Download Introduction - Sree Narayana Gurukulam College of Engineering

SREE NARAYANA GURUKULAM COLLEGE OF ENGINEERING
KOLENCHERY
YEAR: 2015
DEPARTMENT: ELECTRICAL & ELECTRONICS ENGINEERING
SEMESTER: 8
SUBJECT CODE & NAME
EE 010 803 ELECTRICAL SYSTEM DESIGN
FACULTY: NITHIN S. NAIR, ASST. PROFESSOR, EEE DEPT.
1
PART-1 ELECTRICAL MACHINE DESIGN
Introduction
CLASSIFICATION OF DESIGN PROBLEMS
The design of an electrical machine involves solution of many complex and diverse
engineering problems. The design problems may be classified under the following four
headings.




Electromagnetic design
Mechanical design
Thermal design
Dielectric design
Each problems may be solved separately and the result are combined to give
overall solution . Each of these three major problems may be further divided into simple
problems and solutions of individual problem are combined to give the solution of a major
design problem.
The electro magnetic design problem in rotating machines involves the design of stator
& rotor core dimensions, stator & rotor teeth dimensions, air-gap length , stator and rotor
windings. In transformer it is the problem of designing the core and the windings.
The mechanical design in rotating machine involves the design of frame (enclosure),
shaft and bearings. In transformer it is the design of tank (i.e., housing for core and winding
assembly).
The thermal design in rotating machine involves the design of cooling ducts in core and
cooling fans. In case of large machines coolants like air or hydrogen may be forced to circulate
in the ducts and air-gap. In transformer it involves the design of cooling tubes or radiators.
Another important design problem, that may require great attention in the design of
insulations (Dielectric design). Dielectric materials are used to insulate one conductor from
other and also the windings from the core. The dielectric materials are designed to withstand
high voltages stresses. The breakdown of dielectric materials may lead to failure of machine.
STANDARD SPECIFICATIONS
Every country has a standard organisation to fix standard specifications for the
manufactures. The specification are guidelines for the manufactures to produce economic
products without compromising quality. The manufactures who are compiling with the
standards will be issued a certification for their products. The quality of the certified products
will be periodically monitored by the standards organization.
The standard specifications issued for electrical machines includes the followings,
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Standard ratings of machines.
Types of enclosure.
Standard dimensions of conductors to be used.
Method of marking ratings and name plate details.
Performance specifications to be met.
Types of insulation and permissible temperature rise.
Permissible loss and range of efficiency.
Procedure for testing of machine parts and machines.
Auxiliary equipments to be provided.
Cooling methods to be adopted.
In India, the Indian Standard Organisation (ISO) has laid down their
specifications (ISI) for various products. The standards will be amended time to time, in order
to include the latest developments in technology. Recently they have released revised standards
ISO 9002, to comply with international standards.
The name plate of rotating machine has to bear the following details as per ISI
specifications

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




kW or kVA rating of machine
Rated working voltage
Operating speed
Full load current
Class of insulation
Frame size
Manufactures name
Serial number of the product
Some of Indian standard specifications numbers along with year of issue for
electrical machines are listened here.
IS 325-1966
IS 1231-1974
IS 4029-1967
IS 996-1979
IS 1885-1993
IS 9499-1980
IS 7538-1996
:
:
:
:
:
:
:
Specifications for three phase induction motor.
Specification for foot mounted induction motor.
Guide of testing three phase induction motor.
Specifications for single phase ac and universal motor.
Specifications for electric and magnetic circuits.
Conventions concerning electric and magnetic circuits.
Specifications for three phase induction motor for centrifugal pumps
and agricultural applications.
IS 12615-1986
IS 9320-1979
IS 4722-1992
IS 12802-1989
IS 4889-1968
:
:
:
:
:
Specifications for energy efficient induction motor.
Guide for testing dc machines.
Specifications for rotating electrical machines.
Temperature rise measurement of rotating electrical machines.
Method of determination of efficiency of rotating electrical machines.
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IS 13555-1993 : Guide for selection and application of three phase induction motor for
different types of driven equipment.
IS 7132-1973 : Guide for testing synchronous machines.
IS 5422-1996 : Turbine type generators.
IS 7572-1974 : Guide for testing single-phase ac and universal motors.
IS 8789-1996 : Values of performance characteristics for three phase induction
motors.
IS 12066-1986 : Three phase induction motors for machine tools.
IS 1180-1989 : Specifications for outdoor 3-phase distribution transformer upto
100 kVA.( Sealed and Non-sealed type)
IS 2026-1994 : Specifications for power transformers.
IS 11171-1985 : Dry type power transformers.
IS 5142-1969 : Continuously variable voltage auto transformers.
IS 10028-1985 : Code of practice for selection , installation and maintenance of
transformers.
IS 10561-1983 : Application guide for power transformers.
IS 13956-1994 : Testing transformers.
IS 9678-1980 : Methods of measuring temperature rise of electrical equipment.
IS 12063-1987 : Classification of degree of protections provided by enclosures of
electrical equipment.
IS 3855-1966 : Standard dimensions of enameled round copper conductor.
IS 449-1962 : Standard dimensions of enameled round copper conductor ( oleo
resinous enamel ).
IS 1595-1960 : Standard dimensions of enameled round copper conductor.
(synthetic enamel)
IS 1897-1962 : Standard dimensions of bare copper strip.
IS 1666-1961 : Standard dimensions of paper covered rectangular copper conductor for
transformer windings.
IS 2068-1962 : Standard dimensions of cotton covered rectangular copper conductor
windings.
IS 3454-1966 : Standard dimensions of paper covered round conductors used for
transformer windings.
IS 450-1964 : Standard dimensions of cotton covered round conductors used for
transformer windings.
GENERAL DESIGN PROCEDURE
In general any electrical machine has two windings. The dc machine and synchronous
machine has armature and field winding. The induction machine has stator and rotor winding.
The basic principle of operation of all electrical machine is governed by Faraday’s law of
induction. Also in every electrical machine the energy is transferred through the magnetic field.
Hence a general design procedure can be developed for the design of electrical machines.
The general design procedure is to relate the main dimensions of the machine to its
rated power output. An electrical machine is designed to deliver a certain amount of power
called rated power. The rated power output of a machine is defined as the maximum power that
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can be delivered by the machine safely. In dc machine the power rating is expressed in kW and
in ac machine in kVA. In case of motor the output power is expressed in HP.
In electrical machine the core and winding of the machine are together called active
part. ( Because the energy conversion takes place only in the active part of the machine). A
general output equation can be developed for dc machine which relate the power output
volume of active part ( D2 L), speed, magnetic and electric loading. Similarly a general output
equation can be developed for ac machine which relates kVA rating to volume of active part (
D2 L), speed, magnetic and electric loading.
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Module 1
Design of DC machines
CONSTRUCTION
The dc machines used for industrial electric drives have three major parts. They are
field system, armature and commutator. The field system is located on the stationary part of the
machine called stator and consists of main poles, interpoles and frame or yoke.
The main poles are designed to produce the magnetic flux. The interpoles are placed in
between the main poles. They are employed to improve the commutation condition . The frame
provides mechanical support to machine and also serve as a path for flux.
The armature is the rotating part ( or rotor ) of a dc machine and consists of armature
core with slots and armature winding accommodated in slots. The conversion of energy from
mechanical to electrical or vice-versa takes place in armature.
The commutator is mounted on the rotor of a dc machine . the commutator and brush
arrangement works like a mechanical dual converter. In case of generator it rectifies the
induced ac to dc. In case of motor it inverts the dc supply to ac.( In motor the commutator
reverses the current through the armature conductors to get unidirectional torque).
MAIN DIMENSIONS
In rotating machine the active part is cylindrical in shape. The volume of the cylinder is
given by the product of area of cross section and length. If D is the diameter and L is the length
of cylinder, then the volume is given by π D2 L/4. Therefore D and l are specified as main
dimension.
In case of dc machine, D represent the diameter of armature and L represent the length
of armature. In case of ac machine, D represent the inner diameter of stator and L represent the
length of stator core. The fig 1.1 shows the main dimensions of rotating machines
Here
Dr = Diameter of rotator
lg = Length of air-gap
MAGNETIC AND ELECTRIC LOADINGS
Consider a conductor of length L, carrying a current of Iz amperes. If the conductor is
moved in a uniform magnetic field of flux density Bav Wb/m2 then the work done is given by
Work = x B L Iz
Where x is the distance through which the conductor is moved.
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When the conductor is moved through a distance x, the conductor cuts through a
magnetic flux of ф = x BL webers. Hence work done in moving the conductor can be
expressed as,
Work = фIz
In rotating machines the conductors are placed in armature. In one revolution of the
armature, each conductor moves through a total flux of ф webers, where ф is flux per pole and
p is number of poles. If Z is the total number of armature conductors then the work done in
one revolution is given by
Work is p ф x Iz Z
The term pф represent the total flux entering and leaving the armature and so it is called
total magnetic loading (or total flux). The term Iz Z represents the sum of currents in all the
conductors on the armature and so it is called total electric loading (or total current volume or
total ampere conductors on the armature).
Total magnetic loading = pф
Total electric loading = Iz Z
Therefore we can say that the work done in one complete revolution is given by the
product of total magnetic loading ad total electric loading.
The total magnetic loading is defined as the total flux around the armature (or stator
inner) periphery at the air-gap. The total electric loading is defined as the total number of
ampere conductors around the armature (or stator) periphery.
Specific magnetic loading
Each unit area of armature surface is capable of receiving a certain magnetic flux.
Hence the flux per unit area is an important parameter to estimate the intensity of magnetic
loading and it is also criterion to decide the volume of active material. This flux per unit area is
expressed as the average value of the flux density at the armature surface or specific magnetic
loading (by the assuming that the armature is smooth). It is denoted by Bav
The average flux density, Bav is given by the ratio of flux per pole and area under a
pole.
Bav =
Flux per pole
=
.
Pole Pitch x Length of armature
ф
πD x
Flux per pole
=
Area under a pole
=
L
Pф
.
πDL
P
7
Note Pole pitch, τ = πD/p
We can say that the specific magnetic loading is also giving by the ratio of total flux
around the air-gap and the area of flux path at the air-gap.
Bav =
Total flux around the air – gap
Area of flux path at the air – gap
=
Pф
πDL
The typical value of specific magnetic loading for various types of rotating machines
are listed in the table given below.
Specific electric loading
Every section of armature is capable carrying certain amount of current. Hence ampereturn per unit section of armature periphery (circumference) is an important parameter to
estimate the intensity of electric loading and it is also a criterion to decide the volume of active
material . This ampere –turn per unit section of armature periphery is expressed as the specific
electric loading . It is denoted by ac.
The specific electric loading is given by the ratio of total armature ampere conductors
and armature periphery (circumference) at air-gap.
ac =
Total armature ampere conductors
Armature periphery at air – gap
= Iz Z
πD
The typical values of specific electric loading for various types rotating machines are
listed in the table given.
Table: Specific magnetic and electric loadings
Machine
DC machine
Induction motor
Synchronous machine
Turbo alternator
Specific magnetic loading
Bav in Wb/m2
0.4 to 0.8
0.3 to 0.6
0.52 to 0.65
0.52 to 0.65
Specific electric loading
Ac in amp.cond./m
15000 to 50000
5000 to 45000
20000 to 40000
50000 to 75000
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OUTPUT EQUATION
The output of a machine can be expressed in terms of its main dimensions, specific
magnetic and & electric loadings and speed. The equation which relates the power output to D,
L, Bav, ac and n of the machine is known as output equation.
Output equation and Output coefficient of DC machine
From the knowledge of machine theory we get the equation of induced emf in armature
as,
Induced emf in armature, E =
ф ZN
p
= фZnp
60 a
a
In the armature of dc machine the conductors are connected in parallel paths. If a is the
number of parallel paths, then the current through each conductor is given by,
Current through each conductor, Iz = Ia or Ia = aIz
Specific magnetic loading,
Bav = pф = Πdl Bav
ac = IzZ or IzZ = π D ac
πD
In dc generator the electrical power generated in the armature is given by the product of
induced emf and armature current. In case of dc motor the mechanical equivalent of electrical
power in armature is given by the product induced emf (back emf) and armature current
Specific electric loading,
Power developed in armature, Pa = E Ia x 10-3 in kW
On substituting for E and Ia from equations
On substituting for pф and Iz Z we get,
Pa = π DL Bav x π D ac x n x 10-3
= π2 Bav ac x 10-3 x D2 Ln
= Co D2 Ln
where, Co = π2 Bav ac x 10-3
The equation, Pa = Co D2 Ln is called output equation and the terms Co is called output
coefficient. The term D2 L in the output equation is proportional to volume of active part.
Therefore if Co is constant then we can say the power output is directly proportional to the
product of volume of active part and speed.
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Co in terms of maximum gap density,Bg is given by Co = π2 ψ Bg ac x 10-3
Power developed by the armature , Pa is different from the rated power output P, of the
machine. The relationship between the two are
Pa = P/ η for generator
Pa = P for motors
i.e
Pa α Volume of active part x Speed
If Co is varied then power output is directly proportional to the four quantities and they are
Bav, ac, volume of active part and speed.
i.e., Pa α Bav x ac x volume of active part x Speed.
SEPARATION OF D AND L
In dc machine the separation of D and L depends on
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
Pole proportions
Peripheral speed
Moment of inertia
Voltage between adjacent commutator segments
Pole proportions
The dimensions of the machine or decided by the square pole criterion. This states that for a
given flux and cross-section area of pole, the length of mean turn of field winding is minimum,
when the periphery forms a square. This means that the length L must be approximately equal
to pole arc or L=b= ψ τ
. The value of ψ is usually between 0.64 to 0.72 ( the ratio L/ τ
= 0.64 to 0.72 ). However in practice L is slightly greater than pole arc , b and L/ τ is
usually between 0.7 to 0.9.For square pole criterion choose L/ τ
as 0.7.
bp
b
L
τ
ψ
=
=
=
=
=
width of pole body
pole arc
core length
pole pitch = π D/p
ratio of pole arc to
pole pitch = b/ τ
Peripheral speed
The peripheral speed of armature is sometimes a limiting factor to the value of
diameter. The peripheral speed should not exceed about 30 m/sec.
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Moment of inertia
For machines used in control systems, a small moment of inertia is desirable . For low moment
of inertia the diameter should be made as small as possible. Conversely a high inertia machine
may be required for impact load applications and such machines are designed with larger
diameter.
Voltage between adjacent commutator segment
The maximum core length is fixed by the maximum voltage that can be allowed between
adjacent segments. It can be shown that the maximum voltage between adjacent commutator
segments in dc machines is given by equation given below.
The maximum voltage between adjacent segments, Ecm = 2 B gm L Va Tc
where, B gm = Maximum air gap flux density under load conditions.
Tc = Turns per coil
The limiting values of B gm , V a , Ecm , are B gm = 1.2 Wb/m2 , V a 30m/sec.,
Ecm = 30 V.With these limiting values , for Tc = 1, L ≈ 0.4 m. . This is only and indication of
limiting value of core length, but it must be clear that large dc machines should have large
diameters rather than large core lengths.
FACTORS AFFECTING THE SIZE OF ROTATING MACHINES
The factors affecting the size of rotating machines are speed and output coefficient. The
output coefficient in turn depends on specific electric and magnetic loadings.
Speed
The output of a machine is directly proportional to speed. Also the speed is inversely
proportional to volume of active parts.
The output equation, Q= Co D2 L ns
Hence, for same volume with increase in speed the output will increase. For a
given output a high speed machine will have less volume and costs less. Therefore for
reducing the cost highest possible speed may be selected. The maximum speed is limited by
mechanical stresses of the rotating parts.
CHOICE OF SPECIFIC MAGNETIC LOADING
The specific magnetic loading is determined by,
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
Maximum flux density in iron parts of machine
Magnetizing current
Core losses
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Maximum flux density in iron
The maximum flux density in any iron part of machine must be below a certain limiting
value. The maximum flux density occurs in the teeth of the armature (or stator core). [Teeth are
the portion of the core in between slots]
The flux density in the teeth is directly proportional to specific magnetic loading.
Hence the choice of specific magnetic loading should be such that the maximum value of flux
density in the teeth is not exceeded. The maximum value of the flux density in the teeth is
between 1.7 to 2.2 Wb/ m2
Magnetizing Current
The magnetizing current of a machine is directly proportional to mmf. The mmf is
directly proportional to specific magnetic loading. Hence a large value of specific magnetic
loading results in increased values of magnetizing mmf and magnetizing current.
The value of magnetizing current is not usually a serious design consideration in dc
machines. But in induction motors an increased value of magnetizing current results in low
power factor. Hence specific magnetic loading in induction motors is lower than in dc
machines. For synchronous machines the magnetizing current is not so critical and the value of
specific magnetic loading is intermediate between that of dc and induction machines.
Core loss
The core loss in any part of the magnetic circuit is directly proportional to flux density
for which it is going to be designed . The flux density is directly proportional to the specific
magnetic loading . Hence the core loss in a machine varies directly as the specific magnetic
loading. Thus a large value of specific magnetic loading results in increased core loss and
consequently a decreased efficiency and an increased temperature rise.
With a given specific magnetic loading, the core loss increases as the frequency of flux
reversal is increased. This is because the hysteresis loss is directly proportional to the
frequency and eddy current loss is proportional to the square of the frequency. It follows that
for high speed dc machines, or high frequency ac machines, specific magnetic loadings must
be reduced in order to achieve lower iron loss.
CHOICE OF SPECIFIC ELECTRIC LOADING
The choice of specific electric loading depends on
 Permissible temperature rise
 Voltage rating of machine
 Size of machine
 Current density
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Permissible temperature rise
The heat dissipated per unit area of armature is proportional to specific electric loading.
The allowable specific electric loading is fixed by allowable temperature rise and the
cooling coefficient. A high value of ac can be used in a machine when a high temperature rise
is allowed. The maximum allowable temperature rise of a machine is determined by the type of
insulating materials used in it. When better quality insulating materials which can withstand
high temperature rises are used in the machines, increased values of specific electric loading
can be used. This results in reduction in the size of the machine.
A high value of electric loading may be used if the cooling coefficient of the machine is
small. The value of cooling coefficient depends upon the ventilation conditions in the machine.
High speed machines will have better ventilation and so higher value of ac can be used.
Voltage
Let,
Ws = Width of the slot
ds = Depth of the slot
Sf = Slot space factor
ys = Slot pitch
= Current density
The specific electric loading can be related to the above terms by the equation,
ac = ds (Ws / ys ) δ Sf
From the above equation it is clear that the specific electric loading is directly
proportional to slot space factor Sf . In high voltage machines, greater insulation thickness is
required and therefore the space factor for these machines is lower. Hence an increase in
voltage will, in general, necessitate a reduction in specific electric loading ac.
Size of machine
From the above equation it is clear that ac depends on the dimension of the slot. For
large machine the depth of the slot will be greater and so higher values of ac can be used.
Actually if the current density and the slot space factor are assumed constant, the specific
electric loading is proportional to the diameter as slot depth usually depends upon the diameter.
Current density
From the equation, q = ac δ ρ it is clear that a higher value of specific electric loading
can be used in a machine which employs lower current density in its conductors.
(because ac = q/ δ ρ).
Typical values of current density are in the range of 2 to 5 A/mm2.The temperature rise
is usually 40o C (above ambient) for normal applications and cooling co-efficient is between
0.02 and 0.035o C W- m2.
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VARIATION OF OUTPUT AND LOSSES WITH LINEAR DIMENSIONS
Consider two machines of the same type with all their linear dimensions in the ratio x:1
and having the same speed, flux density and current density. Let the machine with linear
dimensions x times be called A and the other machine B.
Output
The output of machine A is x4 times the output of machine B.
Losses
I 2 R loss = Number of conductors x copper loss in each conductor
LIMITATIONS IN DESIGN
The following factors impose limitations on design of electrical machines.
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Saturation
Temperature rise
Stress on insulation
Efficiency
Stress & strain on rotating parts and bearings
Mechanical precision of air-gap
Commutation
Power factor
Specifications
For the maximum utilization of active material the specific magnetic and electric loading can
be kept as high as possible. The value of specific magnetic loading is limited by the saturation
of magnetic materials used in the machine.
The value of specific electric loadings is limited by the allowable temperature rise in
the machine, which in turn depends on the insulating material. If an insulating material is
subjected to temperatures about its limit then its life is drastically reduced.
The huge developed in the electrical machine (due to losses) impose thermal stress on
the insulating material. The operating voltage impose electrical stress on the insulating
materials. The shot circuits currents that may flow in the windings great mechanical stresses on
the insulating materials. The type of insulating materials id decided by the operating
temperature and the size( dimension ) of insulation is decided by the electrical and mechanical
stresses.
The capital and running cost of machine depends on efficiency. If the losses in the
machine are low then the running cost will be less. In order to keep the electric and magnetic
14
losses to low value, the specific, electric, and magnetic loadings should be as low as possible.
This requires larger warn8ing of active materials( iron and copper/ aluminium) which results in
higher capital cost. Hence, for a machine with high value of efficiency the capital cost will be
higher and running cost will be lesser.
The dimension of rotor and central shaft are limited by the mechanical stress and strain
on them. In general the rotor should be stiff and they should not be any significant deflection
due to strain. In high speed machine the rotor slot dimensions as selected such that the
mechanical stresses at the bottom of rotor teeth do not exceed the limit.
The type of bearings to be used in rotating machines are decided by the rotor weight,
external loads, forces due to unbalanced rotors and unbalanced magnetic tool.
In induction motors , in the length of air- gap is kept as small possible for high power
factor. the length of air-gap is mainly limited by the position ( accuracy) of mechanical
fabrication technology. By employing modern CNC machines and dynamic balancing of rotors
smaller air-gaps can be achieved.
The commutations conditions in dc machines limit the maximum power output that can
be delivered by the machine. The maximum power output that can be obtained from a single dc
machine is 10 MW.
In general the power factor should be high in order to reduce the current level for the
same power. For high value of power factor, the specific magnetic loading should be less and
the length of air-gap should as small as possible. These requirements will increase the cost of
machine. ( because when magnetic loading is reduced the volume of active material has to be
increased and for small values of air-gap the fabrication cost will be high).
The specifications imposed by consumers and standards organization are major
limitations in design. Some of the specifications to be met are allowable temperature rise,
losses, efficiency, power factor, voltage rating, torque requirements, etc
SUMMARY OF DESIGN EQUATIONS
1.
2.
3.
4.
5.
6.
Total magnetic loading = pф
Total electric loading = Iz Z
Power developed in the armature of dc machine, Pa = Co D2 L n
Output coefficient of dc machine, Co π2 Bav ac x 10-3
Pole pitch, τ = π D/P
Choice of L / τ for dc machines
 In general, L / τ = 0.7 to 0.9
 For square pole criterion, L / τ = 0.7
15
TUTORIAL PROBLEMS
1.
A 350 kW, 500V, 450 rpm, 6-pole, dc generator 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 loadings.
Given Data
P = 350kW D=0.87m
V = 500 V
L=0.32m
n = 450 / 60 rps
Z = 660
P = 6 Lap wound
Solution
Specific electric loading,
ac = pф
Specific magnetic loading,
Bav = IzZ
πD
The power output of the generator,
P = VI x 10-3 in kW
Full load current , I =
350
500 x 10-3
P
V x 10-3
=
= 700 amps
Neglecting field current, Ia ≈ I,current trough each armature conductor
Specific electric loading, ac =28173 amp.cond./m
Specific magnetic loading, Bav = 0.6929 Wb / m2
Result
Specific electric loading, ac = 28173 amp.cond./m
Specific magnetic loading, Bav = 0.6929 Wb / m2
2.
Determine suitable value for the number of poles depends on the following factors.
1. Frequency of flux reversals should lie between 25 Hz to 50 Hz.
2. The current per brush arm should not exceed 400 A.
3. For reduced cost the highest possible choice of poles should be choosen.
Solution
Speed in rps, n=300 / 60=5 rps
16
Choice of poles can be 10, 12, 14, 16, 18 or 20.
Armature current, Ia
for p = 10, current / brush arm = Ia / 10 =200 amps
and for p = 20, current / brush arm = Ia / 20 =100 amps
For all choice of poles the current limit is not violated. (But for minimum cost we can choose
the maximum number of poles.)
Let number of poles, p = 10
For square pole face, L / τ = 0.7
The power developed in armature, Pa = Co D2 L n
Output coefficient, Co = π2 Bav ac Kws x 10-3
Result
The main dimensions are D and L
The diameter of armature, D = 1.32 m
The length of armature, L = 0.29 m
17
EFFECT OF VENTILATING DUCTS ON RELUCTANCE OF AIR - GAP
When the length of the armature is higher than the diameter or when the length is
greater than 0.1 meter, radial ventilation ducts are provided for better cooling of the core. The
radial ventilating ducts are small gaps of width wd in between the stacks of armature core. The
core is normally divided into stacks of 40 to 80 mm thick, with ventilating ducts of width 10
mm in between two sacks.
The provision of radial ventilating ducts results in contraction of flux in the arial
direction. Due to this the effective axial length of the machine is reduced and this results in an
increase in the reluctance of air-gap.
We can derive an expression for gap contraction factor for ventilating ducts by treating
stacks of laminations as teeth and the ducts as slots.
Contracted or effective axial length, L’ = L – Kcd n d w d
Where, Kcd = carter’s co-efficient for ducts
w d = width of the duct
n d = number of the ducts
The value of Kcd can be obtained from fig 2.7 by using by the ratio in place of ratio.
The gap contraction factor for ducts, Kgd is defined as the ratio of reluctant of air-gap
in machine with armature radial ducts to reluctance of air-gap in machine without armature
radial ducts.
Carter’s coefficient
The Carter’s coefficient is a parameter that can be used to estimate the
contracted or effective slot pitch in case of armatures with open or semienclosed slots. It is the
function of the ratio of slot opening and air-gap length.
The Carter’s coefficient is also used to estimate the effective length of armature when
ducts are employed. In this case it will be a function of wd ,the width of duct.
In electrical machine design the Carter’s coefficient is used to estimate the air-gap
expansion ( or contraction ) factor for slots and ducts. The increase in reluctant of airgap due to slotting and ducts is accounted as additional air-gap, which in turn given by
air-gap expansion factor.
Choice of armature length
The factors to be considered for the choice of armature length are
 Cost
 Ventilation
 Voltage between adjacent commutator segments
18

