Electrical Machine Design Question Bank - As per VTU Syllabus

S D M Education Society’s Shri Dharmasthala Manjunatheshwara Institute of Technology (SDMIT)

Electrical & Electronics Engineering DepartmentUjire – 574240, Karnataka – State, INDIA

QUESTION BANK Class: VI Semester B.E – EEE Year: 2015-16

Subject : 10EE63 Electrical Machine Design (EMD)

PART - A

UNIT – 1 : Principles of Electrical Machines Design1. June / July 2011 : 08 Marks, June / July 2009 : 06 Marks, Dec 2014 / Jan 2015 : 06 Marks

What are the limitations involved in design of electrical machines. OrExplain the limitations in design of electrical machines

2. Dec 2013 / Jan 2014 : 08 Marks, Dec 09 / Jan 2010 : 06 MarksWhat are the major considerations accounted for the good design of electrical machines?

3. June / July 2014 : 10 Marks, Dec 2013 / Jan 2014 : 08 Marks, June / July 2013 : 06 Marks, Dec 2012 : 06 Marks, June 2012 : 12 Marks, Dec 2011 : 07 Marks, June / July 2011 : 08 Marks, Dec 2010 : 08 Marks, Dec 09 / Jan 2010 : 10 Marks, May / June 2010 : 10 Marks,List the desirable properties of insulating materials. Give the classification of insulating materials based on thermal considerations with examples used in each class

4. Dec 2015 / Jan 2016 : 4 MarksWhat are the desirable properties of insulation materials used in electrical machines? Name some insulation materials and state where they are utilized.

5. Dec 09 / Jan 2010 : 04 MarksClassify the magnetic materials in accordance with the value of their relative permeability. Give one example each.

UNIT – 2 : Design of DC Machines1. Dec 2015 / Jan 2016 : 6 Marks, Dec 2014 / Jan 2015 : 06 Marks, Dec 2013 / Jan 2014 : 05 Marks, June 2012 :

08 Marks, Dec 2011 : 08 Marks, June / July 2011 : 08 Marks, Dec 2010 : 08 MarksFrom the first principles derive the output equation of DC Machine

2. Dec 2007 / Jan 2008: 08 Marks, Jan / Feb 2006 : 08 MarksShow that the output of a dc generator with single turn coil is given by the expression S = 0.03 E| v q A / PNWhere E| = average voltage between adjacent commutator segments, v = peripheral velocity of the rotor in m/s, P = Number of Poles, N = Speed in rpm and q = specific electric loading

3. Dec 2014 / Jan 2015 : 05 Marks, Dec 09 / Jan 2010 : 06 MarksProve that in a DC Machine, the volume of the active parts is proportional to torque developed by the machine

4. Dec 2015 / Jan 2016 : 6 MarksDiscuss the factors which influence the selection of i). Number of poles, ii). Number of slots, iii). Air gap of a Machine

5. Dec 2015 / Jan 2016 : 4 MarksDiscuss why the armature core, field poles of a DC Machine are laminated, while yoke is not normally laminated?

6. June / July 2015 : 08 Marks, June / July 2014 : 04 Marks, Dec 2013 / Jan 2014 : 08 Marks, Dec 09 / Jan 2010 : 06 MarksDescribe how specific electric and magnetic loadings play an important role in the design of electrical

Question Bank 2015-16 : – 10EE63 Electrical Machine Design, VI Semester B.E -EEE, SDMIT, Ujire -574240, INDIA Page 1 of 16

machines OrDefine specific loadings and mention the advantages of choice of higher values of specific loadings in the design of any machine

7. Dec 2014 / Jan 2015 : 05 Marks, Dec 2011 : 07 Marks, June / July 2009 : 06 MarksState the factors affecting the value of electrical loading in case of a DC Machine

8. June / July 2014 : 06 Marks, Dec 2013 / Jan 2014 : 04 Marks, June / July 2009 : 06 Marks, Jan / Feb 2006 : 12 Marks, Dec 2013 / Jan 2014 : 04 Marks, June / July 2013: 04 Marks, Dec 2011 : 06 MarksDiscuss the factors that should be given due considerations while selecting number of poles for DC Machine OrMention the various factors that affect the choice of number of poles of a DC Machine Or What are the advantages and disadvantages of large number of poles in DC Machines?

9. June 2012 : 10 MarksGive a step by step procedure of designing a shunt field coil for a DC Machine

10. Dec 09 / Jan 2010 : 05 MarksDiscuss the guiding factors for the choice of number of armature slots in a DC Machine

11. Dec 09 / Jan 2010 : 08 MarksDraw a neat sketch of four pole DC machine indicating the magnetic circuit and different parts involved in it. Also explain with suitable expressions to estimate the total mmf / pole at no load and normal voltage.

12. May / June 2010 : 08 MarksDraw neat diagram showing cross section of DC machine armature slot along with insulation details. Explain the factors considered while fixing up the dimensions of the slot.

13. June / July 2015 : 10 MarksObtain an expression for field ampere turns per meter height of a DC machine in terms of permissible loss, copper space factor and depth of winding.

14. Dec 2014 / Jan 2015 : 08 MarksA 350 kW, 500 V, 415 V, 450 rpm, 6 Pole DC Generator is built with an armature diameter of 0.87 m and core length of 0.32 m. The lap wound armature has 660 conductors. Calculate B arc and ac.

15. Dec 2015 / Jan 2016 : 10 MarksCalculate the diameter and length of armature core of a 70 kW, 240 V, 900 rpm, 4 Pole DC Shunt Generator. The average flux density is 0.7 T and AC/m is 34,000. The ratio of core length to pole pitch is 0.8. Full load armature drop is 9.6 V and field current is 3.0 A.

16. Dec 2014 / Jan 2015 : 10 Marks. Determine the main dimensions, number of poles and length of air gap for a 600 kW, 500 V, 900 rpm generator. Assume : Barc = 0.6 T, (MMF)A4 = 50 % (MMF)Arm , ac / m = 35000, Pole arc / Pole pitch = 0.75, efficiency = 91 %, Kg = 1.15

17. June / July 2014 : 10 Marks, Dec 2013 / Jan 2014 : 07 Marks, June / July 2011 : 12 MarksA Design s required for a 50 kW, 4 Pole, 600 rpm, DC Shunt Generator, the full load terminal voltage being 220 V. If the Bmax = 0.83 Tesla and ac / m = 30,000, calculate suitable dimensions of armature core. Assume that the full load armature voltage drop is 3 % of rated terminal voltage and that the field current is 1.0% of rated full load current. Ratio of pole arc to pole pitch is 0.67.

18. Dec 2007 / Jan 2008: 10 Marks, Jan / Feb 2006 : 12 MarksCalculate the main dimensions of a 20 hp, 1000 rpm, 400 V, dc motor. Bav = 0.37 T, ac = 16000 ac/m. efficiency = 90 %. Choose a suitable number of poles and justify the selection.

19. June / July 2013: 10 MarksA 5 kW, 250 V, 4 Pole, 1500 rpm shunt generator is designed to have a square pole face. The loadings are:Average flux density in the air gap = 0.42 wb / m2, ampere conductors per metre = 15000. Find the main dimensions of the machine. Assume full load efficiency = 87 % and ratio of pole arc to pole pitch = 0.66.

20. Dec 2007 / Jan 2008: 12 Marks. Design a 50 kW, 4 pole, 600 rpm dc shunt generator. The full load terminal voltage being 220 V. if the maximum gap density is 0.83 T and ac = 30000 ac/m. calculate the suitable dimensions of armature core to give a square pole face. Assume that the full load armature drop is 3 % of the rated terminal voltage and the field current is 1 % of the full load current. Ratio of pole arc to pole pitch = 0.67.


