COURSE:
INTRODUCTION TO
ELECTRICAL MACHINES
Prof Elisete Ternes Pereira, PhD
SYNOPSIS

a)
b)
c)
d)
e)
f)
Introducing the Basic types of Electric
Machines
A.C. Motors - Induction and Synchronous Motors
Ideal and Practical Transformers
D.C. Motors and Generators
Self and Separately Exited Motors
Stepper Motors and their characteristics
Assessment of Electric Motors
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
Efficiency
Energy losses
Motor load analysis
Energy efficiency opportunity analysis
Improve power quality
Rewinding
Power factor
Speed control
INTRODUCTION
1
INTRODUCTION
TO
ELECTRICAL MACHINES

Essentially all electric energy is generated in a rotating machine: the
synchronous generator, and most of it is consumed by: electric
motors.
 In many ways, the world’s entire technology is based on these
devices.

The study of the behavior of electric machines is based on three
fundamental principles:
 Ampere’s law,
 Faraday’s law and
 Newton’s Law.
INTRODUCTION

TO
ELECTRICAL MACHINES
Various configurations result and are classified generally by the type
of electrical system to which the machine is connected:

direct current (dc) machines or

alternating current (ac) machines.
INTRODUCTION

TO
ELECTRICAL MACHINES
Machines with a dc supply are further divided into permanent
magnet and wound field types, as shown in Figure 4.1.
INTRODUCTION

ELECTRICAL MACHINES
The wound motors are further classified according to the
connections used:

The field and armature may have separate sources



(separately excited),
they may be connected in parallel

(shunt connected), or
they may be series


TO
(series connected).
(figure follows)
INTRODUCTION



TO
ELECTRICAL MACHINES
AC machines are usually
 single-phase or
 three-phase machines
and may be
 synchronous or
 asynchronous.
See figure next page.
INTRODUCTION

TO
ELECTRICAL MACHINES
Several variations are shown in Figure 4.2.
1.1 - BASIC ELECTROMAGNETIC LAWS:
AMPERE`S LAW AND FARADAY`S LAW

The two principles that describe the electromagnetic
behavior of electric machines are Ampere’s Law and
Faraday’s Law.


These are two of Maxwell’s equations.
Most electric machines operate by attraction or repulsion of
electromagnets and/or permanent magnets.
AMPERE`S LAW


Ampere’s law describes the magnetic field that can be produced
by currents or magnets.
In an electric machine, there will always be at least one set of
coils with currents.

A motor cannot be produced with permanent magnets alone.

 H  d  I enclosed
AMPERE`S LAW


Ampere’s law states that the line integral of the component of the
magnetic field along the path of integration is equal to the current
enclosed by the path.
This is exactly true for static fields and is a very good
approximation for the low-frequency fields dealt with in electric
machines:

 H  d  I enclosed

The right-hand side of the equation represents the current enclosed by
the integration path and is called the magnetomotive force (MMF).

AMPERE`S LAW  H  d  I enclosed




In electric machines, currents are
frequently placed in slots
surrounded by ferromagnetic
teeth.
The MMF corresponding to each
path is the total current enclosed
by the path.
If the slots contain currents that
are approximately sinusoidally
distributed , then the MMF will be
cosinusoidally distributed in space.
In this way, the magnetic field or
flux density in the air gaps of the
machine will often have a
sinusoidal or cosinusoidal
distribution.
An example illustrating the
determination of the MMF is shown
in Figure 4.3, where different
integration paths are shown by dotted
lines.
FARADAY`S LAW

&
EMF
Faraday’s law relates the induced voltage, or electromotive force
(EMF), to the time rate of change of the magnetic flux linkage:
Vind 
d
dt
Magnetic flux

Electric circuit (loop of
conducting material)
I
Vind


For voltage to be induced,
there has to be a variation
in time between the relative
position of the magnetic
flux and the electric circuit
If the electric circuit is closed and current is allowed to flow, the current will produce a magnetic
flux that opposes the increase of the applied flux = Lenz`s law
FARADAY`S LAW
Vind
d

dt
or
 
d  
 E  d    dt  B  dS
where E is the electric field and B
is the magnetic flux density.