Specific magnetic loading
When the length of the core is large, the ratio of inactive copper to active copper
will be small.Hence the machine may cost less. But when the core length is very large then
ventilation of the core will be difficult. The centre portion of the core tends to attain a high
temperature rise and so the core must be ventilated ( or cooled ) by special methods.
An expression for minimum ( or limiting ) value of core length can be derived as shown
below.
For better commutation the voltage between adjacent commutator segments at load is
to be limited to 30V . To achieve this the induced emf in a conductor should not exceed
.
where, Tc = Turns per coil
N c = Number coil between adjacent segments
N c = 1, for simplex lap winding
Nc
= p / 2, for simplex wave winding
p = number of poles
We know that
Induced emf in a conductor = Bav L Va
where, Bav = Specific magnetic loading
L = Length of armature
Va = Peripheral speed
It can be observed that the maximum value of armature core length depends on specific
magnetic loading and peripheral speed.
Choice of armature diameter
The factors to be considered for the choice of armature diameter are
 Peripheral speed
 Pole pitch
 Specific electric loading
 Induced emf power conductor
 Power output
19
Table: Pole pitch
Pole
2
4
6
above 6
Pole pitch mm
Upto 240
240 to 400
350 to 450
450 to 500
In dc machines the peripheral speed lies in the range 15 to 50 m/s. Normally the
peripheral speed should not exceed 30 m / s. If the speed exceeds 30 m / s then special rotor
construction methods have to be employed to prevent the damage due to excessive centrifugal
force.
The diameter should be suitable for accommodating desired number of poles with
normal value of pole pitch. The normal values of pole pitch are given in the above table .
An expression for maximum (or limiting) value of armature diameter can be derived as
shown below.
We know that the induced emf in a conductor should not exceed 7.5 / T c N c volts
for better commutation conditions. Hence an equation for diameter of armature can be derived
in terms of induced emf per conductor.
Induced emf in armature, E = emf per conductor x conductors from parallel path
Power developed in armature, Pa = E Ia x 10-3 in kW
Iit can be observed that the maximum value of armature diameter depends on specific electric
loading and induced emf per conductor.
SELECTION OF NUMBER OF POLES
The number of poles used in dc machine has an import bearing upon the magnetic and
electric circuits. In case of ac machines, number of poles is fixed by the supply frequency and
the speed of the machine. But in the case of Dc machine , any number of poles can be used.
However there is always a very small range of number of poles that gives a design which is
sound from the commercial point of view. The selection of number of poles depends on
 Frequency
 Weigh of iron parts
 Weight of copper
 Length of commutator
 Labour charges
 Flash over and distortion of field form.
The number of poles are chosen such that the frequency lies between 25 to 50 Hz.
20
With large number of poles the flux carried by the yoke reduces. Hence for a given flux, with
large number of poles , area of cross-section of yoke can be reduced, which result in reduction
of iron parts. Also by increasing the number of poles , the weight of iron in the armature core
can be decreased. The overall diameter of the machine decreases as the number of poles is
increased. Therefore form commercial point of view a large number of poles results in reduced
cost.
566
The weight of copper in armature and field winding decreases with increase in number
of poles. With increase in number of poles the length of the commutator reduces and so the
overall length of the machine also reduces. With the increase in number of poles labour
charges will increase.
The use of large number of poles results in increased danger of flash over between
adjacent brush arms. With increase in number of poles there is reduction in distortion of field
from under load conditions.
Advantages of large number of poles
The large number of poles results in reduction of the following
 Weight of armature core and yoke
 Cost of armature and field conductors
 Overall length and diameter of machine
 Length of commutator
 Distortion of field form under load condition
Disadvantages of large number of poles
The large number of poles results in increase of the following
 Frequency of flux reversal
 Labour charges
 Possibility of flash over between brush arms
Guiding factor for choice of number of poles
1.
2.
The frequency should lies between 25 to 50 Hz
The value of current per parallel path is limited to 200 amps, thus the current per
brush should not be more than 400 amps.
Current per parallel path = Ia/p
for lap winding
= Ia /2
for wave winding
Current per brush arm = 2Ia/p
for lap winding
= Ia
for wave winding
where, p= number of poles
21
3.
The armature mmf should not be too large. The normal value of armature mmf per
pole are listed in table 3-2
Table: Armature mmf per pole
4.
Out put in kW
Armature mmfr per pole in AT
Upto 100
100 to 500
500 to 1500
over 1500
5000 or less
5000 to 7500
7500 to 10,000
upto 12,500
If there are more than one choice of number of poles which satisfies the above three
conditions, then choose the largest value for poles. This results in reduction in iron
and copper.
Pole proportions
The cross section of the poles should be circular in order that the length of mean turn of
the field winding is minimum ( since a circle gives minimum periphery for a given area). But
circular poles cannot be laminated , hence the next best alternative is square pole section.
In square section the width of the pole body is equal to the length of the machine. Some
manufactures prefer a square pole face. For square pole face, the pole arc (b) is equal to the
length of the machine . Some manufactures prefer rectangular pole section.
L = bp, for square pole section
L = b, for square pole face
Usually the ration of pole pitch of the ration L / τ is specified.
ψ = b / τ = 0.64 to 0.72 and L / τ= 0.45 to 1.1
LENGTH OF AIR- GAP
In rotating electrical machines a small gap is provided between the rotor and stator to
avoid the friction between the stationary and rotating parts.
A large value of air – gap results in lesser noise, better cooling, reduced pole face
losses, reduced circulating currents and lass distortion of field form. Also larger air-gap results
in higher field mmf which reduces armature reaction.
22
In general, mmf required for air – gap
where Kg = 1.15 = gap contraction factor.
In dc machines the mmf required for air – gap is normally taken as 0.5 to 0.7 times the
armature mmf per pole.
ARMATURE CORE DESIGN
The armature of a dc machine consist of core and winding. The armature core is
cylindrical in shape with slots on the outer periphery of the armature. The core is formed with
circular lamination of thickness 0.5 mm. The winding is placed on the slots in the armature
core. The design of armature core involves the design of main dimensions D&L number of
slots , slot dimensions and depth of core. The design of main dimensions D&L have been
discussed in chapter-1
Number of poles of armature slots
The factors to be considered for selection of number of armature slots are





Slot width ( or pitch)
Cooling armature conductors
Flux pulsations
Commutation
Cost
A large number of slots results in smaller slot pitch and so the width of tooth is also
small. This may lead to difficulty in construction. But large number of slots will lead to
less number of conductors / slot and so the cooling of armature conductors is better.
If the air- gap reluctance per pair of poles is constant then the flux pulsations and
oscillations can be avoided. It can be proved that the air – gap reluctance is constant if
the slots per pole is an integer plus ½.
For sparkles commutation the flux pulsations and oscillation under the interpole must
be avoided. this can be achieved with large number of slots per pole. Infact , the
number of slots in the region between the tips of two adjacent poles should be atleast 3.
From the above calculations we can say the slots per pole should be greater than or
equal to m9, for better commutation. When large number of slots are used the cost of
lamination and the cost of insulation will be high
23
Guiding factors for number of armature slots





The slot pitch should lie between 25 to 35 mm. For small machine it can be 20mm
or even less than 20mm.
The slot loading should not exceed 1500 ampere conductors. Slot loading =
Number of conductors in the slot x current per conductor.
To reduce flux pulsation losses the slot per pole should be an integer plus ½ for lap
winding and slots per pole are should be an integer plus ½ for wave winding.
To avoid sparking the number of slots per pole should have a minimum value of 9.
The slot per pole there is from 9 to 16.In case of small machines it can be 8.
The number of slots selected should be suitable for the type of winding. In case of
simplex lap winding the number of slots should be a multiple of pole pair. In case of
wave winding the number of slots should not be a multiple of pole pair to avoid
dummy coils
Slot dimension
The dimension of the slot are width and depth. Usually the slot area is estimated from
the knowledge of conductor area and slot space factor. The slot space factor lies in the
range of 0.25 to 0.4 and the value depends on the thickness of insulation.
Table: Slot depth
Diamater of armature
Slot depth
m
Mm
0.15
22
0.20
27
0.25
32
0.30
37
0.40
42
0.50
45
After deciding the slot area, the depth of slot is assumed based on the diameter of the armature.
This table can be used as a guide line for choosing the slot depth. Once the depth is finalized
the width can be estimated from the slot area and depth.
The following factors can be considered before finalizing the slot dimensions





Flux density in tooth
flux pulsations
Eddy current loss in conductors
Reactance voltage
Fabrication difficulties
24
The dimensions of the slot and the number of slots will decide the dimensions of the tooth.
The dimensions of the tooth should be chosen such that the flux density in any part of tooth
does not exceed 2.1 Wb / m2 .
The slot opening should be as small as possible in order to reduce flux pulsation losses.
With increase in depth of the slot the eddy current loss in conductors increases, specific
permanents of slot increases,reactance voltage increases and it becomes difficult fabricate
the lamination with narrow width at the roots of teeth.
Depth of armature core
The depth of armature cannot be independently designed, because it depends on the
diameter of armature (D) , innerdiameter of armature (D1) and depth of the slot (ds).
After estimating D, Di and ds the available depth of core dc can calculated . With this
value of dc, the flux density in the core can be estimated and if it does not exceed 1.5
Wb / m2 then the available depth of core is sufficient, otherwise we have to increase
the diameter of the armature D to give sufficient depth for core. The usual value of flux
density in the core is 1.0 to 1.5 Wb / m2 .
DESIGN OF FIELD SYSTEM
The field system consist of poles, pole shoe and field winding. The two types of
field windings are shunt and series field winding. The shunt field winding have large
number of turns made of thin conductors, because the current carried by he is very low.
The series field winding is designed to carry heavy current and so it is made of thick
conductors or strips.
The field coils are former wound , insulated and fixed over the field poles. In
shunt machines, the full winding space along the height of the pole is used to
accommodate shunt field winding. In compound machines , 80% of the winding space
is taken by shunt field and the remaining 20% by series field.
The factors to be considered for the design field winding are




mmf per pole and flux density
Loss dissipated from the surface of field coil
Resistance of the field coil
Current density in the field conductors.
Typical values of various terms used in field design
Bp
ATfi
= 1.2 to 1.7 Wb /m2
= (1.1 to 1.25) AT
25
qf
Sf
df
ρ
δf
=
=
=
=
=
700 W/m2
0.4 to 0.75
30 to 55mm
2 x 10-8 Ω-m
1.2 to 3.5 A / m m2
DESIGN OF SHUNT FIELD WINDING
The design of shunt field winding involves the determination of the following
informations regarding the pole and shunt field winding. A brief discussion about the
estimation of various quantities are presented below.







