€¦ · Web viewerode sengunthar engineering college. perundurai - 638057. department of. electrical and electronics engineering. two mark question with answers. vi semester

ERODE SENGUNTHAR ENGINEERING COLLEGE

PERUNDURAI - 638057

DEPARTMENT OF

ELECTRICAL AND ELECTRONICS ENGINEERING

TWO MARK QUESTION WITH ANSWERS

VI SEMESTER

EE6604- Design of Electrical Machines

Regulation – 2013

Academic Year 2018-2019

EE6604 – DESIGN OF ELECTRICAL MACHINESUNIT 1 – INTRODUCTION

PART AANNA UNIVERSITY TWO MARK QUESTIONS AND ANSWERS

1. What are the major considerations to evolve a good design of electrical machine? (AN) (MAY/JUNE 2013) The major considerations to achieve a good electrical machine are

1. Specific electric loading2. Specific magnetic loading3. Temperature rise4. Efficiency5. Length of air-gap6. Power factor

2. Define specific magnetic loading. (RE) (MAY/JUNE 2012, 2011, NOV/DEC 2011)The specific magnetic loading defined as the average flux density over the air-gap of a

machine. specific magnetic loading = Total flux around the air gap/ Area of flux at the air-gap

= pФ/Лdl

3. Define specific electric loading. (RE) (APR/MAY 2015, 2014, 2013, 2011)The specific electric loading is defined as the number of armature (or) stator ampere

conductors per metre of armature (or) stator periphery at the air-gap. Specific electric loading, ac = Total armature ampere conductors/ Armature periphery at air-gap = Iz Z/лD

4. What is output equation? (RE) (MAY 2014)The equation describing the relationship between output and main dimensions (D and L),

Specific loadings (Bav and ac) and speed (n) of a machine is known as output equation. The output equation of dc machine is, Pa = Co D2 L n, in KW The output equation of ac machine is, Q = Co D2 L ns, in KVA Where Pa = Power developed in armature of dc machine. Q = KVA rating of ac machine. Co = Output coefficient.

5. Define total gap contraction factor, Kg. (RE) (MAY/JUNE 2010, 2011)The total gap Contraction factor, Kg is defined as the ratio of reluctance of slotted armature

with ducts and the reluctance of smooth armature without ducts.

6. What is gap contraction factor for slots? (RE) (DEC 2014)The ratio of reluctance of air gap of slotted armature to reluctance of air-gap of smooth

armature is called gap contraction factor for slots.Kgs = Ys / (Ys – Kcs Ws)

7. Define field form factor. (RE) (MAY/JUNE 2010, 2012)The field form factor, Kr is defined as the ratio of average gap density over the pole pitch to

maximum flux density in the air-gap.

8. Define Stacking factor. (RE)(NOV/DEC 2014)Stacking factor is defined as the ratio of actual length of iron in a stack of assembled core

plates to total axial length of stack.

9. What is real and apparent flux density? (RE)(DEC2012)The real flux density is due to the actual flux through a tooth. The apparent flux density is due

to total flux that has to be pass through the tooth. Since some of the flux passes through slot the real flux density is always less than the apparent or total flux density.

10. What is unbalanced magnetic pull? (RE) (NOV/DEC 2008, MAY/JUNE 2012)The unbalanced magnetic pull is the radial force acting on the rotor due to non uniform air-gap

around armature periphery.

11. What is optimum design? (RE) (MAY/JUNE 2009)The design is said to be optimum when the following are met. (i) Min loss (or ) Max efficiency, (ii) Min cost, (iii) Min Volume, (iv) Min Weight.

12. What are the factors that affect the size of the rotating machine?(AN) (APRIL/MAY 2015, NOV/DEC 2013)The factors which influence the relative values of D and L (size) in DC machines are

Machine proportions Peripheral speed Moment of inertia Voltage between adjacent commutator segments.

13. Write down the classification of magnetic materials.(UN)(APR/MAY 2014, NOV/MAY 2011)All the magnetic materials possess magnetic properties to a greater or a lesser degree. The

properties of materials are characterized by their relative permeability. The magnetic materials are classified as

(i) Diamagnetic materials – Relative permeability(µr) slightly less than unity(ii) Para magnetic materials - Relative permeability(µr) slightly greater than one(iii) Ferro magnetic materials - Relative permeability(µr) is much greater than unity

The Ferro magnetic materials further can be classified as soft magnetic and hard magnetic materials.

14. Mention the main areas of design of electrical machines. (UN)(NOV/DEC 2012)(i) Electrical circuit(ii) Dielectric circuit(iii)Magnetic circuit(iv)Thermal circuit(v)Mechanical circuit

15. What are the important specifications of a DC machine as per IS 4722:1968? (AN)(NOV/DEC 2012)

As per IS 4722:1968 “Specification for rotating electric machinery” the duty types are classified in to eight classes of duty.

16. What are the major considerations in electrical machine design? (AN)(NOV/DEC 2011)

(i) Cost(ii) Durability(iii) Small size and less weight(iv) Wider temperature operating limits.

PART BOTHER POSSIBLE QUESTIONS AND ANSWERS

1. What are the main dimensions of a rotating machine? (AN)The main dimensions of a rotating machine are the armature diameter (or) stator bore, D and

armature (or) stator core length, L

2. What are the factors that decide the choice of specific magnetic loading? (AN) The value of magnetic loading is determined by

1. Maximum flux density in iron parts of machine2. Magnetizing current and 3. Core losses.

3. What are the factors that decide the choice specific electric loading? (AN) The value of specific electric loading is determined by

1. Permissible temperature rise2. Voltage rating of machine3. Size of machine and4. Current density.

4. What is meant by magnetic circuit? (RE) The complete closed path followed by group of lines of magnetic flux is known as magnetic circuit.

Flux = MMF/ Reluctance5. What is meant by magnetic flux? (RE)

The total magnetic lines of force passing through the material is called magnetic flux. It is denoted by Ф.

6. What is meant by MMF? (RE)Magneto Motive Force: The force, which drives the magnetic flux through magnetic circuit, is

called MMF. It’s produced by passing electric current through a wire of no. of turns. It is measured in ampere turns.

7. How is total mmf calculated? (AP)The total mmf required to establish the flux in the magnetic circuit is calculated using the

knowledge of dimensions and configuration of the magnetic circuit. The magnetic circuit is split up into conventional parts, which may be connected in series or parallel. The flux density is calculated in every part and mmf per unit length; ‘at’ is found by consulting ‘B-H’ curves. The summation of mmf in series gives the total mmf.

8. What is meant by Reluctance (S)? (RE)It is the property of magnetic circuit opposes the establishment of magnetic flux in it. The unit

of reluctance is ampere turns/ weber.

9. What is magnetization curve? (RE)

The magnetization curve is a graph showing the relation between the magnetic field intensity,H and the flux density, B of a magnetic material. It is used to estimate the mmf required for flux path in the magnetic material and it is supplied by the manufacturers of stampings or laminations.

10. What is leakage coefficient? (RE)The leakage coefficient is defined as the ratio of total flux to useful flux.

Leakage coefficient, Cl = Total flux / Useful flux

11. What is Carter’s gap co-efficient? (RE)The Carter’s gap coefficient (Kcs) is the ratio of slot width to gap length. The formula which

gives the value of Kcs directly is Kcs = 1 / [ 1+ (5 lg/Ws)]

Where lg – Gap length; Ws = Width of slot

12. What are the difference between leakage flux and fringing flux? (AN)The leakage flux is not useful for energy transfer or conversion. But the fringing flux is useful

flux. The leakage flux flows in the unwanted path. But the fringing flux flows in the magnetic path.

13. What is gap contraction factor for ducts? (RE)The ratio of reluctance of air-gap with ducts to reluctance of air-gap without ducts is known as

gap contraction factor for ducts.Kgd = L / (L – Kcd nd Wd)

14. Write the expression for mmf of air-gap with smooth and slotted armatures. (RE)MMF for air-gap with smooth armature is ATg = 8,00,000 B lg

MMF for air-gap with slotted armature is ATg = 8,00,000 Kg B lg15. List the different types of slots that are used in rotating machines. (RE) The different types of slots are

1. Parallel sided slots with flat bottom2. Parallel sided slots with circular bottom3. Tapered slots with flat bottom4. Tapered slots with circular bottom5. circular slots

16. State the parameters governing slot utilization factor or slot space factor. (UN) The following factors decide the slot utilization factor:

1. Voltage rating2. Thickness of insulation3. Number of conductors per slot4. Area of cross-section of the conductor5. Dimensions of the conductor

16 MARKSPART – B

1. Discuss the advantages of hydrogen cooling. (8) (APRIL/MAY 2015)2. The exciting coil of an electromagnet has a cross section of 120x50mm2 and a length of mean

turn of 0.8m. It dissipates 150W continuously. Its cooling surface is 0.125m2 and specific heat dissipation 25W/m2-oC. Calculate the final steady state temperature rise of the coil surface. Also calculate the hot spot temperature rise of the coil if the thermal resistivity of insulating material used is 8Ωm. The space factor is 0.56. (8)(APRIL/MAY 2015).

