Design of Synchronous Machines MCQs ( Design Of Electrical Machines ) MCQs – Design Of Electrical Machines MCQs

Design of Synchronous Machines MCQs ( Design Of Electrical Machines ) MCQs – Design Of Electrical Machines MCQs

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Design Of Electrical Machines MCQs – Design of Synchronous Machines MCQs ( Design Of Electrical Machines ) MCQs

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Types of Synchronous Machines

1. How many categories can the synchronous motors be divided into?
a) 4
b) 3
c) 5
d) 2
Answer: c
Explanation: The synchronous motors can be divided into 5 types of categories. They are hydro-generators, turbo-alternators, engine driven generators, motors, compensators.


2. How is the hydro-generator driven by?
a) water turbine
b) steam turbine
c) internal combustion engines
d) control of reactive power networks
Answer: a
Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The hydro-generators are also known as water wheel generators.


3. What is the rating of the hydro-generators?
a) 750 MW
b) 1000 MW
c) 20 MW
d) 700 MW
Answer: a
Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The rating of the synchronous generators is 750 MW.


4. What is the speed by which the hydro-generators are driven?
a) 100-1000 rpm
b) 5000 rpm
c) 1500 rpm
d) 3000 rpm
Answer: a
Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The hydro-generators are driven at a speed of 100-1000 rpm.


5. How is the turbo-alternators driven by?
a) water turbines
b) steam turbines
c) engine driven generators
d) compensators
Answer: b
Explanation: The turbo alternators are one category among the synchronous machines. They are driven by the steam turbines.


6. What is the speed of the turbo-alternators?
a) 100-1000 rpm
b) 5000 rpm
c) 1500 rpm
d) 3000 rpm
Answer: d
Explanation: The turbo alternators are one category among the synchronous machines. They are driven by steam turbines. The rating of the turbo-alternators is 3000 rpm.


7. What is the rating of the turbo-alternators?
a) 750 MW
b) 1000 MW
c) 20 MW
d) 700 MW
Answer: b
Explanation: The turbo alternators are one category among the synchronous machines. They are driven by the steam turbines. The speed of the turbo-alternators is 1000 MW.


8. How is the engine driven generators driven by?
a) water turbines
b) steam turbines
c) internal combustion engine
d) compensators
Answer: c
Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings.


9. What is the rating of the engine driven generators?
a) 750 MW
b) 1000 MW
c) 20 MW
d) 700 MW
Answer: c
Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings. These generators have ratings upto 20 MW.


10. What is the speed of the engine driven generators?
a) 100-1000 rpm
b) 5000 rpm
c) 1500 rpm
d) 3000 rpm
Answer: c
Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings. These generators have speed of 1500 rpm.


11. The synchronous motors are cheaper than the induction motors.
a) true
b) false
Answer: a
Explanation: The synchronous motors are cheaper than the induction motors. They have high power lower speed applications and are cheaper when compared to the induction motors.


12. Which among the following are the applications of synchronous motors?
a) compressors
b) blowers
c) fans
d) compressors, fans, blowers
Answer: d
Explanation: They are mainly used in the high power and low speed applications. They include compressors, fans, blowers and low head pumps.


13. What is the application of synchronous compensators?
a) control of real power
b) control of active power
c) control of reactive power
d) control of apparent power
Answer: c
Explanation: The main application of the synchronous compensators is to control the reactive powers. They are used in the power supply networks.


14. What is the rating of the synchronous generators?
a) 750 MVAr
b) 1000 MVAr
c) 100 MVAr
d) 700 MVAr
Answer: c
Explanation: The main application of the synchronous compensators is to control the reactive powers. They are designed for ratings upto 100 MVAr.


15. What is the speed of the engine driven generators?
a) 100-1000 rpm
b) 5000 rpm
c) 1500 rpm
d) 3000 rpm
Answer: d
Explanation: The main application of the synchronous compensators is to control the reactive powers. They are designed to speeds upto 3000 rpm.

Construction of Hydro-Generators – 1

1. How many factors are involved in the construction of hydro-generators?
a) 11
b) 10
c) 9
d) 8
Answer: b
Explanation: The hydro-generators are the synchronous generators that are driven by the water turbines. They are chosen based on 10 different categories.


2. Which factor does the constructional feature of the hydro-generators depend on?
a) speed
b) voltage
c) power
d) current
Answer: a
Explanation: The constructional feature of hydro-generators are basically dependent upon the mechanical considerations. The main mechanical consideration is speed of the machine.


3. What factors does the speed of the machines depend upon?
a) head
b) blades
c) type of turbine used
d) head and type of turbine used
Answer: d
Explanation: The hydro-generators are generally low speed machines. The speed depends upon the head and the type of turbines used.


4. Why is the stator core built up of laminations?
a) to reduce core loss
b) to reduce copper loss
c) to reduce iron loss
d) to reduce eddy current loss
Answer: d
Explanation: The stator core is built up of laminations in order to reduce the eddy current loss. The loss in the laminated core is usually the single largest loss and hence the choice of type and grade of steel is of utmost importance.


5. The modern synchronous machines make use of non-directional cold rolled steel.
a) true
b) false
Answer: a
Explanation: The modern synchronous machines make use of the non-directional cold rolled steel which has electrical characteristics similar to that of the hot rolled steel. But the non-directional cold rolled steel has much improved mechanical characteristics.


6. What is the thickness of the most commonly used grade for stator laminations?
a) 0.5 mm
b) 1 mm
c) 1.5 mm
d) 2 mm
Answer: a
Explanation: The most commonly used grade for the stator laminations is grade 230. The thickness of this most commonly used grade for the stator laminations is 0.5 mm.


7. What is the range of the outside diameter of the stator frame of the large hydro-generator?
a) 3-18 m
b) 2-18 m
c) 3.5-18 m
d) 4-18 m
Answer: c
Explanation: The minimum value for the outside diameter of the stator frame of the large hydro-generators is 3.5 m. The maximum value for the outside diameter of the stator frame of the large hydro-generators is 18 m.


8. How are the stator windings of all synchronous generator connected?
a) star-delta connection
b) star connection
c) star connection with neutral earthed
d) delta connected
Answer: c
Explanation: The stator windings of all the synchronous generator are connected in the star connection with neutral earthing. The main advantage that the winding has to be insulated to earth for the phase voltage and not the line voltage.


9. What among the following is the advantages of the star connection?
a) eliminates single frequency harmonics
b) eliminates double frequency harmonics
c) eliminates triple frequency harmonics
d) eliminates sinusoidal harmonics
Answer: c
Explanation: The hydro-generators stator windings are connected in the star connected with the neutral earthed. It eliminates the triple frequency harmonics from the line voltage.


10. The capacity of the pull out machines used for making the coils limits the pole pitch to less than 0.8 m.
a) true
b) false
Answer: a
Explanation: The capacity of the pull out machines used for making the coils limits the pole pitch to less than 0.8 m. The capacity of the pull out machines used for making coils limits the length of slot portion to about 3 m.


11. What is the main advantage of a winding with multi-turns coils?
a) reduce the choosing of the value of stator slots
b) allows greater flexibility in selecting the value of stator slots
c) increases the flexibility in selecting the number of stator slots
d) has no effect on the number of stator slots
Answer: b
Explanation: The main advantage of a winding with multi-turns coils is that it allows a greater flexibility in selecting the value of stator slots. It in turn helps in choosing the required number of turns per phase.


12. What happens to the current in the windings during the sudden short circuits at the line terminals?
a) the current reduces to 15 times the full load current
b) the current increases to 15 times the full load current
c) the current reduces to 10 times the full load current
d) the current increases to 10 times the full load current
Answer: b
Explanation: The current in the windings during short circuit at the line terminals increases to about 15 times the full load current. It can also increase to higher level depending on the direct axis sub-transient reactance.


13. What happens to the electromagnetic forces during the sudden short circuits at the line terminals?
a) the electromagnetic forces get increased by 250 times the force under normal full load condition
b) the electromagnetic forces get decreased by 250 times the force under normal full load condition
c) the electromagnetic forces get increased by 200 times the force under normal full load condition
d) the electromagnetic forces get decreased by 200 times the force under normal full load condition
Answer: a
Explanation: The electromagnetic forces are directly proportional to the square of the current. It increases by 250 times the force under normal full load conditions.