Dimension of the main field pole
Dimensions of the field coil
Current in the shunt field winding
Resistance of the field coil
Dimensions of the field conductor
Number of turns in the field coil
Losses in the field coil
Dimensions of the main field pole
The dimensions of the rectangular field pole are area of cross – section, length,
width and height of the pole body. For cylindrical poles the diameter has to be
estimated instead of length and width. The area of the pole body can be estimated from
the knowledge of flux per pole, leakage coefficient and flux density in the pole. The
leakage coefficient depends on the power output of the dc machine. A suitable value of
leakage coefficient can be assumed from table given below. The usual range of flux
density in the pole is 1.2 to 1.7 Wb /m2
Table-: Leakage coefficient
Output
in kW
Leakage
co-efficient C
50
100
200
500
1000
1.12 to 1.25
1.11 to 1.22
1.10 to 1.20
1.09 to 1.18
1.08 to 1.16
Flux in the pole body, фp = C ф
Area of pole body, Ap = фp / Bp
When circular poles are employed the area of cross – section will be a circle and so the
diameter of the pole can be estimated from the equation of circle.
26
Ap is the area of the pole and dp is the diameter of the pole body.
When rectangular poles are employed the length of the pole is chosen as 10 to 15mm
less than the length of armature . The reduction in the length of the pole is to permit end play
and to avoid magnetic centering.
Length of the pole, Lp = L – (0.001 to 0.015)
Net iron length of the pole, Lpi = 0.9 Lp
bp is the width of the pole.
The height of the pole body is given by the sum of height of field coil , thickness of
insulation and clearance.
Height of the pole body, hp = hf + Thickness of insulation and clearance
Total height of the pole, hp1 = hp + hs
Dimension of the coil
The field coils are former wound and placed on the pole. The field coils may have
rectangular or circular cross- section, depending on the type of poles. The field coils and their
cross-sections are shown in fig 3.19. The dimension of the field coil are depth, height and
length of mean turn of field coil.
Usually the depth of field coil is assumed and the value depends on the diameter of
armature. A suitable value of depth of field winding can be selected from table 3.9.
The height of field coil depends on the surface required for cooling the coil and number
of turns in the field coil. Hence the weight of the field coil (hf) and the number of terms (T Bf)
cannot be independently designed . Therefore two equations are formed with hf & Tf as
variable and they are solved to estimate hf & Tf . The information of the two equations are
explained in the section “Power loss in the field coil”.
The length of mean turn (Lmt) of field coil can be calculated using the dimension of
pole and depth of field coil.
Table: Depth of field winding
Armature
Depth of
Diameter
Field winding
in m
In mm
0.2
30
0.35
35
0.5
40
0.65
45
27
1.00
50
above 1 m
55
For rectangular field coil
Length of mean turn, Lmt = 2 (L p +b p +2df) ( Refer fig 3.19)
(or) Length of mean turn, Lmt =
where, Lo = Length of outer most turn
Li = Length of inner most turn
For cylindrical field coil,
Length of mean turn, Lmt = π ( dp+ d Lf ) (Refer fig 3.19)
Current in the shunt field winding
The shunt field current can be estimated from the voltage across field coil and the
resistance of field coil. Each pole of a dc machine carries one field and all the field coils are
connected in series to form shunt field winding. Hence the voltage across each field coil is
given by,
Ef is the voltage across each shunt field coil
Normally the voltage across the shunt field winding is equal to 80 to 85% of rated
machine voltage. because in generators 15 to 20 % of rated voltage is absorbed by the field
rheostat provided for voltage regulation. Ijn case of motors this allowance depends on the
range of speed control.
E is the voltage across each shunt field coil,I is the the field current and Rf is the
resistance of each coil field.
Resistance of field coil
The resistance of the field coil can be estimated from the knowledge of resistivity of
copper, length of mean turn of field coil , number fields in turns in field coil and area of cross
section of field conductor.
Length of field coil = Length of mean turn x Number of turns in field coil
Dimension of field conductor
The dimensions of field conductor are area of cross- section and diameter. The area of
the cross section of the field conductor can be estimated from the knowledge of field current (I
28
f)
and current density (δf). The usual range of current density in the field conductor is 1.2 to
3.5 A /m m2
Alternatively the area of cross-section of field conductor can be estimated from
the knowledge of ampere turns, resistivity, length of mean turn and voltage across each field
coil.
We know that , area of cross-section of field conductor, ATf1 = Tf If
Usually the conductors with circular cross-section is used for field winding. Therefore the area
of cross- section is also given by the equation for area of circle.
The conductors used for winding are coated with winding warnish in order to insulate
one turn from the other. The diameter of the conductor included the thickness of insulation is
necessary to estimate the copper space factor.
Copper space factor, Sf = 0.75
Number of turns in the field coil
When the ampere – turns to be developed by the field coil is known the turns can be
estimated from the knowledge of field current.
The field ampere turns is also related to
resistance of field coil and dimensions of field coil, which in turn related to height of field coil.
The height of field coil depends on the cooling surface required for dissipating the heat
developed in the field coil. Hence two equations are formed with Tf & hf as variables and they
are solved to estimate Tf& hf .
Power loss in the field coil
The power loss 9in the field coil is copper loss which depends on resistance and
current. Heat is developed in the field coil due to this loss and the heat is dissipated through the
surface of the coil. If there is no sufficient surface for heat dissipation then heat accumulates,
which may lead to damage ( or burning ) of the coil.
In field coil design the loss dissipated per unit surface area is specified and from which
the required surface area can be estimated. The surface area of field coil depends on length of
mean turn, depth and height of field coil. Usually the length of mean turn is estimated from the
dimensions of the pole, depth of winding is assumed and height of field winding is estimated in
order to provide the required surface area.
The heat can be dissipated from all the four sides of a coil i.e., inner, outer, top and
bottom surface of the coil.
Inner surface area of the field coil = L f ( h f – d f )
Outer surface area of the field coil= Lmt ( h f + d f )
Top surface area of the field coil = Lmt d f
Bottom surface area of the field coil =Lmt d f
29
Total surface area of field coil } S = Lmt (h f -d f ) + L mt dr +Lmdf
= Lmt hf + 2Lmt df = 2Lmt (hf+df)
Permissible copper loss in the field coil ,Qf = S qf
where qf is loss dissipated per unit area.
On substituting for S from equation above we get,
Permissible copper loss, Qr = 2Lmt qr (hr +dr)
Actua copper loss in field coil = I2f Rf = E2f
Rf
On substituting for R4 from equation above we get
Actual copper loss -
E2f
ρ Lmt Tf / af
= E2f af .
ρ Lmt Tf
On equating the permissible copper loss to actual copper loss, we can form an equation with Tf
and hf as variables.
2 Lmt qf (hr + df) = E2f af .
ρ Lmt Tf
Another equation with Tf and hf can be formed from the knowledge of conductor area
in the field coil as shown below
Conductor area in field coil = Number of turns x
Area of cross -section
of field conductor
= Tf af
Also, conductor area in field coil = Copper space factor x Area of cross-section
field coil
= Sf hf df
On equating the equations above we get
Tf af = Sf hf df
30
DESIGN OF SERIES FIELD WINDING
In compound machines, each pole carries a shunt field coil and a series field coil. In
each pole approximately 80% of the height is occupied by shunt field coil and 20% by series
field coil. The design of shunt field coil for compound machines is same as that of a shunt
machine. The series field coil is provided to compensate for the reduction in field mmf due to
armature reaction. The ampere turns to be developed by the series field will be 15 to 25% of
full load armature ampere turn.
In case of series machines each pole carries a series field coil and all the field coils are
connected in series to form series field winding. The ampere turns to be developed by the
series field coil in dc series machine is 1.15 to 1.25 times the full load armature mmf.
Design of series field coil
The stop-by-step procedure for design of series field coil is given below.
Step 1 :
Estimate the ampere turns to be developed by series field coil
Armature ampere turns at full load (per pole)=
armature
Current through x No of
a turn
turns
.
.
Number of poles
= Iz x Z/2
P
= IzZ
2P
For compound machines
Ampere –turns to be developed
by series field coil
ATse = 0.15 to 0.25 x Iz Z
2p
For series machines, Ampere
turns to be developed by
series field coil
ATse = 1.15 to 1.25 x Iz Z
2p
Step- 2 Calculate the number of turns in the series field coil
Number of turns in series
field coil
Tse = ATse (rounded to integer)
Ise
Where Ise = Ia = Current through series field winding at full load
Step- 3 Determine the area of cross-section of the series field conductor
Area of cross- section of series field conductor , ase =
Ise
31
δ
where, δse = Current density in series field conductor
The current density in chosen as 2 to 2.3 A/mm2. For low capacity machines circular
conductors are used and for higher capacity machines rectangular conductor are used.
Step-4 Estimate the dimensions of the field coil.
Conductor area in field coil = Tse ase
Also conductor area in field coil = Copper space factor x Depth of coil
= Sfse hse dse
where, Sfse = Copper space factor for series field coil.
When circular conductors are employed the value of copper space factor is 0.6 to 0.7.
When rectangular conductors are employed, the space factor depends on thickness and type of
insulation.
On equating the two equations for conductor area of field coil we get,
Sfse hse dse = Tse ase
Therefore Height of field coil, hse = Tse ase
Sfse dse
Choose a suitable depth and calculate the height of series field coil using the equation.
In case of compound machine, the total height of pole required to accommodate field
winding will be sum of the height of shunt field coil and series field coil.
Step – 5 Estimate the resistance of series field coil.
Resistance of series field coil = ρLmtse Tse
ase
where lmtse = Length of mean turn of series field coil
The length of mean turn of series field coil can be estimated from the dimensions of the
pole.
Length of mean turn of series field coil = Lmtse = 2 (Lp +bp + 2dse)
32
DESIGN OF COMMUTATOR AND BRUSHES
The commutator and brush arrangement are used to convert the bidirectional internal
current to unidirectional external current viceversa. The current flows through the brush
mounted on the commutator surface. The brushes are located at the magnetic neutral axis
which is midway between two adjacent poles.
When a armature conductor pass through the magnetic neutral axis, the current in the
conductor reverse from the direction to the other. Since the brushes are mounted on magnetic
neutral axis, the coil undergoing current reversal is short circuited by carbon brush. During
this short circuit period, the current must be reduced from its original value to zero and then
built up to an equal value in the opposite direction. This process is called commutation and
the time during which the current reversal takes place is called the time of commutation.
The phenomena of commutation is affected by resistance of the brush, reactance emf
induced by leakage flux and rotational emf induced by armature flux. Based on the factors
effecting the commutation, the process of commutation can be classified into following types.




Resistance commutation
Retarded commutation
Accelerated commutation
Sinusoidal commutation
When the reactance and rotational emfs are eliminated by the compoles, the commutation
is assisted only by the resistance of the brush. This type of commutation is called resistance
commutation. By employing high resistance carbon brushes, a straight line or linear resistance
commutation can be achieved.
When the brushes are placed at geometrical neutral axis, a reactance voltage and rotational
emf are induced in the coil under going commutation. This is due to the shift in magnetic
neutral axis during load conditions. These voltages delays the process of current reversal and
this type of commutation is called delayed commutation. In this case the commutation is
completed before the current can reach its final value and so the current has to jump through
the air in the form of a spark at the trailing edge of brush.
When the brushes are slightly shifted from magnetic neutral axis in the direction of
rotation, the emf induced in the coil undergoing commutation will accelerate the process of
current reversal. This type of commutation is called accelerated commutation. This type of
commutation may give rise to burning of brushes at the leading edge.
When the process of current reversal is delayed at the leading edge and accelerated at the
trailing edge, the commutation is called sinusoidal commutation.
The commutator is cylindrical in shape and placed at one end of the armature. It consists
of a number of copper bars or segments separated from one another by a suitable insulating
material of thickness 0.5 to 1 mm. The number of commutator segments is equal to number of
33
coils in the armature. Each segment is connected to an armature coil and the connections are
made with the help of risers. In large machines the risers are made of copper strips and in
small machines lugs are provided instead of risers.
Te materials used for commutator segments are hard drawn copper or silver copper (0.05%
silver).. The silvered copper can withstand high temperature (350ºC) used for soldering the
risers. The insulators used between commutator segments are made of mica or micanite or
resin bonded asbestos. The different types of materials used for brushes are natural graphite,
hard carbon, electrographite and metal graphite.
A carbon brush placed on the commutator surface and the cross-section commutator are
shown in figure below. The commutator segments are firmly held by V shaped clamping rings
secured by bolts.
Design Formula
In dc machine the number of commutator segments is same as number of coils in
armature. Also the equation for minimum number of coils gives the minimum number of
segments.
Number of coils, C = 1 x uSa
2
Number of commutator segments = C
Minimum number of segments = Ep / 15
Commutator segments pitch, βc = π Dc / C
Choose the diameter of commutator , Dc as 60 to 80% of the diameter of armature, The
commutator diameter is also chosen such that the commutator peripheral speed is limited to
15m/s to 30m/s. The width of commutator segments, βc is greater than or equal to 4mm.
Thickness of brush is selected such that it covers 1 to 3 commutator segments.
Current carried by each brush, Ib = 2I βa/ p for lap winding
= Ia
for wave winding
Total brush area per spindl, Ab = Ib / δb
The number of brush locations are decided by the type of winding. In lap winding the
number of brush locations is equal to number of poles and in wave winding it is always two. In
each locations there may be more than one brush mounted on a spindle.
34
The area of each individual brush should be chosen such that it does not carry more
than 70 A. Hence the number of brushes in a spindle, nb is selected such that each brush does
not carry more than 70 A.
Let ab = Contact area of each brush
nb = Number of brushes per spindle
now, contact area of brushes in a spindle, Ab = n βb ab
Also, ab = wb tb
A = nb wb tb
Usually, the thickness of brush, t = ( 1 to 3 ) x βc
Width of brush, wb =__ Ab__
nbtb
= _ab_
tb
The length of the commutator depends on the space required for mounting the brushes
and to dissipate the heat generated by the commutator losses
Length of the commutator, Lc = nb ( wb + Cb ) + C1 + C2
Cb = clearance between the brushes ( usually 5mm )
C1 = clearance allowed for staggering the brushes ( 10mm for small machine and
30mm for large machine)
C2 = Clearance for allowing end play ( Usually 10 to 25mm).
The losses at the commutator are the brush contact losses and the brush friction losses.
The brush contacts loss depends on materials , condition and equally of commutation obtained.
Hence it is difficult to predetermine accurately the brush contacs losses. The brush contact drop
Vb is independent of load current.
The brush friction loss can be calculated from the formula,
Pbf = μ Pb AB Vc
Pb =
AB =
AB =
AB =
Brush contact pressure pressure on commutator, N / m2
Total contact area of all brushes, m2
pAb for lap winding ;
= μ Coefficient of friction
2Ab for wave winding ;
= Vc Peripheral speed of commutator, m / sec.
35
Table: Properties of brush materials
Type of
material
Brush
contact drop
in V
Current
Density
in A/mm2
Pressure
in kN/ m2
Commutator
speed in
m/sec
Co-efficient
of friction
0.7 – 1.2
0.1
14
50 - 60
0.1 – 0.2
0.7 – 1.8
0.065 - 0.085
14 - 20
20 - 30
0.15 – 0.25`
0.7 – 1.8
0.085 – 0.11
18 – 21
30 – 60
0.1 – 0.2
0.4- 1.7
0.1 - 0.2
18 - 21
20 - 30
0.1 – 0.2
Natural
graphite
Hard
carbon
Electro
graphite
Metal
graphite
TUTORIAL PROBLEMS
1.
Find the main dimension of a 200kW, 250V, 6 pole, 100rpm generator. The maximum
value of flux density in the gap is 0.87 Wb/ m2 and the ampere conductors per meter of
armature periphery are 31000. The ratio of pole arc to pole pitch is 0.67 and the efficiency is
91 percent. Assume the ratio of length of core to pole pitch = 0.75.
Given Data
ψ = 0.67
200kW
N = 1000 rpm
250 V
Bg = 0.87 Wb / m2
6 pole
ac = 31000 amp.cond./m η = 0.91
L / τ = 0.75
Solution
Power developed in armature, Pa
Output coefficient , Co = π2 Bav
=
P = 200 = 219.78 kW
η
0.91
ac x 10-3 = π2 ψ Bg ac x 10-3
= π2 x 0.67 x 0.87 x 31000 x 10-3
= 178.34kW / m3 - rps
Also the power developed in armature, Pa = Co D2 Ln
D2 L =
_ Pa _
Co n
= _ _219.78_______
178.34 x (1000 / 60)
36
= 0.0739m3
Give that L / τ = 0.75
L = 0.75 τ = 0.75 x _ πD _
P
= __0.75 x π__ D
6
= 0.3927 D
Put L = 0.3927 D in the equation for D2 L
D2 L = 0.0739
D2 (0.3927D) = 0.0739
D =
0.0739 1/3 =
0.3927
0.57
;
L = 0.3927 D = 0.3927 x 0.57 = 0.22 m
Result
The diameter of the armature , D = 0.57m
The length of the armature,
L = 0.22 m
2. Find the main dimension and the number of poles of a 37kW, 230V, 1400rpm shunt motor
so that a square pole face is obtained. The average gap density is 0.5 Wb / m2
and the ampere conductors per meter are 22000. The ratio of pole arc to pole pitch is 0.7 and
the full load efficiency is 90 percent.
Given Data
37kW
Square pole face
ψ = 0.7
230 V
Bav = 0.5 Wb / m2
η = 90 %
1400 rpm ac = 22000
Solution
If poles ,p = 2 then f =
pN_ = 2 x 1400 = 23.33Hz
120
120
If poles, P = 4, then f = pN = 4 x 1400 = 46.67Hz
120
120
Power input, Pi = VI x 10-3 kW
37
Also power point, Pi = Power output
Efficiency
= P
η
Load current,
=
I = _____ P ______
η x V x 10-3
37______ = 178.74 A
0.9 x 230 x 10-3
Let, armature current, Ia ≈ P = 37 kW
Also power developed in armature, Pa = Co D2 Ln
D2 L =
Pa
Co n
=
_____ 37__________
108.57 x ( 1400 / 60 )
= 0.0146 m3
Give that , Pole arc = 0.7
Pole pitch
For square pole face, length of armature is equal to pole arc
Pole arc_ =
Pole pitch
_Length _ = L
Pole pitch
τ
L = 0.7 τ = 0.7 x π D
P
= 0.7
= 0.7 x π x D = 0.5498 D
4
Put, L = 0.5498D in the equation for D2 L
D2 L = 0.0146
D2 ( 0.5498D) = 0.0146
D = 0.0146 1/3 = 0.2983m ≈ 0.3m
0.5498
L = 0.5498 D = 0.5498 x 0.3 = 0.165m
Result
Number of poles,
p = 4
38
Diameter of armature, D = 0.3m
Length of armature,
L = 0.165m
3. Calculate the main dimension of a 20HP, 1000rpm, 400V, dc motor. Given that Bav 0.37=
Wb / m2 and ac = 16000 amp. cond. /m. Assume an efficiency of 90%.
Given Data
20 HP Bav = 0.37 Wb / m2
400 V ac = 16000 amp.cond./m
N = 1000 rpm
η = 90%
Solution
Power input, Pi = P = 20 x 0.746 = 16.58 kW
η
0.9
Also power input, Pi = VI x 10-3
Load current, I =
Pi
V x 10-3
= 16.58
= 41.45 A
-3
400 x 10
Armature current Ia ≈ I = 41.45A
Since the armature current is less than 200A, the current per parallel path will not
exceed the upper limit of 200A.
Let p = 2, f = pN = 2 x 1000 = 16.67 Hz
120
120
p = 4, f = pN = 4 x 1000 = 33.33 Hz
120
120
p = 6, f = pN = 6 x 1000 = 50 Hz
120
120
The frequency of flux reversals should lie in the range of 25 to 50 Hz. For
minimum cost the highest possible choice of poles should be chosen.
Hence, the number of poles, p = 6
Output coefficient, Co = π2 Bav ac x 10-3
= π2 x 0.37 x 16000 x 10-3
39
= 58.428 kW / m3 -rps
For dc motor, power
developed in armature
Pa ≈ P = 20HP = 20 x 0.746 = 14.92 kW
Also power developed in armature, Pa = Co D2 Ln
D2 L = Pa = ______14.92________
Con
58.428 x (1000 / 60)
= 0.0153 m3
Let us assume a square pole face, L = 0.7
τ.
L = 0.7τ = 0.7 πD =
P
0.7 x π D = 0.3665D
6
Put L = 0.3665 D in the equation for D2 L.
D2 L = 0.0153
D2 (0.3665D) = 0.0153
D = ( 0.0153 )1/3 = 0.3469 m ≈ 0.35m
0.3665
L = 0.3665D = 0.3665 x 0.35 = 0.128m
Result
Diameter of armature, D = 0.35m
Length of armature, L = 0.128m
4. Design the shunt field winding of a 6 –pole , 440V, dc generator allowing a drop of 15% in
the regulator. The following design data are available. mmf per pole = 7200 AT; mean length
of turn = 1.2m; winding depth = 3.5cm; watts per sq.m. of cooling surface = 650.Calculate the
inner , outer and end surfaces of the cylindrical field coil for cooling. take diameter of the
insulated wire to be 0.4mm greater than the bare wire. Assume 2 micro – ohm – cm as the
resistivity of copper at the working temperature.
Given Data
ATfl = 7200 AT
di = d + 0.4mm
40
Lmt = 1.2m
ρ = μ Ω - cm = 2 x 10-8 Ω - m
df = 3.5 cm = 0.035 m
drop in field regulator = 15%
qf = 650 W / m2
p = 6 ; V = 440V
Solution
Given that drop in field regulator = 15 % of V
Voltage across field winding = 0.85 V
Voltage across each field coil, Ef =
0.85 V = 0.85 x 440 = 62.33 Volts
P
6
Area of cross- section of field conductor } af = AT fl ρ Lmt
Ef
= 7200 x 2 x10-8 x1.2
62.33
= 2.772 x 10-6 m2 = 2.772 mm2
Let the conductors have circular cross – section
Diameter of bare conductor , dc = 4af = 4 x 2.772 = 1.88 mm
π
π
Diameter of insulated conductor, dfci = 1.88 + 0.4 = 2.88 mm
Space for winding, Sf = 0.75 dfc2 = 0.75 x 1.882 = 0.51
dfci
2.28
Cooling surface of the coil, S = 2Lmt (hf + df ) = 2 x 1.2 (hf + 0.035)
= 2.4 hf + 0.084
Permissible loss, Qf = qf S = 650 [ 2.4 hf + 0.084 ] = 1560 hf + 54.6
Actual copper loss in field coil } Qf = I2f Rf = E2f = __ E2f ___ = __ E2f af __
Rf
Tf ρ Lmt
Tf ρ Lmt
af
= 62.332 x 2.772 x 10-6 = 0.45 x 106
Tf x 2 x 10-8 x 1.2
Tf
On equating the permissible loss to actual copper loss, we get
41
1560 hf + 54.6 = 0.45 x 106
Tf
Tf = 0.45 x 106___
1540 hf + 54.6
(or)
Conductor area in field coil = Sf hf df = 0.51 x hf x 0.035 = 0.01785 hf
Also, conductor area = Tf af = Tf x 2.772 x 10- 6
On equating the two equations of conductors area we get,
0.01785 hf = Tf x 2.772 x 10- 6
hf = 2.772 x 10- 6 Tf = 1.553 x 10- 6 Tf
0.01785
Solving for Tf
Tf = _____0.45 x 106 __________ =
1560 x 1.553 x 10-4 Tf + 54.6
0.45 x 106 _
0.242 Tf + 54.6
0.242 T2f + 54.6 Tf - 0.45 106 =0
__________________________
Tf = -54.6 +- √ 54. 62 + 4 x 0.242 x 0.045 x 106 = 1255 ( Taking only positive value)
2 x 0.242
From equation (b), hf = 1.553 x 10- 4 Tf
= 1.553 x 10- 4 x 1255 = 0.195m
Here the field coil is cylindrical and the dimensions of the field coil are shown in fig
E3.11.
Here, Lmt
= π dmf where dmf =mean diameter of field coil
Let
dif
= Inner diameter of field coil
dof
= Outer diameter of field coil
dif
=
dmf
dmf - df
= Lmt
π
and dof = dmf
= 1.2
π
+ df
= 0.382m
dif
= 0.382 – 0.035 = 0.347 m
dof
= 0.382 + 0.035 = 0.417m
42
Length of inner most turn, Lif = π dif = π x 0.347 = 1.09m
Length of outer most turn, Lof = π dof = π x 0.416 = 1.307 m
Inner cooling surface, hf Lif = 0.195 x 1.09 = 0.2126 m2
Outer cooling surface, hf Lof = 0.195 x 1.307 = 0.255 m2
Top cooling surface = π d2 of - π d2 if
4
4
= π (0.416)2
4
- π (0.347)2
4
= 0.0413 m2
Bottom cooling surface = 0.0413 m2
Note: Top and bottom surfaces are end cooling surfaces
Result
Area of cross section of field conductor, af
Diameter of bare conductor, dfc
Diameter of insulated conductor ,dfci
Number turn in filed coil, Tf
Height of field coil, hf
Inner cooling surface of a field coil
Outer cooling surface of a field coil
Top cooling surface of a field coil
Bottom cooling surface of a field coil
= 2.772 mm2
= 1.88 mm
= 2.28 mm
= 1255 turns
= 0.195 m
= 0.2126m2
= 0.255mm2
= 0.0413mm2
= 0.0143mm2
6. Determine the total commutator losses for a 1000kW, 500V, 800 rpm, 10 pole generator.
Given that commutator diameter = 1.0m, current density at brush contact = 75 x 10-3 A / mm2 ,
brush pressure = 14.7 kN / m2 , coefficient of friction = 0.28, brush contact drop = 2.2V.
Given data
p = 100kW
Dc = 1.0m
δb= 75 x 10-3 Amm2 N = 800rpm
p = 10
μ = 0.28
V = 500V
Pb = 14.7 kN / m2
Vb = 2.2 A
Solution
Armature current,
Ia =
P x 1000
V
= 1000 x 1000 = 2000A
500
Brush contact loss, Ia Vb
= 2000 x 2.2 = 4400W
Total brush contact area, AB = 2 Ia = 2 x 2000
δb
75 x 10-3
43
= 53.33 x 10-3 mm2 = 53.33 x 10-3 m2
Commutator peripheral speed, Vc
= π Dc n
= π x 1 x (800 / 60) = 41.89 m / sec
Brush friction loss = π Pa Ab Vc
= 0.28 x 14.7 x 10-3 x 53.33 x 10-3 x 41.89 = 9196 Watts
Total power loss = 4000 + 9196 = 13596 watts
Current per brush area = 2 Ia
P
= 2 x 2000 = 400 A
10
Current per brush is limited to 70A
Number of brushes per spindle, nb
= 400
70
Let number of brushes per spindle, nb = 8
≈6
Current carries by each brush = 400 = 50 amps
8
Area of each brush, ab =
50
= 666.67 mm2
-3
75 x 10
Contact area of each brush arm = nb ab = 8 x 666.67 = 5333.3 mm2
Total brush contact area , AB = 10 x 5333.3 = 53,333 mm2 = 53,333 x 10-6 m2
Brush friction loss = μ Pa Ab Vc = 0.28 x 14.7 x 10-3 x 53,333 x 10-6 x 41.89
= 9196 Watts.
Total loss = 4400 + 9196 = 13.596kW
Result
Total commutator loss = 13.596kW.
44
MODULE-2
DESIGN OF TRANSFORMERS
TRANSFORMER CONSTRUCTION
A transformer consists of two windings coupled through a magnetic medium. The two
windings work at different voltage level. The two windings of the transformer are called high
voltage and low voltage winding. Both windings are wound on a common core. One of the
winding is connected to ac supply and it is called primary. The other winding is connected to
load and it is called secondary. The transformer is used to transfer electricity energy from high
voltage winding to low voltage winding are vice – versa through magnetic field.
The construction of transformer varies greatly, depending on their applications,
winding voltage and current rating and operating frequencies. The two major type of
construction of transformers ( used in transmission and distribution of electric energy) are core
type and shell type. Depending on the application , these transformers can be classified as
distribution transformers and power transformers.
The transformer is extremely important as a component in many different types of
electric circuits, from small – signal electronic circuit to high voltage power transmission
systems.
The most important function performed by transformer are,