21. June / July 2013: 10 MarksDetermine the total commutator losses for a 1000 kW, 500 V, 300 rpm, 10 Pole, dc generator. Diameter of the commutator = 100 cm, current density for the brush = 7.50 A / cm2, brush pressure = 1500 kgf / m2. Assume lap winding, brush contact drop = 2 V. coefficient of friction = 0.20, armature current = F.L current

22. December 2012: 14 MarksA 150 kW, 230 V, 500 rpm, DC shunt motor has square field coil. Find the number of poles, main dimensions and air gap length. Assume Bav = 0.85 T, ac / cm = 290, ratio of width of the pole body ot pole pitch = 0.55, ratio of pole arc to pole pitch = 0.70, efficiency = 91 %, take mmf for air gap = 55 % of armature mmf, gap contraction factor = 1.15

23. June / July 2015 : 08 MarksDetermine the main dimensions, number of poles, number of armature conductors, number of slots, conductors per slot and the size of armature conductors and cross sectional area of armature conductors for A 250 kW, 400 V, 625 A, 600 rpm, lap wound DC compound assuming the following data:Average flux density in the gap = 0.63 T, Specific Electric Loading = 33000 amp conductors / mh ; field and armature copper losses = 5 % of output; ratio of pole arc to pole pitch = 0.70, pole arc = gross length of armature, armature drop = 3 % of terminal voltage, current density = 5 A / mm2, slot pitch = 2.6 cm

24. December 2012: 12 Marks, June / July 2009 : 12 MarksA 250 kW, 500 V, 600 rpm, DC Generator is built with and armature diameter of 75 cm and a core length of 30 cm. The lap connected armature has 720 conductors using the data obtained from this machine, determine the armature diameter, core length, number of armature slots, armature conductors and commutator segments for a 350 kW, 440 V, 720 rpm, 6 Pole DC Generator. Assume a square pole face with ratio of pole arc to pole pitch equal to 0.66. The full load efficiency is 0.91 and to internal voltage drop is 4 % of the rated voltage. The diameter of commutator is 0.70 of the armature diameter. The pitch of commutator segments should not be less than 4 mm. the voltage between adjacent segments should not exceed 15 V at no load.

25. December 2012: 08 MarksDuring the design of armature of a 1000 kW, 500 V, 10 Pole, 300 rpm, DC compound generator following details are obtained: External diameter of commutator = 140 cm; Gross length of the core = 35 cm; Flux / Pole = 0.105 wb. Based on the above details, find out the following details regarding the field system designi. Axial length of the pole, ii. Width of the pole, iii. Height of the pole, iv. Pole arc, v. Depth of the yokeAssume the permissible loss of the cooling surface = 700 W / m2, leakage factor for the pole 1.2, flux density in the pole = 1.6 Tesla, Iron factor for the pole = 0.95, winding type = Lap connected, voltage drop = 10 V, ratio of field ampere turns to the armature ampere turns on full load = 1.2, depth of the winding = 5 cm, copper space factor for the field winding = 0.6, pole shoe thickness at the centre of the pole = 3.6 cm, ratio of pole arc to pitch of the pole = 0.68, flux density in the yoke = 1.5 times gross armature length. Axial length of pole = 1 cm less than gross length of armature.

26. June 2012 : 10 Marks, May / June 2010 : 10 MarksCalculate the diameter and length of armature for a 7.5 kW, 4 Pole, 1000 rpm, 220 V DC Shunt Motor. Given that full load efficiency = 0.83, Maximum flux density = 0.9 wb / m2; ac = 30000 AC/m; field form factor = 0.70; assume that the maximum efficiency occurs at full load and field current is 2.5 % of rated current. The pole face is square.

27. Dec 2011 : 12 MarksDetermine the main dimensions, number of pole and length of air gap of a 600 kW, 500 V, 900 rpm generator. Assume average gap density as 0.60 T and ac = 35000 AC / m. The ratio of pole arc to pole pitch = 0.75, and efiiciency = 91 %. The following are the design constraints: peripheral speed 40 m / sec; frequency of the flux reversal

50 Hz, current / brush 400 A; the armature mmf per pole 7500 AT. The mmf required for the air gap is 50 % of armature mmf and gap contraction factor is 1.15

28. June / July 2011 : 10 Marks, June / July 2009 : 08 MarksThe following are the particulars refer to the shunt field coil for a 440 V, 6 Pole, DC Generator, mmf / pole = 7000 A, depth of winding is 50 mm, length of inner turn = 1.1 m, length of outer turn = 1.4 m, loss radiated from outer surface = 1400 W / m2. Space factor = 0.62, resistivity = 0.02/m/mm2. Calculate diameter of wire, length of coil, number of turns, exciting current. Assume voltage drop of 20 % of terminal voltage across field regulator.


29. Dec 2010 : 12 MarksFind the main dimensions, number of poles and length of air gap of a 1000 kW, 500V, 300 rpm DC Generator. Assume the Bav = 0.7 T, ac /m = 40000, square pole face, ration of pole arc to pitch = 0.70. Assume efficiency = 92 %, and gap contraction factor = 1.15

30. Dec 2010 : 12 MarksCalculate the mmf required for the air gap of a machine having core length = 0.32m including 4 ducts of 10 mm each, pole arc = 0.19 m, slot pitch = 65.40 mm, slot opening = 5 mm, air gap length = 5 mm, flux / pole = 52 mwb. Given carters coefficient is 0.18 for opening / gap = 1 and 0.28 for opening / gap = 2

31. Dec 09 / Jan 2010 : 10 MarksFind the main dimensions and number of poles of a 50 hp, 230 V, 1400 rpm shunts motor so that a square pole face is obtained. Bav = 0.5 T, ac / m = 22000. The ratio of pole arc to pole pitch = 0.70. Assume the efficiency = 90 %. Check the values obtained are within the permissible limits.

32. Dec 09 / Jan 2010 : 07 MarksDesign a shunt field coil of a dc motor from the following data: field ampere turns per pole = 9000, mean length of turn = 1.4 m, depth of coil = 35 x 10-3 m, voltage across the field coil = 40 V, resistivity of the wire = 2.1 x 10-6 ohm – metre, thickness of the insulating varnish = 0.20 mm. Power dissipation from the total surface should not exceed 700 W / m2 .check your design for power dissipation. Take the copper space factor for field coil as 0.65

33. May / June 2010 : 12 MarksCalculate the size of conductor and number of turns for the field coil of a 6 pole, 460 V, dc shunt motor. The coil is to supply an mmf of 4000 A at working temperature. The length of inside turn is 0.74 m, height of the coil = 0.13 m, the coil space factor = 0.52. the permissible loss from external surface alone is 1200 W / m2. Solution should not be attempted by assuming a numerical value for winding depth. Resistivity of conductor = 0.02 Ω / m / mm2. 15 % of the applied voltage is dropped across the field rheostat.

34. June / July 2015 : 10 MarksA 50 HP, 4 Pole, 480 V, 600 rpm shunt motor has a wave wound armature with 770 conductors. The leakage factor for the poles is 1.2. The poles are to be of circular in cross section, and field coils are 70 mm thick and produce an mmf of 10,000 A per pole. The flux density in the poles is 1.5 wb / m2, calculate: i). diameter of poles, ii). Diameter of field wire, iii). Length of field coil, iv). Turns per pole and v). field current.