14. What should be done to the conductors in the overhang of the stator?
a) bracing
b) shaving
c) punching
d) compressing
Answer: a
Explanation: The conductors in the overhang of the stator must be braced. Bracing is the process of raising the mechanical strength of the conductors.


15. How many steel brackets are used along with the support to steel rings?
a) 5
b) 7
c) 4-6
d) 5-9
Answer: c
Explanation: During the process of bracing one or two circular steel rings are used to support the overhang. Along with the steel rings 4-6 steel brackets are also used.

Construction of Hydro-Generators – 2

1. What factor does the rotor body depends upon in the construction of hydro-generators?
a) speed
b) voltage
c) peripheral voltage
d) power
Answer: c
Explanation: In the hydro-generators the salient poles are attached to the rotor body. The type of rotor body used depends in general on the peripheral speed.


2. Which machine makes use of the forged steel construction?
a) low speed
b) high speed
c) very high speed
d) medium speed
Answer: b
Explanation: The forged steel construction is made use for the high speed machines. The earliest construction particularly at relatively low outputs consisted of a body and shaft extension made as single forging.


3. In what type of generator is the thick rolled steel discs made use of?
a) generators running at 400 rpm
b) generators running at 500 rpm
c) generators running at 600 rpm
d) generators running at 600 rpm and above
Answer: d
Explanation: Another type of rotor construction made use of in the hydro-generators is the thick rolled steel discs. It is made use for the generators running at 600 rpm and above.


4. What is the range of the length of the thick rolled sheets used in the rotor design of the hydro generators?
a) 130-180 mm
b) 120-180 mm
c) 150-180 mm
d) 160-190 mm
Answer: b
Explanation: The minimum value of the length of the thick rolled sheets used in the rotor design of the hydro-generators is 120 mm. The maximum value of the length of the thick rolled sheets used in the rotor design of the hydro-generators is 180 mm.


5. The laminations used in the rotor body design is 1.8 mm.
a) true
b) false
Answer: a
Explanation: The laminations used in the rotor body design is 1.8 mm. They are in the form of the overlapping segments tightly bolted.


6. For what peripheral speed of the machine is the segmental rim on fabricated spider used?
a) >120 m per s
b) <120 m per s
c) >130 m per s
d) <130 m per s
Answer: d
Explanation: The segmental rim on fabricated spider is used for the machines whose peripheral speed is upto 130 m per s. The advantage is it is easy to transport and assemble at site.


7. For what peripheral speed is the poles bolted to the yoke?
a) 20 m per s
b) 30 m per s
c) 25 m per s
d) 27 m per s
Answer: c
Explanation: The poles are clamped or fixed to the rim in different ways. In case of the generators the poles are bolted to the yoke when the peripheral speed is upto 25 m per s.


8. What is the range of the peripheral speeds in the water wheel generators?
a) 20 to 80 m per s
b) 20 to 50 m per s
c) 50 to 80 m per s
d) 30 to 70 m per s
Answer: a
Explanation: The minimum peripheral speed in the water wheel generators is 20 m per sec. The maximum peripheral speed in the water wheel generators is 80 m per sec.


9. The length of the mean turn of the winding is smallest with circular poles.
a) true
b) false
Answer: a
Explanation: The circular poles have a lot of advantages. The length of the mean turn of the winding is smallest with circular poles and therefore cost is reduced. The copper losses are also less.


10. Why are damper windings used in the construction of hydro-generators?
a) to control losses
b) to increases efficiency
c) to reduce the oscillations
d) to reduces voltage surges
Answer: c
Explanation: The damper windings are generally used in the construction of hydro-generators. It is generally used to reduce the oscillations in the generator.


11. By how much is the rotor and the turbine runner and the hydraulic thrust requirements more than the dead weight of rotating masses?
a) thrice
b) twice
c) by 4 times
d) by 8 times
Answer: b
Explanation: In case of the vertical shaft generators special features have to be incorporated in the bearing set up. This is because the rotor and the turbine runner and the hydraulic thrust requirements is twice more than the dead weight of rotating masses.


12. How are the bearing cooled in the construction?
a) natural cooling
b) forced cooling
c) air cooling
d) oil cooling
Answer: d
Explanation: Oil is the material which is used for the cooling of the bearings. It is being supplied to the bearings by the pump.


13. By how many minutes is the complete energy of the rotating parts and the machine is brought to rest?
a) 2 minutes
b) 1 minute
c) 3 minutes
d) 4 minutes
Answer: c
Explanation: The brakes are so designed such that the complete energy of the rotating parts and the machine is brought to rest. It is brought to rest by 3 minutes.


14. What material is used to make the material of the pads of the brakes?
a) pads using asbestos
b) pads using metal wires
c) pads using copper
d) pads using the asbestos interlaced with metal wires.
Answer: d
Explanation: The brakes are made to stop the machines immediately to protect the bearings. The brakes have pads using the asbestos interlaced with metal wires.


15. Why are the slip rings made use of?
a) they are used to provide excitation to field windings
b) they are used to provide excitation to the armature windings
c) they are used to reduce the heating effects
d) they are used to reduce the losses
Answer: a
Explanation: The slip rings are the last construction material made use of in the hydro-generators construction. They are used to provide excitation to the field windings.

Construction of Turbo-Alternators

1. How many factors does the construction of the turbo alternators depend upon?
a) 2
b) 4
c) 3
d) 6
Answer: c
Explanation: There are 3 factors which decide the construction of the turbo-alternators. They are i) stator core ii) stator winding iii) rotor.


2. How many poles does the turbo-alternators have and what is the speed of the turbo alternators?
a) 4 poles, 3000 rpm
b) 3 poles, 6000 rpm
c) 2 poles, 5000 rpm
d) 2 poles, 3000 rpm
Answer: d
Explanation: The turbo-alternators are made up of 2 poles. The speed with which the turbo alternators operate are 3000 rpm.


3. What is the relation of the lengths and diameters with the turbo-alternators?
a) long length, long diameters
b) long length, short diameters
c) short lengths, short diameters
d) short lengths, long diameters
Answer: b
Explanation: The turbo-alternators are characterized by the long lengths. The turbo-alternators are also characterized by short diameters.


4. What is the core length and the shaft length of a 500 MW turbo alternator?
a) 5 m, 12 m
b) 6 m, 15 m
c) 3 m, 10 m
d) 4 m, 13 m
Answer: a
Explanation: The core length of a 500 MW turbo alternator is 5 m. The shaft length of the 500 MW turbo alternator is 12 m.


5. What is the outer diameter of the stator core and outer casing of 500 MW turbo alternator?
a) 5 m, 2 m
b) 6 m, 5 m
c) 3 m, 4 m
d) 4 m, 7 m
Answer: c
Explanation: For a 500 MW turbo alternator the outer diameter of the stator core is 3 m. The diameter of the outer casing is 4 m.


6. What type of lamination is used for the stator core of the turbo alternators?
a) stepped
b) smooth
c) interleaved
d) segmental
Answer: d
Explanation: The stator core is made up of segmental laminations. The grain oriented steel laminations are another type of stator laminations made use of.


7. What is the advantage of the grain oriented steel laminations?
a) it lowers the iron loss
b) it lowers the core loss
c) it lowers the heating effects
d) it lowers the harmonics
Answer: b
Explanation: The grain oriented steel laminations are also one type of laminations used for the stator core. They result in the lowering of the core loss.


8. What is the pulsational force produced for 500 MW machine?
a) 70 kN per m
b) 60 kN per m
c) 80 kN per m
d) 90 kN per m
Answer: c
Explanation: The generator designed with low voltage gives high currents which produce high pulsational forces. The pulsational forces can be as high as 80 kN per m.


9. What is the use of the laminated and transposed conductors in turbo alternators?
a) to decrease the harmonics
b) to decrease the heating effects
c) to decrease the iron and core loss
d) to decrease the eddy current loss
Answer: d
Explanation: The presence of large currents requires the use of multi circuit windings. The use of the laminated and transposed conductors in turbo alternators is to reduce the eddy current loss.


