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EEC 233- Electrical machine II

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UNESCO-NIGERIA TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERING TECHNOLOGY
C
A
Ic
B
Ia
Ib
N
ELECTRICAL MACHIENS I I
COURSE CODE: EEC233
YEAR II- SEMESTER III
THEORY
Version 1: December 2008
1
Table of Contents
Chapter 1: Basic Principles of electric machines: ............................... keeW1
1.1 Introduction
.........................................................................................1
1.2 Electro- mechanical energy convertion ........................................... 2
1.3 Alignment devices .................................................................................. 6
1.4 Interaction devices ................................................................................ 7
1.5 Induction devices ................................................................................... 7
1.6 Work Examples ....................................................................................... 8
1.7Electromagnets ...................................................................................... 11
1.8 Faraday's Law ........................................................................................ 12
1.9Lenz's Law ............................................................................................... 13
1.10 Rotating magnetic field ............................................................ Week2
1.11 Synchronous speed: .......................................................................... 14
Chapter 2 Three phase induction motor ....................................... Week3
2.1 Introduction ........................................................................................... 17
2.2 Construction of Induction Motor ...................................................... 18
2.2 The principle of operation of the Induction Motor ............. Week4
2.3 The slip ................................................................................................... 20
2.4 Name Plate............................................................................................. 22
Chapter 3: Synchronous Machine: ....................................................... keeW5
3.1 Introduction ........................................................................................... 18
3.2 Stator construction .............................................................................. 18
3.3 Rotor construction ............................................................................... 18
3.4 Principle of operation of the synchronous generator ................ 20
3.5 Principle of operation of the synchronous motor: ...................... 20
3.6 Excitation Methods .............................................................................. 21
Chapter4:Control & protection of electric motors ............................. keeW6
4.1 Need for Circuit Protection ............................................................... 22
4.2 Types of Overcurrent Protective Devices ............................. Week7
4.2.0 Protection and control devices............................................ Week8
4.2.1 Fuses .................................................................................................... 25
4.2.2 Circuit Breakers ................................................................................ 27
4.3 Control devices ............................................................................. Week9
4.3.1 Relay ................................................................................. 30
4.3.2 Contactors ....................................................................... 31
4.3.3Pushbuttons ............................................................Week10
4.3.4 Selector Switches ........................................................... 35
4.3.5 Indicator Lights .............................................................. 35
Chapter 5: Energy convertion ............................................................. keeW11
5.1 Electro-mechanical energy convertion .......................................... 37
5.1.1Major energy convertion principle ........................................ 38
5.2 Energy convertion ................................................................................ 39
5.3 Linked energy system ............................................................................ 40
5.4 Energy storage ...................................................................................... 41
5.5 Energy ballance ......................................................................... Week12
5..5.1 Block diagram of energy balance equation ......................... 43
5.6 Magnetic field energy and force ........................................... Week13
5.6.1 Magnetic field .............................................................................. 45
5.6.2 Magnetic circuit ......................................................................... 46
5.7 Magnetic field energy............................................................................ 47
5.8 Maxwell stress .................................................................................. 48
5.9 Energy density ........................................................................... Week14
5.10 Faraday's Lenz Law ............................................................................. 49
5.11 Energy Convertion ............................................................................. 50
5.11.1 Constant current ..................................................................... 51
5.11.2 Constant flux ........................................................................... 52
5.11.3 General Condition .................................................................. 53
5.11.4 Diffirencial foam ..................................................................... 54
5.12 Alignment force and torque; single excitation ....................Week15
5.11 Work examples ................................................................................... 56
5.11.3 General Condition
1. Basic Principles of Electric Machines
Week 1
1.1 Introduction
It may be necessary to define what we mean by the term electrical machines. A machine is a
device that does useful work in a predictable way according to some physical laws. It acts as
a transducer, or convertors, accepting an input of energy in one physical from and
transforming it, more or less effectively, into another.
An electromagnetic machine, in the essential conversion process, uses energy in an
intermediate magnetic form. As a motor the machine takes in electrical energy and converts
it into mechanical work, such as driving a machine tool or a lift, or operating a loudspeaker.
An electro-magnetic machine is usually reversible and cab, as a generator, producer
electrical energy form some other kind, such as the mechanical energy of prime- movers or
the a caustic energy of microphones and gramophone pickups.
Electrical energy is versatile and controllable. Its special lie in that can be transfer
continuously and economically from to place (Transmission), made widely available as a
services (distribution), used in conveying intelligence (Telecommunication and data
processing), and applied to indicate an supervise production systems (control,
instrumentation and computation). It is readily converted into sound, light, heat and useful
forms of energy. In particular it is easily converted to or from mechanical energy in the
electromagnetic machines
1
1. Basic Principles of Electric Machines
Week 1
1.2 Electro-mechanical energy conversion
1.2.1 Principal of electrical machines
The operation of electromagnetic mechanical devices can be explained in terms of basic
principals concerned with
i
The development of magneto-mechanical forces and
ii.
The induction of emf (electromotive force) by the rate of change of the linkage.
Thus, electromagnetic energy conversion is based on three bask principles namely (i)
induction (ii) interaction and (iii) alignment
1.
Principle of induction
It is known that when electrons are in motion, they produce a magnetic field. Conversely,
when, a magnetic field embracing a conductor moves relative to the conductor, it produce a
flow of electrons in the conductor.
The phenomenon whereby on e.m.f and hence current (i.e flow of electrons) is induced in
any conductor which is cut across or is cut by a magnetic flux is known as electromagnetic
induction
The induced emf E is given by
(a)
E = NdØ OR (B)
e = Bluw ……..volts
dt
Where N = number of turns of the coil
Ø = flux in webers linking the coil
T = time in seconds
2
1. Basic Principles of Electric Machines
Week 1
This is the equation for the induced emf when the magnetic flux moves relatively to the
conductor.
In equation 1.1(b), B = flux density in wb/m2
L = effective length of conductor in metre , U = velocity of the conductor in m/s
And this is the equation for the induced emf when the conductor moves relatively to the flux.
The induction principle is employed in devices such as induction motors generators,
transformers, controlling instrument etc.
1.2.2 Sketchmatic explanation of the induction principle
Fig:1.1a. Voltage & Current induced in the secondary Fig: 1.1b Conductor stationary, while the
circuit due to flux linkage with the primary winding.
field moves (current will be induced on the
galvanometer)
Fig: 1.1c Conductor moves, while the field stationary (current will be induced on the
galvanometer)
3
1. Basic Principles of Electric Machines
Week 1
Fig. 1.1 (a) shows an irobn- cored solenoid with a permanent magnetic place adjascent ot it. If the
magnet‟s position is changed from position CD to position AB, the flux linking with the
coils of the solenoid will change, leading to an induced emf in the coil which can be
detected by the sensitive galvanometer G. in this arrangement, the conductor (coil) i
stationary whilst the filed (magnet) moves as in alternators i.e a.c generators.
Fig 1.1 (b) shows a filed arrangement that is stationary while the conduct a-b is free to move about
the vertical axis. An emf, detectable by galvanometer G, will be induced in the conductor as
it cuts through the flux through the flux. This principle is employed in the construction of
d.c. generator,.
F ig 1.1(c) when a coil (Ni) is made to carry an alternative current (ii) it produces an alternative flux
(g). if a second coil (N2) is now placed in a region whereby the alternative flux produced by
the first coil links with the second coil, an emf ( usually of the some frequency) will be
induced in the second coil. This is the principle of the transformer and the induction motors
2. Principle of interaction
An electric current flowing in a direction making an angle (preferably a right-angle) with a
magnetic filed produced by another current ( or a magnet) experience a force fe, the relative
direction being shown in Fig 1.2.
Fig: 1.2 Principle of interaction.
4
1. Basic Principles of Electric Machines
Week 1
The force, fe, arises from the interaction of the flux (created by the current l‟ flowing in the
conductor with the flux produced by a second current or magnet. Since lines of flux do not
cross, the two fluxes will realign. Resulting in a stronger fie ld one side. The conductor and
weaker filed on t6he other side. The conductor then tends to move from the region of
stronger field to the region of weaker filed. Employed in electric motors.
3. Principle of alignment
A pieces of ferromagnetic materials in a magnetic field experience of force urging it towards
a region where the field is stronger, or tending to align it so as to shorten the magnetic flux
path as shown in fig: 1.3
Fig: 1.3a Moving coil meter
Fig: 1.3b The force „fe‟ on shaped high
permeability pieces in a field
Fig: 1.3c Polar attraction & repulsion on separately magnetized bodies
5
1. Basic Principles of Electric Machines
1.3
(a)
Lifting magnet
(f)
(b)
Relay
(g)