35. June / July 2013: 10 MarksAn electromagnetic coil has an internal diameter of 0.8 m and external diameter of 0.4 m. its height is 0.20 m. the outside cylindrical surface of the coil can dissipate 1000 W / m2. Calculate the total mmf per coil if the voltage applied across the coil is 50 V. Assume the space factor to be 0.60 and resistivity of the wire to be 0.02 Ω / m / mm2

36. June / July 2014 : 10 MarksDuring the design of armature of a 1000 kW, 500 V, 10 Pole, 300 rpm, DC Compound Generator, following information has been obtained:External diameter of armature = 1.4 m, Gross core length = 0.35 m, flux per pole = 0.105 wb.Based on the above information, find out the following details regarding the design of field system : i). Axial length of the pole, ii). Width of the Pole, iii). Height of the Pole.Permissible loss per square meter of the cooling surface may be assumed as 700 W / m2. Assume leakage factor of the pole Kl = 1.2, flux density in the pole BP = 1.6 Tesla, Iron factor ki = 0.95, voltage drop as 2 % of rated voltage, ATf = 1.2 ATa , Copper space factor Sf = 0.60 and depth of the winding as 0.05 m and thickness of the pole shoe = 4 cm

37. Dec 2015 / Jan 2016 : 10 Marks, Dec 2007 / Jan 2008: 10 MarksA shunt field coil has to develop an mmf of 9000 AT, the voltage drop in he field coil = 40 V, resistivity of round wire = 0.021 Ω / m / mm2. The depth of the winding = 35 mm, length of mean turn = 1.4 m. design a coil so that the power dissipated = 700 W / m2 of the total coil surface. Take the diameter of insulated wire to be 0.2 mm greater than bare wire.


UNIT - 3 & 4: Design of Transformers ( Single and Three Phase)


1. Dec 2014 / Jan 2015 : 06 Marks, June / July 2014 : 4 Marks, Dec 2011 : 10 Marks, Dec 2010 : 10 Marks, Jan / Feb 2006: 08 MarksDerive the output equation of a 3 phase core type transformer, and deduce the expression for volt / turn.

2. Dec 2015 / Jan 2016 : 10 MarksShow that the output of a 3 phase core type transformer is 5.23 f B m H d2 HW x 10-3 kVA, where ‘ f ’ is the supply frequency, Bm - maximum value of flux density in wb / m2, d – effective diameter of the core, H is magnetic potential gradient in amperes / meter, Hw – Height of the Window in meters.

3. Dec 2014 / Jan 2015 : 04 MarksDiscuss the factors involved in optimum design of transformer

4. Dec 2014 / Jan 2015 : 10 Marks, June / July 2014 : 10 Marks, Dec 2013 / Jan 2014 : 10 Marks, June 2012 : 10 Marks, Dec 2010 : 10 Marks, May / June 2010 : 12 Marks, Dec 2007 / Jan 2008: 10 Marks, Jan / Feb 2006: 08 Marks, June / July 2015 : 10 MarksDerive an expression for the total leakage reactance of a transformer, referred to the primary

5. Dec 2015 / Jan 2016 : 08 Marks, Dec 2014 / Jan 2015 : 10 Marks, Dec 2007 / Jan 2008: 10 MarksA 1Φ, 440 V, 50 Hz transformer is built I form of stampings having a relative permeability of 1000. The length of the flux path is 2.5 m, the area of cross section of the core = 2.5 x 10-3 m2 and the primary winding has 800 turns. Estimate the maximum flux and the no load current of the transformer. Assume: loss at i) working flux density = 2.6 w / kg; ii) Iron weight = 7.8 x 103 kg/m3; iii) stacking factor = 0.90

6. Dec 2014 / Jan 2015 : 10 MarksA 250 kVA, 6600 /400V, 3 Φ, core type transformer has a total loss of 4800 W at full load. The transformer tank is 1.25 m height and 1m x 0.5m in plan. Design a suitable scheme for tubes, if the average temperature rise is to be limited to 350C . the diameter of the tubes is 50 mm and are spaced 75 mm apart. The average height of the tubes = 1.05m.

7. June / July 2014 : 04 Marks, June 2012 : 5 Marks, June / July 2015 : 10 MarksDerive the expression for the no load current of the transformer.

OrExplain the procedure to determine the no load current of the transformer.

8. June / July 2014 : 10 Marks, June / July 2009 : 10 Marks, June / July 2015 : 10 MarksCalculate: i). net cross section of the core, ii). Gross area of the core, iii). Core dimensions, iv). Window area, v). Dimensions of the window, for a 200 kVA, 6600/250V, 50 Hz, single phase shell type oil immersed transformer based on the following design parameters:Window Space Factor Kw = 0.28, Bm = 1.1 Tesla, Average current density = 2.2 A / mm2, stacking factor = 0.90, window proportion- = 2.5 :1, Rectangular core proportion = 1.8 :1. Net cross section of copper in the window is 0.20 time the net cross section of iron in the core.

9. Dec 2015 / Jan 2016 : 08 MarksA 3 phase, 50 Hz oil cooled core type transformer has the following dimensions: Distance between the adjacent core centers is 0.2 m, height of the window is 0.24 m. Diameter of circumscribing circle is 0.14 m. Flux density in the core is 1.25 T and the current density in the conductors is 2.5 A / mm 2. Estimate the kVA rating, Assume a window space factor of 0.20 and a core area factor of 0.56, core is 2 stepped.

10. June / July 2014 : 12 MarksA 300 kVA, 1100/440V, 50 HZ, 3 phase, delta / star, core type oil immersed, self cooled transformer gave the following results during the design calculations of magnetic frame and windings: Centre to centre distance between the cores = 36 cm, Height of the window = 44 cm, Height of the yoke = 17 cm, total weight of the magnetic frame = 700 kg, average specific iron loss = 2.1 W / kg, outer diameter of the HV winding = 35 cm, Resistance of LV winding per phase = 0.0047 ohms, Resistance of HV winding per phase = 9.74 ohms. Based on the above design data, calculate the following: i. The dimensions of the tank with clearance; ∆L = 8 cm; ∆W = 10 cm; ∆H = 45 cm, ii. The temperature rise of the transformer with plain tank, iii. Number of cooling tubes, if the temperature rise not to exceed 350C. Assuming diameter of cooling tube as 5 cm and length of cooling tube = 95 cm

11. Dec 2013 / Jan 2014 : 05 Marks, Dec 09 / Jan 2010 : 10 Marks, June / July 2009 : 10 Marks, Dec 2007 / Jan 2008: 5 Marks Prove that EMF/turn of a single phase transformer is K √Q, where Q is the output kVA rating / phase

12. Dec 2015 / Jan 2016 : 04 Marks


For a constant volume of conductors in a transformer, show that for a minimum copper loss, current densities in the windings must be equal.

13. Dec 2013 / Jan 2014 : 05 Marks Show that ratio of net core area to area of circumscribing circle in a 2 stepped core of transformer is 0.71

14. Dec 2013 / Jan 2014 : 10 MarksCalculate the active and reactive components of no load current of 400V, 50 Hz, single phase transformer having the following particulars: stacking factor = 0.9, density = 7.8 x 10-3 kg / m3 , length of mean flux path = 2.2 m, gross iron section = 10 x 10-2 m2, primary turns = 200, joints equivalent to 0.2 mm air gap, use the following data:

Bm wb/m2 0.9 1.0 1.2 1.3 1.4Mmf AT/m 130 210 420 660 1300Iron loss W / kg 0.8 1.3 1.9 2.4 2.9

15. Dec 2013 / Jan 2014 : 10 Marks The tank of the 1250 kVA natural oil cooled transformer has the following dimensions length, width and height as 1.55 m, 0.65 m and 1.85 m respectively. The full load loss is 13.1 kW. Find the number of cooling tubes. Assume heat dissipation due to radiation and convection as 12.50 W/m2 (6+6.5) respectively. Improvement in convection due to provision of tubes = 40 %; temperature limitation = 400C, length of the tubes = 1m and diameter of the tubes = 5 cm. neglect top and bottom surfaces of cooling.

