Electromech. Energy. Conv

7/29/2019 Electromech. Energy. Conv.

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Electromechanical Energy

Conversion

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Converts electrical energy into mechanical energy

or

Converts mechanical energy into electrical energy.

Electromechanical energy conversion device:

Electromechanical energy conversions use a

magnetic field as the medium of energy

conversion

Introduction

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IntroductionThree categories of electromechanical energy conversion

devices:

z

Transducers (for measurement and control)- small motionTransform the signals of different forms. Examples:

microphones, sensors and speakers.

z Force producing devices (translational force)- limited

mechanical motion.

Produce forces mostly for linear motion drives, Example

Actuators - relays, solenoids and electromagnets.

z Continuous energy conversion equipment.Operate in rotating mode. Examples: motors and

generators.


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Energy Conversion Process

The principle of conservation of energy:

Energy can neither be created nor destroyed. It

can only be changed from one form to another.Therefore total energy in a system is constant


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An electromechanical converter system has three

essential parts:

An electrical system (electric circuits such as windings)

A magnetic system (magnetic field in the magnetic cores and air gaps)

A mechanical system (mechanically movable parts such as a rotor in an

electrical machine).


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EM Energy Conversion: Analogy

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Field Energy

ElectricalEnergy

(input)

Mechanical

Energy

(output)

Thermal

Energy(losses)


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ElectromechanicalSystem

Electrical System Magnetic System Mechanical System

Voltages andCurrents

Magnetic Flux Position, Speedand Acceleration

Circuit Equations

(KVL and KCL)

Force/Torque Eqns

(Newtons Law)Froce/Torque

emf

Concept of electromechanical system modeling


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The energy transfer equation is as follows:

+

+

=

lossesEnergy

fieldmagnetic

inenergystored

inIncrease

output

energy

Mechanical

sourcesfrom

inputenergy

Electrical

Electrical system Magnetic system Mechanical

system

Electrical

loss

Field loss Mechanical

loss

P mech


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Electrical energyinput from sources

resis tan ce loss

=

Mechanical energyoutput + friction

and windage loss

+

Increase instored field

energy + core loss

The energy balance can therefore be written as:

For the lossless magnetic energy storage system in differential form,

dWe = i d = differential change in electric energy input

dWm =fm dx = differential change in mechanical energy outputdWf= differential change in magnetic stored energy

fmedWdWdW +=


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Here e is the voltage induced in the electric terminals by changing

magnetic stored energy.

Together with Faradays law for induced voltage, form the basis forthe energy method.

fme dWdWdteidW +==

ididtdt

ddW

dt

dedt;eidW

e

e

==

==We can write


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Singly-excited

System

Energy, Coenergy

and Force or Torque


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Energy in Magnetic SystemConsider the electromechanical system below:

Schematic of an electromagnetic relay

Axiallength(perpendiculartopage)=l


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Energy in Magnetic SystemThe mechanical force fm is defined as acting from the

relay upon the external mechanical system and the

differential mechanical energy output of the relay is

Then, substitution dWe

= id , gives

dWf= id fm dx

Value of Wfis uniquely specified by the valuesof and x, since the magnetic energy storage

system is lossless.

dWm = fm dx


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Wf = id

Energy in Magnetic System

i

dWf = id

d

dWf = differential change in magnetic stored energy


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Energy and CoenergyThe -i characteristics of an electromagnetic systemdepends on the air-gap length and B-H characteristics

of the magnetic material.

For a larger air-gap

length the characteristic

is essentially linear. The

characteristic becomes

non linear as the air-gaplength decreases.


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Energy and Coenergy

For a particular value of air-gap length, thefield energyis represented by the

red area between axis and -i characteristic. The blue area between i axisand - i characteristic is known as the coenergy


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Energy and CoenergyThe coenergy is defined as

From the figure of - i characteristic,

Wf + W

f= i

Note that Wf > Wf if the - i characteristic is nonlinear and Wf = Wf if it is linear.

diWi

0

'

f =

The quantity of coenergy has no physical significance.However, it can be used to derive expressions for force

(torque) developed in an electromagnetic system


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Determination of Force from

Energy

The magnetic stored energy Wfis a state

function, determined uniquely by theindependent state variables and x. This is

shown explicitly by

dWf(, x) = id fm dx


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EnergyFor any function of two independent variables

F(x1,x2), the total differential equation ofFwith respect to the two state variables x1 and x2can be written

dF(x1,x2) =F(x1,x2)

x1 x2

dx1+ F(x1,x2)

x2 x1

dx2


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Energy

Therefore, for the total differential of Wf

dWf(,x) =

Wf(,x)

xd+

Wf(,x)

x dx

And we know that

dWf(,x) = id fm dx


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EnergyBy matching both equations, the current:

i = Wf(,x) x

where the partial derivative is taken while

holding x constant and the mechanical force:

fm = W

f(,x)

x where the partial derivative is taken while

holding constant.