Changing voltage and current level in an electronic system
matching source and load impedances for maximum power transfer in electronic
and control circuitry.
Electrical isolation
Core type transformer
In core type transformer , the magnetic core is built of laminations to form a rectangular
frame and the windings are arranged concentrically with each other around the legs or limbs.
The top and bottom horizontal portion of the core are called yoke. The yokes connect the two
lings and have a cross section area equal to are greater than that of limbs.
Each limb carries one half of primary and secondary. The two windings are closely
coupled together to reduce the leakage reactance. The low voltage winding is wound near the
core and high voltage winding is wound over low voltage winding away from core in order to
reduce the amount of insulating materials required.
45
Shell type transformer
In shell type transformer the windings are put around the central limb and the flux path
is completed through two side limbs. The central limb carries total mutual flux while the side
limbs forming a part of a parallel magnetic circuit carry half the total flux. The cross sectional
area of the central limb is twice that of each side limbs.
Table 4.1 : Comparison of core and shell type transformer
Core type
1. Easy in design and construction
2.Has low mechanical strength due to non
– bracing of windings
3. Reduction of leakage reactance is not
easily possible
4. The assembly can be easily dismantled
for repair work.
5. Better heat dissipation from windings.
6. Has longer mean length of core and
shorter mean length of coil turn. hence
best suited for EHV ( Extra High
Voltage) requirements.
Shell type
1. Comparatively complex
2. high mechanical strength
3. Reduction of leakage reactance is
highly possible.
4. It cannot be easily dismantled
for repair work.
5. Heat is not easily dissipated from
windings since it is surrounded by core.
6. it is not suitable for EHV ( Extra High
Voltage) requirements.
Distribution transformer
Transformer upto 200kVA ( or 500kVA) are used to step down distribution voltage to a
standard service voltage or from transmission voltage to distribution voltage are known as
distribution transformers. They are kept in operation all the 24 hours a day whether they are
carrying any load or not.
The load on the distribution transformer varies from time to time and the transformer
will be on no-load most of the time. Hence in distribution transformer the copper loss ( which
depends on load) will be more when compared to core loss ( which occurs as long as
transformer is in operation).Hence distribution transformer are designed with less iron loss and
designed to have the maximum efficiency at a load much lesser than full load. Also it should
have good regulation to maintain the variation of supply voltage within limits and so it is
designed with small value of leakage reactance
Power transformer
The transformers used in sub- stations and generating stations are called power
transformer. They have rating above 200kVA. Usually a substation will have number of
transformers working in parallel. During heavy load periods all the transformers are put in
operation and during light load periods some transformers are disconnected. Therfore the
power transformers should be designed to have maximum efficiency at or near full load. Power
transformers are designed to have considerably greater leakage reactance that is possible in
46
distribution transformers in order to limit the fault current. In the case of power transformers
inherent voltage regulation is less important than the current limiting effect of higher leakage
reactance.
OUTPUT EQUATION OF SINGLE PHASE TRANSFORMER
The equation which relates the rate kVA output of transformer to the area of core and
window is called output equation. In transformers the output kVA depends on flux density and
ampere turns. The flux density is related to core area and the
ampere – turns is related to window area.
The low voltage winding is placed nearer to the core in order to reduce the insulation
requirement. The space inside the core is called window and it is the space available for
accommodating the primary and secondary winding. The window area is shared between the
winding and their insulations.
The induced emf in a transformer, E =Emf per turn,
the window in single phase transformer contains one primary and one secondary
windings. The window space factor Kw is the ratio of conductor area in window to total area of
window.
Window space factor, Kw = Conductor area in window
Total area of window
= Ac
Aw
Conductor area in window, Ac = Kw Aw
The current density δ is same in both the windings.
Current density, δ =
Ip
ap
= Is
as
Area of cross - section of primary conductor, ap = Ip
δ
Area of cross – section of secondary conductor, as = Is
δ
If we neglect magnetizing mmf then primary ampere turns is equal to secondary ampere
turns.
Ampere turns, AT = Ip Tp = Is Ts
Total copper area in window
Ac = Copper area of
+ Copper area of
47
primary winding
= Number of primary
turns x area of
cross – section of
primary conductor
= Tp Ap + Ts as = Tp
= 1 (Tp Ip + Ts Is ) = 1
δ
δ
Ip + Ts Is
δ
δ
secondary winding
Number of secondary
+ turns x area of
cross – section of
secondary conductor
(ap = Ip and as = Is )
δ
δ
= ( AT + AT) ( AT = Ip Tp = Is Ts )
= 2AT
δ
On equating the 2 equations,
Kw Aw = 2 AT / δ
Ampere turns, AT = 1 Kw Aw δ
2
The kVA rating of single phase transformer is given by,
kVA rating, Q = Vp Ip x 10-3 ≈ Ep Ip x10-3 (Ep ≈ Vp)
= Ep Tp Ip x 10-3
Tp
(Et = Ep and AT = Tp Ip )
Tp
= Et AT x 10-3
On substituting Et and AT we get,
Q = 4.44 fфm Kw Aw δ x 10-3
2
= 2.22 fфm Kw Aw δ x 10-3
( B m = фm )
Ai
= 2.22 f Bm Ai Kw Aw δ x 10-3
The equation is the output equation of single phase transformer.
48
OUTPUT EQUATION OF THREE PHASE TRANSFORMER
The cross-section has three limbs and two windows. Each limbs carries the low voltage
and high voltage winding of a phase.
The induced emf per phase, E = 4.44 fфm T volts
Emf per turn, Et = E = 4.44 fфm volts
T
In case of three phase transformer , each window has two primary and two secondary
windings. The window space factor Kw is the ratio of conductor area in window to total area of
window,
Kw = Conductor area in window
Total area of window
= Ac
Aw
Note: In three phase transformers , one window is considered.
The current density δ is same in both the windings.
Current density, δ =
Ip = Is
ap
As
where, Ip = Primary current per phase
Is = Secondary current per phase
Are of cross – section of primary conductor, ap = Ip
δ
Area of cross – section of secondary conductor, as = Is
δ
If we neglect magnetizing mmf then primary ampere turns per phase is equal to secondary
ampere turns per phase.
Ampere turns, AT = Ip Tp = Is Ts
Total copper
area in window = Ac = 2 x Number of
2 x number of
primary turns x area + secondary tuns x area
of cross – section of
of cross – section of
primary conductor
secondary conductor
= 2 Tp ap + 2 Tp as
49
= 2 Tp Ip + 2 Ts Is
δ
δ
( ap = Ip and as = Ip )
δ
δ
= 2 (Tp Ip + Ts Is )
δ
= 2 ( AT +AT )
δ
( AT= Ip Tp = Is Ts )
= 4 AT
δ
On equating the 2 eqns. we get,
4 AT = Kw Aw
δ
Ampere – turn, AT = Kw Aw δ
4
The kVA rating of three phase transformer is given by,
kVA rating, Q = 3 x Volt – ampere per phase x 10-3 = 3 Vp Ip x 10-3
= 3 Ep Ip x 10-3
( 3 Ep ≈ Vp )
= 3 x Ep x Tp Ip x 10-3 ( Et = Ep and AT = Tp Ip )
Tp
Tp
= 3 Et AT x 10-3
On substituting Et and AT we get,
Q = 3 x 4.44 fфm x Kw Aw δ x 10-3
4
= 3 x 33 fфm x Kw Aw δ x 10-3 ( Bm = фm )
Ai
= 3.33 fBm Ai Kw Aw δ x 10-3
The equation (4.24) is the output equation of three phase transformer.
50
Emf per turn
The transformer design starts with selection of an appropriate value of emf per turn.
Hence an equation for emf per turn can be developed by relating output kVA, magnetic and
electric loading. In transformer the ratio of specific magnetic and electric loading is specified
rather than actual value of specific loadings.
Let, ratio of specific magnetic
and electric loading
= γ = фm
AT
The volt ampere per phase a transformer is given by the product of voltage and current
per phase. Considering the primary voltage and current per phase we can write,
kVA per phase, Q = Vp Ip x 10-3
= 4.44 fфm Tp Ip x 10-3 (Vp ≈ Ep = 4.44 fфm Tp
)
= 4.44 fфm ATp x 10-3 (Tp Ip = AT )
= 4.44 fфm фm x 10-3 ( AT = фm )
r
r
фm2 = ___ Qr _____
4 x 44f x 10-3
we can say that the emf per turn is directly proportional to K. The value of K depends
on the type , service condition and the method of construction of transformer. The value of K
for different types of transformers are listed in table
Table : Values of K for different types of transformer
Transformer type
Single phase shell type
Single phase core type
Three phase shell type
Three phase core type, distribution transformer
Three phase core type, power transformer
K
1.0 to 1.2
0.75 to 0.85
1.3
0.45
0.6 to 0.7
Note : Q is kVA rating for single phase transformer and Q is kVA per phase for three phase
transformer.
51
DESIGN OF CORES
For core type transformer the cross-section may be rectangular, square or stepped.When
circular coils are required for distribution and power transformers, the square and stepped cores
are used.
For shell type transformer the cross-section may be rectangular. When rectangular
cores are used the coils are also rectangular in shape. The rectangular cores is suitable for small
and low voltage transformers. In core type transformer with rectangular cores, the ratio of
depth to width of the core is 1.4 to 2. In shell type transformers with rectangular cores the
width of the central limb is 2 to 3 times the depth of the core .
The excessive leakage fluxes produced during short circuit and over loads, develop
serve mechanical stresses on the coils. on circular coils these forces are radial and there is no
tendency for the coil to change its shape. But the rectangular and square coils the forces are
perpendicular to the conductors and tends to deform the shape of coil. Hence circular coils are
employed in high voltage and high capacity transformers.
In square cores the diameter of the circumscribing circle is larger than the diameter of
stepped cores of same area of cross – section. Thus when stepped cores are used the length of
mean turn of winding is reduced with consequent reduction in both cost of copper and copper
loss. However with larger number of steps a large number of different sizes of laminations
have to be used. This results in higher labour charges for shearing and assembling different
types of laminations.
Square core
Let d= diameter of circumscribing circle
Also, d = diagonal of the square core
a = side of square
d is the diameter of circumscribing circle
side of square, a = d
Gross core area, Agi = area of square = a2 = 0.5d2
Let stacking factor, Sf = 0.9
Net core area, Ai = stacking factor x Gross area core
= 0.9 x 0.5 d2 = 0.45 d2
Note : The gross core area is the area including insulation area and net core area is the
area of iron alone excluding insulation area.
52
Area of circumscribing circle = π d2
4
The ratio,
Net core area__________
Area of circumscribing circle
=
The ratio,
Gross core area__________
Area of circumscribing circle
=
0.45 d2
(π / 4) c
= 0.58
0.5 d2
= 0.64
2
(π / 4) d
Another useful ratio for the design of transformer core is core area factor. It is the ratio
of net core area and square of the circumscribing circle
Core area factor, =
Net core area__________
Square of circumscribing circle
= Ai =
d2
0.45 d2
d2
= 0.45
Two stepped core or cruciform core
In stepped core the dimension of the step should be chosen, such as to a occupy
maximum area within a circle. the dimension of the two step to give maximum area for the
core in the given area of circle are determined as follows [ Also for given diameter of circle]
Let a = Length of the rectangle
b = Breadth of the circle
d = Diameter of the circumscribing circle
Also, d = Diagonal of the rectangle
θ = Angle between the diagonal and length of the rectangle
The maximum core area for a given d is obtained when θ is maximum value. Hence
differentiate Agi with respect to θ and to equate to zero to solve for maximum value of θ.
cos θ = a ;
b
a = d cos θ
sin θ = b ;
d
b = d cos θ
The two stepped core can be divided into three rectangles gives the gross core area.
53
Gross core area, Agi = ab +
a-b
2
a – b b = ab + 2(a – b) b
2
2
b+
= ab + ab - b2 = 2ab - b2
On substituting for a and b we get,
Agi = 2(d cos θ ) (d sinθ) – ( d sinθ) 2 = 2 d2 cosθ sinθ - d2 sin2 θ
= d2 ( 2 sinθ cos θ - sin2 θ ) = d2 (sin2θ - sin2 θ)
= d2 sin2θ - d2 sin2 θ
To get maximum value of θ , differentiate Agi with respect to θ , and equate to zero,
i.e., d Agi =0
dθ
On differentiating the equation with respect to θ we get,
_d__ Agi = d2 cos 2θ x 2 - d2 2 sinθ cos θ
dθ
put _d__ Agi = 0
dθ
d2 cos 2θ x 2 - d2 2 sinθ cos θ = 0
d2 2 sinθ cos θ = d2 cos 2θ x 2
2θ = tan-1 2
d 2 sinθ = d cos 2θ x 2
2
2
sin2θ
cos2θ
= 2
tan2θ
=2
θ= 1
2
tan-1 2 = 31.72o
When θ = 31.72o, the dimension of the core (a & b) will give the maximum area for core for a
specified d.
When θ = 31.45o ,
a = d cos θ
= d cos31.72o
= 0.85d
b = d sinθ
= d sin 31.72o
= 0.53d
54
On substituting the above value of a & b in the equation we get,
Gross core – area, Agi = 2ab - b2 = 2(0.85d) (0.53d) - (0.53 d)2 = 0.618 d2
Let stacking factor, Sf = 0.9
net core-area, Ai = Stacking factor x Gross core area
= 0.9 x 0.618 d2 = 0.56 d2
The ratio, _ Net core area__________ = 0.56 d2 = 0.71
Area of circumscribing circle
(π / 4)d2
The ratio, _ Gross core area__________
Area of circumscribing circle
= 0.618 d2 = 0.79
(π / 4)d2
Another useful ratio for the design of transformer core is core area factor . It is the ratio
of net core area and square of the circumscribing circle
Core area factor,Kc = _ Net core area__________
square of circumscribing circle
= Ai = 0.56 d2 = 0.56
d2
d2
Multi – stepped cores
We can prove that the area of circumscribing circle is more effectively utilized by
increasing the number of steps. The most economical dimensions of various steps for a multi
stepped core can be calculated as shown for cruciform ( or two stepped ) core. The results are
tabulated in the table .
Table : Ratio of area of core and circumscribing circle
Ratio
______
Agi ____________
Area of circumscribing circle
Square
core
0.64
0.58
0.45
Cruciform
core
0.79
0.71
0.56
3-stepped
core
0.84
0.75
0.6
4-stepped
core
0.87
0.78
0.62
_________
Ai ___________
Area of circumscribing circle
Core area factor, Kc = Ai / d2
55
Check the flux density in the core
The flux density decides the area of cross-section of core and core loss. Higher value of
flux density results in smaller core area, lesser cost, reduction in length of mean turn of
winding, higher iron loss and large magnetizing current.
The choice of flux density depends on the service condition (i.e., distribution or
transmission)and the material used for laminations of the core. The laminations made with cold
rolled silicon steel can work with higher flux densities than the laminations made with hot
rolled silicon steel. Usually the distribution transformers will have low flux density to achieve
lesser iron loss.
When hot rolled silicon steel is used for laminations the following values can be used
for maximum flux density (Bm).
Bm = 1.1 to 1.4 Wb / m2
-
For distribution transformers
Bm = 1.2 to 1.5 Wb / m2
-
For power transformers
When cold rolled silicon steel is used for lamination the following values can be used
for maximum flux density (Bm).
Bm = 1.55 Wb / m2
- For transformers with voltage rating upto 132kV
Bm = 1.6 Wb / m2
- For transformers with voltage rating 132kV to 275 kV
Bm = 1.7 Wb / m2
- For transformers with voltage rating 275kV to 400kV
OVERALL DIMENSIONS OF THE TRANSFORMER
The main dimensions of the transformer are Height of window (Hw) and width of
window (Ww). The other important dimensions of the transformer are width of largest stamping
(a), diameter of circumscribing circle (d), distance between core centres (D), height of yoke
(H), depth of yoke(D), overall height of transformer frame (H) and over all width of
transformer frame (W).
DESIGN OF WINDING
The transformer has one high voltage winding and the low voltage winding. the design
of winding involves the determination of number of turns and area of cross – section of the
conductor used for winding. the number of turn is estimated using voltage rating and emf per
turns ( or by using ampere – turns and rated current ). The area of cross – section is estimated
using rated current and current density.
56
Usually the number of turns of low voltage winding is estimated first using the given
data and it is corrected to nearest integer. Then the number of turns of high voltage winding are
chosen to satisfy the voltage rating of the transformer.
Number of turns in low
voltage winding
where, VLV
ILV
TLV
or AT
ILV
= Rated voltage of low voltage winding
= Rated current of low voltage winding
Number of turns in high
voltage winding
where, VHV
= VLV
Et
THV = TLV x VHV
VLV
= Rated voltage of high voltage winding
Note : In step-up transformer, TLV = Tp , VLV = Vp , THV = TS and VHV = Vs
In step down transformer, THV = TP , VHV = VP , TLV = TS and VLV = VS
Rated current in a winding =
kVA per phase x 10-3
Voltage rating of the winding
The area of cross section of primary and secondary winding conductors are estimated
by assuming a current density. The choice of current density depends on the allowable
temperature rise, copper loss and method of cooling. The range of current density foe various
types of transformers are given below.
δ = 1.1 to 2.2 A / mm2
- For distribution transformers
δ = 1.1 to 2.2 A / mm2
- For small power transformers with self oil cooling
δ = 2.2 to 3.2 A / mm2
- For large power transformers with self oil cooling
or air – blast
δ = 2.2 to 3.2 A / mm2
- For large power transformers with forced
circulation of oil or with water cooling coils.
Area of cross – section of
primary winding conductor
ap
= Ip
δ
Area of cross – section of
secondary winding conductor
as
= Is
δ
57
Note : In transformers same current density is assumed for primary and secondary.
COOLING OF TRANSFORMERS
The losses developed in the transformer cores and windings are converted in to thermal
energy and cause heating of corresponding transformer parts. The heat dissipation in
transformer occurs by Conduction, Convention and Radiation. The paths of heat flow in
transformer are the following