This law states that the voltage induced in a loop is equal to the time
rate of change of the flux linking the loop.

The negative sign indicates that the voltage is induced such that the
current would oppose the change in flux linkage.

The change in flux linkage can be caused by a change in flux density
and/or a change in geometry.
MAGNETIC FORCE

 
dF  I (d  x B)
or

F

I
I

The change in

F
BASIC CONCEPTS - REVIEW
1. MAGNETIC CIRCUITS
CONCEPTS REVIEW
MAGNETIC CIRCUITS

Lets consider first, the most basic ideal circuit:

Some relevant parameters:
 m mrm0 >> m0
Ac
 lc
i
N

CONCEPTS REVIEW
MAGNETIC CIRCUITS
Ampère`s law applied over the typical
mean-length (lc) results in:

 
 H  d   it
c

for magnetic circuits
H  Ni
The Magnetic Flux can be written as a
function of B:
 
   B  da
for magnetic circuits
  BAc

Substituting we find:
S


B
 d s  0
  BA  m r mo HA 
s

The Magnetic Flux Density, B, in
terms of the Magnetic Field, H, is:


B  m r mo H
Ni

 m m
r o


A 
CONCEPTS REVIEW
MAGNETIC CIRCUITS

This found equation:



Ni

 m m
r
o

fmm  Ni

A 

But,

Then, for magnetic circuits:
Can be written in terms of
the `Magnetomotive
Force`:

fmm
m
So that:
 
fmm  Ni   H  d 
fmm  Ni  H c  c
CONCEPTS REVIEW
MAGNETIC CIRCUITS

Also, if this equation:

Ni

 m m
r
o



A 

is equal to this:
fmm

m
Then, the `Magnetic Reluctance` is given by:
m 
c
m r m 0 Ac
BASIC CONCEPTS - REVIEW
2. TRANSFORMERS PRINCIPLES
CONCEPTS REVIEW
TRANSFORMER BASICS

Conceptual Schematic – Ideal Transformer

Ideal circuit = no loses.
CONCEPTS REVIEW
TRANSFORMER BASICS

FIRST: a voltage source is connected in the primary side
and the secondary side is an open circuit;


we want to find the voltage induced in the open
secondary coil
When the primary is energized: current in the primary coil
 magnetic flux in the core.
CONCEPTS REVIEW
TRANSFORMER BASICS

Flux generated by current 1 in coil 1:
11  N1  (t )

By Faraday`s law, the induced voltage is:
d11
dt

Since there is no losses, the induced voltage is exactly the
same as the applied voltage in coil 1:
d11
d (t )
v1 (t ) 
 N1
dt
dt
CONCEPTS REVIEW
TRANSFORMER BASICS

The flux in coil 2, that was generated by current 1:
21  N2  (t )

And so, the voltage induced in coil 2 is given by the
equation:
d21
d (t )
v2 (t ) 
 N2
dt
dt
CONCEPTS REVIEW
TRANSFORMER BASICS

From the voltage equations:
v1 (t ) 

d11
d (t )
 N1
dt
dt
v2 (t ) 
We get the Transformer Ratio Equation:
v1 (t ) N1

v2 (t ) N 2
d21
d (t )
 N2
dt
dt
CONCEPTS REVIEW
TRANSFORMER BASICS

SECOND: there is now a load connected to the secondary
coil, so i2(t) can flow.