16. June / July 2013 : 06 Marks, Why the core of the transformer is stepped? Why the width of the central limb of shell type transformer is taken double? Define iron space factor

17. June / July 2013 : 12 MarksDetermine the dimensions of core and yoke for a 200 kVA, 50 Hz, 1 phase core type transformer. A cruciform core is used with distance between adjacent limbs equal to 1.6 times the width of core laminations. Assume voltage per turn = 14 V, Bm = 1.1 Tesla, Kw = 0.32, current density = 3 A / mm2 and stacking factor = 0.90. The net iron area = 0.56 d2, where d = diameter of the circumscribing circle. Also width of the largest stamping = 0.85 d. Assume CRGO steel

18. June / July 2013 : 10 MarksCompute the active and reactive components of no load current of 5500/440 V, 50 Hz, 1 phase core type transformer with the following particulars:Mean length of the magnetic path = 300 cm, gross cross sectional area = 150 cm2, Bm = 1.5 Tesla, Specific core loss = 2.5 W / kg, Ampere turns / cm at 1.15 Tesla = 7.0, specific gravity of core material = 7.5 gm / cm 3. The effect of joins is equivalent to an air gap of 1mm in the magnetic circuit.

19. June / July 2013 : 10 Marks, May / June 2010 : 08 MarksThe tank of a 300 kVA, 3 phase oil immersed self cooled transformer is 145 cm in height and 55 cm x 115 cm in plan. The full load loss is 5.5 kW. Find the number of cooling tubes necessary to limit the temperature rise to 350C. the tubes are 5 cm in diameter and have an average length of 100 cm. neglect heat dissipation from top and bottom surfaces. Tank surface dissipates heat due to convection at the rate of 6.5 W / m2 / 0C and due to radiation at the rate of 6 W / m2 / 0C. Heat dissipated from the tube surface by convection improves by 35 %.

20. Dec 2012 : 08 Marks, June 2012 : 5 MarksDerive the e m f equation of a transformer, then show that volts / turn = K √ kVA. Also explain the factors to be taken into account while selecting the value of constant K.

21. Dec 2012 : 12 Marks, June 2012 : 10 MarksDetermine the main dimensions of the core and number of turns and area of C.S of the conductors of the primary and secondary of a 125 kVA, 6600 / 460 V, 50 Hz, single phase core type distribution transformer. Maximum flux density in the core = 1.2 wb / m2. Current density = 250 A / cm2. Assume cruciform for the assembled core allowing 8% for the insulation between laminations. Take yoke cross section as 15 % greater than the core. Window. Net cross section of copper is 0.225 times the net cross section of iron in the core. Window space factor = 0.30. Draw a neat sketch to a suitable scale, show the values.

22. Dec 2012 : 08 MarksCalculate the overall dimensions, number of turns and cross section of the conductors for a 200 kVA, 6600 / 440 V, delta – star, 50 Hz, 3 – phase core type transformer. Given the following data: e m f / turn = 10 volts, maximum flux density in the core = 1.3 T, current density = 2.5 A / mm2, window space factor = 0.30, overall height = over all width. Stacking factor = 0.90, use 3 stepped core. Draw a sketch showing details.


23. Dec 2012 : 08 MarksDesign a suitable cooling tank with cooling tubes for a 500 kVA, 6600 / 400 V, 50 Hz, 3 phase, oil immersed natural cooled transformer with the following data: temperature rise allowed for transformer tank is < 600C. Dimensions of the transformer are 100 cm height, 96 cm in length, width = 47 cm, total losses = 7 kW. Allowable temperature rise for tank walls = 350C. Cooling tube diameter = 5 cm, spaced 7.5 cm, length of cooling tubes = 90 % of height of the tank. Take clearances i) along length = 14 cm, ii) along width = 13 cm, iii) along height = 40 cm. determine the number of tubes required and draw a sketch showing their possible arrangement.

24. June 2012 : 8 MarksThe full load efficiency of a 300 kVA, 1 Φ core type transformer is 98.2 % at UPF. Design the number of cooling tubes necessary if the temperature rise is 350C. tank area may be assumed as 4.92 m2. Assume the tube diameters as 5 cm and average length of 105 cm. heat dissipation may be assumed as 12. 5 W / m2 / 0C

25. Dec 2011 : 10 MarksCalculate the dimensions of the core, the number of turns and the cross sections of the conductors for a 100 KVA, 2300 / 400 V, 50 Hz. Single phase shell type transformer, assuming: ratio of magnetic and electric loading 480 X 108; maximum flux density 1.1 T; current density 2.2 X 106 A / m2; window space factor 0.30; ratio of depth of stacked core to width of central limb 2.6; ratio of height to width of window 2.5; stacking factor 0.90

26. June / July 2011 : 10 MarksDerive the output equation of single phase transformer

27. June / July 2011 : 12 MarksFind the main dimensions of a core and window for a 500 kVA, 6600 / 400V, 50 Hz, single phase core type oil immersed self cooled distribution transformer. Assume the flux density as 1.2 T and current density of 2.75 A / mm2, window space factor 0.32; volt / turn = 16.8; use cruciform core section, height of the window is 3 times its width. Also find number of turns and cross sectional area of primary and secondary winding.

28. June / July 2011 : 8 MarksCalculate the no load current for 11000 / 400 V, 50 Hz, single phase core type transformer. If mean length of magnetic path is 300 cm, gross iron area of the core is 150 cm2, maximum flux density is 1.2 T, core loss / kg of iron is 3.3 W, AT/m of the transformer iron is 800; density of iron is 7.5 g /c.c, stacking factor 0.95, and joints are equivalent 0.1 mm of air gap.

29. Dec 2010 : 10 MarksA 100 KVA, 2000 / 400 V, 50 Hz. Single phase shell type transformer has the following particulars: maximum flux density 1.1 T; current density 2.2 X 106 A / m2; window space factor 0.30; ratio of depth of stacked core to width of central limb 2.6; ratio of height to width of window 2.5; stacking factor 0.90

30. Dec 09 / Jan 2010 : 10 MarksDetermine the dimensions of core and yoke for a 100 kVA, 50 Hz, 1 phase core type transformer. A square type core is used with distance between adjacent limbs equal to 1.6 times the width of core laminations. Assume voltage per turn = 14 V, Bm = 1.1 Tesla, Kw = 0.32, current density = 3 A / mm2 and stacking factor = 0.90. flux density in the yoke to be 80 % of flux density in core

31. May / June 2010 : 06 Marks, June / July 2015 : 10 MarksExplain the design of tubes for transformer tank. Give their general arrangement diagram

32. May / June 2010 : 07 MarksDetermine the dimensions for core and window for a 5 KVA, 50 Hz, single phase core type transformer. A rectangle core is used with long side twice as long as short side. The window height is 3 times its width. Volt / turn = 1.8 V; window space factor 0.2, current density 1.8 A / mm2, flux density 1 T. Draw a propionate sketch and indicate the dimensions.

33. May / June 2010 : 08 Marks, June / July 2009 : 10 MarksA 15000 kVA, 33 / 6.6 kV, 3 phase star-delta core type transformer has the following data: area of cross section of core limit = 0.15 m2, Area cross section of yoke = 0.18 m2 ; length of flux path in each limit 2.3 m; each yoke 1.6 m; number of turns in h v winding = 450; AT/m in core leg is 540 A /m and in yoke is 260 A/m as obtained from magnetization curves. Loss per kg in iron is 2.25 W /kg in limit and 1.4 W / kg in yoke. Density of iron is 7.8 g / c.c. estimate the no load current per phase.

34. Dec 2007 / Jan 2008: 5 MarksShow that the losses in a transformer are proportional to the cube of its linear dimensions.


