electric machinery - جامعة نزوى · copyright © the mcgraw-hill companies, inc. permission...

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3-0

PowerPoint Slides to accompany

Electric Machinery Sixth Edition

A.E. Fitzgerald

Charles Kingsley, Jr.

Stephen D. Umans

Chapter 3

Electromechanical-Energy-

Conversion Principles


3-1

3.1 FORCES AND TORQUES IN MAGNETIC FIELD SYSTEMS

)( BvEF q

Lorentz Force Law:

For many charged particle

)( BvEF v N/m3 ( coulombs/m3)

vJ Current density A/m2 BJF v

AJI Current A BIF N/m


3-2

Example 3.1: A nonmagnetic motor containing a single-turn coil is placed in a uniform magnetic field of magnitude B0, as shown in Fig. 3.2. The coil sides are at radius R and the wire carries current I as indicated. Find the θ-directed torque as a function of rotor position α when I=10 A, B0=0.02 T and R=0.05 m. Assume that the rotor is of length l=0.3 m.


3-3

Very few problems can be solved using Lorentz force, where current-carrying elements and simple structures exist.

Most electromechanical-energy-conversion devices contain magnetic material and forces can not be calculated from Lorentz force.

Thus, We will use ENERGY METHOD based on conservation of energy.


3-4

Electrical terminals: e and i

Mechanical terminals: ffld and x

Losses separated from energy storage mechanism

Interaction through magnetic stored energy


3-5

dt

dxfie

dt

Wdfld

fld

Time rate of change of Wfld (field energy) equals to the difference of input electrical power and output mechanical power for lossless systems.

or

dxfdiWd fldfld

Force can be solved as a function of flux linkage λ and position x.


3-6

heatto

converted

Energy

fieldmagneticin

storedenergy

inIncrease

output

energy

Mechanical

sources

electricfrom

inputEnergy

fldmechelec WdWdWd

3.2 ENERGY BALANCE

Energy neither created nor destroyed, it only changes the form.

Energy balance equation is written for motor action below

For lossless magnetic-energy-storage system

energystoredmagneticinchangealDifferenti:

outputenergymechanicalalDifferenti:

inputenergyelectricalalDifferenti:

fld

mech

elec

Wd

Wd

Wd


3-7

3.3 ENERGY IN SINGLY-EXCITED MAGNETIC FIELD SYSTEMS

Schematic of an electromagnetic relay. Figure 3.4


3-8

ixL )(

The magnetic circuit can be described by an inductance which is a function of the geometry and permeability of the magnetic material.

When air-gap exist in most cases Rgap>>Rcore and energy storage occurs in the gap.

Magnetic nonlinearity and core losses neglected in practical devices.

Flux linkage and current linearly related.

Energy equation

Wfld uniquely specified by the value of λ and x. Thus, λ and x are called STATE VARIABLES.

dxfdiWd fldfld


3-9

Integration paths for Wfld. Figure 3.5

Magnetic stored energy Wfld uniquely determined by λ and x regardless of how they are brought to their final values.

bpath

fld

apath

fldfld WdWdW22

OR magnetic stored energy:

dVBdHWV

B

fld

0

0

0

000 ),(),(

dxixW fld


3-10

Example 3.2:The relay shown on the figure is made of infinitely-permeable magnetic material with a movable plunger, also of infinitely-permeable material. The height of the plunger is much greater than the air-gap length (h>>g). Calculate the magnetic stored energy Wfld as a function of plunger position (0<x<d) for N=1000 turns, g=2 mm, d=0.15 m, l=0.1 m, and i=10 A.


3-11

3.4 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROM ENERGY

Consider any state function F(x1, x2), the total differential of F with respect to the two variables x1 and x2

2

2

1

1

21

12

),( xdx

Fxd

x

FxxFd

xx

Similarly, for energy function Wfld(λ, x)

xdx

Wd

WxWd

fld

x

fld

fld

),(

dxfdixWd fldfld ),(


3-12

x

fldWi

x

Wf

fld

fld

fldfld Tf

Once we know the energy, current and more importantly force can be calculated.

