lecture notes on electrical technology –r15

lecture notes on electrical technology –r15
LECTURE NOTES ON ELECTRICAL TECHNOLOGY –R15
LECTURE NOTES
ON
ELECTRICAL TECHNOLOGY
B.Tech ECE
II YEAR I SEMESTER
(JNTUA-R15)
Mrs.S.JAREENA
ASSISTANT PROFESSOR
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
CHADALAWADA RAMANAMMA ENGINEERING COLLEGE
CHADALAWADA NAGAR, RENIGUNTA ROAD, TIRUPATI (A.P) - 517506
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LECTURE NOTES ON ELECTRICAL TECHNOLOGY –R15
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY ANANTAPUR
II B.Tech I-Sem (E.C.E)
T
Tu
C
3
1
3
(15A02306) ELECTRICAL TECHNOLOGY
Objective:
Electrical Technology contains Single phase transformers, Induction motors, DC generators and
motors which are widely used in industry are covered and their performance aspects will be
studied.
UNIT- I DC GENERATORS
D.C. Generators – Principle of Operation – Constructional Features – E. M.F Equation–
Numerical Problems – Methods of Excitation – Separately Excited and Self Excited Generators –
Build-Up of E.M.F - Critical Field Resistance and Critical Speed - Load Characteristics of Shunt,
Series and Compound Generators- Applications
UNIT – II D.C. MOTORS
D.C Motors – Principle of Operation – Back E.M.F. –Torque Equation – Characteristics and
Application of Shunt, Series and Compound Motors-Speed Control of D.C. Motors: Armature
Voltage and Field Flux Control Methods. Three Point Starter-Losses – Constant & Variable
Losses – Calculation of Efficiency - Swinburne’s Test.
UNIT-III
SINGLE PHASE TRANSFORMERS
Single Phase Transformers - Constructional Details- Emf Equation - Operation on No Load and
on Load - Phasor Diagrams-Equivalent Circuit - Losses and Efficiency-Regulation-OC and SC
Tests – Sumpner’s Test - Predetermination of Efficiency and Regulation.
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UNIT-IV 3-PHASE INDUCTION MOTORS
Polyphase Induction Motors-Construction Details of Cage and Wound Rotor Machines- Principle of Operation – Slip- Rotor Emf and Rotor Frequency - Torque Equation- Torque Slip
Characteristics.
UNIT – V SYNCHRONOUS MACHINES
Principle And Constructional Features of Salient Pole and Round Rotor Machines – E.M.F
Equation- Voltage Regulation by Synchronous Impedance Method- Theory of Operation of
Synchronous Motor.
OUTCOME:
After going through this course the student gets a thorough knowledge on DC Motors &
Generators, Transformers and Induction motors with which he/she can able to apply the above
conceptual things to real-world problems and applications.
TEXT BOOKS:
1. Electric Machines –by I.J.Nagrath & D.P.Kothari,Tata Mc Graw Hill, 7th Edition.2005
2. Basic Electrical Engineering –By T.K.Nagasarkar and M.S. Sukhija Oxford University Press.
REFERENCE BOOKS:
1. Electrical and Electronic Technology, Hughes, Pearson Education.
2. Electrical Machines, P. S. Bimbhra, Khanna Publishers, 2011.
3. Basic Electrical Engineering, 2nd Edition, V.N. Mittle and Aravind Mittal, Mc Graw hill Education,
2006.
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UNIT-I
DC GENERATORS
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Generators
There are two types of generators, one is ac generator and other is dc generator. Whatever may
be the types of generators, it always converts mechanical power to electrical power. An ac
generator produces alternating power.
A DC generator produces direct power. Both of these generators produce electrical power, based
on same fundamental principle of Faraday's law of electromagnetic induction. According to these
law, when an conductor moves in a magnetic field it cuts magnetic lines force, due to which an
emf is induced in the conductor. The magnitude of this induced emf depends upon the rate of
change of flux (magnetic line force) linkage with the conductor. This emf will cause an current
to flow if the conductor circuit is closed. Hence the most basic two essential parts of a generator
are
1. a magnetic field
2. conductors which move inside that magnetic field.
Constructional Features
A DC generator has the following parts
1. Yoke
2. Pole of generator
3. Field winding
4. Armature of DC generator
5. Brushes of generator
6. Bearing
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Fig. A Cut Away View Of Practical DC Generator
Yoke of DC Generator
Yoke of DC generator serves two purposes,
1. It holds the magnetic pole cores of the generator and acts as cover of the generator.
2. It carries the magnetic field flux.
In small generator, yoke are made of cast iron. Cast iron is cheaper in cost but heavier than steel.
But for large construction of DC generator, where weight of the machine is concerned, lighter
cast steel or rolled steel is preferable for constructing yoke of DC generator. Normally larger
yokes are formed by rounding a rectangular steel slab and the edges are welded together at the
bottom. Then feet, terminal box and hangers are welded to the outer periphery of the yoke frame.
Armature Core of DC Generator
The purpose of armature core is to hold the armature winding and provide low reluctance path
for the flux through the armature from N pole to S pole. Although a DC generator provides direct
current but induced current in the armature is alternating in nature. That is why, cylindrical or
drum shaped armature core is build up of circular laminated sheet. In every circular lamination,
slots are either die - cut or punched on the outer periphery and the key way is located on the
inner periphery as shown. Air ducts are also punched of cut on each lamination for circulation of
air through the core for providing better cooling.
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Armature Winding of DC Generator
Armature winding are generally formed wound. These are first wound in the form of flat
rectangular coils and are then pulled into their proper shape in a coil puller. Various conductors
of the coils are insulated from each other. The conductors are placed in the armature slots, which
are lined with tough insulating material. This slot insulation is folded over above the armature
conductors placed in it and secured in place by special hard wooden or fiber wedges.
Commutator of DC Generator
The commutator plays a vital role in dc generator. It collects current from armature and sends it
to the load as direct current. It actually takes alternating current from armature and converts it to
direct current and then send it to external load. It is cylindrical structured and is build up of
wedge - shaped segments of high conductivity, hard drawn or drop forged copper. Each segment
is insulated from the shaft by means of insulated commutator segment shown below. Each
commutator segment is connected with corresponding armature conductor through segment riser
or lug.
Brushes of DC Generator
The brushes are made of carbon. These are rectangular block shaped. The only function of these
carbon brushes of DC generator is to collect current from commutator segments. The brushes are
housed in the rectangular box shaped brush holder. As shown in figure, the brush face is placed
on the commutator segment with attached to the brush holder.
Bearing of DC Generator
For small machine, ball bearing is used and for heavy duty dc generator, roller bearing is used.
The bearing must always be lubricated properly for smooth operation and long life of generator.
Emf equation for dc generator
The derivation of EMF equation for DC generator has two parts:
1. Induced EMF of one conductor
2. Induced EMF of the generator
Derivation for Induced EMF of One Armature Conductor
For one revolution of the conductor,
Let,
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Φ = Flux produced by each pole in weber (Wb) and
P = number of poles in the DC generator. therefore,
Total flux produced by all the poles = ø*p
And,
Time taken to complete one revolution = 60/N
Where,
N = speed of the armature conductor in rpm.
Now,according to Faraday’s law of induction, the induced emf of the armature conductor is
denoted by “e” which is equal to rate of cutting the flux.
Therefore,
Induced emf of one conductor is
Derivation for Induced EMF for DC Generator
Let us suppose there are Z total numbers of conductor in a generator, and arranged in such a
manner that all parallel paths are always in series.
Here,
Z = total numbers of conductor
A = number of parallel paths
Then,
Z/A = number of conductors connected in series
We know that induced emf in each path is same across the line
Therefore,
Induced emf of DC generator
E = emf of one conductor × number of conductor connected in series.
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Induced emf of DC generator is
Simple wave wound generator
Numbers of parallel paths are only 2 = A
Therefore,
Induced emf for wave type of winding generator is
Simple lap-wound generator
Here, number of parallel paths is equal to number of conductors in one path i.e. P = A
Therefore,
Induced emf for lap-wound generator is
Methods Of Excitation
An electric generator or electric motor consists of a rotor spinning in a magnetic field. The
magnetic field may be produced by permanent magnets or by field coils. In the case of a machine
with field coils, a current must flow in the coils to generate the field, otherwise no power is
transferred to or from the rotor. The process of generating a magnetic field by means of an
electric current is called excitation.
For a machine using field coils, which is most large generators, the field current must be
supplied, otherwise the generator will be useless. Thus it is important to have a reliable supply.
Although the output of a generator can be used once it starts up, it is also critical to be able to
start the generators reliably. In any case, it is important to be able to control the field since this
will maintain the system voltage.
Types of excitation
(1)seperately excited generator.
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(2)self excited generator.
self generator is classified into 3 types.
1.shunt generator.
2.series generator.
3.compound generator.
compoud generator is again classified into 2 types.
1.short shunt generator.
2.long shunt generator.
Separately excited generators.
These kind of generators has provided field exciter terminals which are external DC
voltage source is supplies to produce separately magnetic field winding (shunt field) for
magnetize of the generator as illustrated in figure as below.
Self excited field generators.
This type of generator has produced a magnetic field by itself without DC sources from an
external. The electromotive force that produced by generator at armature winding is supply to a
field winding (shunt field) instead of DC source from outside of the generator. Therefore, field
winding is necessary connected to the armature winding. They may be further classified as
a) Shunt generator.
This generator, shunt field winding and armature winding are connected in parallel
through commutator and carbon brush as illustrated in the figure below.
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b) Series generator
The field winding and armature winding is connected in series. There is different from shunt
motor due to field winding is directly connected to the electric applications (load). Therefore,
field winding conductor must be sized enough to carry the load current consumption and the
basic circuit as illustrated below.
Series generator
c) Compound generator
The compound generator has provided with magnetic field in combine with excitation of shunt
and series field winding, the shunt field has many turns of fine wire and caries of a small current,
while the series field winding provided with a few turns of heavy wire since it is in series with an
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armature winding and caries the load current. There are two kinds of compound generator as
illustrated in figure 5 and 6.
A short-shunt compound generator
Characteristic of separately excited generator
The generated electromotive force (EMF) is proportional to both of a magnetic density of flux
per pole and the speed of the armature rotated as expression by the relation as following.
Eg = κ φ n
Where
K = Constant for a specific machine
φ = The density of flux per pole
n = Speed of the armature rotation
Eg = Generator voltage
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By holding the armature speed (n) at a constant value it can show that generator voltage (Eg) is
directly proportional to the magnetic flux density. Which, flux density is proportionately to the
amount of field current (If). The relation of field current and generate voltage as impressed by
figure .
From the figure when the field current (If) is become zero a small generate voltage is produce
due to a residual magnetism.
As the field current increases cause to increase generated voltage linearly up to the knee of the
magnetization curve. Beyond this point by increasing the field current still further causes
saturation of the magnetic structure.
Generator voltage (Eg) is also directly to the armature speed. The formula and a magnetization
curve can be both impressed about this relation.
Where
Eg = Generator voltage or the value of EMF at speed n
Eg' = Generator voltage or the value of EMF at speed n’
n = Speed of the generator armature ( n’ ≠ n )
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Example 1:
The open circuit terminal voltage versus the field current for a separately excited
DC generator with provided the following test data at revolving speed 1400 rpm as show
by the table1 below.
Magnetic curve for example 3.1
Solution
Curve (a) in figure 8 shows the characteristic at revolving speed 1400 rpm
obtained by the data as show in table 1. To obtain the characteristic at 1000 rpm, is made
of the relation as Eg = Kφn
For instance, at a field current of 0.4 Amp the terminal voltage is 114 volts, when the speed is
reached to 1400 rpm and kept its field current constant at this value, the open circuit voltage at
1000 rpm becomes.
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Voltage Regulation
When we add load on the generator, the terminal voltage will decrease due to
(a) The armature winding resistance is mainly of armature resistance. It is cause directly
decrease in terminal voltage as following relation.
Vt = Eg - Ia Ra
Where,
Vt = Terminal or output voltage
Ia = Armature current or load current
Ra = Armature resistance
(a) Load characteristic of
(b) Circuit diagram
a separately excited DC generator
The decrease in magnetic flux due to armature reaction. The armature current establishes a
magneto motive force (MMF), which it distorts to main flux, and makes result in weakened flux.
We can put inter-pole between main field poles to reduce the armature reaction.
To have some measure by how much the terminal voltage change from no-load condition and on
load condition, which is called “voltage regulation”.
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Where Vnl = No-load terminal voltage Vfl = Full-load terminal voltage
Remark:
A separately excited generator has disadvantage of requiring an external DC source. It is
therefore used only where a wide range of terminal voltage required.
Example 2
The separately excited generator of example 1 is driven at revolving speed 1000 rpm and the
field current is adjusted to 0.6 Amp. If the armature circuit resistance is 0.28 ohm, plot the output
voltage as the load current is varied from 0 to 60 Amp. Neglect armature reaction effects. If the
full-load current is 60 Amp, what is the voltage regulation?
Solution
From example 1, Eg = 153 volts when the field current is 0.6 Amp, which is the open circuit
terminal voltage. When the generator is loaded, the terminal voltage is decreased by internal
voltage drop,
namely.
Vt = Eg - Ia Ra
For a load current of, say 40 Amp.
Vt = 153 - (40 × 0.28) = 141.80 Volts.
This calculation is for a number of load currents and the external characteristic can
be plotted as show in fig. 10 at full load the terminal voltage.
Vt = 153 - (60 × 0.28) = 136.20 Volts.
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Calculated load characteristic of an example 3.2
Critical Field Resistance And Critical Speed
The critical field resistance is the maximum field circuit resistance for a given speed with
which the shunt generator would excite. The shunt generator will build up voltage only if field
circuit resistance is less than critical field resistance. It is a tangent to the open circuit
characteristics of the generator at a given speed.
Critical resistance
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Suppose a shunt generator has built up voltage at a certain speed. Now if the speed of the
prime mover is reduced without changing Rf, the developed voltage will be less as because the
O.C.C at lower speed will come down (refer to figure). If speed is further reduced to a certain
critical speed (ncr), the present field resistance line will become tangential to the O.C.C at ncr.
For any speed below ncr, no voltage built up is possible in a shunt generator.
Critical Speed
Load characteristics
Self excited DC shunt generator
A shunt generator has its shunt field winding connected in parallel with the armature so that the
machine provides it own excitation. For voltage to build up, there must be some residual
magnetism in the field poles. There will be a small voltage (Er) generated.
(a) Shunt generator circuit
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(b) load characteristic of shunt generator
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If the connection of the field and armature winding are such that the weak main pole flux aids to
the residual flux, the induced voltage will become larger. Thus more voltage applied to the main
field pole and cause to the terminal voltage increase rapidly to a large value.
When we add load on the generator, the terminal voltage will decrease due to.
a) The armature winding resistance
b) The armature reaction
c) The weakened flux due to the connection of the generator to aids or oppose to the residual
flux.
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Circuit diagram for the solution of example 3
b) For a load of 20 kw when the terminal voltage is 135 volts, therefore the load current
Series Generator
The field winding of a series generator is connect in series with the armature winding. Since it
carries the load current, the series field winding consists of only a few turns of thick wire. At noload, the generator voltage is small due to residual field flux only. When a load is added, the flux
increase, and so does the generated voltage.
(a) Circuit diagram of series generator
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(b) load characteristics
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Figure shows the load characteristic of a series generator driven at a certain speed. The dash line
indicated the generated EMF of the same machine with the armature opencircuited and the field
separated excited. The different between the two curves is simply the voltage drop (IR) in the
series field and armature winding.
V t = Eg - Ia Ra + Rf
Where
Rf = The series field winding resistance
Ra = The armature winding resistance
The series generators are obviously not suited for applications requiring good voltage regulation.
Therefore, they have been used very little and only in special applications for example, as
voltage booster. The generator is placed in series with a supply line. When the current
consumption is increase, the generated voltage of the series machine goes up because the
magnetic field current is increases.
Compound generator
The compound generator has both a shunt and a series winding. The series field winding usually
wound on the top of a shunt field. The two winding are usually connected such that their ampereturns act in the same direction. As such the generator is said to be cumulatively compound.
Simple circuit for compound generator
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(a) Curve s is represent the terminal voltage characteristic of shunt field winding alone.
Under-compound, this condition the addition of series field winding too short it is cause the
terminal voltage no rise to certain value and reduce while increasing in load current.
(b) Flat compound by increasing the number of a series field turns. It is cause to rise up in
terminal voltage and when no-load and full load condition a terminal voltage is made nearly
same value or equal.
(c) Over-compound, if the number of series field turns is more than necessary to compensated of
the reduce voltage. In this case while a full load condition a terminal voltage is higher than a
no-load voltage. Therefore over-compound generator may use where load is at some distance
from generator. Voltage drop in the line has compensated by used of an over-compound
generator.
(d) If a reversing the polarity of the series field occur this cause to the relation between series
field and shunt field, the field will oppose to each other more and more as the load current
increase. Therefore terminal voltage will drop, such generator is said to be a differentially
compound.
The compound generator are used more extensively than the other type of dc generator because
its design to have a wide variety of terminal voltage characteristics.
Machine Efficiency
The efficiency of any machine is the ratio of the ratio of the output power to the input power.
The input power is provided by the prime mover to drive the generator. Because part of the
energy delivered to the generator is converted into heat, it represents wasted energy. These losses
are generally minimized in the design stage; however, some of these losses are unavoidable.
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Losses of generator
The losses of generators may be classified as
1) Copper losses
The copper losses are present because of the resistance of the windings. Currents flowing
through these windings create ohmic losses. The windings that may be present in addition to
the (I2 R ) armature winding are the field windings, inter-pole and compensate windings.
2) Iron losses
As the armature rotates in the magnetic field, the iron parts of the armature as well as the
conductors cut the magnetic flux. Since iron is a good conductor of electricity, the EMF s
induced in the iron parts courses to flow through these parts. These are the eddy currents.
Another loss occurring in the iron is due to the Hysteresis loss is present in the armature core.
3) Other rotational losses consist of
3.1 bearing friction loss
3.2 friction of the rushes riding on the commutator
3.3 windage losses
Windage losses are those associated with overcoming air friction in setting up circulation
currents of air inside the machine for cooling purposes. These losses are usually very small.
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Applications Of Dc Generators
Applications of Separately Excited DC Generators
These types of DC generators are generally more expensive than self-excited DC generators
because of their requirement of separate excitation source. Because of that their applications are
restricted. They are generally used where the use of self-excited generators are unsatisfactory.
1. Because of their ability of giving wide range of voltage output, they are generally used for
testing purpose in the laboratories.
2. Separately excited generators operate in a stable condition with any variation in field
