Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) MCQs – Design Of Electrical Machines MCQs
Latest Design Of Electrical Machines MCQs
By practicing these MCQs of Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to “Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Design Of Electrical Machines to approach the success. Tab this page to check “Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.
Design Of Electrical Machines MCQs – Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) MCQs
The most occurred mcqs of Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) in past papers. Past papers of Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) Mcqs. Past papers of Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) Mcqs. The Important series of Aspects of Design of Mechanical Parts MCQs ( Design Of Electrical Machines ) Mcqs are given below:
Design of Shaft
1. What is the consideration for the determination of the diameter of shaft?
a) stiffness
b) voltage
c) current
d) rigidity
Answer: c
Explanation: The main aspect for the design of the diameter of the shaft is the stiffness. The diameter of the shaft depends on the stiffness of the machine.
2. What is the meaning of stiffness?
a) ability to transmit the power
b) ability to withstand the weight of the rotor
c) ability to withstand unbalanced magnetic pull
d) ability to withstand the weight of rotor and unbalanced magnetic pull
Answer: d
Explanation: The diameter of shaft for an electrical machine is determined by considerations of stiffness. The stiffness is the ability to withstand the weight of rotor and unbalanced magnetic pull.
3. What should be the first property of the shaft design?
a) the shaft design should be such that the shaft must have enough corrosion resistance
b) the shaft design should be such that the shaft must have enough mechanical strength
c) the shaft design should be such that the shaft has enough tensile strength
d) the shaft design should be able to withstand the voltage fluctuations
Answer: b
Explanation: The shaft design should be such that the shaft must have enough mechanical strength. The strength should be such that it should withstand all loads without causing much residual strain.
4. What is the second property of the shaft design?
a) the shaft design should be such that it has high rigidity
b) the shaft design should be such that it should have high tensile strength
c) the shaft design should be such that it should have high corrosion resistance
d) the shaft design should be such that it should withstand voltage fluctuations
Answer: a
Explanation: The second property of the shaft design such that it should have high rigidity. The rigidity should be such that the deflection of shaft under operation of machine does not reach such a dangerous value as to cause the rotor to touch the stator.
5. The critical speeds of rotation should be different from running speed of machine.
a) true
b) false
Answer: a
Explanation: The critical speed relation is the third property of the shaft design. The critical speeds of rotation should be different from the running speed of the machine.
6. What is the formula of the diameter of the shaft?
a) diameter of the shaft = 5.5 + (output in watt/rps)1/3 mm
b) diameter of the shaft = 5.5 – (output in watt/rps)1/3 mm
c) diameter of the shaft = 5.5 * (output in watt/rps)1/3 mm
d) diameter of the shaft = 5.5 / (output in watt/rps)1/3 mm
Answer: c
Explanation: The output is first calculated from the operation of the machine. Next, the tachogenerator is used to calculate the speed of the machine and on the substitution of the values gives the diameter of the shaft.
7. What is the relation of the diameter of the shaft in the bearings to the diameter under the armature?
a) diameter of the shaft is very much greater than the diameter under the armature
b) diameter of the shaft is greater than the diameter under the armature
c) diameter of the shaft is equal to the diameter under the armature
d) diameter of the shaft is lesser than the diameter under the armature
Answer: d
Explanation: There are certain rules in the design of the shaft. The diameter of the shaft in the bearings is less than the diameter under the armature.
8. What happens when the diameter under armature is 150 mm or above?
a) diameter of the shaft in bearing is 100 mm smaller than the maximum diameter
b) diameter of the shaft in bearing is 90 mm smaller than the maximum diameter
c) diameter of the shaft in bearing is 70 mm smaller than the maximum diameter
d) diameter of the shaft in bearing is 50 mm smaller than the maximum diameter
Answer: d
Explanation: The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter of the shaft in bearing is 50 mm smaller than the maximum diameter.