3. Describe the classification of insulating materials used for electrical machines. (8) (APRIL/MAY 2015).

4. The temperature rise of a transformer is 25oC after one hour and 37.5oC after two hours of starting from cold conditions. Calculate its final steady state temperature rise and the heating time constant. If its temperature falls from the final steady value to 40 oC in 1.5 hour when disconnected, calculate its cooling time constant. The ambient temperature is 30 oC. (APRIL/MAY 2015).

5. What are the limitations in the design of electrical apparatus? Explain them. (8) (DEC 2014)6. Explain the modern trends in design of electrical machines. (8) (DEC 2014)7. Derive the equation of temperature rise of a machine when it is run under steady load

conditions starting from cold conditions. (8) (DEC 2014, MAY/JUNE 2013) 8. What are the main groups of Electrical conducting materials? Describe the properties and

applications of those materials. (16) (APR/MAY 2014, MAY/JUNE 2013)(NOV/DEC 2011)9. Explain in detail the various cooling methods of electrical machines. (16) (APR/MAY 2014)10. Discuss the choice of specific magnetic loading. (8) (NOV/DEC 2013, MAY/JUNE 2012)11. Discuss the choice of specific electric loading. (8) (NOV/DEC 2013, MAY/JUNE 2012)12. Calculate the specific electric and magnetic loading of 100 HP, 3000V, 3 phase, 50 Hz, 8 pole,

star connected flame proof induction motor having stator core length 0.5 m and stator bore 0.66 m. Turns/phase =286. Assume full load efficiency = 0.938 and best power factor = 0.86. (8) (NOV/DEC 2013)

13. A 600 rpm, 50 Hz, 10000V, 3 phase, Synchronous generator has the following design data: B av

= 0.48 Wb/m2, current density = 2.7 A/mm2, slot space factor = 0.35, number of slots = 144, slot size = 120x20mm, D = 1.92m, L= 0.4m. Determine the kVA rating of the machine. (8) (NOV/DEC 2013)

14. (i)State and explain various classes of insulating materials, employed in electrical machines, according to temperature limits.(10) (NOV/DEC 2012)(ii) A field coil has a cross section of 100 x 50 mm2 and its length of mean turn is 1 m. estimate the hot spot temperature above that of the outer surface of the coil if the total loss in the coil is 120W. Assume: space factor = 0.56. Thermal resistivity of insulating material = 8Ωm. (6)(NOV/DEC 2012)

15. (i) State and explain the general factors that influence the choice of specific electric and magnetic loadings for rotating machines. (12)(ii) Explain briefly about the rating of electrical machines. (6) (NOV/DEC 2012)

16. A 350kW, 500V, 450 rpm, 6 pole DC generator is build with an armature diameter of 0.87m and core length is 0.32m. The lap wound armature has 660 conductors. Calculate the specific electric and magnetic loadings.(8) (MAY/JUNE 2012)

17. A 20 HP, 440 V, 4 pole, 50 Hz, 3 phase induction motor is built with a stator bore of 0.25m and core length of 0.16m. The specific electric loading is 23000 ampere conductors per metre. Find the specific magnetic loading of the machine. Assume full load efficiency of 84% and a power factor of 0.82. (8) (MAY/JUNE 2012)

18. Describe any two methods used for determination of motor rating for variable load drives with suitable diagrams. (NOV/DEC 2011)

OTHER POSSIBLE QUESTIONS1. Discuss in detail about the cooling methods adopted in transformer. (AN) 2. What are the advantages of hydrogen as a better cooling medium for turbo alternators? (AN) 3. Discuss about the various types of thermal ratings of the electrical machines.(AN)4. Discuss about the various Insulating materials and their grades. (AN)5. Explain in detail about the MMF calculation for tapered Teeth. (EV)6. Discuss in detail about the real and apparent flux densities. (AN) 7. Write about specific magnetic loading and specific electric loading. (AP) 8. Explain about the cooling of Turbo alternators. (EV)9. What are the direct and indirect cooling methods used in electrical machines. (AN) 10. Derive an equation for the slot leakage reactance. (CR) 11. Discuss in detail about the unbalance magnetic pull. (AN) 12. Explain about the air gap reluctances in different types of armature slots. (EV)13. Derive the relationship between real and apparent flux densities. (CR)14. Obtain the expression for leakage reactance of a single phase core type transformer.(UN)

Bloom’s Level Key WordRemembering REUnderstanding UNApplying APAnalyzing ANEvaluating EVCreating CR

UNIT 2. DC MACHINESPART A

ANNA UNIVERSITY TWO MARK QUESTIONS AND ANSWERS

1. What is meant by peripheral speed of armature? (RE) (NOV/DEC 2013, MAY/JUNE 2012) The distance travel by the armature per unit time is called as peripheral speed.

Va = П Dn m/secn = speed in r.p.sD = diameter of armature in m

2. What are the factors to be considered for the selection of number of poles in dc machine? (AN)(MAY/JUNE 2014, 2013)(NOV/DEC 2011) The factors to be considered for the selection of number of pole in dc machine are frequency of flux reversals, current/ brush arm and armature mmf per pole.

3. Why square pole is preferred? (AP) (APRIL/MAY 2015, NOV/DEC 2013) If the cross-section of the pole body is square then the length of the mean turn of field winding Is minimum. Hence to reduce the copper requirement a square cross-section is preferred for the Poles of dc machine.

4. What is equalizer connection? (RE) (APRIL/MAY 2011)

The equalizer connections are low resistance copper conductors employed in lap winding to equalize the induced emfs in parallel paths. The difference in induced emf in parallel paths is due to slight unsymmetry in flux per pole in dc machines.

5. What are the factors to be considered for estimating the length of air-gap in dc machine? (AP) (MAY/JUNE 2012) The factors to be considered for estimating the length of air-gap are armature reaction, cooling, iron losses, distortion of field form and noise.

6. How the effects of armature reaction can be reduced? (RE) (APR/MAY 2011) The effects of armature reaction can be reduced by

1. increasing the length of air-gap at pole tips2. increasing the reluctance at pole tips3. Providing compensating winding and interpoles.

7. What are the factors that influence the choice of commutator diameter? (AN)(MAY 2014,NOV/DEC 2012)

1. Peripheral speed2. The peripheral voltage gradient should be limited to 3 V/mm3. Number of coils in the armature

8. What are dummy coils? (UN) ( NOV/DEC 2008) The coils which are placed in armature slot for mechanical balance but not connected electrically to the armature winding are called dummy coils.

9. Define copper space factor of the coil. (RE)(APRIL/MAY 2015)It is defined as the ratio of conductor area to the cross section area of the coil.

10. Why DC motors are preferred in general? (EV)(NOV/DEC 2014)DC motors are mostly preferred for the following advantages as follows:

Constant speed applications Variable speed applications No harmonic interference High starting torque

11. Mention the factors governing the length of armature core in a dc machine. (RE)(NOV/DEC 2014)

The factors which influence the relative values of D and L in dc machines are Machine proportions Peripheral speed Moment of inertia Voltage between adjacent commutator segments.

12. What is meant by magnetic circuit calculations? (RE)(NOV/DEC 2013)The magnetic circuit is split up into convenient parts which may be connected in series or

parallel. The flux density is calculated in every part and mmf per unit length ‘at’ is found by consulting B-H (or) B-at curves.

13. Write down the output equation of a dc machine. (UN)(MAY/JUNE 2013)

The equation describing the relationship between output and main dimensions (D and L), Specific loadings (Bav and ac) and speed (n) of a machine is known as output equation. The output equation of dc machine is, Pa = Co D2 L n, in KW Where Pa = Power developed in armature of dc machine. Co = Output coefficient.14. Name any two methods to reduce armature reaction.(UN) (NOV/DEC 2011)

Increase in length of air gap at pole tips Increasing reluctance of pole tips Compensating winding Interpoles

15. What is slot loading? (RE) (MAY/JUNE 2011)It is defined as the product of current through each conductor and conductors per slot.

Slot loading = Iz Zs


1. What are the effects of armature reaction? (AN) The various effects of armature reaction are reduction in induced emf, increase in iron losses, delayed commutation, sparking and ring firing.

2. How the polarities of interpole are decided? (RE) The polarity of the interpole must be that of the main pole just ahead (in the direction of rotation) for a generator and just behind (in the direction of rotation) for a motor.

