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
ELECTRICAL ENGINEERING PROGRAM COURSE CONTENT
Subject
Machine II
Year
2
Semester
3
Course Code
EEC233
Credit Hours
3
Theoretical
1
Practical
2
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
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
3
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
4
5.12 Alignment force and torque; single excitation ....................Week15
5.11 Work examples ................................................................................... 56
5.11.3 General Condition
5
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
convert 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
6
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
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
7
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)
8
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.
9
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
(a)
Lifting magnet
(f)
Actuator
10
1.3
(b)
Relay
(g)
Electromagnetic pump
( c)
Telephone
(h)
Loudspeaker
(d)
Moving-iron indicator (i)
Moving –coil indicator
(e)
Reluctance motor
(k)
Industrial rotating machine
Alignment devices
(a)
The lifting magnet: Attract ferromagnetic loads such as beams, plates, and
scrap-iron.
(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.
1.4
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
11
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)
(n)
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
12
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
=
900
1 second
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
t
13
1/60
=
0.012 – 0.01667 = 0.72v
=
0.72v
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 106
wb.
:. Average emf induced in coil
= (change in flux x No of turns) = NdØ
Time taken.
Dt
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.
14
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.
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
15
0
Time1
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.
16
N
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
17
Week 2
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 three-phase 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.
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
18
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°.
19
10
5
-5
-10
1
Point1
Point4
2
3
4
5
6
7
Point2
Point3
Point 5
Point6
Point7
5
4. Protection & control of electric motors
Week 8
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
4. Protection & control of electric motors
Week 8
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
7
4. Protection & control of electric motors
Week 8
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
 Squirrel cage type
 Wound rotor
2.2.2.1 Squirrel cage rotor:
8
4. Protection & control of electric motors
Week 8
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:
9
4. Protection & control of electric motors
Week 8
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
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
components from the environment. A cooling fan is
10
Shaft
4. Protection & control of electric motors
Week 8
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.
11
4. Protection & control of electric motors
Figure2.7: Sectional view of Induction Motor
12
Week 8
4. Protection & control of electric motors
Week 8
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
13
4. Protection & control of electric motors
Week 8
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
Hertz
Rated supply frequency
Frame
External dimensions based on NEMA Regulations
14
4. Protection & control of electric motors
Week 8
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 2.2:Explanation of the codes used on AC motor nameplates
15
4. Protection & control of electric motors
Week 8
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
Metal frame
Laminated iron
core with slots
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 alternator,
Insulated copper
bars are placed in
the slots to form
the three-phase
winding
which is itself excited by the pilot exciter. In
externally fed fields, the source can be a shaft-driven
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).
16
4. Protection & control of electric motors
Week 8
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
17
4. Protection & control of electric motors
Week 8
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 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
18
4. Protection & control of electric motors
Week 8
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
19
4. Protection & control of electric motors
Week 8
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.
20
4. Protection & control of electric motors
Week 8
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.
Short Circuit
conductor
Insulation
Figure4. 3: Short circuit fault between two condu
21
4. Protection & control of electric motors
Week 8
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,
22
4. Protection & control of electric motors
Week 8
is easy to replace, totally 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
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
23
Figure 4.5: A HBC fuse
4. Protection & control of electric motors
Week 8
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
circuit
breakers
generally
employ
a
bimetal
strip
to
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
24
4. Protection & control of electric motors
Week 8
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).
25
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
26
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
To Power Circuit
Relay Open
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
Figure4.11: a relay
controlling switch is open, so the relay is de-energised and
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.
27
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
de-energized, the magnetic field collapses, and the movable contacts open under
spring pressure.
28
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
4.3.2.2 Contactors and Overload Relays
29
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
30
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.
.
31
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.
32
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
33
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 coverted 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 as arrangement, has obvious limitations. It may also be
34
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 energy-conversion
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
=
10-8
R
=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
Conversely, if e 12V,
= 1 x RT2 = 4W
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)
5.3
Linked energy systems
35
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 short-circuited, 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 given stimulus, such as the occurrence of a transmission- line fault or
starting of a large motor.
5.4 Energy storage
36
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.
37
Week 12
5.5 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
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
38
corresponding symbols these relations are total energy wf + ws + w
2.2(a)
Energy change dwe + dwm = dwf + dws + dw
Energy rate Pe + pm
= dwf + dws
dt
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 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
39
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 wellestablished principles and laws summarized in the following table.
Part of system
Quantities
Principles
Electrical
Voltage, current
Faraday-Lenz and
Kirchhoff laws
Conversion
E.M.F
current
magnetic Magneto- mechanical
40
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
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
41
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.
42
Week 13
5.6 Magnetic field energy and forces
In order to be able to analyise mathematically the electromechnical 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.
43
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
cross- sectional 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
Which means that H = Ub = 800000B
44
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 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.
45
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.
.
46
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 airgap, 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 (m3 is [HdB, and for and values Bi and H1 the final energy density (shaded area) is
clearly ½ B1H1 . 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.
47
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]
48
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 against the positive direction of x) of the right-hand member’
49
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.
(2)
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
50
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
(4)
∆wf,
∆x
=
fm
2.28
∆wf,
=
mm
2.29
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
2.30
=
torgue, mm
2.31
=
dwf
dx
dwf
dQ
51
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.
52
Solution
The mmf (F) =
NI 400 X 1.6
Area fine B =UOH
Then F = 640
:.
=
640 A.t HX 21g
=>
BX2lg
Uo
640 X 4π X 190-7
2 x 10-3
B =640 x UO
2lg
= 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
A dynamic phonograph pickup for vertical recording the effective length of
conductor in moving coil is
L
=
20turns =20cm =0.2m
53
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
Example4.
54
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.
:.
E
=
80 x 0.015 x 2 x 1500
60
=
60 volts
Example 5.
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
55
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.
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
56
And rotational loss
=
IAV
=
300w p =iv :.i = p1
v
At full-load of 5KW, IA =5000W
=
20A
And the armature copper loss
=
IA2 Ra =202 x1 =400w
The energy balance equation gives
Mech. Energy
:input
converted
+
field elect
input output
=
increase
energy
Field
energy
stored
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%
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
57