9. What happens in the case of the small shafts?
a) the diameter in the bearings should be about 1/3 of the maximum diameter
b) the diameter in the bearing should be about 2/3 of the maximum diameter
c) the diameter in the bearing should be about 2/5 of the maximum diameter
d) the diameter in the bearing should be about 1/5 of the maximum diameter
Answer: b
Explanation: The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter in the bearing should be 2/3 of the maximum diameter.
Bearings
1. What bearing is made use of in the horizontal shaft machines?
a) plain bearing
b) thrust bearing
c) push bearing
d) throw bearing
Answer: a
Explanation: The plain bearing is made use of in the horizontal shaft machine. The thrust bearing are used for vertical shaft machine.
2. What is the plain bearing used?
a) sleeve bearings
b) anti friction bearing
c) sleeve or anti friction bearing
d) sleeve and anti friction bearing
Answer: c
Explanation: The plain bearing can be the sleeve bearings made use of. It can also be friction bearings used which is either ball or roller bearings.
3. In horizontal shaft machines, the forces acting in which direction is prominent?
a) circular
b) vertical
c) horizontal
d) radial
Answer: d
Explanation: Plain bearings are made use of for the horizontal shaft machines. The forces acting in the radial direction are most prominent.
4. In vertical shaft machines, which load is taken up by the thrust bearings?
a) perpendicular load acting upwards
b) perpendicular load acting downwards
c) axial load acting upwards
d) axial load acting downwards
Answer: d
Explanation: The thrust bearings are made use of for the vertical shaft machines. In vertical shaft machines, the axial load acting downwards is taken up by the thrust bearings.
5. How are the radial loads caused?
a) dynamic unbalance of rotor
b) unbalanced magnetic pull of rotor
c) dynamic unbalance of rotor or unbalanced magnetic pull of rotor
d) dynamic unbalance of rotor and unbalanced magnetic pull of rotor
Answer: c
Explanation: The radial loads are caused by the dynamic unbalance of rotor. It can also be caused by the unbalanced magnetic pull of rotor.
6. What is the additional setup provided to the simple thrust bearing when it is not able to take up radial loads?
a) ball bearing
b) gear bearing
c) steel bearing
d) guide bearing
Answer: d
Explanation: Sometimes the simple thrust bearing wont be able to pull the radial loads. Therefore an additional bearing called guide bearing is provided.
7. How many guide bearings are used along with the simple thrust bearings to pick up radial loads?
a) 2
b) 3
c) 2 or 3
d) 4
Answer: c
Explanation: The simple thrust bearing cannot pull the radial load and hence guide bearings are used in addition to it. Usually 2-3 guide bearings are used in vertical shaft machines depending upon the load.
8. Where are the bearings for the horizontal shaft placed?
a) outside the machine
b) in the end shields of the machine
c) outside the machine or in the end shields of the machine
d) outside the machine and in the end shields of the machine
Answer: c
Explanation: The bearings for the horizontal shaft machines are being placed outside the machine. It can also be placed along the end shields of the machine.
9. What is the other name of the bearings outside the machine?
a) ball bearing
b) pedestal bearing
c) guide bearing
d) gear bearing
Answer: b
Explanation: The bearings for the horizontal shaft machines are placed outside the machine or in the end shields of the machine. The other name for the bearings outside the machine is pedestal bearing.
10. The phosphor bronze sleeve bearings are used for the small electrical machines.
a) true
b) false
Answer: a
Explanation: The phosphor bronze sleeve bearings are used for the small electrical machines. The diameter of the machines are between 50-60 mm.
11. Anti-friction bearings are lubricated by charcoal.
a) true
b) false
Answer: b
Explanation: Anti-friction bearings are one type of bearing used to take radial loads. They are always lubricated by grease.
Frames of D.C. & A.C. Machines
1. What is the work of the frame of dc machines?
a) to reduce the voltage
b) to reduce the flux
c) to carry the flux
d) to carry the current
Answer: c
Explanation: The main function of the frame of dc machines is to carry the flux. Thus the frame must be large enough to carry flux.
