DESIGN OF ELECTRICAL MACHINES

PART-A

UNIT I

1 What are the basic components required for the rotating machines? (AU MAY

2017-R2008)

Ans: The basic components of all electromagnetic apparatus are the field and armature

windings supported by dielectric or insulation, cooling system and mechanical parts.

2 Mention the different types of duties and ratings of rotating electric machinery (AU

MAY 2017-R2008)

Ans: i) Continuous duty ii) Fluctuating duty and iii) Short time and intermittent duty.

3 What are the electrical properties of insulating materials? (AU MAY 2017)

Ans: High dielectric strength, high insulating resistance with low dielectric loss, good mechanical strength, good thermal conductivity and high degree of thermal stability.

4 Mention the different types of duties of electrical machines (AU MAY 2017) (AU

MAY 2016)

Ans: 1. continuous duty

2. short time duty (T‹‹ Th )

3. intermittent periodic duty

4. intermittent periodic duty with starting

5. intermittent periodic duty with starting and braking

6. continuous duty with intermittent periodic loading

7. continuous duty with starting and braking

8. continuous duty with periodic speed changes

5 Define Specific Electric Loading (AU NOV 2016) (AU MAY 2015) (AU MAY 2014)

(AU MAY 2013) (AU MAY 2011)

Ans: Specific Electric Loading is defined as the ratio of total number of ampere

conductors and the armature periphery at the air gap.

6 Mention the factors that affect the size of rotating machines. (AU NOV 2016) (AU

MAY 2015) (AU NOV 2013)

Ans: 1.Speed 2. Output coefficient 3. Specific Electric Loading

4. Specific Magnetic Loading

7 Define Specific Magnetic Loading (AU MAY 2016)

Ans: Specific Magnetic Loading is defined as the ratio of total flux around the air gap and

the area of flux path at the air gap

8 Define gap concentration factor for slots? (AU NOV 2014)

Ans: The gap contraction factor for slots, Kgs is defined as the ratio of reluctance of air-gap

of slotted armature to reluctance of air-gap of smooth armature.

Kgs = Reluctance of air - gap of slotted armature/ Reluctance of air-gap of smooth armature

Kgss=yss/(yss-Kcs*wo)

9 Define stacking factor (AU NOV 2014)

Ans: Stacking factor is defined as the ratio of actual length of iron in stacks of assembled

core plates to total axial length of stack.

10 Write down the classification of magnetic materials (AU MAY 2014)

Ans: Based on relative permeability 1. Ferromagnetic materials 2.Paramagnetic

materials & 3.Diamagnetic materials

11 Define peripheral speed and write the expression (AU NOV 2013) (AU MAY 2012)

Ans: Peripheral speed is the translational speed that may exist at the surface of rotor,

while it is rotating.

Va = πDns

12 What are the major considerations in electrical machine design? (AU MAY 2013)

Ans: The factors to be considered while designing the electrical machines are 1. Cost

2. Durability 3. Compliance with the performance specification and consumer requirement.

13 Mention the main areas of design of electrical machines (AU NOV 2012)

Ans: 1. Electromagnetic design 2. Mechanical design 3. Thermal design 4. Dielectric

design.

14 What are the important specifications as per IS 4722:1968 (AU NOV 2012)

Ans: IS: 4722-1968: specification for Rotating Electrical machinery:

1. continuous duty

2. short time duty (T‹‹ Th )

3. intermittent periodic duty

4. intermittent periodic duty with starting

5. intermittent periodic duty with starting and braking

6. continuous duty with intermittent periodic loading

7. continuous duty with starting and braking

8. continuous duty with periodic speed changes

15 What is meant by Magnetic Loading (AU MAY 2012)

Ans: The total magnetic load is defined as the total flux around the armature periphery and is

given by pϕ Weber’s

16 How materials are classified according to their degree of magnetism? (AU MAY

2011)

Ans: Based on relative permeability 1. Ferromagnetic materials 2.Paramagnetic

materials & 3.Diamagnetic materials

Based on degree of magnetism: Soft magnetic materials and Hard magnetic materials

17 Compare radial and axial cooling of machines (AU NOV 2010)

Ans:

S No Radial cooling Axial cooling

1 For cooling of rotors For cooling of stators

2 Air is forced only through

sub slots

Air is forced through

airgap from both ends

18 What are the effects of using open slots and ducts on magnetic circuit of a machine?

(AU NOV 2010)

Ans: 1. Flux will be crowded over the teeth, 2 reluctance of slotted armature with ducts

will be increased.

19 List the advantages of hydrogen cooling (AU MAY 2010)

Ans:

Advantages of hydrogen cooling:

Compared with air, hydrogen has the following properties:-

i) (1/14) th

density thereby the windage losses and noise reduced

ii) 14 times specific heat and 1.5 times heat transfer leading to improved cooling

iii) 7 times thermal conductivity resulting in reduced temperature gradient

iv) reduced corona effect v) will not support combustion so long as the hyd /air mixture

exceeds 3/1.

20 What is meant by subadvanced magnetic pull? (AU MAY 2010)

Ans: Sub advanced magnetic pull is the radial force acting on the rotor due to non

uniform air gap around armature periphery

UNIT II

1. Identify the role of main poles and interpoles in DC machine (AU MAY 2017-

R2008)

Ans: Poles are joined to the yoke with the help of bolts or welding. They carry field

winding and pole shoes are fastened to them. Pole shoes serve two purposes; (i) they

support field coils and (ii) spread out the flux in air gap uniformly.

Interpoles are designed in DC motors to overcome the effects of the armature reactance

and the self-induction of the machine.

2. Write any two guiding factors for the choice of number of poles (AU MAY 2017-

R2008) (AU NOV 2016) (AU MAY 2014) (AU MAY 2013) (AU NOV 2011)

Ans: 1) Frequency 2) Weight of iron parts 3) Weight of copper parts 4) Length of

commutator 5) Labour charges 6) Flash over and distortion of field form.

3. Distinguish between real and apparent flux densities in the tooth section of slot (AU

MAY 2017) (AU NOV 2012)

Ans:

S No Apparent flux density (Bapp) Real flux density (Breal)

1 Bapp =

Breal =

2 Higher than Breal Lesser than Bapp

4. Write the expression for brush friction losses (AU MAY 2017)

Ans: Brush friction loss =Wbf = μ ρb p Ab Vc

μ = coefficient of friction (0.1 - 0.3) ; ρb = brush pressure (10 – 15 kN/ m2)

5. Write the expression for output coefficient of DC machines. (AU NOV 2016)

Ans: Co =π2 Bav ac x10-3

Co - output coefficient, Bav = Specific Magnetic Loading ac = Specific Electric Loading

6. What is real and apparent flux density (AU MAY 2016)

Ans: The real flux density is due to the actual flux through a tooth. The apparent flux

density is due to total flux 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. B

real=Actual flux in the tooth/tooth area B app=Total flux in the slot pitch/tooth area

7. Define field form factor. (AU MAY 2016)

Ans: Field form factor is the ratio of average gap density over the pole pitch to

maximum flux density in the air gap.

8. Give the reason why square poles are preferred. (AU MAY 2015) (AU NOV 2013)

Ans: To reduce copper requirements.

9. Define copper space factor of a coil (AU MAY 2015)

Ans: The copper space factor of a coil is defined as the ratio of conductor area and the

area of the cross section of the coil.

Copper space factor = Conductor area / Area of cross section of the coil

Conductor area = Number of turns x area of cross section of conductor

10. Why DC motors are preferred in general (AU NOV 2014)

Ans: DC motors have almost linear characteristics with their speed of rotation being

determined by the current flowing through the motor winding, so mostly preferred.

11. Mention the factors governing the length of armature core in a dc machine (AU

NOV 2014)

Ans: 1.Flux density in tooth. 2.Flux pulsations. 3.Eddy current loss in conductors.

4.Reactance voltage. 5.Fabrication difficulties.

12. Mention the factors to be considered in the design of commutator of a DC machine.

(AU MAY 2014)

Ans: The factor influencing commutator design are No of commutator segments, commutator

diameter, length of commutator, commutator losses. What is meant by magnetic circuit

calculations? (AU NOV 2013)

Ans: Calculation of reluctance, flux, flux density and mmf for various sections of

magnetc circuit.