For a system with rotating mechanical terminal

x

dTdiWd fldfld ),(

),(fld

fld

WT


3-13

Example 3.4:The magnetic circuit below consists of a single-coil stator and an oval rotor. Because the air-gap is nonuniform, the coil inductance varies with rotor angular position, measured between the magnetic axis of the stator coil and the major axis of the rotor, as

)2(cos)( 20 LLL

where where L0=10.6 mH and L2=2.7 mH. Note the second-harmonic variation of inductance with rotor angle θ.


3-14

3.5 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROM COENERGY

dxfdixWd fldfld ),(

Mathematically manipulated to define a new state function known as the COENERGY, from which force can be obtained directly as a function of current.

),(),( xWixiW fldfld

dxfdixiWd fldfld ),(

Note that energy and coenergy equal for linear systems.


3-15

dxfdixiWd fldfld ),(

xdx

Wid

i

WxiWd

i

fld

x

fld

fld

),(

x

fld

i

xiW

),(

i

fld

fldx

xiWf

),(

i

fld idxixiW0

),(),(


3-16

dVdHBWV

H

fld

0

0

dVdHBWV

H

H

fld

c

0

In field-theory terms, for soft magnetic materals (B=0 when H=0)

For permanent magnet materials (B=0 when H=Hc)


3-17

Effect of x on the energy and coenergy of a singly-excited device: (a) change of energy with held constant; (b) change of coenergy with i held constant. Figure 3.11


3-18

Example 3.5: For the relay below, find the force on the plunger as a function of x when the coil is driven by a controller which produces a current as a function of x of the form

A)( 0

d

xIxi


3-19

Example 3.6: The magnetic circuit in the figure is made of high-permeability electrical steel. The rotor is free to turn about a vertical axis. The dimensions are shown in the figure.

a) Derive an expression for the torque acting on the rotor in terms of the dimensions and the magnetic field in the two air gaps. Assume the reluctance of the steel to be negligible and neglect the effects of fringing.

b) The maximum flux density in the overlapping portions of the air gaps is to be limited to approximately 1.65 T to avoid excessive saturation of the steel. Compute the maximum torque for r1=2.5 cm, h=1.8 cm, and g=3 mm.


3-20

3.6 MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS

Many electromechanical devices have multiple electrical terminals.


3-21

Integration path to obtain Wfld(10, 20

, 0).

dTdidiWd fldfld 221121 ),,(

01

0

02

00

0

102211

0

20212021 ),,(),,0(),,(

didiW fld

For magnetically linear systems

2121111 iLiL

2221212 iLiL

D

LLi 2121221

D

LLi 2111212

21122211 LLLLD

USING ENERGY FUNCTON:


3-22

01

002

00

0

1

0

20121022

0

2

0

2011021

)(

)()(

)(

)(),,(

d

D

LLd

D

LW fld

000000 21

0

01221

0

02222

0

011021

)(

)(

)(2

)(

)(2

)(),,(

D

L

D

L

D

LW fld

21,

fld

fld

WT


3-23

dTdidiiiWd fldfld 221121 ),,(

01

0

02

00

0

102211

0

20212021 ),,(),,0(),,(

ii

fld diiiidiiiiiW

000000 2101222

02221

011021 )(

2

)(

2

)(),,( iiLi

Li

LiiW fld

USING COENERGY FUNCTON:

2121111 iLiL

2221212 iLiL

21, ii

fld

fld

WT


3-24

d

Ldii

d

Ldi

d

LdiT fld

)()(

2

)(

2

1221

222211

21

nnnnn

n

n

n i

i

i

LLL

LLL

LLL

2

1

21

22221

11211

2

1

ILλ )(

For a general n electrical terminal

ILI )(2

1T

fldW IL

Iθd

dT T

fld

)(

2

1


3-25

Example 3.7: In the figure, the inductances in henrys are given as L11=(3+cos 2θ)x10-3; L12=0.3 cos θ; L22=30+10 cos 2θ. Find and plot the torque Tfld(θ) for current i1=0.8 A and i2=0.01 A.


3-26

310)sin4.22sin64.1( fldT


3-27

3.7 FORCES AND TORQUES IN SYSTEMS WITH PERMANENT MAGNETS

Special case must be taken when dealing with hard magnetic material because magnetic flux density is zero when H=Hc not when H=0.

•Consider fictitious winding

•In normal operation, the fictitious winding carries NO current

•Current in the winding can be adjusted to zero out the field produced by permanent magnet in order to achieve the “zero force” starting point.