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Energy: Linear SystemFor a linear magnetic system for which =L(x)i:

Wf(,x) = i(,x)d=0

L(x) d=0

12 2

L(x)

and the force,fm can be found directly:

fm = Wf(,x)

x =

x

1

2

2

L(x)

=2

2L(x)2

dL(x)

dx


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CoenergyBy expanding dWf

(i,x):

and, from the previous result:

dW f'(i,x) = di + fm dx

dWf

'(i,x) =

Wf' (i,x)

ix

di +Wf

' (i,x)

xi

dx


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CoenergyBy matching both equations, :

= Wf'

(i,x)i

x


holding x constant and the mechanical force:

fm =W

f

'(i,x)

xi


holding i constant.


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Coenergy: Linear SystemFor a linear magnetic system for which =L(x)i:

Wf' (i,x) = (i,x)di =

0

i

L(x)idi =0

i

L(x) i2

2

and the force,fm can be found directly:

fm =Wf

'(i,x)

xi

=

xL(x)

i2

2

i

=i2

2

dL(x)

dx


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Determination of Torque from

CoenergyFor a system with a rotating mechanical terminal,

the mechanical terminal variables become the

angular displacement and the torque T.

Therefore, equation for the torque:

T=Wf

'(i,)

i

where the partial derivative is taken whileholding constant.


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Determination of Force Using

Energy or Coenergy?

The selection of energy or coenergy as thefunction to find the force is purely a matter of

convenience.

They both give the same result, but one or the

other may be simpler analytically, depending on

the desired result and characteristics of the

system being analyzed.


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Direction of Force Developed1. By using energy function:

fm = +W

f' (i,x)

xi

The negative sign shows that the force acts in a

direction to decrease the magnetic field stored

energy at constant flux.

2. By using coenergy function:

The positive sign emphasizes that the force acts

in a direction to increase the coenergy at

constant current.

fm = Wf(,x)

x


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Direction of Force Developed

3. By using inductance function:

The positive sign emphasizes that the force actsin a direction to increase the inductance at

constant current.

fm = +i2

2

dL(x)

dx i


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B-H Curve and Energy Density

In a magnetic circuit having a substantial air

gap g, and high permeability of the iron core,

nearly all the stored energy resides in the gap.Therefore, in most of the cases we just need

to consider the energy stored in the gap. The

magnetic stored energy,

diWf = 0

NHgi = and NAdBNABdNdd === )()( in which

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Therefore, dBHAgdBNA

N

HgW

BB

f

==

00

However, Ag is volume of the air gap. Dividing

both sides of the above equation by the volume

Ag results in

dBHAg

W

w

Bf

f

== 0

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wf is known as energy density.

dBHwB

f = 0where is energy per unit volume

The area between the B-Hcurve and B axis

represents the energy

density in the air gap.H

B

wf

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B-H Curve and Energy DensityIn the same manner,

dHBwH

f

=

0

' is coenergy per unit volume.

The area between the B-Hcurve and H axis represents

the coenergy density in the

air gap.H

Bwf

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B-H Curve and Energy DensityFor a linear magnetic circuit, B = H or H =B/, energy density:

2

2

00BdBBdBHw

BB

f === and coenergy density:

2

2

00

' HHdHBdHw

HH

f

===

In this case, it is obvious that wf = wf.

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Example 3.1 PC SenThe dimensions of the relay system are shown in

figure below. The magnetic core is made of cast

steel whose B-H characteristic is shown in Figure

1.7 (pg.6). The coil has 300 turns, and the coilresistance is 6 ohms. For a fixed air-gap length lg =

4 mm, a dc source is connected to the coil to

produce a flux density of 1.1 Tesla in the air-gap.

Calculate

(a)The voltage of the dc source.