From the internal most heated spots of a given part ( core or winding ) to their outer
surface in contact with the oil.
From the outer surface of a transformer part to the oil that cools it.
From the oil to the walls of a cooler, eg. wall of tank.
From the walls of the cooler to the cooling medium air or water.
In the path 1 mentioned above heat is transformed by conduction. In the path 2 and 3
mentioned above heat is transferred by convention of the oil. In path 4 the heat is dissipated by
both convection and radiation.
The various methods of cooling transformers are









Air natural
Air blast
Oil natural
Oil natural air – forced
Oil natural – water forced
Forced circulation of oil
Oil forced – air natural
Oil forced – air forced
Oil forced – water forced
The choice of cooling method depends upon the size, type of application and
type of condition obtaining at the site where the transformer in installed.
Air – natural is used for transformers up to 1.5 MVA. Since cooling by air is not so
effectives and proves in sufficient for transformers of medium sizes, oil is used as a coolant.
Oil is used for almost all transformers except for the transformers used for special
applications. Both plain walled and corrugated walled tanks are used in oil cooled transformer.
In oil – natural air-forced method the oil circulating under natural head transfers heat to
tank walls. The air is blown through the hollow space to cool the transformer. In oil natural –
water forced method, copper cooling coils are mounted above the transformer core but below
the surface of oil. Water is circulated through the cooling coils to cool the transformer.
In oil forced – air natural method of cooling, oil is circulated through the transformer
with the help of pump and cooled in a heat exchanger by natural circulation of air. In oil forced
58
– air forced method, oil is cooled in external heat exchanger uasing air blast produce by fans.
In oil forced - water forced method, heated oil is cooled in a water heat exchanger. In this
method pressure of oil is kept higher than that of water to avoid leakage of oil.
Natural cooling is suitable up to 10 MVA. The forced oil and air circulation or
employed for transformers of capacities 30 MVA and upwards. The forced oil and water is
used for transformers designed for power plants.
Transformer oil as a cooling medium
For the transformer oil, the specific heat dissipation due to convection of oil is given
by
λconv 40.3[ θ ] 1/ 4 W / m2 -o C
H
where θ = Temperature difference of the surface relative to oil , o C
H = Height of dissipating surface, m
The average working temperature of oil is 50 o C 60 o C. for θ = 20 o C and H = 0.5
to 1m, λconv = 80 to 100 W / m2 -o C . The corresponding figure for convention of air is 8 W
/ m2 -o C .The corresponding figure for convection due to oil is more than 10 times than that
of air.
Temperature rise in plain walled tanks
The transformer core and winding assembly is placed inside a container called tank.
The walls of the tank dissipate heat by both radiation and convection. For a temperature rise of
40 o C above the ambient temperature of 20 o C, the specific heat dissipation are as follows.
1. Specific heat dissipation due to radiation, λrod = 6 W / m2 -o C
2. Specific heat dissipation due to convection , λcon = 6.5 W / m2 -o C
The total specific heat dissipation in plain walled tanks is 12.5 W / m2 -o C
The temperature rise, θ = _____
Total loss___________ = Pi + Pc
Specific heat
Heat dissipation
12.5 St
dissipation
x surface of the tank
where,
Pi = Iron loss
Pc = Copper loss
59
St = Heat dissipation surface of the tank
The heat dissipation surface of the tank is given by total area of vertical sides plus one
half area of the top cover. If the oil is in contact with the cover then the total heat dissipating
surface of the tank is given by total area of vertical sides plus full area of the top cover. The
area of bottom of the tank is given by total area of vertical sides plus full area of the top cover.
The area of bottom of the rank should be neglected as it has very little cooling effect.
For transformer of low capacity the plain walled tanks have sufficient surface to keep
the temperature rise within the limits. But for transformer of large output, the plain walled
tanks are not sufficient to dissipate the losses. This is because, the volume and hence losses
increase as cube of linear dimensions, while the dissipating surface increase as the square of
linear dimensions. Thus an increase in rating results in an increase in loss to be dissipated per
unit area giving a high temperature rise. Modern oil immersed power transformers with natural
oil cooling and a plain tank may be provided for output not exceeding 20 to 30 KVA.
Transformer rated for longer outputs much be provided with means to improve the
conditions of heat dissipation. This may be achieved by providing corrugations cooling tubes
and radiators.
DESIGN OF TANK WITH COOLING TUBES
The transformers are provided with cooling tubes to increase the heat dissipating area.
The tubes are mounted on the vertical sides of the transformer tank. But the increase in
dissipation of heat is not proportional to increase the area, because the tubes would screen,
(conceals) some of the tanks surface preventing radiations from the screened surface. On the
other- hand the tubes will improve the circulation of oil. This improves the dissipation of loss
by convention. The circulation of oil is due to more effective pressure heads produce by
columns of oil in tubes.
the improvement in loss dissipation by conviction is equivalent to loss dissipated by
35% of tube surface area.Hence to baccount for th9is improvement in dissipation of loss by
convection and additional 35 % tube area is added to actual tube surface area or the specific
heat dissipation due to convection is taken as 35 % more than that without tubes.
Let, The dissipation surface of tank = St
The dissipation surface of tubes = X St
Loss dissipated by surface of the
tank by radiation and convection = ( 6+6.5)St = 12.5 St
Loss dissipated by
tubes by convection = 6.5 x 135 x
100
Total loss dissipated by
X St = 8.8 X St
60
walls and tubes
= 12.5 St + 8.8 X St = (12.5 + 8.8 X) St
Actual total area of tank
walls and tubes
= St + X St = St (1 + X)
Loss dissipated per m2 of
dissipated surface
= Total loss dissipated
Total area
= St (125 + 8.8X)
St ( 1 + X)
= 12.5 + 8.8X
1 +X
Temperature rise in
transformer with cooling tubes
θ=
_ Total loss __
Loss dissipated
Total loss, Ploss = Pi + Pc
where,
Pi = Iron loss
Pc = Copper loss
θ = ____ Pi + Pc ____
St ( 12.5 + 8.8X )
From the 2 equations,
(or) 12.5 + 8.8X = Pi + Pc _
θ St
X = [ Pi + Pc _ _ 12.5 ]
θ St
_1__
8.8
Total area of cooling tubes = X St
On substituting for X we get,
Total area of cooling tubes =
_ 1__ [ Pi + Pc _ _ 12.5 ] St
8.8
θ St
= [ _ 1__
8.8
Let
Pi + Pc _ _ 12.5
θ St
St ]
= Length of the tube
61
dt = Diameter of the tube
surface area of each tube = π dt
(surface area of a cylinder)
Total number of tubes, nt
= Total area of tubes
Area of each tube
nt = _____1______ [ Pi + Pc _
8.8 π dt
θ
_ 12.5 St ]
The standard diameter of the cooling tubes is 50mm and the length of the tube depends
on the height of the tank.The tubes are arranged with a center to center spacing of 75mm.
The dimensions of the tank are decided by the dimensions of the transformer frame and
clearance required on all the sides.
Let
C1
C2
C3
C4
Doc
= Clearance between winding and tank along with the width.
= Clearance between the winiding and the tank along the length
= Clearance between the transformer frame and the tank at the bottom.
= Clearance between the transformer frame and the tank at the top.
= Outer diameter of the coil
Width of the tank, WT = 2D + Doc + 2 C1 ( for 3 – phase)
= D + Doc+ 2 C1 ( for 1 – phase)
Length of the tank, LT = Doc + 2 C2
Height of the tank, HT = H + C3 + C4
The clearance eon the sides depends on voltage and power rating of the winding. The
clearance at the top depends on the oil height above the assembled transformer and the space
for mounting the terminals and tap changing gear. The clearance at the bottom depends on the
space required for mounting the transformer frame inside the tank. The typical values of the
clearances are listed in the table.
Table: Clearances between transformer frame and tank
Voltage
kVA rating
Clearance in mm
C1
C2
C3
C4
62
Upto 11kV
Upto 11kV
11 kV to 33kV
11kV to 33kV
< 1000 kVA
1000 to 5000 kVA
< 1000 kVA
1000 to 5000 kVA
40
70
75
85
50
90
100
125
75
100
75
100
375
400
450
475
ESTIMATION OF NO LOAD CURRENT OF TRANSFORMER
The no – load current of a transformer has two components . They are magnetizing
component and loss component. The magnetizing current depend on the mmf required to
establish the desired flux. The loss component of no – load current depends on the iron loss.
No – Load current of single phase transformer
Total length of core = 2 lc
Total length of yoke = 2 ly
Here, lc = Hw = Height of window
and ly = Hw
= width of window
mmf of core = mmf per meter for maximum flux density in core x Total length of core
= aty x2 lc = 2 atc 2 lc
mmf for yoke = mmf per meter for maximum flux density in yoke x mmf for joints
= aty x2 ly = 2 aty ly
Maximum value of magnetizing current = ATo / Tp
If the magnetizing current is sinusoidal then,
__
rms value of magnetizing current, Im = ATo / √ 2 Tp
Note: Since maximum value of flux is used in the design of transformer, the current given by
equation (4.53) will be maximum value.
When the magnetizing current is non – sinusoidal , the peak factor
__
should be used in place of √ 2 .
63
 Im = ATo / Kpk Tp
The value of and
are taken from B –H curves for transformer steel. The joints in a
magnetic circuit may be taken as short air gaps in parallel with iron paths.
The loss component of no – load current, Il = Pi / Vp
where, Pi = Iron loss in W
Vp = Terminal voltage of primary winding
The iron losses are circulated by finding the weight of cores and yokes. The loss per kg
of iron is taken from the loss curves given by the manufacturer of transformer laminations.
_______
No – load current, Io = √ I2m + I2l
No – load current of three – phase transformer
Total length of core = 3 lc
Total length of yoke =2 ly
Here, lc = Hw = Height of window
ly = 2Ww + d
where, Ww = Width of window
d = diameter of circumscribing circle
mmf for core = mmf per meter for maximum flux density in core x Total length of core
= atc x 3 lc = 3 atc lc
mmf for yoke = mmf per meter for maximum flux density in yoke x Total length of yoke
= aty x 2 ly = 3 aty ly
Total magnetizing mmf
required for the transformer } = mmf for core + mmf for yoke + mmf for joints
Total magnetizing mmf
64
required per phase
= ATo = 3 atc lc + 2 aty ly + mmf for joints
3
Maximum value of magnetizing
current per phase
= ATo
Tp
If magnetizing current is sinusoidal then,
____
rms value of magnetizing current per phase, Im = ATo √ 2Tp
If the magnetizing current is non – sinusoidal then Kpk should be used in the place of
__
√2.
The value of atc and aty are taken from B-H curves.
Let, Pi = Total iron loss for the three phases,  Pi = 3 Vp Il
 Loss component of no – load current, Il = Pi / 3 Vp
_______
Hence, no –load current per phase, Io = √ I2m + I2l
Alternate formula of magnetizing current
Magnetizing current = Magnetizing mmf per meter x length of flux path in iron
Now a curve can be plotted between and magnetizing volt ampere per kg using the B –
H curve of the material. Usually the manufactures will supply the magnetizing characteristics.
From the characteristics the magnetizing current can be calculated
TUTORIAL PROBLEMS
1. Estimate the main dimensions including winding conductor area of a 3 – phase , - Y core
type transformer rated at 300kVA, 50Hz. A suitable core with 3 – steps having a
circumscribing circle of 0.25m diameter and a leg spacing of 0.4m is available. Emf per turn =
(stacking factor).
Solution
Let 440V side be secondary and 6600V be primary. Here the secondary is star
connected and primary is delta connected.
65
For 3 stepped core, the ratio
Gross core area,
= 0.84 x area of circumscribing cirle
= 0.84 x 0.049 = 0.041
Net core area,
= 0.9 x 0.041 = 0.0369 = 0.037 x 10
Given that, leg spacing = 0.45m
Width of window,
= leg spacing = 0.45m
Result
Number of primary turns per phase,
Number of secondary turns per phase,
= 780
= 30
2. The tank of 1250kVA, natural oil cooled transformer had the dimensions length, width and
height as 0.65 x 1.55 x 1.85m respectively. The full load loss = 13.1 kW, loss dissipation due
to radiations =
, loss dissipation due to convection =
, improvement in
convection due to provision of tubes = 40%, temperature rise = , length of each tube = 1m,
diameter of tube = 50mm. Find the number of tubes for this transformer. Neglect the top and
bottom surface of the tank as regards the cooling.
Given Data
kVA = 1250
Tank dimension = 0.65 x 1.55 x 1.85m
Solution
Length = 0.65m
Width = 1.55m
height = 1.85m
The diameter of the tube is 50mm and the standard distance between the tubes is half
of the diameter and so, let distance between tubes = 25mm.
The width of the tank is 1550mm. If we leave an edge spacing of 62.5mm on their sides
then we can arrange 20 tubes widthwise with a spacing of 75mm between centers of tubes. On
length we can arrange 8 tubes with same spacing as that of width wise tubes. But one row is
not sufficient to accommodate the required 158 cooling tubes. Hence three rows of cooling
66
tubes are provided on both lengthwise and widthwise. The plan of the cooling tubes is shown
in fig 4.7.1
Result
The total number of tubes provided = 160
They are arranged as 3 row on widthwise with each row consisting of 20, 19 & 20 tubes
and 3 rows on length wise each row consisting of 8.7 & 6 tubes.
67
MODULE 3
DESIGN OF SYNCHRONOUS MOTOR
MAIN DIMENSIONS OF ROTATING MACHNES
In rotating machine the active part is cylindrical in shape. The volume of the cylinder is
given by the product of area of cross section and length. If D is the diameter and L is the length
of cylinder, then the volume is given by π D2 L/4. Therefore D and l are specified as main
dimension.
In case of dc machine, D represent the diameter of armature and L represent the length
of armature. In case of ac machine, D represent the inner diameter of stator and L represent the
length of stator core. The fig 1.1 shows the main dimensions of rotating machines
Here
Dr = Diameter of rotator
lg = Length of air-gap
MAGNETIC AND ELECTRIC LOADINGS
Consider a conductor of length L, carrying a current of l z amperes. If the conductor is
moved in a uniform magnetic field of flux density Bav Wb/m2 then the work done is given by
Work = x B L lz
Where x is the distance through which the conductor is moved.
When the conductor is moved through a distance x, the conductor cuts through a
magnetic flux of ф = x BL webers. Hence work done in moving the conductor can be
expressed as,
Work = ф lz
In rotating machines the conductors are placed in armature. In one revolution of the
armature, each conductor moves through a total flux of ф webers, where ф is flux per pole and
p is number of poles. If Z is the total number of armature conductors then the work done in
one revolution is given by
Work is p ф x lz Z
The term pф represent the total flux entering and leaving the armature and so it is called
total magnetic loading (or total flux). The term lz Z represents the sum of currents in all the
conductors on the armature and so it is called total electric loading (or total current volume or
total ampere conductors on the armature).
68
Total magnetic loading = pф
Total electric loading = lz Z
Therefore we can say that the work done in one complete revolution is given by the
product of total magnetic loading ad total electric loading.
The total magnetic loading is defined as the total flux around the armature (or stator
inner) periphery at the air-gap. The total electric loading is defined as the total number of
ampere conductors around the armature (or stator) periphery.
Specific magnetic loading
Each unit area of armature surface is capable of receiving a certain magnetic flux.
Hence the flux per unit area is an important parameter to estimate the intensity of magnetic
loading and it is also criterion to decide the volume of active material. This flux per unit area is
expressed as the average value of the flux density at the armature surface or specific magnetic
loading (by the assuming that the armature is smooth). It is denoted by Bav
The average flux density, Bav is given by the ratio of flux per pole and area under a
pole.
Bav =
Flux per pole
=
Area under a pole
=
ф
πD x
фFlux per pole
.
Pole Pitch x Length of armature
..
Pф
L
πD
.
P
Note Pole pitch, τ = πD/P
We can say that the specific magnetic loading is also giving by the ratio of total flux
around the air-gap and the area of flux path at the air-gap.
Bav =
Total flux around the air – gap
Area of flux path at the air – gap
=
Pф
πDL
Specific electric loading
Every section of armature is capable carrying certain amount of current. Hence ampereturn per unit section of armature periphery (circumference) is an important parameter to
estimate the intensity of electric loading and it is also a criterion to decide the volume of active
material . This ampere –turn per unit section of armature periphery is expressed as the specific
electric loading . It is denoted by ac.
69
The specific electric loading is given by the ratio of total armature ampere conductors
and armature periphery (circumference) at air-gap.
ac =
Total armature ampere conductors
Armature periphery at air – gap
= lz Z
πD
OUTPUT EQUATION
The output of a machine can be expressed in terms of its main dimensions, specific
magnetic and & electric loadings and speed. The equation which relates the power output to D,
L, Bav, ac. and n of the machine is known as output equation.
Output equation and output coefficient of AC machine
The equation of induced emf, frequency, current through each conductor and total
number of armature conductors of an ac machine are given below. These equations are
obtained from the knowledge of machine theory.
Induced emf per phase, Eph = 4.44 f ф T ph Kws
The frequency of induced emf, f =
Current through each conductor,Iz = Iph / a
where, Iph = Current per phase
and
a = Number of parallel circuits or paths per phase.
Total number of armature conductors, Z = Number of phases x 2Tph
= 3 x 2Tph = 6Tph
Specific magnetic loading= Bav
Specific electric loading= ac
,
pф = π DL Bav
Iz Z = π D ac
.
Consider a 3-phase machine having one circuit (one parallel path) per phase. The
voltampererating of one phase is given by the product of voltage per phase and current per
phase. hence the kVA rating of 3-phase ac machine can be written as shown in equation (1.30).
kVA rating of 3-phase machine, Q = 3 E ph I ph x 10-3
On substituting for E ph , I ph we get,
Q = 3 x 4.44 ф T ph Kws x Iz x 10-3 =
70
= 6.66 p ns ф T ph Kws Iz x 10-3
= 1.11 x pф
x
Iz 6 T ph x ns x Kws x 10-3
On substituting for 6 T ph we get,
Q = 1.11 x pф x Iz Z x ns x Kws x 10-3
On substituting for pф and Iz Z we get,
Q = 1.11 x π DL Bav x π D ac x ns x Kws x 10-3
= 1.11 π2 Bav ac Kws x 10-3 x D2 L ns
= 11 Bav ac Kws x 10-3 x D2 L ns
= Co D2 L ns
where, Co = 11 Bav ac Kws x 10-3
The equation , Q = Co D2 L ns is called output equation and Co is called output coefficient.
The term D2 L in the output equation is proportional to volume of active part. Therefore, if Co
is constant then we can say that the kVA rating is directly proportional to the product of
volume of active part and speed.
i.e., Q α Volume of active part x Speed
If Co is varied then power output is directly proportional to the four quantities : Bav , ac,
volume of active part and speed.
i.e., Q α Bav x ac x Volume of active part x Speed
SEPARATION OF D AND L
We can say that the output of a rotating electrical machine is directly proportional to the
term D2 L. The active part of the rotating electrical machine is cylindrical in shape and the
volume of cylinder is π D2 L. Hence the term D2 L is related to volume of active part. The
separation of D and L refers to the selection of an appropriate value of D and L for a given
volume of active part.
For a given volume of active part there are various choice of D and L. The choice of
D&L and the ratio between them depends on a number of factors in various types of rotating
machines. In the following sections a brief discussion about various factors that decides the
choice of D and L in various types of rotating machines are presented. In general a ratio of L/ τ
or L/D is assumed where τ = pole pitch = π D/p. Using the output equation and from the
knowledge of kW or kVA rating, specific loading and speed, the value of D2 L are estimated.
71
SEPARATION OF D AND L FOR SYNCHRONOUS MACHINES
In synchronous machines the separation of D and L depends on