3-28 Integration path for calculating Wfld (if = 0, x ) in the permanent magnet system of Fig. 3.17. Figure 3.18

dxfdixiWd fldffffld ),(

bpath

fld

apath

fldffld WdWdxiW11

),0(

0

00

0),(),(),0(

fI

fff

x

fffldffld dixixdxIifxiW

0

0

),(),0(

fI

fffffld dixixiW

If0 is the current to zero-out the field.


3-29

Figure 3.19

Example 3.8: The magnetic circuit is excited by a samarium-cobalt permanent magnet and includes a movable plunger. Also shown is the fictitous winding of Nf turns carrying a current if which is included here for the sake of the analysis. The dimensions are: Wm=2 cm, Wg=3 cm, W0=2 cm, d=2 cm, g0=0.2 cm, and D=3 cm.

a) Find an expression for the coenergy of the system as a function of plunger position x,

b) Find an expression for the force on the plunger as a function of x,

c) Calculate the force at x=0 and x=0.5 cm.


3-30

A different solution for permanent magnet circuits:

dHA e

cR

dd

NiA eeq

R

)(

dHNi ceq)(


3-31

Example 3.9: Figure shows an actuator consisting of an infinitely-permeable yoke and plunger, excited by a section of NdFeB magnet and an excitation winding of N1=1500 turns. The dimensions are: W=4 cm, W1=4.5 cm, D=3.5 cm, d=8 mm, and g0=1 mm.

a) Find x-directed force on the plunger when the current in the excitation winding is zero and x=3 mm.

b) Calculate the current in the excitation winding required to reduce the plunger force to zero.


3-32

3.8 DYNAMIC EQUATIONS

We are interested in the operation of complete electromechanical system and not just of the electromechanical energy conversion system around which it is built.

For Electrical Terminal:

td

xd

xd

xLdi

td

idxLiRv

)()(0

For multiple-excited system, we will have similar equation for each terminal


3-33

)( 0xxKfK

td

xdBfD

2

2

td

xdMfM

For Mechanical Terminal:

Spring:

Damper:

Mass:

K

B

M

002

2

)( fxxKdt

xdB

dt

xdMf fld

x

M

B

K

fldf

0f

: Spring constant (N/m)

: Damping constant (N.s/m)

: Mass (kg)


3-34

td

txd

xd

xLdi

td

tidxLtiRtv

)()()()()()(0

))(),(())(()()(

)( 02

2

0 txtifxtxKdt

txdB

dt

txdMtf fld

Dynamic Equations (Electrical and Mechanical Equations Together):

xd

xLdif fld

)(

2

2

These equations completely specify the behavior of electromechanical device. Solution of these equations will describe the position x and the current i at any time t in the system.


3-35

)( 0 KTK

td

dBTF

2

2

td

dJTJ

For Rotational Mechanical Terminal:

Torsional Spring:

Friction:

Inertia:

K

B

J

002

2

)( TKdt

dB

dt

dJT fld

: Torsional Spring constant (N.m/rad)

: Friction constant (N.m.s/rad)

: Inertia constant (kg.m2/rad)


3-36

Example 3.10: Figure shows in cross section a cylindrical solenoid magnet in which the cylindrical plunger of mass M moves vertically in brass quide rings of thickness g and mean diameter d. The permeability of brass is µ0. The plunger is supported by a spring with K constant. Its unstretched length is l0. A mechanical load force ft is applied to the plunger from the mechanical system connected to it. Assume that frictional force is linearly proportional to the velocity with coefficient B. The coil has N turns and resistance R. Its terminal voltage is vt and its current i. Derive the dynamic equations of motion of the electromechanical system.


3-37

EXTRA Example: A two poles VR machine is shown in figure. Stator and rotor has infinite permeability.

a) Find gap cross-sectional area as a function of θ.

b) Find the inductance for the machine.

c) Write down the dynamic equations.

d) Solve the dynamic equations to find the position of rotor as a function of time initially starting from θ0=25 degrees.

0Stator Axes

Rotor Axes

r

Numerical Values:

N=100 turns, g=0.0005 m, d=0.1 m, r=0.04 m, J=0.05, B=0.02, θ0=30, R=0.5 ohm, E=10 Volt.

electric machinery - جامعة نزوى · copyright © the mcgraw-hill companies, inc. permission...

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