60 + Interview Questions in Aspects Of Design Of Mechanical Parts in Design Of Electrical Machines Page 1 InterviewSolution

1.	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 bearingI got this question by my school principal while I was bunking the class.My enquiry is from Bearings in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct option is (d) guide bearing Easy explanation: SOMETIMES the SIMPLE thrust bearing wont be able to PULL the radial LOADS. Therefore an ADDITIONAL bearing called guide bearing is provided.

Discussion

2.	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 rotorThe question was asked in my homework.This intriguing question originated from Bearings in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right CHOICE is (c) DYNAMIC unbalance of ROTOR or unbalanced MAGNETIC pull of rotor To ELABORATE: The radial loads are caused by the dynamic unbalance of rotor. It can also be caused by the unbalanced magnetic pull of rotor.

Discussion

3.	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 / 2This question was posed to me in a national level competition.I want to ask this question from Design of Fan in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» RIGHT option is (a) maximum AIR passing = 2 * VOLUME of air passing per second Easiest 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.

Discussion

4.	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 fanThe question was asked in my homework.The query is from Design of Fan in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct ANSWER is (c) NUMBER of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of fan To ELABORATE: 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.

Discussion

5.	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 utilizationI got this question in unit test.Question is taken from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT ANSWER is (d) width of fan = area of OUTLET opening / 2.88 * outside diameter * COEFFICIENT of utilization To elaborate: 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.

Discussion

6.	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 speedThis question was posed to me by my college director while I was bunking the class.The origin of the question is Design of Fan in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct OPTION is (a) area of outlet opening = maximum AIR passing PER second / 0.42 * peripheral speed Best explanation: The maximum air passage per second ALONG with the peripheral speed is calculated. On substitution the area of outlet opening is OBTAINED.

Discussion

7.	What is the range of the difference of air temperature at inlet and outlet?(a) 11-15^0C(b) 10-13^0C(c) 12-16^0C(d) 14-18^0CI had been asked this question in examination.Origin of the question is Design of Fan in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (c) 12-16^0C For explanation: The difference of AIR temperatures at inlet and OUTLET has a MINIMUM value of 12^0C. The difference of air temperatures at inlet and outlet has a MAXIMUM value of 16^0C.

Discussion

8.	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 outletThe question was posed to me in quiz.Asked question is from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer»

Discussion

9.	How many steps are required in the design of the fan?(a) 7(b) 8(c) 9(d) 6The question was posed to me in examination.My enquiry is from Design of Fan in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct option is (a) 7 The 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.

Discussion

10.	How many data are required for the design of fan?(a) 3(b) 4(c) 5(d) 6I had been asked this question by my college professor while I was bunking the class.My question is taken from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right choice is (c) 5 The EXPLANATION is: There are 3 DATA REQUIRED in the design of fan. They are outside diameter of fan, VOLUME of air, hydrodynamic resistance.

Discussion

11.	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 coefficientsI have been asked this question during an internship interview.The question is from Design of Fan topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT OPTION is (d) area of cross section and hydrodynamic coefficients Easy explanation: The hydrodynamic RESISTANCE is CALCULATED from the hydrodynamic coefficients. It is ALSO calculated from the area of cross section.

Discussion

12.	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^-3I got this question in final exam.Asked question is from Design of Fan in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» RIGHT OPTION is (a) 12-20 * 10^-3 The 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.

Discussion

13.	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^-3I had been asked this question in a national level competition.Query is from Design of Fan in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct choice is (b) 30 * 10^-3 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.

Discussion

14.	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^-3This question was posed to me by my college professor while I was bunking the class.This interesting question is from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (b) 40-60 * 10^-3 The best I can explain: 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.

Discussion

15.	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 pathsThe question was posed to me in an online interview.Question is from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct answer is (d) sharp or projecting inlet EDGES, inlet corners, variations in cross-sections of air PATHS To EXPLAIN: 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.