excitation. Because of this property they are used as supply source of DC motors, whose
speeds are to be controlled for various applications. Example- Ward Leonard Systems of
speed control.
Applications of Shunt Wound DC Generators
The application of shunt generators are very much restricted for its dropping voltage
characteristic. They are used to supply power to the apparatus situated very close to its position.
These type of DC generators generally give constant terminal voltage for small distance
operation with the help of field regulators from no load to full load.
1. They are used for general lighting.
2. They are used to charge battery because they can be made to give constant output voltage.
3. They are used for giving the excitation to the alternators.
4. They are also used for small power supply.
Applications of Series Wound DC Generators
These types of generators are restricted for the use of power supply because of their
increasing terminal voltage characteristic with the increase in load current from no load to full
load. We can clearly see this characteristic from the characteristic curve of series wound
generator. They give constant current in the dropping portion of the characteristic curve. For this
property they can be used as constant current source and employed for various applications.
1.They are used for supplying field excitation current in DC locomotives for regenerative
breaking.
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2.This types of generators are used as boosters to compensate the voltage drop in the feeder in
various types of distribution systems such as railway service.
3.In series arc lightening this type of generators are mainly used.
Applications of Compound Wound DC Generators
Among various types of DC generators, the compound wound DC generators are most
widely used because of its compensating property. We can get desired terminal voltage by
compensating the drop due to armature reaction and ohmic drop in the in the line. Such
generators have various applications.
1. Cumulative compound wound generators are generally used lighting, power supply purpose
and for heavy power services because of their constant voltage property. They are mainly
made over compounded.
2.
Cumulative compound wound generators are also used for driving a motor.
3. For small distance operation, such as power supply for hotels, offices, homes and lodges,
the flat compounded generators are generally used.
4. The differential compound wound generators, because of their large demagnetization
armature reaction, are used for arc welding where huge voltage drop and constant current is
required.
At present time the applications of DC generators become very limited because of technical
and economic reasons. Now a days the electric power is mainly generated in the form of
alternating current with the help of various power electronics devices.
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UNIT-II
DC MOTORS
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Direct Current Motor (DC motor)
DC motor is similar to dc generator; in fact the same machine can act as motor or generator. The
only difference is that in a generator the EMF is greater than terminal voltage, whereas in motor
the generated voltage EMF is less than terminal voltage. Thus the power flow is reversed, that is
the motor converts electrical energy into mechanical energy. That is the reverse process of
generator.
DC motors are highly versatile machines. For example, dc motors are better suited fore many
processes that demand a high degree of flexibility in the control of speed and torque. The dc
motor can provided high starting torque as well as high decelerating torque for application
requiring quick stop or reversals.
DC motors are suited in speed control with over wide range is easily to achieve compare with
others electromechanical.
DC Motor Basic Principles
(a) Energy Conversion
If electrical energy is supplied to a conductor lying perpendicular to a magnetic field,
the interaction of current flowing in the conductor and the magnetic field will produce
mechanical force (and therefore, mechanical energy).
(b) Value of Mechanical Force
There are two conditions which are necessary to produce a force on the conductor. The
conductor must be carrying current, and must be within a magnetic field. When these two
conditions exist, a force will be applied to the conductor, which will attempt to move the
conductor in a direction perpendicular to the magnetic field. This is the basic theory by which
all DC motors operate.
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The force exerted upon the conductor can be expressed as follows.
F = B i l Newton
(1)
where B is the density of the magnetic field, l is the length of conductor, and i the value of
current flowing in the conductor. The direction of motion can be found using Fleming’s Left
Hand Rule.
Fleming’s Left Hand Rule
The first finger points in the direction of the magnetic field (first - field), which goes
from the North pole to the South pole. The second finger points in the direction of the current in
the wire (second - current). The thumb then points in the direction the wire is thrust or pushed
while in the magnetic field (thumb - torque or thrust).
Principle of operation
Consider a coil in a magnetic field of flux density B (figure ). When the two ends of the
coil are connected across a DC voltage source, current I flows through it. A force is exerted on
the coil as a result of the interaction of magnetic field and electric current. The force on the two
sides of the coil is such that the coil starts to move in the direction of force.
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Fig.1. Torque production in a DC motor
In an actual DC motor, several such coils are wound on the rotor, all of which experience
force, resulting in rotation. The greater the current in the wire, or the greater the magnetic field,
the faster the wire moves because of the greater force created.
At the same time this torque is being produced, the conductors are moving in a magnetic
field. At /dt) as shown indifferent positions, the flux linked with it changes, which causes an
emf to be induced (e = d figure 5. This voltage is in opposition to the voltage that causes current
flow through the conductor and is referred to as a counter-voltage or back emf.
Fig.2. Induced voltage in the armature winding of DC motor
The value of current flowing through the armature is dependent upon the difference between the
applied voltage and this counter-voltage. The current due to this counter-voltage tends to oppose
the very cause for its production according to Lenz’s law. It results in the rotor slowing down.
Eventually, the rotor slows just enough so that the force created by the magnetic field (F = Bil)
equals the load force applied on the shaft. Then the system moves at constant velocity.
Construction
DC motors consist of one set of coils, called armature winding, inside another set of coils or a set
of permanent magnets, called the stator. Applying a voltage to the coils produces a torque in the
armature, resulting in motion.
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Stator
The stator is the stationary outside part of a motor.
 The stator of a permanent magnet dc motor is composed of two or more permanent magnet
pole pieces.
 The magnetic field can alternatively be created by an electromagnet. In this case, a DC coil
(field winding) is wound around a magnetic material that forms part of the stator.
Rotor
The rotor is the inner part which rotates.
 The rotor is composed of windings (called armature windings) which are connected to the
external circuit through a mechanical commutator.
Both stator and rotor are made of
ferromagnetic materials. The two are separated by air-gap.
Winding
A winding is made up of series or parallel connection of coils.
 Armature winding - The winding through which the voltage is applied or induced.
 Field winding - The winding through which a current is passed to produce flux
(for the electromagnet)
 Windings are usually made of copper.
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Torque Developed
The turning or twisting moment of a force about an axis is called torque. It is measured by the
product of the force and the radius at which this force acts.
Consider a pulley of radius meter acted upon by a circumferential force of newton which causes
it to rotate at rpm.
Armature torque of a motor
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Shaft torque
Induced Counter-voltage (Back emf):
Due to the rotation of this coil in the magnetic field, the flux linked with it changes at different
positions, which causes an emf to be induced (refer to figure 2).
The induced emf in a single coil, e = døc/dt
Since the flux linking the coil, c = Sin
Induced voltage , e =  cost
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Note that equation (4) gives the emf induced in one coil. As there are several coils wound all
around the rotor, each with a different emf depending on the amount of flux change through it,
the total emf can be obtained by summing up the individual emfs.
The total emf induced in the motor by several such coils wound on the rotor can be obtained by
integrating equation (4), and expressed as:
Eb = K m
(5)
where K is an armature constant, and is related to the geometry and magnetic properties of the
motor, and m is the speed of rotation.
The electrical power generated by the machine is given by:
Pdev = Eb Ia = Km Ia
(6)
DC Motor Equivalent circuit
The schematic diagram for a DC motor is shown below. A DC motor has two distinct
circuits: Field circuit and armature circuit. The input is electrical power and the output is
mechanical power. In this equivalent circuit, the field winding is supplied from a separate DC
voltage source of voltage Vf. Rf and Lf represent the resistance and inductance of the field
winding. The current If produced in the winding establishes the magnetic field necessary for
motor operation. In the armature (rotor) circuit, VT is the voltage applied across the motor
terminals, Ia is the current flowing in the armature circuit, Ra is the resistance of the armature
winding, and Eb is the total voltage induced in the armature.
Fig.4. DC motor representation
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Counter EMF in DC motor
When voltage is applied to dc motor, current will flow into the positive brush through the
commutator into the armature winding. The motor armature winding is identical to the generator
armature winding. Thus the conductors on the north field poles are carry current in one direction,
while all conductors on the south field poles carry the current in opposite direction. When the
armature carry current it will produce a magnetic field around the conductor of it own which
interact with the main field. It is cause to the force developed on all conductors and tending to
turn the armature.
The armature conductors continually cut through this resultant field. So that voltages are
generated in the same conductors that experience force action. When operating the motor is
simultaneously acting as generator. Naturally motor action is stronger than generator action.
Although the counter EMF is opposite with the supplied voltage, but it cannot exceed to applied
voltage. The counter EMF is serves to limit the current in an armature winding. The armature
current will be limited to the value just sufficient to take care of the developed power needed to
drive the load.
In the case of no load is connected to the shaft. The counter EMF will almost equal to the applied
voltage. The power develops by the armature in this case is just the power needed to overcome
the rotational losses. It’s mean that the armature current IA is controlled and limited by counter
EMF therefore
Where:
VL = Line voltage across the armature winding
Ra = Resistance of the armature winding
Ea = Induced EMF or generated voltage
Ia = Armature current
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Since, EA is induced or generated voltage it is depend on the flux per pole and the speed of the
armature rotate (n) in rpm.
Therefore
Ea=Kφn
Where:
K = the constant value depending on armature winding and number of pole of
machine.
φ = Rotation of the armature
Where:
Z = Total number of conductor in the armature winding
a = Number of parallel circuit in the armature winding between positive and negative brushes.
For wave wound armature “a” = 2
Lap wound armature “a” = P
Example
A dc motor operated at 1500 rpm when drawing 20 amps from 220 volts supply, if the armature
resistance is 0.2 ohms. Calculate the no load speed assumed Ia = 0 amp (This amount to
assuming the brushes and rotation loss are negligible)
Solution
When load condition Ia = 20 amps.
Ea = VL – Ia Ra = 220 – 20 (0.2) = 216 Volts.
And
Ea = k φ n
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216 = Kφ ×1500
Kφ = 216/1500
= 0.144
At no load condition Ia = 0 Amp.
Ea = VL = 220 Volts.
Hence
Ea = kφn
n =220/ kφ
=220/ 0.144
= 1528 rpm.
Mechanical power develop in dc motor (Pd)
Let,
Pd = Mechanical power develop
T = Torque exerted on the armature
Pd = ωT
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DC Machine Classification
DC Machines can be classified according to the electrical connections of the armature
winding and the field windings. The different ways in which these windings are connected lead
to machines operating with different characteristics. The field winding can be either self-excited
or separately-excited, that is, the terminals of the winding can be connected across the input
voltage terminals or fed from a separate voltage source (as in the previous section). Further, in
self-excited motors, the field winding can be connected either in series or in parallel with the
armature winding. These different types of connections give rise to very different types of
machines, as we will study in this section.
(a) Separately excited machines
 The armature and field winding are electrically separate from each other.
 The field winding is excited by a separate DC source.
Fig.5. Separately excited dc motor
The voltage and power equations for this machine are same as those derived in the previous
section. Note that the total input power = Vf If + VT Ia
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(b)Self excited machines
In these machines, instead of a separate voltage source, the field winding is connected
across the main voltage terminals.
1.Shunt machine
 The armature and field winding are connected in parallel.
 The armature voltage and field voltage are the same.
Fig.6. shunt motor
Total current drawn from the supply, IL = If + Ia
Total input power = VT IL
Voltage, current and power equations are given in equations (7), (8) and (9).
2.Series DC machine
 The field winding and armature winding are connected in series.
 The field winding carries the same current as the armature winding.
A series wound motor is also called a universal motor. It is universal in the sense that it will run
equally well using either an ac or a dc voltage source.
Reversing the polarity of both the stator and the rotor cancel out. Thus the motor will always
rotate the same direction regardless of the voltage polarity.
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Fig.7.Series Motor
Compound DC machine
If both series and shunt field windings are used, the motor is said to be compounded. In a
compound machine, the series field winding is connected in series with the armature, and the
shunt field winding is connected in parallel. Two types of arrangements are possible in
compound motors:
Cumulative compounding - If the magnetic fluxes produced by both series and shunt field
windings are in the same direction (i.e., additive), the machine is called cumulative compound.
Differential compounding - If the two fluxes are in opposition, the machine is differential
compound.
In both these types, the connection can be either short shunt or long shunt.
Speed control of DC motor
Many applications require the speed of a motor to be varied over a wide range. One of
the most attractive features of DC motors in comparison with AC motors is the ease with which
their speed can be varied.
We know that the back emf for a separately excited DC motor:.
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(7)
From the above equation, it is evident that the speed can be varied by using any of the following
methods:
 Armature voltage control (By varying VT)
 Field Control (By Varying )
 Armature resistance control (By varying Ra)
Armature voltage control
This method is usually applicable to separately excited DC motors. In this method of speed
control, Ra and  are kept constant.
In normal operation, the drop across the armature resistance is small compared to Eb and
therefore: Eb VT
Since, Eb = Kø ωm
Angular speed can be expressed as:
ωm= VT/ Kø
(8)
From this equation, If flux is kept constant, the speed changes linearly with VT.
 As the terminal voltage is increased, the speed increases and vice versa.
 The relationship between speed and applied voltage is shown in figure 8. This method provides
smooth variation of speed control.
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Fig.8.Variation of speed with applied voltage
Field Control , ( )
In this method of speed control, Ra and VT remain fixed.
Therefore, from equation (7):
mI/
Assuming magnetic linearity,  If
(OR) m  I/IF
(9)
i.e., Speed can be controlled by varying field current If.
 The field current can be changed by varying an adjustable rheostat in the field circuit (as shown
in figure 9).
 By increasing the value of total field resistance, field current can be reduced, and therefore
speed can be increased.
Fig.9.
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The relationship between the field winding current and angular speed is shown in figure 10
.
Fig.10: Variation of speed with field current
Armature Resistance Control
The voltage across the armature can be varied by inserting a variable resistance in series with the
armature circuit.
Fig.11. Armature resistance method for speed control
From speed-torque characteristics , we know that:
For a load of constant torque VT and  are kept constant, as the armature resistance Ra is
increased, speed decreases. As the actual resistance of the armature winding is fixed for a given
motor, the overall resistance in the armature circuit can be increased by inserting an additional
variable resistance in series with the armature. The variation if speed with respect to change in
this external resistance is shown in figure 12. This method provides smooth control of speed..
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Fig. 12: Variation of speed with external armature resistance
DC Shunt Motor speed control
All three methods described above can be used for controlling the speed of DC Shunt Motors.
Series Motor speed control
The speed is usually controlled by changing an external resistance in series with the armature.
The other two methods described above are not applicable to DC series motor speed control.
Applications of dc motors
Types of Losses in a DC Machines
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The losses can be divided into three types in a dc machine (Generator or Motor). They are
1. Copper losses
2. Iron or core losses and
3. Mechanical losses.
All these losses seem as heat and therefore increase the temperature of the machine. Further the
efficiency of the machine will reduce.
1. Copper Losses:
This loss generally occurs due to current in the various windings on of the machine. The
different winding losses are;
Armature copper loss = I2a Ra
Shunt field copper loss = I2shRsh
Series field copper loss = I2se Rse
Note: There’s additionally brush contact loss attributable to brush contact resistance (i.e.,
resistance in the middle of the surface of brush and commutator). This loss is mostly enclosed in
armature copper loss.
2. Iron Losses
This loss occurs within the armature of a d.c. machine and are attributable to the rotation of
armature within the magnetic field of the poles. They’re of 2 sorts viz.,
(i) Hysteresis loss
(ii) eddy current loss.
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Hysteresis loss:
Hysteresis loss happens in the armature winding of the d.c. machine since any given part of the
armature is exposed to magnetic field of reverses as it passes underneath sequence poles. The
above fig shows the 2 pole DC machine of rotating armature. Consider a tiny low piece ab of the
armature winding. Once the piece ab is underneath N-pole, the magnetic lines pass from a to b.
Half a revolution well along, identical piece of iron is underneath S-pole and magnetic lines pass
from b to a in order that magnetism within the iron is overturned. So as to reverse constantly the
molecular magnets within the armature core, particular quantity of power must be spent that is
named hysteresis loss. It’s given by Steinmetz formula.
The steinmetz formula is
Hysteresis loss Ph= ηB16max fV watts
Where,
η = Steinmetz hysteresis co-efficient
Bmax = Maximum flux Density in armature winding
F = Frequency of magnetic reversals
= NP/120 (N is in RPM)
V = Volume of armature in m3
If you want to cut back this loss in a d.c. machine, armature core is created of such materials that
have an lesser value of Steinmetz hysteresis co-efficient e.g., silicon steel.
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Eddy current loss:
In addition to the voltages evoked within the armature conductors, some of other voltages
evoked within the armature core. These voltages turn out current currents within the coil core as
shown in Fig. These are referred to as eddy currents and power loss attributable to their flow is
named eddy current loss. This loss seems as heat that increases the temperature of the machine
and efficiency will decrease.
If never-ending cast-iron core is employed, the resistance to eddy current path is tiny attributable
to massive cross-sectional space of the core. Consequently, the magnitude of eddy current and
therefore eddy current loss are massive. The magnitudes of eddy current are often decreased by
creating core resistance as high as sensible. The core resistances are often greatly exaggerated by
making the core of skinny, spherical iron sheets referred to as lamination's shown in the fig. The
lamination's are insulated from one another with a layer of varnish. The insulating layer features
a high resistance, thus only small amount of current flows from one lamination to the opposite.
Also, as a result of every lamination is extremely skinny, the resistance to current passing over
the breadth of a lamination is additionally quite massive. Therefore laminating a core will
increase the core resistance that drops the eddy current and therefore the eddy current loss.
Eddy Current loss Pe=KeB2maxf2t2V Watts
Where, ke = constant
Bmax = Maximum flux density in wb/m2
T = Thickness of lamination in m
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V = Volume of core in m3
Note: Constant (Ke) depend upon the resistance of core and system of unit used.
It may well be noted that eddy current loss be subject to upon the sq. of lamination thickness. For
this reason, lamination thickness ought to be unbroken as tiny as potential.
3.Mechanical Loss
These losses are attributable to friction and windage.