Electromagnetic pump
( c)
Telephone
(h)
Loudspeaker
(d)
Moving-iron indicator (i)
Moving –coil indicator
(e)
Reluctance motor
(k)
Week 1
Actuator
Industrial rotating machine
Alignment devices
(a)
The lifting magnet: Attract ferromagnetic loads such as beams, plates, and scrapiron.
(b)
The relay: the coil current causes the armature to be attracted towards the cover
against a spring load: Millions of such relays do useful work in automatic telephone
exchanges, traffic light installation and simple control systems,.
(c)
The telephone receivers: has a ferromagnetic diaphragm attracted by a permanent
magnet, the field is caused to fluctuated by the speech currents in the coil, so varying
the deflection of the claptrap and producing sound waves in the air.
(d)
The moving- iron indicator, uses the force between the fixed and moving irons to
deflect a pointed against a spring.
(e)
The Reluctance motor- the forces urge a displaced rotor in alignment with the
magnetized stator.
(f)
The actuator-the current-carrying coil “suck” a displaced ferromagnetic plunger into
a position of symmetry: this is a useful and forceful device.
6
1. Basic Principles of Electric Machines
1.4
Week 1
Interaction devices
(g)
Electromagnetic pump: current passed through a conducting liquid in an enclosed
channel forces the liquid to move by interaction with a magnetic cross field; liquid
sodium-potassium or lithium can be pumped in thy way for the extraction of heat
from a nuclear reactor.
(h)
Loudspeaker: alternating current in the coil flow in the radial magnet filed of the port
magnet,. And the consequent movement of the attached diaphragm sets up sound
waves. This is the same essential arrangement as a “generator‟ of mechanical
vibrations
(i)
Moving –coil indicator-current (normally direct) in the coil of the indirect develops a
force in the radial permanent- magnet filed to move pointed against a control spring.
(k)
Industrial rotating machines: Current caused to flow in conductors the surface of a
rotor, mounted within a magnetic stator develop interaction forces tending to turn the
rotor.
1.5
Induced voltage devices
Recalling that a conductor moving or cutting magnetic lines of flux or that the flux
moves relative to the conductor will proan induced voltage, the following devices
employ the induced voltage arrangement.
(i)
The transformer- an alternating current flowing in the primary coil (winding) set up
an alternating flux that links with the secondary coil inducing a voltage in the latter.
(m)
The generatopr-usually constructed like (k) but with the rotor mechanical energy (via
the prime –mover) will have emf induced in the stator coils. ( the stator is slotted to
house conductors)
7
1. Basic Principles of Electric Machines
(n)
Week 1
The induction motor- the stator usually carried one or htree –phawindings whilst the
rotor may have a similar arrangement of coil as the stator or just carry or alirmium
bars. Electrical energy supplied to the stator windings produced a rotating magnetic
field with cuts the rotor conductors and hence induced voltages in them. A complete
rotor circult will have current flowing in the rotor conductors (caused by the induced
voltage) and by interaction forces produced motion of the rotor.
1.6 Work Examples
Examples 1
A conductor carries a current of 800 A at right- angle to magnetic field having a density of
0.5wb/m2 calculated the force on a metre length of the conductor.
Solution
The force F is given by
F = Bli
= 0.5 X 1 X 800
= 400N
Example 2
A four –pole generator has a magnetic flux of 12 mnb /pole calculated the average value of
the emf generated in one of the armature conductors while
it is moving through the
magnetic flux of one pole, if armature is driven at 900 r.p.m
Solution
When a conductor moves through the magnetic field of one pole, it cuts a magnetic flux of
12 x 10-3 wb.
Time taken for a conductor to move through one revolution
=
60
=
1 second
8
1. Basic Principles of Electric Machines
900
Week 1
15
Since the machine has 4 poles, time taken for a conductor to move through the field of one
pole = ¼ x 1/15 = 1/60s
:. Average emf generated in one conductor rate of change of flux
=
Ø
=
12 x 10-3
=
0.012 – 0.01667 = 0.72v
=
0.72v
1/60
t
Example 4
A magnetic flux of 400 uwb passing through a coil of 1200 turns is reversed in 0.1s calculate
the average emf induced in the coil.
Solution
The magnetic flux has to decrease form 400 uwb to zero and then increase to 400wwb in the
reverse direction, hence the increase of flux is 400 (-400-400) uwb = -800 x 10-6 wb.
:. Average emf induced in coil
= (change in flux x No of turns) = NdØ
Time taken.
Dt
9
1. Basic Principles of Electric Machines
Week 1
1.7 Electromagnets
Anything with an electrical current running through it has a magnetic field. Figure1 shows different
sources of magnetic field.
Figure1.4: Different sources of magnetic field
The most common forms of electromagnets are the Solenoids .When the wire is shaped into a coil
as shown in Figure1.1, all the individual flux lines produced by each section of wire join together to
form one large magnetic field around the total coil.
As with the permanent magnet, these flux lines leave the north of the coil and re-enter the coil at its
south pole. The magnetic field of a wire coil is much greater and more localized than the magnetic
field around the plain conductor before being formed into a coil. This magnetic field around the coil
can be strengthened even more by placing a core of iron or similar metal in the center of the core.
The metal core presents less resistance to the lines of flux than the air, thereby causing the field
strength to increase. (This is exactly how a stator coil is made; a coil of wire with a steel core.) The
advantage of a magnetic field which is produced by a current carrying coil of wire is that when the
current is reversed in direction the poles of the magnetic-as shown in Figure1.2- field will switch
positions since the lines of flux have changed direction. Without this magnetic phenomenon
existing, the AC motor as we know it today would not exist.
10
1. Basic Principles of Electric Machines
Week 1
Iron Core
N
Magnetic field Lines
S
The Current
Battery
S
N
Figure1.5: Reversing the polarity of the solenoid by reversing the current direction
1.8 Faraday's Law
Faradays law states whenever the magnetic flux linked with a circuit changes, an e.m.f. is always
induced in it, or Whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor.
The phenomenon of inducing a current by changing the magnetic field in a coil of wire is known as
electromagnetic induction.
Figure1.3 shows an electromagnet which is connected to an AC power source. Another
electromagnet is placed above it. The second electromagnet is in a separate circuit. There is no
physical connection between the two circuits. Voltage and current are zero in both circuits at Time1.
At Time2 voltage and current are increasing in the bottom circuit
11
1. Basic Principles of Electric Machines
0
Time1
Week 1
0
0
Time2
Time3
Figure 1.6: Experment showing the electromagnetic induction phenomena
A magnetic field builds up in the bottom electromagnet. Lines of flux from the magnetic field
building up in the bottom electromagnet cut across the top electromagnet. A voltage is induced in
the top electromagnet and current flows through it. At Time 3 current flow has reached its peak.
Maximum current is flowing in both circuits. The magnetic field around the coil continues to build
up and collapse as the alternating current continues to increase and decrease. As the magnetic field
moves through space, moving out from the coil as it builds up and back towards the coil as it
collapses, lines of flux cut across the top coil. As current flows in the top electromagnet it creates its
own magnetic field.
1.9 Lenz's Law
Lenz's law enables us to determine the direction of the induced current: "The direction of the
induced current is such as to oppose the change causing it." The Figure 1.4a shows the north pole of
a bar magnet approaching a solenoid. According to Lenz's law, the current which is thereby
generated in the coil must cause an effect which opposes the approaching magnetic field.
12
1. Basic Principles of Electric Machines
N
Week 1
N
Iron Core
Ammeter
a
N
N S
b
Figure1.7: Experiment demonstrating Lenz's law
This is achieved if the direction of the induced current creates a north pole at the end of the
solenoid closest to the approaching magnet, as the induced north pole tends to repel the approaching
north pole. The Figure1.4b shows the north pole of a bar magnet withdrawing from a solenoid.
According to Lenz's law, the current which is thereby generated in the coil must cause an effect
which opposes the departing magnetic field. This is achieved if the direction of the induced current
creates a south pole at the end of the solenoid closest to the departing magnet, as the induced south
pole tends to attract the departing north pole
13
1.
Basic Principles of Electric Machines
Week Two
1.4 Rotating magnetic field
The three-phase induction motor also operates on the principle of a rotating magnetic
field. The following discussion shows how the stator windings can be connected to a threephase ac input and have a resultant magnetic field that rotates.
Figure 1.5 shows how the three phases are tied together in a Y-connected stator. The dot in
each diagram indicates the common point of the Y-connection. You can see that the
individual phase windings are equally spaced around the stator. This places the windings
120° apart.
C
A
B
Ic
Ia
Ib
N
Figure 1.5:- Three-phase, Y-connected stator.
Using the left-hand rule the electromagnetic polarity of the poles can be determined at any
given instant.
5
1.
Basic Principles of Electric Machines
Week Two
The results of this analysis are shown for voltage points 1 through 7 in figure 2. At point 1,
the magnetic field in coils A is maximum with polarities as shown. At the same time,
negative voltages are being felt in the B and C windings. These create weaker magnetic
fields, which tend to aid the A field. At point 2, maximum negative voltage is being felt in
the C windings. This creates a strong magnetic field which, in turn, is aided by the weaker
fields in A and B. As each point on the voltage graph is analyzed, it can be seen that the
resultant magnetic field is rotating in a clockwise direction. When the three-phase voltage
completes one full cycle (point 7), the magnetic field has rotated through360°.
6
1.
Basic Principles of Electric Machines
Week Two
10
5
-5
-10
1
Point1
Point4
2
3
4
5
6
7
Point2
Point3
Point 5
Point6
Point7
5
1. Basic Principles of Electric Machines
Week 2
1.5 Synchronous speed:
The speed of the rotating magnetic field is referred to as synchronous speed (Ns). Synchronous
speed is equal to 120 times the frequency (f), divided by the number of poles (P).
Ns 
120 f
p
If the frequency of the applied power supply for the two-pole stator used in the previous example is
50 Hz, synchronous speed is 3000 RPM.
Ns 
120  50
 3000 RPM
2
The synchronous speed decreases as the number of poles increase. The following table shows the
synchronous speed at 50 Hz for the corresponding number of poles.
No of poles
Synchronous speed
2
3000rpm
4
1500rpm
6
1000rpm
8
750rpm
Table 1.1: Different speeds for different number of poles
6
2. Three phase induction motor
Week 3
2.1 Introduction
Three-phase AC induction motors are widely used in industrial and commercial applications. They
are classified either as squirrel cage or wound-rotor motors.
These motors are self-starting and use no capacitor, start winding, centrifugal switch or other
starting device.
They produce medium to high degrees of starting torque. The power capabilities and efficiency in
these motors range from medium to high compared to their single-phase counterparts.
Popular applications include grinders, lathes, drill presses, pumps, compressors, conveyors, also
printing equipment, farm equipment, electronic cooling and other mechanical duty applications.

