COURSE:
INTRODUCTION TO ELECTRICAL MACHINESProf Elisete Ternes Pereira, PhD
SYNOPSIS Introducing the Basic types of Electric
Machines
a) A.C. Motors - Induction and Synchronous Motorsb) Ideal and Practical Transformersc) D.C. Motors and Generatorsd) Self and Separately Exited Motorse) Stepper Motors and their characteristicsf) Assessment of Electric Motors
i. Efficiencyii. Energy lossesiii. Motor load analysisiv. Energy efficiency opportunity analysisv. Improve power qualityvi. Rewindingvii. Power factorviii. Speed control
INTRODUCTION1
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 TO 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 (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.
enclosedIdH
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:
The right-hand side of the equation represents the current enclosed by the integration path and is called the magnetomotive force (MMF).
enclosedIdH
AMPERE`S LAW
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.
enclosedIdH
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:
dt
dVind
indV
I
Electric circuit (loop of conducting material)
Magnetic flux
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
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.
SdBdt
ddE
dt
dVind
where E is the electric field and B is the magnetic flux density.
or
MAGNETIC FORCE
The change in
or
)( BxdIFd
I
I
F
F
BASIC CONCEPTS - REVIEW
1. MAGNETIC CIRCUITS
CONCEPTS REVIEWMAGNETIC CIRCUITS
Lets consider first, the most basic ideal circuit:
Some relevant parameters:
=m mrm0 >> m0 Ac N
lc i
CONCEPTS REVIEWMAGNETIC CIRCUITS
Ampère`s law applied over the typical mean-length (lc) results in:
NiH
A
iNHABA
or
or
tcidH
0 sdBs
S
adB
cBA
HB or
The Magnetic Flux can be written as a function of B:
The Magnetic Flux Density, B, in terms of the Magnetic Field, H, is:
for magnetic circuits
for magnetic circuits Substituting we find:
CONCEPTS REVIEWMAGNETIC CIRCUITS
A
iN
or
This found equation:
m
fmm
Can be written in terms of the `Magnetomotive Force`:
So that:
Nifmm
dHNifmm
But,
Then, for magnetic circuits:
ccHNifmm
CONCEPTS REVIEWMAGNETIC CIRCUITS
A
iN
or
Also, if this equation:
m
fmm
Then, the `Magnetic Reluctance` is given by:
is equal to this:
cr
cm A0
BASIC CONCEPTS - REVIEW
2. TRANSFORMERS PRINCIPLES
CONCEPTS REVIEWTRANSFORMER BASICS
Conceptual Schematic – Ideal Transformer
Ideal circuit = no loses.
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 REVIEWTRANSFORMER BASICS
Flux generated by current 1 in coil 1:
By Faraday`s law, the induced voltage is:
Since there is no losses, the induced voltage is exactly the same as the applied voltage in coil 1:
)(111 tN
dt
d 11
dt
tdN
dt
dtv
)()( 1
111
CONCEPTS REVIEWTRANSFORMER BASICS
The flux in coil 2, that was generated by current 1:
And so, the voltage induced in coil 2 is given by the equation:
)(221 tN
dt
tdN
dt
dtv
)()( 2
212
CONCEPTS REVIEWTRANSFORMER BASICS
From the voltage equations:
We get the Transformer Ratio Equation:
dt
tdN
dt
dtv
)()( 2
212
dt
tdN
dt
dtv
)()( 1
111
2
1
2
1
)(
)(
N
N
tv
tv
CONCEPTS REVIEWTRANSFORMER 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:
)()()( 2211 tiNtiNtH
CONCEPTS REVIEWTRANSFORMER BASICS
For this expression:
When i2(t) equal to zero:
The flux in coil 1, produced by current 1, is:
But, and
So:
)()()( 2211 tiNtiNtH
)(
)( 11 tiNtH
)()( 111 tNt
AtHNt )()( 111
AtBt )()( )()( tHtB
CONCEPTS REVIEWTRANSFORMER BASICS
Substituting
Into:
We get the expression of the flux in coil 1 produced by current 1:
where L1 is the self inductance of coil 1; in this case given by:
)(
)( 11 tiNtH
AtHNt )()( 111
AN
L2
11
)()()(
)( 111
2111
111 tiLtiAN
AtiN
Nt
CONCEPTS REVIEWTRANSFORMER BASICS
For the general case, when i2 0
and:
)()(
)( 2211 tiNtiNtH
AtiN
AtiN
AtHAtBt
)()()()()( 2211
CONCEPTS REVIEWTRANSFORMER BASICS
The flux in coil 1, produced by both currents, i1 and i2, is given by:
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:
In a similar we may find the flux in coil 2:
)()()()( 221
1
21
11 tiANN
tiAN
tNt
)()()( 2111 tMitiLt
)()( 22 tNt )()()( 2212 tiLtiMt
CONCEPTS REVIEWTRANSFORMER BASICS
The induced voltages in each coil are, then:
Given that: and
dt
tdiM
dt
tdiLtv
)()()( 21
11
dt
tdiL
dt
tdiMtv
)()()( 2
21
2
dt
tdtv
)()(
)(
)(ti
tL
CONCEPTS REVIEWTRANSFORMER BASICS
Another equation very much used in transformers design and analysis is the following:
This, however, is not an exact equation and can only be used when the magnetic permeability of the nucleus can be considered infinite.
1
2
2
1
N
N
i
i
1
2
2
1
)(
)(
N
N
ti
ti
CONCEPTS REVIEWTRANSFORMER BASICS
or
This relation comes from Ampere`s law, that for this case is:
When we assume a very large r so that H 0. In this case:
1
2
2
1
N
N
i
i
)()()( 2211 tHtiNtiN
)()( 2211 tiNtiN
CONCEPTS REVIEWTRANSFORMER BASICS
and
The negative sign indicates that the currents produce magnetic fields with opposite polarities.
