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‹‹ T h ) 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.
58
Embed
DESIGN OF ELECTRICAL MACHINES PART-A UNIT I 1 2017-R2008)
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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)
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.
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.