10. What is the voltage range for large turbo-alternators?
a) 15-20 kV
b) 18-23 kV
c) 20-23 kV
d) 20-25 kV
Answer: d
Explanation: The generation voltage normally used are 15 kV for 100-200 MW machines. The voltage range for the large turbo-alternators are 20-25 kV.


11. The overhang has to be highly reinforced in turbo alternators.
a) true
b) false
Answer: a
Explanation: The overhang has to be highly reinforced in turbo alternators. The overhang is distributed over a large number of sections which eliminates the concentration of conductors in the overhang.


12. What is the use of the slot in the rotor?
a) for inserting the field windings
b) for inserting the armature windings
c) for securing the field windings
d) for inserting and securing the field windings
Answer: d
Explanation: The rotors slots are used for inserting the field windings. The rotor slots are also used for securing the field windings also.


13. How many types are the rotor slots distinguished into?
a) 2
b) 3
c) 4
d) 5
Answer: a
Explanation: There are 2 types of rotor slots in the turbo-alternators. They are i) radial sort ii) parallel slot rotors.


14. The rotor is slotted for one-third of its periphery.
a) true
b) false
Answer: b
Explanation: The rotor is slotted for only two-thirds of its periphery. It is because to reduce the flux pulsations and the cost factors.


15. What is the mechanical strength of the end bells used?
a) 1120 MN per m2
b) 1130 MN per m2
c) 1150 MN per m2
d) 1140 MN per m2
Answer: c
Explanation: The end bells are made of a non-magnetic austenitic steel in order to reduce the leakage flux. The end bells have high mechanical strength of about 1150 MN per m2.

 Armature Design – 1

1. How many factors does the design of armature of synchronous machines depend upon?
a) 2
b) 4
c) 3
d) 5
Answer: d
Explanation: There are 5 factors that decide the design of the armature of synchronous machines. They are i) single or double layer winding, ii) number of armature slots, iii) coil span, iv) turns per phase, v) conductor section.


2. When are the double layer bar windings made use of during the armature design?
a) large values of flux per pole and small number of turns per phase
b) large values of flux per pole and large number of turns per phase
c) small values of flux per pole and small number of turns per phase
d) small values of flux per pole and large number of turns per phase
Answer: a
Explanation: The double layer bar windings are made use of when there is large values of flux per poles. The double layer bar windings are made use of when there are small number of turns per phase.


3. Which type of machines have a large number of poles per phase?
a) high voltage machines and machines with high value of flux per pole
b) high voltage machines and machines with small value of flux per pole
c) small voltage machines and machines with high value of flux per pole
d) small voltage machines and machines with low value of flux per pole
Answer: b
Explanation: The high voltage machines have a large number of poles per phase. The small value of flux per pole per phase also results in large number of poles per phase.


4. Which among the following makes double layer windings advantageous than the single layer windings?
a) ease in the manufacture of coils and lower cost of winding
b) less number of coils are required as spare in the case of winding repairs
c) fractional slot windings can be employed
d) ease in the manufacture of coils and lower cost of winding, fractional slot windings can be employed, less number of coils are required as spare in the case of winding repairs
Answer: d
Explanation: The double layer windings have an advantage over single layer windings because of ease in the manufacture of coils and lower cost of winding, less number of coils are required as spare in the case of winding repairs, fractional slot windings can be employed.


5. The single layer windings have higher efficiency and quieter operation because of narrow slot openings.
a) true
b) false
Answer: a
Explanation: The single layer winding has high efficiency and quiet operation because of narrow slot openings. They also have higher space factor owing to the absence of inter layer separator.


6. When is the double layer bar or wave windings made use of?
a) when single turns coils are necessary as with turbo alternators and unipolar low voltage machines
b) when single turns coils are necessary as with turbo alternators and bipolar low voltage machines
c) when single turns coils are necessary as with turbo alternators and multipolar low voltage machines
d) when double turns coils are necessary as with turbo alternators and unipolar low voltage machines
Answer: c
Explanation: The double layer bar or wave windings are used when the single turns coils are necessary. They are also made use of with the multipolar low voltage machines.


7. How many factors are related in the selection of the armature slots?
a) 5
b) 6
c) 7
d) 4
Answer: b
Explanation: There are 6 factors associated with the selection of the armature slots. They are i) Balanced windings ii) Cost iii) Hot spot temperature iv) Leakage reactance v) Tooth ripples vi) flux density in iron.


8. How is the number of armature slots associated with the armature windings?
a) number of slots should be such that unbalanced winding is obtained
b) number of slots should be such that balanced winding is obtained
c) number of slots should be so low as possible
d) number of slots should be high as possible
Answer: b
Explanation: The number of slots should be such that balanced winding is obtained. Balanced windings should be obtained because it may lead to losses and heating effects.


9. How is the number of armature slots associated with the cost factor?
a) small number of slots leads to less cost
b) small number of slots leads to high cost
c) large number of slots leads to high cost
d) large number of slots leads to low cost
Answer: a
Explanation: The smaller the number of slots, the less will be the cost. This is because there are fewer coils to wind, form insulate, place into slots, and connect.


10. How is the number of armature slots associated with the hotspot temperature?
a) small number of slots leads to less hotspot temperature
b) small number of slots leads to high hotspot temperature
c) large number of slots leads to high hotspot temperature
d) large number of slots leads to low hotspot temperature
Answer: b
Explanation: The small number of slots leads to an increase in hotspot temperature. The small number of slots results in bunching of conductors leaving small space for circulation of air.


11. How is the number of armature slots associated with the leakage reactance?
a) small number of slots leads to less leakage reactance
b) small number of slots leads to high leakage reactance
c) large number of slots leads to high leakage reactance
d) large number of slots leads to low leakage reactance
Answer: a
Explanation: When the number of slots is small the leakage flux is increased. As the leakage flux is increased, the leakage reactance is increased owing to conductors lying near each other.


12. How is the number of armature slots associated with the tooth ripples?
a) tooth ripples are increased, if the number of slots are increased
b) tooth ripples are decreased, if the number of slots are increased
c) tooth ripples are increased, if the number of slots are decreased
d) tooth ripples are decreased, if the number of slots are decreased
Answer: b
Explanation: The tooth ripples in the field form and the consequent pulsation losses in pole face decrease if a large number of slots are used. Also, the waveform of generated voltage is free from ripples.


13. How is the number of armature slots associated with the flux densities in iron?
a) tooth ripples are increased, if the number of slots are increased
b) tooth ripples are decreased, if the number of slots are increased
c) tooth ripples are increased, if the number of slots are decreased
d) tooth ripples are decreased, if the number of slots are decreased
Answer: a
Explanation: The large number of slots a greater space is taken up by the insulation. This results in the narrower teeth giving flux densities which may go beyond acceptable limits.


14. The value of slot pitch depends upon the voltage of the machine.
a) true
b) false
Answer: a
Explanation: The value of the slot pitch serves as a guide when choosing the number of armature slots. The value of the slot pitch depends upon the voltage of the machine.


15. What is the value of the slot pitch for the low voltage machines?
a) slot pitch < 25 mm
b) slot pitch = 25 mm
c) slot pitch less than equal to 25 mm
d) slot pitch greater than equal to 25 mm
Answer: c
Explanation: The value of the slot pitch depends upon the voltage of the machine. The slot pitch is less than equal to 25 mm for low voltage machines.

Armature Design – 2

1. What is the range for the stator slot pitch for the large hydro-electric generators?
a) 50-60 mm
b) 50-70 mm
c) 50-80 mm
d) 50-90 mm
Answer: d
Explanation: The minimum value for the stator slot pitch for the large hydro-electric generators is 50 mm. The maximum value for the stator slot pitch for the large hydro-electric generators is 90 mm.


2. When is the range of the number of slots per pole per phase in the salient pole machines?
a) 2-3
b) 3-4
c) 2-4
d) 2-6
Answer: c
Explanation: The minimum value of the number of slots per pole per phase is 2. The maximum value of the number of slots per pole per phase is 4.


3. Fractional windings are invariably used in synchronous machines.
a) true
b) false
Answer: a
Explanation: Fractional slot windings reduces the distribution factor for higher harmonics thus reducing their corresponding generated emfs and making the voltage nearly sinusoidally. Fractional slot windings are invariably used in synchronous machines.