2. Why is the length of the yoke made larger?
a) to protect the armature windings
b) to cover the armature windings
c) to cover the field windings
d) to cover and protect the field windings
Answer: d
Explanation: The length of the yoke is usually made larger than the pole cores. It is because to cover and protect the field windings.
3. What is the formula for the depth of the yoke?
a) depth of yoke = thickness/2
b) depth of yoke = thickness
c) depth of yoke = 2*thickness
d) depth of yoke = 3*thickness
Answer: b
Explanation: The depth of yoke is equal to the thickness of the yoke. It is calculated to give the required cross-section for the magnetic circuit.
4. In large machines, the thickness is relatively larger to the diameter.
a) true
b) false
Answer: b
Explanation: The thickness is used in the calculation of the depth of the yoke. In large machines, the thickness is relatively smaller to the diameter.
5. What is the formula in order to check the rigidity?
a) moment of inertia ≥ (weight of magnetic frame * radius2 * 10-6) / 225
b) moment of inertia ≤ (weight of magnetic frame * radius2 * 10-6) / 225
c) moment of inertia = (weight of magnetic frame * radius2 * 10-6) / 225
d) moment of inertia < (weight of magnetic frame * radius2 * 10-6) / 225
Answer: a
Explanation: The moment of inertia, the weight of magnetic frame and the radius is calculated first. The machine is highly rigid if the moment of inertia is greater than or equal to the product of weight of magnetic frame and square of radius divided by 225.
6. What is the formula for the thickness of the ac machines?
a) thickness = 40 * inner diameter of frame/12
b) thickness = 40 + inner diameter of frame/12
c) thickness = 40 – inner diameter of frame/12
d) thickness = 40 * inner diameter of frame*12
Answer: a
Explanation: The thickness of the ac machines depend upon the inner diameter of the frame. On obtaining the inner diameter of frame and on substitution gives the thickness of ac machines.
7. What is the formula for the breadth of the ac machine?
a) breadth = 6 + 0.01 * inner diameter of frame
b) breadth = 6 – 0.01 * inner diameter of frame
c) breadth = 6 * 0.01 * inner diameter of frame
d) breadth = 6 / 0.01 * inner diameter of frame
Answer: a
Explanation: The breadth of the ac machines also depends upon the inner diameter of the frame. On substituting the values the breadth is calculated.
8. What is the formula for the checking of rigidity of induction machines?
a) moment of inertia ≥ radius / length of stator core * 90
b) moment of inertia ≥ radius * length of stator core * 90
c) moment of inertia ≥ radius * length of stator core / 90
d) moment of inertia ≤ radius / length of stator core * 90
Answer: c
Explanation: The radius and length of the stator core along with the moment of inertia is calculated. If the moment of inertia is greater than or equal to the product of length and radius divided by 90, the machine is more rigid.
9. What is the formula for the radius at the centre of gravity?
a) radius at the centre of gravity = inner diameter1.5/6.3
b) radius at the centre of gravity = inner diameter2/6.3
c) radius at the centre of gravity = outer diameter1.5/6.3
d) radius at the centre of gravity = outer diameter2/6.3
Answer: c
Explanation: The outer diameter of stator core is first calculated. On substituting the values the radius at the centre of gravity is obtained.
Design Of Single Phase Induction Motors MCQs
10. What is the formula of the centrifugal force?
a) centrifugal force = weight of revolving body * 39.43 * speed2 * radius of circular path
b) centrifugal force = weight of revolving body / 39.43 * speed2 * radius of circular path
c) centrifugal force = weight of revolving body * 39.43 / speed2 * radius of circular path
d) centrifugal force = weight of revolving body * 39.43 * speed2 / radius of circular path
Answer: a
Explanation: The weight of revolving body, speed, radius of circular path is calculated. On substitution the centrifugal force is obtained.
