course: introduction to electrical machines prof elisete ternes pereira, phd

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.


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.


Machines with a dc supply are further divided into permanent magnet and wound field types, as shown in Figure 4.1.


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)


AC machines are usually single-phase or three-phase machines

and may be synchronous or asynchronous.

See figure next page.


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


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:


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


A

iN

or

Also, if this equation:

m

fmm

Then, the `Magnetic Reluctance` is given by:

is equal to this:

cr

cm A0


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.


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


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


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


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


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


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


For the general case, when i2 0

and:

)()(

)( 2211 tiNtiNtH

AtiN

AtiN

AtHAtBt

)()()()()( 2211


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


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


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


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


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


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



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


t

dtivw0

.

dt

dv

t N

ididtdt

dw

0 0

0

The processes of energizing the winding is seeing in the figure:


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:


Linear System

0 N 0

i

Stored ENERGY

Co-ENERGY

orLi

imm '

25,05,0' iLimm

iL


In linear systems:


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


It can also be shown that the Energy per Volume Unit is:


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.


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.


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 .


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


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:


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


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:


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

course: introduction to electrical machines prof elisete ternes pereira, phd

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