Simple and rugged construction

Low cost and minimum maintenance

High reliability and sufficiently high efficiency since there is no losses in brush contacts
or mechanical friction

Needs no extra starting motor and need not be synchronized

Need only one source of power
2.2 Construction of Induction Motor:
An Induction motor has basically two parts, Stator and Rotor. Also some of other parts were
acknowledged in the following section
1
2. Three phase induction motor
Week 3
2.2.1 Stator construction:
Figure2.1: Diagram construction of the stator
The stator is made up of several thin laminations of aluminum or cast iron. They are punched and
clamped together to form a hollow cylinder (stator core) with slots as shown in Figure 2.1. Coils of
insulated wires are inserted into these slots.
The iron core on the figure has paper liner insulation placed in some of the slots.
Each grouping of coils, together with the core it surrounds, forms an electromagnet (a pair of poles)
on the application of AC supply. The number of poles of an AC induction motor depends on the
internal connection of the stator windings. The stator windings are connected directly to the power
source. Internally they are connected in such a way, that on applying AC supply, a rotating magnetic
field is created
2.2.2 Rotor construction:
There are two main types
2
2. Three phase induction motor


Week 3
Squirrel cage type
Wound rotor
2.2.2.1 Squirrel cage rotor:
This rotor has a laminated iron core with slots(Figure2.2), and is mounted on a shaft. Aluminum
bars are molded in the slots and the bars are short circuited with two end rings. The bars are skewed
on a small rotor to reduce audible noise. Fins are placed on the ring that shorts the bars. These fins
also work as a fan and improve cooling.
Most motors use the squirrel-cage rotor because There are no commutators, slip rings or brushes.
Hence this is a most rugged and maintenance-free construction.
Rotor bars) slightly skewed(
End ring
Figure2.2:Squriell cage rotor
2.2.2.2 Wound rotor:
3
2. Three phase induction motor
Week 3
Wound Rotor
Brush
Slip Rings
External Rotor
Resistances
Figure 2.3: Schemtic diagram showing Induction motor, wound rotor type
The wound rotor or slip-ring induction motor differs from the squirrel-cage motor only in the rotor
winding. The rotor winding consists of insulated coils, grouped to form definite polar areas of
magnetic force having the same number of poles as the stator. The ends of these coils are brought
out to slip-rings. By means of brushes, a variable resistance is placed across the rotor winding
(Fig.2.3). By varying this resistance, the speed and torque of the motor is varied. The wound rotor
motor is an excellent motor for use on applications that require an adjustable-varying speed (an
adjustable speed that varies with load) and high starting torque.
Terminals
Slip rings
Bearings
Fan
Laminated core
Shaft
Coils
Figure 2.4:Wound rotor
2.2.3 Enclosure
Enclosure or the frame (Figure2.5) ,its main
application to hold the parts together. also it Helps
with heat dissipation. In some cases, protects internal
4
2. Three phase induction motor
Week 3
components from the environment. A cooling fan is attached to the shaft at the left-hand side. This
fan blows air over the ribbed stator frame
Figure2.5:Induction Motor enclosure
2.2.3
Bearings:
There are two main types, the sleeve bearings and ball bearings .Ball (Roller) Bearings
(Figure2.6a) Support shaft in any position. Its Grease lubricated and required no maintenance
The Sleeve Bearings(figure2.6b) are Standard on most motors. They are only used with horizontal
shafts and its oil lubricated.
(a)
Figure2.6: a)Ball Bearings
b)Sleeve bearings
(b)
b
2.2.5 Conduit Box
Point of connection of electrical power to the motor’s stator windings.
2.2.6 Eye Bolt
Used to lift heavy motors with a hoist or crane to prevent motor damage, as it can be seen in the
following figure.
5
2. Three phase induction motor
Week 3
Figure2.7: Sectional view of Induction Motor
6
2. Three phase induction motor
Week 4
2.2 The principle of operation of the Induction Motor:
The three-phase current with which the motor is supplied establishes a rotating magnetic
field in the stator. This rotating magnetic field cuts the conductors in the rotor inducing
voltages and causing currents to flow. These currents set up an opposite polarity field in the
rotor(Lenz's law). The attraction between these opposite stator and rotor fields produces the
torque which causes the rotor to rotate. This simply is how the squirrel-cage motor works
Figure2.8:The magnetic field created in the stator and the rooted in the squirrel cage induction motor
2.3 The Slip:
There must be a relative difference in speed between the rotor and the rotating magnetic
field. If the rotor and the rotating magnetic field were turning at the same speed no relative
motion would exist between the two, therefore no lines of flux would be cut, and no voltage
would be induced in the rotor. The difference in speed is called slip. Slip is necessary to
produce torque. Slip is dependent on load. An increase in load will cause the rotor to slow
down or increase slip. A decrease in load will cause the rotor to speed up or decrease slip.
Slip is expressed as a percentage and can be determined with the following formula.
Slip (%) 
( Ns  Nr ) 100
Ns
1
2. Three phase induction motor
Week 4
Where Nr is the actual speed of the rotor
For Example, a four-pole motor operated at 60Hz has a synchronous speed (Ns) of
1800 RPM. If the rotor speed at full load is 1765 RPM (Nr), then the slip will be
calculated
Slip (%) 
(1800  1765) 100
 1.9%
1800
2.4 Name Plate:
It is essential that all motors have nameplates with certain information useful in the
identification of the type of motor, The following Table explain the indication of each code
used on the shown nameplate
Name Of Manufacturer
ORD. No.
IN123456789
HIGH EFFICIENCY
FRAME
286T
H.P.
42
SERVICE
FACTOR
1.10
3PH
AMPS
40
VOLTS
415
Y
R.P.M
1790
HERTZ
60
4POLE
DUTY
CONT
DATE
01/15/2003
TYPE
CLASS INSUL.
F
NEMA
Design
B
NEMA
NOM. EFF.
95
Address Of Manufacturer
Figure2.9:Typical Nameplate of an AC induction motor
Term
Description
Volts
Rated Supply voltage
HP
Rated motor output
Amps
Rated full load current
RPM
Rated full load speed of the motor
2
2. Three phase induction motor
Week 4
Hertz
Rated supply frequency
Frame
External dimensions based on NEMA Regulations
Duty
Motor load condition ,either its continuities load, short time,etc
Date
Date of manufacturing
Class Insulation
Specifies the max. limit temperature of the winding
NEMA Design
Types of NEMA design, A,B,C etc
Service factor
Factor by which the motor can be overloaded beyond the full load.
NEMA Nom
Efficiency
Motor efficiency at rated load
PH
Number of phases
Pole
Number of poles
Motor safety standard
Y
The connection either star or delta
Table 2.1:Explanation of the codes used on AC motor nameplates
3
3. Synchronous machine
Week 5
3.1 Introduction:
Synchronous motors are motors that always run at the same speed regardless of load.
Synchronous motors are somewhat more complex than squirrel-cage and wound rotor motors and,
hence, are more expensive. There is no slip in a synchronous motor, that is, the rotor always moves
at exactly the same speed as the rotating stator field.. The machine consists of three main parts:

Stator, which carries the three phase winding,

Rotor, with one DC winding or permanent magnets

Slip rings or excitation machine (exciter) (in case of electrical excitation).
Synchronous motors are used whenever exact speed must be maintained or for power factor
correction. Synchronous motors are more expensive than other types for the lower horsepower
ratings, but may possibly be more economical for 100 hp and larger ratings.
3.2 Stator construction:
The stator of a synchronous generator holds a three-phase winding where the individual phase
windings are distributed 120° apart in space and is sometimes called the armature winding. The
stator must be made of laminated iron sheets in order to reduce eddy currents.
3.3 Rotor construction:
The rotor holds a field winding,which is magnetized
by a DC current( the field current).
The rotating
field winding can be energized through a set of slip
rings and brushes (external excitation), or from a
diode-bridge mounted on the rotor (self-excited). The
rectifier-bridge
is
fed
from
a
shaft-mounted
Metal frame
Laminated iron
core with slots
Insulated copper
bars are placed in
the slots to form
the three-phase
winding
alternator, which is itself excited by the pilot exciter.
In externally fed fields, the source can be a shaftdriven dc generator, a separately excited dc generator, or a solid-state rectifier. Several variations to
these arrangements exist. There are two types of rotors:

Salient-pole rotor (Fig.2) for low-speed machines (e.g.hydro-generators)

Cylindrical rotor (Fig.3) for high-speed machines (e.g. turbo-generators).
1
3. Synchronous machine
Week 5
3-Phase Stator Winding
Rotor Field
Winding
Brushes
-
+
Slip Rings
Cylindrical
Pole Rotor
Field current
a- Schematic diagram showing a cylindrical rotor of a synchronous machine
Steel
retaining
ring
Shaft
Shaft
Wedges
DCcurrent
current
DC
terminals
terminals
b- cylindrical rotor of a synchronous machine
Figure3.1: cylindrical rotor of a synchronous machine
2
3. Synchronous machine
Week 5
3-Phase Stator Winding
Rotor Field
Winding
Brushes
-
+
Slip Rings
Salient Pole
Rotor
Field current
a- Schematic diagram showing a salient-pole rotor of a synchronous machine
Slip
rings
Pole
Fan
DC excitation
winding
b- salient-pole rotor of a synchronous machine
Figure3.2: Salient-Pole rotor of a synchronous machine
3.4 Principle of operation of the synchronous generator:
When the 3phase rotor is rotated (by an external prime-mover) the rotating magnetic
flux(induced by DC current) induces voltages in the stator windings. These voltages are sinusoidal
with a magnitude that depends on the field current, and also differ by 120° in time and have a
frequency determined by the angular velocity of the rotation.
3.5 Principle of operation of the synchronous motor:
The stator is supplied with three phase supply in order to develop a rotating magnetic field. Also
the rotor is supplied with DC supply to produce constant
magnetic field. As a result of the interaction of these two fields, the rotor will start to move.
However, the synchronous motor is not self started. Consequently, it usually equipped with squirrel
cage windings that mounted on the pole faces of the synchronous motor rotor. These rotor windings
3
3. Synchronous machine
Week 5
are frequently referred to as damper or amortisseur windings. Thus, the synchronous motor starts as
an induction motor. When the motor accelerates to near synchronizing speed (about 95%
synchronous speed), DC current is introduced into the rotor field windings. This current creates
constant polarity poles in the rotor, causing the motor to operate at synchronous speed as the rotor
poles "lock" onto the rotating AC stator poles.
3.6 Excitation Methods
Two methods are commonly utilized for the application of the direct current (DC) field current to
the rotor of a synchronous motor.

Brush-type systems apply the output of a separate DC generator (exciter) to the slip rings of the
rotor.

Brushless excitation systems utilize an integral exciter and rotating rectifier assembly that
eliminates the need for need for brushes and slip rings.
3.7 Method of Synchronization
There are three basic method of synchronizing two or more machine:
. Bright lamp method
. Dark lamp method
. automatic method
4
3. Protection & control of electric motors
Week 6
4.1 Need for Circuit Protection
Current flow in a conductor always generates heat(Figure1). The greater the current flow, the
hotter the conductor. Excess heat is damaging to electrical components. For that reason, conductors
have a rated continuous current carrying capacity or ampacity. Overcurrent protection devices, such
as circuit breakers, are used to protect conductors from excessive current flow. These protective
devices are designed to keep the flow of current in a circuit at a safe level to prevent the circuit
conductors from overheating.
Normal Current Flow
Excessive current Flow
Figure4. 1:The effect of current flowing in conductors
overcurrent is defined as any current in excess of the rated current of equipment of a conductor. It
may result from overload, short circuit, or ground fault
Overloads: An overload occurs when too many devices are operated on a single circuit, or a piece of
electrical equipment is made to work harder than it is designed for. For example, a motor rated for
10 amps may draw 20, 30, or more amps in an overload condition.
1
3. Protection & control of electric motors
Week 6
Good Insulation
Damaged Insulation
Figure4. 2: Insulation of electric conductors
Conductor Insulation Motors, of course, are not the only devices that require circuit protection for
an overload condition. Every circuit requires some form of protection against overcurrent. Heat is
one of the major causes of insulation failure of any electrical component. High levels of heat can
cause the insulation to breakdown and deteriorated, exposing conductors (Figure4.2).
Short Circuits When two bare conductors touch, a short circuit occurs (Figure4.3).
When a short circuit occurs, resistance drops to almost zero. Short circuit current can be thousands
of times higher than normal operating current.
The heat generated by this current will cause extensive damage to connected equipment and
conductors. This dangerous current must be interrupted immediately when a short circuit occurs.
2
3. Protection & control of electric motors
Short Circuit
Week 6
conductor
Insulation
Figure4. 3: Short circuit fault between two condu
3
4.Protection & control of electric motors
Week 7
4.2 Types of Overcurrent Protective Devices
Circuit protection would be unnecessary if overloads and short circuits could be eliminated.
Unfortunately, overloads and short circuits do occur. To protect a circuit against these currents, a
protective device must determine when a fault condition develops and automatically disconnect the
electrical equipment from the voltage source. An overcurrent protection device must be able to
recognize the difference between overcurrents and short circuits and respond in the proper way.
Slight overcurrents can be allowed to continue for some period of time, but as the current magnitude
increases, the protection device must open faster. Short circuits must be interrupted instantly.
Several devices are available to accomplish this.
4.2.1 Fuses
A fuse is a one-shot device (Figure1). The heat produced by overcurrent causes the current
carrying element to melt open, disconnecting the load from the source voltage. There are three types
of fuses, namely