PART - B

UNIT – 5 & 6 : Design of Induction Motors1. Dec 2014 / Jan 2015 : 05 Marks, June 2012: 10 Marks, Dec2013 / Jan 2014 : 08 Marks, June/July 2011: 10

Marks, May/June10: 05 Marks, Jan / Feb06: 10Marks, Dec 2015 / Jan 2016 : 04 Marks, June / July 2014 : 10 Marks, June / July 2015 : 10 MarksDiscus the various factors considered when estimating length of air gap of a 3-phase induction motor. Give the expression used in calculations of length of air gap.

OrExplain the factors which influence the length of air gap of 3phase induction motor & write few empherical formulae for the length of air gap.

OrMention the factors that are affected by the length of the air gap in 3phase induction motors. List the advantages & disadvantages of larger air gap in 3phase induction motors.

OrDiscuss the various factors that influence the choice of length of the air gap of an induction motor.

2. Dec 2012: 06 MarksWrite a brief note on choice of i) air gap flux density & ii) Specific electric loading. Bring out the advantages &disadvantages in the case of induction motor.

3. Dec 2010: 12 Marks, Dec07/Jan08: 12MarksDiscus the various factors which influence the selection of air gap, stator & rotor slots in an induction motor.

4. Dec 2015 / Jan 2016 : 04 MarksDiscuss the advantages of skewing the rotor slots in an induction motor.

5. Dec 2014 / Jan 2015 : 05 MarksBriefly give the design procedure stator teeth and core of three phase induction motor

6. Dec 2014 / Jan 2015 : 08MarksDiscuss briefly the design of end ring and rotor bar of a squirrel cage phase induction motor

7. June / July 2015 : 10 MarksExplain the step by step procedure of designing squirrel cage rotor of an induction motor.

8. June / July 2014 : 08 MarksWhat is meant by the terms ‘crawling’ & ‘cogging’ in case of a three phase induction motors? What steps would you take in the design procedure, so as to minimize these tendencies?

9. Dec 09/Jan10: 10 Marks, May/June10: 05 MarksDiscuss the various considerations to be taken into account while selecting the no. of rotor slots in squirrel cage induction motors.

10. Dec 2015 / Jan 2016 : 04 MarksWhat are the factors to be considered while designing the rotor of a slip ring induction motor?

11. Dec2013 / Jan 2014 : 05 Marks, Jun/July 2013: 10 Marks, Dec 09/Jan10: 10 Marks, Dec07/Jan08: 05MarksWhat are the usual values of specific loading?, With usual notations derive the output equation of a three phase induction motor.

12. Dec2013 / Jan 2014 : 07 MarksWrite the design procedure of slip ring induction motor rotor.

13. Dec 2015 / Jan 2016 : 4 MarksDiscuss the factors to be considered while choosing the number of slots for the rotor of an induction motor

14. Dec 2011: 10 Marks, May/June10, June/July09: 10 MarksDerive the output equation of a 3phase induction motor & explain factors which influence the choice of specific loadings.

15. Dec 2011: 10 MarksDerive the expression for rotor bar current & end ring current.

16. Jun/July 2013: 05 MarksMention the types of leakage reactance’s & draw the phasor diagram of an induction motor neglecting the loss component of no load current and the leakage reactance of rotor winding.


17. Dec 2010: 10 Marks, With equations, explain the estimation of No-load current of a three phase induction motor.

18. May / june 2010: 06 MarksDiscus the factors that contribute to the production of noise in induction motors.

19. May / June 2010: 06 MarksDiscuss in detail the design of a rotor of a single phase induction motor of split phase type including design of winding

20. Dec 2014 / Jan 2015 : 12 Marks, Jan / Feb06: 10MarksA 11 kW, three phase, 6 pole, 50 Hz, 220 V, Y – connected induction motor has 54 slots, each containing 9 conductors. Calculate the values of bar and end ring currents. The number of stator bars is 64. The machine has an efficiency of 0.86 and p.f of 0.85. The rotor mmf may be assumed to be 85 % of stator mmf. Also find the end ring sections if the current density is 5 A / mm2

21. Dec 2015 / Jan 2016 : 12 MarksA 30 HP, 3 phase, 440 V, 960 rpm, 50 Hz delta connected induction motor has a specific electric loading of 25000 AC/m and specific magnetic loading of 0.46 T. The full load efficiency is 86 %, pf is 0.87, pole pitch core length = 1. Find following : i) Stator core dimensions, ii) Number of Stator Slots and number of turns in the stator winding.

22. Dec 2014 / Jan 2015 : 10 Marks, June / July 2015 : 10 MarksDetermine the main dimensions, turns/phase, number of slots, conductor cross section and slot area of a 250 hp, three phase, 50 Hz, 400 V, 1410 rpm, slip ring induction motor. Assume: i). Barc = 0.5 T, ii). Ac/m = 30000 A, iii) Efficiency = 0.90, iv). P.f = 0.90, v). Winding factor = 0.955, vi). Current density = 3.5 A / mm2, vii). Slot space factor = 0.40, viii). Core length / Pole pitch = 1.2. Assume machine is delta connected

23. June / July 2014 : 10 MarksCalculate the following design information for a 30KW, 440V, 3-Phase, 6pole, 50 Hz delta connected, cage induction motor. i) Main dimensions of stator frame ii) No of turns per phase in state winding iii) No of stator slots iv) No of conductors per slot.The various data required for the design calculations are:Specific magnetic loading, Bav = 0.48 Tesla, Specific electric loading, q = 26000 AC/m, Full load efficiency = 0.88, Full load p.f, cosØ = 0.86, Winding factor Kw = 0.95, Slots per pole per phase = 3, Assume for best

power factor operation D=0.135 P√L24. June / July 2014 : 12 Marks

During the design of 3-phase, 5KW,400V,50Hz,4 pole squirrel cage induction motor designed for star-delta starting, following data have been obtained: Rotor diameter = 0.14m, Core length = 0.11m,turns per phase on stator = 360,air gap length = 0.45mm, Coater’s gap contraction coefficient = 1.25, Assuming that the ampere turns required for the iron parts are about 32% of that needs for the gap, calculate the magnetizing current drawn by the motor & comment, upon its value.Assume winding factor, Kw = 0.955, efficient = 85% & p.f = 0.85

25. Dec2013 / Jan 2014 : 08 Marks, A 1.1 kW,3phase,50Hz 1500rpm delta connected induction motor has stator bore D = 0.15m & core length L = 0.06m. Estimate the main dimensions of a 3.7kW,3phase 50Hz,1000rpm delta connected motor having the same loadings as the previous one. The efficiency & power factor are also same. Assume the same L/τ ratio.

26. Dec2013 / Jan 2014 : 12 MarksEstimate the stator core dimensions, number of stator slots & number of stator conductor per slot for a 100KW,3300V,50Hz,12pole star connected slip ring induction motor Assume: average gap density = 0.4wb/m2; ampere conductors/m = 25000; efficiency = 0.9, power factor = 0.9. The slot loading should not exceed 500 ampere conductors.

27. Jun/July 2013: 10 MarksFind the main dimension, number of stator turns, size of conductors & number of slots of a 5 HP,400V,3phase,50Hz,4pole, squirrel cage induction motor using delta starter. Assume the following data: Average flux density in the air gap = 0.46 T, Amp. Conductor/m at armature periphery = 22 × 103

Full load efficiency = 83%, Full load p.f = 0.84(lag), Winding factor = 0.955, current density = 4 A/mm2

Number of slots/pole/phase = 3, Take L/τ = 1.5

28. Jun/July 2013: 10 Marks. A 3phase,11 kW,440V,50Hz, 6pole delta connected squirrel cage induction motor has 54 stator slots each containing 28 conductors. Calculate the value of bar & end ring currents. The number of rotor slots is 57.