3. What is the effect of interpole on main pole? (RE) In case of generator the interpole will magnetize the leading edge and demagnetize the trailing edge of main pole. In case of motor the interpoles will demagnetize the leading edge and magnetize the trailing edge of main pole.

4. What are the different types of commutation? AN) The different types of commutation are:

1. Resistance commutation 2. Retarded commutation 3. Accelerated commutation 4. Sinusoidal commutation

16 MARKSPART – B

1. Explain the various steps involved in the design of shunt field winding (armature winding) of dc machine. (16) (APRIL/MAY 2015, 2014,NOV/DEC 2013,2011)

2. (i)Determine the air gap length of a DC machine from the following particulars: gross length of core = 0.2m; number of ducts = 1 and is 10mm wide; slot pitch = 25mm; slot width = 10mm; carter’s coefficient for slots and ducts = 0.32; gap density at pole centre = 0.7 Wb/m2; field

mmf per pole = 3900AT, mmf required for iron parts of magnetic circuit = 800 AT. (8) (APRIL/MAY 2015).

(ii)A 5kW, 4 pole, 250V, 1500 rpm shunt generator is designed to have a square pole face. The loadings are: average flux density in the gap = 0.42 Wb/m2 and ampere conductors per metre = 15000. Find the main dimensions of the machine. Assume full load efficiency = 0.87 and ratio of pole arc to pole pitch = 0.66. (8) (APRIL/MAY 2015).

3. Calculate 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 = 65.4 mm; slot opening = 5 mm; flux per pole = 52mWb. Given carter’s coefficient is 0.18 for opening/gap = 1, and is 0.28 for opening/gap = 2. (8)(NOV/DEC 2014, 2013).

4. (i)Derive the output equation of a DC machine. (8) (NOV/DEC 2014, 2012).(ii)Calculate the diameter and length of armature for a 7.5kW, 4 pole, 1000 rpm, 220V shunt motor. Given full load efficiency = 0.83. Maximum gap flux density = 0.9 Wb/m2, specific electric loading = 30000 ampere conductors/metre; field form factor 0.7. Assume that the maximum efficiency occurs at full load and the field current is 2.5% of rated current. The pole face is square. (8) (DEC 2014)

5. (i)Explain the various factors that are affected by the selection of poles in a DC machine. (8)

(ii) Determine the diameter and length of armature core for a 55kW, 110V, 1000 rpm, 4 pole shunt generator, assuming specific electric and magnetic loadings of 26000 ampere conductors per metre and 0.5 Wb / m2 respectively. The pole arc should be about 70% of pole pitch and length of core about 1.1 times the pole arc. Allow 10 A for the field current and assume a voltage drop of 4 V for the armature circuit. Specify the winding to be used and also determine suitable values for the number of armature conductors and slots. (16)(NOV/DEC 2014, 2013) (MAY/JUNE 2012)

6. A design is required for a 50kW, 4 pole, 600 rpm, DC shunt generator, the full load terminal voltage being 220V. If the maximum gap density is 0.83 Wb/ m2 and the armature ampere conductors per metre are 30,000. Calculate suitable dimensions of armature core to give a square pole face. Assume that the full load armature voltage drop is 3% of the rated terminal voltage, and that the field current is 1% of rated full load current. Ratio of pole arc to pole pitch is 0.67. (16) (APR/MAY 2014)(NOV/DEC 2011

7. Explain the various steps involved in the design of commutator and brushes of a dc machine. (16) (MAY/JUNE 2013)

8. Calculate the diameter and length of armature for a 7.5 kW, 4 pole, 1000 rpm, 220V shunt motor. Given: full load efficiency = 0.83; maximum gap flux density = 0.9 Wb/m2, specific electric loading = 30000 ampere conductors per metre; field form factor = 0.7. Assume that the maximum efficiency occurs at full load and the field current is 2.5 percent of rated current. The pole face is square. (16) (MAY/JUNE 2013)

9. Determine the mmf required for the air gap of a DC machine having open slots, given the following particulars:Slot pitch = 4.3 cm; slot opening = 2.1 cm; gross length of core; = 48 cm; pole arc = 18 cm; air gap length = 0.6 cm; flux per pole = 0.056 weber.There are 8 ventilating ducts each 1.2 cm wide.

Ratio: Slot opening/ Gap length 1 2 3 3.5 4.0Carter’s coefficient 0.15 0.28 0.37 0.41 0.43

The above data may be used for ducts also.10. Derive the relationship between real and apparent flux densities. (CR) (8)(MAY/JUNE 2012)11. A 500kV, 4 pole, 1500 rpm shunt generator is designed to have a square pole face. The

loadings are: average flux density in the gap = 0.42 Wb/m2 and ampere conductor per metre = 15000. Find the main dimensions of the machine. Assume full load efficiency 0.87 and ratio of pole arc to pole pitch = 0.66. (8) (MAY/JUNE 2012)

OTHER POSSIBLE QUESTIONS

1. A 4 pole, 25 HP, 500 V, 600 rpm series motor has an efficiency of 82%. The pole faces are square and the ratio of pole arc to pole pitch is 0.67. Take Bav = 0.58 Wb/m2 and ac = 17000 amp.cond/m. Obtain the main dimensions of the core. (EV) (12)

2. Enumerate the procedure for shunt field design. (AP) (4)3. A 4 pole, 400 V, 960 rpm, shunt motor has an armature of 0.3 m in diameter and 0.2 m in length.

The commutator diameter is 0.22 m. Give full details of a suitable winding including the number of slots, number of commutator segments and number of conductors in each slot for an average flux density of approximately 0.55wb/m2Wb / m2 in the air gap. (EV) (12)

4. Find the main dimension and number of poles of a 37 kW, 230V, 1400 rpm, shunt motor so that a square pole face is obtained. The average gap density is 0.5 wb/m2 and the ampere conductors / meter are 22000. The ratio of pole arc to pole pitch is 0.7 and the full load efficiency is 90%. (EV)

(4)5. Derive the output equation of a DC machine. (AN) (10) 6. Distinguish between lap winding and wave winding. (CR) (6)7. Determine the total commutator losses for a 1000kw, 500V, 800rpm, 10 poles generator. Given

that commutator diameter is 1.0m, current density at brush contact =75x10-3 A/mm2 brush pressure = 14.7kv/ m2, co efficiency offriction = 0.28, brush contact drop= 2.2V. (EV) (10)

8. Discuss the significance of specific loadings in the design of DC machines. (CR) (6)9. Design a suitable commutator for a 350kw, 600rpm, 440V, 6poles DC generator having an

armature diameter of 0.75m. The number of slots is 288. Assume suitable values where it’s necessary? (CR) (10)

10. Discuss the choice of poles and speed in DC machine. (CR)(6)

11. List out the procedure involved in design of shunt field winding and series field winding? (RE) 12. Find the main dimension and number of poles of a 37 kW, 230V, 1400 rpm, shunt generator so that a square pole face is obtained. The average gap density is 0.7 wb/m2 and the ampere conductors / meter are 25000. The ratio of pole arc to pole pitch is 0.7 and the full load efficiency is 90 %. (RE) (6)13. Determine the shunt field winding of a 6-pole, 440V, dc generator allowing a drop of 15 % in the

regulator. The following design date are available, mmf per pole = 7200 AT; mean length of turn = 1.2 m; winding depth = 3.5 cm; watts per sq.cm. of cooling surface = 650. Calculate the inner, outer and end surfaces of the cylindrical field coil for cooling. Take diameter of the insulated wire to be 0.4 mm greater than the bare wire. Assume 2 micro – ohm / cm as the resistivity of copper at the working temperature. (CR)

14. Determine the diameter and length of armature core for a 55kW, 110V, 1000 rpm,4 pole shunt generator, assuming specific electric and magnetic loadings of 26000 amp.cond. / m and 0.5 Wb / m2 respectively. The pole arc should be about 70% of pole pitch and length of core about 1.1 times the pole arc. Allow 10 ampere for the field current and assume a voltage drop of 4

volts for the armature circuit. Specify the winding to be used and also determine suitable values for the number of armature conductors and slots. (CR) (16)


UNIT 3.TRANSFORMERSPART A


1. What is window space factor? (RE) (MAY/JUNE 2013) The window space factor is defined as the ratio of copper area in window to total area of window.

2. Difference between core type and shell type transformer. (AN) (NOV/DEC 2013)

CORE TYPE SHELL TYPE1. Core is surrounded by the winding.2. Construction is simple.3. Repair is easy.4. High capacity machine.