13. Write down the output equation of DC machine (AU MAY 2013)

Ans: Pa = E Ia x 10 -3

= (p/A)(ΦZN/60) Iz A 10 -3

= {π2 Bave ac x10

-3 }

D

2 L n = Co D

2 L n

where Co = output coefficient

Generator: Pa = (P/η) – (FW & Iron losses) ; Motor : Pa = P + (FW & Iron losses)

14. Mention the factors that influence the choice of commutator diameter. (AU NOV

2012)

Ans: The factors which influences the choice of commutator diameter are (i) The

peripheral speed (ii) The peripheral voltage gradient should be limited to 3 V/mm

(iii)Number of coils in the armature.

15. Define field form factor (AU MAY 2012)

Ans: Field form factor is the ratio of average gap density over the pole pitch to maximum

flux density in the air gap.

16. Explain unbalanced magnetic pull (AU MAY 2012)

Ans: Unbalanced magnetic pull is the radial force acting on the rotor due to non uniform

air gap around armature periphery

17. Show how specific electric loading and specific magnetic loading are interdependent

(AU NOV 2011)

Ans: Output of a dc machine is proportional to the product of their Bav and ac,

Pa = (Bav x ac)

For a particular out, Bav and ac are interdependent. i.e., if one is chosen higher, the other

value of other has to be lower.

18. What is slot loading?(AU MAY 2011)

Ans: The slot loading is the number of ampere conductor per slot. This value should not

exceed 1500 A. ( Iz Z < 1500 A) ( Zs – No of conductors/slot)

19. State any two methods to reduce armature reaction. (AU MAY 2011) (AU NOV

2010)

Ans: 1. Compensating windings are provided to neutralize the effects of armature

reaction.

2. By increasing reluctance of pole tips, the distorting effects armature reaction can be

reduced. 3. By increasing the length of air gap at pole tips.

20. List the advantages of lap winding in a DC machine (AU NOV 2010)

Ans: The total number of brushes is equal to the number of poles. It provides large

current and low voltage. Hence it is used for large machines.

UNIT III

1. List the advantages and disadvantages of three phase transformers (AU MAY 2017)

(R2008)

Ans:

Advantages i) A 3 phase transformer is lighter, occupies lesser space, cheaper and more efficient than a

bank of 1- phase transformers.

ii) In case of 3- phase transformers, than is only one unit to install and operate. Hence the

installation and operational costs are smaller for 3 – phase units.

Disadvantages i) Greater cost of standby Units

ii) Increased cost and inconvenience of repairs.

iii) In Single Phase transformer ( three Single Phase Transformer) failure of one

transformer, the other two, Single Phase Transformer still supply the power, while it is

not possible in case of failing a Three Phase Transformer.

2. What is meant by core earthing? (AU MAY 2017) (R2008)

Ans: Core earthing doing for the purpose of operator protection.incase of any insulation

leakage or damage, the operator will get shock hazard.

3. Define window space factor. (AU MAY 2017) (AU NOV 2016) (AU MAY 2013) (AU

NOV 2010)

Ans: The window space factor is defined as the ratio of copper area in window to total

area of window.

Kw =

4. How magnetic curves are used for calculating the no-load current of a transformer

(AU MAY 2017)

Ans: i) IM (Magnetizing/reactive /wattless) current. It magnetizes core

ii) Iw (Coreloss/ active/wattful ) current. It supplies hysteresis and eddy current

loss and negligible I²R loss

5. Mention the methods by which heat dissipation occurs in a transformer (AU NOV

2016) (AU MAY 2014) (AU NOV 2011)

Ans: i) Conduction

ii) Radiation

iii) Convection (Natural and Artificial)

6. Why the area of yoke is usually 15-20% more than that of core? (AU MAY 2016)

Ans: By keeping yoke area 15-20% higher ,the flux density in the yoke is reduced,

resulting into reduction in iron losses for yoke. The reduced core area results working

flux density and needed to increase the number of turns.

7. Why the efficiency of transformer is so high? (AU MAY 2016)

Ans: Transformers operate at higher efficiency when compared to other electrical

machines.This is due to the absence of mechanical losses which is due to the absence of

moving parts

8. List the advantages of stepped cores (AU MAY 2015)

Ans: For same area of cross section the stepped cores will have lesser diameter of the

circumscribing circle than square cores. This results in length of mean turn of the winding

with consequent reduction in both cost of copper and copper loss.

9. Explain why circular coils are preferred in transformers? (AU MAY 2015) (AU

NOV 2012)

Ans: The circular coils are preferred in transformer because the rectangular coil makes

use more length of copper for the same number of turns as compared to circular core.

10. Mention why stepped cores are used in transformers. (AU NOV 2014)

Ans: The stepped core are generally used for transformer because of the following reason LV

and HV coils are circular, for better utilization of space, for reducing the mean length of LV

and HV turns and resulting in saving of copper material

11. Why are the cores of large transformers are built-up of circular cross section. (AU

NOV 2014) (AU NOV 2013 )

Ans: The cross-section of the cores of large transformers are circular because the section

has the smallest perimeter for the given area and therefore requires less copper than

rectangular or square section.

12. Define voltage regulation (AU MAY 2014 ) (AU NOV 2011)

Ans: It is defined as change in voltage from no load voltage to full load voltage when the full

load voltage is thrown off.

% Voltage regulation of transformer =

Where V0 = open circuit voltage Vf = full load voltage

13. Differentiate core and shell type transformers. (AU NOV 2013 )

Ans:

S No Core type transformers Shell type transformers

1 In core, the windings surround the core

In shell type, the core surrounds the windings

2 Easier assembly and insulation of winding

Comparatively difficult

3 Reduction of leakage reactance is not easily possible

Reduction of leakage reactance is highly possible

4 Low mechanical strength High mechanical strength

14. Mention the cooling methods used for dry type transformers. (AU MAY 2013)

Ans: i) Air natural ii) Air blast

15. What are the factors to be considered for selecting cooling methods of a transformer

(AU NOV 2012)

Ans: The choice of method of cooling of a transformer depends on the size, type of application

and types of conditions obtaining at the site where the transformer is installed.

16. What are the merits of shell type over core type transformer? (AU MAY 2012)

Ans: Leakage reactance can be reduced to any desired value. They can produce greater

withstanding forces under short circuit conditions as the windings are surrounded and

supported by the core. The cooling is better in core than in winding.

17. Define stacking factor (AU MAY 2012)

Ans: Stacking factor is the ratio of Area of crosssection of iron in core to

Area of crosssection of iron in core

Sf =

Stacking factor (iron space factor) = 0.9

18. Give the relationship between emf per turn and KVA rating of transformer (AU

MAY 2011 )

Ans: Et = k

K =

19. What are the factors affecting the choice of flux density of core in the transformer?

(AU MAY 2011)

Ans: 1. Core area, 2.Length of mean turn of windings, 3.Iron loss, 4. Magnetizing current, 5.

Cost of iron, 6. Cost of conductor, 7. Temperature gradient across core, 8.Harmonics.

20. What are the advantages of having circular coil in a transformer? (AU NOV 2010)

Ans: The circular coils are preferred in transformer because the rectangular coil makes use

more length of copper for the same number of turns as compared to circular core.

UNIT IV

1. Define cogging (AU MAY 2017) (R2008)

Ans: When the rotor teeth and stator teeth face each other, the reluctance of the magnetic

path is minimum, that is why the rotor tends to remain fixed. This phenomenon is called

cogging or magnetic locking of induction motor.

2. State the effect of change of number of poles in induction motor (AU MAY 2017)

(R2008)

Ans: Changing the number of poles in a machine gives a set of discrete operating speeds.

3. State the rules for selecting rotor slots of squirrel cage machines. (AU MAY 2017)

Ans: Rules for selecting rotor slots of squirrel cage machines

Number of stator slots should not be equal to rotor slots satisfactory results are obtained

when Sr is 15 to 30% larger or smaller than Ss.

The difference (Ss - Sr) should not be equal to + or - p, + or – 2p or + or – 5 p to avoid

synchronous cusps.

The difference (Ss - Sr) should not be equal to + or - 1, + or – 2 , + or – (p+1) or + or –

(p+2) to avoid noise and vibrations.

4. What are the ranges of efficiency and power factor in an induction motor (AU MAY

2017)

Ans: 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 specification says that the product of efficiency and power factor shall be in the

range of 0.83 to 0.88.