4. What is the relation between coil span and harmonics?
a) low coil span decreases harmonics to less amount
b) low coil span decreases the harmonics drastically
c) high coil span decreases the harmonics drastically
d) high coil span decreases the harmonics by small amount
Answer: b
Explanation: The coil span is kept low in order to decreases the harmonics. The advantage of having lower coil spans is that it reduces the harmonics drastically.


5. The coil span should be 8.33 percent of pole pitch to obtain the maximum reduction of harmonics.
a) true
b) false
Answer: a
Explanation: The coil span adjustment indirectly affects the harmonics reduction. The coil span should be minimum of about 8.33 percent of pole pitch to obtain the maximum reduction of harmonics.


6. When is the formula for the flux per pole?
a) flux per pole = average magnetic field * pole pitch * length of the core
b) flux per pole = average magnetic field / pole pitch * length of the core
c) flux per pole = average magnetic field * pole pitch / length of the core
d) flux per pole = 1 / average magnetic field * pole pitch * length of the core
Answer: a
Explanation: To obtain the value of flux per pole first the average magnetic field is obtained. Then the pole pitch and the length of the core is obtained to obtain the flux per pole.


7. What is the formula for the turns per phase in the armature design?
a) turns per phase = voltage per phase * parallel paths per phase / 4.44 * flux density * frequency * winding space factor
b) turns per phase = voltage per phase / parallel paths per phase * 4.44 * flux density * frequency * winding space factor
c) turns per phase = voltage per phase * parallel paths per phase * 4.44 * flux density * frequency * winding space factor
d) turns per phase = voltage per phase * parallel paths per phase * 4.44 * flux density / frequency * winding space factor
Answer: a
Explanation: For obtaining the turns per phase, the voltage per phase is obtained along with the parallel paths per phase. Next the winding space factor is calculated and the substitution in the formula gives the turns per phase.


8. What is the formula for current in each conductor?
a) current in each conductor = kVA * 103 * 3 * voltage per phase
b) current in each conductor = kVA / 103 * 3 * voltage per phase
c) current in each conductor = kVA * 103 / 3 * voltage per phase
d) current in each conductor = kVA * 103 * 3 / voltage per phase
Answer: c
Explanation: The kVA output is first obtained from the operation of the machine. Next the voltage per phase is calculated to obtain the current in each conductor.


9. What is the permissible current density in the armature conductors?
a) 3-4 A per mm2
b) 3-6 A per mm2
c) 4-6 A per mm2
d) 3-5 A per mm2
Answer: d
Explanation: The minimum permissible value of the current density in the armature conductors is 3 A per mm2. The maximum allowed value of the current density in the armature conductors is 6 A per mm2.


10. What is the formula for the area of cross section of armature conductors?
a) area of cross section = current per conductor * current density in the armature conductors
b) area of cross section = current per conductor + current density in the armature conductors
c) area of cross section = current per conductor – current density in the armature conductors
d) area of cross section = current per conductor / current density in the armature conductors
Answer: d
Explanation: For obtaining the area of the cross section the current per conductor is calculated. Next the current density is calculated and the ratio of both gives the current density of the area of cross section.

Design of Field Winding – 1

1. What type of coils are made use of for machines with small number of poles?
a) iron wound coils
b) wire wound coils
c) rectangular coils
d) square coils
Answer: b
Explanation: Different type of coils are being made use of for the machines as per the quantity of number of poles. For small number of poles the wire wound coils are made use of.


2. What type of strips is made use of for field coils of small alternators?
a) wood covered rectangular strips
b) bare copper strips
c) glass covered rectangular strips
d) iron strips
Answer: c
Explanation: The glass covered rectangular strips are made use of for the field coils of small alternator. The bare copper strips are being used are made use of for the field coils of large alternator.

 

Thermal Design Aspects of Electrical Machines & D.C. Machines MCQs




3. What should be the maximum width of the edge conductors used in the large alternators?
a) 6 mm
b) 5 mm
c) 4 mm
d) 3 mm
Answer: a
Explanation: The field coils of the large alternators use strip on edge winding wherein the bare copper strips are insulated from each other by interturn insulation. The width of the edge conductors does not exceed 6 mm.


4. For machines with Class B insulation, how many layers of inter turn insulation is made use of and what is the distance between the layers?
a) 4, 0.18 mm
b) 3, 0.25 mm
c) 2, 018 mm
d) 2, 0.25 mm
Answer: c
Explanation: The machines using Class B insulation makes use of 2 layers of inter turn insulation. The distance between the insulation is given to be 0.18 mm.


5. What material is the paper strips stuck with?
a) synthetic resin varnish
b) shellac
c) synthetic resin varnish and shellac
d) synthetic resin varnish or shellac
Answer: d
Explanation: The paper strips used in the Class B insulation machines are stuck on with shellac. They are also made of the synthetic resin varnish materials.


6. What is the thickness of the flanges and what material is used in the flanges?
a) 10 mm thick, resins
b) 10 mm thick, asbestos
c) 15 mm thick, asbestos
d) 10 mm thick, bakelized asbestos
Answer: d
Explanation: The flanges used in the Class B insulation is made up of 10 mm thickness. The flanges are made up of the bakelized asbestos.


7. Current is passed simultaneously through the conductors to raise the temperature of the field coil.
a) true
b) false
Answer: a
Explanation: The current is passed simultaneously through the conductors to raise the temperature of the field coil. The temperature should be high enough so that polymerization of interturn insulation is complete.


8. During the pressing and consolidation by how much is the thickness of the interturn insulation reduced to?
a) 0.36 mm to 0.26 mm
b) 0.36 mm to 0.25 mm
c) 0.30 mm to 0.25 mm
d) 0.32 mm to 0.25 mm
Answer: a
Explanation: The thickness of the interturn insulation before the pressing and consolidation is 0.36 mm. After the process of pressing and consolidation the thickness of the interturn insulation is reduced to 0.26 mm.


9. How many layers does the machine with Class F insulation consists of?
a) 2
b) 3
c) 4
d) 5
Answer: b
Explanation: The machines with Class B insulation consists of 2 layers. The machines with Class F insulation consists of 3 layers.


10. What is the thickness of the layers of Class F insulation and what material is layers made of?
a) 0.18 mm, asbestos paper
b) 0.10 mm, asbestos paper
c) 0.18 mm, thick epoxy treated asbestos paper
d) 0.10 mm, thick epoxy treated asbestos paper
Answer: c
Explanation: The thickness of the layers is 0.18 mm. The layers are made up of thick epoxy treated asbestos paper.


11. What is the lamination material of the pole body and the thickness of the pole body insulation?
a) epoxy resin, 5 mm thick
b) epoxy resin. 4 mm thick
c) asbestos, 4 mm thick
d) asbestos 5 mm thick
Answer: b
Explanation: The pole body insulation is made up of the epoxy resin laminations. The thickness of the pole body insulation is 4 mm thick.


12. What is the range of the pressure under which the field coils are consolidated?
a) 4-10 MN per m2
b) 3-10 MN per m2
c) 4-12 MN per m2
d) 4-15 MN per m2
Answer: c
Explanation: The minimum pressure under which the field coils are consolidated is 4 MN per m2. The maximum value of the pressure under which the field coils are 12 MN per m2.


13. What is the range of the exciter voltage in the field coils?
a) 50-100 V
b) 150-300 V
c) 200-400 V
d) 50-400 V
Answer: d
Explanation: The minimum value of the exciter voltage across the field coils is 50 V. The maximum value of the exciter voltage is 400 V.


14. The field winding should be designed for a voltage from 15-20% less than the exciter voltage.
a) true
b) false
Answer: a
Explanation: The field winding should be designed for a voltage from 15-20% less than the exciter voltage. This is because to allow for the drop in voltage between field and exciter and to allow for variations in the reluctance of the magnetic field.


15. What is the formula for the voltage across each field coil?
a) voltage across each field coil = (0.8-0.85)*exciter voltage/number of poles
b) voltage across each field coil = (0.8-0.85)*exciter voltage*number of poles
c) voltage across each field coil = (0.8-0.85)/exciter voltage*number of poles
d) voltage across each field coil = (0.8-0.85)/exciter voltage/number of poles
Answer: a
Explanation: The exciter voltage value is first set up for the particular machine. Then with the number of poles, the voltage across each field coil is calculated.