Pole proportions
Number of poles
Peripheral speed
Short circuit ratio
Pole proportion
In salient pole synchronous machines the choice of diameter (D) depends on the type of
pole and the permissible peripheral speed. The Two-types of poles used in salient pole
machines are round poles and rectangular poles.
When round poles are used the ratio of L/ τ is between 0.6 to 0.7, where τ = pole pitch π
D/p. When rectangular poles are employed the ratio of L/ τ is between 1 to 5.
Peripheral speed
For the large high speed machines D is fixed by the limiting peripheral speed. The
output equation for synchronous machine can be expressed in terms of peripheral speed (Va) as
shown below.
The output equation is, Q = Co D2 L ns
where, Co = 11 Bav ac Kws x 10-3
But V a = π Dn2
On substituting the expression for Co and D in the output equation.
We can say that an increase in machine rating will necessitate an increase in Va which
in turn is achieved by increasing D, until the maximum permissible peripheral speed is
reached. Once this happens , the value of D cannot be increased further and the only way to get
increased output is to increase the length L. In general the value of D is calculated using the
limiting value of peripheral speed Va for cylindrical rotor synchronous machines. Then using
the values of D2 L and D, the value of L is estimated.
Number of poles
The small diameter and large number of poles results in the small pole pitch and so less
space for field coils. Hence a large diameter is advisable for machines having large number of
poles. The empirical relationship used to decide the number of poles is
τ / L = 0.5 + (6/p)
72
Short circuit ratio
A major factor influencing the design of synchronous machines is their short circuit
ratio (SCR).
Field current required to produce rated voltage on open circuit
SCR=
Field current required to produce rated current at short circuit
Higher values of SCR results in higher stability limit and a low value of inherit
regulation . For high values of SCR, the length of core (L) should be large.
Output coefficient
The output of the machine is directly proportional to output coefficient. Also the
volume of active parts is inversely proportional to output coefficient.
The output equation, Q = Co D2 L ns
If Q is constant, then
Hence higher value of Co results in higher output. With high Co , the volume of active
part decreases and the machine costs less. The output coefficient Co depends on specific
loadings Bav and ac. Hence for higher Co , higher specific loadings are choosen
Co = 11 Bav ac Kws x 10-3
The high values of specific loadings may affect some of the performance characteristics
of the machine like temperature rise, efficiency, power factor, commutation conditions, etc. So,
specific loadings are chosen such that they give best performance and minimum cost.
DESIGN OF THREE PHASE INDUCTION MOTOR
CONSTRUCTION
The two major parts of three phase induction motor are stator and rotor. The stator
consists of core and winding. The stator core is made of laminated sheet steel of thickness
0.5mm. The stator core internal diameter and the length are the main dimensions of induction
motor.
The two different types of rotor of three phase induction motors are squirrel cage rotor
and wound rotor ( or slip ring rotor). In squirrel cage construction the rotor consists of core,
copper of aluminium bars and end-rings. The end-rings are provided to short – circuit the rotor
bars at both the ends. The rotor core is made of laminated sheet steel with thickness 0.5mm.
The aluminium bars and the end – rings are casted directly over the rotor core. When copper is
employed, the rotor bars are inserted on the slots from the end of rotor and the end rings are
joined to them by bracing.
The wound rotor consists of a core, winding, slip-ring and brushes. The rotor core is
made of laminations and it carries a three – phase winding. One end of each phase are
73
connected to from a star point. The other ends of each phase are connected to three slip – rings.
The slip – rings are mounted on the rotor shaft and they are insulated from the rotor and from
each other. Carbon brushes are mounted over the slip – rings which offer the facility of
connecting the external resistances to rotor winding.
MAIN DIMENSIONS
The main dimension of induction motor are the diameter of stator bore, D and the
length of stator core, L. The product D2 L is determined from input kVA, specific electric and
magnetic loadings. The separation of D and L, from the product D2 L, depends on the ratio L /
τ where τ ( = π D / p), is the pole pitch. In induction motors most of the operating characteristic
are decided by L / τ ratio of the motor. The ratio core length to the pole pitch (L / τ) for various
design features are listed below.
For minimum cost,
For good power factor ,
For good efficiency,
For good over all design
= L / τ = 1.5 to 2
= L / τ = 1.0 to 1.25
= L / τ = 1.5
=L/τ=1
Generally L / τ lies between 0.6 to 2. It can be shown that, for best power factor the pole pitch
τ is given by the equation,
τ=
0.18 L
The diameter of stator bore and hence the diameter of rotor is also limited by peripheral
speed. Standard constructions are employed for peripheral speeds upto
60 m/s. For higher peripheral speeds up to 75 m / s, special constructions methods should be
employed for rotor which results in higher cost. For normal design, the diameter should be so
chosen that the peripheral speed does not exceed about 30 m /s.
The stator is provided with radial ventilating ducts if the core length exceeds 125mm.
The width of each duct is about 8 to 10mm.
SEPARATION OF D AND L FOR INDUCTION MOTORS
The operating characteristics of an induction motor are mainly influenced by the ratio
L/ τ. The ratio of L/ τ for various design features are listed below.
For minimum cost,
For good power factor,
For good efficiency,
For good overall design,
L/ τ = 1.5 to 2
L/ τ = 1.0 to 1.25
L/ τ = 1.5
L/ τ = 1.0
Generally L/ τ lies between 0.6 to 2.
Output equation and output coefficient of AC machine
74
The equation of induced emf, frequency, current through each conductor and total
number of armature conductors of an ac machine are given below. These equations are
obtained from the knowledge of machine theory.
Induced emf per phase, Eph = 4.44 f ф T ph Kws
f is the frequency of induced emf
Current through each conductor,Iz = Iph / a
where, Iph = Current per phase
and
a = Number of parallel circuits or paths per phase.
Total number of armature conductors, Z = Number of phases x 2Tph
= 3 x 2Tph = 6Tph
Specific magnetic loading, Bav
pф = π DL Bav
Specific electric loading, ac
Iz Z = π D ac
.
Consider a 3-phase machine having one circuit (one parallel path) per phase. The
voltampererating of one phase is given by the product of voltage per phase and current per
phase. hence the kVA rating of 3-phase ac machine can be written
kVA rating of 3-phase machine, Q = 3 E ph I ph x 10-3
On substituting for E ph , I ph we get,
Q = 3 x 4.44 ф T ph Kws x Iz x 10-3 = 3 x 4.44 x
= 6.66 p ns ф T ph Kws Iz x 10-3
= 1.11 x pф
x
Iz 6 T ph x ns x Kws x 10-3
On substituting for 6 T ph from equation we get,
Q = 1.11 x pф x Iz Z x ns x Kws x 10-3
On substituting for pф and Iz Z we get,
Q = 1.11 x π DL Bav x π D ac x ns x Kws x 10-3
= 1.11 π2 Bav ac Kws x 10-3 x D2 L ns
= 11 Bav ac Kws x 10-3 x D2 L ns
= Co D2 L ns
where, Co = 11 Bav ac Kws x 10-3
75
The equation , Q = Co D2 L ns is called output equation and Co is called output coefficient.
The term D2 L in the output equation is proportional to volume of active part. Therefore, if Co
is constant then we can say that the kVA rating is directly proportional to the product of
volume of active part and speed.
i.e., Q α Volume of active part x Speed
If Co is varied then power output is directly proportional to the four quantities : Bav , ac,
volume of active part and speed.
i.e., Q α Bav x ac x Volume of active part x Speed
Efficiency & power factor
For squirrel cage induction motors the efficiencies varies from 0.72 to 0.91 and power
factor varies from 0.66 to 0.9. Fro slip ring induction motors the efficiency varies from 0.84 to
0.91 and power factor varies from 0.7 to 0.92
CHOICE OF SPECIFIC LOADINGS
For the design of induction motor, the horse power rating, speed, power factor and
efficiency are specified. Therefore, in order to calculate the value of D2 L, we must evaluate
the out put coefficient. The value of out put coefficient depends upon the choice of specific
electric loading (ac) and specific magnetic loading ( Bav).
The choice of specific electric loading depends on copper loss, temperature rise,
voltage rating and overload capacity. it also depends upon size of motor and cooling. It varies
between 5000 to 45000 amp. cond. / m depending on the factors mentioned above.
The choice of specific electric loading depends on power factor, iron loss and over load
capacity. For 50 Hz machines of normal design the value of Bav lies between 0.3 to 0.6 Wb /
m2 . For machines used in cranes, rolling mills, etc., where a large over load capacity is
required, a value of 0.65 Wb / m2 may be used.
Choice of specific electric loading
A large value of ac results in higher copper losses and higher temperature rise. For
machines with high voltage rating smaller values of ac should be preferred. Since for high
voltage machines the space required for insulation is large.
For high overload capacity , lower values of ac should be selected. Since large values
of ac results in large number of turns per phase, leakage reactance will be high. Large values of
leakage reactance results in reduced overload capacity.
76
Choice of specific magnetic loading
With large values of Bav , the magnetizing current will be high, which result in poor
power factor. However, in education motors the flux density in the air gap should be such that
there is no saturation in any part of the magnetic circuit.
A large value of Bav results in increased iron loss and decreased efficiency. With higher
values of Bav , higher values of overload capacity can be obtained. Since the higher Bav
provides, large values of flux per pole, the turns per phase will be less and so the leakage
reactance will be less. Lower value of leakage reactance results in higher over load capacity.
MAIN DIMENSIONS
The main dimension of induction motor are the diameter of stator bore, D and the
length of stator core, L. The product D2 L is determined from input kVA, specific electric and
magnetic loadings. The separation of D and L, from the product D2 L, depends on the ratio L /
τ where τ ( = π D / p), is the pole pitch. In induction motors most of the operating characteristic
are decided by L / τ ratio of the motor. The ratio core length to the pole pitch (L / τ) for various
design features are listed below.
For minimum cost,
For good power factor ,
For good efficiency,
For good over all design
= L / τ = 1.5 to 2
= L / τ = 1.0 to 1.25
= L / τ = 1.5
=L/τ=1
Generally L / τ lies between 0.6 to 2. It can be shown that, for best power factor the pole pitch
τ is given by the equation,
τ=
0.18 L
The diameter of stator bore and hence the diameter of rotor is also limited by peripheral
speed. Standard constructions are employed for peripheral speeds upto
60 m/s. For higher peripheral speeds up to 75 m / s, special constructions methods should be
employed for rotor which results in higher cost. For normal design, the diameter should be so
chosen that the peripheral speed does not exceed about 30 m /s.
The stator is provided with radial ventilating ducts if the core length exceeds 125mm.
The width of each duct is about 8 to 10mm.
STATOR WINDING
For small motor upto 5HP, single layer winding like mush winding, whole coil
concentric winding and bifurcated concentric winding are employed. For large capacity
machines , double layer windings( either lap or wave winding) are employed with diamond
shaped coils.
77
The stator winding can be designed for either star or delta depending on the running
condition.
Stator turns per phase
The turns per phase Ts , can be estimated from stator phase motor and maximum flux in
the core. The maximum flux in the core can be estimated from Bav ,D2 L and p.
Specific magnetic loading, Bav = P ф
π DL
фm = Bav π DL
P
Stator phase voltage, Es = 4.44 Kws fфm Ts
Here, ф = фm
Stator turns per phase, Ts = _____ Ts____
4.44 Kws fфm
Length of mean turn
The approximate length of mean turn of the winding on induction motor stators for use
on voltage upto 650 V may be calculated from the following empirical relationship.
Length of mean turn of stator Lmts = 2L + 2.3τ + 0.24
Here the value of L and τ are expressed in m.
Stator conductors
The area of cross section of stator conductors can be estimated from the knowledge of
current density, kVA rating of the machine and stator phase voltage. The current density in the
stator winding is usually lies between 3 to 5 A/ mm2
kVA rating of 3 – phase machine, Q =3 ES x 10-3
where,
Es = Stator phase voltage
Is = Stator phase current
 Is = ____Q_______
3 ES x 10-3
Let δs = Current density in stator conductor
as = Area of cross – section of state conductor
ds = Diameter of stator conductor
78
as = Is
and
δs
ds = √ 4 as
π
( since as = πd2s /4 )
Round conductors are used for small diameters. If the diameter is more than 2 or
3mm then bar or strip conductors are used.
STATOR CORE
The stator core is made of laminations of thickness 0.5mm. The design of stator core
involves selection of number of slots, estimation of dimensions of teeth and depth of stator
core.
Stator slots
The different types of slots used in induction motor are open slots and semi enclosed
slots. The shape of the slots have an important effect upon the operating performance of the
motor as well as the problem of installing the winding.
When open slots are used the winding coils can be formed and fully insulated before
installing and also it is easier to replace the individual coils.Another advantage of open slots is
that their use avoids excessive
slot leakage their by reducing the leakage reactance.
When semienclosed slots are used , the coils must be tapped and insulated after their
placed in the slots. The advantages of semi enclosed slots are less air gap contraction factor
giving a small value of magnetizing current, low tooth pulsation loss and much quieter
operation (less noise). Semi enclosed slots are mostly preferred for induction motors.
In small motors where round conductors are used, the tapered slot with parallel sided
tooth arrangement is useful as it give the maximum slot area for a particular tooth density. In
large and medium size machines were strip conductors are preferred, parallel sided slots with
tapered teeth are used.
Choice of stator slot
The number of stator slots depends on tooth pulsations loss, leakage reactance,
ventilation, magnetizing current, iron loss and cost. In general the number of slots to be
selected to give and integral number of slots should be between 15 to 25mm. For semi enclosed
slots the slot pitch may be less than 15mm.
The stator slot pitch, YSS =
Let Ss = total stator slots
Gap surface
Total number of stator slots
;
πD = Gap surface
79
then,Ss =
πD
Yss
Total number of stator conductors = No. of phases x Conductors per phase
= 3 x 2Ts = 6 Ts
where, Ts = Turns per phase
Conductors per slot,
Zss = Total stator conductors
Number of stator slots
=
6Ts
Ss
Note : Zss must be even for double layer winding
Guide lines for selecting stator slots
Step - 1
:
The stator slot pitch various from 15mm to 25mm.
Calculate the range of stator slots using the equation.
Stator slots, Ss = π D / Yss .
Minimum number of slots are obtained when Yss = 25mm.
Maximum number of slots are obtained when Yss = 15mm.
Step -2
:
Stator slots should be multiple of q where q is slots per
pole per phase. For q = 2, 3, 4, etc., Calculate stator slots.
Ss = Number of phases x poles x q
Step – 3
:
Select the choices of stator slots which are common
between the values obtained in step 1 & 2.
Step – 4
:
The best choice of stator slot is given by the value of slot
in the list obtained from step 3 & satisfying the slot loading.
Slot loading - IzZss
where, Iz = Current through a conductor
Zss = Conductors per slot
Note : Try to choose lesser number of slots which results in reduced labour & less amount of
copper.
Area of stator slot
Approximate area of each slot
=
Copper area per slot
Space factor
= ____ Zss x as______
Space factor
80
The space factors vary from 0.25 to 0.4. High voltage machines have lower space
factors due to large thickness of insulation. After obtaining the area of the slot, the dimensions
of the slot should be adjusted. The slot should not be too wide to give a thin tooth. The width
of the slot should be so adjusted such that the mean flux density in the tooth lies between 1.3 to
1.7 Wb / m2 . The width of tooth should not be too large as it results in narrow and deep slots.
The deepest slots give a large value of leakage reactance. in general the ratio of slot depth to
width should be between 3 & 6.
Stator teeth
The dimensions of the slot determine the value of flux density in the teeth. A high value
of flux density in the teeth is not desirable, as it leads to higher iron loss and a greater
magnetizing mmf. The maximum value of Bs ( the mean flux density of stator tooth ) should
not exceed 1.7 Wb / m2.
Minimum teeth area per pole = (фm / 1.7)
Also, Teeth area per pole = number of slots per pole x net iron length x width of tooth.
= (Ss / p) x Li x Wts
When teeth area per pole is minimum, the width of tooth will be minimum
( фm /1.7 ) = (Ss / p) x Li x Wts min
The minimum width of
stator tooth
=
Wts min = ____ фm _____
1.7 (Ss / p) Li
The minimum width of stator tooth is either near the gap surface or at one – third height
of tooth from slot opening. A check per minimum tooth width using the above equation should
be applied before finally deciding the dimensions of stator slot.
Depth of stator core
The cross – section of stator core is shown in fig 5.3. The depth of stator core depends
on the flux density in the core. The flux density in the stator core lies between 1.2 to 1.5 Wb /
m2 .
The flux passing through the stator core is half of the flux per pole
 Flux in the stator core = фm
2
Let Bcs = Flux density in stator core
81
Area of stator core =
_Flux through core _________
Flux density in the stator core
= фm
2 Bcs
Also, Area of stator core = length x depth = Li x dcs
On equating the above two equations for area of stator aore we get,
Li dcs= фm
2 Bcs
 Depth of core, dcs
= __ фm __
2 Bcs Li
Outer diameter of stator core, Do = D + 2 ( Depth of stator slots + Depth of core)
= 2 (dss+ dcs)
LENGTH OF AIR GAP
The length of air gap in induction motor is decided by considering the following
factors.






Power factor
Pulsation loss
Cooling
Over- load capacity
Unbalanced magnetic pull
Noise
Power factor
The mmf required to send the flux through air gap in proportional to the product of flux
density and the length of air gap. Even with very small densities, the mmf required for air –
gap is much more than that for the rest of the magnetic circuit. Therefore , it is the length of air
gap that primarly determines the magnetizing current drawn by the machine.
Overload capacity
The length of air gap affects the value of zig zag leakage reactance which forms a large
part of total leakage reactance. If the length of the air gap is large then the
zig zag leakage flux will be less and so the leakage reactance will be less. With lesser value of
leakage reactance the overload capacity increases. Hence, greater is the length of air gap,
greater is the overload capacity.
82
Pulsation loss
With larger length of air gap, the variation of reluctance due to slotting is small. The
tooth pulsation loss, which is produced due to variation in reluctance of the air gap, is reduced
accordingly. Therefore , the pulsation loss is less with large air gaps.
Unbalanced magnetic pull
If the length of the air gap is small, then even a small reflection or eccentricity of the
shaft would produce a large irregularity in the length of air gap. It is responsible for production
of large unbalanced magnetic pull which has the tendency to bend the shaft still more at a place
where it is already bend resulting in fouling of rotor with stator. If the length of air gap of a
machine is large , a small eccentricity would not be able to produce noticeable unbalanced
magnetic pull.
Cooling
If the length of air gap is large, the cylindrical surfaces of rotor and stator are separated
by a large distance. This would affect better facilities for cooling at the gap surfaces especially
when a fan is fitted for circulation of air.
Cooling
The principal cause of noise in induction motors is the variation of reluctance of the
path of the zig zag leakage flux. To ensure that the noise produced will not be objectionable, it
is necessary to make the zig zag leakage as small as possible. This can be done by increasing
the length of the air gap.
From the above, we conclude that the length of air gap in a deduction machine should
be as small as mechanically possible in order to keep down the magnetizing current and to
improve the power factor. This is a major consideration . But if a higher overload capacity ,
better cooling, reduction in noise are reduction in unbalanced magnetic pull is important, then
large air gap lengths should be used.
Relation for calculation of length of air gap ( lg )
The following empirical formula can be used to calculate the length of air gap ( lg ).
Table: Length of air - gap
D
In m
lg
in mm
83
0.15
0.20
0.25
0.30
0.45
0.55
0.65
0.80
4.
0.35
0.50
0.60
0.70
1.3
1.8
2.5
4.0
1.
For small induction motor,
____
lg = 0.2 + 2√ DL in mm
2.
Alternate formula for small induction motors,
lg = 0.125 + 0.35D + L + 0.015 Va in mm
3.
Another formula for general use,
lg = 0.2 + D in mm
For machines with journal bearings,
__
= 1.6 √D - 0.25 in mm
Note : In all the above formulae the value of D and L are expressed in meter and Va in m /
sec.
Typical values of length of air gap for 4 – pole machines in relation to the main
dimension D are listed in above table .
CHOICE OF ROTOR SLOTS
With certain combinations of stator and rotor slots, the following problems may
develop in the induction motor.