5. List the advantages of using open slots (AU NOV 2016) (AU NOV 2014) (AU NOV

2013 )

Ans: The winding coils can be formed and fully insulated before installing and also it is easier to replace the individual coils.

It avoids excessive slot leakage thereby reducing the leakage reactance.

6. Why induction motor is called as rotating transformer? (AU NOV 2016)

Ans: 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

7. What are the factors to be considered for choice of specific electric loading (AU

MAY 2016)

Ans: The factors to be considered for the choice of specific electric loading are

• Copper loss

• Temperature rise

• Voltage rating

• Synchronous reactance

• Stray load losses

8. How the induction motor can be designed for best power factor? (AU MAY 2016)

(AU MAY 2015) (AU NOV 2013 ) (AU NOV 2012)

Ans: For best power factor the pole pitch, τ is chosen such that τ

9. Where mush winding is used? (AU MAY 2015)

Ans: This winding is very commonly used for small induction motors having circular

conductors.

10. Why fractional slot winding is not used for induction motor? (AU NOV 2014)

Ans: Fractional slot windings tend to create non-uniform flux density distributions in

air gap. Hence, these windings are not used in induction motor.

11. Write down the output equation for 3-phase Induction Motor (AU MAY 2014 )

Ans: Output, Q = 1.11 x PΦ x 3Iph Zph x ns Kw η cos Φ x 10-3

kW

12. Define stator slot pitch (AU MAY 2014 ) (AU NOV 2011)

Ans: The stator slot pitch is defined as the distance between cemtres of two adjacent

stator slots in linear scale measurement.

Yss =

13. Write down the equation for output coefficient in an Induction Motor. (AU MAY

2013) (AU NOV 2011) (AU NOV 2011)

Ans: Co = (11 Bav q Kw η cos Φ x 10-3

)

14. What is meant by ideal short circuit current. (AU MAY 2013)

Ans: Ideal short circuit current.is defined as the current drawn by the motor at standstill

if its resistace is neglected.

Isct =

=

15. Name the different types of fluxes associated with three phase induction motors (AU

NOV 2012)

Ans: The different leakage fluxes are:

Stator and rotor slot leakage flux

Stator and rotor zig-zag leakage flux

Stator and rotor overchange leakage flux

Belt (or) differential leakage flux

16. What are the advantages and disadvantages of large air-gap length, in induction

motor? (AU MAY 2012)

Ans: Advantages A large air-gap length results in higher overload capacity, better cooling,

reduction in noise and reduction in unbalanced magnetic pull.

Disadvantages

The disadvantage of large air-gap length is that it results in high value of magnetizing

current

17. What are the factors to be considered for selection of number of slots in induction motor

stator (AU MAY 2012)

Ans:

18. Write down the functions of end-rings in the rotor of a cage induction motor (AU

MAY 2009 )

Ans: End rings are provided to short circuit the rotor bars at both the ends.

19. Define dispersion coefficient and give its significance in an induction motor (AU NOV

2009) (AU NOV 2010)

Ans:

20. How crawling can be prevented by design in an induction motor? (AU MAY 2011)

Ans:

21. Give the methods to reduce harmonic torques in an induction motor(AU NOV 2010)

Ans: The methods used for reduction or elimination of harmonic torques are chording, integral slot winding, skewing and increasing the length of air-gap.

UNIT V

1. What is run away speed? (AU MAY 2017-R2008) (AU NOV 2013 ) (AU MAY

2012) (AU MAY 2011) (AU NOV 2010)

Ans: The runaway speed is defined as the speed which the prime mover should have, if it is suddenly unloaded, when working at its rated load.

2. Classify the types of synchronous machines (AU MAY 2017) (R2008)

Ans: Based on construction the synchronous machines may be classified as, 1. Salient pole machines. 2. Cylindrical rotor machines.

3. What are the factors that are affected due to SCR? (AU MAY 2017)

Ans:

4. State three important features of turboalternators (AU MAY 2017) (AU NOV 2012)

Ans: 1. The rotors of turbo alternators have large axial length and small diameters.

2. Damping torque is provided by the rotor itself and so there is no necessity for additional

damper winding.

3. They are suitable for high speed operations and so number of poles is usually 2 or 4.

5. State the factors for separation of D and L for cylindrical rotor machine (AU NOV

2016)

Ans: The factors to be considered are

1. Peripheral speed

2. Number of poles

3. Short circuit ratio

6. List the factors that govern the design of field system of alternator (AU MAY 2016)

Ans:

7. What is short circuit ratio? (AU MAY 2016) (AU MAY 2014 ) (AU MAY 2013) (AU

NOV 2012) (AU NOV 2011)

Ans: The short circuit ratio (SCR) of a synchromous machine is define as the ratio of

field current required to produce rated voltage or open circuit to field current required

to circulate rated current at short circuit.

8. How is cylindrical pole different from salient pole in a synchronous machine? (AU

MAY 2015) (AU MAY 2009)

Ans: i) Cylindrical pole are non, projecting pole whereas the salient pole machines are

projecting pole.

ii) Cylindrical rotor construction is used for turbo alternators which are driven by high

speed steam or gas turbines where as salient pole construction is used for generators

driven by hydraulic turbine since these turbines 'operate at relatively low speeds.

9. How the dimensions of induction generator differ from that of an induction motor (AU

MAY 2015)

Ans:

10. How is the efficiency of an alternator affected by load power factor? (AU NOV

2014)

Ans: At Unity Power factor : The effect of armature reaction is minimum, hence

merely distorts the main flux.

At Lagging Power factor: The armature mmf opposes the main field, results in a

weakened field and a low generated voltage.

At Unity Power factor: The armature mmf aids the main field, results in a higher

generated voltage.

11. Mention the factors to be considered for the selection of number of armature slots.

(AU NOV 2014)

Ans: i) Balanced winding ii) Cost iii) Leakage reactance iv) Tooth ripples

12. Define Specific Magnetic Loading of synchronous machines (AU MAY 2014 ) (AU

NOV 2011)

Ans: Specific Magnetic Loading is defined as the ratio of total flux around the air gap and

the area of flux path at the air gap

Bav =

13. What are the limiting factors for the diameter of synchronous machine? (AU NOV

2013 )

Ans: The limiting factor of synchronous machine is the peripheral speed. The limiting

value of peripheral speed is 175 m/s for cylindrical and 80 m/s for salient pole machines

14. Explain how the value of SCR affects the design of alternator (AU MAY 2012)

Ans: For high stability and low regulation value of SCR should be high which

requires large air gap. When length of air gap is large, mmf requirement will be high and

so field system will be large. Hence the machine will be costlier.

15. Give the need for damper winding in synchronous machine. (AU MAY 2011)

Ans: To prevent hunting.

16. What is the effect of SCR on synchronous machine performance?

Ans: 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 so the

field system will be large. Hence the machine will be costlier.

17. Mention the factors to be considered for the design of field system in alternator.

Ans: The factors that govern the design of field system of the alternators are

1. Number of poles and voltage across each field winding

2. Amp-turn per pole

3. Copper loss in the field coil

4. Dissipating surface of field coil

5. Specific loss dissipation and allowable temperature rise.

18. Why semi- closed slots are generally preferred for the stator of induction motors?

Ans:

19. Why choice of high specific loading in the design of synchronous generators loads to

poor Voltage regulation?

Ans:

20. Write the expressions for length of air-gap in salient pole synchronous machine?

Ans:

PART-B

UNIT 1 1. Describe any two methods used for determination of motor rating for variable load drives

with suitable diagrams . (AU MAY 2017) RATING OF MACHINES:

DETERMINATION OF MOTOR RATING: Different methods to calculate the proper rating of motors for variable load drives.

method of variable losses

equivalent current method

equivalent torque method

equivalent power method

Methods of average losses:

The method consists of finding average losses Qav in the motor when it operates according to the

given load diagram. These losses are compared with Qnom. The method of average losses

presuppose that when Qav = Qnom , the motor will operate without temperature rise going

above the permissible for the particular class of insulation.

θper=Qnom/Sλ

and therefore θm=Qav/Sλ

θm= θper=Qav/Sλ=Qnom/Sλ

2. Discuss in details the various ratings and duties of electric machines (8). (AU MAY

2008) TYPES OF DUTIES AND RATING OF MACHINES: Various types of duty as per IS:4722-1968 “Specification for rotating machines”.