Design of Field Winding – 2

1. What is the formula for the winding height in the design of the field windings?
a) winding height = height of the pole – height of shoe + space taken by the spool, flanges, etc
b) winding height = height of the pole + height of shoe + space taken by the spool, flanges, etc
c) winding height = height of the pole + height of shoe – space taken by the spool, flanges, etc
d) winding height = height of the pole – height of shoe – space taken by the spool, flanges, etc
Answer: d
Explanation: The height of the pole and the height f the shoe is calculated. Next the space taken by the spool, flanges is calculated and the winding height is obtained.


2. What is the approximate value of the space taken by spools, flanges, etc?
a) 15 mm
b) 10 mm
c) 12 mm
d) 20 mm
Answer: d
Explanation: The value of the space taken by spools, flanges are required for the calculation of the winding height. The approximate value of the space taken by spools, flanges, etc. is 20 mm.


3. What is the winding depth for the pole pitch of 0.1 mm?
a) 25 mm
b) 35 mm
c) 45 mm
d) 50 mm
Answer: a
Explanation: The winding depth is 25 mm for the pole pitch of 0.1 mm. The winding depth is 35 mm for the pole pitch of 0.2 mm. The winding depth is 45 mm for the pole pitch is 0.3 mm.


4. What is the formula for the voltage across each field coil?
a) voltage across each field coil = field current * resistance of each field at 75°C
b) voltage across each field coil = field current / resistance of each field at 75°C
c) voltage across each field coil = field current + resistance of each field at 75°C
d) voltage across each field coil = field current – resistance of each field at 75°C
Answer: a
Explanation: The field current is first calculated from the machine. Next the resistance value of each field at 75°C is calculated and this gives the voltage across each field coil.


5. What is the range of the current density in the field conductors?
a) 3 to 5 A per mm2
b) 3 to 4 A per mm2
c) 4 to 5 A per mm2
d) 3 to 6 A per mm2
Answer: b
Explanation: The minimum range of the current density in the field conductors is 3 A per mm2. The maximum value of the current density in the field conductors is 4 A per mm2.


6. What is the formula for the field current of the synchronous machines?
a) field current = current density * area of conductors
b) field current = current density / area of conductors
c) field current = current density – area of conductors
d) field current = current density + area of conductors
Answer: a
Explanation: For the calculation of the field current, first the current density is calculated. Next, the area of conductors is calculated and the field current is calculated.


7. What is the formula for the number of field turns of the field windings?
a) number of field turns = field mmf per pole at full load * field current
b) number of field turns = field mmf per pole at full load / field current
c) number of field turns = field mmf per pole at full load + field current
d) number of field turns = field mmf per pole at full load – field current
Answer: b
Explanation: The field mmf per pole at full load is calculated from the voltage across each field coil. Next, the field current is calculated and from these values the number of field turns is calculated.


8. What is the relation between winding space and the depth?
a) winding space is directly proportional to the depth
b) winding space is indirectly proportional to the depth
c) winding space is directly proportional to the square of the depth
d) winding space is indirectly proportional to the square of the depth
Answer: b
Explanation: The winding space is indirectly proportional to the depth. If the winding space is less, then the depth is increased.


9. What is the formula of the resistance of the winding is calculated at 75°C?
a) resistance of the winding = (Number of field turns * pole proportion * length of mean turns of the coil) / area of the field conductors
b) resistance of the winding = (Number of field turns * pole proportion * length of mean turns of the coil) * area of the field conductors
c) resistance of the winding = (Number of field turns / pole proportion * length of mean turns of the coil) / area of the field conductors
d) resistance of the winding = (Number of field turns * pole proportion / length of mean turns of the coil) / area of the field conductors
Answer: a
Explanation: The number of field turns is calculated along with the pole proportion. The length of mean turns of the coil and the area of the field conductors is calculated and on substituting the values the resistance of the winding is obtained.


10. What is the formula of the dissipating surface of the coil?
a) dissipating surface of the coil = 2*length of mean turns of the coil*(winding height * diameter of winding)
b) dissipating surface of the coil = 2*length of mean turns of the coil*(winding height / diameter of winding)
c) dissipating surface of the coil = 2*length of mean turns of the coil*(winding height + diameter of winding)
d) dissipating surface of the coil = 2*length of mean turns of the coil/(winding height * diameter of winding)
Answer: c
Explanation: First the length of the mean turns of the coil is calculated. Then the winding height and the diameter of the winding is calculated to obtain the dissipating surface of the coil.


11. What is the formula for the cooling co-efficient to the rotating field coils?
a) cooling coefficient of rotating field coils = 0.05 to 0.08 / 1 + armature voltage
b) cooling coefficient of rotating field coils = 0.05 to 0.08 / 1 – armature voltage
c) cooling coefficient of rotating field coils = 0.08 to 0.12 / 1 + armature voltage
d) cooling coefficient of rotating field coils = 0.08 to 0.12 / 1 – armature voltage
Answer: c
Explanation: The armature voltage is first calculated for the calculation of the cooling coefficient of rotating field coils. The cooling coefficient is used to calculate the temperature rise.


12. What is the formula for the temperature rise in the design of field windings?
a) temperature rise = 1 / copper loss in each field coil at 75°C * cooling coefficient of rotating field coils * dissipating surface of the coil
b) temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils * dissipating surface of the coil
c) temperature rise = copper loss in each field coil at 75°C / cooling coefficient of rotating field coils * dissipating surface of the coil
d) temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils / dissipating surface of the coil
Answer: d
Explanation: The copper loss in each field coil is first calculated using its formula. Next, the cooling coefficient of rotating field coils is calculated. Finally dissipating surface of the coil is calculated and this gives the temperature rise.


13. If the temperature increases beyond the acceptable limits the depth of the winding should be decreased.
a) true
b) false
Answer: b
Explanation: The temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils / dissipating surface of the coil is calculated. If the temperature rise crosses the specified limits, the depth of the winding is increased.


14. The increase in the depth of the winding increases the heat dissipating surface.
a) true
b) false
Answer: a
Explanation: The increase in the depth of the windings increase the heat dissipating surface. The increase in the heat dissipation decreases the temperature rise.


15. What is the minimum clearance between adjacent field coils and pole drawing?
a) 14 mm
b) 15 mm
c) 13 mm
d) 12 mm
Answer: b
Explanation: The final step in the design of field windings is the checking of the clearance between the adjacent field coils and pole drawing. The minimum value of clearance should be 15 mm.

Design of Rotor – 1

1. How many factors does the design of rotor of synchronous machines depend upon?
a) 2
b) 3
c) 4
d) 5
Answer: c
Explanation: There are 4 factors which are associated with the design of rotor in the synchronous machines. They are height of pole, design of damper windings, height of pole shoe, pole profile drawing.


2. What is the formula for the flux in pole body?
a) flux in pole body = leakage coefficient * useful flux per pole
b) flux in pole body = leakage coefficient / useful flux per pole
c) flux in pole body = leakage coefficient – useful flux per pole
d) flux in pole body = leakage coefficient + useful flux per pole
Answer: a
Explanation: The leakage coefficient is obtained first from its formula. Next, the value of useful flux per pole is calculated and this gives the flux in pole body value.


3. What is the range of the permissible values of the flux densities in pole body?
a) 1.4-1.7 Wb per m2
b) 1.5-1.7 Wb per m2
c) 1.4-1.6 Wb per m2
d) 1.5-1.6 Wb per m2
Answer: b
Explanation: The minimum value of the flux density in the pole body is given to be 1.5 Wb per m2.The maximum permissible value of the flux density in the pole body is given to be 1.7 Wb per m2.


4. What is the range of the leakage coefficient in the pole body?
a) 1.1 to 1.2
b) 1.00 to 1.5
c) 1.15 to 1.2
d) 0.75 to 2.3
Answer: c
Explanation: The minimum value of the leakage coefficient in the pole body is 1.15. The maximum value of the leakage coefficient in the pole body is 1.2.