Bracing of Rotor Windings
1. What is the use of the wire bands of rotor?
a) used for bracing the rotor windings
b) used for circulating the current in the rotor windings
c) used for the encircling of the rotor windings
d) used for the protecting the rotor windings
Answer: a
Explanation: Bands used on the rotors of electrical machines are intended for bracing the rotor windings. This is done against their shift in the radial direction under action of centrifugal forces.
2. Where are the wire bands placed?
a) active portions of rotor conductors
b) inactive portions of rotor conductors
c) active or inactive portions of rotor conductors
d) active and inactive portions of the rotor conductors
Answer: d
Explanation: The wire bands are placed on the active portions of the rotor conductors. They are also placed in the inactive portions of the rotor conductors.
3. What are the factors on which the sizes of bands placed on depend?
a) length of air gap
b) method of cooling of armatures
c) length of air gap and method of cooling of armatures
d) method of cooling of armatures or length of air gap
Answer: c
Explanation: The sizes of bands placed on the active portions of the conductors depend upon the length of air gap. They also depend upon the method of cooling of armatures.
4. In what machines are the wire bands along the active length of windings placed?
a) dc or ac machines
b) dc and ac machines
c) dc machines
d) ac machines
Answer: c
Explanation: Wire bands are generally placed on both the active and inactive portions of rotor conductors. The wire bands along the active length of windings are placed along the dc machines.
5. What is the range of the width of each band that should not be exceeded?
a) 10-15 mm
b) 15-20 mm
c) 20-25 mm
d) 18-23 mm
Answer: b
Explanation: Bands placed along the active length of windings are housed in the ring slots. The width of each band should not exceed 15 to 20 mm.
6. What is the maximum value above which the total width of the bands should not exceed?
a) 25-35% of the axial length of armature core
b) 30-35% of the axial length of armature core
c) 25-30% of the axial length of armature core
d) 35-40% of the axial length of armature core
Answer: a
Explanation: The total width should not exceed 25% of the axial length of the armature core. The total width should not exceed maximum of 35% of the axial length of the armature core.
7. What is the formula for the breadth of the ring slot?
a) breadth of the ring slot = (number of turns in a band + 1)*diameter of band wire – 2*constant
b) breadth of the ring slot = (number of turns in a band + 1)*diameter of band wire + 2*constant
c) breadth of the ring slot = (number of turns in a band + 1)*diameter of band wire * 2*constant
d) breadth of the ring slot = (number of turns in a band + 1)*diameter of band wire / 2*constant
Answer: b
Explanation: The number of turns in a band is first calculated along with the diameter of band wire. The value of constant is just fixed and on substitution gives the breadth of the ring slot.
8. What is the value of the constant used in the calculation of the breadth of the ring slot for the diameter of band wire < 1.5 mm?
a) 1 mm
b) 1.5 mm
c) 2 mm
d) 3 mm
Answer: a
Explanation: The value of the constant used in the calculation of the breadth of the ring slot is 1 mm for the diameter of band wire < 1.5 mm. The value of the constant used in the calculation of the breadth of the ring slot is 1.5 mm for the diameter of band wire > 1.5 mm.
9. What is the maximum width of the bands placed on the end windings of induction machines and high speed dc machines?
a) 30 mm
b) 35 mm
c) 40 mm
d) 45 mm
Answer: c
Explanation: The maximum width are obtained for the bands placed on the end windings of induction machines and high speed dc machines. The maximum width is 40 mm.
10. What is the diameter of the wire bands made of tin, steel or bronze wire?
a) 2 mm
b) 1 mm
c) 4 mm
d) 3 mm
Answer: d
Explanation: The wire bands are generally made up of tin, steel or bronze wires. The diameter of those wire bands are 3 mm.
11. What is the function of the bands when it is placed on overhang?
a) used to reduce the centrifugal forces
b) used to increase the centrifugal forces
c) used to balance the centrifugal forces
d) used to withstand the centrifugal forces
Answer: d
Explanation: The bands when placed on overhead only are used to withstand the centrifugal forces. The centrifugal forces are due to the weight of the overhang.