Continuous duty(S1)

Short time duty(S2)

Intermittent periodic duty(S3)

Intermittent periodic duty with starting(S4)

Intermittent periodic duty with starting and braking(S5)

Continuous duty with intermittent periodic loading(S6)

Continuous duty with starting and braking(S7)

Continuous duty with periodic speed changes(S8)

3. Write down the properties and application of insulating materials.(8) (AU NOV 2008)

Insulating Materials: Insulating materials are used to provide an electrical insulation

between parts at different potentials. Design of electrical machine is limited by the

restriction imposed by the insulating materials.

Properties of insulating materials: The proper selection of an insulating material for a

particular condition needs the knowledge of their electrical and mechanical

properties. The three fundamentals electrical qualities of insulating materials of great

importance to the designer of electrical machine or equipment are,

insulation resistance or resistivity

electric strength or dielectric strength

dielectric loss angle

In addition to the above, other properties such as mechanical strength, heat

resistance, etc.,

An ideal insulating materials must have the following properties,

high insulating resistance

high dielectric strength

low dielectric loss and low dielectric loss angle

no attraction for moisture

good heat conductivity

sufficient mechanical strength to withstand vibration and bending

solid materials should have a high melting or softening point.

liquid material should not evaporate.

Classification of Insulation Systems

Class A insulation consists of materials such as cotton, silk and paper when suitably

impregnated or coated or when immersed in a dielectric liquid such as oil. Other

materials or combinations of materials may be included in this class if by experience or

tests they can be shown to be capable of operation at the Class A temperature.

Maximum allowed temperature: (IEC60034-1 & NEMA MG1-12.43): 105C, 221F.

Class B insulation consists of materials or combinations of materials such as mica, glass

fibre, asbestos, etc., with suitable bonding, impregnating or coating substances (beware

a few older applications used asbestos). Other materials or combination of materials, not

necessarily inorganic, may be included in this class, if by experience or tests they can be

shown to be capable of operation at the class B temperature.

Maximum allowed temperature: (IEC60034-1 & NEMA MG1-12.43): 130C, 266F.

Class C insulation consists of materials or combinations of materials such as mica,

porcelain, glass, quartz with or without an inorganic binder (beware a few older

applications used asbestos). Other materials or combinations of materials may be

included in this class, if by experience or tests they can be shown to be capable of

operation at temperatures above the Class H limit. Specific materials or combinations of

materials in this class will have a temperature limit, which is dependent upon their

physical, chemical and electrical properties.

Maximum allowed temperature: (IEC60034-1 only): >180C, 356F.

Class E insulation consists of materials or combinations of materials, which by experience

or tests can be shown to be capable of operation at Class E temperature (materials

possessing a degree of thermal stability allowing them to be operated at a temperature

15 Centigrade degrees higher than Class A materials).

Maximum allowed temperature: (IEC60034-1 only): 120C, 248F.

Class F insulation consists of materials or combinations of materials such as mica, glass

fibre, asbestos, etc., with suitable bonding, impregnating or coating substances, as well

as other materials or combinations of materials, not necessarily inorganic, which by

experience or tests can be shown to be capable of operation at the Class F temperature

(materials possessing a degree of thermal stability allowing them to be operated at a

temperature 25 Centigrade degrees higher than Class B materials).

Maximum allowed temperature: (IEC60034-1 & NEMA MG1-12.43): 155C, 311F.

Class H insulation consists of materials such as silicone elastomer and combinations of

materials such as mica, glass fibre, asbestos etc., with suitable bonding, impregnating or

coating substances such as appropriate silicone resins. Other materials or combinations

of materials may be included in this class if by experience or tests they can be shown to

be capable of operation at the Class H temperature.

Maximum allowed temperature: (IEC60034-1 & NEMA MG1-12.43): 180C, 356F.

Application of insulating materials

Cables and transmission lines:

Insulating material is generally used as a protective coating on electrical conductor and cables.

Cable cores which touch each other should be separated and insulated by means of insulation

coating on each core, e.g. polyethylene, cross linked polyethylene-XLPE, polyvinyl chloride-

PVC, Teflon, silicone etc. Hanging disk insulators (bushings) are used in high voltage

transmission bare cables where they are supported by electrical poles. Bushings are made from

glass, porcelain, or composite polymer materials.

The transformer insulation:

(a)conductor or turn-to-turn insulation,

(b) coil-to-coil insulation,

(c)low voltage coil-to-earth insulation,

(d) high voltage coil-to-low voltage coil insulation, and

(e) high voltage coil-to-ground insulation

Transformer oil

provides the required dielectric strength and insulation

cools the transformer by circulating itself through the core and the coil structure.

should be in the liquid state over the complete operating range of temperatures between -

40°C and+50°C

gets oxidized when exposed to oxygen at high temperatures

formation of peroxides, water, organic acids and sludge.

chemical deterioration of the paper insulation and the metal parts of the

transformer.

sludge being heavy, reduces the heat transfer capabilities of the oil, and also

forms as a heat insulating layer on thecoil structure, the core and the tank walls.

the effects of oxidation are minimized by designing them such that access to oxygen itself

is limited.

sealed transformers

filling with nitrogen gas,

providing oxygen absorbers like activated clay or alumina

arc discharge inside a transformer decomposes the oil and causes explosions

Application in Rotating machines

low voltage machines: up to 6,600 V , class E or F insulation

high voltage machines: 6,600 V and up, Class F insulation

machines above 22 kV rating are not built except under special conditions

Application in Circuit Breakers

Low voltage breakers

use synthetic resin moldings to carry the metallic parts.

for higher temperatures ceramic parts are used.

if the arc is likely to come into contact with molded parts, melanine or some

special kind of alkyd resins are used because of their greater arc resistance.

High voltage breakers: air circuit breakers and oil circuit breakers.

many insulating fluids are suitable for arc extinction

the choice depends on the rating and type of the circuit breaker.

commonly used insulating fluids:

atmospheric air,

compressed air,

high vacuum,

SF6

oil (transformer oil) (interrupts the arc)

Askarels produce large quantities of toxic and corrosive products

4. Describe the properties and applications of electrical conducting materials. (AU MAY 2013)

Conducting materials Commonly used conducting materials are copper and aluminum. Some of the desirable

properties a good conductor should possess are listed below.

1. Low value of resistivity or high conductivity

2. Low value of temperature coefficient of resistance

3. High tensile strength

4. High melting point

5. High resistance to corrosion

6. Allow brazing, soldering or welding so that the joints are reliable

7. Highly malleable and ductile

8. Durable and cheap by cost

For the same resistance and length, cross-sectional area of aluminum is 61% larger than

that of the copper conductor and almost 50% lighter than copper.

Though the aluminum reduces the cost of small capacity transformers, it increases the size

and cost of large capacity transformers. Aluminum is being much used now a days only

because copper is expensive and not easily available. Aluminum is almost 50% cheaper

than Copper and not much superior to copper.

Cables and wiring

Low- and medium-voltage cables

Transformers

Busbars

5. Explain the choice of specific magnetic loading (8) (AU MAY 2012) Choice of Specific Magnetic loadings

Following are the factors which influences the performance of the machine.

(i) Iron loss: A high value of flux density in the air gap leads to higher value of flux in the

iron parts of the machine which results in increased iron losses and reduced efficiency.

(ii) Voltage: When the machine is designed for higher voltage space occupied by the

insulation becomes more thus making the teeth smaller and hence higher flux density in

teeth and core.

(iii) Transient short circuit current: A high value of gap density results in decrease in

leakage reactance and hence increased value of armature current under short circuit

conditions.

(iv) Stability: The maximum power output of a machine under steady state condition is

indirectly proportional to synchronous reactance. If higher value of flux density is used it

leads to smaller number of turns per phase in armature winding. This results in reduced

value of leakage reactance and hence increased value of power and hence increased

steady state stability.

(v) Parallel operation: The satisfactory parallel operation of synchronous generators

depends on the synchronizing power. Higher the synchronizing power higher will be the

ability of the machine to operate in synchronism. The synchronizing power is inversely

proportional to the synchronous reactance and hence the machines designed with higher

value air gap flux density will have better ability to operate in parallel with other

machines.

6. Explain the choice of specific electric loading (8) (AU NOV 2013) Choice of Specific Electrical and Magnetic loadings

Following are the some of the factors which influence the choice of specific electric

loadings.

