Alternator / Generator MCQs with Explanatory Answers)
Alternator / Generator MCQs with Explanatory Answers
1. The rating of Alternator / Generator may be expressed in ___________
- kW
- kVA
- kVAR
- HP
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Answer: 2. kVA
Explanation: The power √3 V_{L} I_{L }Cos φ delivered by the alternator for the same value of current, depends upon p.f. (Power Factor=Cos φ) of the load. But the alternator conductors are calculated for a definite current and the insulation at magnetic system are designed for a definite voltage independent of p.f. (Cos φ) of the load. For this reason apparent power measured in kVA is regarded as the rated power of the alternator.
2. Most of the Alternators have rotating field and stationary armature.
- True
- False
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Answer: ( 1 )
Most of the alternators have rotating field and stationary armature. This is because of some of the advantages as given below
- The stationary armature can be easily insulated
- Stationary armature windings are safe from vibration and centrifugal forces and can be braced against high electromagnetic force
- Number of slip rings required to provide direct current to the rotating field is less compared to number of slip rings required to provide direct current to the rotating armature.
3. Salient pole type rotors are designed for alternators with many number of poles and Cylindrical rotors are designed for alternators with two to four poles.
- True
- False
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Answer: ( 1 )
Salient pole rotors are basically used in low and medium speed Alternators. As speed is low, the desired frequency is obtained by having larger number of poles. Hence these machines have large diameters and short axial lengths.
Figure 1: Salient Pole Vs. Cylindrical Rotors
On the other hand, cylindrical rotors are used for turbo alternators running at high speed as 3600 RPM or 1800 RPM.
4. Alternators are also called Synchronous generators because
- They must run at a constant speed irrespective of desired frequency.
- They must run at synchronous speed to give the desired frequency
- They must run at a speed which varies with frequency.
- They must run at a speed equal to the desired frequency.
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Answer:
Let us consider an Alternator with P number of poles on the rotor revolving with N revolutions per minute speed. When a pair of poles pass through a conductor, EMF is induced in the conductor. The polarity of the induced EMF depends upon the polarity of the magnetic poles which cut through the conductor.
Number of cycles per revolution = P/2
Number of revolutions per second = N/60
Therefore, Number of cycles per second = PN /120
However, number of cycles per second is the frequency of generated voltage. Hence, frequency, f = PN /120
Since number of poles is usually fixed for an alternator, the speed of the rotor must be in synchronization with the desired speed. Hence Alternators are also called as Synchronous Generators.
5. For a short pitch coil, the pitch factor is————
- Less than 1
- Equal to 1
- More than 1
- Zero
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Answer: ( 1 )
Pole pitch is the angular distance between two poles on a machine. It is always 180 degrees electrical regardless of the number of poles in the machine. However, two sides of a coil are not always corresponding points under the adjacent poles and hence the coil has a span or pitch less than 180 degrees. This is known as a short pitch coil. The pitch factor, K_{c} is the ratio of the voltage generated in the short pitch coil to the voltage generated in the full pitch coil. Since EMF generated in the coil is always less than the EMF generated in the pole or full pitched coil. Hence for a short pitch coil, K_{c }is less than 1.