Discussion

16.	What is the formula for the total head produced?(a) total head produced = ∑ coefficient of hydrodynamic resistance + volume of air passing per second^2(b) total head produced = ∑ coefficient of hydrodynamic resistance – volume of air passing per second^2(c) total head produced = ∑ coefficient of hydrodynamic resistance * volume of air passing per second^2(d) total head produced = ∑ coefficient of hydrodynamic resistance / volume of air passing per second^2This question was posed to me in an interview for internship.Question is from Design of Fan in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer»

Discussion

17.	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 m^2(b) 600 NM per m^2(c) 650 NM per m^2(d) 700 NM per m^2This question was addressed to me in an internship interview.My doubt stems from Bracing of Rotor Windings topic in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT answer is (b) 600 NM per m^2 For EXPLANATION I would SAY: The value of permissible stress for STEEL wire for the diameter of branding wire of 0.5-1.2 mm is 600 NM per m^2. The value of permissible stress for steel wire for the diameter of branding wire of 1.5-2 mm is 570 NM per m^2.

Discussion

18.	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 passing^2(b) head of air inside the machine = hydrodynamic resistance + volume of air passing^2(c) head of air inside the machine = hydrodynamic resistance – volume of air passing^2(d) head of air inside the machine = hydrodynamic resistance / volume of air passing^2I got this question in my homework.This interesting question is from Design of Fan topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT choice is (a) head of air inside the MACHINE = HYDRODYNAMIC resistance * volume of air passing^2 Best 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.

Discussion

19.	What is the value of permissible stress for bronze wire for the diameter of branding wire of 1 mm?(a) 350 NM per m^2(b) 250 NM per m^2(c) 300 NM per m^2(d) 450 NM per m^2The question was posed to me in an online interview.I need to ask this question from Bracing of Rotor Windings in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct answer is (a) 350 NM per m^2 The explanation: The permissible stress for bronze wire for the DIAMETER of branding wire of 1 MM is 350 NM per m^2. The permissible stress for bronze wire for the diameter of branding wire of 1.5 mm is 300NM per m^2.

Discussion

20.	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 wiresThis question was posed to me in a job interview.This interesting question is from Bracing of Rotor Windings in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT choice is (d) mean DIAMETER at the position of centre of GRAVITY = Inner diameter – diameter of STATOR wires Best 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.

Discussion

21.	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 forcesThe question was asked at a job interview.My question is based upon Bracing of Rotor Windings topic in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right choice is (d) used to withstand the centrifugal forces To explain: 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.

Discussion

22.	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 forcesI had been asked this question in my homework.The above asked question is from Bracing of Rotor Windings in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct choice is (d) USED to withstand the CENTRIFUGAL forces To explain I would say: 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.

Discussion

23.	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 mmThis question was addressed to me at a job interview.The doubt is from Bracing of Rotor Windings topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right answer is (d) 3 mm To EXPLAIN: The wire bands are GENERALLY MADE up of TIN, steel or bronze wires. The DIAMETER of those wire bands are 3 mm.

Discussion

24.	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 mmThis question was addressed to me in an interview for internship.Query is from Bracing of Rotor Windings topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right option is (c) 40 mm For explanation I would say: 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.

Discussion

25.	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 mmThis question was posed to me in exam.I'm obligated to ask this question of Bracing of Rotor Windings in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right choice is (a) 1 mm Best 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.

Discussion

26.	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 – 2constant(b) breadth of the ring slot = (number of turns in a band + 1)diameter of band wire + 2constant(c) breadth of the ring slot = (number of turns in a band + 1)diameter of band wire 2constant(d) breadth of the ring slot = (number of turns in a band + 1)diameter of band wire / 2*constantI got this question in an interview for internship.My enquiry is from Bracing of Rotor Windings in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct CHOICE is (b) breadth of the ring slot = (number of turns in a band + 1)diameter of band WIRE + 2constant For 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.

Discussion

27.	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 mmThis question was posed to me in an interview for job.This intriguing question originated from Bracing of Rotor Windings topic in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct choice is (B) 15-20 mm For explanation I would say: 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.

Discussion

28.	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 coreThe question was posed to me by my school principal while I was bunking the class.I want to ask this question from Bracing of Rotor Windings topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct option is (a) 25-35% of the axial LENGTH of ARMATURE CORE The best 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.

Discussion

29.	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 gapI had been asked this question by my college director while I was bunking the class.I'm obligated to ask this question of Bracing of Rotor Windings in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right answer is (c) length of AIR GAP and method of cooling of ARMATURES The 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.