(i) Copper loss: Higher the value of q larger will be the number of armature of conductors

which results in higher copper loss. This will result in higher temperature rise and

reduction in efficiency.

(ii) Voltage: A higher value of q can be used for low voltage machines since the space

required for the insulation will be smaller.

(iii) Synchronous reactance: High value of q leads to higher value of leakage reactance

and armature reaction and hence higher value of synchronous reactance. Such machines

will have poor voltage regulation, lower value of current under short circuit condition

and low value of steady state stability limit and small value of synchronizing power. (iv) Stray load losses: With increase of q stray load losses will increase. Values of specific

magnetic and specific electric loading can be selected from Design Data Hand Book for

salient and non salient pole machines.

7. Explain in detail the various cooling methods of electrical machines. (AU MAY 2014)

8. A field coil has a heat dissipating surface of 0.15 m2 and length of mean turn 1 m. It dissipates

loss of 150 W, the emissivity being 34 W/m2_°C. Estimate the final steady temperature rise of

the coil and its time constant if the cross section of the coil is 100*50 mm2. Specific heat of

copper is 390 J/kg° C. The space factor is 0.56. Copper weighs 8900 kg/m3. (8) (AU MAY 2011) 9. Derive the equation of temperature rise of a machine when it runs under steady load

conditions starting from cold conditions. (AU NOV 2014 )

10. Calculate the specific magnetic loading of 100HP, 300V, 3phase, 50 Hz, 8 pole, star connected flame

proof induction motor having stator core length=0.5m and stator bore=0.66m, turns/phase=286.

Assume full load efficiency as 0.938 and power factor as 0.86(AU NOV 2013)

11. Describe the methods of measurement of temperature rise in various parts of an electrical

machine. . (AU MAY 2013)

UNIT 2

1. Derive the output equation of a DC machine. (8)(AU NOV 2014)

Output equation relates the output and main dimensions of the machine. Actually it

relates the power developed in the armature and main dimensions.

E : EMF induced or back EMF

Ia : armature current

Φ:Average value of flux / pole

Z : Total number of armature conductors

N : Speed in rpm

n : Speed in rps

P : Number of poles

A : number of armature paths or circuits

D : Diameter of the armature

L : Length of the armature core

Power developed in the armature in kW, Pa= E Ia x 10-3

=(φZNP/60 A)× Ia× 10-3

=(Pφ)×(IaZ/A)×N x 10-3

/60

=(Pφ)×(IaZ/A)×n x 10-3

....... (1)

The term P φ represents the total flux and is called the magnetic loading. Magnetic

Loading per unit area of the armature surface is called the specific magnetic loading or average

value of the flux density in the air gap Bav. That is,

Bav = Pφ /π DL Wb/m2 or tesla denoted by T

Therefore Pφ = Bav π DL ................ (2)

The term (Ia Z/A) represents the total ampere-conductors on the armature and is called the

electric loading. Electric loading/unit length of armature periphery is called the specific

electric loading q. That is,

q= IaZ / π A D Amp-cond / m

Therefore Ia Z/A = q π D ............ (3)

Substitution of equations 2 and 3 in 1, leads to

kW = Bav π DL × q π D × (n × 10-3

)

= B q D2 L n

= π2Bav q C0 D

2 L N 10

-3

Where C0= π2Bav q 10

-3 is called the output coefficeint of the DC machine and is equal to 1.64

x 10-4 Bq.

Therefore D2 L = (Kw/1.64 × 10-4 B q N) m

3

The above equation is called the output equation. The D2L product represents the size of

the machine or volume of iron used. In order that the maximum output is obtained /kg of iron

used,

D2L product must be as less as possible. For this, the values of q and Bav must be high.

(Ia/A can be written as Iz, which can be used in the above derivation to keep it simpler)

2. Find the dimensions of a 200kW, 250V, 6 pole, 1000 rpm DC generator. The maximum

value of flux density in air gap is 0.87wb/m2 and the ampere conductors per metre length

of armature periphery are 31000. Ratio of pole arc to pole pitch is 0.67 and efficiency

91%. Assume that the ratio of core length to pole pitch = 0.75. (AU MAY 2008)

3. Explain the effects of choice of number of poles in a DC Machine on

(1) Frequency of flux reversal (2) Weight of iron (3) Weight of copper and (4) Length of

commutator. (8)(AU MAY 2011)

1. Frequency

As the number of poles increases, frequency of the induced emf f = PN/120 increases, core loss

in the armature increases and therefore efficiency of the machine decreases.

2. Weight of the iron used for the yoke

Since the flux carried by the yoke is approximately f /2 and the total flux f T = pf is a constant

for a given machine, flux density in the yoke,

By =

=

=

By

of poles increases, Ay and hence the weight of iron used for the yoke reduces.

3. Weight of overhang copper

For a given active length of the coil, overhang µ pole pitch p D/P goes on reducing as the

number of poles increases. As the overhang length reduces, the weight of the inactive

copper used at the overhang also reduces.

4. Length of the commutator

Since each brush arm collects the current from every two parallel paths, current / brush arm

= 2 Ia / A and the cross sectional area of the brush / arm

reduces as the number of poles increases.

As Ab = tbwbnb and tb is generally held constant from the commutation point of view, wbnb

reduces as Ab reduces. Hence the length of the commutator

Lc = (wbnb + clearances) reduces as Ab reduces or the number of poles increases.

wb – width of the brush, tb – thickness of the brush, nb – number of brushes per spindle

4. Discuss various methods to determine mmf required for teeth of an Electric Machine.

(8)(AU MAY 2011)

The calculation of MMF for producing flux in the Teeth of the machine is difficult because :

i) the teeth are tapered when parallel sided slots are used and this results in variation in

the flux density over the depth of the tooth.

ii) the slots provide another parallel path for the flux flow, the teeth are normally worked

in saturation and hence µr becomes low.

Following methods are usually employed for the calculation of MMF required for the

tapered teeth:-

i) Graphical method :ATt = Mean ord. x lt

Mean ord. is the mean ord. of “at” variation with tooth depth.

ii) Simpson’s rule :at mean = (at1 +4 at2 + at3)/6 A/m

iii) Bt1/3 method : ATt = at1/3 x lt ,

where at1/3 = MMF for corresponding to B at 1/3 rd

height from the narrow end.

5. Explain the various steps involved in the design of shunt field winding of a dc machine

(AU NOV 2013) Area of each pole (Ap) = Flux in the pole body / Flux density = cl φ / Bp

where cl = leakage coefficient

Width of the pole = Ap /Li ; where Li = net iron length = 0.9 L

Height of the pole(hf) chosen based on the MMF to be provided by the pole at full-load.

( ATf at full-load) /(ATArm. at full-load) =1.0 to 1.25

(to overcome armature reaction)

Copper loss in the field winding = If2 Rf

= (δfaf)2

{ρLmt Tf}/af = δf2

ρLmt Tf af

= δf2

ρLmt Sf hfdf ; ………………….(1)

where df = depth of the pole winding. Tf = no.of turns in each field coil.

Permissible loss = S qf = 2Lmt hf qf ..(2)

_______

Equating (1) and (2) ; δf =104

√ qf / Sf df MMF per metre of the field winding = ATf/ hf = If Tf

/hf = δf (af Tf )/ hf

= δf Sf hfdf/hf = δf Sf df = Sqrt {2 qf Sf df /ρ}

6. Determine the diameter and length of armature core for 55 kW, 110V, 1000rpm, 4 pole

shunt generator assuming specific electric and magnetic loadings of 26000 ampere

cond/meter and 0.5 wb/m2 repectively. The pole arc should be about 70% of pole pitch

and length of core about 1.1 times the pole arc. Allow 10A for the field current and

assume a voltage drop of 4V for armature circuit. (AU NOV 2014).