1. Winding is surrounded by the core.2. Construction is difficult.3. Repair is difficult.4. Low capacity machine.

3. Why circular coils are preferred in transformers? (RE) (APR/MAY 2015, NOV/DEC 2012) (Or) Why are the cores of large transformers built-up of circular cross section? (NOV/DEC 2013, 2014) The excessive leakage flux produced during short circuit and over loads develops severe mechanical stresses on the coils. On circular coils these forces are radial and there is no tendency for the coil to change its shape. But on rectangular coils the forces are perpendicular to the conductors and tend to deform the coil in circular form.

4. List the different types of windings used in core type transformers. (RE) (APRIL/MAY 2011) The different types of windings employed in core type transformers are

1. Cylindrical winding 5. Cross-over winding2. Helical winding 6. Disc and continuous disc winding3. Double helical winding 7. Aluminium coil winding4. Multi-layer helical winding.

5. What are the factors to be considered for selecting the cooling method of a transformer? (AN)(MAY/JUNE 2013, NOV/DEC 2012) The choice of cooling method depends on KVA rating of transformer, size, application and the site condition where it has to be installed.

6. How the heat dissipates in a transformer? (AP)(MAY 2011)

Or What are the methods by which heat dissipation occurs in a transformer? (APR/MAY 2014)(NOV/DEC 2011) The heat dissipation in a transformer occurs by conduction, Convection and Radiation.

7. Why stepped cores are used in transformers? (AN) (DEC 2014)When stepped cores are used, the diameters of the circumscribing circle is minimum for a

given area of the core, which helps in reducing the length of mean turn of the winding with consequent reduction in both cost of copper and copper loss.

8. What are the advantages of stepped cores? (APRIL/MAY 2015)If the number of steps increased in the transformer core construction results in diameter of

circumscribing circle over the core will be less. Hence, there is a saving in copper for the windings.

9. Define the term: ‘Voltage Regulation’. (APR/MAY 2014)(NOV/DEC 2011)The regulation of a transformer is defined as the ratio of change in secondary terminal voltage

between no load and full load conditions to the secondary no load voltage.

Regulation = No load secondary voltage – Full load secondary voltage

Secondary no load voltage

10. What are the cooling methods used for dry type transformers. (RE) (MAY/JUNE 2013 The different methods of cooling of transformers are,

1. Air natural (AN)2. Air Blast (AB)

11. Write the advantages of shell type transformer over core type transformer.(RE) (MAY/JUNE 2012)

The shell type transformers have greater capability of withstanding forces produced under short circuit conditions.

In shell type, the windings can be easily subdivided by using sandwich coils. It is possible to reduce the leakage reactance to any desired value. In case of shell type transformers, the core is exposed and therefore cooling is better in

core than in windings.

12. What is meant by stacking factor? (RE)(NOV/DEC 2014, MAY/JUNE 2012)Stacking factor is defined as the ratio of actual length of iron in a stack of assembled core

plates to total axial length of stack.

13. Give the relationship between emf per turn and kVA rating in a transformer. (UN)(MAY/JUNE 2011)

EMF per turn, Vt = Q SQRTof kVAWhere Q = SQRT of 4.44fx103r

14. What are the factors affecting the choice of flux density of core in a transformer?(AN) (NOV/DEC 2012)

The factors affecting the choice of flux density of core in a transformer No load current drawn Saturation limit of the magnetic materials Iron losses in the magnetic material

Temperature rise of the transformerPART B

OTHER POSSIBLE QUESTIONS AND ANSWERS

1. What is the range of efficiency of a transformer? (RE) The efficiency of a commercial transformer will be in the range of 94% to 99%. Among the available electrical machines the transformers have the highest efficiency.

2. What is transformer bank? (RE) A transformer bank consists of three independent single phase transformers with their primary and secondary windings connected either in star or in delta.

3. What is the cause of noise in transformer? (RE)1. Mechanical forces developed during working2. Loosening of stampings in the core3. Expansion and contraction of oil level

4. Difference between distribution and power transformer. (AN)

Distribution transformer Power transformer1. Power rating <= 200 KVA.2. used for distribution purposes.3. Energy efficiency is good.4. Regulation is low

1. Power rating > 200 KVA.2. used for transmission purposes.3. Power efficiency is good.4. Regulation is high.

5. What is the range of flux density in transformers? (RE) When hot rolled silicon steel laminations are used the maximum flux densities are in the range of 1.1 to 1.35 Wb/m2 for distribution transformers and 1.25 to 1.45 Wb/m2 for power transformers. When cold rolled silicon steel is employed the flux density may go upto 1.7 Wb/m2

Generally low voltage rating machines are designed to work with low values of flux densities.

6. What are the factors to be considered to choose the type of winding for a core type transformer? (AN) The type of winding used for core type transformers defends on the following factors

1. Current density 4. Surge voltage2. Short circuit current 5. Impedance3. Temperature rise 6. Transport facilities.

7. What is tertiary winding? (RE) Some three phase transformers may have a third winding called tertiary winding apart from

primary and secondary. It is also called auxiliary winding (or) stabilizing winding.

8. How the tertiary winding is connected? Why? (RE) The testing winding is normally connected in delta. When the tertiary is connected in delta, the unbalance in phase voltage during unsymmetrical faults in primary or secondary is compensated by the circulating currents flowing in the closed delta.9. List the different methods of cooling of transformers. (RE) The different methods of cooling of transformers are,

3. Air natural (AN)4. Air Blast (AB)

5. Oil Natural (ON)6. Oil Natural Air Forced (ONAF)7. Oil Natural Water Forced (ONWF)8. Oil Forced (OF)9. Oil Forced Air natural (OFAN)10. Oil Forced Air Forced (OFAF)11. Oil Forced Water Forced (OFWF)

10. How the heat dissipates in a transformer? (AP) The heat dissipation in a transformer occurs by conduction, Convection and Radiation.

11. Why transformer oil is used as a cooling medium? (AP) When transformer oil is used as a coolant the heat dissipation by convection is 10 times more than the convection due to air. Hence the transformer oil is used as coolant. Specific heat dissipation by convection due to air = 8 W/m2-o C Specific heat dissipation by convection due to oil = 80 to 100 W/m2-o C.

12. How is leakage reactance of winding estimated? (AN)It is estimated by primarily estimating the distribution of leakage flux and the resulting flux

leakage of the primary and the secondary windings. The distribution of the leakage flux depends upon the geometrical configuration of the coils and the neighboring iron masses and also on the permeability of the iron.

13. List the assumptions made for calculation of leakage flux and leakage reactance. (RE)1. The primary and secondary windings have an equal axial length2. The flux paths are parallel to the windings along the axial height3. Primary windings mmf is equal to secondary winding mmf4. Half of the leakage flux in the duct links with each winding.

14. In transformers, why the low voltage winding is placed near the core? (AP)The winding and core are both made of metals and so insulation has to be placed in between

them. The thickness of insulation depends on the voltage rating of the winding. In order to reduce the insulation requirement the low voltage winding is placed near the core.

15. What are the disadvantages of stepped cores? (EV)With large number of steps a large number of different sizes of laminations have to be used.

This results in higher labour charges for shearing and assembling different types of laminations.

16. What are the salient features of distribution transformer? (AN)1. The distribution transformers will have low iron loss and higher value of copper loss.2. The capacity of transformers will be up to 500 KVA3. The transformers will have plain walled tanks are provided with cooling tubes or radiators4. The leakage reactance and regulation will be low.

17. What types of forces acts on the coils of a transformer in the event of a short circuit on a transformer? (AN)

During short circuit conditions the radial forces will be acting on the coil, which is due to short circuit currents.18. How the leakage reactance of a transformer is reduced? ((UN)

In transformer the leakage reactance is reduced by interleaving the high voltage, and low voltage winding.

19. What is the range of current densities used in the design of transformer winding? (UN) The choice of current density depends on the allowable temperature rise, copper loss and method of cooling. The range of current density for various types of transformers is given below:

1. δ = 1.1 to 2.2 A/mm2 – for distribution transformers2. δ = 1.1 to 2.2 A/mm2 – for small power transformers with self oil cooling3. δ = 2.2 to 3.2 A/mm2 – for large power transformers with self oil cooling4. δ = 5.4 to 6.2 A/mm2 – for large power transformers with forced circulation of oil

PART – B1. Derive the output equation of a single phase transformer in terms of core and window area. (8)

(APRIL/MAY 2015) (NOV/DEC 2014,NOV/DEC 2013)(MAY/JUNE 2012)2. A 3 phase, 50Hz, single phase core type transformer has the following dimensions: distance

between core centres = 0.3 m, height of window = 0.35m, diameter of circumscribing circle = 0.2m. The flux density in the core is 1.3 Wb/m2 and the current density in the conductors 2.5 A/mm2. Estimate kVA rating if window space factor = 0.2 and a core area factor = 0.56. The core is 2 stepped.(8) (APRIL/MAY 2015).