5. What is the formula for the area of cross-section of pole body for rectangular poles?
a) area of cross section of pole body = 0.98 * axial length of the pole * breadth of the pole
b) area of cross section of pole body = 0.98 / axial length of the pole * breadth of the pole
c) area of cross section of pole body = 0.98 * axial length of the pole / breadth of the pole
d) area of cross section of pole body = 1/0.98 * axial length of the pole * breadth of the pole
Answer: a
Explanation: The axial length of the pole and the breadth of the pole are calculated. Next by multiplying the two values with the stacking factor, we get the area of cross section of pole body.


6. What is the formula for the copper area of the field windings?
a) copper area = full load field mmf * current density in the field winding
b) copper area = full load field mmf / current density in the field winding
c) copper area = full load field mmf + current density in the field winding
d) copper area = full load field mmf – current density in the field winding
Answer: b
Explanation: For the calculation of the copper area, first the current density in the field winding is calculated. Next the full load field mmf is calculated and the ratio gives the copper area of field windings.


7. What is the formula for the total space required for the winding?
a) total space = copper area + space factor
b) total space = copper area – space factor
c) total space = copper area / space factor
d) total space = copper area * space factor
Answer: c
Explanation: The copper area is calculated from its respective formula. Then the space factor is calculated and the ratio gives the value of total space.


8. What is the value of space factor for the strip on edge winding?
a) 0.8-0.9
b) 0.4
c) 0.65
d) 0.75
Answer: a
Explanation: The space factor for the strip on edge winding is 0.8-0.9. The space factor for small round wires is 0.4 and for large round wires it is 0.65. The space factor for large rectangular conductors is 0.75.


9. What is the formula for the height of winding?
a) height of winding = total winding area / depth of winding
b) height of winding = total winding area * depth of winding
c) height of winding = total winding area + depth of winding
d) height of winding = total winding area – depth of winding
Answer: a
Explanation: The total winding area is first calculated. Next the depth of the winding is calculated. The ratio of both gives the height of winding.


10. What is the formula for the radial length of the pole shoe?
a) radial length of the pole shoe = height of winding – height of pole shoe – 0.02
b) radial length of the pole shoe = height of winding + height of pole shoe – 0.02
c) radial length of the pole shoe = height of winding – height of pole shoe + 0.02
d) radial length of the pole shoe = height of winding + height of pole shoe + 0.02
Answer: d
Explanation: First the height of the winding is calculated from its formula. Next the height of pole shoe is calculated. Both the values are added with 0.02 to give the radial length of the pole shoe.


11. What is the formula for the height of pole body?
a) height of pole body = height of the winding + 0.02
b) height of pole body = height of the winding * 0.02
c) height of pole body = height of the winding – 0.02
d) height of pole body = height of the winding / 0.02
Answer: a
Explanation: The height of the pole body is one of the design factors in the design of rotor. It is obtained by adding the value of the height of winding with 0.02, which is the approximate space occupied by flanges.


12. What is the range of the ratio of radial length of pole to pole pitch?
a) 0.3-1
b) 0.3-1.5
c) 0.7-1
d) 0.7-1.5
Answer: b
Explanation: The minimum value of the ratio of radial length of pole to pole pitch is given to be 0.3. The maximum value of the ratio of radial length of pole to pole pitch is given to be 1.5.


13. The damper windings are made use of in synchronous generators to reduce the oscillations and to prevent hunting.
a) true
b) false
Answer: a
Explanation: The purpose of the damper windings is to reduce the oscillations and to prevent the hunting in synchronous generators. Next, the damper windings are used to suppress the negative sequence field in the synchronous generator.


14. The mmf of the damper windings depends on the pole pitch value.
a) true
b) false
Answer: a
Explanation: The mmf of the damper windings depends on the pole pitch value. The value for the mmf of the damper windings = 0.143 * specific electric loading * pole pitch.


15. What is the formula for the area per pole of damper pass provided?
a) area per pole of damper pass = 0.2 * specific electric loading * pole pitch * current density in damper bars
b) area per pole of damper pass = 0.2 * specific electric loading * pole pitch / current density in damper bars
c) area per pole of damper pass = 0.2 * specific electric loading – pole pitch / current density in damper bars
d) area per pole of damper pass = 0.2 + specific electric loading * pole pitch / current density in damper bars
Answer: b
Explanation: The specific electric loading and the pole pitch is calculated first. Next the current density in damper bars is next calculated. Substituting in the above formula gives the area per pole of damper pass provided.

Design of Rotor – 2

1. What is the range of current density in the damper bars?
a) 3-4 A per mm2
b) 3-5 A per mm2
c) 3-6 A per mm2
d) 4-6 A per mm2
Answer: a
Explanation: The minimum value of the current density in the damper bars is given to be 3 A per mm2. The maximum value of the current density in the damper bars is given to be 4 A per mm2.


2. What is the percentage of the damper windings slot pitch with respect to stator slot pitch?
a) 30%
b) 40%
c) 20%
d) 60%
Answer: c
Explanation: The slot pitch of damper windings is 20% of the stator slot pitch. This is because to reduce the current induced in damper windings by tooth ripples.


3. What is the formula for the pole arc?
a) pole arc = number of bars per pole * stator slot pitch * 0.8
b) pole arc = number of bars per pole / stator slot pitch * 0.8
c) pole arc = number of bars per pole * stator slot pitch / 0.8
d) pole arc = 1 / number of bars per pole * stator slot pitch * 0.8
Answer: a
Explanation: The number of bars per pole is calculated along with the stator slot pitch. Next, all the values are multiplied by 0.8 and the pole arc value is obtained.


4. What is the formula for the length of each damper bar for small machines?
a) length of each damper bar = 1.1 * axial length
b) length of each damper bar = axial length + 0.1
c) length of each damper bar = axial length – 0.1
d) length of each damper bar = 1.1 / axial length
Answer: a
Explanation: Length of each damper bar = 1.1 * axial length is the formula for the length of each damper bar for small machines. length of each damper bar = axial length + 0.1 is the formula for the length of each damper bar for large machines.


5. What is the formula for the area of cross-section of each damper bar?
a) area of cross section of each damper bar = total area of bars per pole – number of damper bars per pole
b) area of cross section of each damper bar = total area of bars per pole + number of damper bars per pole
c) area of cross section of each damper bar = total area of bars per pole / number of damper bars per pole
d) area of cross section of each damper bar = total area of bars per pole * number of damper bars per pole
Answer: c
Explanation: The total area of bars per pole is first obtained. Next the number of damper bars per pole is calculated and the ratio gives the area of cross section of each damper bar.


6. What is the formula of the area of each ring short-circuiting the bars?
a) area of each ring short-circuiting the bars = (0.7-0.9) * area of damper bar
b) area of each ring short-circuiting the bars = (0.8-1) * area of damper bar
c) area of each ring short-circuiting the bars = (0.7-1) * area of damper bar
d) area of each ring short-circuiting the bars = (0.8-0.9) * area of damper bar
Answer: b
Explanation: First the area of damper bars is calculated first. Next the area of damper bars is multiplied by a value between 0.8-1 to obtain the area of each ring short-circuiting the bars.


7. Given : total area = 473 mm2 and Number of bars = 8 for a rotor design, what is the value of area of each damper bar?
a) 59 mm2
b) 455 mm2
c) 475 mm2
d) 3784 mm2
Answer: a
Explanation: Area of each bar = total area / number of bars
Area of each bar = 473/8 = 59 mm2.


8. What is the formula for the height of pole shoe sufficient to accommodate the damper windings?
a) height of pole shoe = diameter of damper bars
b) height of pole shoe = 2 * diameter of damper bars
c) height of pole shoe = diameter of damper bars/2
d) height of pole shoe = 3 * diameter of damper bars/2
Answer: b
Explanation: First the diameter of the damper bars is calculated. It is then multiplied by 2 to get the height of pole shoe.


9. Pole profile drawing helps in obtaining the various dimensions of the pole.
a) true
b) false
Answer: a
Explanation: The pole profile drawing is essential to obtain the design characteristics of the rotor. It helps in the process of obtaining the various dimensions of the pole.