Discussion

30.	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 machinesI have been asked this question during an online exam.Question is from Bracing of Rotor Windings in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT CHOICE is (c) dc machines 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.

Discussion

31.	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 windingsI had been asked this question in examination.I'd like to ask this question from Bracing of Rotor Windings in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT choice is (a) used for bracing the rotor windings The 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.

Discussion

32.	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 conductorsI got this question during an online interview.Question is from Bracing of Rotor Windings topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right CHOICE is (d) active and INACTIVE portions of the rotor conductors To explain I WOULD say: 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.

Discussion

33.	What is the formula of the centrifugal force?(a) centrifugal force = weight of revolving body * 39.43 * speed^2 * radius of circular path(b) centrifugal force = weight of revolving body / 39.43 * speed^2 * radius of circular path(c) centrifugal force = weight of revolving body * 39.43 / speed^2 * radius of circular path(d) centrifugal force = weight of revolving body * 39.43 * speed^2 / radius of circular pathThis question was addressed to me during an internship interview.I'd like to ask this question from Frames of D.C. & A.C. Machines topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT option is (a) centrifugal force = weight of revolving body * 39.43 * SPEED^2 * radius of circular path The explanation: The weight of revolving body, speed, radius of circular path is CALCULATED. On SUBSTITUTION the centrifugal force is obtained.

Discussion

34.	What is the formula for the radius at the centre of gravity?(a) radius at the centre of gravity = inner diameter^1.5/6.3(b) radius at the centre of gravity = inner diameter^2/6.3(c) radius at the centre of gravity = outer diameter^1.5/6.3(d) radius at the centre of gravity = outer diameter^2/6.3The question was asked at a job interview.This intriguing question originated from Frames of D.C. & A.C. Machines in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct choice is (c) RADIUS at the centre of GRAVITY = outer DIAMETER^1.5/6.3 To elaborate: The outer diameter of stator CORE is first CALCULATED. On substituting the values the radius at the centre of gravity is obtained.

Discussion

35.	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 frameThe question was asked by my school teacher while I was bunking the class.The origin of the question is Frames of D.C. & A.C. Machines in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct option is (a) breadth = 6 + 0.01 * inner diameter of frame The EXPLANATION: The breadth of the AC MACHINES also depends UPON the inner diameter of the frame. On substituting the values the breadth is calculated.

Discussion

36.	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 * 90This question was addressed to me in final exam.This intriguing question comes from Frames of D.C. & A.C. Machines in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right choice is (c) moment of inertia ≥ radius * length of stator core / 90 For explanation I WOULD SAY: 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.

Discussion

37.	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*12This question was posed to me in final exam.I'd like to ask this question from Frames of D.C. & A.C. Machines in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The CORRECT answer is (a) thickness = 40 * inner DIAMETER of frame/12 To ELABORATE: 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.

Discussion

38.	What is the formula in order to check the rigidity?(a) moment of inertia ≥ (weight of magnetic frame * radius^2 * 10^-6) / 225(b) moment of inertia ≤ (weight of magnetic frame * radius^2 * 10^-6) / 225(c) moment of inertia = (weight of magnetic frame * radius^2 * 10^-6) / 225(d) moment of inertia < (weight of magnetic frame * radius^2 * 10^-6) / 225I had been asked this question in homework.This question is from Frames of D.C. & A.C. Machines in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» CORRECT answer is (a) moment of inertia ≥ (weight of magnetic frame * RADIUS^2 * 10^-6) / 225 Easy 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.

Discussion

39.	In large machines, the thickness is relatively larger to the diameter.(a) true(b) falseI have been asked this question in unit test.This interesting question is from Frames of D.C. & A.C. Machines topic in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (b) false For 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.

Discussion

40.	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 = 2thickness(d) depth of yoke = 3thicknessThe question was asked by my college professor while I was bunking the class.Query is from Frames of D.C. & A.C. Machines in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct option is (B) DEPTH of YOKE = thickness 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.

Discussion

41.	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 windingsThe question was asked in examination.My doubt stems from Frames of D.C. & A.C. Machines topic in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» RIGHT option is (d) to cover and protect the FIELD WINDINGS To elaborate: The length of the YOKE is usually made larger than the pole cores. It is because to cover and protect the field windings.