Friction loss occurs due to the friction in bearing, brushes etc.

windage loss occurs due to the air friction of rotating coil.
These losses rely on the speed of the machine. Except for a given speed, they're much constant.
Constant and Variable Losses
The losses in a d.c. machine is also further classified into (i) constant losses (ii) variable losses.
Constant losses
Those losses in a d.c. generator that stay constant at all loads are referred to as constant losses.
The constant losses in a very d.c. generator are:
(a)iron losses
(b)mechanical losses
(c)shunt field losses
Variable losses
Those losses in a d.c. generator that differ with load are referred to as variable losses. The
variable losses in a very d.c. generator are:
Copper loss in armature winding (I2Ra)
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Copper loss in series field winding (I2seRse)
Total losses = Constant losses + Variable losses.
Generally this copper loss is constant for shunt and compound generators.
Three point starter
The figure above shows that typical representation diagram of a 3 point starter for DC shunt
motors with its protective devices. It contains 3 terminals namely L, Z, & A; hence named 3
point starter. The starter is made up of of starting resistances divided into many section and
which are connected in series within the armature. The each tapping point on the starting
resistances is carried out to a no. of studs. The starter 3 terminals L,Z & A are connected to the
positive terminal of line, shunt field and armature terminal of motor respectively. The remaining
terminal of the shunt and armature are connected to the negative line terminal. The No volt coil
release is connected in series with field winding. The handle one end is connected to the L
terminal by means of over load release coil. Then another end of handle travels against the
twisting spring & make touching base with every single stud in the course of starting operation,
tripping out the starting resistance as it moves above every stud in clockwise.
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Armature Reaction in DC Motor
Working:

Initially the DC supply is turned on with the handle is in OFF position.

Now the handle is moved towards clockwise direction to the 1st stud. Once it contacts with
the 1st stud, immediately the shunt field coil is connected to the supply, however the entire
starting resistances is injected with armature circuit in series.