12. What is the function of the bands when they are distributed along the axial length of armature?
a) used to reduce the centrifugal forces
b) used to increase the centrifugal forces
c) used to decrease the centrifugal forces
d) used to withstand the centrifugal forces
Answer: d
Explanation: The bands are distributed along the axial length of the armature and they withstand the centrifugal forces. The centrifugal forces are due to the weight of both the active and inactive parts of armature.
13. What is the formula of the mean diameter at the position of centre of gravity?
a) mean diameter at the position of centre of gravity = Inner diameter + diameter of stator wires
b) mean diameter at the position of centre of gravity = Inner diameter * diameter of stator wires
c) mean diameter at the position of centre of gravity = Inner diameter / diameter of stator wires
d) mean diameter at the position of centre of gravity = Inner diameter – diameter of stator wires
Answer: d
Explanation: The inner diameter and the diameter of the stator wires is first calculated. Then on substitution gives the mean diameter at the position of centre of gravity.
14. What is the value of permissible stress for bronze wire for the diameter of branding wire of 1 mm?
a) 350 NM per m2
b) 250 NM per m2
c) 300 NM per m2
d) 450 NM per m2
Answer: a
Explanation: The permissible stress for bronze wire for the diameter of branding wire of 1 mm is 350 NM per m2. The permissible stress for bronze wire for the diameter of branding wire of 1.5 mm is 300 NM per m2.
15. What is the value of permissible stress for steel wire for the diameter of branding wire of 0.5-1.2 mm?
a) 570 NM per m2
b) 600 NM per m2
c) 650 NM per m2
d) 700 NM per m2
Answer: b
Explanation: The value of permissible stress for steel wire for the diameter of branding wire of 0.5-1.2 mm is 600 NM per m2. The value of permissible stress for steel wire for the diameter of branding wire of 1.5-2 mm is 570 NM per m2.
Design of Fan
1. What is the formula for the fundamental relationship for the design of the ventilation system?
a) head of air inside the machine = hydrodynamic resistance * volume of air passing2
b) head of air inside the machine = hydrodynamic resistance + volume of air passing2
c) head of air inside the machine = hydrodynamic resistance – volume of air passing2
d) head of air inside the machine = hydrodynamic resistance / volume of air passing2
Answer: a
Explanation: First the hydrodynamic resistance is calculated along with the volume of air passing. On substitution in the formula gives the head of air inside the machine.
2. What is the formula for the total head produced?
a) total head produced = ∑ coefficient of hydrodynamic resistance + volume of air passing per second2
b) total head produced = ∑ coefficient of hydrodynamic resistance – volume of air passing per second2
c) total head produced = ∑ coefficient of hydrodynamic resistance * volume of air passing per second2
d) total head produced = ∑ coefficient of hydrodynamic resistance / volume of air passing per second2
Answer: c
Explanation: The various coefficient of hydrodynamic resistance is calculated along with the volume of air passing per second. On substituting the various values and on addition gives the total head produced.
3. What are the ventilating parts in the ventilating circuits?
a) sharp or projecting inlet edges
b) inlet corners
c) variations in cross-sections of air paths
d) sharp or projecting inlet edges, inlet corners, variations in cross-sections of air paths
Answer: d
Explanation: There are various ventilation parts provided in the ventilation circuits. They are sharp or projecting inlet edges, inlet corners, variations in cross-sections of air paths.
4.What is the range of the coefficients of hydrodynamic resistances for the protruding edges at inlet?
a) 40-50 * 10-3
b) 40-60 * 10-3
c) 30-50 * 10-3
d) 30-40 * 10-3
Answer: b
Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For protruding edges the range is about 40-60 * 10-3.
5. What is the range of the coefficients of hydrodynamic resistances for the rectangular edges at inlet?
a) 10-20 * 10-3
b) 30 * 10-3
c) 20-30 * 10-3
d) 20-25 * 10-3
Answer: b
Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For rectangular edges the range is about 30 * 10-3.