7. A design is required for a 50KW, 4 pole, 600rpm, 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 metre are 30,000. Calculate suitable dimensions of armature core to

give a square pole face. Assume full load armature voltage drop 3% of the rated terminal

voltage and field current 1% of rated full load current, ratio of pole arc to pole pitch is

0.67 (8) (CO-2)(AU MAY 2009) (AU NOV 2009) (AU MAY 2010) (AU NOV 2011)

(AU MAY 2014)

Power developed in the armature in kW=EIa ×10-3

E = V+voltage drop in the armature circuit=220+0.03×220=226.6V

Ia = IL+ ISh

IL =

=

= 227.3A

Ia = 227.3 + (0.01×227.3)=229.6A

kW = 226.6 ×229.6×10-3

=52

D2L =

≈ 0.031m-3

For a square pole face, L=ψ t = 0.67(πD/4) 0.53 D

3= 0.031 m

3

D3=

m

3

D = 0.39 m

L = 0.53 x 0.39

L = 0.21 m

8. Explain the various factors that are affected by the selection of number of poles in a dc

machine (AU NOV 2014)

9. Determine the apparent flux density in teeth of a DC machine if the real flux density in

teeth is 2.15 Wb/m2, slot pitch is 28 mm, slot width is 10 mm, gross core length is 0.35

m, no. of ventilating ducts is 4 each 10 mm wide. Magnetizing force corresponding to

flux density of 2.15 Wb/m2 is 55000 AT/m and iron stacking factor is 0.9.

(8)(AU MAY 2011)

10. A 5 kw, 250 V, 4 pole, 1500 rpm DC Shunt Generator is designed to have a square pole

face. The specific magnetic loading and specific electric loadings are 0.42 Wb/m2 and

15000 AC/m respectively. Find the main dimensions of the machine. Assume full load

efficiency= 0.87 and pole arc to pole pitch ratio is 0.66. (8)(AU MAY 2011 (CO-2)

11. A 350kW, 500 V, 450 rpm, 6 pole DC machine is built with an armature diameter of

0.87m and core length of 0.32m. the lap wound armature has 660 conductors. Calculate

the specific electric and magnetic loading (8) (AU MAY 2012)

12. Calculate the mmf required for air gap of machine having core length = 0.32m, including

4 ducts of 10mm each, pole arc=0.19m, slot pitch =65.4mm, slot opening=5mm, air gap

length=5mm, flux per pole=52mwb. Given carter’s coefficient is 0.18 for opening/gap=1

and 0.28 for opening/gap=2. (AU NOV 2013)

13. Determine the no. of poles armature diameter, core length for the preliminary design of a

500KW, 400V 600rpm dc shunt generator. Assuming an average flux density in air gap

of 0.7 wb/m2 and specific electric loading of 38400 ac/m. take the core length/pole arc =

1.1.

UNIT 3

1. Derive the output equation of a three phase transformer. (8)(AU MAY 2011) (AU

MAY 2015)

Each window contains two primary and two secondary windings

Ac =2 (Tpap + Tsas ) = 4 AT / δ

→ AT = Kw Aw δ/4

Rating in kVA = Q = 3Vp Ip x 10 -3

= 3Ep Ip x 10 -3

=3 Et ( Tp Ip )x 10 -3

= 3 x 4.44 fΦm ( Kw Aw δ/4 ) 10 -3

where Φm = Bm Ai

= 3.33 fΦm ( Kw Aw δ ) 10 -3

Using the output equation it can also be shown that

_____

E t = K √ kVA

where K =√ 4.44 f r 10 3 ;

r = Φm / AT

r is a constant for transformer of a given type ,service and method of connection,

since Φm determines the core section and AT fixes the total copper area.

2. Explain the design of transformer tank with cooling tubes (AU MAY 2013)

The specific heat dissipation due to convection of oil

= λconv = 40.3 (θ /H) ¼

W/m2- oC ; = temp difference of the surface relative

to the oil and H = height of the dissipating surface.

Experimentally found that a plain tank surface dissipates 6.0 W/m2-oC by radiation

and 6.5 W/m2-oc by convection (.for a temp rise of 40

oC above an ambient temp of 20

oc). Thus a total of 12.5 W/m

2-

oC is taken.

The temp rise θ = total loss/ (λ St) = (Pi + Pc) / (λ St)

Where St = heat dissipating surface area of the tank.

For small transformers , plain walled tank is enough to dissipate the losses. As the

rating of the transformer increases, the volume increases as he cube of the linier

dimensions but the heat dissipating surface area increases only as square of the linier

dimensions. So above certain rating, plain tank becomes inadequate to dissipate losses

and the area is increased by providing tubes. For larger ratings forced air cooling is

used.

If tubing is provided, the oil circulation is improved due to the head of the oil, and

this causes an additional dissipation by convection of about 35 % .

Let x St be the area of the cooling tubes. Then

Loss dissipated by the tank surface = 12.5 St W/0C

Loss dissipated by the tubes (1.35 x 6.5) x St W/0 C = 8.8 x St W/

0 C

Total loss dissipated by the tank and oil tubes

= (12.5 St + 8.8 x St ) W/0C

Hence θ = ( Pi + Pc)/ (12.5 St + 8.8 x St ) →Total tube area x St

= (1/8.8) [ {(Pi + Pc)/ θ} – 12.5 St ]

The number of tubes = nt = Total tube area /(π dt lt )

The arrangement of the tubes on tank side walls should be made uniformly with a

spacing of usually 75 mm. Examples of calculation of nt and the arrangement of the

tubes should be studied.

3. Derive the output equation of a single phase transformer in terms of core and window

area (AU NOV 2014)

4. A 250kVA, 6600/400V, 3phase core type transformer has a total loss of 4800W at full

load. The transformer tank is 1.25m high and 1m x 0.5m in plan. Design suitable tubes if

the average temperature rise is to be limited to 35°C. The diameters of the tubes are

50mm and are spaced 75mm from each other. The average height of tubes is 1.02m.

Specific heat dissipation due to radiation and convection is respectively 6 and 6.5

W/m2°C. Assume that the convection is improved by 35 percent due to provision of

tubes. (AU NOV 2013 ) (AU MAY 2011 ) (AU NOV 2014 ) (AU NOV 2013 )

5. Determine the dimensions of the 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 1.1Wb/m2, window space factor 0.32, current density 3A/mm

2 and 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. (AU MAY 2012 )

(AU NOV 2014 )

6. A single phase 400V, 50Hz transformer is built from stampings having a relative

permeability of 1000. The length of flux path is 2.5m. The maximum flux density in core

is 1 wb/m2, weight of core is 43.8kg and the primary winding has 800 turns. The iron

loss at working flux density is 2.6 W/kg. Find the no load current of the transformer. (AU

MAY 2013)

7. The ratio of flux to full load mmf in a 400 kVA, 50 Hz single phase core type transformer

is 2.4*10–6

. Calculate the net iron area and the window area of the transformer if the

maximum flux density in the core is 1.3 Wb/m2, Current density is 2.7 A/mm2 and

window space factor is 0.26. Also calculate the full load mmf. (8)(AU MAY 2011)

8. Calculate the core and windows area required for a 1000 kVA, 6600/400 V, 50 Hz, single

phase core type transformer. Assume a maximum flux density of 1.25 wb/m2 and current

density of 2.5A/mm2.

Voltage per turn = 30V. window space factor =0.32. (8) (AU MAY

2010)

9. Calculate approximate overall dimensions for a 200kVA, 6600/400V, 50Hz, 3phase core

type transformer. The following data may be assumed; emf per turn=10V; maximum flux

density=1.3Wb/m2. Current density= 2.5A/mm

2. 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 = 0.6d2; where ‘d’ is the diameter of

circumscribing circle (AU NOV 2012 )

10. Describe the methods of cooling of transformers (AU MAY 2013)

11. Determine the man dimensions of 5kVA, 11000/400volts, 50Hz, single phase core type

transformer having the following data:

The net conductor are in window is 0.6 times the net cross section area of iron in

core. The core is of square cross section, maximum flux density is 1 wb/m2. Current

density= 1.4A/m m2.

Window space factor 0.2. height of window is 3 times its width.

(AU MAY 2008) 12. Determine the man diameter of core and window for a 5kVA, 50Hz, single phase core

type transformer. A rectangle core is used with long side as twice as short side. Height of

window is 3 times its width. Voltage per turn is 1.8V, space factor 0.2, Current density=

1.8A/m m2..

maximum flux density is 1 wb/m2.

(AU NOV 2013)

UNIT 4

1. Derive the output equation of AC machine in terms of its main dimensions (8) (MAY

2012)

Output Equation is the mathematical expression which gives the relation between the

various physical and electrical parameters of the electrical machine.