10. The pole shoe drawing is completed by fixing the height of pole shoe.
a) true
b) false
Answer: a
Explanation: The pole shoe surface can be drawn with the armature surface being fixed. The pole shoe drawing is completed by fixing the height of pole shoe.

Losses and Temperature Rise

1. How many types of losses are present in synchronous machines?
a) 7
b) 3
c) 4
d) 5
Answer: a
Explanation: There are 7 losses in the synchronous machines. They are i) iron loss due to main field, ii) iron loss due to parasitic field, iii) I2R loss in the armature winding, iv) eddy current loss in armature conductors, v) stray load loss, vi) loss in field windings, vii) friction and windage loss.


2. What is the classification of the iron loss due to the main field?
a) hysteresis loss
b) eddy current loss
c) hysteresis loss or eddy current loss
d) hysteresis loss and eddy current loss
Answer: d
Explanation: The iron loss due to main field is due to the hysteresis loss. The eddy current loss also contribute to the iron losses due to main field.


3. What are the factors the pole face loss depends upon?
a) slot opening
b) air gap length
c) number of slots and speed of machines
d) slot opening, air gap length, number of slots and speed of machines
Answer: d
Explanation: The pole face loss depends upon the slot opening and the air gap length. It also depends on the number of slots and speed of machines.


4. What is the range of the pole face loss in the synchronous machines?
a) 40-60 % of iron loss
b) 20-60 % of iron loss
c) 25-70 % of iron loss
d) 40-80 % of iron loss
Answer: c
Explanation: The pole face loss has a minimum value of 25% of the iron loss. The pole face loss has a maximum value of 70% of the iron loss.


5. What is the formula for the copper loss in the synchronous machine?
a) copper loss per phase = current per phase * dc resistance
b) copper loss per phase = current per phase2 * dc resistance2
c) copper loss per phase = current per phase2 * dc resistance
d) copper loss per phase = current per phase * dc resistance2
Answer: c
Explanation: First the current per phase is calculated and the value is squared. Next the dc resistance is calculated and the sum of the square of the current per phase and dc resistance gives the copper loss per phase.


6. What is the formula for the total eddy current loss in conductors?
a) total copper loss = 3 * average value of the eddy current constant * current per phase2 * dc resistance
b) total copper loss = 3 / average value of the eddy current constant * current per phase2 * dc resistance
c) total copper loss = 3 * average value of the eddy current constant / current per phase2 * dc resistance
d) total copper loss = 3 * average value of the eddy current constant * current per phase2 / dc resistance
Answer: a
Explanation: The average value of the eddy current constant is obtained. Next the I2R loss values are calculated and multiplying with 3 gives the total copper loss.


7. What is the cause of the stray load losses in the synchronous machine?
a) stray field
b) stray armature
c) stray field and stray armature
d) stray field or stray armature
Answer: a
Explanation: The stray load loss occurs due to stray fields. They are formed when the machine is being loaded.


8. What is the voltage drop in the carbon and graphite brushes?
a) 1 V
b) 0.3 V
c) 0.6 V
d) 0.75 V
Answer: a
Explanation: The voltage drop in the carbon and graphite brushes is 1 V. The voltage drop in the brushes containing metal is 0.3 V.


9. What factors does the friction and windage loss depend upon?
a) construction of the machine
b) speed of the machine
c) rating of the machine
d) construction, speed, rating of the machine
Answer: d
Explanation: This loss consists of the bearing friction and rotor windage loss. The loss depends upon the type of construction, speed and ratings of the machines.


10. What is the reduction in the total friction loss with the hydrogen cooling?
a) 0.3-0.5 % of kVA rating
b) 0.2-0.3 % of kVA rating
c) 0.3-0.4 % of kVA rating
d) 0.3-0.6 % of kVA rating
Answer: c
Explanation: The friction loss depends upon the type of construction, speed and ratings of the machines. The hydrogen cooling reduces the total friction loss by 0.3-0.4% of the kVA rating.


11. What is the formula to obtain the temperature rise of the surface?
a) temperature rise of the surface = Surface area * cooling coefficient * dissipating surface
b) temperature rise of the surface = Surface area / cooling coefficient * dissipating surface
c) temperature rise of the surface = Surface area * cooling coefficient / dissipating surface
d) temperature rise of the surface =1 / Surface area * cooling coefficient * dissipating surface
Answer: c
Explanation: The surface area is first calculated from its formula. Next, the cooling coefficient and the dissipating surface are obtained and on substitution gives the temperature rise of the surface.


12. What factor/s does the cooling coefficient depend upon?
a) speed of the cooling medium
b) configuration of the surface
c) speed of the machine and configuration of the surface
d) speed of the machine or configuration of the surface
Answer: c
Explanation: The cooling coefficient depends upon the speed of the machine. The cooling coefficient also depends upon the configuration of the surface.


13. The value of the cooling coefficient varies from 0.025 to 0.04 in the back of the stator core.
a) true
b) false
Answer: a
Explanation: The cooling coefficient value is required in the calculation of the temperature rise of the surface. The value varies from 0.025 to 0.04 for the back of the stator core.


14. The peripheral speed is the armature peripheral speed in the stationary field coils.
a) true
b) false
Answer: a
Explanation: There are various peripheral speeds in various parts of the machine. In the stationary field coils the peripheral speed is nothing but the armature peripheral speed.


15. What all factors does the heat to be dissipated by cooling surfaces depend upon?
a) hysteresis loss
b) eddy current loss
c) heating loss
d) hysteresis, eddy and heating losses
Answer: d
Explanation: The heat to be dissipated by the cooling surface of the armature core would consist of the hysteresis loss and the eddy current loss. It also consists of the heating loss or the I2R in the active part of the armature.

Synchronous Machines Output Equation

1. What is the formula for output equations in synchronous machines?
a) kVA output = output coefficient * diameter2 * length * synchronous speed
b) kVA output = output coefficient / diameter2 * length * synchronous speed
c) kVA output = output coefficient * diameter2 / length * synchronous speed
d) kVA output = output coefficient * diameter2 * length / synchronous speed
Answer: a
Explanation: The output equation is found out by first calculating the output coefficient. Next, the diameter and length are obtained, and the synchronous speed is calculated using the tacho-generator to obtain the kVA output.


2. What is the formula of the output coefficient?
a) output coefficient = 11 * specific magnetic loading / specific electrical loading * winding space factor * 10-3
b) output coefficient = 11 / specific magnetic loading * specific electrical loading * winding space factor * 10-3
c) output coefficient = 11 * specific magnetic loading * specific electrical loading * winding space factor * 10-3
d) output coefficient = 11 * specific magnetic loading * specific electrical loading / winding space factor * 10-3
Answer: c
Explanation: The output coefficient is one of the terms required in the calculation of the output of the machine. The specific magnetic and electrical loading terms are first calculated along with the winding space factor.


3. What is the formula for the output equation with respect to the peripheral speed?
a) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor * 10-3 * peripheral speed2 *Length * synchronous speed
b) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor * 10-3 * peripheral speed2 *Length / synchronous speed
c) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor / 10-3 * peripheral speed2 *Length * synchronous speed
d) output = 1.110 / specific magnetic loading * specific electrical loading * winding space factor * 10-3 * peripheral speed2 *Length * synchronous speed
Answer: b
Explanation: The output equation with respect to the peripheral speed depends on the square of the peripheral speed of the machine. It doesn’t consist of the diameter term in the output equation.


4. How many factors does the choice of specific magnetic loading depend upon?
a) 4
b) 2
c) 5
d) 8
Answer: c
Explanation: The choice of specific magnetic loading depends upon 5 factors basically. They are a) Iron Loss, b) Voltage, c) Transient short circuit current, d) Stability, e) Parallel Operation.


5. How is the iron loss related with the choice of specific magnetic loading?
a) choice of magnetic loading is directly proportional to the iron loss
b) choice of magnetic loading is indirectly proportional to the iron loss
c) choice of magnetic loading is directly proportional to the square of the iron loss
d) choice of magnetic loading is indirectly proportional to the square of the iron loss
Answer: a
Explanation: The choice of specific magnetic loading is directly proportional to the iron loss. The iron loss increases with the increase in the air gap density.