Another equation extensively employed in the design of transformers is the following:
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:
wAN
VB pico
pico1
)sen()(1 wtVtv pico
CONCEPTS REVIEWTRANSFORMER BASICS
With primary voltage:
The flux then will be:
Such that:
Resulting in,
wAN
V
AB picopico
pico1
)sen()(1 wtVtv pico
)cos()( wtt pico
)sen()cos()sen( 11 wtwNwtdt
dNwtV picopicopico
CONCEPTS REVIEWTRANSFORMER BASICS
BASIC CONCEPTS - REVIEW
2. ELECTROMECHANICAL ENERGY-CONVERSION
ENERGY and FORCE
Energy storage in a system of current conductors
Most of the important applications of electromagnetic fields are based in the capacity to store energy.
JJThe instantaneous input power given by the source is:
So, the input energy is:
CONCEPTS REVIEWENERGY CONVERSION
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.
ivP .
t
dtivw0
.
Input energy:
Faraday`s law:
Input energy:
where is the linkage flux
This integral equation gives the total energy stored in the system
CONCEPTS REVIEWENERGY CONVERSION
t
dtivw0
.
dt
dv
t N
ididtdt
dw
0 0
0
The processes of energizing the winding is seeing in the figure:
CONCEPTS REVIEWENERGY CONVERSION
t N
ididtdt
dw
0 0
0
0 N 0
i0
i
Stored ENERGY
Co-ENERGY
The area above the curve is numerically equal to the Stored Energy.
There is no physical correspondence to the area below the curve, but it is called Co-Energy.
If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY
In linear systems:
CONCEPTS REVIEWENERGY CONVERSION
Linear System
0 N 0
i
Stored ENERGY
Co-ENERGY
orLi
imm '
25,05,0' iLimm
iL
If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY
In linear systems:
CONCEPTS REVIEWENERGY CONVERSION
0 N 0
i0
i
Stored ENERGY
Co-ENERGY
Non-Linear System Linear System
0 N 0
i
Stored ENERGY
Co-ENERGY
222
0 2
1
2
)(
2
0
ooo
m LiL
Li
Ld
Lwenergystored
t N
ididtdt
dw
0 0
0
If the system is LINEAR the ENERGY is EQUAL to the CO-ENERGY
It can also be shown that the Energy per Volume Unit is:
CONCEPTS REVIEWENERGY CONVERSION
Linear System
0 N 0
i
Stored ENERGY
Co-ENERGY
22
2
1
2Li
Lwm
)/(22
1 32
2 mJB
Hwm
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:
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 REVIEWENERGY CONVERSION
Coil CoreSpring
Mass
xElectric Source
i
Fm
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:
Since: , a small energy variation is given by:
CONCEPTS REVIEWENERGY CONVERTION
)( dxFidd mm
imm '
iddidd mm '
By substitution we arrive in the following equation:
The magnetic force can, then, be found, as a function of current (i) and position (x or ) - by the equation:
CONCEPTS REVIEWENERGY CONVERTION
dxFddi mm '
x
xiF m
m
),('
The magnetic force can, then, be found, as a function of current (i) and position (x or ) - by the equation:
In lienar systems: ; for this case we can write the equations:
CONCEPTS REVIEWMAGNETIC FORCE
x
xiF m
m
),('
),(' i
T m
x
xiLiFm
),(5,0 2
),(
5,0 2 iLiT
2, 5,0 Lim
for rotating systemsor
or for rotating systems
Alternatively, we can obtain the force as a function of flux () and position (x or ), when and x/ are chosen as independent variables:
And for linear systems:
CONCEPTS REVIEWMAGNETIC FORCE
x
xF m
m
,
,mmT
x
xF m
m
,
,mmT
or
or
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 REVIEWSTEP 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.
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 m-mm.
CONCEPTS REVIEWSTEP 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
So, the rotor moves clockwise by an angle of: 90º - 60º = 30º
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 .
CONCEPTS REVIEWSTEP MOTOR
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.
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
CONCEPTS REVIEWSTEP MOTOR
Step motors may be easily electronically controlled .
They may be operated at low speeds and admit acceleration without difficulty.
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:
CONCEPTS REVIEWMOTORS
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 :
t t t
m dtidt
ddti
dt
ddtiviv
0 0 0
22
11
2211 )(
t t
m idid0 0
2211
The linkage flux in coil 1 is partialy due to current i1 and partialy to currente i2 and is given by:
CONCEPTS REVIEWMOTORS
Where L1 is the self inductance of coil 1 and M is the mutual inductance
2111 MiiL
Similarly, the linkage flux in coil 2 is given by:
1222 MiiL
CONCEPTS REVIEWMOTORS
Similarly, the linkage flux in coil 2 is given by:
So that,
t t I I
diiMdiiLMiiLdidi0 0 0 0
21111211111
1 2
)(
2
0
212
110 11 5,0I
tdiiMILdi
1
0
12222
0
22 5,0It
diiMILdi
Substituting:
CONCEPTS REVIEWMOTORS
in this previously given equation:
2
0
212
110 11 5,0I
tdiiMILdi
1
0
12222
0
22 5,0It
diiMILdi
)(5,05,0 21222
211 iidMILILm
t t
m idid0 0
2211
We obtain:
21222
211 5,05,0 IMIILILm
and
or
In a linear system , so the torque in the rotor is obtained:mm ,
d
dMII
d
dLI
d
dLIT m
m 2122
212
1 5,05,0'
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