Discussion

42.	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 currentThe question was posed to me by my college director while I was bunking the class.The origin of the question is Frames of D.C. & A.C. Machines topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Right OPTION is (c) to carry the FLUX To explain: The MAIN FUNCTION of the frame of dc machines is to carry the flux. Thus the frame must be large ENOUGH to carry flux.

Discussion

43.	Anti-friction bearings are lubricated by charcoal.(a) true(b) falseThe question was posed to me in examination.My question is from Bearings topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct answer is (b) false For explanation: Anti-friction bearings are one TYPE of bearing used to TAKE RADIAL loads. They are always LUBRICATED by grease.

Discussion

44.	The phosphor bronze sleeve bearings are used for the small electrical machines.(a) true(b) falseThe question was asked in final exam.Question is from Bearings in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (a) true The BEST explanation: The phosphor bronze SLEEVE bearings are USED for the small electrical machines. The diameter of the machines are between 50-60mm.

Discussion

45.	What is the other name of the bearings outside the machine?(a) ball bearing(b) pedestal bearing(c) guide bearing(d) gear bearingI have been asked this question during an interview.This intriguing question originated from Bearings in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (b) PEDESTAL BEARING To explain I WOULD say: 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.

Discussion

46.	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 machineI have been asked this question in an international level competition.My question is taken from Bearings topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» Correct ANSWER is (c) outside the machine or in the end SHIELDS of the machine Best 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.

Discussion

47.	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 downwardsThis question was posed to me during an interview.The above asked question is from Bearings topic in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» The correct OPTION is (d) axial load ACTING downwards To elaborate: 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.

Discussion

48.	In horizontal shaft machines, the forces acting in which direction is prominent?(a) circular(b) vertical(c) horizontal(d) radialThe question was posed to me during an interview.I need to ask this question from Bearings in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» CORRECT choice is (d) radial The best EXPLANATION: Plain bearings are made USE of for the HORIZONTAL shaft machines. The forces ACTING in the radial direction are most prominent.

Discussion

49.	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) 4The question was asked in my homework.I'm obligated to ask this question of Bearings topic in section Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» RIGHT answer is (C) 2 or 3 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.

Discussion

50.	What is the plain bearing used?(a) sleeve bearings(b) anti friction bearing(c) sleeve or anti friction bearing(d) sleeve and anti friction bearingThe question was asked in a national level competition.Enquiry is from Bearings topic in division Aspects of Design of Mechanical Parts of Design of Electrical Machines
Answer» RIGHT OPTION is (c) sleeve or ANTI friction bearing For 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.

Discussion

Explore topic-wise InterviewSolutions in .

What is the range of the difference of air temperature at inlet and outlet?(a) 11-15^0C(b) 10-13^0C(c) 12-16^0C(d) 14-18^0CI had been asked this question in examination.Origin of the question is Design of Fan in division Aspects of Design of Mechanical Parts of Design of Electrical Machines

How many steps are required in the design of the fan?(a) 7(b) 8(c) 9(d) 6The question was posed to me in examination.My enquiry is from Design of Fan in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines

How many data are required for the design of fan?(a) 3(b) 4(c) 5(d) 6I had been asked this question by my college professor while I was bunking the class.My question is taken from Design of Fan topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines

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 mmThis question was addressed to me at a job interview.The doubt is from Bracing of Rotor Windings topic in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines

In large machines, the thickness is relatively larger to the diameter.(a) true(b) falseI have been asked this question in unit test.This interesting question is from Frames of D.C. & A.C. Machines topic in division Aspects of Design of Mechanical Parts of Design of Electrical Machines

Anti-friction bearings are lubricated by charcoal.(a) true(b) falseThe question was posed to me in examination.My question is from Bearings topic in portion Aspects of Design of Mechanical Parts of Design of Electrical Machines

The phosphor bronze sleeve bearings are used for the small electrical machines.(a) true(b) falseThe question was asked in final exam.Question is from Bearings in chapter Aspects of Design of Mechanical Parts of Design of Electrical Machines

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) 4The question was asked in my homework.I'm obligated to ask this question of Bearings topic in section Aspects of Design of Mechanical Parts of Design of Electrical Machines