The machine has an efficient of 86% & a power factor of 0.85. The rotor mmf is 80% of stator mmf.

29. Jun/July 2013: 05MarksDetermine the leakage permeance per meter length of rectangular semi enclosed slots having the following dimension in mm: Slot width = 10 , Slot opening = 4 , Height of conductor portion = 25, Height above conductor but below wedge = 1 , Wedge height = 3 , Lip height = 1.5

30. Dec 2012: 14 MarksDetermine the man dimension , no of slots are of C.S conductor, slot area of a 150 KW, 3 – phase, 50Hz, 400V,1400rpm slip ring induction motor. Take flux density in the air gap = 0.5 Tesla, ampere conductor/cm = 300, efficiency as 90%, p.f=0.9, stator current density = 3.5A/mm2 , slot space factor = 0.4, the ratio of core length to pole pitch = 1.2, machine is delta connected, take slots\pole\phase = 4

31. Dec 2012: 12 MarksA 15 KW, 400V, 3phase, 50Hz, 6pole induction motor has a diameter of 30cms & length of the core is 12cm. The no of stator slot=72 with conductors per slot=20. The stator is delta connected calculate the magnetizing current per phase if the length of the air gap is 0.55mm. The gap contraction factor is 1.2, the mmf for iron parts is 30% that for a air gap, coil span = 11slots,find the no load current & the power factor if the windage & friction losses are 250 watts & iron losses are 850 watts, winding phase spread = 60degree.

32. Dec 2012: 08 MarksFor a 5HP, 400V, 3phase, 50Hz, 4pole squirrel cage induction motor to be started by a star-delta starter, find length of air gap, no. of rotor slots, rotor bars & end ring currents, copper loss in bars & end ring currents, rotor end ring current density = 5A/mm2 ,end ring mean diameter = 7.5cm. Take stator core diameter = 10.5cm, stator core length = 12.5cm,no of stator slots = 36,conductos per slot = 104, efficiency = 0.8 & p.f = 0.83, rotor bar length = 10.5cm, rotor bar current density = 5A/mm2, receptivity of copper = 0.021Ωm×mm2.

33. June 2012: 10 Marks, Estimate the stator core dimensions, number of stator slots, and number of stator conductors per slot for a 100 kW, 3300 V , 50Hz, 12 pole star connected slip ring induction motor. Take the average gap density of 0.4 Wb/m2 , ac/m, 25000, efficiency=90%, power factor=0.9 and winding factor =0.96. choose the main dimensions to give best p.f the slot loading must not 500 ac

34. June 2012: 08 Marks, June/July 2011: 08 Marks, A 3phase, 4pole, 50Hz induction motor has 24 stator slots & 28 rotor slots. Prove that tendency to run as synchronous motor at a speed of 214.3 rpm

35. June 2012: 12 Marks, Dec 2015 / Jan 2016 : 12 MarksA 90kW, 500V, 50Hz, 3phase, 8pole induction motor has a star connected stator winding accommodated in 63 slots with 6conductors/slot. If the slip ring voltage on open circuit is to be above 400v. Find a suitable rotor winding statingi) No. of slots ii)No. of conductor/slot iii)Coil span iv) Approximate full load current/phase in rotorAssume efficiency = 0.9, p.f = 0.86

36. Dec 2011: 10 MarksDetermine the main dimensions, no. of turns/ph., no. of slots, conductor section & slot are of a 150kW, 3phase 50Hz, 400V, 1410 rpm, slip ring induction motor. Assume average flux density as 0.57, specific electric loading as 30,000AC/m, efficiency = 0.9, p.f = 0.9, current density = 3.5 Amp/mm2, slot space factor = 0.4, machine winding is delta connected & slot loading is not to exceed 750Amp conductor.

37. Dec 2011: 10 MarksCalculate the equivalent resistance of rotor per phase with respect to stator, the current in the each bar & end ring and total rotor copper loss for a 415V,50hz, 4pole, 3phase induction motor having the following data :Stator :- slots = 48; conductors/slot = 35; current in each conductor = 10Amps.Rotor :- slots = 57; length of each bar = 0.12m; area of each bar = (9.5 X 5.5)mm2; mean diameter of end ring = 0.2m, area of each end ring = 76mm2; resistivity of copper = 0.02 ohm/m/mm2; full load p.f = 0.85

38. June/July 2011: 12 Marks, Dec 09/Jan10: 10 MarksA 15kW, 3phase, 6pole, 50Hz cage motor has the following data : D = 0.32m; l = 0.1 No. of stator slots = 54; No. of conductors/slot = 24; current in each conductor 17.5A; Full load pf = 0.85(lag). Design a suitable cage rotor giving number of rotor section of each rotor bar & end ring & effective resistance of the rotor. Give the load speed as 950rpm. Resistivity of copper as 0.02Ω mm2/m.

39. Dec 2010: 08 MarksSelect dimensions from the following range for a 25 HP, 400V, 3phase,6pole,50Hz induction motor. The


mean gap density is not to exceed 0.45 wb/m2 & specific electric loading is not to exceed 25000 Ampere conductors/m. Calculate also the turns per phase for the stator winding. The product of efficiency & p.f may be taken as 0.72, & the motor must be suitable for Y-Δ startingStator bore in m 0.25 0.3 0.36Core length in m 0.10 0.12 0.19Core length in m 0.14 0.16 0.18, Assume Kw = 0.955.

40. Dec 2010: 10 MarksCalculate the equivalent resistance of rotor per phase referred to stator from the following data of a 400V,3phase, 4pole, 50Hz cage motor. Stator slots 48,conductor/slot = 30, rotor slots = 53, one bar per rotor slot, length of each bar 12cm, area of bar 0.6cm2, mean diameter of end rings 18cm, area of ring 1.5 cm2 full pitch winding with 60 phase spread for stator. Specific resistance is 0.021Ω/m.mm2

41. Dec 09/Jan10: 10 MarksDetermine the main dimensions, number of radial ventilating ducts, no. of stator slots & no.of turns per phase of a 3.7kW, 400V, 3phase, 4pole, 50Hz cage induction motor, to be started by a star-delta starter. Assume Bav = 0.45wb/m2; ac/m = 23000, efficiency = 0.85 & p.f =0.84lagging.

42. May/June10: 10 Marks. A 15kW 440V,4pole,50Hz,3phase,induction motor is built with a stator bore diameter of 0.25m & a core length of 0.16m. The specific electric loading is 23000AC/m. Using the data obtained from this machine, determine the core dimensions, no. of stator slots & no. of stator conductors for a 11kW, 460V, 6pole, 50Hz, 3phase induction motor. Assume a full load efficiency of 84% & a p.f for each machine. The winding factor is 0.955

43. May/June10: 10 MarksDesign a cage motor for a 40HP, 3phase,400V,50Hz,6pole delta connected induction motor having the following data :- Full load efficiency = 87%, Full load p.f = 0.85, stator bore diameter = 0.33m, Gross core length = 0.17m, No. of stator slots = 54, conductor slot = 14. Assume any missing data suitably.

44. June/July09: 10 MarksDetermine the diameter of stator bore & core length of a 50kW, 415 volts, 3phase, 6pole star connected induction motor for which the specific electric & magnetic loadings are 32000 Ac/m & 0.51wb/m2 respectively. The machine is having an efficiency of 90% & the p. f is 0.91. Design the machine for best p. f. Estimate the no. of stator conductors required for a winding in which the conductors are connected in two parallel paths. Also find the no. of slots & also suitable no. of conductors/slots.