As the handle moved gradually towards the final stud, so that the starting resistance is cut out
step by step in armature circuit. And finally the handle is detained magnetically by the No
volt coil release since it is energized by the filed winding.

In case if the shunt field winding excitation is cut out by accident or else the supply is
interrupted then the no volt coil release gets demagnetized and handle returned back to the
original position under the influence of spring.
Note: If we were not used No volt coil release; then if the supply is cut off the handle would
remain in the same position, causing an extreme current in armature.

If any fault occurs on motor or overload, it will draw extreme current from the source. This
current raise the ampere turns of OLR coil (over load relay) and pull the armature Coil, in
consequence short circuiting the NVR coil (No volt relay coil). The NVR coil gets
demagnetized and handle comes to the rest position under the influence of spring. Therefore
the motor disconnected from the supply automatically.
Characteristic of DC Shunt Motor
Disadvantage:
In point starter, no volt relay coil is connected in series with field circuit; hence it carries shunt
current in the field. When the speed control of DC motor through field regulator, it may be
weakened the shunt field current to such extent the no volt coil release might not in a position to
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hold the starter handle in ON position. This might the motor disconnected from the source when
it is not anticipated. This can be overcome by using the point starter.
Swinburne’s Test for DC Machines
In this technique, the DC Generator or DC Motor is run as a motor at no load; with that losses of
the DC machines are determined. When the losses of DC machine well-known, then we can find
the efficiency of a DC machine in advance at any desired load. In DC machines this test is
applicable only throughout the flux is constant at all load (DC Shunt machine and DC
Compound Machine). This test maintains of two steps;
Determination of Hot Resistance of Windings:
The resistance of armature windings and shunt field windings are measured with the help of a
battery, ammeter and voltmeter. Since these armature and shunt filed resistances are measured
while the DC machine is cold, it should be transformed to values equivalent to the temperature at
which the DC machine would work at full load. These values are measured generally when the
room temperature increases above 40oC. Take on the hot resistance of armature winding and
shunt field winding be Ra and Rsh correspondingly.
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
Condition for maximum Efficiency in DC Machine
Determination of Constant Losses:
On no load the DC machine run as a motor with the supply voltage is varied to the normal rated
voltage. With the use of the field regulator R the motor speed is varied to run the rated speed
which is shown in the figure.
Let
V = Supply Voltage
I0 = No load current read by A1
Ish = Shunt Field current ready by A2
No load armature current Iao = I0 – Ish
No load Input power to motor = VI0
No load Input power to motor = VIa0
= V (I0 – Ish)
As the output power is nil, the no loads input power to the armature provides Iron loss, armature
copper loss, friction loss and windage loss.
Constant loss Wc = Input power to Motor – Armature copper loss
Wc = VI0 – (I0 – Ish2Ra)
As the constant losses are identified, the efficiency of the DC machine at any loads can be
determined. Suppose it is desired to determine the DC machine efficiency at no load current.
Then,
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Armature current Ia = I-Ish (For Motoring)
Ia = I+Ish (For Generating)
To find the Efficiency when running as a motor:
Input power to motor = VI
Armature copper loss =Ia2Ra = (I-Ish2Ra)
Constant Loss = Wc
Total Loss = (I-Ish2Ra)+Wc
Motor Efficiency η = (Input power – Losses)/ Input
η = {VI – (I-Ish2Ra)} / VI

Condition for maximum Efficiency in DC Machine
To find the Efficiency when running as a Generator:
Output Power of Generator = VI
Armature copper loss =Ia2Ra = (I+Ish2Ra)
Constant Loss = Wc
Total Loss = (I+Ish2Ra)+Wc
Motor Efficiency η= Output power/ (Output power + Losses)
η = VI / {VI + (I+Ish2Ra) + Wc}
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Merits:

Since this test is no load test, power required is less. Hence the cost is economic.

The efficiency of the machine can be found very easily, because the constant losses are
well known.

This test is appropriate.
Demerits:

When the DC machine is loaded, this test does not deliberate the stray load loss that
occurs.

Using this method we cannot check the DC machine performances at full load.
Example: It does not indicate the commutation performance is satisfactory on full load and
cannot indicate the specified limit of temperature rise.