Semi-enclosed (Rewireable) fuse

Cartridge fuses

High Breaking Capacity(HBC)
Fuse Cap
Good Element
Glass or Ceramic
Body
Open
Element
Figure 4.4: Plug fuse
4.2.1.1 Cartridge
The cartridge type have fuses which look similar to those you
would
find in a standard household plug. This type is improvement of
the
rewirable fuse type. It is main advantages, is easy to replace,
totally
1
4.Protection & control of electric motors
Week 7
enclosed and its current rating is very accurate
Figure 4.5: A cartridge fuse and its holder
4.2.1.2 HBC
HBC stands for "high blow current (sometimes described as HRC = high rupture current). HBC
fuses are designed not to explode when failing under currents
many times their normal working current (e.g. 1500 amps in
a 10
amp circuit). They are therefore to be preferred for the
protection of main voltage circuits where the power source
may
be capable of providing very high currents. HBC types can
usually be recognized by being sand filled though they may
have
a thick ceramic body.
4.2.1.3 Semi-enclosed(Rewireable) fuses
Figure 4.5: A HBC fuse
As the name indicates, the rewireable type have a fuse wire held at both
ends by a small retaining screw. Once the fuse is blown, the fuse wire is
the only pieces to be replaced. It is cheap, but replacing a wrong size of
element can cause catastrophic consequences.
Figure 4.6: Rewireable fuses
2
4. Protection & control of electric motors
Week 8
Figure 4.7:Miniature circuit breakers with different poles
The problem with fuses is they only work once. Every time you blow a fuse, you
have to replace it with a new one. A circuit breaker(Figure 4.7) does the same thing
as a fuse .It opens a circuit as soon as current climbs to unsafe levels ,but you can
use it over and over again.
The basic circuit breaker consists of a simple switch,(see figure 4.8) connected to
either a bimetallic strip or an electromagnet. The diagram below shows a typical
electromagnet design.
Figure4.8: Cut view of a miniature circuit breaker
27
4. Protection & control of electric motors
circuit
breakers
generally
employ
a
bimetal
strip
to
Week 8
sense
overload
conditions(Figure4.9b). When sufficient overcurrent flows through the circuit
breaker’s current path, heat build up causes the bimetal strip to bend. After bending
a predetermined distance the bimetal strip makes contact with the tripper bar
activating the trip mechanism.
A bimetal strip is made of two dissimilar metals bonded together(Figure4.8). The
two metals have different thermal expansion characteristics, so the bimetal bends
when heated. As current rises, heat also rises. The hotter the bimetal becomes the
more it bends, until the mechanism is released.
Material 1
}
Bi-metal
Material2
Heat source
Figure 4.8:The effect of heat on a bimetal strip
Short circuit protection is accomplished with an electromagnet (Figure4.8a). The
electromagnet is connected in series with the overload bimetal strip. During
normal current flow, or an overload, the magnetic field created by the
electromagnet is not strong enough to attract the armature. When a short circuit
current flows in the circuit, the magnetic field caused by the electromagnet attracts
the electromagnet’s armature. The armature hits the tripper bar rotating it up and to
the right. This releases the trip mechanism and operating mechanism, opening the
contacts. Once the circuit breaker is tripped current no longer flows through the
electromagnet and the armature is released. (See figure 4.9).
28
4.Protection & control of electric motors
Week 8
Magnetic Circuit Breaker(a)
Magnetic
coil
Current in
Magnetic
coil
Current in
Electrical
contacts
Electrical
contacts
Current
out
Current
out
Spring
Spring
Latching
mechanisim
Latching
mechanisim
Overload Conditions
Normal Conditions
Thermal-magnetic Circuit Breaker(b)
Magnetic
coil
Magnetic
coil
Current in
Electrical
contacts
Electrical
contacts
Current in
Current out
Spring
Current out
Latching
mechanisim
Spring
Normal Conditions
Latching
mechanisim
Overload Conditions
Figure4.9: The principle of operation of circuit breakers
29
4.Protection & control of electric motors
Week 9
4.3 Control devices
4.3.1 Relay:
Iron Core
Iron Core
Switch
Switch
Relay Coil
Relay Coil
Battery
Contacts
Contacts
To Power Circuit
Relay Open
To Power Circuit
Relay Close
Figure4.10:The principle of operation of the relay
The relay is a remotely controlled switch. In the diagram above, a power circuit contains a
switch which is opened and closed by operation of a
relay. The relay is activated by a magnetic core which is
energised when a controlling switch is closed. As the
core is energised, it lifts and closes a pair of contacts in a
second circuit - usually a power circuit. The current
required for the relay is usually much lower than that used
for the power circuit so it can be provided by a battery
In the left hand of figure4.10,the diagram shows the
controlling switch is open, so the relay is de-energised and
Figure4.11: a relay
the
power circuit contacts are open. If the controlling switch is
closed, as in the right hand diagram, the relay is therefore energised and its core magnet lifts
to close the contacts in the power circuit.
1
4.Protection & control of electric motors
Week 9
4.3.2 Contactors:
(a)
( b)
Figure4.12: a)Construction of a contactor ,b) A contactor
Figure4.12a shows the interior of a basic contactor. There are two circuits involved with the
operation of a contactor, the control circuit and the power circuit. The control circuit is
connected to the coil of an electromagnet, and the power circuit is connected to the
stationary contacts.
When the control circuit supplies power to the coil, a magnetic field is produced,
magnetizing the electromagnet. The magnetic field attracts the armature to the magnet,
which, in turn, closes the contacts. With the contacts closed, current flows through the power
circuit from the line to the load. Figure(4.13)
When current no longer flows through the control circuit, the electromagnet's coil is deenergized, the magnetic field collapses, and the movable contacts open under spring
pressure.
2
4.Protection & control of electric motors
Week 9
Figure4.13:Principle of operation of the contactor
4.3.2.1 Overload relays
Overload relays (Figure4.14)are designed to meet the special protective needs of motor
control circuits. Overload relays allow harmless temporary overloads that occur when a
motor starts.
Overload relays trip and disconnect power to the motor if an overload condition persists.
Overload relays can be reset after the overload condition has been corrected.
Figure4.14: Overload relay
3
4.Protection & control of electric motors
Week 9
4.3.2.2 Contactors and Overload Relays
Contactors are used to control power in a variety of applications. When used in motorcontrol applications, contactors can only start and stop the motors. Contactors cannot sense
when the motor is being overloaded and provide no overload protection.
Most motor applications require overload protection, although some smaller motor, such as
household garbage disposals, have overload protection built into the motor. Where overload
protection is required, overload relays (such as the one shown here) provide such protection.
4.3.2.3 Motor Starter
Contactors and overload relays are separate control devices. When a contactor is combined
with an overload relay, it is called a motor starter.Figure4.15
4
4.Protection & control of electric motors
Week 10
4.3.3Pushbuttons
A pushbutton is a control device used to manually open and close a set of contacts. Pushbuttons may
be illuminated or non-illuminated and are available in a variety of configurations and actuator
colors.
Figure 4.16: Different types of pushbuttons
i) Normally Open Pushbuttons
Pushbuttons are used in control circuits to perform various functions such as starting and stopping a
motor. A typical pushbutton uses an operating plunger, a return spring, and one set of contacts.
This illustration shows a pushbutton with normally open contacts. Pressing the button causes the
contacts to close (figure4.17). This pushbutton has momentary contacts which means that the
contacts will open when the pushbutton is released.
ii) Normally Closed Pushbuttons
Pushbuttons with normally closed contacts, such as the one shown here, are also used in control
circuits. The contacts remain in the closed position allowing current to flow through them until the
pushbutton is pressed(figure4.17). Pressing the pushbutton opens the contacts and interrupts current
flow. The pushbutton shown here also has momentary contacts; however, normally open and
normally closed pushbuttons with maintained contacts are also available.
.
1
4.Protection & control of electric motors
Week 10
Normally Open Pushbutton
Normally Close Pushbutton
Figure 4.17: The mechanism of the pushbuttons
4.3.4 Selector Switches
Selector switches are another means to manually open and close contacts and are commonly used
to select one of two or more circuit possibilities.
Selector switches may be maintained, spring return, or key operated and are available in twoposition, three-position, and four-position types.
The basic difference between a pushbutton and a selector switch is the operator mechanism. A
selector switch operator mechanism is rotated to open and close contacts.
2
4.Protection & control of electric motors
Week 10
Figure 4.18: Different types of selector switches
4.3.5 Indicator Lights:
Indicator lights, often referred to as pilot lights, provide a visual indication of a circuit's operating
condition. An indicator light may be wired to turn on for any predetermined condition. Indicator
lights are available in round designs with 16 mm, 22 mm, or 30 mm mounting diameters as well as
in square designs.
Figure 4.19: Different types of push buttons switch
3
5. energy conversion
Week 11
5.1 Electro-mechanical energy conversion
Energy is converted to electrical form because of the advantages listed in the introductory