3. A 250 kVA, 6600 / 400 V, 3-phase core type transformer has a total loss of 4800W on full load. The transformer tank is 1.25 m in height and 1 m x 0.5 m in plan. Design a suitable scheme for cooling tubes if the average temperature rise is to be limited to 35°C. The diameter of the tube is 50 mm and is spaced 75 mm from each other. The average height of the tube is 1.05 m. Specific heat dissipation due to radiation and convection is respectively 6 and 6.5 W/m2-oC.Assume that convection is improved by 35% due to provision of tubes.(16)(APRIL/MAY 2015) (NOV/DEC 2014, 2013)(MAY/JUNE 2012).

4. Determine the dimensions of core and yoke for a 200kVA, 50Hz single 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 14V, maximum flux density 1.1 Wb / m2, window space factor 0.32, current density 3 A/mm2, stacking factor =0.9. The net iron area is 0.56d2 in a cruciform core where d is the diameter of circumscribing circle. Also the width of largest stamping is 0.85d.(16) (NOV/DEC 2014)(MAY/JUNE 2012)

5. Develop the output equation for a single phase as well as three phase transformer. (16) (APR/MAY 2014)

6. A 6600V, 60 Hz single phase transformer has a core of sheet steel. The net iron cross sectional area is 22.6 x 10-3 m2, the mean length is 2.23 m and there are four lap joints. Each lap joint takes ¼ times as much reactive mmf as is required per metre of core. If Bm = 1.1 Wb / m2, determine (i) The number of turns on the 6600V winding and(ii) The no load currentAssume an amplitude factor of 1.52 and that for given flux density, mmf per metre = 232 A/m. specific loss = 1.76 W/kg. Specific gravity of plates = 7.5. (16) (APR/MAY 2014)(NOV/DEC 2011)

7. Determine the diameter of core and window for a 5 kVA, 50 Hz, 1 phase, core type transformer. A rectangle core is used with long side twice as short side. The window height is 3 times the width. Voltage per turn = 1.8V, space factor = 0.2, δ = 1.8 A/mm2, Bm = 1 Wb/mm2. (8) (NOV/DEC 2013)

8. Describe the methods of cooling of transformers. (16) (MAY/JUNE 2013)(NOV/DEC 2011)9. A single phase 400V 50 Hz transformer is built from 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 is 2.5 x 10 -3

m2 and the primary winding has 800 turns. Estimate the maximum flux and no load current of the transformer. The iron loss at the working flux density is 2.6 W/kg. iron weights 7.8x103

kg/m3. Stacking factor is 0.9. (16) (MAY/JUNE 2013)

10. Calculate approximate overall dimensions for a 200kVA, 6600/440V, 50Hz, 3 phase core type transformer. The following data may be assumed:Emf/turn = 10V; maximum flux density = 1.3 Wb/m2, current density = 2.5 A/mm2; window space factor = 0.3; overall height = overall width; stacking factor = 0.9. Use a 3 stepped core. For a 3 stepped core, width of largest stamping = 0.9d and net iron area = 0.6 d 2, where ‘d’ is the diameter of circumscribing circle. (16) (NOV/DEC 2012)

11. (i)A single phase, 400V, 50 Hz, transformer is built from stampings having a relative permeability of 1000. The length of the flux path is 2.5 m, the maximum flux density in the core is 1 Wb/m2, the weight of core is 43.8kg and the primary winding has 800 turns. The iron loss at the working flux density is 2.6 W/kg. Find the no load current of the transformer. (6)(ii)Explain the design of transformer tank with cooling tubes. (10) (NOV/DEC 2012)

OTHER POSSIBLE QUESTIONS1. Estimate the main dimensions including winding conductor area of a 3 phase delta-star core

type transformer rated at 300KVA, 6600/400 V, 50 Hz. A suitable core with 3 steps having a circumscribing circle of 0.25m diameter and leg spacing of 0.4 m is available. EMF / turn = 8.5 V, δ = 2.5A/mm2, kw=0.28 and Sf=0.9 (stacking factor). (CR) (10)

2. Derive the output equation of a single phase transformer. (AN)(6)

3. Determine the main dimensions of the core, the number of turns, the cross sectional area of conductors in primary and secondary windings of a 100 kVA, 2200/ 480 V, 1-phase, core type transformer, to operate at a frequency of 50 Hz, by assuming the following data. Approximate volt per turn = 7.5 volt. Maximum flux density = 1.2 Wb / m2. Ratio of effective cross – sectional area of core to square of diameter of circumscribing circle is 0.6. Ratio of height to width of window is 2. Window space factor = 0.28. Current density = 2.5 A/mm2. (CR)

(10)4. Explain how to estimate the no-load current of a three phase transformer. (EV) (6)5. Describe about the effect of frequency on Iron losses. (UN) (6)6. Derive the voltage per turn equation of a transformer. (AN) (8)7. Discuss about the various methods of cooling of power transformer. (CR)

(8)8. Determine the core and yoke dimensions for a 250 kVA, 50Hz, single phase, core type

transformer, Emf per turn = 12 V, the window space factor = 0.33, current density = 3A / mm2 and Bmax = 1.1 T. The distance between the centers of the square section core is twice the width of the core. (CR) (6)

9. Calculate the dimensions of the core, the number of turns and cross sectional area of conductors in the primary and secondary windings of a 250 kVA, 6600 / 400 V, 50 Hz, single phase shell type transformer. Ratio of magnetic to electric loadings = 560 x 10-8, Bm = 1.1 T, δ = 2.5 A / mm2, Kw = 0.32, Depth of stacked core / width of central limb = 2.6; height of window / width of window = 2.0. (EV)

(10)10. A 375 kVA, single phase core type transformer operating on 6.6 kV / 415V is to be designed

with approximately 7.5V per turn and a flux density of 1.1 T. Design a suitable core section and yoke section using two sizes of stampings. The width of smaller stampings may be approximately 0.62 times the larger stampings. State the assumptions made. . (CR)

(6) 11. The tank of a 500 kVA, 50Hz, 1-phase, core type transformer is 1.05 x 0.62 x 1.6 m high. The

mean temperature rise is limited to 35°C. The loss dissipating surface of tank is 5.34 m2. Total loss is 5325 W. Find the area of tubes and number of tubes needed. (RE) (10)

12. The tank of 1250 kVA, natural oil cooled transformer has the dimensions length, width and height as 0.65 x 1.55 x 1.85 m respectively. The full load loss = 13.1 kW, loss dissipation due to radiations = 6 W / m2-°C, loss dissipation due to convection = 6.5 W / m2°C, improvement in convection due to provision of tubes = 40%, temperature rise = 40°C, length of each tube = 1m, diameter of tube = 50mm. Find the number of tubes for this transformer. Neglect the top and bottom surface of the tank as regards the cooling. (EV) (12)


UNIT 4. THREE PHASE INDUCTION MOTORPART A


1. Write the expression for output equation and output coefficient of induction motor. (UN)(MAY/JUNE 2013) (APR/MAY 2014) (NOV/DEC 2011) The equation for input KVA is considered as output equation in induction motor. The input KVA, Q = Co D2 L ns in KVA

Output coefficient, Co = 11 KWS Bav ac * 10-3 in KVA/m3-rps.

2. Why fractional slot winding is not used for induction motor? (RE) (DEC 2014)Windings with fractional number of slots per pole phase create asymmetrical mmf distribution

around the air gap and favor the creation of noise in the motor. Therefore, fractional windings are not used in induction motor starter.

3. What are the factors to be considered for estimating the length of air-gap in induction motor? (CR) (MAY/JUNE 2012) The following factors are to be considered for estimating the length of air-gap.

1. Power factor 4. Unbalanced magnetic pull2. Overload capacity 5. Cooling3. Pulsation loss 6. Noise

4. What is crawling and cogging? (RE) (APR/MAY 2011) Crawling is a phenomenon in which the induction motor runs at a speed lesser than sub synchronous speed.

Cogging is a phenomenon in which the induction motor.

5. How the induction motor can be designed for best power factor?(RE) (APRIL/MAY 2015, Nov/Dec 2013, 2012)

Total permeance of leakage flux path of an induction motor is given by

^ = A L/Ʈ +BƮ

^ = A pL/πD +BπD/p

For the best power factor, leakage reactance is minimum in terms of minimum permeance of leakage flux paths

d^/ dD = 0 -ApL/πD2 + Bπ/p = 0

D2 = (A/B) x (p2/π2) x L D = Square root of (A/B) x (p/π) x Square root of L D = (Square root of 0.18) x (p/π) x Square root of L)

πD/p = (Square root of 0.18) x (Square root of L) Ʈ =( Square root of 0.18 x L)

6. Where mush winding is used? (AP)(APRIL/MAY 2015)Two types of winding used in induction motors are single layer mush winding or double layer

lap winding. For small motors mush winding is suitable.