(b)The stored field energy.

lg 5 cm

5 cm

10 cm

5 cm

10 cm

Depth =10 cm


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Example 3.2 PC Sen

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The -i relationship for an electromagneticsystem is given by

which is valid for the limits 0 < i < 4 A and 3 < g >g). Calculate

a) The magnetic storage energy Wfas a function of plunger position ( 0< x


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Example 3The magnetic circuit shown is made of high-permeability

electrical steel. Assume the reluctance of steel -- infinity.Derive the expression for the torque acting on the rotor .

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Example 4

The magnetic circuit below consists of a single coil stator and

an oval rotor. Because of the air-gap is non uniform, the coil

inductance varies with the rotor angular position.

Given coil inductance L() = Lo + L2cos2, where Lo= 10.6mH and L2= 2.7 mH.

Find torque as a function of for a coil current of 2 A.

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Doubly-excited

Systems

Energy, Coenergyand Force or Torque

R t ti M hi


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Rotating Machines

z Most of the energy converters, particularly thehigher-power ones, produce rotational motion.

z The essential part of a rotating electromagneticsystem is shown in the figure.

z The fixed part is called the stator,

the moving part is called the rotor.

z The rotor is mounted on a shaftand is free to rotate between

the poles of the stator

z Let consider general case where

both stator & rotor have windings

carrying current ( is and ir )

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Rotating Machines


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Rotating Machinesz Assume general case, both stator and rotor have winding carrying

currents (non-uniform air gap silent pole rotor)

z The system stored field energy, Wf can be evaluated byestablishing the stator current is and rotor current ir and let

system static, i.e. no mechanical output

Stator and rotor

flux linkage isexpressed in

terms of

inductances L

(which depends

on position rotor

angle , L()

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Reluctance Torque It is caused by thetendency of the induced pole to align with

excited pole such that the minimum reluctance

is produced. At least one or both of thewinding must be excited.

Alignment Torque It is caused by a tendencyof the excited rotor to align with excited stator

so as to maximize the mutual inductance.

Both winding must be excited.

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C li d i l M hi


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Cylindrical Machines

z Reluctance machines are simple in construction,

but torque developed in these machines is small.

z Cylindrical machines, although more complex in

construction, produce larger torques.

z Most electrical machines are of the cylindrical

type.

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Cylindrical Machinesz A cross sectional view of an elementary

two pole cylindrical rotating machineis (uniform air gap) shown.

z The stator and rotor windings are placedon two slots.

z In the actual machine the windingsare distributed over several slots.

z

If the effects of the slots are neglected,the reluctance of the magnetic path isindependent of the position of the rotor.

z Assumed Lss and Lrr are constant (i.e no

reluctance torque produced).z Alignment torque is caused by the

tendency of the excited rotor to alignwith the excited stator, depends on

mutual inductance10/23/2012 54

C li d i l hi


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Cylindrical machines

z Torque produced

z Mutual inductance

z Currents

z Rotor position

WhereM = peak value of mutual inductance

= the angle between magnetic axis ofthe stator and rotor windings

m = angular velocity of rotor

siniMid

dMcosii

d

dLiiT rsrs

srrs ===

Tm when =90o

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z Torque in general varies sinusoidally with time

z Average value of each term is zero unless thecoefficient of t is zero

)tsin()tcos(tMcosIIT mrsrmsm ++=

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z Non zero average torque exists/develop only if

z Synchronous machine

z Single phase machine

z Pulsating torque

z Polyphase machine minimize pulsating torque

z Not self starting (m = 0 Tavg = 0

Case 1:

Machine develop torque

if sum or difference of

the angular speed of thestator and rotor current

Wr =0 Idc at rotor

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z Asynchronous machines

z Single phase machine

z Pulsating torquez Not self starting

z Polyphase machine minimize pulsating torque and self starting10/23/2012 58

Example


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Example

z In a electromagnetic system, the rotor has no winding

(i.e. we have a reluctance motor) and the inductance

of the stator as a function of the rotor position is

Lss = L0 + L2 cos 2. The stator current is is= Ism sin t

(a) Obtain an expression for the torque acting on the rotor

(b) Let = mt+ , where m is the angular velocity of therotor and is the rotor position at t = 0. Find thecondition for the non-zero average torque and obtain

the expression for the average torque.

Sen pg 111

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Example 5


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Example 5

In a doubly excited rotating actuator shown in figure

below, the stator inductances are given as L11= (3+cos2)mH, L12 = 0.3cos, and the rotor impedance is L22 =

30+10cos2. Find the developed torque in the system fori1=0.8A and i2 = 0.01 A.

Fgrd pg 140

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Electromech. Energy. Conv