Using this test we cannot determine the accurate efficiency of DC machine, because iron
loss at actual loads is greater than at no load. This is primarily owed to armature
reaction interfere with field.
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UNIT-III
SINGLE PHASE TRANSFORMERS
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TRANSFORMERS
The transformer is a device that transfers electrical energy from one electrical circuit to
another electrical circuit. The two circuits may be operating at different voltage levels but always
work at the same frequency. Basically transformer is an electro-magnetic energy conversion
device. It is commonly used in electrical power system and distribution systems.
SINGLE PHASE TRANSFORMERS
INTRODUCTION
In its simplest form a single-phase transformer consists of two windings, wound on an iron
core one of the windings is connected to an ac source of supply f. The source supplies a current
to this winding (called primary winding) which in turn produces a flux in the iron core. This flux
is alternating in nature (Refer Figure 4.1). If the supplied voltage has a frequency f, the flux in
the core also alternates at a frequency f. the alternating flux linking with the second winding,
induces a voltage E2 in the second winding (called secondary winding). [Note that this
alternating flux linking with primary winding will also induce a voltage in the primary winding,
denoted as E1. Applied voltage V1 is very nearly equal to E1]. If the number of turns in the
primary and secondary windings is N1 and N2 respectively, we shall see later in this unit that
.
The load is connected across the secondary winding, between the terminals a1, a2. Thus, the load
can be supplied at a voltage higher or lower than the supply voltage, depending upon the ratio
N1/N2
.
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When a load is connected across the secondary winding it carries a current I2, called load
current. The primary current correspondingly increases to provide for the load current, in
addition to the small no load current. The transfer of power from the primary side (or source) to
the secondary side (or load) is through the mutual flux and core. There is no direct electrical
connection between the primary and secondary sides.
In an actual transformer, when the iron core carries alternating flux, there is a power loss in the
core called core loss, iron loss or no load loss. Further, the primary and secondary windings have
a resistance, and the currents in primary and secondary windings give rise to I 2 R losses in
transformer windings, also called copper losses. The losses lead to production of heat in the
transformers, and a consequent temperature rise. Therefore, in transformer, cooling methods are
adopted to ensure that the temperature remains within limit so that no damage is done to
windings’ insulation and material.
In the Figure 4.1 of a single-phase transformer, the primary winding has been shown connected
to a source of constant sinusoidal voltage of frequency f Hz and the secondary terminals are kept
open. The primary winding of N1 turns draws a small amount of alternating current of
instantaneous value i0, called the exciting current. This current establishes flux φ in the core (+ve
direction marked on diagram). The strong coupling enables all of the flux φ to be confined to the
core (i.e. there is no leakage of flux).
CONSTRUCTION OF A TRANSFORMER
There are two basic parts of a transformer:
1. Magnetic core
2. Winding or coils
MAGNETIC CORE:
The core of a transformer is either square or rectangular in size. It is further divided in two
parts. The vertical portion on which the coils are bound is called limb, while the top and bottom
horizontal portion is called yoke of the core as shown in fig. 2.
Core is made up of laminations. Because of laminated type of construction, eddy current
losses get minimized. Generally high grade silicon steel laminations (0.3 to 0.5 mm thick) are used.
These laminations are insulated from each other by using insulation like varnish. All laminations are
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varnished. Laminations are overlapped so that to avoid the airgap at the joints. For this generally ‗L‘
shaped or ‗I‘ shaped laminations are used which are shown in the fig. 3 below.
WINDING:
There are two windings, which are wound on the two limbs of the core, which are
insulated from each other and from the limbs as shown in fig. 4. The windings are made up of
copper, so that, they possess a very small resistance. The winding which is connected to the load
is called secondary winding and the winding which is connected to the supply is called primary
winding. The primary winding has N1 number of turns and the secondary windings have N2
number of turns.
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Fig. 4 Single Phase Transformer
PRINCIPLE OF OPERATION OF A SINGLE PHASE TRANSFORMER
A single phase transformer works on the principle of mutual induction between two magnetically
coupled coils. When the primary winding is connected to an alternating voltage of r.m.s value,
V1 volts, an alternating current flows through the primary winding and setup an alternating flux
ϕ in the material of the core. This alternating flux ϕ, links not only the primary windings but also
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the secondary windings. Therefore, an e.m.f e1 is induced in the primary winding and an e.m.f e2
is induced in the secondary winding, e1 and e2 are given:
If the induced e.m.f is e1 and e2 are represented by their rms values E1 and E2 respectively, then
K is known as the transformation ratio of the transformer. When a load is connected to the
secondary winding, a current I2 flows through the load, V2 is the terminal voltage across the
load. As the power transfered from the primary winding to the secondary winding is same,
Power input to the primary winding = Power output from the secondary winding.
The directions of emf‘s E1 and E2 induced in the primary and secondary windings are such that,
they always oppose the primary applied voltage V1.
EMF Equation of a transformer:
Consider a transformer having,
N1 =Primary turns
N2 = Secondary turns
Φm = Maximum flux in the core
Φm = Bm × A webers
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f= frequency of ac input in hertz (Hz)
The flux in the core will vary sinusoidal as shown in figure, so that it increases from zero to
maximum “ϕm” in one quarter of the cycle i.e, 1/4f second.
i.e, E1 =4.44fφm×N1 = 4.44fBm×A×N1
Similarly;
E2= 4.44 f φm × N2 = 4.44 f Bm × A × N2
Transformation Ratio:
(1) Voltage Transformation Ratio
(2) Current Transformation Ratio
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Voltage Transformation Ratio:
Voltage transformation ratio can be defined as the ratio of the secondary voltage to the primary
voltage denoted by K.
Current Transformation Ratio:
Consider an ideal transformer and we have the input voltampere is equal to output voltampere.
Mathematically, Input Voltampere = Output Voltampere
TRANSFORMER ON NO-LOAD
Theory of Transformer On No-load, and Having No Winding Resistance and No Leakage
Reactance of Transformer
Let us consider one electrical transformer with only core losses, which means, it has only
core losses but no copper loss and no leakage reactance of transformer. When an alternating
source is applied in the primary, the source will supply the current for magnetizing the core of
transformer.
But this current is not the actual magnetizing current, it is little bit greater than actual
magnetizing current. Actually, total current supplied from the source has two components, one is
magnetizing current which is merely utilized for magnetizing the core and other component of
the source current is consumed for compensating the core losses in transformer. Because of this
core loss component, the source current in transformer on no-load condition supplied from the
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source as source current is not exactly at 90° lags of supply voltage, but it lags behind an angle θ
is less than 90°. If total current supplied from source is Io, it will have one component in phase
with supply voltage V1 and this component of the current Iw is core loss component. This
component is taken in phase with source voltage, because it is associated with active or working
losses in transformer. Other component of the source current is denoted as Iμ. This component
produces the alternating magnetic flux in the core, so it is watt-less; means it is reactive part of
the transformer source current. Hence Iμ will be in quadrature with V1 and in phase with
alternating flux Φ.
Hence, total primary current in transformer on no-load condition can be represented as
Now you have seen how simple is to explain the theory of transformer in no-load.
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TRANSFORMER ON LOAD
Theory of Transformer On Load But Having No Winding Resistance and Leakage
Reactance
Now we will examine the behavior of above said transformer on load, that means load is
connected to the secondary terminals. Consider, transformer having core loss but no copper loss
and leakage reactance. Whenever load is connected to the secondary winding, load current will
start to flow through the load as well as secondary winding. This load current solely depends
upon the characteristics of the load and also upon secondary voltage of the transformer. This
current is called secondary current or load current, here it is denoted as I2. As I2 is flowing
through the secondary, a self mmf in secondary winding will be produced. Here it is N2I2, where,
N2 is the number of turns of the secondary winding of transformer.
This mmf or magneto motive force in the secondary winding produces flux φ2. This φ2 will
oppose the main magnetizing flux and momentarily weakens the main flux and tries to reduce
primary self induced emf E1. If E1 falls down below the primary source voltage V1, there will be
an extra current flowing from source to primary winding. This extra primary current I2′ produces
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extra flux φ′ in the core which will neutralize the secondary counter flux φ 2. Hence the main
magnetizing flux of core, Φ remains unchanged irrespective of load.
So total current, this transformer draws from source can be divided into two components, first
one is utilized for magnetizing the core and compensating the core loss i.e. Io. It is no-load
component of the primary current. Second one is utilized for compensating the counter flux of
the secondary winding. It is known as load component of the primary current. Hence total no
load primary current I1 of a electrical power transformer having no winding resistance and
leakage reactance can be represented as follows
Where θ2 is the angle between Secondary Voltage and Secondary Current of transformer. Now
we will proceed one further step toward more practical aspect of a transformer.
Transformer On Load, With Resistive Winding, But No Leakage Reactance
Now, consider the winding resistance of transformer but no leakage reactance. So far we have
discussed about the transformer which has ideal windings, means winding with no resistance and
leakage reactance, but now we will consider one transformer which has internal resistance in the
winding but no leakage reactance. As the windings are resistive, there would be a voltage drop in
the windings.
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We have proved earlier that, total primary current from the source on load is I1. The voltage drop
in the primary winding with resistance, R1 is R1I1. Obviously, induced emf across primary
winding E1, is not exactly equal to source voltage V1. E1 is less than V1 by voltage drop I1R1.
Again in the case of secondary, the voltage induced across the secondary winding, E2 does not
totally appear across the load since it also drops by an amount I2R2, where R2 is the secondary
winding resistance and I2 is secondary current or load current.
Similarly, voltage equation of the secondary side of the transformer will be
Theory of Transformer On Load, With Resistance As Well As Leakage Reactance in
Transformer Windings
Now we will consider the condition, when there is leakage reactance of transformer as well as
winding resistance of transformer.
Let leakage reactances of primary and secondary windings of the transformer are X1 and X2
respectively.
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Hence total impedance of primary and secondary winding of transformer with resistance R1 and
R2 respectively, can be represented as,
We have already established the voltage equation of a transformer on load, with only
resistances in the windings, where voltage drops in the windings occur only due to resistive
voltage drop. But when we consider leakage reactances of transformer windings, voltage drop
occurs in the winding not only because of resistance, it is because of impedance of transformer
windings. Hence, actual voltage equation of a transformer can easily be determined by just
replacing resistances R1 & R2 in the previously established voltage equations by Z1 and Z2.
Therefore, the voltage equations are,
Resistance drops are in the direction of current vector but, reactive drop will be perpendicular to
the current vector as shown in the above vector diagram of transformer.
Equivalent Circuit of Transformer
Equivalent impedance of transformer is essential to be calculated because the electrical power
transformer is an electrical power system equipment for estimating different parameters of
electrical power system which may be required to calculate total internal impedance of an
electrical power transformer, viewing from primary side or secondary side as per requirement.
This calculation requires equivalent circuit of transformer referred to primary or equivalent
circuit of transformer referred to secondary sides respectively. Percentage impedance is also very
essential parameter of transformer. Special attention is to be given to this parameter during
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installing a transformer in an existing electrical power system. Percentage impedance of different
power transformers should be properly matched during parallel operation of power transformers.
The percentage impedance can be derived from equivalent impedance of transformer so, it can
be said that equivalent circuit of transformer is also required during calculation of % impedance.
Equivalent Circuit of Transformer Referred to Primary
For drawing equivalent circuit of transformer referred to primary, first we have to establish
general equivalent circuit of transformer then, we will modify it for referring from primary side.
For doing this, first we need to recall the complete vector diagram of a transformer which is
shown in the figure below
.
Let us consider the transformation ratio be,
In the figure above, the applied voltage to the primary is V1 and voltage across the primary
winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the primary is
partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding. The voltage
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appeared across winding is countered by primary induced emf E1. So voltage equation of this
portion of the transformer can be written as,
The equivalent circuit for that equation can be drawn as below,
From the vector diagram above, it is found that the total primary current I1 has two components,
one is no - load component Io and the other is load component I2′. As this primary current has
two components or branches, so there must be a parallel path with primary winding of
transformer. This parallel path of current is known as excitation branch of equivalent circuit of
transformer. The resistive and reactive branches of the excitation circuit can be represented as
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The load component I2′ flows through the primary winding of transformer and induced voltage
across the winding is E1 as shown in the figure right. This induced voltage E1 transforms to
secondary and it is E2 and load component of primary current I2′ is transformed to secondary as
secondary current I2. Current of secondary is I2. So the voltage E2 across secondary winding is
partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears across load. The load voltage is V2.
The complete equivalent circuit of transformer is shown below.
Now if we see the voltage drop in secondary from primary side, then it would be ′K′ times
greater and would be written as K.Z2.I2. Again I2′.N1 = I2.N2
Therefore,
From above equation, secondary impedance of transformer referred to primary is,
So, the complete equivalent circuit of transformer referred to primary is shown in the figure
below,
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Approximate Equivalent Circuit of Transformer
Since Io is very small compared to I1, it is less than 5% of full load primary current, Io changes
the voltage drop insignificantly. Hence, it is good approximation to ignore the excitation circuit
in approximate equivalent circuit of transformer. The winding resistance and reactance being in
series can now be combined into equivalent resistance and reactance of transformer, referred to
any particular side. In this case it is side 1 or primary side.
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Equivalent Circuit of Transformer Referred to Secondary
In similar way, approximate equivalent circuit of transformer referred to secondary can be
drawn.
Where equivalent impedance of transformer referred to secondary, can be derived as
Losses in Transformer:
Losses of transformer are divided mainly into two types:
1. Iron Loss
2. Copper Losses
IRON LOSS:
This is the power loss that occurs in the iron part. This loss is due to the alternating frequency of
the emf. Iron loss in further classified into two other losses.
a) Eddy current loss
b) Hysterisis loss
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a) Eddy Current Loss:
This power loss is due to the alternating flux linking the core, which will induced an emf in the
core called the eddy emf, due to which a current called the eddy current is being circulated in the
core. As there is some resistance in the core with this eddy current circulation converts into heat
called the eddy current power loss. Eddy current loss is proportional to the square of the supply
frequency.
b) Hysterisis Loss:
This is the loss in the iron core, due to the magnetic reversal of the flux in the core, which results
in the form of heat in the core. This loss is directly proportional to the supply frequency.
Eddy current loss can be minimized by using the core made of thin sheets of silicon steel
material, and each lamination is coated with varnish insulation to suppress the path of the eddy
currents. Hysterisis loss can be minimized by using the core material having high permeability.
COPPER LOSS:
This is the power loss that occurs in the primary and secondary coils when the transformer is on
load. This power is wasted in the form of heat due to the resistance of the coils. This loss is
proportional to the sequence of the load hence it is called the Variable loss where as the Iron loss
is called as the Constant loss as the supply voltage and frequency are constants
EFFICIENCY:
It is the ratio of the output power to the input power of a transformer
Input = Output + Total losses
= Output + Iron loss + Copper loss
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where,
Wcopper is the copper loss at full load
Wcopper = I2R watts
CONDITION FOR MAXIMUM EFFICIENCY:
In general for the efficiency to be maximum for any device the losses must be minimum.
Between the iron and copper losses the iron loss is the fixed loss and the copper loss is the
variable loss. When these two losses are equal and also minimum the efficiency will be
maximum.
Therefore the condition for maximum efficiency in a transformer is
Copper loss = Iron loss
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O.C. and S.C. Tests on Single Phase Transformer
The efficiency and regulation of a transformer on any load condition and at any power
factor condition can be predetermined by indirect loading method. In this method, the actual load
is not used on transformer. But the equivalent circuit parameters of a transformer are determined
by conducting two tests on a transformer which are,
1. Open circuit test (O.C Test)
2. Short circuit test (S.C.Test)
The parameters calculated from these test results are effective in determining the
regulation and efficiency of a transformer at any load and power factor condition, without
actually loading the transformer. The advantage of this method is that without much power loss
the tests can be performed and results can be obtained. Let us discuss in detail how to perform
these tests and how to use the results to calculate equivalent circuit parameters.
Open Circuit Test (O.C. Test)
The experimental circuit to conduct O.C test is shown in the Fig. 1.
Fig 1. Experimental circuit for O.C. test
The transformer primary is connected to a.c. supply through ammeter, wattmeter and
variac. The secondary of transformer is kept open. Usually low voltage side is used as primary
and high voltage side as secondary to conduct O.C test.
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The primary is excited by rated voltage, which is adjusted precisely with the help of a
variac. The wattmeter measures input power. The ammeter measures input current. The
voltemeter gives the value of rated primary voltage applied at rated frequency.
Sometimes a voltmeter may be connected across secondary to measure secondary voltage
which is V2 = E2 when primary is supplied with rated voltage. As voltmeter resistance is very
high, though voltmeter is connected, secondary is treated to be open circuit as voltmeter current
is always negligibly small.
When the primary voltage is adjusted to its rated value with the help of variac, readings of
ammeter and wattmeter are to be recorded.
Let,
Vo = Rated voltage
Wo = Input power
Io = Input current = no load current
As transformer secondary is open, it is on no load. So current drawn by the primary is no
load current Io. The two components of this no load current are,
Im = Io sin Φo
Ic = Io cos Φo
where cos Φo = No load power factor
And hence power input can be written as,
Wo = Vo Io cos Φo
The phasor diagram is shown in the Fig.
Fig.
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As secondary is open, I2 = 0. Thus its reflected current on primary is also zero. So we have
primary current I1 =Io. The transformer no load current is always very small, hardly 2 to 4 % of
its full load value. As I2 = 0, secondary copper losses are zero. And I1 = Io is very low hence
copper losses on primary are also very very low. Thus the total copper losses in O.C. test are
negligibly small. As against this the input voltage is rated at rated frequency hence flux density