45. June/July09: 10 MarksA 15 kW, 420V, 50Hz, 3phase induction motor with a synchronous speed of 1000rpm has a delta connected stator winding accommodated in 54 slots with 26 conductors per slot. If the slip ring voltage on open circuit is to be about 350V, design a suitable rotor winding giving the details of i) no. of slots ii) no. of conductors/slot iii) approximate full load rotor current/phase iv) size of the rotor slot. Assume efficiency & p. f as 0.91

46. Dec07/Jan08: 08MarksA 8pole, 500V, dc shunt generator, with all field coils connected in series requires a mmf of 5000At/pole. The poles are of rectangular dimensions 120×25mm2. Determine i)the area of cross section of the wire ii) No. of turns iii) MMF supplied by the field. A conductor of round cross section is used. Resistivity is 0.02Ω mm2/m & the insulation of the wire increases the diameter by 0.2mm. Allow a voltage drop of 50V in the field regulator.

47. Dec 07/Jan 08: 15MarksDetermine the main dimensions, turns/phase, no. of slots, conductor area & slot area in an induction motor, rated for 250 HP, 400V, 3phase, 1410rpm(slip ring induction motor). Assume B average = 0.5Tesla, specific electric loading 30,000AC/m. Efficiency & p. f = 0.9, winding factor = 0.955, current density = 3.5A/mm2, slot space factor = 0.4, Ratio of core length to pole pitch = 1.2. The machine is delta connected.

48. JanFeb06: 10MarksCalculate the following design information for a 30kW, 440V, 3phase,6pole 50Hz delta connected squirrel cage induction motor : i) Main dimensions of stator frame ii) No. of turns per phase on stator winding iii) No. of stator slots iv) No. of conductor per slot. Assume B average = 0.48Tesla; q = 26000 ac/m ; Full load efficiency = 88% ; Full load p. f = 86%. Assume winding factor = 0.955.

49. Jan / Feb06: 10MarksA 5HP, 4pole, 3phase induction motor works on a 50hz, 400 volts mains. It is designed for star-delta starting & has the following design data : Rotor diameter = 14cm ; Gross core length = 11.5cm; Turns per phase on stator = 360 ; Air gap length = 0.4mm; Winding factor = 0.955; Iron factor = 0.95; Carters gap – contraction


coefficient = 1.25. Assume that the ampere-turns required for the iron parts are about 30% of that required for the gap. Calculate the magnetizing current.

50. June / July 2015 : 10 MarksDesign a wound rotor for a 3 phase, 850 kW, 6600 V, 50 Hz, 12 Pole, induction motor with full load efficiency of 92 % and power factor of 0.91, based on the following information’s:Gross length of the stator = 45 cm, internal diameter of the stator = 122 cm, number of stator slots = 144, number of conductors per slot = 10, number of rotor slots per pole per phase = 3.50, voltage between the slip rings = 600 V, current density = 5 A / mm2, The machine is star connected.

UNIT - 7 & 8 : Design of Synchronous Machines1. Dec 2014 / Jan 2015 : 06 Marks

Discuss the factors effecting choice of magnetic-loading in case of a synchronous machine.

2. June / July 2014 : 04 MarksCompare round-poles v/s rectangular poles.

3. Dec 2014 / Jan 2015: 06 Marks, Dec2013/Jan2014 : 08marks, JuneJuly2013: 10marks, JanFeb2006 :10Marks, Dec 2011 : 08marks, May / June 2010 : 10Marks, June / July 2014 : 06Marks, Dec 2012 : 04marks, June / July 2011:10 marks, Dec 09 / Jan10 :10 marks, Junjuly2009 :10marks, Dec 2015 / Jan 2016 : 08 Marks, Dec07/ Jan08 :08MarksWhat is short circuit ratio in a synchronous machine? How does it influence the design of alternators ?

OrExplain the term ‘short circuit ratio’ as applied in synchronous machines. How the value of short circuit ratio does affect the design of alternators?

OrDefine SCR of a Synchronous machine. Discuss its effect on performance of the machine?

4. Dec 2014 / Jan 2015: 08 MarksWhat are steps involved , in design of field windings of a synchronous machine

5. Dec 2014 / Jan 2015: 06Marks, Dec 2015 / Jan 2016 : 05 Marks, Dec2013/Jan2014 : 10marks,June/ july2011 : 06Marks, Dec 2011:06marks, Jun / July2009 :10marksDiscuss any five factors to be considered in selection of no. of slots in synchronous machines.

OrExplain the factor to be considered while selecting suitable no. of armature slots in synchronous machine.

OrWhat are the factors that effect selection of armature slots?

6. June / July 2014 ,2013: 08Marks, 10Marks, June / July 2013: 08Marks, 10Marks, June/July 2011: 10 Marks, Dec 2015 / Jan 2016 : 05 Marks, Dec2010 : 10marks, June / July 2015 : 10 MarksFrom the first principles, derive output equation 3phase alternator in terms of specific loadings, diameter & length of the stator core. What are the usual values of specific loadings?

Or

Derive the output equation of a synchronous machine. And show that HP = Input KVA X n X cosθ -------------------------------- 0.746

7. June / July2013: 10marksExplain the choice of Bav & ac in synchronous machine.

8. June / July 2014 : 06MarksIt is advisable to have field system rotating & armature stationary for large synchronous machines. Justify the above statement

9. June / July 2014 : 06Marks, Dec 2011 : 08marks, Jun / July 2011: 06marksEnumerate the advantages & disadvantages of providing a large air gap in synchronous machines.

10. Dec 2012 : 02marksWhat are the advantages of arranging stator core in packets of an alternator ?

11. Dec2010 : 08 Marks, June / July 2015 : 10 MarksExplain in steps, the design of field winding of a salient pole synchronous machine.


12. May/ June 2010 :06marks. With usual notations, derive an expression for AT per meter height of the field winding of a salient pole synchronous machine.

13. Dec 2014 / Jan 2015 : 10 Marks. Determine the main dimensions of 1000KVA, 50Hz, 3phase, 375rpm alternator. The average air-gap density is 0.55Wm/m2, ampere conductor per meter is 28000. Use rectangular poles of assume ratio of core-length to pole pitch as 2. Maximum permissible peripheral speed is 50m/s. The runaway speed is 1.8 times the synchronous speed. Assume kw = 0.955

14. June / July 2014 : 06Marks. Calculate the stator core dimensions for a 10 MVA, 11KV, 50Hz, 3phase, 2pole turbo alternator, based on the following information. Specific magnetic loading Bav= 0.63 Tesla , specific electric loading q = 48,000 amp. cond./mt , limiting peripheral speed, v = 120m/sec, length of air gap = lg = 2.0cm, Stator wdg. Factor, Kw = 0.955

15. June / July 2014 : 08Marks. A 500KVA, 33KV, 50Hz, 600rpm, 3phase salient pole alternator has 180 turns per phase. Estimate the length of air gap if the average flux density is 0.54Wb/mt2. The ratio of pole arc to pole pitch, 0.65, the short circuit ratio,1.2 the gap construction factor 1.15 & the winding factor, 0.955. The mmf required for gap is 80% of no load field mmf & the winding factor 0.955

16. Dec2013/Jan2014 : 12marksThe following is the design data available for a 1250KVA, 3phase,50Hz,3300V,star connected,300rpm salient pole alternator stator bore D = 19m; stator core length L = 0.335m; pole arc/pole pitch = 0.66; turns per phase = 150; single layer concentric winding with 5 conductors/slot; short circuit ratio = 1.2. Assume winding factor = 0.96 Calculate i) Specific magnetic loading ii) Armature mmf per pole iii) Gap density over pole arc iv) Air gap length. Given mmf required for air gap is 0.88 of a no load field mmf & gap contraction factor is 1.15.