7. List the advantages of using open slots. (AP)(NOV/DEC 2014,2013)When open slots are used, the winding coils can be formed and insulated before they are

inserted in the slots and also it is easier to replace the individual coils.

8. Define stator slot pitch. (RE)(APR/MAY 2014)(NOV/DEC 2011)Normally, the stator slot pitch is defined as the distance between centres of two adjacent stator

slots in linear scale measurement.

Stator slot pitch, Yss = πD S

s9. What is meant by an ideal short circuit current? (RE)(MAY/JUNE 2013)

It is measured based on following condition as the current drawn by the motor at standstill neglecting its resistance. For good power factor, the value of short circuit current should be large and it results in low leakage reactance.

10. Name the different types of leakage fluxes associated with 3 phase induction motors. (UN) (NOV/DEC 2012)

The different types of leakage fluxes associated with 3 phase induction motors are as follows:(i) Slot leakage flux(ii) Zig- zag leakage flux(iii) Overhang leakage flux(iv) Harmonic (or) differential leakage flux.

11. What are the advantages and disadvantages of large air-gap length in induction motor? (UN) (MAY/JUNE 2012)Advantage

A large air-gap length results in higher overload capacity, better cooling, reduction on noise and reduction in unbalanced magnetic pull.Disadvantage

The disadvantage of large air-gap length is that it results in high value of magnetizing current.12. What are the factors to be considered for selecting the number of slots in induction machine stator? (UN)(NOV/DEC 2011)

The number of stator slots selected is based on the following factors:(i) Tooth pulsation loss(ii) Leakage reactance

(iii) Mechanical difficulties(iv) Magnetizing current and iron loss(v) Cost

13. How crawling can be prevented by design in an induction motor? (UN)(APR/MAY 2011)The crawling can be prevented by selecting the rotor slots of cage type induction motor could

not be exceeding stator slots by about 15-30%.

14. Define dispersion coefficient of an induction motor. (RE)(APR/MAY 2011)It is defined as the ratio of magnetizing current to ideal short circuit current.


1. What are the main dimensions of induction motor? (AN)The main dimensions of induction motor are stator core internal diameter and stator core

length.

2. What are the different types of induction motor? How they differ from each other? (AN)The two different types of induction motor are squirrel cage and slip ring type. The stator is

identical for both types but they differ in the construction of rotor.1. The squirrel cage rotor has copper or aluminium bars mounted on rotor slots and short circuited

at both ends by end rings.2. The slip-ring rotor carries a three phase winding. One end of each phase is connected to a slip

ring and other ends are star connected.

3. Why wound rotor construction is adopted? (UN) The wound rotor has the facility of increasing the rotor resistance through slip rings. Higher

values of rotor resistance are needed during starting to get a high value of starting torque.

4. What is rotating transformer? (RE) The principle of operation of induction motor is similar to that a transformer. The stator winding is equivalent to primary of a transformer and the rotor winding is equivalent to short circuited secondary of a transformer. In transformer the secondary is fixed but in induction motor it is allowed to rotate.

5. Why rotating machines with aluminium armature coils have increased leakage reactance? (UN)

Aluminium coils in armature require more space for accommodation of conductors. Large size slots are designed. Hence with large size slots the value of leakage reactance increases.

6. Why choice of high specific loading in the design of synchronous generators loads to poor voltage regulation? (UN)

High value of specific electric loading will mean more number of turns per phase. This will cause high value of leakage reactance and poor voltage regulation.

7. List the advantages and disadvantages of using closed type of rotor slot in squirrel cage induction motor. (RE) Advantages

1. Low reluctances2. Less magnetizing current

3. Quitter operation4. Large leakage reactance and so starting current is limited

Disadvantages1. Reduced over load capacity

8. Name the methods used for reducing harmonic torques. (RE)1. Chording2. Integral slot winding3. Skewing and4. Increasing the length of air gap

9. What is full pitch and short pitch or chording? (RE) When the coil span is equal to pole pitch (180 deg electrical), the winding is called full pitched winding. When the coil span is less tha pole pitch (180 deg electrical), the winding is called short pitched or chorded.

10. What are the ranges of efficiency and power factor in induction motor? (AP) Squirrel cage motors Efficiency = 0.72 to 0.91

Power factor = 0.66 to 0.9 Slip ring motors

Efficiency = 0.84 to 0.91 Power factor = 0.7 to 0.92 The ISI specifications says that the product of efficiency and power factor shall be in the range of 0.83 to 0.88

11. What are the ranges of specific magnetic loading and electric loading in induction motor? (AP) Specific magnetic loading = 0.3 to 0.6 Wb/ m2

Specific electric loading = 5000 to 45000 amp.cond/m.

12. What are the factors to be considered for the choice of specific magnetic loading? (UN) The choice of specific magnetic loading depends on

1. power factor2. iron loss3. over load capacity4.

13. What are the factors to be considered for the choice of specific electric loading? (RE) The choice of specific electric loading depends on

1. temperature rise2. voltage rating3. Size and Cost of the machine4. overload capacity

14. What is integral slot winding and fractional slot winding? (RE) In integral slot winding the total number of slots is chosen such that the slots per pole is an integer. The integer should also be a multiple of number of phases.

In fractional slot winding the total number of slots is chosen such that the slots per pole is not an integer.

16. What is Skewing? (RE)Skewing is twisting either the stator or the rotor core. The motor noise, vibrations, cogging and

Synchronous cusps can be reduced or eliminated by skewing either the stator or the rotor.

17. What are the special features of the cage rotor of induction machine? (RE)1. The cage rotor can adopt itself for any no. of phases and poles2. It is cheaper and rugged3. It is suitable for any type of starting method except using rotor resistance starter.

18. What is the condition for obtaining the maximum torque in case of 3 phase induction motor? (RE)The maximum torque occurs in induction motor when rotor reactance is equal to rotor

resistance.

19. What are the main dimensions of induction motor? (RE)1. Stator core internal diameter2. Stator core length.

20. What is harmonic induction torque and harmonic synchronous torque? (RE)Harmonic induction torques is torques produced by harmonic fields due to stator winding and

slots.Harmonics synchronous torques is torques produced by the combined effect of same order of

stator and rotor harmonic fields.

21. What is the condition for obtaining the maximum torque in case of 3- phase induction motor? (RE)

The maximum torque occurs in induction motor when rotor reactance is equal to rotor Resistance.

16 MARKSPART –B

1. Write short notes on :(i) Design of rotor bars and slots(ii) Design of end rings. (16)(APRIL/MAY 2015, NOV/DEC 2013)

2. A 15kW, 440V, 4 pole, 50 Hz, 3 phase induction motor is built with a stator bore 0.25m and a core length of 0.16m. The specific electric loading is 23000 ampere conductors per metre. Using the data of this machine, determine the core dimensions, number of stator slots and number of stator conductors for a 11kW, 460V, 6 pole, 50Hz motor. Assume a full load efficiency of 84 % and power factor of 0.82 for each machine. The winding factor is 0.955.(16) (APRIL/MAY 2015).

3. (i)Derive the output equation of ac machine in terms of the main dimensions. (8) (NOV/DEC 2014, 2013)(MAY/JUNE 2012)(ii)Find the main dimensions of a 15kW, three phase 400V, 50Hz, 2810 rpm squirrel cage induction motor having an efficiency of 0.88 % and full load power factor of 0.9. Specific magnetic loading is 0.5 Wb/m2 and specific electric loading = 25000 A/m. Take rotor peripheral speed as approximately 20m/sec at synchronous speed. (8) (NOV/DEC 2014)

4. Estimate the stator core dimensions, number of stator slots and number of stator conductors per slot for a 100kW, 3300V, 50Hz, 12 pole star connected slip ring induction motor . Assume :Average gap density = 0.4 Wb/m2

Conductors per metre = 25000 A/mEfficiency = 0.9; power factor – 0.9 and winding factor = 0.96.Choose main dimension to give best power factor. The slot loading should not exceeding 500 ampere conductors. (16) (NOV/DEC 2014, MAY/JUNE 2013).