part of the note. It is seldom available or used in electrical form, but converted into electrical
form at the input to a system and back to non-electrical form at the output of a system. A
typical example is the processing of energy from and hydro generating plant. It is converted
into electrical form at the power plant. Transmitted through transmission lines and
distribution lines, and converted to mechanical energy in an electric motor are the point use.
A second example is in the conversion of the energy in sound pressure waves, and the
transmission in electrical form from the taker to the listener in a telephone system. Few more
energy conversion principles will be mentioned.
5.1.1 Major energy coversion principles
Energy conversion between electrical and non- electrical forms includes
5.2
(i)
Electrochemical eg battery
(ii)
Electrothermal eg. Thermocouple
(iii)
Photo electrical eg photo cell
Energy coversion
Theoretically, only a sourceless current is needed to develop a mechanical force
magnetically. But in a machine the production of force is hardly enough: something must
move in order to do useful work done demands a corresponding energy supply form
somewhere.
In a device energized only by a permanent magnet, the only energy source is the magnet
itself. If the displacable part of the machine moves under force and does work, this can only
be at the expense of the field energy of the permanent magnet, which must decrease. Such an
5. energy conversion
Week 11
arrangement has obvious limitations. It may also be inconvenient a permanent magnet”
lifting magnet, for example would not e capable of releasing its load. Where the magnetic
affected by the movement is produced by a current circuit, changes of field energy have to
be supplied electrically from a source. This implies the appearance in the circuit of an
electromotive force e, which, with the current I, represents the delivery or absorption by the
source of energy at the rate ei consider the elementary system of fig 2.1 A sources of voltage
is connected to a device (e.g a secondary battery or a machine) in which the energyconversion process results in the appearance of an e.f.m.
The effective resistance of the circuit is represented by R. if
current flows into the circuit
form the positive terminal of the source, and the input power p = Vi Rl + el, has the
direction as shown at (a). However¸ if e >, the current reverses and we can now call it –e. the
power input from the now p = v (-i) = Rl2 + e (-l), which is negative, i.e it is an output from
the device into the source, as at (b). to illustrate this simple but fundamental point, suppose
that v = 10v d.c nad R = 1-2. then if e = 8vd.c.
The current o = v-e
R
=
10-8
=2A
1
And the source provides an input power p = 10 x 2 = 20w The converting device accepts 8
x 2 W as a motor And power loss due to plissipation
= 1 x RT2 = 4W Conversely, if e
12V, The current again is 10 -12 -2A (I.e reversed) The device produces 12 x 2 = 24 W as
a generator of which RT2 22 x 1 = 4W is dissipated in R, and 10 x 20W is delivery as an
output to the source. In the case of the electromagnetic machine, the relationship between
the emf and the magnetic field is obtained from the faraday induction law (which had been
mentioned in 1.1)
5. energy conversion
5.3
Week 11
Linked energy systems
An electromechanical machine forms a coverting link between an electrical energy system (
such as a main power –supply network) and a mechanical one (such as a prime- mover or a
train). In action a machine is not an isolated things, but has a behavious strongly influenced
by its terminal systems. A relay, for instance, will be affected if its operating battery
becomes discharged; a loudspeaker will behave very differently it enclosed in an evacuated
vessels with the air loading thus removed; a hydro electric generator, suddenly shortcircuited, will react severely on the turbine and pipe-line.
A machine can, of course, be studies initially in
isolation, but the engineering interest
begins in fact when the complete linked system is considered. Again, the steady-state
behavious is informative up to a point, but operation in responses to change – i.e, the
transient responses is fro move important and fundamental.
Fig 2.1 Electro–mechanical linked energy system
System analysis can be complicated. Fig 2.2 shows diagrammatically a typical electric
supply system feeding a mechanical load through an electromechanical machine. In some
cases we might simplify the analysis by assuming, say, that the terminal voltage and
frequency of the machine were constant. This is good enough if the machine is a small
contactor but if it is a 25MW motor the effects of its behaviour reach for back through even
an extensive supply system. Methods are available for evaluating such a complex for any
5. energy conversion
Week 11
given stimulus, such as the occurrence of a transmission- line fault or starting of a large
motor.
5.4 Energy storage
We now consider how a flux is established and energy is stored in simple toroidal magnetic
Circuit of cross sectional area A, path length L, and of material of constant permeability u,
The flux is to be established by a current i, in a uniformly wound coil of N-turns. In order
To concentrate on energy storage we neglect the coil resistance. With i initially zero, let a
Voltage V, be applied to the coil terminals, what happen thereafter depends on Faraday’s
Law of electromagnetic induction.
5. energy conversion
5.5
Week 12
Energy balance
Fig: 2.1 Electromechanical machine conventions
A machine accepts energy in a variety of forms from its attached terminal systems. By
conversion we take energy input as positive, so that an output is regarded as a negative input.
The machine internally electrical energy- mechanical energy is a motor mechanical energy to
electrical energy is a generator converts some energy, stores some, and dissipates the rest:
these energies are positive if they increase with time. As the prime object of a machine is
conversion to useful output, one of the terminal inputs will normally be negative. Recalling
the principle of conservation of energy which states that energy is neither created nor
destroyed and combining it with the laws of electric and magnetic fields, electric circuits and
Newtonian mechanics, the energy balance can be expressed as:Total terminal energy input internal energy + Dissipation 2.1 for an electromechanical
machine using a magnetic field as the means of conversion, the balance can be stated in more
specific terms as electrical energy input + mechanical energy input
= stored magnetic –field energy + stored mechanical energy + Dissipation
5. energy conversion
Week 12
Reckoned from an initial condition of zero energy, w = o.A comparable relation must apply
to energy changes dw, and also to energy rate dw/dt i.e to power, P. in corresponding
symbols these relations are total energy wf + ws + w
Energy change dwe + dwm = dwf + dws + dw
Energy rate Pe + pm
= dwf + dws
dt
2.2(a)
2.2(b)
+
p 2.2(c)
dt
The rates of change of stored field energy wf and stored mechanical energy, ws, are left in
differential form because there is always a practical limit to storage. A magnetic field can not
grow in strength indefinitely when ferromagnetic materials is employed; and if the kinetic
energy in a flywheel is continually increased, the speed must rise and the wheel may burst
under centrifugal force.
We shall now examine the electromechanical machine in more detail with fig 2.3. The
machine links an electric source of voltages supplying a current; and a mechanical sources
represented by a bar moving to positive directions, thus both vi and fmu are inputs ( The
mechanical source could alternatively be a shaft rotated at angular speed wr by a tongue mm
to give an input power mnwr ). The electrical end of the machine is precisely that of fig 2.1
(a), with opposing v. the mechanical end has the magnetically developed force fe opposing
fm > fm it can reverse speed w so that the mechanical system is driven and absorbs a
mechanical output.
The behavior can now be summarized. With the machine operating in the steady state as a
motor, the applied voltage u drive +I against e to give a total electrical power input pe =
u(+e), of which the part ei is converted. The outcome of conversion is the force fe which
drives the bar against fm to develop the mechanical input pm = fm (-u) which, being
5. energy conversion
Week 12
negative, is actually an output. With the machine as a generator; the bar is driven at speed u
by the force fm to provide the mechanical input pm = fm ( +u), as a result which e now
exceeds u and reverse the current to provide the negative electrical input (i.e output (i.e
output) pe = u (-i) the sum of the inputs (pe +pm) must be rate of rise of internal energy
storage plus the rate of energy dissipation.
A real electromagnetic machine has fairly obvious points of attachment (e.g the electrical
terminals and the shaft) by which it is connected to the electrical and mechanical sources to
form a link between them. But it is very to concentrate source to from link between them.
But it is very convenient attention on the conversion region enclosed by the chain- dotted
line in fig 2.3, for it contains only the essential quantities e and i,. U and fe. Various losses,
and the mechanical storage, are excluded so that attention can be directed on to the physical
process if useful energy conversion by electromagnetic means outside the conversion region
we can account for conduction and core losses associated with the electrical end and
represented rough by the resistance R in fig 2.3, and friction and similar losses on the
mechanical side. It is to be noted that the externally applied force fm is not necessarily equal
to –fe because there may be force-absorbing components of inertial and elasticity in the
mechanical working parts of the machine itself, as well as internal friction.