In an induction motor the output equation can be obtained as follows

Consider an ‘m’ phase machine, with usual notations

Output Q in kW = Input x efficiency

Input to motor = mVph Iph cos Φ x 10-3

kW

For a 3 Φ machine m = 3

Input to motor = 3Vph Iph cos Φ x 10-3

kW

Assuming Vph = Eph, Vph = Eph = 4.44 f Φ TphKw

= 2.22 f ΦZphKw

f = PNS/120 = Pns/2,

Output = 3 x 2.22 x Pns/2 x ΦZphKw Iph η cos Φ x 10-3

kW

Output = 1.11 x PΦ x 3Iph Zph x ns Kw η cos Φ x 10-3

kW

PΦ = BavπDL, and 3Iph Zph/ πD = q

Output to motor = 1.11 x BavπDL x πDq x ns Kw η cos Φ x 10-3

kW

Q = (1.11 π2 Bav q Kw η cos Φ x 10

-3) D

2L ns kW

Q = (11 Bav q Kw η cos Φ x 10-3

) D2L ns kW

Therefore Output Q = Co D2L ns kW

where Co = (11 Bav q Kw η cos Φ x 10-3

)

Vph = phase voltage ; Iph = phase current

Zph = no of conductors/phase

Tph = no of turns/phase

Ns = Synchronous speed in rpm

ns = synchronous speed in rps

p = no of poles, q = Specific electric loading

Φ = air gap flux/pole; Bav = Average flux density

kw = winding factor

η = efficiency

cosΦ= power factor

D = Diameter of the stator,

L = Gross core length

Co = Output coefficient

2. Describe the steps involved in the design of end rings. (AU MAY 2014)

Squirrel cage rotor consists of a set of copper or aluminum bars installed into the

slots, which are connected to an end-ring at each end of the rotor. The construction of this

type of rotor along with windings resembles a ‘squirrel cage’. Aluminum rotor bars are

usually die-cast into the rotor slots, which results in a very rugged construction. Even

though the aluminum rotor bars are in direct contact with the steel laminations, practically

all the rotor current flows through the aluminum bars and not in the lamination

Most of the induction motor are squirrel cage type. These are having the

advantage of rugged and simple in construction and comparatively cheaper. However

they have the disadvantage of lower starting torque. In this type, the rotor consists of bars

of copper or aluminum accommodated in rotor slots.

Selection of number of rotor slots:

The number of rotor slots may be selected using the following guide lines.

(i) To avoid cogging and crawling: (a)Ss≠ Sr (b) Ss – Sr= ±3P

(ii) To avoid synchronous hooks &cusps in torque speed characteristics Ss - Sr ≠ ±P, ±2P,

±5P.

(iii) To noisy operation Ss - Sr ≠ ±1, ±2, (±P ±1), (±P ±2)

Rotor Bar Current:

Bar current in the rotor of a squirrel cage induction motor may be determined by

comparing the mmf developed in rotor and stator. Hence the current per rotor bar is given

by Ib = ( Kws x Ss x Z's ) x I'r / ( Kwr x Sr x Z'r ) ;

where Kws – winding factor for the stator,

Ss – number of stator slots,

Z's – number of conductors / stator slots,

Kwr – winding factor for the rotor,

Sr – number of rotor slots,

Z'r – number of conductors / rotor slots and

I'r – equivalent rotor current in terms of stator current and is given by

I'r = 0.85 Is where is stator current per phase.

Copper loss in rotor bars:

Knowing the length of the rotor bars and resistance of the rotor bars cu losses in

the rotor bars can be calculated. Length of rotor bar lb = L + allowance for skewing

Rotor bar resistance = 0.021 x lb / Ab

Copper loss in rotor bars = Ib x rb x number of rotor bars.

End Ring Current:

All the rotor bars are short circuited by connecting them to the end rings at both

the end rings. The rotating magnetic field produced will induce an emf in the rotor bars

which will be sinusoidal over one pole pitch. As the rotor is a short circuited body, there

will be current flow because of this EMF induced. The distribution of current and end

rings are as shown in Fig. below. Referring to the figure considering the bars under one

pole pitch, half of the number of bars and the end ring carry the current in one direction

and the other half in the opposite direction. Thus the maximum end ring current may be

taken as the sum of the average current in half of the number of bars under one pole.

Maximum end ring current Ie(max) = ½ ( Number rotor bars / pole) Ib(av)

= ½ x Sr/P x Ib/1.11

Hence rms value of Ie = 1/2 x Sr/P x Ib/1.11

= 1/ x Sr/P x Ib/1.11

Maximum value of end ring current Ie(max) = (2/π) (Sr/2P) (Ib(max))

Since the bus bar current varies sinusoidally,

Ib(max)=

Ie(max) = (2/π) (Sr/2P) ( )

Ie =

=

(Sr/2P) ( )

=

Area of end ring:

Knowing the end ring current and assuming suitable value for the current density

in the end rings cross section for the end ring can be calculated as

Area of each end ring Ae = Ie / δe mm2, current density in the end ring may be

assume as 4.5 to 7.5 amp/mm2.

Copper loss in End Rings:

Mean diameter of the end ring (Dme) is assumed as 4 to 6 cms less than that of

the rotor. Mean length of the current path in end ring can be calculated as lme = Dme.

The resistance of the end ring can be calculated as re = 0.021 x lme / Ae

Total copper loss in end rings = 2 x Ie 2 x re

Equivalent Rotor Resistance:

Knowing the total copper losses in the rotor circuit and the

equivalent rotor current equivalent rotor resistance can be calculated as follows.

Equivalent rotor resistance r'

r = Total rotor copper loss / 3 x (Ir' )2

3. Estimate the stator core dimensions and the total number of stator conductor for a 3ф,

100kW, 3300V, 50Hz, 12 pole star connected slip ring induction motor. Assume: average

gap density=0.4Wb/m2, conductors per metre = 25000 A/m, efficiency = 0.9, power

factor = 0.9, winding factor = 0.96. choose the main dimension to give best power factor.

(AU MAY 2014)

Given: P=100kW

VL=3300V

F=50Hz

p=12

Bav=0.4Wb/m2

ac = 25000 A/m

η=0.9

cosф=0.9

Kws=0.96

Solution:

Q = Co D2L ns kVA

where Co = (11 Bav ac Kw η cos Φ x 10-3

)

Q=P/ (η Cosф)

=100/(0.9*0.9)

=123.45kVA

ns = 2f/P

=(2*50)/12

= 8.33 rps

Q = (11 Bav ac Kw η cos Φ x 10-3

) D2L ns

Q = (11*0.4*25000*0.96*0.9*0.9* 10-3

)* D2L *8.33

123.45 = (11*0.4*25000*0.96*0.9*0.9* 10-3

)* D2L *8.33

D2L =

D2L = 0.1408

For best power factor,

τ =

τ2=0.18 L

[

]

2 = 0.18L

D2 = 2.626L

but, D2L = 0.1408

Simplifying, we get D = 0.779m & L = 0.236m

Stator pitch Yss lies between 15 to 25 mm

Yss=

Ss=

If Yss = 15mm, Ss = [π*0.778 ] / [15*10-3

]

= 163

If Yss = 25mm, Ss = [π*0.778 ] / [25*10-3

]

= 98

So, the number of slots varies between 98 and 163

Ss = m*p*qs

= 3*12*qs

qs 2 3 4 5

Ss 72 108 144 180

108 & 144 lies between specified range. Hence Ss=108 or 144

To finalize, slot loading is calculated. Slot loading should be less than 500A

Slot loading = Iz * Zss

Iz = ILS =

Iz=21.599A

Zss = Zs / Ss

Zs=2*m*Tph = 6 Tph

Es = 4.44 f ф kWS Tph

Tph=Es / 4.44 f ф kWS

Bav = pф / πDL

Ф = Bav πDL / p

= (0.4*π*0.0188*0.231) / 12

0.0188wb

Es = 3300 / = 1905.25V

Tph=Es / 4.44 f ф kWS

= 1905.25 / (4.44*50*0.0188*0.96)

= 475.52

≈ 476turns

Zs = 6 Tph

= 6*476

= 2856

Slot loading = Iz * Zs / Ss

If Ss = 108, Slot loading = Iz * Zs / Ss

= 21.59 * 2856/108

= 570.93 > 500, condition not satisfied

If Ss = 144, Slot loading = Iz * Zs / Ss

= 21.59 * 2856/144

= 428.2 < 500, condition satisfied.