5. Describe the steps involved in the design of end rings. (16) (APR/MAY 2014)6. Estimate the stator core dimensions, number of stator slots and number of stator conductors per

slot for a 100kW, 3300V, 50 Hz, 12 pole star connected slip ring induction motor. Assume: average gap density = 0.4 Wb/m2, conductors/metre = 25000 A/m, efficiency = 0.9 and winding factor = 0.96. Choose main dimensions to give best power factor. The slot loading should not exceed 500 ampere conductors. (16) (APR/MAY 2014)(NOV/DEC 2011)

7. Describe the effect of dispersion coefficient due to the following factors in an induction motors:(i) Overload capacity(ii) Air gap length(iii) Number of poles and (iv) Frequency

8. (i) State and explain the factors to be considered when estimating the length of air gap of a 3 phase induction motor. (10)(ii)A 11 kW, 3 phase, 6 pole, 50 Hz, 220V, star connected induction motor has 54 stator slots, each containing 9 conductors. Calculate the current in rotor bars. The number of rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of 0.85. the rotor mmf may be assumed as 85% of stator mmf. (8) (NOV/DEC 2012)

9. A 15kW, 400V, 3 phase, 50Hz, 60Hz, 6 pole induction motor has a diameter of 0.3m and the core length of 0.12m. the number of stator slots is 72 with 20 conductors per slot. The stator is delta connected. Calculate the value of magnetizing current per phase if the length of air gap is 0.55m. the gap contraction factor is 1.2. Assume the mmf required for the iron parts to be 35% of the air gap mmf. Coil span = 11 slots. (16) (NOV/DEC 2012)

10. Determine the approximate diameter and length of stator core, the number of stator slots and the number of stator conductors for a 11kW, 400V, 3 phase, 4 pole, 1425 rpm, delta connected induction motor. Bav = 0.45 Wb/m2, ac = 23000 amp.conductors/m, full load efficiency = 0.85, power factor = 0.88, L/Ʈ = 1. The stator employs a double layer winding. (MAY/JUNE 2012)

11. Estimate the main dimension, air gap length, stator slots, slots / phase and cross sectional area of stator and rotor conductors for three phase, 15HP, 400V, 6 pole, 50Hz, 975 rpm induction motor. The motor is suitable for star – delta starting. Bav = 0.45 wb/m2. ac = 20000 AC/m. L / τ = 0.85. η = 0.9 , P.F = 0.85. (EV) (MAY/JUNE 2012)

12. Describe the steps involved in the design of magnetizing current for an induction motor from design data. (NOV/DEC 2011)

OTHER POSSIBLE QUESTIONS1. Estimate the main dimension, air gap length, stator slots, slots / phase and cross sectional area of

stator and rotor conductors for three phase, 15HP, 400V, 6 pole, 50Hz, 975 rpm induction motor. The motor is suitable for star – delta starting. Bav = 0.45 wb/m2. ac = 20000 AC/m. L / τ = 0.85. η = 0.9 , P.F = 0.85. (EV) (16)

2. A 15 kW, three phase, 6 pole, 50 Hz, squirrel cage induction motor has the following data, stator bore dia = 0.32m, axial length of stator core = 0.125 m, number of stator slots = 54, number of conductor / stator slot = 24, current in each stator conductor = 17.5 A, full load P.F = 0.85 lag. Design a suitable cage rotor giving number of rotor slots section of each bar and section

of each ring. The full speed is to be 950 rpm, use copper for rotor bar and end ring conductor. Resistivity of copper is 0.02 Ωm. (EV) (16)

3. A 90 kW, 500V, 50 Hz, three phase, 8 pole induction motor has a star connected stator winding accommodated is 63 slots with a 6 conductors / slot. If slip ring voltage, an open circuit is to be about 400V at no load find suitable rotor winding. Calculate number of rotor slots, number conductors / slot, coil span, number of slots per pole. P.F = 0.9 and the efficiency is 0.85. (EV)

(16) 4. Determine the approximate diameter and length of stator core, the number of stator slots and

the number of conductors for a 20 kW, 400V, 3 phase, 4pole, 1200rpm, delta connected induction motor. Bav =0.5T, η = 0.82, ac = 26,000 amp.cond/m, power factor = 0.8, L/τ = 1, double layer stator winding. (EV) (16)

5. Estimate the main dimensions, air-gap length, stator slots, stator turns per phase and cross sectional area of stator and rotor conductors for 3 phase, 110 kW, 3300V, 50 Hz, 10 poles, 600 rpm, Y connected induction motor, Bav = 0.48 Wb/m2, ac = 28,000 amp.cond/m, L/τ = 1.25, η = 0.9, power factor = 0.86. (EV) (16)

6. Design a cage rotor for a 18.8HP, 3phase, 440V, 50Hz, 1000rpm, induction motor having full load efficiency of 0.86, power factor = 0.86, D=0.25m, L=0.14m, Zss/Ss= 54. Assume missing data if any. (CR) (16)


UNIT 5. SYNCHRONOUS MACHINES

PART AANNA UNIVERSITY TWO MARK QUESTIONS AND ANSWERS

1. What is runaway speed? (RE) (NOV/DEC 2013, MAY/JUNE 2012, APR/MAY 2011) The runaway speed is defined as the speed which the prime mover would have, if it is suddenly unloaded, when working at its rated load.

2. What are the factors to be considered for the choice of specific magnetic loading? (RE)(MAY/JUNE 2013)(NOV/DEC 2011) The factors to be considered for the choice of specific magnetic loading are

1. Iron loss 4. Stability 2. Voltage rating 5. Parallel operation 3. Transient short circuit current

3. Define Short Circuit Ratio of a synchronous machine (SCR)? (RE)(APRIL/MAY 2008, MAY/JUNE 2009, 2012, 2013 AND 2014)(NOV/DEC 2011)

The Short Circuit Ratio (SCR) is defined as the ratio of field current required to produce rated voltage on open circuit to field current required to circulate rated current at short circuit.

It is also given by the reciprocal of synchronous reactance, Xd in p.u (per unit). For turbo=alternators SCR is normally between 0.5 to 0.7. For salient pole alternator SCR varies from 1.0 to 1.5.

4. Distinguish between cylindrical pole and salient pole construction. (AN)(APRIL/MAY 2015)S.No. Salient pole construction Cylindrical pole construction

1. Poles are projecting out from the surface Un slotted portion of the cylinder acts as pole hence poles are non-projecting

2. Diameter is high and axial length is small Smaller diameter and large axial length3. Air gap is non-uniform Air gap is uniform4. Mechanically weak Mechanically strong5. Preferred for low speed alternators Used in high speed alternators6. Separate damper winding is needed It is not necessary7. Prime movers used are hydraulic turbines

and IC enginesThe prime movers used are steam turbines and electric motors.

6. How the dimensions of induction generator differ from that of an induction motor?(AN)(APRIL/MAY 2015)S.No. Induction motor Induction generator

1. Single stator winding are used Double stator winding are used. One for motoring action and other generating action

2. Diameter is high and axial length is small Smaller diameter and large axial length3. Air gap is non-uniform Air gap is uniform4. Mechanically strong Mechanically weak

7. How is the efficiency of an alternator affected by load power factor? (AP) (DEC 2014)The efficiency of an alternator depends not only on KVA output but also on power factor of the

load. For a given load, efficiency is maximum at unity power factor and decreases as the power factor falls.

8. List the factors to be considered for the choice of number of slots in synchronous machine. (AN) (DEC 2014)

Or Mention the factors to be considered for the selection of number of armature slots. The factors to be considered for the choice of number of slots are

1. Balanced winding 4. Leakage reactance2. Cost 5. Tooth losses3. Hot spot temperature in winding 6. Tooth flux density.

9. Define specific magnetic loading of a synchronous machine. (RE)(APR/MAY 2014)The specific magnetic loading defined as the average flux density over the air-gap of a

machine. Specific magnetic loading = Total flux around the air gap/ Area of flux at the air-gap

= pФ/Лdl

10. What is the limiting factor for the diameter of synchronous machine? (RE)(NOV/DEC 2013)The diameter of stator in synchronous machine is limited by selecting suitable value of

peripheral speed. Maximum permissible speed is 175 m/s for cylindrical rotor type and 80 m/s for salient pole rotor type.

11. State the importance features of turbo alternator rotors. ((UN)(NOV/DEC 2012)The rotor winding is not concentrated but it is distributed in slots. The slot pitch is selected

such that undesirable harmonics are not introduced in the flux density curve.12. How the value of SCR affects the design of alternator? (AP)(MAY/JUNE 2012)

(i) Voltage regulation (ii) Stability (iii) Short circuit current (iv) Parallel operation (v) Self excitation.

For high stability and low regulation, the value of SCR should be high, which requires large air gap. When the length of air gap is large, the mmf requirement will be high and so the field system will be large. Hence the machine will be costlier. 13. Give the need for damper winding in synchronous machine. (RE)(MAY/JUNE 2011)

Use of damper winding on synchronous generator(i) Suppress the negative sequence field(ii) Damp out the oscillations during hunting

Use of damper winding in synchronous motor(i) Start the motor(ii) Develops damping power when the machine starts hunting


1. Name the two types of synchronous machines. (RE) Based on construction the synchronous machines may be classified as,

1. Salient pole machines2. Cylindrical rotor machines.3.

2. Define critical speed. (RE)The rotor of an alternator rotates with prime mover speed. The rotor core is structure which has

certain mass and property of elasticity. The rotor core is designed corresponding to natural frequency is called critical speed.