6. What is the range of the coefficients of hydrodynamic resistances for the rounded edges at inlet?
a) 12-20 * 10-3
b) 10-20 * 10-3
c) 15-20 * 10-3
d) 12-30 * 10-3
Answer: a
Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For rounded edges the range is about 12-20 * 10-3.
7. What factor/factors are required to evaluate the hydrodynamic resistance?
a) area of cross section
b) hydrodynamic coefficients
c) area of cross section or hydrodynamic coefficients
d) area of cross section and hydrodynamic coefficients
Answer: d
Explanation: The hydrodynamic resistance is calculated from the hydrodynamic coefficients. It is also calculated from the area of cross section.
8. How many data are required for the design of fan?
a) 3
b) 4
c) 5
d) 6
Answer: c
Explanation: There are 3 data required in the design of fan. They are outside diameter of fan, volume of air, hydrodynamic resistance.
9. How many steps are required in the design of the fan?
a) 7
b) 8
c) 9
d) 6
Answer: a
Explanation: There are 7 steps involved in the design of fan. They are maximum air passing per second, peripheral speed of fan, width of fan, static head of fan under rated duty, inside diameter of fan, number of blades, power input of fan.
10. What is the formula for the volume of air?
a) volume of air = 0.9 * losses in kW / difference of air temperature at inlet and outlet
b) volume of air = 0.9 * losses in kW * difference of air temperature at inlet and outlet
c) volume of air = 0.9 / losses in kW * difference of air temperature at inlet and outlet
d) volume of air = 1 / 0.9 * losses in kW * difference of air temperature at inlet and outlet
Answer: a
Explanation: The losses in kW is calculated along with the difference of air temperatures at inlet and outlet. On substitution the volume of air can be obtained.
11. What is the range of the difference of air temperature at inlet and outlet?
a) 11-150C
b) 10-130C
c) 12-160C
d) 14-180C
Answer: c
Explanation: The difference of air temperatures at inlet and outlet has a minimum value of 120C. The difference of air temperatures at inlet and outlet has a maximum value of 160C.
12. What is the formula for the area of outlet opening?
a) area of outlet opening = maximum air passing per second / 0.42 * peripheral speed
b) area of outlet opening = maximum air passing per second * 0.42 * peripheral speed
c) area of outlet opening = maximum air passing per second * 0.42 / peripheral speed
d) area of outlet opening = 1/maximum air passing per second * 0.42 * peripheral speed
Answer: a
Explanation: The maximum air passage per second along with the peripheral speed is calculated. On substitution the area of outlet opening is obtained.
13. What is the formula of the width of fan?
a) width of fan = area of outlet opening * 2.88 * outside diameter */ coefficient of utilization
b) width of fan = 1/ area of outlet opening * 2.88 * outside diameter * coefficient of utilization
c) width of fan = area of outlet opening * 2.88 * outside diameter * coefficient of utilization
d) width of fan = area of outlet opening / 2.88 * outside diameter * coefficient of utilization
Answer: d
Explanation: The area of outlet opening and the outside diameter is calculated. After fixing the coefficient of utilization and on substituting the width of fan is obtained.
14. What is the formula for the number of blades?
a) number of blades = 3.14 * outside diameter * (1.25 – 1.5)* width of fan
b) number of blades = 3.14 / outside diameter * (1.25 – 1.5)* width of fan
c) number of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of fan
d) number of blades = 3.14 * outside diameter * (1.25 – 1.5) / width of fan
Answer: c
Explanation: The outside diameter and the width of fan is first calculated. Then the range is fixed according to the diameter and on substitution gives the number of blades.
15. What is the formula for the maximum air passing per second at maximum efficiency?
a) maximum air passing = 2 * volume of air passing per second
b) maximum air passing = volume of air passing per second
c) maximum air passing = 2 / volume of air passing per second
d) maximum air passing = volume of air passing per second / 2
Answer: a
Explanation: The volume of air passing per second is first calculated and multiplying it by 2 gives the maximum air passing per second. The maximum air passing per second is used in the calculation of the area of the outlet opening.