and to its rotative speed. An individual coil of winding generates a full cycle of a c. voltage each time it is swept by a pair of magnetic poles.  The relation between the generated frequency from cycles per pole pair to a machine basis is given as

$f = \dfrac{{P\omega }}{{4\pi }} \times \dfrac{{Cycle}}{{\sec }}$

Where

P = Number of Poles
ω = Rotative speed in rad/sec

Hence the number of poles connected the to the alternator is

$\begin{array}{l}P = \dfrac{{4\pi f}}{\omega }\\\\P = \dfrac{{4\pi \times 60}}{{14.347}}\\\\P = 53\end{array}$

 

Ques 68. The stator of an alternator rarely uses ______ slots.

  1. Wide-open type
  2. Semi-closed type
  3. Closed type
  4. None of the above.

Answer 3. Closed Type

Explanation:-

Constructionally the staler for both types of alternators is identical. The stator is made of laminations of special magnetic iron or steel alloy and can accommodate armature conductors. The whole structure is fitted in a frame. The frame may be of cast iron or welded steel plate. T

he laminations in the stator are insulated from each other with paper or varnish depending Upon the size of the machine. The stampings also have openings that make axial and radial ventilating ducts to provide efficient cooling.

Slots provided on the stator core are mainly of three types 

  1. Open type
  2. Semi-closed type.
  3. Fully closed type

Open type:- Open type slots are commonly used because the coils can be formed, wound and insulated prior to being placed in the slots. These slots also provide the facility in removal and replacement maintenance of defective coil. However, these type of slots has the disadvantage of distributing the air gap flux into branches which tends to produce ripples in the emf wave.

Semi-closed type:- The semi-closed type slots are better in this respect but do not permit the use of pre-wound coils and lack ventilation as well as pose difficulty in maintenance.

Closed type:  The wholly closed type slots do not distribute the air gap flux but

  1. They tend to increase
  2. The armature conductors have to be threaded through, thereby increasing the initial labor and cost of winding
  3. They present a complicated problem of end connections. Hence, they are rarely used.

 

Ques 69. The advantages of having the revolving field of the alternator are

  1. Ease of construction
  2. Better Insulation to the armature
  3. Reduce Rotor weight
  4. All of the above

Answer. 4. All of the above

Explanation:-

 FIELD AND ARMATURE CONFIGURATIONS

There are two arrangements of fields and armatures:

  1. Revolving armature and the stationary field
  2. Revolving field and stationary armature.

 ADVANTAGES OF ROTATING FIELD IN AN ALTERNATOR

In large alternators, the rotating field arrangement is usually forward due to the following advantages.

  1. Ease of Construction: Armature winding of large alternators is complex, the connection, and bracing of the armature windings can be easily made for the stationary stator.
  2. The number of Slip Rings: If the armature is made rotating, the number of slip rings required for power transfer from the armature to the external circuit is at least three. Also, heavy current flows through brush and slip rings cause problems and require more maintenance in large alternators. Insulation required for slip rings from the rotating shaft is difficult with the rotating armature system.
  3. High voltage generation:- Voltages can be generated as high as 11,000 and 13,800 V. These values can be reached because the stationary armature windings do not undergo vibration and centrifugal stresses.
  4. High current Rating:-  Alternators can have relatively high current ratings. Such ratings are possible because the output of the alternator is taken directly from the stator windings through heavy, well-insulated cables to the external circuit. Neither slip rings nor a commutator is used.
  5. Better Insulation to Armature: Insulation arrangement of armature windings can easily be made from the core on the stator.
  6. Reduced Rotor Weight and Rotor Inertia:  Since the field system is placed on the rotor, hence the insulation requirement is less (for low dc voltage). Also, rotational inertia is less. It takes lesser time to gain full speed.
  7. Improved Ventilation Arrangement: The cooling can be provided by enlarging the stator core with radial ducts. Water cooling is much easier if the armature is housed in the stator.

Hence in almost all of the alternators, the armature is housed in the stator while the dc field system is placed in the rotor.

 

Ques 70. The a.c. the armature winding of an alternator operates at _______ the field winding.

  1. The same voltage as
  2. Much lesser voltage than
  3. Much higher voltage than
  4. None of the above

Answer. 3. Much Higher voltage than

Explanation:- 

AC is used everywhere, the generation level of a.c voltage may be higher as 11 kV to 33 kV. This gets induced in the armature. For the stationary armature, a large space can be provided to accommodate a large number of conductors and the insulation.

The rotor carries a field winding which is supplied with d.c. through two slip rings by a separate d.c source. The field voltage is usually in the range between 100 and 250 V. The amount of power delivered to the field circuit is relatively small.

Since the voltage applied to the rotating field is low voltage DC, the problem of ARC over at the slip ring is not encountered.

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