The motor may refuse start.
The motor may crawl at some subsynchronous speed.
Severe vibrations are developed and so the noise will be excessive.
The above affects are due to harmonic magnetic fields developed in the machine.
The harmonic fields are due to winding, sloting, saturation and irregularities in air gap.
The harmonic fields are superposed upon the fundamental sine wave field and induce
emfs in the rotor windings and thus circulate harmonic currents. Theses harmonic currents, in
turn, interact with the harmonic fields to produce harmonic torques.
Infact the harmonic fields may be thought of as separate low power motors that are
direct coupled to the same shaft as the fundamental. Therefore, the net motor torque is equal to
the sum of the torque due to the fundamental and the torques produced by group of harmonic
fields.
the space harmonic fields have more poles than the fundamental and therefore have
lower synchronous speeds. Some of these fields revolve in the forward direction and some in
the backward direction. As motor speeds above their respective values, the forward rotating
harmonic fields produce breaking torques. The backward rotating harmonic fields produce
braking torque at all speeds. In addition the harmonic fields are responsible for increase in
stray load losses and increased motor heating.
84
The squirrel cage motor will circulate currents due to any harmonic emf produce by the
gap flux except that had a wave length equal to the pitch of the bars. The effect of space
harmonic fields produced by windings are greatly intensified by slotting. The slots introduces
steps in the mmf wave and produces further harmonics and also modulates the gap flux. hence
the choice of rotor slots is particularly important in the case of squirrel cage machines. Any bad
combination of stator and rotor slots may result in awkward behaviour.
Harmonic induction torques
A three phase winding carrying sinusoidal currents produces harmonics of the order n =
6N ±1 , where N is an integar. The movement of the harmonic is with are against the direction
of rotation depending upon the sign ( + means with the rotation and – means against the
rotation). The number of poles for the nth harmonic is n times the number of poles of the
fundamental. Hence the synchronous speed of nth harmonic is
1 / nth of the synchronous speed of fundamental.
For N = 1, the winding produces forward rotating 7th harmonic and backward rotating
th
5 harmonic.
For N=2, the winding produces forward rotating 13th harmonic and backward rotating
th
11 harmonic.
In induction motors only 5th and 7th harmonic are more pronounced and produces dips
in the torque – speed characteristics. With the certain combinations of rotor and stator slots the
dip due to 7th harmonic may became very pronounced. The 7th harmonic produces dip at 1/7th
synchronous speed. the 5th harmonic produces dip at – 1/5th synchronous speed.
If 7th harmonic is more pronounced, the machine may crawl at a speed little lesser than
1/7th synchronous speed.
If the mechanical load on the shaft requires a constant load torque and if the torque
developed by the rotor is below this load torque then the motor cannot accelerate upto its full
speed but continuous to run at a speed little lower than 1/7th synchronous speed. This condition
of the motor is called crawling.
Stator slots produces harmonics of the order 6A± 1 in a 3- phase machine, where A is
an integer. For a 4 pole, 36 slot machine with q = 3 and when A=1, forward rotating 19th
harmonic and back ward rotating 17th harmonic are produced. This causes dip in the speed –
torque characteristics at 1/19th and 1/17th synchronous speeds respectively.
The effect of production of dips may be augmented by rotor slotting
Corresponding to the above 4 pole,36 slot stator if we choose 76 rotor slots, there would be one
rotor bar corresponding to every 19th harmonic pole. Thus the 19th harmonic torques would
be very large and the rotor would vibrate considerably. Therefore it is necessary to avoid
values of rotor slots exceeding stator slots by about 15 to 30%.
Harmonic synchronous torque
If the stator and rotor harmonics are of the same order then the torque will be
alternatively in opposite directions as they move past each other. But if their speeds happen to
coincide they will lock together and if sufficiently powerful giving rise to a synchronous
torque. In such a case the motor would crawl at constant subsynchronous speed.
85
The difference in harmonic induction and synchronous torques is that the operating
speed changes slightly for the former but is constant for the later, for a small variation of the
shaft load.
The stator produces harmonies of the order n = 6q ±1 = 2 (Ss / p) ± 1 for (A =1). These
harmonics revolve at a speed 1/n of synchronous speed with respect of stator.
The rotor slotting produces harmonics field of the order n’= 2 (Ss / p) ± 1 for (A =1)
and they revolve at a speed 1/n’ that of the fundamental.
The speeds would be equal if, 2 (Ss / p) ± 1=2 (Sr / p) ± 1
one of the possibilities for this to happen, is when Ss = Sr.
When the number of rotor slots is equal to the number if stator slots, the speeds of all
the harmonics produced by the stator slotting coincide with the speed of the corresponding
rotor harmonics. Thus harmonics of every order would try to exert synchronous torques at their
corresponding synchronous speeds and the machine would refuse to start. This is known as
cogging.
The other possibility for equal stator and rotor harmonics speeds are, when
Ss – Sr = ± P.
Thus inorder to avoid synchronous cusps the difference of stator an rotor slots should
not be equal to ± P or a multiple of p. ( synchronous cusps are the synchronous torques
produced due to harmonic synchronous speeds , due to synchronous cusps the machine will
crawl.)
SUMMARY OF DESIGN EQUATIONS
1.Power developed in the armature of dc machine, Pa = Co D2 L n
2. kVA rating of 3-phases ac machine, Q = Co D2 L ns
3. Output coefficient of ac machine, Co =11 Bav ac x 10-3
4. Pole pitch, τ = π D/P
5. Choice of L / τ for induction motors
 For minimum cost, L / τ=1.5 to 2
 For good power factor, L / τ =1.0 to 1.25
 For good efficiency, L / τ = 1.5
 For good overall design, L / τ =1.0
 For best power factor, τ =
6. Choice of L / τ for salient pole synchronous machines
a. When round poles are employed, L / τ = 0.6 to 0.7
b. When rectangular poles are employed, L / τ = 1 to 5
7. In cylindrical rotor synchronous machines,
 Stator inner diameter, D = Dr + 2
86
PART-2 ELECTRICAL SYSTEM DESIGN
MODULE-4
Electrical system design is the design of electrical systems. This can be as simple as a flashlight
cell connected through two wires to a light bulb or as involved as the space shuttle. Electrical
systems are groups of electrical components connected to carry out some operation. Often the
systems are combined with other systems. They might be subsystems of larger systems and have
subsystems of their own. For example a subway rapid transit electrical system is composed of the
wayside electrical power supply, wayside control system, and the electrical systems of each transit
car. Each transit car’s electrical system is a subsystem of the subway system. Inside of each transit
car there are also subsystems, such as the car climate control system.
Design
The following would be appropriate for the design of a moderate to large electrical system.
1. A specification document is written. It probably would have been written by the
customer. The specification document states in plain language and numerical detail
what the customer expects. If it is well written, it will be used as a reference throughout
the electrical system design.
2. A functional specification (design) document that goes into more technical details may
be created. It uses the specification document as its basis. Here calculations may be
used or referenced to support design decisions.
3. Functional diagrams may be made. These use block diagrams indicating information
and electrical power flow from component to component. They are similar to the
functional flow block diagrams used with computer programs.
4. Schematic diagrams showing the electrical interconnections between the components
are made. They may not show all the conductors and termination points. Except for
one-line diagrams, this should show all the circuit nodes. One-line diagrams represent
the three or four conductors of three-phase power circuits with one line.
5. Wiring diagrams are sometimes made. These show and name the termination points and
names of each conductor. In some systems enough information can be put onto the
schematics so that wiring diagrams are not needed.
6. Physically smaller systems that are built many times may use a cable harness. A fullsized to-scale wiring diagram can be made of a cable harness. This wiring diagram can
then be laid on a peg board and used to guide the construction of more cable harnesses.
Harnesses can be inserted into to their equipment as an assembly. Cable harnesses that
are reused many times, like automobile wiring harnesses, are created with automated
machinery.
87
7. A wire list is made in spreadsheet or list format. It shows the electrical assembly people
what wires are to be connected and to where. When it is printed out on paper, it is easy
for the assembly people to check off conductors as they are connected. The wire list
contains at a minimum each wire name, terminal name, and wire model number or
gage. It may also contain the wire termination device model numbers, voltage classes,
conductor class (high-voltage, medium voltage, or control wiring), etc.
ELECTRICAL ESTIMATION AND COASTING
1. Introduction
Before any electrical project is initiated, it is essential to list out the materials
required and compute the cost involved for completion of that work. Thus estimation
consists of two parts; (a) preparing list of various items involved and (b) calculating
the cost of materials and labour cost involved for executing the work. The quantity
and specification of various materials required for installation work written in a
tabular form is called schedule of materials.
2. Graphical symbols for diagram
In engineering drawing it is common practice to employ graphical symbols to
represent various components. In order to get the same meaning to every one who
reads the drawing, symbols are standardized by Bureau of Indian Standards (BIS). As
far as possible these symbols are agreed with the convention adopted by the
International Electro Technical Commission. An important criterion in the selection
of symbol is that, as far as possible, they should be self explanatory and easy to
draw. IS 2032 gives a list of standard symbols.
88
89
90
91
92
93
3. Standard values of voltages
For the sake of completeness, all the standard values of voltages given in IS: 585 –
1962.
• Single phase, two wire system – The standard voltage shall be 240 V
• Three phase system o 415 V, 3.3 kV, 6.6 kV, 11 kV, 22 kV, 33 kV, 66 kV, 110 kV, 132 kV, 220
kV and 400 kV.
• The standard dc voltage shall be 220 / 440 V
4. Voltage limits for AC system
The voltage at any point of the system under normal conditions shall not depart from
the declared voltage by more than the values given below;
• 6% in the case of low (250V or less) or medium (251 to 650 V) voltage
• 6% in the higher side or 9% on the lower side in case of High voltage (651 V to
33 kV)
• 12.5% in case of Extra High voltage ( above 33 kV)
5. Distance from Electric Lines
No building shall be allowed to be erected or re- erected, or any additions or
alterations made to the existing building unless the following minimum clearances
are provided from the over head electric supply lines.
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Note:- For extra high voltage lines apart from the minimum clearance indicated, a
vertical and horizontal clearance of 3.0 m from every additional 33 kV or part thereof
shall be provided.
6. Wiring Installations
A major portion of the fixed installation design in a building relates to wiring
installation. The essential design and constructional requirements for electrical
wiring installations are as follows.
(6.1) Fittings and Accessories
• A ceiling rose or any other attachment shall not be used on a circuit, the
voltage of which normally exceeds 250 V.
• Each 15 A socket outlet provided in building for the use of domestic appliances
such as AC, water cooler etc.
• Each socket outlet shall be controlled by a switch which shall preferably be
located immediately adjacent thereto or combined therewith.
• Ordinary socket outlet may be fixed at any convenient place at a height above
20 cm from the floor level. In a situation where the socket outlet is accessible
to children, socket outlet which automatically gets screened by the withdrawal
of plug is preferable.
• In an earthed system of supply, a socket outlet with plug shall be three pin
types with third terminal connected to earth.
• All lamps unless otherwise required and suitably protected, shall be hung at a
height of not less than 2.5 m above floor level.
Unless otherwise specified, the clearance between the bottom most point of the
ceiling fan and the floor shall be not less than 2.4 m. the minimum clearance
between the ceiling and the plane of the blade shall be not less than 30 cm.
(6.2) Reception and Distribution of Main Supply
• There shall be circuit breaker or a linked switch with fuse on each live
conductor of the supply mains at the point of entry. The main switch shall be
easily accessible and shall be situated near to the termination of service line.
• Branch distribution board shall be provided with a fuse or a miniature circuit
breaker (MCB) or both of adequate rating / setting.
• Light and fans may be wired on a common circuit. Such sub-circuit shall not
have more than a total of 10 points of light, fan and 5 A socket outlets. The
load of such circuit shall be restricted to 800 Watts. Power sub-circuit shall be
designed according to the load but in no case shall there be more than two 15
A outlets on each sub-circuit.
• The load on any low voltage sub circuit shall not exceed 3000 Watts. In case of
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new installation, all circuits and sub-circuits shall be designed by making a
provision of 20% increase in load due to any future modification.
• The distribution fuse board shall be located as near as possible to the centre of
the load. These shall be fixed in suitable stanchion or wall and shall not be
more than 2 m from the floor level.
• All conductors shall be of copper or aluminium. Conductor for final sub-circuit
of fan and light wiring shall have a nominal cross sectional area not less than 1
Sq. mm copper and 1.5 Sq. mm aluminium. The cross sectional area for power
wiring shall be not less than 2.5 Sq. mm copper, 4 Sq. mm aluminium. The
minimum cross sectional area of conductors of flexible cord shall be 0.5 Sq.
mm copper.
(6.3) Conduit wiring
• Rigid non-metallic conduits are used for surface, recessed and concealed
conduit wiring. Conductors of ac supply and dc supply shall be bunched in
separate conduits. The numbers of insulated cables that may be drawn into
the conduit are given in table.
Maximum permissible number of 1.1 kV grade single core cables that may be
drawn into rigid non metallic conduits
Conduit shall be fixed by saddles secured to suitable wood plugs or other plugs
with screws at an interval of not more than 60 cm. whenever necessary, bends or
diversions may be achieved by bending the conduits or by employing normal
bends, inspection bends, inspection boxes, elbows or similar fittings.
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• Energy meters shall be installed at a height where it is convenient to note the
meter reading; it should preferably not be installed at a height not less than 1
m from the ground.
(8.6) Sequence to be followed in carrying out the estimate
1. Wiring layout: Prepare building plan on a suitable scale and mark
electrical points, switch boards, main board, meter board, distribution
board etc. on the plan using specified symbols. The path of wiring
showing connection to each point is marked by a little thick line.
2. Calculation of total connected load: The total connected load and hence
the total current is calculated for deciding the cable size, rating of main
switch board and distribution board.
3. Selection of Main Switch: Once the connected load is calculated, the
main switch can be conveniently selected from the available standard
switch list.
(3.1) List of standard Iron Clad main switches for domestic purpose:
a) DPIC (Double Pole Iron Clad) main switch: 5,15 or 30 A, 250V or DPMCB
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(Double Pole Miniature Circuit Breaker): 5, 10, 16, 32 and 63 A, 250 V
b) TPIC (Triple Pole Iron Clad) main switch: 30, 60, 100, 200 A, 500 V or
TPMCB (Triple Pole Miniature Circuit Breaker): 16, 32 and 63 A, 500 V,
beyond this TPMCCB (Triple Pole Molded Case Circuit Breaker): 100,
200, 300 and 500 A, 660 V
c) TPN main switch: 30, 60, 100, 200, 300 A, 500 V or TPNMCB: 16, 32,
63A, 500 V, beyond this TPNMCCB: 100, 200, 300, 500 A, 660 V.
4. Selection of Main Distribution Board: The Main Distribution Board is a
fuse box or MCB box where different sub-circuits are terminated.
Numbers of sub-circuits are decided based on the total connected load or
total number of points.
5. Assumptions: the conditions which are not specified in the question may
be assumed conveniently. Eg:- location of main switch board, switch
boards, height of building(if not specified)
6. Calculation of length of conduit: To avoid duplicity in calculating the
length of conduit pipe, this may be calculated in three stages. (a) The
conduit installed from switch board up to horizontal run (HR) including
from main switch or DB to HR. (b) The conduit on walls running parallel
to the floor ie, the HR below ceiling. (c) The conduit installed between HR
and ceiling, along ceiling and ceiling to last point on HR.
The total length of conduit is calculated by adding the length of conduit
obtained from the three stages and including 10% wastage.
7. Calculation of length of phase wire and neutral wire: The phase wire and
neutral wire is calculated sub-circuit wise. Once it is calculated, wastage
of 15% is included.
8. Calculation of length of earth wire: The earth wire is run along the
conduit. The calculations are carried out in length but it is converted in
to weight while preparing material table.
9. Preparing Material Table: The material table should be prepared with
complete specification of each item.
(8.7) Current rating of copper conductor single core cables
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(8.8) Selection, rating and installation of equipments on the main switch board
Eg 1:- There are 4 light/ power sub-circuits in an installation of a house wiring. One
of them is a sub-circuit for 15 a socket. Draw the single line diagram showing cutout,
meter, main switch, main distribution board and other equipment. Make your own
assumptions for number of electrical points in each sub-circuit and find out the
rating of main switch and distribution board.
ANS:
Load in sub-circuit No. 1
Light point = 2 x 60 = 120 W
Fan point = 2 x 60 = 120 W
5 A socket = 4 x 100 = 400 W
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100
101
Distribution Inside Large Buildings
In large buildings the type of distribution depends on the building
type, dimension, the length of supply cables, and the loads. The
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distribution system can be divided in to:
The vertical supply system (rising mains).
The horizontal supply (distribution at each floor level).
In most cases a high voltage supply and transformer substation is
required. Normally HV switchgear and substation transformers are
installed at ground floor ( or basement ).
However, often there are appliances with large power demand
installed on the top floors (converters and motors for lifts,
airconditioning
equipment and electric kitchens).
As it is desirable to brining the high voltage supply as close as
possible to the load centers, transformers are installed at the top floor,
or
if required, additional ones are installed on one of the intermediate
floors.
In such cases transformers with non-inflammable insulation and
cooling
are used.
The arrangement of the rising mains depends on the size and shape of
the building and suitable size of shafts for installing cables and bus
ducts
must be provided in coordination with the building architect.
The vertical supply system are implemented in several ways, some of
which are :
Single Rising Main
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Applications :Where high supply security is not important.
Advantages :a) The different loads of individual floors are balanced out.
b) Only a small main L.V board is required.
c) Simple in construction and operation.
Disadvantages :Low supply security (a fault in the rising mains effect all floors).
104
105
Applications :High rise building with high load concentration.
Advantages :Easier mounting.
Smaller size for rising mains.
Disadvantages :A fault in any rising mains effect several floors (relatively low
security).
Loads are balanced only within each group.
Larger power distribution board.
Individual Floor Supply
Applications :In high rise buildings were stories are let separately (metering is at
central point at ground floor).
Advantages :a) Smaller size of cables can be used (easy installation).
b) In the case of a fault in arising main, only one story is effected.
Disadvantages:a) Different loading of the individual floors can not be balanced out.
b) The rising main must be rated for the peak load of each floor.
c) Uneconomical – large number of cables and the size of the rising
main shaft is quite large.
d) Large low voltage distribution board with numerous circuits.
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Applications :In large buildings when relatively higher security is required.
Advantages :a) Higher power supply security ( in the event of a fault, it is possible
to switch off the faulty part and leave the majority of the building
operational )
b) A small low voltage distribution board is required.
c) The differing loading of individual floor are balanced out ( smaller
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sizes for rising mains )
Power distribution in an industry
The power distribution in an industry has different levels
• Main Switch Board (MSB) level
• Sub Switch Board (SSB) level
• Distribution Board (DB) level
DB is the last element before the loads. But large loads are directly connected to SSB
or MSB.
DB / DFB (Distribution Fuse Board) / FDB (Fuse Distribution Board)
� Usually even numbers of ways are used in DBs (2, 4, 6, 8, 10 and 12). As per
IS the maximum number of ways is limited to 12.
Eg:- 12 way 3 ph DB = 4 x 12 = 48 cable connection including neutral.
� Usual current rating of DB s are : 16A, 32A and 63A
� 63A, 12 way DB s are not common. Since maximum input current = 63 x
12 = 700A, which is not possible to handle by a DB. Hence 63A DB is 2 ways or
4 ways.
Motor loads up to 20 hp are fed from DB s of various rating.
� All DBs have isolator or SFU as incomer switch. But in some case this is
avoided if the switch board supplying to the DB is within 3m from the DB
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In a designed system 20% spare outlets are kept for future expansion. ie, in
each DB, 1 or 2 outlets shall be kept as spares.
Selection of rating of incomer isolator/SFU and incomer feeder size
In any system, all the connected loads will not be put on simultaneously. This
reduces the maximum demand from simply computing by adding all connected loads.
The maximum demand is expressed through a factor called ‘Diversity Factor,
Diversity Factor (DF) = Sumof connected load
Simultaneousmax.demand(MD)
>1
� From the requirement data, the details of connected load on each DB are
known to us. For spare outlets, an average of other outlets can be assumed.
� If the DF is known, we can find the maximum current requirement of the DB to
feed all loads including spares. Instead of furnishing the DF, a usual practice is
specifying MD. A commonly accepted and safe value of DF is 1.5. this value can
be assumed for each DB
� If motor loads are connected, for selection of isolator / SFU, the starting
current has to be taken in to account rather than continuous current.
Eg:- 5 hp - 5Nos and 10 hp - 2Nos motors are connected to a DB
Total connected load = 45hp
MD= 45 30
1.5
= hp
Corresponding maximum current is 30 x 1.4 = 42 A. This current is the
continuous maximum current
� When motors are started we have to account the starting inrush current of
large motor in the down stream. Starting current of DOL starting motor is 2.5
times the rated current and for assisted starting (star delta), it is 1.5 times the
rated current.
So the MD calculation in the above case is as follows:
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� One 10 hp (one higher rating) kept aside
� Now only MD of 20 hp is existing
� Its maximum current = 20 x 1.4 = 28 A
� For one 10 hp alone, maximum current = 2.5 x (10 x 1.4) = 35 A
� Therefore MD of the DB = 28 + 35 = 63 A
� ie, incoming feeder, isolator/SFU of the DB can be rated to 63 A
Grading or Discrimination between Feeder Fuse and DB Fuse
The feeder to a DB will be fed from an SSB or MSB. This feeder will be protected by
the HRC fuse in the SSB or MSB. It is necessary that the feeder protective fuse
should not blow off before the motor protective fuse in the DB. This is achieved by
proper grading between the fuses. The fuse of SSB/MSB is denoted as major fuse
and that of DB is termed as minor fuse. For achieving grading the ratio between
major and minor fuses shall be 2:1 or more
� Feeder cable is selected by considering the 20% excess of the MD of DB. Also
major fuse rating should match with the cable selection.
� If the cable length exceeds 75 to 100mtr, the voltage drop condition should be
taken in to account. The voltage drop in the feeder should not be more than 3%
in the maximum demand condition.
Eg 1:- 50 hp, 415 V, 3 ph Induction motor use PVCAPVC 150 m cable.
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Feeder cable is selected by considering the 20% excess of the MD of DB. Also
major fuse rating should match with the cable selection.