Hence, Zs = 2856

Zss = 20

Ss = 144

4. Determine the main dimensions , no: of radial ventilating ducts, no: of stator slots and no:

of turns/phase of a 3.7 Kw, 400V, 3 phase, 4 pole , 50 Hz, squirrel cage IM to be started

by star-delta starter. Assume average flux density in the air-gap is 0.45 Wb/mm2.

Ampere-conductors per metre = 23,000, efficiency =85% and power factor =0.84,

winding factor =0.955 , stacking factor=0.9. Since the machine is rated at 3.7 Kw, 4pole

are sold at competitive price and therefore choose the main dimensions to give a cheap

design. (AU MAY 2010) (AU NOV 2011 )

Given: P = 3.7 kW

VL = 400V

p = 4

f = 50Hz

Bav= 0.45Wb/m2

ac = 23000 A/m

η = 0.85

cosф = 0.84

Kws =0.96

ki = 0.9

Solution:

Q = P/ (η Cosф)

= 3.7 / (0.85*0.84)

= 5.18 kVA

ns = 2f/P

=(2*50)/4

= 25 rps

Q = (11 Bav ac Kw η cos Φ x 10-3

) D2L ns

Q = (11*0.45*23000*0.96*0.85*0.84 x 10-3

) D2L 25

5.18 = (11*0.45*23000*0.96*0.85*0.84 x 10-3

) D2L 25

D2L =

D2L = 1.91*10

-3 m

3

Wkt, L/τ = 1.5 to 2

Taking L/τ = 1.5

= 1.5

=

L/D = 1.178

L = 1.178 D

D2

(1.178 D )= 1.91*10-3

m3

D3

=

D = 0.12m

L = 0.13m

Pole pitch, τ =

= (π * 0.12) / 4

Length of coil is 0.13m

Radial duct = 10mm

Li = 0.9 (0.13 - 0.01)

Li = 0.108m

Turns per phase, фm = Bav * L * τ

= 0.45 * 0.13 * 0.094

фm = 5.5 * 10-3

wb

Es = 400V

Es = 4.44 f ф kWS Ts

400 = 4.44 * 50 * 5.5 * 10-3

* 0.955 * Ts

Ts = 343 turns

Slots per pole per phase, qs = 3

Stator slots, Ss = m*p*qs

= 3 * 4 * 3

= 36

Slot pitch Yss =

= (π * 0.12) / 36

= 0.01047m

=10.47 mm

Number of stator slots = 6Ts

= 6*343

= 2058

Zss = Zs / Ss = 2058 / 36 = 57

Actual turns per phase, Ts = (36*57) / (2*3)

5. Describe the steps involved in the design of magnetizing current for an induction motor

from design data (AU NOV 2011 )

Magnetising current of an induction motor is responsible for producing the required

amount of flux in the different parts of the machine. Hence this current can be

calculated from all the magnetic circuit of the machine. The ampere turns for all

the magnetic circuit such as stator core, stator teeth, air gap, rotor core and rotor

teeth gives the total ampere turns required for the magnetic circuit. Based on the

total ampere turns of the magnetic circuit the magnetizing current can be

calculated as

Where, p – no of pairs of poles,

AT30 – Total ampere turns of the magnetic circuit at 600 from the centre of the

pole,

Tph – Number of stator turns per phase.

AT60 = ATg + ATts + ATtr + ATcs + ATcr

ATg = mmf of air gap,

ATts = mmf of stator teeth,

ATtr = mmf of rotor teeth,

ATcs = mmf of stator core,

ATcr = mmf of rotor core

mmf for airgap, Bg 60 = 1.36 x AT60

mmf for airgap, = 8,00,000 x Bav x kg x ag

where, kg = kgss x kgsr

= {Gap contraction factor for stator slots} x {Gap contraction factor for

rotor slots}

ag = Area of air gap, m2

mmf of stator teeth :

Flux density at height of tooth from narrow end

Bt 1/3 =

фm

x Li x

= width of stator at 1/3

rd from narrow end

= π (D + 2dss/3)

- Wss SL

The calculation of mmf for stator teeth is based on Bts60.

Bts60 = 1.36 x Bts 1/3

mmf required for stator teeth, ATts = at ts x dss

where, at ts = mmf/metre for stator teeth

dss = depth of stator slots (m)

6. Determine the main dimensions , no: of radial ventilating ducts, no: of stator slots and no: of

turns/phase of a 3.7 Kw, 400V, 3 phase, 4 pole , 50 Hz, squirrel cage IM to be started by star-

delta starter. Assume average flux density in the air-gap is 0.45 Wb/mm2.

Ampere-conductors per

metre = 23,000, efficiency =85% and power factor =0.84, winding factor =0.955 , stacking

factor=0.9. Since the machine is rated at 3.7 Kw, 4pole are sold at competitive price and therefore

choose the main dimensions to give a cheap design. (AU MAY 2011 ) (CO1)

7. Determine the approximate diameter and length of stator core, the number of stator slots and the

number of stator conductors for a 11kW, 400V, 3phase, 4 pole, 1425rpm, delta connected

induction motor. Bav = 0.42 Wb/m2,ac=23000 amp.cond/m, full load efficiency =0.85, power

factor =0.88, L/=1. The stator employs a double layer winding. (AU MAY 2012) (CO3)

PART-B

UNIT 5 1. Describe procedure to design the field winding of a synchronous machine (8) (MAY 2012)

2. State and explain the factors considered for the selection of number of slots (AU NOV 2012)

Derive the expression for length of air gap of a synchronous motor

3. Derive the output equation of an AC machine. (8)(MAY 2011) (CO5)

Output Equation is the mathematical expression which gives the relation between the

various physical and electrical parameters of the electrical machine.

In an induction motor the output equation can be obtained as follows

Consider an ‘m’ phase machine, with usual notations

Output Q in kW = Input x efficiency

Input to motor = mVph Iph cos Φ x 10-3

kW

For a 3 Φ machine m = 3

Input to motor = 3Vph Iph cos Φ x 10-3

kW

Assuming Vph = Eph, Vph = Eph = 4.44 f Φ TphKw

= 2.22 f ΦZphKw

f = PNS/120 = Pns/2,

Output = 3 x 2.22 x Pns/2 x ΦZphKw Iph η cos Φ x 10-3

kW

Output = 1.11 x PΦ x 3Iph Zph x ns Kw η cos Φ x 10-3

kW

PΦ = BavπDL, and 3Iph Zph/ πD = q

Output to motor = 1.11 x BavπDL x πDq x ns Kw η cos Φ x 10-3

kW

Q = (1.11 π2 Bav q Kw η cos Φ x 10

-3) D

2L ns kW

Q = (11 Bav q Kw η cos Φ x 10-3

) D2L ns kW

Therefore Output Q = Co D2L ns kW

where Co = (11 Bav q Kw η cos Φ x 10-3

)

Vph = phase voltage ; Iph = phase current

Zph = no of conductors/phase

Tph = no of turns/phase

Ns = Synchronous speed in rpm

ns = synchronous speed in rps

p = no of poles, q = Specific electric loading

Φ = air gap flux/pole; Bav = Average flux density

kw = winding factor

η = efficiency

cosΦ= power factor

D = Diameter of the stator,

L = Gross core length

Co = Output coefficient

(AU NOV 2012)

4. For a 250kVA, 1100V, 12pole,500 rpm,3 phase alternator. Determine air-gap diameter,

core length, number of stator conductors, number of stator slots and cross section of stator

conductors. Assuming average gap density as 0.6 Wb/m2, specific electric loading of

30000 amp.cond/m and L/ τ = 0.85 (AU MAY 2012)

5. Determine the main dimensions of a 100 kVA, 50 Hz, three phase 375 rpm alternator. The

average air gap flux density is 0.55 Wb/m2and ampere conductors per metre is 28000.

Given that τ / L must be between 1 to 5. The maximum permissible peripheral speed is 50

m/sec. The runaway speed is 1.8 times synchronous speed. (8) (MAY 2011)

6. Discuss the step by step procedure to design the rotor of a synchronous machine (AU May

2009)

7. Determine for a 500 KVA, 6600V,50Hz,500 rpm, Y-connected 3-ф salient pole alternator

(i) The dia. at the air gap

(ii) Core length for Sq.pole face

(iii) No. of stator slots

(iv) No. Of stator conductors for double layer winding

Assume: Sp.magnetic loading=0.6T

Sp.electric loading =30,000Ac/m

Winding factor=0.955 (AU MAY 2008)

8. 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 / ּז 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

times synchronous speed.