in the core is at its maximum value. Hence iron losses are at rated voltage. As output power is
zero and copper losses are very low, the total input power is used to supply iron losses. This
power is measured by the wattmeter i.e. Wo. Hence the wattmeter in O.C. test gives iron losses
which remain constant for all the loads.
...
Wo = Pi = Iron losses
Calculations : We know that,
Wo = Vo Io cos Φ
cos Φo = Wo /(Vo Io ) = no load power factor
Once cos Φo is known we can obtain,
Ic = Io cos Φo
and
Im = Io sin Φo
Once Ic and Im are known we can determine exciting circuit parameters as,
Ro = Vo /Ic Ω
and
Xo = Vo /Im Ω
Key Point : The no load power factor cos Φo is very low hence wattmeter used must be low
power factor type otherwise there might be error in the results. If the meters are connected on
secondary and primary is kept open then from O.C. test we get Ro'and Xo' with which we can
obtain Ro and Xo knowing the transformation ratio K.
Short Circuit Test (S.C. Test)
In this test, primary is connected to a.c. supply through variac, ammeter and voltmeter as
shown in the Fig. 3.
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Experimental circuit for O.C. test
The secondary is short circuited with the help of thick copper wire or solid link. As high voltage
side is always low current side, it is convenient to connect high voltage side to supply and
shorting the low voltage side.
As secondary is shorted, its resistance is very very small and on rated voltage it may draw
very large current. Such large current can cause overheating and burning of the transformer. To
limit this short circuit current, primary is supplied with low voltage which is just enough to cause
rated current to flow through primary which can be observed on an ammeter. The low voltage
can be adjusted with the help of variac. Hence this test is also called low voltage test or reduced
voltage test. The wattmeter reading as well as voltmeter, ammeter readings are recorded.
Now the current flowing through the windings are rated current hence the total copper loss is full
load copper loss. Now the voltage supplied is low which is a small fraction of the rated voltage.
The iron losses are function of applied voltage. So the iron losses in reduced voltage test are very
small. Hence the wattmeter reading is the power loss which is equal to full load copper losses as
iron losses are very low.
...
Wsc = (Pcu) F.L. = Full load copper loss
Calculations : From S.C. test readings we can write,
Wsc = Vsc Isc cos Φsc
...
cos Φsc = Vsc Isc /Wsc = short circuit power factor
Wsc = Isc2 R1e = copper loss
...
R1e =Wsc /Isc2
while
Z1e =Vsc /Isc = √(R1e2 + X1e2)
...
X1e = √(Z1e2 - R1e2)
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Thus we get the equivalent circuit parameters R1e, X1e and Z1e. Knowing the transformation
ratio K, the equivalent circuit parameters referred to secondary also can be obtained.
Important Note : If the transformer is step up transformer, its primary is L.V. while secondary
is H.V. winding. In S.C. test, supply is given to H.V. winding and L.V is shorted. In such case
we connect meters on H.V. side which is transformer secondary through for S.C. test purpose
H.V side acts as primary. In such case the parameters calculated from S.C. test readings are
referred to secondary which are R2e, Z2e and X2e. So before doing calculations it is necessary to
find out where the readings are recorded on transformer primary or secondary and accordingly
the parameters are to be determined. In step down transformer, primary is high voltage itself to
which supply is given in S.C. test. So in such case test results give us parameters referred to
primary i.e. R1e, Z1e and X1e.
Key point : In short, if meters are connected to primary of transformer in S.C. test, calculations
give us R1e and Z1e if meters are connected to secondary of transformer in S.C. test calculations
give us R2e and Z2e.
Calculation of Efficiency from O.C. and S.C. Tests
We know that,
From O.C. test, Wo = Pi
From S.C. test, Wsc = (Pcu) F.L.
Thus for any p.f. cos Φ2 the efficiency can be predetermined. Similarly at any load which is
fraction of full load then also efficiency can be predetermined as,
where
n = fraction of full load
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where
I2= n (I2) F.L.
Calculation of Regulation
From S.C. test we get the equivalent circuit parameters referred to primary or secondary.
The rated voltages V1, V2 and rated currents (I1) F.L. and (I2) F.L. are known for the given
transformer. Hence the regulation can be determined as,
where I1, I2 are rated currents for full load regulation.
For any other load the currents I1, I2 must be changed by fraction n.
...
I1, I2 at any other load = n (I1) F.L., n (I2) F.L.
Key Point : Thus regulation at any load and any power factor can be predetermined, without
actually loading the transformer.
Example 1 : A 5 KVA, 500/250 V, 50 Hz, single phase transformer gave the following readings,
O.C. Test : 500 V, 1 A, 50 W (L.V. side open)
S.C. Test : 25 V, 10 A, 60 W (L.V. side shorted)
Determine : i) The efficiency on full load, 0.8 lagging p.f.
ii) The voltage regulation on full load, 0.8 leading p.f.
iii) The efficiency on 60% of full load, 0.8 leading p.f.
iv) Draw the equivalent circuit referred to primary and insert all the values in it.
Solution : In both the tests, meters are on H.V. side which is primary of the transformer. Hence
the parameters obtained from test results will be referred to primary.
From O.C. test,
Vo = 500 V,
Io = 1 A,
Wo= 50 W
...
cos Φo = Wo/Vo Io = 50/(500x1) = 0.1
...
Ic = Io cos = 1 x 0.1 = 0.1 A
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and
Im = Io sin Φo = 1 x 0.9949 = 0.9949 A
...
Ro =Vo /Ic = 500/0.1 = 5000 Ω
and
Xo = Vo/Im = 500/0.9949 = 502.52 Ω
and
Wo = Pi= iron losses = 50 W
From S.C. test,
Vsc = 25 V, Isc = 10 A, Wsc = 60 W
...
R1e = Wsc /Isc2 = 60/(10)2 = 0.6 Ω
Z1e = Vsc /Isc = 25/10 = 2.5 Ω
...
X1e= √(2.52 - 0.62) = 2.4269 Ω
(I1) F.L. = VA rating/V1
= (5 x 103 )/500 = 10 A
and
Isc = (I1) F.L.
...
Wsc = (Pcu) F.L. = 60 W
i) η on full load, cos = 0.8 lagging
ii) Regulation on full load, cos Φ2 = 0.8 leading
= - 1.95 %
iii) For 60% of full load, n = 0.6 and cos Φ2 = 0.8 leading]
...
Pcu = copper loss on new load = n2 x (Pcu) F.L.
= (0.6)2 x 60 = 21.6 W
= 97.103 %
iv) The equivalent circuit referred to primary is shown in the Fig. 4.
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Fig. 4
Example 2 : The open circuit and short circuit tests on a 10 KVA, 125/250 V, 50 Hz, single
phase transformer gave the following results :
O.C. test : 125 V, 0.6 A, 50 W (on L.V. side)
S.C. test : 15 V, 30 A. 100 W (on H.V. side)
Calculate : i) copper loss on full load
ii) full load efficiency at 0.8 leading p.f.
iii) half load efficiency at 0.8 leading p.f.
iv) regulation at full load, 0.9 leading p.f.
Solution : From O.C. test we can weite,
Wo = Pi = 50 W = Iron loss
From S.C. test we can find the parameters of equivalent circuit. Now S.C. test is conducted
on H.V. side i.e. meters are on H.V. side which is transformer secondary. Hence parameters from
S.C. test results will be referred to secondary.
Vsc = 15 V, Isc = 30 A, Wsc = 100 W
...
R2e = Wsc/(Isc)2 =10/(30)2 = 0.111Ω
Z1e = Vsc /Isc = 15/30 = 0.5 Ω
...
X2e = √(Z2e2 - R2e2) = 0.4875 Ω
i) Copper loss on full load
(I2) F.L. = VA rating/V2 = (10 x 103)/250 = 40 A
In short circuit test, Isc = 30 A and not equal to full load value 40 A.
Hence Wsc does not give copper loss on full load
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...
Wsc = Pcu at 30 A = 100 W
Now
Pcu α I2
( Pcu at 30 A)/( Pcu at 40 A) = (30/40) 2
100/( Pcu at 40 A) = 900/1600
Pcu at 40 A = 177.78 W
...
(Pcu) F.L. = 177.78 W
ii) Full load η , cos Φ2 = 0.8
iii) Half load η , cos Φ2 = 0.8
n = 0.5 as half load,
(I2) H.L. = 0.5 x 40 = 20
= 97.69%
iv) Regulation at full load, cos Φ = 0.9 leading
= -1.8015%
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Sumpner's Test Or Back-To-Back Test On Transformer
Sumpner's test or back to back test on transformer is another method for
determining transformer
efficiency,
voltage
regulation
and
heating
under
loaded
conditions. Short circuit and open circuit tests on transformer can give us parameters
of equivalent circuit of transformer, but they cannot help us in finding the heating information.
Unlike O.C. and S.C. tests, actual loading is simulated in Sumpner's test. Thus the Sumpner's test
give more accurate results of regulation and efficiency than O.C. and S.C. tests.
Sumpner's Test
Sumpner's test or back to back test can be employed only when two identical transformers are
available. Both transformers are connected to supply such that one transformer is loaded on
another. Primaries of the two identical transformers are connected in parallel across a supply.
Secondaries are connected in series such that emf's of them are opposite to each other. Another
low voltage supply is connected in series with secondaries to get the readings, as shown in the
circuit diagram shown below.
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In above diagram, T1 and T2 are identical transformers. Secondaries of them are connected in
voltage opposition, i.e. EEF and EGH. Both the emf's cancel each other, as transformers are
identical. In this case, as per superposition theorem, no current flows through secondary. And
thus the no load test is simulated. The current drawn from V1 is 2I0, where I0 is equal to no load
current of each transformer. Thus input power measured by wattmeter W1 is equal to iron losses
of both transformers.
i.e. iron loss per transformer Pi = W1/2.
Now, a small voltage V2 is injected into secondary with the help of a low voltage transformer.
The voltage V2 is adjusted so that, the rated current I2 flows through the secondary. In this case,
both primaries and secondaries carry rated current. Thus short circuit test is simulated and
wattmeter W2 shows total full load copper losses of both transformers.
i.e. copper loss per transformer PCu = W2/2.
From above test results, the full load efficiency of each transformer can be given as –
Predetermination of Voltage Regulation
Modern power systems operate at some standard voltages. The equipments working on these
systems are therefore given input voltages at these standard values, within certain agreed
tolerance limits. In many applications this voltage itself may not be good enough for obtaining
the best operating condition for the loads. A transformer is interposed in between the load and
the supply terminals in such cases. There are additional drops inside the transformer due to the
load currents. While input voltage is the responsibility of the supply provider, the voltage at the
load is the one which the user has to worry about.
If undue voltage drop is permitted to occur inside the transformer the load voltage becomes too
low and affects its performance. It is therefore necessary to quantify the drop that takes place
inside a transformer when certain load current, at any power factor, is drawn from its output
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leads. This drop is termed as the voltage regulation and is expressed as a ratio of the terminal
voltage (the absolute value per se is not too important).
The voltage regulation can be defined in two ways - Regulation Down and Regulation up. These
two definitions differ only in the reference voltage as can be seen below. Regulation down: This
is defined as ‖ the change in terminal voltage when a load current at any power factor is applied,
expressed as a fraction of the no-load terminal voltage.
Expressed in symbolic form we have,
Where,
Vnl is the no-load terminal voltage.
Vl is load voltage.
Normally full load regulation is of interest as the part load regulation is going to be lower.
This definition is more commonly used in the case of alternators and power systems as the userend voltage is guaranteed by the power supply provider. He has to generate proper no-load
voltage at the generating station to provide the user the voltage he has asked for. In the
expressions for the regulation, only the numerical differences of the voltages are taken and not
vector differences.
In the case of transformers both definitions result in more or less the same value for the
regulation as the transformer impedance is very low and the power factor of operation is quite
high. The power factor of the load is defined with respect to the terminal voltage on load. Hence
a convenient starting point is the load voltage. Also the full load output voltage is taken from the
name plate. Hence regulation up has some advantage when it comes to its application. Fig. 23
shows the phasor diagram of operation of the transformer under loaded condition. The no-load
current I0 is neglected in view of the large magnitude of I‘2.
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Fig. Regulation of Transformer
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Fig. Variation of full load regulation with power factor
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Predetermination of Efficiency
Transformers which are connected to the power supplies and loads and are in operation
are required to handle load current and power as per the requirements of the load. An unloaded
transformer draws only the magnetization current on the primary side, the secondary current
being zero. As the load is increased the primary and secondary currents increase as per the load
requirements. The volt amperes and wattage handled by the transformer also increases. Due to
the presence of no load losses and I2R losses in the windings certain amount of electrical energy
gets dissipated as heat inside the transformer. This gives rise to the concept of efficiency.
Efficiency of power equipment is defined at any load as the ratio of the power output to
the power input. Putting in the form of an expression,
While the efficiency tells us the fraction of the input power delivered to the load, the
deficiency focuses our attention on losses taking place inside transformer. As a matter of fact the
losses heat up machine. The temperature rise decides the rating of the equipment.
The temperature rise of the machine is a function of heat generated the structural
configuration, method of cooling and type of loading (or duty cycle of load). The peak
temperature attained directly affects the life of the insulations of the machine for any class of
insulation.
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Fig. Efficiency
A typical curve for the variation of efficiency as a function of output is given in Fig. The losses
that take place inside the machine expressed as a fraction of the input is sometimes termed as
deficiency. Except in the case of an ideal machine, a certain fraction of the input power gets lost
inside the machine while handling the power. Thus the value for the efficiency is always less
than one. In the case of a.c. machines the rating is expressed in terms of apparent power. It is
nothing but the product of the applied voltage and the current drawn. The actual power delivered
is a function of the power factor at which this current is drawn. As the reactive power shuttles
between the source and the load and has a zero average value over a cycle of the supply wave it
does not have any direct effect on the efficiency. The reactive power however increases the
current handled by the machine and the losses resulting from it. Therefore the losses that take
place inside a transformer at any given load play a vital role in determining the efficiency. The
losses taking place inside a transformer can be enumerated as below:
1. Primary copper loss
2. Secondary copper loss
3. Iron loss
4. Dielectric loss
5. Stray load loss
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These are explained in sequence below.
Primary and secondary copper losses take place in the respective winding resistances due to the
flow of the current in them
The primary and secondary resistances differ from their d.c. values due to skin effect and
the temperature rise of the windings. While the average temperature rise can be approximately
used, the skin effect is harder to get analytically. The short circuit test gives the value of Re
taking into account the skin effect.
The iron losses contain two components - Hysteresis loss and Eddy current loss. The
Hysteresis loss is a function of the material used for the core.
For constant voltage and constant frequency operation this can be taken to be constant. The eddy
current loss in the core arises because of the induced emf in the steel lamination sheets and the
eddies of current formed due to it. This again produces a power loss Pe in the lamination.
where t is the thickness of the steel lamination used. As the lamination thickness is much
smaller than the depth of penetration of the field, the eddy current loss can be reduced by
reducing the thickness of the lamination. Present day laminations are of 0.25 mm thickness and
are capable of operation at 2 Tesla. These reduce the eddy current losses in the core. This loss
also remains constant due to constant voltage and frequency of operation. The sum of hysteresis
and eddy current losses can be obtained by the open circuit test.
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The dielectric losses take place in the insulation of the transformer due to the large electric
stress. In the case of low voltage transformers this can be neglected. For constant voltage
operation this can be assumed to be a constant.
The stray load losses arise out of the leakage fluxes of the transformer. These leakage fluxes
link the metallic structural parts, tank etc. and produce eddy current losses in them. Thus they
take place ‘all round‘ the transformer instead of a definite place , hence the name ‘stray‘. Also
the leakage flux is directly proportional to the load current unlike the mutual flux which is
proportional to the applied voltage. Hence this loss is called ‘stray load‘ loss. This can also be
estimated experimentally. It can be modeled by another resistance in the series branch in the
equivalent circuit. The stray load losses are very low in air-cored transformers due to the absence
of the metallic tank
Thus, the different losses fall in to two categories Constant losses (mainly voltage dependant)
and Variable losses (current dependant). The expression for the efficiency of the transformer
operating at a fractional load x of its rating, at a load power factor of Ө2, can be written as
Here S in the volt ampere rating of the transformer (V‘2 I‘2 at full load), Pconst being constant
losses and Pvar the variable losses at full load.
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UNIT-IV
3-PHASE INDUCTION MOTORS
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4.1. Three Phase Induction Motor
The most common type of AC motor being used throughout the work today is the
"Induction Motor". Applications of three-phase induction motors of size varying from half a
kilowatt to thousands of kilowatts are numerous. They are found everywhere from a small
workshop to a large manufacturing industry.
The advantages of three-phase AC induction motor are listed below:
• Simple design
• Rugged construction
• Reliable operation
• Low initial cost
• Easy operation and simple maintenance
• Simple control gear for starting and speed control
• High efficiency.
Induction motor is originated in the year 1891 with crude construction (The induction machine
principle was invented by NIKOLA TESLA in 1888.). Then an improved construction with
distributed stator windings and a cage rotor was built.
The slip ring rotor was developed after a decade or so. Since then a lot of improvement has taken
place on the design of these two types of induction motors. Lot of research work has been carried
out to improve its power factor and to achieve suitable methods of speed control.
4.2 Types and Construction of Three Phase Induction Motor
Three phase induction motors are constructed into two major types:
1. Squirrel cage Induction Motors
2. Slip ring Induction Motors
4.2.1 Squirrel cage Induction Motors
(a) Stator Construction
The induction motor stator resembles the stator of a revolving field, three phase alternator. The
stator or the stationary part consists of three phase winding held in place in the slots of a
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laminated steel core which is enclosed and supported by a cast iron or a steel frame as shown in
Fig: 4.1(a).
The phase windings are placed 120 electrical degrees apart and may be connected in either star
or delta externally, for which six leads are brought out to a terminal box mounted on the frame of
the motor.When the stator is energized from a three phase voltage it will produce a rotating
magnetic field in the stator core.
Fig.4.1.
(b) Rotor Construction
The rotor of the squirrel cage motor shown in Fig: 4.1(b) contains no windings. Instead it is a
cylindrical core constructed of steel laminations with conductor bars mounted parallel to the
shaft and embedded near the surface of the rotor core.
These conductor bars are short circuited by an end rings at both end of the rotor core. In large
machines, these conductor bars and the end rings are made up of copper with the bars brazed or
welded to the end rings shown in Fig: 4.1(b).In small machines the conductor bars and end rings
are sometimes made of aluminium with the bars and rings cast in as part of the rotor core.
Actually the entire construction (bars and end-rings) resembles a squirrel cage, from which the
name is derived.
The rotor or rotating part is not connected electrically to the power supply but has voltage
induced in it by transformer action from the stator. For this reason, the stator is sometimes called
the primary and the rotor is referred to as the secondary of the motor since the motor operates on
the principle of induction and as the construction of the rotor with the bars and end rings
resembles a squirrel cage, the squirrel cage induction motor is used.
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The rotor bars are not insulated from the rotor core because they are made of metals having less
Resistance than the core. The induced current will flow mainly in them. Also the rotor bars are
usually not quite parallel to the rotor shaft but are mounted in a slightly skewed position. This
feature tends to produce a more uniform rotor field and torque. Also it helps to reduce some of
the internal magnetic noise when the motor is running.
(c) End Shields
The function of the two end shields is to support the rotor shaft. They are fitted with bearings and
Attached to the stator frame with the help of studs or bolts attention.
4.2.2 Slip ring Induction Motors
(a) Stator Construction
The construction of the slip ring induction motor is exactly similar to the construction of squirrel
cage induction motor. There is no difference between squirrel cage and slip ring motors.
(b) Rotor Construction