The machine has new been reduced to an analyzable form. Its behaviors under specified
conditions involves the forces and movement of the mechanical parts, the voltages and
current at the electrical terminals and processes of energy conversion and storage and
dissipation going on inside. Evaluation is based on the well-established principles and laws
summarized in the following table.
5. energy conversion
Week 12
Part of system
Quantities
Principles
Electrical
Voltage, current
Faraday-Lenz and
Kirchhoff laws
Conversion
E.M.F
current
magnetic Magneto- mechanical
field,. Force, displacement
Principles
induction
and
Force, displacement speed
thermal laws Newton law
Mechanical
5.5.1 Block diagram for energy balance equation
The energy balance equation is given by equation 2.2 as electrical energy input mechanical
energy input
= stored magnetic-=field energy + stored mechanical energy + dissipation
The dissipation (energy lossess) arise from three main causes
(ii)
Part of electrical energy is converted directly to heat in the resistance of current path.
(ii)
Part of mechanical energy developed with the device is absorbed in friction ad
windage and converted to heat.
(iii)
Part of the energy absorbed by the coupling field is converted to heat in magnetic
core losses (for magnetic coupling ) or dielectric loss) for electric coupling).
if we associate the various losses with the corresponding energies, equation 2.2 be written as
Electrical energy
Input minus
Resistance losess
mechanical energy
= Output plus friction
and windage losses
increase in energy stored
+
in the coupling field
plus associated losses
5. energy conversion
Week 12
Equation 2.3 is obtained ( for a motor) with the mechanical energy transferred to the R.H.S
of the equality sign and neglecting the energy mechanical stored
energy ( for a machine
without a flywheel and neglecting the mass of the shaft). If there is a flywheel, the stored
mechanical energy is 1/2mu2 or ½ mr2w2
Where m
=
mass of flywheel
V
=
linear velocity of rotating wheel
w
=
angular velocity of rotating wheel
r
=
radius of flywheel
Equation 2.3 may be represented in the form of a block diagram as shown in fig 2.4
Fig 2.4 General representation of electromagnetic energy conversion. Fro a generator action, the
positions of the electrical system and that or the mechanical system will be interchanged.
5. energy conversion
5.6
Week 13
Magnetic field energy and forces
In order to be able to analyse mathematically the electromechanical system that is
completely described by the energy balance equation, we need to be able to determine
qualitatively the energy of the magnetic field and the associated force.
5.6.1 Magnetic field
A magnetic field is a region of space in which certain physical effects occurs in particular
the development of mechanical force. A pictorial model of the field can be made by drawing
closed loops of magnetic flux, such that their direction and spacing at any point are a
measure of the flux density. The magnetic circuit in the present context is composed partly
of ferromagnetic material such as iron, and partly of an airgap. The iron serves to “guide” the
flux in a desired path; the
airgap is necessary to make useful magnetic effects readily
accessible.
The lines in a flux plot have no real existence. In a given region a magnetic field may change
direction, become weaker in some place and stronger in others.
5.6.2 Magnetic circuit n/a
Engineers look upon magnetic flux (Weber) as produced by electric current. A current I
develops around any path that links it a magneto motive force (M.M.F) F = I (ampere). The
effect of a current can be multiplied. By coiling the electric circuit into N turns so that
around a path linking all N turns the m.m.f is N times as great, giving F =- ampere- turn.
The m.m.f is distributed along the path, to give along a path element of length dx the
magnetic field intensity h (ampere-turn/ metre). The summation of Hdx around a single loop
closed with F i.e F = Hdx = m.m.d.
5. energy conversion
Week 13
At any point, H gives rise to a flux density B = NH (tesla or Weber/m2) our Henry/meter]
Flux summation of the flux density over the area available to the flux path given the total
flux [ i.e Ø i.e. BA. (Weber).-2.6 where A is the are of flux path.
The „ law of the magnetic circuit relates the total flux Ø to
the mmf f through the
expression.
Ø = F =F Comparable to the law of electric circuit 2.7
S
I
=
Where s
V
v.g (ohm‟s law)
R
=
=
the total reluctance (ampere-turn per weber]
And = 1/s = total permeance [ weber per ampere-turn]
For a path- length x of materials of absolute permeability U, and having a uniform crosssectional area A over which the density B is everywhere the same, the mmf f require = N x x
= Hx…………..2.8
From equation 2.6, 2.7 and 2.8, F Øs F = mmf
And the reluctance of the path S =
f/Ø = Hx
2.9
= x……….2.10
And the 1/s UA/x
For a succession of parts , x, y, z
F = fx + fy + f2 +
and S = SX + SY + SZ +
2.11
2.12
If, however the parts are in parallel and share the flux
F = fx =fy =fz and
For fields in ferromagnetic materials U is very much greater, and the relative permeability Nr
=u =Uo 4 /107 1/80000
5. energy conversion
Week 13
Which means that H = Ub = 800000B
For field ferromagnetic materials U is very much greater,
Ad the relative permeability Ur = U
Uo
Since, usually Ur is large, then it is convenient ( it simplies analysis) to assume that the
whole mmf is required for the excitation of the air gap i.e the whole of the field energy is
stored in the air-gap
5.7
Magnetic field energy
With the assumption that the magnetic filed energy is concentrated within the air gap. It
becomes easy to calculated the magnetic field energy.
A magnet attract on iron bar. If the iron bar is light enough and the magnet filed is enough,
the bar will be seen to move up to get attached to the magnet. The movement of the bar
signifies that work is done, since the iron bar has mass and covered some distance
(work done = force x distance). This means that the space that the file occupies (the field
region) can demonstrated or has on attribute of force. And hence, the filed region must
process some energy. If can be easily noticed that the force is strong when the air gap is short
but rapidly diminishes as the air gap length is increased.
5.8 Maxwell stress
Fig 2.5 maxwell forces
Maxwell formulated the concept that the forces is transmitted across the gap between a pair
of magnetized surface as a result of two stresses. If at a point in the gap the flux density is B
and the corresponding field intensity is H =B/U,. then there is a tensile stress of magnitude
5. energy conversion
Week 13
1.2 BH along the direction of a flux line and a compressive stress ½ BH along all directions
at right angles to a flux line.
Fig 2.5 shows two iron bars forming part of magnetic circuit when, as at (a), the polar
surfaces are close together, the flux is mainly concentrated between the surfaces. The density
B is large, and so therefore is H, and ½ BH represented a strong tensile force of attraction
between the faces. Not all the flux is useful; some, of the leakage flux, exists at the sides of
each bar. Flux crossing the boundary between air and a high permeable materials must enter
or leave the boundary between air and a high permeable materials must enter or leave the
boundary almost at right angles, so that the tensile stress due to faces. All the comprehensive
stresses balance out by symmetry.
In case (b), the greater reluctance of the long air gap reduces the total flux, the useful flux
density of the pole faces is smaller while the leakage flux is much greater hence the forces of
attraction between the pole faces is much less than in case (a)
In most practical applications, the air gap is small enough to enable us assumes a uniform
flux density over the polar area. i.e in the air gap.
.
5. energy conversion
Week 14
5.9 ENERGY DENSITY
The Maxwell stress concept is another way of saying that the energy to the value
½ BH is stored in a unity cube of the space occupied by = magnetic, thus ½ BH is the energy density
[weber (metre2) x (amper/metre] = volt –second x ampere/cubic metre
= Joule /cubic metre.
Fig 2.6 magnetic energy.
Consider an air gap, initially unmagnetized. Apply a magnetic force to the gap, an increase of H
from zero causes the flux density B = μoH to increase proportionately, Fig 2.6(o). The energy (m 3
is [HdB, and for and values Bi and H1 the final energy density (shaded area) is clearly ½ B 1H1 . The
some summation applies to a filed set up by in a ferromagnetic matter, with similar result Fig. 2.6
(b0, if the permeability U is constant; but for the same and density B1 a much smaller magnetizing
force Hii is need and much less energy is stored.
If the ferromagnetic materials is subjected to saturation, the stored energy is as shown in Fig 2.6(c0
and is calculated by piece-wise approximation to composite area of DOAD plus area of trapezium
AB,CD.
5. energy conversion
Week 14
5.10 FARADAY- LENZ LAW
When the flux 4 associated with an electric increase in time at by amount d4, an emf, e = de/dt
appears in the circuit. The minus sign implies that the direction of the emf is such that a current
produced by it in the circuit opposes the change d4.
Flux-linkage 4 (weber- turn] is the product of a magnetic flux and the number of turns through
which it passes in the same direction. Since the current is proportional the flux,
then flux likage Ų = NQ
Since we are neglecting the coil resistance, then around the electric circuit loop formed by the
voltage source and the N turns of coil on the toroid, the KVL gives
u = +Dq/DT =-E. 2.17
The instantaneous electric power input to the coil
P =vi = (dw/dt)i
The total energy required to establish from zero a flux Q1 and a linkage Q1, (corresponding to a
current mmf F1 =Ni) is
Wf = (t pdt = (4 idQ =(Q fdQ
Since the core of the
toroid has constant permeability.
Which is represented by the shaded area in Fig, 2. 6 (d). this magnetically stored energy can be