We want to find the new induced voltage.
By applying Ampere`s law to the circuit, using the line of
average path/length , we get the following expression:
H (t )   N1 i1 (t )  N2 i2 (t )
CONCEPTS REVIEW
TRANSFORMER BASICS

For this expression:
H (t )   N1 i1 (t )  N2 i2 (t )

When i2(t) equal to zero:

The flux in coil 1, produced by current 1, is:

But,

So:
 (t )  B(t ) A
and
H (t ) 
N1 i1 (t )

B (t )  m H (t )
11(t )  N1m H (t ) A
11(t )  N1  (t )
CONCEPTS REVIEW
TRANSFORMER BASICS

Substituting H (t ) 
N1 i1 (t )



Into:
11(t )  N1m H (t ) A
We get the expression of the flux in coil 1 produced by
current 1:
N1 i1 (t )
N12 A
11 (t )  N1m
Am
i1 (t )  L1 i1 (t )


where L1 is the self inductance of coil 1; in this case given by:
N12 A
L1  m

CONCEPTS REVIEW
TRANSFORMER BASICS

For the general case, when i2  0
N1 i1 (t ) N 2 i2 (t )
H (t ) 




and:
N1i1 (t )
N 2i2 (t )
 (t )  B(t ) A  mH (t ) A  m
A m
A


CONCEPTS REVIEW
TRANSFORMER BASICS

The flux in coil 1, produced by both currents, i1 and i2, is
given by:
 mN12 A 
 mN N A 
 i1 (t )   1 2  i2 (t )
1 (t )  N1 (t )  



  

The first term in parenthesis is the self-inductance of coil 1,
the second term is the mutual inductance between coils 1
and 2; then:
1 (t )  L1i1 (t )  Mi2 (t )

In a similar we may find the flux in coil 2:
2 (t )  N2 (t )
2 (t )  M i1 (t )  L2 i2 (t )
CONCEPTS REVIEW
TRANSFORMER BASICS

The induced voltages in each coil are, then:
v1 (t )  L1
di1 (t )
di (t )
M 2
dt
dt
v2 (t )  M

di1 (t )
di (t )
 L2 2
dt
dt
d (t )
Given that: v(t ) 
and
dt
L   (t )
i (t )
CONCEPTS REVIEW
TRANSFORMER BASICS

Another equation very much used in transformers design
and analysis is the following:
i1 (t ) N 2

i2 (t ) N1

or
i1 N 2

i2 N 1
This, however, is not an exact equation and can only be
used when the magnetic permeability of the nucleus can be
considered infinite.
CONCEPTS REVIEW
TRANSFORMER BASICS

This relation comes from Ampere`s law, that for this case
is:
N1i1 (t )  N2i2 (t )  H (t )

When we assume a very large mr so that H  0. In this case:
N1i1 (t )   N2i2 (t )
The negative sign indicates that
the currents produce magnetic
fields with opposite polarities.

and
i1 N 2

i2 N1
CONCEPTS REVIEW
TRANSFORMER BASICS

Another equation extensively employed in the design of
transformers is the following:
B pico 


V pico
N1 wA
It is called “the Design Equation” and it encounters many
practical usage.
To deduce it we assume a sinusoidal voltage applied to the
primary side when the secondary is open:
v1 (t )  V pico sen( wt )
CONCEPTS REVIEW
TRANSFORMER BASICS

With primary voltage: v1 (t )  V pico sen( wt )

The flux then will be:

Such that:
 (t )   pico cos( wt )
V pico sen( wt )  N 1

Resulting in,
B pico 
d
 pico cos( wt )  N1 w pico sen( wt )
dt
 pico
A

V pico
N1wA
BASIC CONCEPTS - REVIEW
2. ELECTROMECHANICAL ENERGY-CONVERSION
ENERGY and FORCE
CONCEPTS REVIEW
ENERGY CONVERSION

Energy storage in a system of current conductors

Most of the important applications of electromagnetic fields are based in
the capacity to store energy.
In this ideal magnetic circuito the
energy must be stored in the
system of conductors of current,
made of a N turns winding and by
currente i.