The rotor of the slip ring induction motor is also cylindrical or constructed of lamination.

Squirrel cage motors have a rotor with short circuited bars whereas slip ring motors have
wound rotors having "three windings" each connected in star.

The winding is made of copper wire. The terminals of the rotor windings of the slip ring
motors are brought out through slip rings which are in contact with stationary brushes as
shown in Fig.4.2.
Fig.4.2
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The Advantages of the Slipring Motor
• It has susceptibility to speed control by regulating rotor resistance.
• High starting torque of 200 to 250% of full load value.
• Low starting current of the order of 250 to 350% of the full load current.
Hence slip ring motors are used where one or more of the above requirements are to be met.
4.2.3 Comparison of Squirrel Cage and Slip Ring Motor
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4.3 Principle of Operation
The operation of a 3-phase induction motor is based upon the application of Faraday Law and the
Lorentz force on a conductor. The behaviour can readily be understood by means of the
following example.
Consider a series of conductors of length l, whose extremities are short-circuited by two bars A
and B (Fig.3.3 a). A permanent magnet placed above this conducting ladder, moves rapidly to
the right at a speed v, so that its magnetic field B sweeps across the conductors. The following
sequence of events then takes place.
1. A voltage E = Blv is induced in each conductor while it is being cut by the flux (Faraday law).
2. The induced voltage immediately produces a current I, which flows down the conductor
underneath the pole face, through the end-bars, and back through the other conductors.
3. Because the current carrying conductor lies in the magnetic field of the permanent magnet, it
experiences a mechanical force (Lorentz force).
4. The force always acts in a direction to drag the conductor along with the magnetic field. If the
conducting ladder is free to move, it will accelerate toward the right. However, as it picks up
speed, the conductors will be cut less rapidly by the moving magnet, with the result that the
induced voltage E and the current I will diminish. Consequently, the force acting on the
conductors wilt also decreases. If the ladder were to move at the same speed as the magnetic
field, the induced voltage E, the current I, and the force dragging the ladder along would all
become zero.
In an induction motor the ladder is closed upon itself to form a squirrel-cage (Fig.3.3b) and the
moving magnet is replaced by a rotating field. The field is produced by the 3-phase currents that
flow in the stator windings.
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Fig.4.3
SLIP
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Frequency of rotor current
Example
A 3-phase, 460 V, 100 hp, 60 Hz, four-pole
induction machine delivers rated output power at a slip of 0.05.
Determine the:
(a) Synchronous speed and motor speed.
(b) Speed of the rotating air gap field.
(c) Frequency of the rotor circuit.
(d) Slip rpm.
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(e) Speed of the rotor field relative to the (i) rotor structure. (ii)
Stator structure. (iii) Stator rotating field.
(f) Rotor induced voltage at the operating speed, if the
stator-to-rotor turns ratio is 1 : 0.5.
Solution
4.4 Rotating Magnetic Field and Induced Voltages
Consider a simple stator having 6 salient poles, each of which carries a coil having 5 turns
(Fig.4.4). Coils that are diametrically opposite are connected in series by means of three jumpers
that respectively connect terminals a-a, b-b, and c-c. This creates three identical sets of windings
AN, BN, CN, which are mechanically spaced at 120 degrees to each other. The two coils in each
winding produce magneto motive forces that act in the same direction.
The three sets of windings are connected in wye, thus forming a common neutral N. Owing to
the perfectly symmetrical arrangement, the line to neutral impedances are identical. In other
words, as regards terminals A, B, C, the windings constitute a balanced 3-phase system.
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For a two-pole machine, rotating in the air gap, the magnetic field (i.e., flux density) being
sinusoidally distributed with the peak along the centre of the magnetic poles. The result is
illustrated in Fig.4.5. The rotating field will induce voltages in the phase coils aa', bb', and cc'.
Expressions for the induced voltages can be obtained by using Faraday laws of induction.
Fig: 4.4 Elementary stator having terminals A, B, C connected to a 3-phase source (not shown).
Currents flowing from line to neutral are considered to be positive.
Fig: 4.5 Air gap flux density distribution.
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Let us consider that the phase coils are full-pitch coils of N turns (the coil sides of each phase are
180 electrical degrees apart as shown in Fig.4.5). It is obvious that as the rotating field moves (or
the magnetic poles rotate) the flux linkage of a coil will vary. The flux linkage for coil aa' will be
maximum.
Hence,
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Where f is the frequency in hertz. Above equation has the same form as that for the induced
voltage in transformers. However, ØP represents the flux per pole of the machine.
The above equation also shows the rms voltage per phase. The N is the total number of series
turns per phase with the turns forming a concentrated full-pitch winding. In an actual AC
machine each phase winding is distributed in a number of slots for better use of the iron and
copper and to improve the waveform. For such a distributed winding, the EMF induced in
various coils placed in different slots are not in time phase, and therefore the phasor sum of the
EMF is less than their numerical sum when they are connected in series for the phase winding. A
reduction factor KW, called the winding factor, must therefore be applied. For most three-phase
machine windings KW is about 0.85 to 0.95.
Therefore, for a distributed phase winding, the rms voltage per phase is
Erms = 4.44fNphφpKW
Where Nph is the number of turns in series per phase.
TORQUE EQUATION
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Starting torque
The torque developed by the motor at the instant of starting is called starting torque.
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Condition for maximum torque
Torque under running condition
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Condition for maximum torque under running condition
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Fig. Effect of rotor resistance on torque-speed characteristic
TORQUE – SPEED CHARACTERISTICS
For small values of slip s, the torque is directly proportional to s.
For large values of slip s, the torque is inversely proportional to s.
Fig. Complete torque-speed characteristic of a three phase induction machine
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Unit-V
SYNCHRONOUS MACHINES
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Synchronous machines
The synchronous motor and induction motor are the most widely used types of AC
motor. The difference between the two types is that the synchronous motor rotates in exact
synchronism with the line frequency. The synchronous motor does not rely on current induction
to produce the rotor's magnetic field. By contrast, the induction motor requires "slip", the rotor
must rotate slightly slower than the AC current alternations, to induce current in the rotor
winding. Small synchronous motors are used in timing applications such as in synchronous
clocks, timers in appliances, tape recorders and precision servomechanisms in which the motor
must operate at a precise speed; speed accuracy is that of the power line frequency, which is
carefully controlled in large interconnected grid systems.
These machines are commonly used in analog electric clocks, timers and other devices
where correct time is required. In high-horsepower industrial sizes, the synchronous motor
provides two important functions. First, it is a highly efficient means of converting AC energy to
work. Second, it can operate at leading or unity power factor and thereby provide power-factor
correction.
Synchronous motors fall under the more general category of synchronous machines which also
includes the synchronous generator. Generator action will be observed if the field poles are
"driven ahead of the resultant air-gap flux by the forward motion of the prime
mover". Motor action will be observed if the field poles are "dragged behind the resultant air-gap
flux by the retarding torque of a shaft load".
There are mainly two types of rotor used in construction of alternator,
1. Salient pole type.
2. Cylindrical rotor type.
CONSTRUCTION OF SALIENT POLE ROTOR MACHINES
The construction of a synchronous motor(with salient pole rotor) is as shown in the figure at left.
Just like any other motor, it consists of a stator and a rotor. The stator core is constructed with
thin silicon lamination and insulated by a surface coating, to minimize the eddy current and
hysteresis losses. The stator has axial slots inside, in which three phase stator winding is placed.
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The stator is wound with a three phase winding for a specific number of poles equal to the rotor
poles.
The rotor in synchronous motors is mostly of salient pole type. DC supply is given to the rotor
winding via slip-rings. The direct current excites the rotor winding and creates electromagnetic
poles. In some cases permanent magnets can also be used. The figure above illustrates
the construction of a synchronous motor very briefly.
Fig. Salient pole synchronous machinr
Working principle of salient pole synchronous machine
The stator is wound for the similar number of poles as that of rotor, and fed with three phase AC
supply. The 3 phase AC supply produces rotating magnetic field in stator. The rotor winding is
fed with DC supply which magnetizes the rotor. Consider a two pole synchronous machine as
shown in figure below.
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
Now, the stator poles are revolving with synchronous speed (lets say clockwise). If the
rotor position is such that, N pole of the rotor is near the N pole of the stator (as shown in
first schematic of above figure), then the poles of the stator and rotor will repel each other,
and the torque produced will be anticlockwise.