6. For a single layer, 36 slot, four pole, four phase stator winding, the value of distribution factor is——–
- 0.5
- 0.96
- 0.4
- 0.2
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Answer: ( 2 )
Given: Number of slots = 36
Number of phases = 4
Number of slots per phase = 36/4 = 12
Number of poles = 4
Number of poles per slot per pole, m = 12/4 = 3
Slot angle, α = 360/36 = 10 degree mechanical = 20 degree electrical
Substituting the corresponding values, we get: K_{D} = 0.5/0.52 = 0.96
7. For a three-phase star connected alternator driven at 800 RPM, required to generate a line to line voltage of 430 Volts at 60Hz, if the stator winding is 2 slots per phase per pole and each coil has 4 turns, the value of useful flux per pole is———
- 0.035 Wb
- 0.001 Wb
- 0.027 Wb
- 0.03 Wb
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Answer: ( 3 )
Given:
Number of phase = 3
Speed of the rotor, N = 800 RPM
Line to line voltage, V_{L} = 430 Volts
Therefore, Phase voltage, V_{ph} = V_{L }/ √3 = 248.26 Volts
Frequency of generated voltage, f = 60Hz
Therefore, number of poles, P = 120f /N = 9
Number of slots per pole per phase, m = 2
Total number of stator slots = 2 x Number of poles x Number of phase = 48
Therefore, slot angle, α = 360/48 = 7-1/2 degree mechanical = 33.75 degree electrical
Hence, Distribution Factor, K_{D} = sin [ ( 2 x 33.75 ) / 2 ] / 2sin ( 33.75 / 2 ) = 0.55 / 0.58 = 0.95
Pitch Factor, K_{c} = 1 ( Full pitch coil )
Number of slots per phase = 2 x 9 = 18
Number of turns per coil = 4
Number of turns per phase, T_{ph} = 4 x 9 = 36
Hence, RMS value of induced EMF, E_{Ph} = 4.44 K_{c}K_{D}T_{ph}φf
As induced EMF is the generated phase voltage
Hence, flux per pole, φ = 248.26 / 261.072 = 0.027 Wb
8. For a three-phase, sixteen pole star connected alternator with 144 slots and 12 conductors per slot, if flux per pole is 0.03 Wb and speed is 400 RPM, the line induced EMF and generated frequency is —————– respectively
- 3366.29 Volts, 50 Hz
- 2000 Volts, 60 Hz
- 3366.29 Volts, 53.33 Hz
- 3400 Volts, 50 Hz
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Given:
Number of phase = 3
Number of poles, P = 16
Number of slots = 144
Number of slots per pole = 9
Speed of rotor = 400 RPM
Number of slots per pole per phase, m = 3
Slot angle, α = 360/144 degree mechanical = 20 degree electrical
Distribution Factor, K_{D }= 0.5 / 0.52 = 0.96
Number of turns per phase, T_{ph} = ( 144 x 12 ) / ( 2 x 3 ) = 288
Assuming full pitch coil, K_{c} = 1
Flux per pole, φ = 0.03 Wb
Generated frequency, f = NP/120 = 53.33 Hz
Hence, induced EMF, E_{ph} = 4.44K_{c }K_{D}fT _{ph}φ = 1943.53 Volts
The line voltage, V_{L} = √3 E_{ph} = 3366.29 Volts
9. For an Alternator connected to a load, the terminal voltage per phase will———-
- Be equal to the induced EMF
- Vary with the load
- Constant irrespective of the load
- None of these
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When the load connected to the Alternator varies during the latter’s operation, the generated voltage also varies. This is because of the voltage drops in armature resistance R_{a}, armature leakage reactance, X_{L} and voltage drop due to armature reactance, X_{a}. The armature reaction is due to the action of armature current flux and field flux.
10. For a three phase, star connected alternator rated 1500 KVA, 13000 V, if the armature effective resistance is 1.6 Ohms and the synchronous reactance is 32 Ohms respectively, the percentage regulation for a load of 1275kW at 0.8 leading power factor is————–
- 12%
- 13.20%
- 14.5%
- 20%
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Answer: ( 2 )
Given:
Effective resistance, R_{a} = 1.6 Ohms
Synchronous Reactance, X_{s} = 32 Ohms
Line Voltage, V_{L} = 13000 Volts
Power Factor, cosφ = 0.8
Hence, sinφ = 0.6
Three phase power = 1275 kW = √3 V_{L}I_{L}cosφ
Therefore, line current, I_{L} = 70.78 Amperes
Phase voltage, V_{ph} = V_{L }/ √3 = 7505.55 Volts
No load voltage for leading power factor of 0.8 is, E_{p}^{2} = (V_{p}cosφ + I_{L}R_{a})^{2} + (-V_{P}sinφ + I_{L}X_{s})^{2 }, or E_{P} = 6514.32 Volts
Therefore, Percentage regulation = ((E_{P} – V_{ph})/E_{P}) x 100 = -13.20%
11. Alternators cannot be connected in parallel with one another
- True
- False
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Answer: ( 1 )
Alternators can be connected in parallel with one another with common bus bars, a process known as Synchronization.
KVA and thank you so much
Welcome
Kva.. ya que se determina es la potencia aparente a suministrar.
What is difference between real power and apparent power ?
Sir, where are the remaining MCQs