ًًThe instantaneous input power given by the source is: P  v.i
So, the input energy is:
t
w   v. i dt
0
CONCEPTS REVIEW
ENERGY CONVERSION
t

Input energy:
w   v. i dt
0

Faraday`s law:

Input energy:
v
d
dt
d
w
idt 
dt
0
t
0  N
 id 
0
where  is the linkage flux
This integral equation gives the total energy stored in the system
CONCEPTS REVIEW
ENERGY CONVERSION
d
w
idt 
dt
0
t

0  N
 id 
0
The processes of energizing the winding is seeing in the figure:

The area above the curve 
is numerically equal to the
Stored Energy.
0  N 0
Stored
ENERGY
There is no physical 
correspondence to the area
below the curve, but it is
called Co-Energy.
Co-ENERGY
i0
i
CONCEPTS REVIEW
ENERGY CONVERSION

If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY

Linear System
0  N 0
m   'm  i
Stored
ENERGY
Co-ENERGY
i

  Li
or
In linear systems:
m   'm  0,5i  0,5L i 2

L
i
d
w
idt 
dt
0
CONCEPTS REVIEW
ENERGY CONVERSION

t
If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY


Non-Linear System
Linear System
0  N 0
0  N 0
Stored
ENERGY
Stored
ENERGY
Co-ENERGY
Co-ENERGY
i0

i
i
In linear systems:
0

2o
( Lio ) 2 1 2
stored energy  wm   d 

 Lio
L
2L
2L
2
0
0  N
 id 
0
CONCEPTS REVIEW
ENERGY CONVERSION

If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY

Linear System
0  N 0
2
1 2
wm 
 Li
2L 2
Stored
ENERGY
Co-ENERGY
i

It can also be shown that the Energy per Volume Unit is:
1
B2
2
wm  mH 
2
2m
( J / m3 )
CONCEPTS REVIEW
ENERGY CONVERSION

Force

Lets now consider a magneto-mechanic arrangement, to see the
exchange of energy between the magnetic field and the mechanic
system, and how the magnetic force can be derived:
Coil
Spring
Core
Fm
Mass
i
Electric
Source


x
When the current flows in the coil the magnetic flux will produce a
force on the iron-magnetic core pulling it to the coil nucleus.
This is how the interaction occurs.
CONCEPTS REVIEW
ENERGY CONVERTION

The force and the magnetic flux are depended of current and
position, that is: (i,x), Fm(i,x)

Or, it is equally true to state that, the force and the current are
depended of flux and position, that is: i( ,x), Fm( ,x)

The law of energy conservation requires that any variation in the
magnetic energy stored in the magnetic circuit should be balanced,
either by a variation in the input energy from the voltage source or by
a variation of energy in the mechanical system; the following
equation describes this requirement:
dm  id  ( Fm dx)

Since:
m   'm  i
, a small energy variation is given by:
dm  d 'm  di  id
CONCEPTS REVIEW
ENERGY CONVERTION

By substitution we arrive in the following equation:
di  d 'm   Fm dx

The magnetic force can, then, be found, as a function of current (i) and
position (x or ) - by the equation:
 'm (i, x)
Fm 
x
CONCEPTS REVIEW
MAGNETIC FORCE

The magnetic force can, then, be found, as a function of current (i) and position (x or
) - by the equation:
Fm 

 'm (i, x)
x
In lienar systems:
or
m,  0,5Li 2
 L(i, x) 
Fm  0,5i 2 

 x 
 'm (i, )
T

for rotating systems
; for this case we can write the equations:
or
 L(i,  ) 
T  0,5i 2 





for rotating systems
CONCEPTS REVIEW
MAGNETIC FORCE

Alternatively, we can obtain the force as a function of flux () and position (x or ),
when  and x/ are chosen as independent variables:
   , x  
Fm   m


x



or
 m  ,   
Tm  




And for linear systems:
 m  , x  
Fm  

x


or
   ,   
Tm   m





CONCEPTS REVIEW
STEP MOTOR
Now lets consider a machine with six poles in the stator (armature), arranged in
three groups (phases) a-aa, b-bb, c-cc.