6. How is the voltage related with the air gap density?
a) air gap density is directly proportional to the voltage
b) air gap density is indirectly proportional to the voltage
c) air gap density is directly proportional to the square of the voltage
d) air gap density is indirectly proportional to the square of the voltage
Answer: b
Explanation: The air gap density is indirectly proportional to the voltage. High voltage machine should have low air gap density, to avoid excessive values of flux density in the teeth and core.


7. How is the transient short circuit current related with the air gap density?
a) air gap density is directly proportional to the short circuit current
b) air gap density is indirectly proportional to the short circuit current
c) air gap density is directly proportional to the square of the short circuit current
d) air gap density is directly proportional to the square of the short circuit current
Answer: a
Explanation: The air gap density is directly proportional to the short circuit current. The air gap density should be kept low in order to reduce the initial electromagnetic forces under short circuit condition.


8. How is the steady state stability related with the air gap density?
a) air gap density is directly proportional to the steady state stability
b) air gap density is indirectly proportional to the steady state stability
c) air gap density is directly proportional to the square of the steady state stability
d) air gap density is directly proportional to the square of the steady state stability
Answer: a
Explanation: The air gap density is directly proportional to the steady state stability. The steady state stability is improved if the air gap density is high.


9. The machines having high air gap density operates poorly when connected in synchronism.
a) true
b) false
Answer: b
Explanation: The machines having high air gap density allows high amount of synchronizing power. Thus the machines having high air gap density provides high synchronism.


10. What is the range of the air gap density for salient pole machines?
a) 0.52-0.65 Wb per m2
b) 0.5-0.6 Wb per m2
c) 0.54-0.65 Wb per m2
d) 0.44-0.65 Wb per m2
Answer: a
Explanation: The range of air gap density for salient pole machines is 0.52-0.65 Wb per m2. The range of air gap density for turbo-alternators is 0.54-0.65 Wb per m2.


11. How many factors influence the choice of specific electric loading?
a) 2
b) 3
c) 4
d) 5
Answer: c
Explanation: There are 4 factors that influence the choice of specific electric loading. They are a) Copper loss and temperature rise, b) voltage, c) synchronous reactance, d) stray load loss.


12. How is the specific electric loading related to copper losses and temperature rise?
a) high specific electric loading gives high copper losses and high temperature rise
b) high specific electric loading gives low copper losses and high temperature rise
c) high specific electric loading gives high copper losses and low temperature rise
d) high specific electric loading gives low copper losses and low temperature rise
Answer: a
Explanation: The specific electric loading is directly proportional to the copper losses and the temperature rise. The high specific electric loading gives high copper losses and high temperature rise.


13. High value of the specific electric loading can be used for low voltage machines.
a) true
b) false
Answer: a
Explanation: High value of specific electric loading can be used for low voltage machines. This is because the space required for insulation is small.


14. How is the specific electric loading related to the synchronous reactance of the machines?
a) specific electric loading is high, leakage reactance is high, giving low synchronous reactance
b) specific electric loading is high, leakage reactance is low, giving low synchronous reactance
c) specific electric loading is high, leakage reactance is high, giving high synchronous reactance
d) specific electric loading is low, leakage reactance is high, giving high synchronous reactance
Answer: c
Explanation: The specific electric loading is directly proportional to the synchronous reactance. If the specific electric loading is high, the synchronous reactance becomes high.


15. What is the value of specific electric loading for the salient pole alternators?
a) 20,000-40,000 A per m
b) 50,000-75,000 A per m
c) 25,000-40,000 A per m
d) 20,000-45,000 A per m
Answer: a
Explanation: The value of specific electric loading for the salient pole alternators is 20,000-40,000 A per m. The value of specific electric loading for the turbo alternators is 50,000-75,000 A per m.

Slot Dimensions

1. What is the range of the flux density in the teeth at no load?
a) 1.7-1.8 Wb per mm2
b) 1.3-1.5 Wb per mm2
c) 1.3-1.6 Wb per mm2
d) 1.4-1.6 Wb per mm2
Answer: a
Explanation: The minimum value of the flux density in the teeth is given to be 1.7 Wb per mm2. The maximum value of the flux density in the teeth is given to be 1.8 A per mm2.


2. What is the formula for the minimum width of the tooth?
a) minimum width of tooth = flux * pole proportion * (number of stator slots / number of poles) * length * 1.8
b) minimum width of tooth = flux / pole proportion * (number of stator slots / number of poles) * length * 1.8
c) minimum width of tooth = flux * pole proportion / (number of stator slots / number of poles) * length * 1.8
d) minimum width of tooth = flux * pole proportion * (number of stator slots / number of poles) * length / 1.8
Answer: b
Explanation: The flux value, pole proportion and the length values are first obtained. Then the ratio of the number of stator slots to number of poles is obtained and on substitution gives the minimum width of tooth.


3. Name the slots that are commonly used.
a) parallel sided
b) square sided
c) rectangular
d) circular
Answer: a
Explanation: The most commonly used type of slots are parallel sided. The other type of slots may be used for the purposes required.


4. How is the teeth and the minimum width designed in the machines?
a) teeth is tapered and minimum width is across the medium
b) teeth is sharpened and minimum width occurs across the air gap
c) teeth is widened and minimum width occurs across the air gap
d) teeth is reduced and minimum width occurs across the medium
Answer: a
Explanation: Parallel sided slots are made use of. Hence, the teeth is tapered and their minimum width occurs at the air gap surface.


5. What is the formula for the maximum permissible width of slot?
a) maximum permissible width = slot pitch * minimum width of the teeth
b) maximum permissible width = slot pitch + minimum width of the teeth
c) maximum permissible width = slot pitch / minimum width of the teeth
d) maximum permissible width = slot pitch – minimum width of the teeth
Answer: d
Explanation: The slot pitch is first calculated with its respective formula. Next, the minimum width of the teeth is calculated and the difference between both gives the maximum permissible width.


6. By how much should the depth of slot not exceed the width?
a) two times
b) three times
c) four times
d) six times
Answer: b
Explanation: The depth of the slot depends upon the width of the slot. The depth should not exceed three times the width of the slot.


7. Why are slot made deeper in the machine?
a) to increase the short circuit current
b) to reduce the short circuit current
c) to increase the open circuit current
d) to reduce the open circuit current
Answer: b
Explanation: The slots used in the machine are basically deeper slots. The slots are made deeper even more to increase the leakage reactance and to limit the short circuit current.


8. What is the formula for the height of length of mean turn of armature?
a) length of mean turn = 2*length + 2.5*pole pitch + 0.06 kV + 0.2
b) length of mean turn = 2*length + 2*pole pitch + 0.06 kV + 0.2
c) length of mean turn = 2*length + 2.5*pole pitch – 0.06 kV – 0.2
d) length of mean turn = 2*length – 2.5*pole pitch – 0.06 kV – 0.2
Answer: a
Explanation: The length of the slots is obtained along with the pole pitch. The output kV is calculated and on substituting we get the length of mean turn.


9. The flux density in the armature core of salient pole machines lies between 1-1.2 Wb per m2.
a) true
b) false
Answer: a
Explanation: The value of depth of core can be calculated by assuming a suitable value of flux density. The value of the flux density varies from 1-1.2 Wb per m2.


10. What is the formula for the depth of armature core?
a) depth of armature core = flux / length of the iron core * flux density
b) depth of armature core = flux * length of the iron core * flux density
c) depth of armature core = flux / 2 * length of the iron core * flux density
d) depth of armature core = flux * 2 * length of the iron core * flux density
Answer: c
Explanation: The flux value is calculated along with the length of the iron core. Next, the suitable flux density is chosen and the depth of armature core is calculated.


11. What is the formula for the outer diameter of the stator?
a) outer diameter = inner diameter + depth of the slots + depth of armature core
b) outer diameter = inner diameter + 2*depth of the slots + depth of armature core
c) outer diameter = inner diameter + 2*(depth of the slots + depth of armature core)
d) outer diameter = inner diameter + depth of the slots + 2*depth of armature core
Answer: c
Explanation: The depth of the armature core is calculated and the depth of the slots is also calculated. The inner diameter is calculated and substituting the outer diameter is obtained.

Design of Synchronous Machines MCQs ( Design Of Electrical Machines ) MCQs – Design Of Electrical Machines MCQs

 

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