� If the cable length exceeds 75 to 100mtr, the voltage drop condition should be
taken in to account. The voltage drop in the feeder should not be more than 3%
in the maximum demand condition.
Design of incomer SFU, Cable size and Bus bar rating for SSBs and MSBs
Switch boards in general are power distribution centers with
SFUs/MCCBs/ACBs/OCBs for controlling outlets and incomer. Unlike DBs, switch
boards are specified by its total current carrying capacity or incomer current rating.
Where as in DBs current rating of the outlet is the specified rating. Standard switch
board ratings are 100 A, 200 A, 400 A, 800 A, 1200 A, 1600 A, 2000 A, 2500 A and
3200 A. If the incomer supply is controlled with an SFU, the switch board is called
switch fuse controlled board and if the incomer is ACB/ OCB controlled, it is called
breaker controlled board.
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A switch board having three sections
� Outlet control gears
� Bus chamber
� Incomer control gear
The outlet switch, fuse and cable rating are decided by the load that has to be
handled through that feeder. If the number of loads is more, SSB is required,
which is installed almost at the load centers. In smaller set up SSB may not be
necessary and MSB will be the only switch board.
Consider the setup:-
For 63 A and 100 A respective rating of switch and fuse are
available.
� For 80 A, 100 A switch with 80 A fuse may be used, since 80 A
switch is not available.
� For 40 hp motor with star-delta starter
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• Starting current = 40 x 1.4 x 1.5 = 84 A
Therefore 100 A switch and fuse are used
� Spare is taken as 100 A
� Total out going fuse rating = 63 x 2 + 100 x 3 + 80 = 506 A
� DF of 2 is assumed
� The MD will be = (506/2) = 253 A
� Taking 20 % extra, the maximum current requirement = 303.6 A
say, 300 A
Hence the incomer switch and fuse shall have a rating of 300
A is used. If 300A switch is not available, 400A switch
with 300A fuse can be used. The incomer cable is also rated
for 300A.
� But in this case, instead of 56A (40 x 1.4) continuous current of
40 hp motor, we have taken 84 A and fuse of 100A. Considering
all these, a practical and most economical selection is 250A
incomer.
� Since the incomer fuse is 250A, any fuse on the outlet greater
than 125A will grade with 250A. Here maximum fuse rating is
100A and grading is automatically satisfied.
Next step is finding Bus bar size.
� Bus bar materials are:
• Aluminum or Aluminium alloy – working current density, 0.8
A/ Sq.mm
• Copper – working current density, 1.2 A/ Sq.mm
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• For the above set up:� 250/0.8 = 312.5 Sq. mm
� For neutral bus bar, half the size of phase bus
bar size is sufficient.
� ie, 40 x 8 mm or 50 x 6 mm Al bus bar may be
used for phases and 20 x 8 mm or 25 x 6 mm
for neutral.
Or
� 31 x 6 mm Cu bus bar may be used for phases
and 31 x 3 mm for neutral.
� For small switch boards the distance between the bus supports
will be 50 cms.
� If DF is not given, we can assume, DF as 2 for all switch boards.
� The term ampacity is some times used to denote the maximum
current rating of the feeders. If DF is not clearly known, the total
ampacity of outlet feeders shall not be more than two times the
ampacity of the incomer feeder.
� The feeder cables need to be selected for the fuse used in the SFU.
• Eg:- when we want 125A feeder, the fuse and cable
corresponding to 125A. But the switch may be 200A, since
above 100A, only 200A switch is available.
� Standard switch ratings are: 32A, 63A, 100A, 200A, 250A, 400A,
630A AND 800A. Some manufactures makes 125A and 320A also.
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the outlet switches is rated to 200A. But the scheme is correct. Though the
switch is rated to 200A, the fuse is only 125A, which will grade with the
incomer 250A.
� There is no lower limit for the outlet of fuses, except those are imposed by
practical consideration of mounting. ie, it may not be possible to mount a 5A
fuse in a 32A switch. But there is lower limit for outlet switch rating.
� Maximum outlet fuse rating is 1
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� When the incomer of a switch board is controlled with an SFU
2
of incomer fuse rating. There
is no upper limit for switch rating except that is imposed by
economic consideration.
� Minimum outlet switch rating is 1 to 1
10 12
of incomer fuse
rating. There is no lower limit for fuse rating other than
availability and mounting possibility.
� When the incomer of a switch board is controlled with a Breaker, the
maximum current rating of outlet fuse should not be more than 1
3
of
incomer rating (setting of CB) and minimum outlet switch rating shall not
be less than 1
5
of the breaker rating.
CBs are available in the ranges of 400A, 800A, 1200A, 1600A,
2400A, 3200A and 3600A
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MODULE-5
Earthing
Earthing or grounding means connecting all parts of the apparatus (other than live
part) to the general mass of earth by wire of negligible resistance. This ensures that
all parts of the equipment other than live part shall be at earth potential (ie, zero
potential) so that the operator shall be at earth potential at all the time, thus will
avoid shock to the operator. The neutral of the supply system is also solidly earthed
to ensure its potential equal to zero.
Earthing shall generally be carried out in accordance with the requirement of Indian
Electricity Rule 1956, particularly IE Rules 32, 51, 61, 62, 67, 69, 88(2) and 90.
• All medium voltage equipment shall be earthed two separate and distinct
connections with earth through an earth electrode. In the case of high and
extra high voltage the neutral point shall be earthed by not less than two
separate and distinct connections.
• Each earth system shall be so devised that the testing of individual earth
electrode is possible. It is recommended that the value of any earth system
resistance shall not be more than 5 Ω, unless otherwise specified.
• Under ordinary conditions of soil, use of copper, iron or mild steel electrodes is
recommended. In direct current system, however due to corrosive action, it is
recommended to use only copper electrode. Use similar materials for earth
electrode and earth conductors to avoid corrosion.
(7.1) Design data on earth electrode
Standard earth electrodes are;
(a) Road and pipe electrodes, (b) Strip or conductor electrodes, (c)
Plate electrodes, and (d) Cable sheaths.
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(7.2) Design of earth electrode
Earth resistivity, ρ = 2πSR Ω-m , where S = distance between successive
electrode in m, R = earth megger reading in Ω
Permissible current density for 3 sec;
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119
(7.2) Specification
• The earth rod shall be situated at a distance not less than 1.5 m from the
building whose installation being earthed
• The size of the continuity conductor shall be 2.9 mm2 (14 SWG) or half of the
installation conductor size.
• The permissible value of earth resistance is,
o Large Power Station - 0.5 Ω
o Major Power Station - 1 Ω
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121
122
Substations
On the basis of design substations may be classified in to
(a)Outdoor type
i. Pole mounted (single stout pole/ H-type/ 4-pole structure
employed for transformers of 25 kVA, 100 kVA and above 100 kVA)
ii. Foundation mounted (For transformers above 250 kVA and voltage
of 33 kV and above)
(b) Indoor type (In this the substation apparatus are installed within the
building)
(10.1) Outdoor substation
When transformers are installed out door, certain clearances must be maintained.
• Clearance between supplier’s and consumer’s structure should not be less than 3
meters. This is for maintaining the minimum sectional clearance of 206 m at 11 kV.
• Supplier’s and consumer’s structure shall be braced together when the clearance
between them is 5 m or less.
• The ground clearance of the live parts of CTPT unit shall not be less than 3.7 m.
• Phase to phase clearance at the AB switch shall be 915 mm
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• Phase to earth clearance at the AB switch shall be 610 mm. It is the clearance
between
the operating rode of the AB switch and the jumpers of 11 kV down conductors
• The supported length of 11 kV jumpers shall be limited to 1.5 m for standard
conductors
and 2.44 m for solid conductors (No. 2 or No. 0 SWG copper).
• Where there is a cable end box with open terminations, the clearance of the live
pars
to ground shall not be less than 305 m
• The ground clearance of ht parts, usually 11 kV at the transformer bushings shall not
be
less than 2.75 m.
• The ground clearance of AB switch handle shall be between 1 and 1.2 m
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Indoor Substation
Indoor substation of 11 kV/415 V are usually installed at industrial areas and other load
areas
like multistoried buildings, telephone exchange etc. Substation building is constructed
for
installing transformer, HT and LT panel etc. Room size should be sufficient, so as to give
adequate clearance between wall and various equipments. Suitable ventilation for
entry of
fresh air at the bottom of transformer room and exit of hot air at top on opposite sides
are
necessary. The installation of transformer should that the cable boxes are on the sides
and not
facing the door.
The OH line terminates on a DP structure outside the indoor substation. All protection
accessories such as AB switch, LA and DO fuse are installed in the DP structure. CT PT
unit is
installed for connecting metering device. Supply to HT side of transformer is brought
through
UG cable. Both sides of the transformer are protected by suitable capacity CB.
Adequate fire
fitting equipment shall be provided at easily accessible positions. Danger notice
board should
be provided on the HV and MV equipments.
125
Key diagram of a 11 kV/415 V indoor substation
126
Constructional Details of 33/11 kV and 11/0.4 kV
Distribution Substations
The power supply to the city of Baghdad is provided basically from
two main substations of 400/132 kV (Baghdad East and Baghdad
west),
which in turn supply many substations of 132/33 kV (or 132/33/11 kV)
distributed geographically throughout the city. A schematic diagram
showing the structure of this system is given in Fig.1. These 132/33 kV
substations in turn provide power supply to a very large number of
33/11 kV substations as shown in Fig.2. Most of these 33/11 kV
substations are equipped with two transformers of 31.5 MVA each (or
2x16 MVA) as shown in Fig.3. To the 11 kV busbars of these
substations (which is divided into two sections) several 11 kV feeders
127
(up to fourteen) are connected using underground cables and/or
overhead line systems.
Each circuit branch is served by a draw-out circuit breaker mostly
by using one of these types :
SF6 Circuit Breaker
Vacuum Circuit breaker
Minimum-Oil Circuit Breaker
There are about 1150 feeders operating at 11 kV in the city
Baghdad alone with a total length of 6845 km.
Equipment layout of a typical such substation is shown in Fig.4.
Also equipment layout and some constructional details of these
substations is shown in Fig.5.
A small 11/0.4 kV transformer size 250 kVA is connected to the 11
kV busbars and used to provide power supply to all substation
axillaries.
The substation axillaries include providing power supply to emergency
lighting, airconditioning, various socket outlets, battery chargers, and
all
other control and monitoring equipment.
Each 11kV feeder provides the supply to large number of 11/0.4 kV
distribution transformers installed using one of the following systems:
i. Pole-mounted transformers supplied directly from 11 kV overhead
lines through manually operated fused switches. The transformer size
in this system is mostly 250 kVA (in few cases in areas with low load
density the 100 kVA size is used).
ii. Compact type unit substations installed usually at street
pavements, in industrial, residential, and commercial areas. . These
substations are provided with three compartments:
High Voltage Compartment
It is placed at one side of the substation, and has an independent
access through a double-sided door with a specially designed lock. It
has the capacity to hold up to three cubicles.
Transformer Compartment
This compartment occupies the middle of the substation and
128
designed to accommodate standard transformers of sizes up to 630
kVA (in some cases up to 1000 kVA). The cover of this
compartment is removable to enable transformer installation on site.
Low Voltage Compartments.
The low voltage compartment is placed at the other side of the
substation and it is provided with all required protective and control
devices. To the low voltage busbars several outgoing 0.4 kV feeders
are installed. Each low voltage feeder provides the power supply to
various numbers of consumers. A single line diagram of a typical
such substation is shown in Fig.6.
The compact type unit substations has the following advantages :
a) Reduction in civil engineering work (only a small excavation is
required)
b) Can be easily transported by a truck due to its small size
c) Remarkable reduction in the installation cost (all internal
connections are made at the factory)
d) Minimum space requirement
e) Adaptation to any application using different standard schemes.
f) Designed for operation outdoors ( weatherproof )
g) High operation safety for both the operator and the equipment
h) Reduction in the maintenance cost as compared with open installations
i) Special sandwich construction of walls prevent quick and direct
heating of equipment caused by direct sunshine conditions.
iii. Privately owned substations installed at consumer's premises in
building basements or in conventional brick-wall rooms. Transformer
sizes used in this case vary from 100 to 1000 kVA and in accordance
with load size.
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NEON-SIGN INSTALLATIONS
Neon signs and field-installed skeleton tubing installations require a transformer or power
supply to step up the voltage to a high level that will cause ignition of connected neon tubing.
This transformer can be considered the “heart beat” of the neon sign or outline lighting system.
Recent articles in the IAEI News have been written to provide information relative to neon
signs and neon installations. This article will visit some of the general requirements one must
be aware of to properly install or inspect a neon transformer or power supply. Keep in mind
that installation conditions can vary from site to site. There are some important factors and
concerns involved with this type of installation that require close attention. High voltages
require increased focus on insulation, spacing, and types of wiring methods and products used.
Specified minimum clearances for high voltage conductors must be maintained.
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Listing and Marking Requirements
Figure 1. Listing marks and SCGFP marks
A good starting point is the identification and listing marks required on the transformer or
power supply. These marks help give an indication that the products have been evaluated by an
electrical testing laboratory and can assist in determining if the product is being used properly.
Neon transformers and power supplies are required to be listed and suitable for an intended
use. If the transformer or supply is remote, it will be field-installed, and the secondary wiring
to the sign copy will also have to be field-installed. Conformance assessment and approval
responsibility of the field-installed transformer and secondary wiring rests with the authority
having jurisdiction. This applies not only to field-installed systems but also to listed section
signs. If the sign is a listed section sign, verify that each section of the copy (sign) and related
power supply or transformer(s) bears a listed section sign mark. Listed section signs, what is
covered under that listing, and the field wiring inspection were reviewed in detail in the
May/June issue of the IAEI News. [Figure 1]
Location of the transformer installation can have an impact on the type of transformer
enclosure required. Transformers installed in wet locations, such as on a roof, or on the inside
of a parapet wall of a building must be suitable for those locations or be installed in suitable
enclosures that provide adequate environmental protection [NEC 600.21(C)]. There are
transformers and listed transformer enclosures specifically for this purpose. Transformers and
power supplies generally are required to have secondary-circuit ground-fault protection and
they must be so marked. This is a marking requirement of the Code and the applicable product
standard [UL 2161; 600.23(F)]. Other important markings on the transformer are the input and
output (secondary) voltages and the current ratings.
Location and Accessibility
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Figure 2. Neon transformers and power supplies are permitted above suspended ceilings where
not supported by the ceiling grid system and appropriate access is provided to the transformer
or supply
Locating the transformer or power supply is an inherent part of effective design using
ingenuity. The location of the transformers is critical in the tube loading and selection process.
The Code addresses this generally in requiring the transformer to be as near to the lamps as
practicable to keep the secondary conductors as short as possible [600.21(B)]. Maximum
lengths are specified in the Code for neon secondary circuits, which will be covered in the next
article in this series.
Another important aspect of the location is accessibility. The transformer must be accessible
for the installer, the inspector, and service personnel. Unlike the general requirement for
standard building power transformers, which are required to be readily accessible, the neon
transformer or power supply only has to be accessible. Definitions can help one establish the
key difference between the two terms. Basically, the neon transformer can be located in areas
where it could be accessed by ladders or other portable means. Service and repair personnel see
the benefits of properly located transformers and power supplies and endure the hardships of
installations that do not provide proper access or adequate working spaces for this type of
equipment.
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Figure 3. Adequate working space at remote neon transformers and power supplies is required
The NEC includes requirements for adequate working space about electrical equipment that is
likely to require examination, adjustment, servicing, or maintenance while energized.
Minimum working space is generally 750 mm (30 in.) in width, and not less than 900 mm (3
ft) from the front of the equipment enclosure. It is interesting to note that within Article 600
specific working space is required to be provided at each ballast, transformer, or power supply
or its enclosure where it is not installed in a sign. Minimum space to be provided is not less
than 900 mm (3 ft) x 900 mm (3 ft) x 900 mm (3 ft). Where located in attics or soffits, an
access opening or door at least 900 mm (3 ft) x 600 mm (2 ft) is required to gain access (see
figure 3). This is often the case in storefronts and above show windows where the neon or sign
is installed to the front or fascia of the building. Sometimes a single access door to these spaces
is acceptable, but where the transformers or power supplies are remote from the access
opening, suitable passageway and walkways are required. This requirement of the Code is far
easier to comply with in new construction than in existing buildings, but keep in mind the NEC
does not differentiate here. Electrical safety and safety of the worker is important here [NEC
600.21(D) and (E)].
Another popular location for remote neon transformers and power supplies is above suspended
ceilings. The ceiling grid cannot be depended upon for support, which means transformers and
power supplies must be securely fastened to the structure in another fashion. Cord- and plugconnected transformers or power supplies are not permitted above suspended ceilings (see
figure 2).
Grounding and Bonding
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Figure 4. Connections of primary branch-circuit equipment grounding conductor and
secondary circuit bonding conductor at transformer enclosure
Understanding grounding and bonding is essential to safe neon installations. Grounding the
equipment as required establishes a reference to earth on the metal equipment associated with
the sign or outline lighting system. This is accomplished when the equipment grounding
conductor with the branch circuit is connected to the transformer or power supply. All metal
parts associated with the neon sign or outline lighting must also be bonded together and
connected back to the transformer enclosure where the equipment grounding conductor of the
branch circuit is connected. This accomplishes two tasks. It puts all metallic parts associated
with the sign at the same potential and bonds the metallic parts and components to the
equipment grounding conductor of the branch circuit feeding the transformer. This, in turn,
puts the bonded metal parts at earth potential (ground reference). Ungrounded metal parts or
metal parts that are not bonded and are located in close proximities to high voltage secondary
circuits can start to rise in potential owing to capacitance coupling effects. This can lead to
potential shock hazards and fire hazards that can result from tracking, etc. A properly grounded
and bonded system minimizes these potential hazards and enhances overall electrical safety.
[Figure 4]
Proper Loading
The Code addresses the loading of neon transformers in a general nature only. This is left as a
primary responsibility of the installer to ensure that the design of the system is such that the
right amount of tubing is being supplied by the appropriate transformer or power supply.
Basically the length of tubing, the diameter and type of gas used have to be married to the
appropriate transformer so that a continuous overcurrent beyond the design loading of the
transformer will not result. Transformer manufacturers can provide the tube footage charts for
use in design of these systems. Once again, this should be done in the design and planning
stages of a neon system installation [NEC 600.41(A)]. There are occasions where the inspector
should verify that the proper transformer is being used. A good indication that transformers are
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underloaded or overloaded is a pattern of service callbacks to keep replacing transformers that
continue to fail.
Connections to Transformer
Figure 5. Transformer terminals indicating specific connection terminals for the grounded
(neutral) conductor and the ungrounded (hot) conductor
Connections are important parts of any electrical installation. Generally, if failure is going to
happen, it usually starts at connection points or terminations. It is rare for failures to start in the
middle of a conductor between terminals, but that is is not to say it cannot happen. When the
transformer is installed one of the last steps prior to testing the system is checking the
connections at the transformer. Primary branch-circuit connections and secondary connections
are required. In transformers within integral enclosures, there are usually separate
compartments for primary and secondary connections. The integrity of the terminal
connections is important. Try to avoid loose connections or connections with devices not
intended for such use as they will eventually spawn failure in the system. Where the
transformer is field-installed with a suitable enclosure, both secondary and primary conductors
are in the same enclosure and attention must be focused on the minimum spacing required
between them. A minimum spacing of 38 mm (1 ½ in.) between each secondary conductor and
all other objects including the branch-circuit conductors is required [NEC 600.32(E)]. Another
item to verify is the minimum length of insulation on the conductor where it emerges from the
raceway into the enclosure. In damp and wet locations a minimum distance of 100 mm (4 in.)
is necessary, and a minimum of 65 mm (2 ½ in.), in dry locations. Suitable secondary
conductors should be used. Another important point to verify is that proper polarity is
maintained for the primary connection to the transformer as many of the secondary-circuit
ground-fault protection units are polarity sensitive (see figure 5). Many are marked to indicate
the terminal specifically for the grounded conductor, and some transformers are also sensitive
to being connected to a multiwire branch circuit. A multiwire branch circuit is one that
includes two or three circuits that share a common neutral conductor (see the definition of
multiwire branch circuit in Article 100). It is important to follow the manufacturer’s
installation instructions and recommendations.
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Figure 6. Transformer checklist diagram from Neon Lighting, page 230
As with any electrical installation, there are unique situations or conditions that will be
encountered in the field that are not specifically addressed in this article. General requirements
and more common elements for field-installed transformers and electronic power supplies have
been provided. Also included in this article is a helpful diagram and checklist for these types of
installations derived from Neon Lighting, IAEI’s new book. [Figure 6] The checklist serves as
a basis for installers and inspectors and might not be totally inclusive.
Safe electric signs, neon, and outline lighting systems are the desire of the installer and the
inspector. Working together and effective communication are the keys to attaining safe neon
installations. Education is the key to understanding these systems and installations and to
becoming more capable of properly applying Code rules to them. The next article in this series
provides a review of the secondary conductor installations and wiring methods from the
transformer output hubs to the tubing electrodes.
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