3. List the factors to be considered for the choice of specific electric loading? (RE) The factors to be considered for the choice of specific electric loading are

1. Copper loss 4. Synchronous reactance2. Temperature rise 5. Stray load losses3. Voltage rating

4. What are the advantages of large air-gap in synchronous machines? (RE) The advantages of large air-gap are

1. Reduction in armature reaction2. Small value of regulation3. Higher value of stability4. A higher synchronizing power which makes the machine less sensitive to load variation.5. Better cooling6. Lower tooth pulsation loss7. Loss noise8. Smaller unbalanced magnetic pull.

5. Write the expressions for length of air-gap in salient pole synchronous machine? (UN) Length of air-gap

lg = (A Tfo)/(Bg Kg * 10-6 )

(or) lg = (A Ta SCR Kf)/ (Bav Kg *106 )

6. What is the limiting factor for the diameter of synchronous machine? (RE) The limiting factor for the diameter of synchronous machine is the peripheral speed. The limiting value of peripheral speed is 175 m/sec for cylindrical rotor machines and 80 m/sec for salient pole machine.

7. How are iron and friction losses of an alternator measured? (UN) Iron and friction losses of an alternator can be measured by coupling the alternator to a suitable calibrated dc motor and driving it at synchronous speed with normal excitation. Then,Iron and Friction losses = output of motor in watts.

8. What is limiting factor for the diameter of synchronous machine? (RE)The limiting factor for the diameter of synchronous machine is the peripheral speed. The

limiting value of peripheral speed is 175 m/sec for cylindrical rotor machines and 80 m/sec for salient pole machines.

9. What is the range of rotor current density? (RE)Rotor current density ranges from about 2.5 A/mm2 for conventionally cooled machines.

However, in modern direct cooled generators, the rotor current density may be as high as 9.5 to 14 A/mm2.10. Mention the factors that govern the design of field system of alternator. (RE) The following factors to be considered for the design of field system in alternator:

1. Amp-turn per pole2. Copper loss in field coil3. Dissipating surface of field coil4. Number of poles and voltage across each field coil

11. What are the prime movers used for a) salient pole alternator b) Non-salient pole alternator? (AN)

The prime movers used for salient pole alternators are water wheels like keplan turbine. Francis turbine, Pelton wheel and diesel or petrol engines.

The prime movers used for non-salient pole alternators are steam turbines and gas turbines.

12. State the factors for separation of D and L for cylindrical rotor machine. (UN) The separation of D and L in cylindrical rotor machine depends on the following factors:

1. Peripheral speed2. Number of poles3. Short circuit ratio (SCR)4.

13. Salient pole alternators are not suitable for high speeds. Why? (EV) The salient pole rotors cannot withstand the mechanical stresses developed at high speeds. The projecting poles may be damaged due to mechanical stresses.

14. How the value of SCR affects the design of alternator? (AP)For high stability and low regulation, the value of SCR should be high, which requires large air

gap. When the length of air gap is large, the mmf requirement will be high and so the field system will be large. Hence the machine will be costlier.

PART – B1. Describe the construction of turbo alternator with neat sketch.(8)(APRIL/MAY 2015)2. For a 250kVA, 1100V, 12 pole, 500 rpm, 3 phase alternator, determine core diameter and core

length. Assume average gap density as 0.6Wb/m2 and specific electric loading of 30,000 ampere conductors per metre, L/Ʈ = 1.5. (8)(APRIL/MAY 2015)(MAY/JUNE 2012).

3. Determine the output coefficient for a 1500kVA, 2200V, 3 phase, 10 pole, 50Hz, star connected alternator with sinusoidal flux distribution. The winding has 60o phase spread and full pitch coils, ac = 30000 amp.cond/m, Bav = 0.6 Wb/m2. If the peripheral speed of the rotor must not exceed 1000 m/sec and the ratio of pole pitch to core length is to be between 0.6 and 1, find D and L, assume an air gap length of 6 mm. Find also the approximate number of stator conductors. (16) (APRIL/MAY 2015).

4. Find the main dimensions of a 25000kVA, 187.5 rpm, 50 Hz, 3 phase, 3 kV salient pole synchronous generator. The generator is to be a vertical water wheel type. The specific magnetic loading is 0.6 Wb/m2 and the specific electric loading is 34000A/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 2 times the normal speed. (16) (NOV/DEC 2014, 2013, 2012).

5. Find the main dimensions of 100MVA, 11kV, 50 Hz, 150 rpm, three phase water wheel generator. The average gap density = 0.65 Wb/m2 and ampere conductors per metre is 40000 A/m. The peripheral speed should not exceed 65m/s at normal running speed in order to limit runaway peripheral speed. (16) (NOV/DEC 2014)

6. Explain the step by step procedure for the design of field winding of synchronous machine. (16) (APR/MAY 2014, MAY/JUNE 2013, NOV/DEC 2012,2011)

7. Determine a suitable number of slots and conductors per slot, for the stator winding of a 3 phase, 3300V, 50 Hz, 300 rpm alternator. The diameter 2.3 and the axial length of core is 0.35m. The maximum flux density in the airgap should be approximately 0.9 Wb/m2. Assume sinusoidal flux distribution. Use single layer winding and star connection for stator. (16)(APR/MAY 2014, MAY/JUNE 2013)

8. Explain the armature winding and rotor design of turbo alternator. (16) (NOV/DEC 2013)9. State and explain the factors to be considered for the selection of armature slots in an

alternator. (8) (NOV/DEC 2012)10. Derive the expression for length of air gap of a synchronous machine. (8) (MAY/JUNE 2012)11. Determine the main dimensions of a 75000kVA, 13.8kV, 50 Hz, 62.5 rpm, 3 phase, star

connected alternator. Also find the number of stator slots, conductors per slot, conductor area and workout winding details. The peripheral speed should be about the 40m/s. Assume, average gap density = 0.65 Wb/m2, ampere conductors per metre = 40000 and current density 4A/mm2. (MAY/JUNE 2012)

12. A 1000 kVA, 3300V, 50Hz, 300 rpm, three phase alternator has 180 slots with 5 conductors / slot, single layer winding with full pitch coil is used. The winding is star connected with one circuit / phase. Determine specific electric loading and magnetic loading, if stator core is 0.2 m and core length = 0.4 m. The machine has 60o phase spread. (EV)

OTHER POSSIBLE QUESTIONS1. Determine the main dimension for 1000 kVA, 50 Hz, three phase, 375 rpm alternator.

The average air gap flux density = 0.55 wb/m2 and ampere conductors / m = 28000. Use rectangular pole. Assume a suitable value for L / τ in order that bolted on pole Construction is used for which machine permissible peripheral speed is 50 m/s. The runway speed is 1:8 time’s synchronous speed. (EV) (16)

2. Find main dimension of 100 MVA, 11 KV, 50 Hz, 150 rpm, three phase water wheel generator. The average gap density = 0.65 wb/m2 and ampere conductors / m are 40000. The peripheral

speed should not exceed 65 m/s at normal running speed in order to limit runaway peripheral speed. (CR) (16)

3. Determine suitable number of slots conductors / slot for stator winding of three phase, 3300V, 50 Hz, 300 rpm alternator, the diameter is 2.3m and axial length of core = 0.35 m. Maximum flux density in air gap should be approximately 0.9 wb / m2. Assume sinusoidal flux distribution use single layer winding and star connection for stator. (EV) (16)

4. Determine for 500kVA, 6600V, 20Hz, 500 rpm and connected three phase salient pole machine diameter, core length for square pole face number of stator slots and number of stator conductors for double layer winding. Assume specific magnetic loading = 0.68 tesla, ac = 30000 AC/m and Kws = 0.955. (EV) (16)

5. A 1000 kVA, 3300V, 50Hz, 300 rpm, three phase alternator has 180 slots with 5 conductors / slot, single layer winding with full pitch coil is used. The winding is star connected with one circuit / phase. Determine specific electric loading and magnetic loading, if stator core is 0.2 m and core length = 0.4 m. The machine has 60o phase spread. (EV) (16)

6. Determine for a 15 MVA, 11kV, 50 Hz, 2pole, star connected turbo alternator (i) air- gap diameter, (ii) core length, (iii) number of stator conductors, from the given data Bav= 0.55 Wb/m2, ac = 36000 amp.cond/m, δ = 5A/mm2, synchronous speed ns = 50 rps, Kws = 0.98, peripheral speed = 160 m/s. (EV) (16)


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