The stator poles are rotating with synchronous speed, and they rotate around very fast and
interchange their position. But at this very soon, rotor can not rotate with the same angle
(due to inertia), and the next position will be likely the second schematic in above figure. In
this case, poles of the stator will attract the poles of rotor, and the torque produced will be
clockwise.
Fig. Salient pole rotoe

Hence, the rotor will undergo to a rapidly reversing torque, and the motor will not start.
But, if the rotor is rotated upto the synchronous speed of the stator by means of an external force
(in the direction of revolving field of the stator), and the rotor field is excited near the
synchronous speed, the poles of stator will keep attracting the opposite poles of the rotor (as the
rotor is also, now, rotating with it and the position of the poles will be similar throughout the
cycle). Now, the rotor will undergo unidirectional torque. The opposite poles of the stator and
rotor will get locked with each other, and the rotor will rotate at the synchronous speed.
The salient features of pole field structure has the following special feature1. They have a large horizontal diameter compared to a shorter axial length.
2. The pole shoes covers only about 2/3rd of pole pitch.
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3. Poles are laminated to reduce eddy current loss.
4. The salient pole type motor is generally used for low speed operations of around 100 to 400
rpm, and they are used in power stations with hydraulic turbines or diesel engines.
Construction of round (or) cylindrical rotor synchronous machine
Fig. Round (or) cylindrical rotor
The cylindrical rotor is generally used for very high speed operation and are employed in steam
turbine driven alternators like turbo generators. The cylindrical rotor type machine has uniform
length in all directions, giving a cylindrical shape to the rotor thus providing uniform flux cutting
in all directions. The rotor in this case consists of a smooth solid steel cylinder, having a number
of slots along its outer periphery for hosing the field coils.
The cylindrical rotor alternators are generally designed for 2-pole type giving very high speed of
Ns = (120 × f)/P = (120 × 50) / 2 = 3000 rpm. Or 4-pole type running at a speed of Ns = (120 × f)
/ P = (120 × 50) / 4 = 1500 rpm. Where f is the frequency of 50 Hz.
The a cylindrical rotor synchronous generator does not have any projections coming out from the
surface of the rotor, rather central polar area are provided with slots for housing the field
windings as we can see from the diagram above. The field coils are so arranged around these
poles that flux density is maximum on the polar central line and gradually falls away as we move
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out towards the periphery. The cylindrical rotor type machine gives better balance and quieteroperation along with lesser windage losses.
Synchronous speed
Example
A 3-phase, 12-pole (6-pole-pair) synchronous motor is operating at an AC supply frequency of
50 Hz. The number of poles per phase is 12/3 = 4, so the synchronous speed is:
EMF Equation
Consider following
Φ= flux per pole in wb
P = Number of poles
Ns = Synchronous speed in rpm
f = frequency of induced emf in Hz
Z = total number of stator conductors
Zph = conductors per phase connected in series
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Tph = Number of turns per phase
Assuming concentrated winding, considering one conductor placed in a slot
According to Faraday's Law electromagnetic induction,
The average value of emf induced per conductor in one revolution
eavg = dΦ /dt
eavg = Change of Flux in one revolution/ Time taken for one revolution
Change of Flux in one revolution = p x Φ
Time taken for one revolution = 60/Ns seconds.
Hence eavg = (p x Φ ) / ( 60/Ns) = p x Φ x Ns / 60
We know f = PNs /120
hence PNs /60 = 2f
Hence eavg = 2 Φ f volts
Hence average emf per turn = 2 x 2 Φ f volts = 4Φf volts
If there are Tph, number of turns per phase connected in series, then average emf induced in Tph
turns is
Eph,avg = Tph x eavg = 4 f Φ Tph volts
Hence RMS value of emf induced E = 1.11 x Eph, avg
= 1.11 x 4 Φ f Tph volts
= 4.44 f Φ Tph volts
Eph,avg= 4.44 f Φ Tph volts
This is the general emf equation for the machine having concentrated and full pitched winding.In
practice, alternators will have short pitched winding and hence coil span will not
be 180o(degrees), but on or two slots short than the full pitch.
If we assume effect of
Kd= Distribution factor
Kc or KP = Cos α/2
Eph,avg= 4.44Kc Kd f Φ Tph volts
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This is the actual available voltage equation of an alternator per phase.If alternator or AC
Generator is Star Connected as usually the case, then the Line Voltage is √3 times the phase
voltage.
Voltage regulation by synchronous impedance method
This method is also called E.M.F. method of determining the regulation. The method requires
following data to calculate the regulation.
1. The armature resistance per phase (Ra).
2. Open circuit characteristics which is the graph of open circuit voltage against the field current.
This is possible by conducting open circuit test on the alternator.
3. Short circuit characteristics which is the graph of short circuit current against field current.
This is possible by conducting short circuit test on the alternator.
Let us see, the circuit diagram to perform open circuit as well as short circuit test on the
alternator. The alternator is coupled to a prime mover capable of driving the alternator at its
synchronous speed. The armature is connected to the terminals of a switch. The other terminals
of the switch are short circuited through an ammeter. The voltmeter is connected across the lines
to measure the open circuit voltage of the alternator.
The field winding is connected to a suitable d.c. supply with rheostat connected in series.
The field excitation i.e. field current can be varied with the help of this rheostat. The circuit
diagram is shown in the Fig.
Fig. Circuit diagram for open circuit and short circuit test on alternator
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Open Circuit Test
Procedure to conduct this test is as follows :
i) Start the prime mover and adjust the speed to the synchronous speed of the alternator.
ii) Keeping rheostat in the field circuit maximum, switch on the d.c. supply.
iii) The T.P.S.T switch in the armature circuit is kept open.
iv) With the help of rheostat, field current is varied from its minimum value to the rated value.
Due to this, flux increasing the induced e.m.f. Hence voltmeter reading, which is measuring line
value of open circuit voltage increases. For various values of field current, voltmeter readings are
observed.
Note : This is called open circuit characteristics of the alternator, called O.C.C. This is shown in
the Fig.
Fig. O.C.C. and S.C.C. of an alternator
Short Circuit Test
After completing the open circuit test observation, the field rheostat is brought to maximum
position, reducing field current to a minimum value. The T.P.S.T switch is closed. As ammeter
has negligible resistance, the armature gets short circuited. Then the field excitation is gradually
increased till full load current is obtained through armature winding. This can be observed on the
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ammeter connected in the armature circuit. The graph of short circuit armature current against
field current is plotted from the observation table of short circuit test. This graph is called short
circuit characteristics, S.C.C. This is also shown in the Fig.
The S.C.C. is a straight line graph passing through the origin while O.C.C. resembles B-H
curve of a magnetic material.
Note : As S.C.C. is straight line graph, only one reading corresponding to full load armature
current along with the origin is sufficient to draw the straight line.
Determination of From O.C.C. and S.C.C.
The synchronous impedance of the alternator changes as load condition changes. O.C.C. and
S.C.C. can be used to determine Zs for any load and load p.f. conditions.
In short circuit test, external load impedance is zero. The short circuit armature current is
circulated against the impedance of the armature winding which is Zs. The voltage responsible
for driving this short circuit current is internally induced e.m.f. This can be shown in the
equivalent circuit drawn in the Fig. 5.3.
Fig. Equivalent circuit on short circuit
From the equivalent circuit we can write,
Zs = Eph/ Iasc
Now value of Iasc is known, which can observed on the alternator. But internally induced
e.m.f. can not be observed under short circuit condition. The voltmeter connected will read zero
which is voltage across short circuit. To determine Zs it is necessary to determine value of E
which is driving Iasc against Zs.
Now internally induced e.m.f. is proportional to the flux i.e. field current If.
Eph α Φ α If
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So if the terminal of the alternator are opened without disturbing If which was present at the
time of short circuited condition, internally induced e.m.f. will remain same as Eph. But now
current will be zero. Under this condition equivalent circuit will become as shown in the Fig.
Fig.
It is clear now from the equivalent circuit that as Ia = 0 the voltmeter reading (Voc)ph will be
equal to internally induced e.m.f. (Eph).
This is what we are interested in obtaining to calculate value of Zs. So expression for Zs can be
modified as,
So O.C.C. and S.C.C. can be effectively to calculate Zs.
The value of Zs is different for different values of If as the graph of O.C.C. is non linear in
nature.
So suppose Zs at full load is required then,
Iasc = full load current.
From S.C.C. determine If required to drive this full load short circuit Ia. This is equal to
'OA', as shown in the Fig.2.
Now for this value of If, (Voc)ph can be obtained from O.C.C. Extend kine from point A, till
it meets O.C.C. at point C. The corresponding (Voc)ph value is available at point D.
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(Voc)ph = OD
While (Iasc)ph = OE
at full load
General steps to determine Zs at any load condition are :
i) Determine the value of (Iasc)ph for corresponding load condition. This can be determined from
known full load current of the alternator. For half load, it is half of the full load value and so on.
ii) S.C.C. gives relation between (Iasc)ph and If. So for (Iasc)ph required, determine the
corresponding value of If from S.C.C.
iii) Now for this same value of If, extend the line on O.C.C. to get the value of (Voc)ph. This is
(Voc)ph for same If, required to drive the selected (Iasc)ph.
iv) The ratio of (Voc)ph and (Iasc)ph, for the same excitation gives the value of Zs at any load
conditions.
The graph of synchronous impedance against excitation current is also shown in the Fig. 2.
Regulation Calculations
From O.C.C. and S.C.C., Zs can be determined for any load condition.
The armature resistance per phase (Ra) can be measured by different methods. One of the
method is applying d.c. known voltage across the two terminals and measuring current. So value
of Ra per phase is known.
So synchronous reactance per phase can be determined.
No load induced e.m.f. per phase, Eph can be determined by the mathematical expression
derived earlier.
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where
Vph = Phase value of rated voltage
Ia = Phase value of current depending on the load condition
cosΦ = p.f. of load
Positive sign for lagging power factor while negative sign for leading power factor, Ra and
Xs values are known from the various tests performed.
The regulation then can be determined by using formula,
Advantages and Limitations of Synchronous Impedance Method
The main advantages of this method is the value of synchronous impedance Zsfor any load
condition can be calculated. Hence regulation of the alternator at any load condition and load
power factor can be determined. Actual load need not be connected to the alternator and hence
method can be used for very high capacity alternators.
The main limitation of this method is that the method gives large values of synchronous
reactance. This leads to high values of percentage regulation than the actual results. Hence this
method is called pessimistic method.
Theory of operation of synchronous motor
Electrical motor in general is an electro-mechanical device that converts energy from electrical
domain to mechanical domain. Based on the type of input we have classified it into single phase
and 3 phase motors. Among 3 phase induction motors and synchronous motors are more widely
used. When a 3 phase electric conductors are placed in a certain geometrical positions (In certain
angle from one another) there is an electrical field generate. Now the rotating magnetic field
rotates at a certain speed, that speed is called synchronous speed. Now if an electromagnet is
present in this rotating magnetic field, the electromagnet is magnetically locked with this rotating
magnetic field and rotates with same speed of rotating field.
Synchronous motors is called so because the speed of the rotor of this motor is same as the
rotating magnetic field. It is basically a fixed speed motor because it has only one speed, which
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is synchronous speed and therefore no intermediate speed is there or in other words it’s in
synchronism with the supply frequency. Synchronous speed is given by
Construction of Synchronous Motor
Normally it's construction is almost similar to that of a 3 phase induction motor, except the fact
that the rotor is given dc supply, the reason of which is explained later. Now, let us first go
through the basic construction of this type of motor.
From the above picture, it is clear that how this type of motors are designed. The stator is given
is given three phase supply and the rotor is given dc supply.
Main Features of Synchronous Motors
1. Synchronous motors are inherently not self starting. They require some external means to
bring their speed close to synchronous speed to before they are synchronized.
2. The speed of operation of is in synchronism with the supply frequency and hence for
constant supply frequency they behave as constant speed motor irrespective of load
condition
3. This motor has the unique characteristics of operating under any electrical power factor.
This makes it being used in electrical power factor improvement.
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Principle of Operation Synchronous Motor
Synchronous motor is a doubly excited machine i.e two electrical inputs are provided to it.
It’s stator winding which consists of a 3 phase winding is provided with 3 phase supply and rotor
is provided with DC supply. The 3 phase stator winding carrying 3 phase currents produces 3
phase rotating magnetic flux. The rotor carrying DC supply also produces a constant flux.
Considering the frequency to be 50 Hz, from the above relation we can see that the 3 phase
rotating flux rotates about 3000 revolution in 1 min or 50 revolutions in 1 sec. At a particular
instant rotor and stator poles might be of same polarity (N-N or S-S) causing repulsive force on
rotor and the very next second it will be N-S causing attractive force. But due to inertia of the
rotor, it is unable to rotate in any direction due to attractive or repulsive force and remain in
standstill condition. Hence it is not self starting. To overcome this inertia, rotor is initially fed
some mechanical input which rotates it in same direction as magnetic field to a speed very close
to synchronous speed. After some time magnetic locking occurs and the synchronous motor
rotates in synchronism with the frequency.
Methods of Starting of Synchronous Motor
1. Synchronous motors are mechanically coupled with another motor. It could be either 3
phase induction motor or DC shunt motor. DC excitation is not fed initially. It is rotated at
speed very close to its synchronous speed and after that DC excitation is given. After some
time when magnetic locking takes place supply to the external motor is cut off.
2. Damper winding : In case, synchronous motor is of salient pole type, additional winding is
placed in rotor pole face. Initially when rotor is standstill, relative speed between damper
winding and rotating air gap flux in large and an emf is induced in it which produces the
required starting torque. As speed approaches synchronous speed , emf and torque is
reduced and finally when magnetic locking takes place, torque also reduces to zero. Hence
in this case synchronous is first run as three phase induction motor using additional winding
and finally it is synchronized with the frequency.
Applications of synchronous motors
Synchronous motors are especially useful in applications requiring precise speed and/or position
control.
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
Speed is independent of the load over the operating range of the motor.

Speed and position may be accurately controlled using open loop controls, e.g. stepper
motors.

Low-power applications include positioning machines, where high precision is required,
and robot actuators.

They will hold their position when a DC current is applied to both the stator and the rotor
windings.

A clock driven by a synchronous motor is in principle as accurate as the line frequency of its
power source. (Although small frequency drifts will occur over any given several hours, grid
operators actively adjust line frequency in later periods to compensate, thereby keeping
motor-driven clocks accurate (see Utility frequency#Stability).)

Record player turntables

Increased efficiency in low-speed applications (e.g. ball mills)
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