17. Dec 2015 / Jan2016 : 10 Marks. Design suitable values of diameter and length of a 75 MVA, 11 kV, 50 Hz, 3000 rpm, 3 Phase Star connected alternator. Also determine the value of flux, conductors / slot, number of turns / phase and size of armature conductors: Given Average gap density = 0.6 T, Ampere Conductors / meter = 50000, Peripheral Speed = 180 m / s, Winding factor = 0.95, Current density = 6 A / mm2

18. Dec2013/Jan2014 : 10marksFind the main dimension of a 2500 KVA; 187.5 rpm 50Hz; 3phase salient pole synchronous generator. The generator is to be vertical wheel type. Specific magnetic loading is 0.6Wb/m2 and specific loading is 34000 ac/m. Use circular poles ratio of core length to the pole pitch = 0.65 Specify the type of pole construction used if the runaway speed is about two times the normal speed. Assume Kw = 0.955

19. June / July2013: 10marksObtain the suitable values of diameter & core length for a 1500 kVA,3300V,3phase delta connected, 10pole alternator which has specific magnetic loading 0.51 T & specific electric loading 34,000 A/m. The pole pitch to core length is 0.8. Assume winding factor as 0.955, frequency 50Hz

20. Dec 2012: 14marks. For the preliminary design of a 1250 KVA,3.3 KV 50 Hz, 300rpm 3phase star connected salient pole alternator. Determine the dimension of the stator slot. Also sketch the slot showing the arrangement of conductors & insulation details. Assume specific magnetic loading = 0.58 T, specific electric loading = 330AC/cm, core length = 1.1 x (pole pitch) , field form factor = 0.66, stacking factor = 0.96, current density = 4A/mm2, maximum flux density in the teeth is limited 1.8 T, winding type = full pitched with 60° phase spread/slot/pole/phase = 3, slot loading should not exceed 1500 A, no. of ventilating ducts = 4, width of ventilating ducts = 1cm

21. Dec 2011: 12 Marks. Determine for a 500 KVA, 6600V, 12 poles, 500 rpm, 3 phase alternator, suitable values for (1) the diameter at air gap (II) the core length (iii) the number of stator conductors (iv) the number of stator slots. Assume a stator connected stator winding a specific magnetic loading of 0.6Wb/m2

& specific electeic loading of 30,000AC/m. Assume ratio length : pole pitch = 1.5

22. Dec 2011:08marks. A 500 kVA, 3.3KV, 50Hz, 600 rpm, 3 phase salient pole alternator has 180turns/phase. Estimate the length of air gap if the average flux density = 0.575 Wb/m2; pole arc/pole pitch = 0.66; short circuit ratio = 1.2; gap contraction factor = 1.15. The mmf required for the gap is 82% of the no load field mmf. KW = 0.955.

23. JuneJuly2011:10 Marks, June / July 2015 : 10 MarksDuring the design of stator for a 3phase, 7.5KVA, 6.6KV, 50Hz,3000rpm turbo alternator, following information have been obtained. Internal diameter of stator = 0.75m, Gross core length = 0.9m, No. of stator/slots/pole/phase = 7, Sectional area of stator conductor = 190mm2, No. of conductors per slot = 4. Based on above data calculatei) Flux/pole ii)Specific loading iii) Current density for stator winding


24. JuneJuly2011:08marksFind the main dimensions of a 100MVA, 11KV, 50Hz, 150rpm, 3phase water generators. Given Bav = 0.65Wb/m2; ac/m = 40000. The peripheral speed should not exceed 65 m/sec at normal running sped in order to limit the runway speed. Assume Kw = 0.95

25. Dec2010 : 10marksDetermine the suitable no. of slots & conductors/slot, for the stator winding of a 3phase 3300V, 50Hz, 300rpm alternator. The diameter is 230 cm & axial length of core is 35cm, the maximum flux density in air gap should be approximately 0.9wb/m2. Assume the sinusoidal flux distribution. Use single layer winding & star connection for stator.

26. Dec2010 : 12marksThe following is the design data available for a 1250 KVA, 3phase 50Hz, 3300V,star connected, 300rpm alternator of salient pole type. Stator bore D =1.9m, stator bore length, L = 0.335mt, pole arc/pole pitch = 0.66, turns/phase = 150, single layer concentric winding with 5 conductors per slot, short circuit ratio = 1.2. Assume that the distribution of gap flux is rectangular region. Calculate :i) Specific magnetic loading ii)Armature mmf per pole iii)Gap density over pole arc iv)Air gap lengthmmf required for air gap is 0.88 of No-load field mmf & the gap contraction factor is 1.15.

27. Dec09/Jan10 :10marksFind the main dimensions of a 100 MVA, 11KV, 50Hz, 150 rpm, 3phase water wheel generator. Given thatBav = 0.65wb/m2 & ac/m are 40000. The peripheral speed should not exceed 65m/s at normal running speed in order to runway speed.

28. Mayjun2010 :10marksFor a 125 MVA, 6.6kV,50Hz 3phase, 3000rpm star connected turbo alternator, calculate bore diameter, core length, no. of slots & turns per phase using the following data :Average flux density in the air gap = 0.55wb/m2, Ampere conductors/m = 57,000, Length of air gap = 3cms, peripheral velocity <150m/sec.

29. Junjuly2009 :10marksA 15kW, 420V, 3phase alternator has 180 slots with 5 conductors/slot. Single layer winding with full pitched coils are used. The winding is star connected with one circuit per phase. The stator bore diameter is 0.2m & the core length is 0.4m. The winding is having a 60 degree phase spread, with full pitched coils.

30. Junjuly2009 :10marksThe following particulars refer to a 1250KVA, 3phase, 3300V, 50Hz star connected alternator running at 300rpm. Stator bore diameter = 1900cms, core length = 33.5cms, no. of ventilating ducts = 4 with 1cm width each, number slots = 180, size of slots = 13.5mm×53mm, air gap flux density = 0.576wb/m2. Ampere conductors per cm of periphery = 331. Calculate i) Flux density in stator teeth ii)Cross section of the conductor iii)No. of conductors per slots.

31. Dec07/Jan08 :12MarksObtain the main dimensions of a 500KVA , 6600V 50Hz, 12pole star connected salient pole alternator giving the following details. I) Internal diameter ii) Length of the machine No. of conductors/phase.B = 0.56T, specific electric loading = 26,000 AC/meter winding factor = 0.96

32. Jan / Feb 06 :10MarksFind the main dimensions of a 2500KVA, 187.5 rpm 50Hz, 3phase, salient pole synchronous generator. The generator is to be vertical water wheel type. The specific magnetic loading is 34,000 AC/m, use circular poles with ratio of core length to pole pitch = 0.65. Specify the type of pole construction used if the runaway speed is about two times the normal speed.

33. Dec2015/Jan2016 : 12 MarksThe field coils of a salient pole alternator are wound with a single layer winding of bare copper strip of 30 mm deep, with separating insulation of 0.15 mm thick. Determine a suitable winding length, number of turns and thickness of conductors to develop an mmf of 1200 AT with potential difference of 5 Volts per coil and with a loss of 1200 W / m2 of total coil surface. The mean length of turn is 1.2 m and resistivity of copper is 0.021 Ω / m / mm2

34. June / July 2015 : 10 Marks. Design the field coil of a 3 phase, 16 poles, 50 Hz, salient pole alternator, based on the following design information: Diameter of stator at the gap surface = 1.0 m, gross length of stator core = 0.3 m, section of the pole body = 0.5 m x 0.3 m, height of pole = 0.15 m, ampere turns per pole


= 6500, exciter voltage = 110 V, Assume 30 volts as reserve; depth of filed coil = 0.03 m, insulation of pole = 0.01 m, current density = 2.6 A / mm2.


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Electrical Machine Design Question Bank - As per VTU Syllabus