assumed to be uniformly distributed through the active volume Al of the core. Then because
Q1 = QIN =NB,A and Nli =Fi =H
il
The energy density = ½ 4ili/Al =1/2 Bi Hi
The total magnetic energy can be stated in several ways
[f =1/24I =1/2QF =1/2Q2S =1/2 F2S =1/2Q / 1/2F2 =1/2Li2]
5. energy conversion
Week 14
Also, since Q BA and F =Hl and H =B/U
Then wf =1/2 QF =1/2 BAHL =1/2AL
U
But al = volume
Wf = Vol. B2
2U
Any expression for the energy of the field wf in equation 2.32 and 2.23 may be employed
depending on the parameters given.
5.11 ENERGY CONVERSION
Fig. 2.7 Energy change with position
Fig 2.7 (a) shows airgap region and existing coil of a magnetic circuit, the ferromagnetic
core of which has a high. Permeability, the plane parallel polar faces. Have an area A and are
spaced x apart. The n-turn coil carrying current I magnetic the system. The problem is to find
the magnetic force of alteraction between the polar faces.
In the comparable system Fig 2.7 (b), with a rotatable part (rotor) port (stator), the problem is
to find the tongue.
Insight into the inteplay of energy can be obtained from a study of finite mobvements, say
from an initial position (1) to afinal direction position (2). At (a0. this movement -∆x (i.e
5. energy conversion
Week 14
against the positive direction of x) of the right-hand member’ are ( b) it is a rotation -∆Q of
the rotor. The static 4/I relations for the two positions are shown in Fig 2.7 © differ because
the gap reluctance for (2) is less than for (1). Clearly the filed energy will differ too. How it
changes depends on the conditions wholly in the gap, and the effect of coil resistance will
initially be ignored.
(1)
CONSTANT CURRENT
Let the
current be held constant at is throughout, as shown at Fig 2.7 (d). for
position the linkage is 41 and the filed energy is ½ Fig 4. to reach position (2) a
linkage 42 with constant current, an electrical energy input + ∆we =(Ų2-Ų1 ) i0 must
be fig 2.7 (d). the current sources. This is represented by the hatched area at Fig
2.7(d0. Now, the increase in field energy is ∆WF =1/2 (Ų2-Ų1)io, which, comparing
the expressions or the hatched areas at (d), is only one-half of∆we. What has
happened to the other half?
Writing the energy balance and excluding the loss and mechanical storage terms:
∆we + ∆wf
i.e (42-41)io + ∆wm =1/2 (Ų2-Ų1)io
Hence ∆wm =1/2 (Ų2-Ų1) io]
As 41 the mechanical input is negative: it is in fact an output work
(force x displacement). A precisely similar consideration gives for the rotary case (b)
the output work (torgue x angular displacement). For constant current, therefore, the
source provides ∆we, of which one-half is taken as energy into the filed and the
other half is converted into mechanical energy output.
5. energy conversion
(2)
Week 14
CONSTANT FLUX
Let the flux be kept constant so that linkage is always . the condition implies that the
current fall from Li to i1 to compensate for the rise in permeance in Fig (e). there is
no electrical energy transfer between the source and system for with constant linkage
there is no induced e.m.f to be balance. Hence ∆we =o. but there is change of filed
energy ∆wf =1/2Ų0 (l2-l1) which is negative because l2 <l1. the energy balance is
O +∆wm =1/2 Ų0 (l2 –l1)
Hence ∆wm =fm (-∆x) =1/2 Ų0 (l2 –l1)
For constant flux, therefore, the mechanical work done comes from an equal
reduction in filed energy.
(3)
GENERAL CONDITION
In a practical device neither of condition (1) and (2) is likely to apply consistently.
The transition will follow some arbitrary contour, such as that in Fig 2.7 (f), with
changes in both 4 and i. the energy balance is then some combination of cases (1) and
(2). The change in field energy is the shaded area ∆wf, and this will correspond, as
before in magnitude to the mechanical energy ∆wm even, if owing to saturation
effect, the Q/I relation is non- linear, they are ∆wf can still be found by graphical
integration. Ni any case,
The mean force
The mean torgue
=
∆wf,
∆x
=
fm
2.28
∆wf,
=
mm
2.29
5. energy conversion
(4)
Week 14
DIFFENTIAL FORM
IF ∆ AND ∆Q are reduced to the infinitesimal differentials dx and dq, the force and
torgue are obtained for a single position x or Q. Then
Force, fm
=
torgue, mm
=
dwf
dx
dwf
dQ
2.30
2.31
5. energy conversion
Week 15
5.12 ALIGNMENT FORCE AND TORQUE: SINGLE EXCITATION
Fig: 2.8 Reluctance motor
The reluctance motor shown in figure above depend on the tendency of the rotor to
Align itself magnetically with the stator. A flux plot for the machine shows that,
provided there is ad equate ove4rlop, all the active flux and all the field energy can be
assumed to occupy the overlap regions. The active gap volume changes with angle Q
between the two magnetic exes, and the torgue is d e f/dQ eqtn 2.31 using equation
2.22 a basis. An angular increase Dq others the active volume of each gap by - rlg
dQ and the gap energy by -1/2 B2 Lr Dq/Uo
the torgue for the two poles is consequently
me
=
dwe
Dq
=
B2Lrlg
uo
2.24
=
4π x10-7 (const.)
The minus sign indicating that the force acts to reduce Q.
Example 1
With the rotor dimentions shown and a coil of 400 turns carrying 1.6A, calculated the
torgue acting on the rotor.
5. energy conversion
Week 15
Solution
The mmf (F) =
NI 400 X 1.6
Area fine B =UOH
Then F = 640
:.
=
640 A.t HX 21g
=>
BX2lg
Uo
B =640 x UO
2lg
640 X 4π X 190-7
2 x 10-3
= 0.40 tesla
0.43
i.e The flux density in each 1mm gap B = 0.40T.
:.
me
=
-0.42 X 0.025 X0.03 X0.001 X107 = 0.0955N-M
-B2l/g4 π
Example 2
The 4-l characteristics of a magnetic circuit are frequently described with straight
segments as shown below. The act is considered linear up to pt. a and in saturation
from a to 5 find the field energy
Example 3
A dynamic phonograph pickup consists of a 20-turn coil length of each coil =1cm)
moving normal to a field of B=0.2t. if the maximum allowable amplitude is 0.02mm,
calculated the output voltage at 100-HZ and at 100Hz
Solution
5. energy conversion
Week 15
A dynamic phonograph pickup for vertical recording the effective
conductor in moving coil is
L
=
20turns =20cm =0.2m
For a sinusoidal displacement of amplitude 2 x 10-5 m =0.02mm
X (t)
=Asinwt = 2x 10-5 sinwt
The velocity at 100HZ is d x /dt or
U
=dx =2 x10-5 x 2π x 103conwt m/s.
Neglecting internal impedance, the output voltage is
V
=
e =Blu =0.2 x 0.2 x 4π x 10-2 coswt colt
The r.m.s value of output voltage is V = vmax = 0.2 x0.2 x 4 π x10-2
2
2
= 0.0036v =3.6mv
(ii)
The velocity at 100HZ is
U
=
dx =2 x 100-5 x 2 π x 102 coswt m/s
And output voltage (neglecting internal impedance) is
V
=e bLu = 0.2 x 0.2 4 π x 10-3 coswt volt
And r.m.s value of output voltage
=
0.2 x 0.2 x 4 π x 10-3
2
=
0.00036v =0.36mv
length of
5. energy conversion
Week 15
Example4.
In a d.c machine, shown above, the armature is wound on a laminated iron cylinder
15cm long and 15cm in diameter. The N and S role faces are 15cm long (into the
paper) and 10cm along the circumference the average flux density in the air gap
under the pole faces is 1T. If there are 80 conductors in series between the brushes
and the machine turning at N =1500rpm, calculated the no-load terminal voltage
Solution
For Ns conductor (no coils) in series, the average emf is
E =NS ∆Ø wb =Ns Ø wb P poles
∆t s
pole
rev
=
Ns Øpn
n
60
rev
sec
No 2 conductor in sec)
volts
Where the flux per pole Ø = BA = 1 x0. 15 x 0.1 = 0.015 wb.
:.
Example 5.
E
=
80 x 0.015 x 2 x 1500
60
=
60 volts
5. energy conversion
Week 15
A magnetic circuit is completed through a soft –iron rotor as shown in the figure
above. Assuming (1) all the reluctance of the magnetic circuit is in the air gaps of
length L
(ii)
There is no fringing so the effective area of each gap is the area
Derive on expression for the torgue as a function of angular position.
Solution
The total reluctance for two air gaps in series is
R
=
2l
UOA
(since Ur =1 for air) R = F =1L
Q A
=
2L
Urwq
For an N-turn coil, the inductance is
L
=
NØ = NF
N2I
I
IR
IR
=
N2 trwØ
2L
Since torgue =1/2 I2 dL = Uo N2rw
dØ
2L
The torgue is independent of Ø under the assume conditions.
Example 6
(a) Wiring diagram
(b)Steady state model
A commutator machine, with the wiring diagram and steady-state model showed
above, is rated 5KW , 250V, 2000rmp. The armature resistance RA is 1. Drive from
the electrical and at 2000rmp, the no-load powder input to the armature is IA = 1.2A
at 250V with the field winding (RE =250) excited by IF =1A. Calculate the efficiency
of this machine.
5. energy conversion
Week 15
Solution
In fig (a), input power IF 2RF = 12 X 250 =250W is required to provide the necessary
magnetic flux. This is power lost and it appears as heat.
In no-load steady operation, there is no output and no change is all loss
[The armature cooper loss at no load is negligible (1.22 x 1=1.44w) and most of the
input power at no- load goes to supply air, bearing brush friction, eddy cumenty and
hystersis losses] the losses are associated with flux changes in the rotting armature
core, are dependent on speed but are nearly independent of load.
Hence, field loss
= IF2RF
=
250W
IAV
=
At full-load of 5KW, IA =5000W
=
20A
And the armature copper loss
=
IA2 Ra =202 x1 =400w
And rotational loss
=
300w p =iv :.i = p1
v
The energy balance equation gives
Mech. Energy
:input
+
field elect
input output
=
increase
energy
Field
energy
stored converted
to heat
i.e
Mech. input
+
250-5000
= 0
field arm
less less
rotat
less
Mech. input + 250 -5000 = 0 + 250 +300 + 400
(field input is all loss and appears o n both sides)
Mech. Input = 500 + 300 + 400 =5700W.
Hence efficiency
=
output
Input
=
elect output
mech. output + elect input
=
5000
5700 + 250
=
0.84. or 84%
5. energy conversion
Week 15
Example 7
In the relay shown above, the contacts are held open by the spring excerting a force
of 0.1N . The gap length is 4mm when the contacts are open
and 1mm when
closed. The coil of 5000 turns would on core 1cm2 in cross- section. Assuming
1)
All reluctance is in a uniform air gap