Coils are wounded for the three phases but, for clarity's sake only the coils in phase
a-aa are indicated.


The rotor in this example has
four poles, as shown.

The idea is to review qualitatively
the behavior of this system after
the energizing of each phase
sequentially
CONCEPTS REVIEW
STEP MOTOR

When coil a-aa is energized, the rotor searches for a position of minimum reluctance,
corresponding, in this case, to the alignment of the rotor in the position: a-aa with I-II

Then, the current in coil a-aa is
interrupted and coil b-bb is energized

The position of minimum reluctance now
is reached when b-bb is aligned with mmm.

So, the rotor moves clockwise by an angle of: 90º 60º = 30º
CONCEPTS REVIEW
STEP MOTOR

As the windings become energized sequentially, one at each time, following from a-aa
 b-bb  c-cc, the rotor moves clockwise in steps of 30º .

This is a very useful and widely
employed machine, known as the
“Step Motor”.

If the windings are energized in the
sequence a-aa  c-cc  b-bb, the
rotor will turn anticlockwise.

The speed of the rotation is determined by the rate the current is switched from one
winding to the next.
CONCEPTS REVIEW
STEP MOTOR

Consider now the case where windings a-aa
and b-bb are energized simultaneously.

The position of minimum reluctance is not
reached by the alignment of a-aa with I-II or by
aligning b-bb with m-mm.

In fact, the rotor will stop in a position of partial
alignment between poles a-I and b-m.

This corresponds to a 15º rotation

Step motors may be easily electronically controlled .

They may be operated at low speeds and admit acceleration
without difficulty.
CONCEPTS REVIEW
MOTORS

Consider now the case where windings are
present in both, the stator and the rotor part of
the machine

This is a more practical case.

The energy stored in such systems can be
described by the equation:
d
d
m   (v1i1  v2i2 )dt   1 i1dt   2 i2 dt
dt
dt
0
0
0
t
t


t
t
t
0
0
m   d1i1   d2i2
1 and 2 are the total linkage flux in coils 1 and 2.
The linkage flux in coil 1 is partialy due to curren i1
and partialy to currente i2 :
CONCEPTS REVIEW
MOTORS

The linkage flux in coil 1 is partialy due to
current i1 and partialy to currente i2 and is given
by:
1  L1i1  Mi2
Where L1 is the self inductance of coil 1 and M is
the mutual inductance

Similarly, the linkage flux in coil 2 is given by:
2  L2i2  Mi1
CONCEPTS REVIEW
MOTORS

So that,
t
t
 i d   i d ( L i
1
1
1
0
11

0
I2
0
0
 Mi2 )  L1  i1di1  M  i1di2
0
t
I1
I2
i1d1  0,5L1I12  M  i1di2
0

Similarly, the linkage flux in coil 2 is given by:
t
I1
0
0
2
i
d


0
,
5
L
I
 M  i2 di1
2
2
2
2

CONCEPTS REVIEW
MOTORS

Substituting:
t
 i d
0
1
1
I2
 0,5L I  M  i1di2 and
2
1 1
0

I1
t
 i d
2
2
 0,5L I  M  i2 di1
2
2 2
0
0
in this previously given equation:
t
t
0
0
m   d1i1   d2i2

We obtain:
m  0,5L1I12  0,5L2 I 22  M  d (i1i2 )

In a linear system
m,  m
Tm 
or
m  0,5L1 I12  0,5L2 I 22  MI1 I 2
, so the torque in the rotor is obtained:
 'm
dL
dL
dM
 0,5I12 1  0,5I 22 2  I1 I 2

d
d
d

The presented developments (equations+ideas)
are useful in the study of the behavior of electrical
machines, and are used in the study of
electromechanical energy conversion.
Prof. Elisete Ternes Pereira, 2010