Lecture 21Lecture 21 Magg,netic Circuits, Materials 21.pdf · Lecture 21Lecture 21 Magg,netic...

Lecture 21Lecture 21Magnetic Circuits, Materialsg ,

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Magnetic Circuits and TransformersMagnetic Circuits and Transformers

1. Understand magnetic fields and their interactionswith moving charges.

2 Use the right hand rule to determine the direction2. Use the right-hand rule to determine the directionof the magnetic field around a current-carryingwire or coilwire or coil.


3. Calculate forces on moving charges and current carrying wires due to magnetic fieldscarrying wires due to magnetic fields.

4 Calculate the voltage induced in a coil by a4. Calculate the voltage induced in a coil by a changing magnetic flux or in a conductor cutting through a magnetic fieldcutting through a magnetic field.

5 Use Lenz’s law to determine the polarities of5. Use Lenz s law to determine the polarities of induced voltages.


6. Apply magnetic-circuit concepts to determine the magnetic fields in practical devices.

7. Determine the inductance and mutual inductance of coils given their physical parameters.

8. Understand hysteresis, saturation, core loss, d dd i dand eddy currents in cores composed of

magnetic materials such as iron.


9. Understand ideal transformers and solve circuits that include transformers.

10. Use the equivalent circuits of real transformers to determine their regulations gand power efficiencies.


Magnetic Field Linesg

Magnetic fields The fluxMagnetic fields can be visualized as lines of flux h f l d

The flux density vector B is tangent to h li f flthat form closed

pathsthe lines of flux

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.density flux MagneticB =

Magnetic Fieldsg

• Magnetic flux lines form closed paths that g pare close together where the field is strong and farther apart where the field is weak.p

• Flux lines leave the north-seeking end of a gmagnet and enter the south-seeking end.

• When placed in a magnetic field, a compass indicates north in the direction of the flux


lines.

Right-Hand Ruleg


Forces on Charges Moving in g gMagnetic Fields

Buf ×= quq

( )θsinquBf =


Forces on Current-Carrying Wires

Blf ×=ddqd

y g

ddqdt

q

Bl ×= ddtdq

Bl ×= idForce on straight wire of length l in a constant

( )θiilBf

Force on straight wire of length l in a constant magnetic field


( )θsinilBf =

Force on a Current Carrying Wirey g

l 1Ai

ml101

==

TBAi5.0

10=

90=θ o

NTmAilBf 5)5.0)(1)(10()sin( === θ


Flux Linkages and Faraday’s LLaw

Magnetic flux passing through a surface area A:

φ d⋅= ∫ ABA∫

For a constant magnetic flux density perpendicular

BA=to the surface:

The flux linking a coil with N turns:


φλ N=

Faraday’s Lawy

Faraday’s law of magnetic induction:

dtde λ

=dt

The voltage induced in a coil whenever its gflux linkages are changing. Changes occur from:

• Magnetic field changing in time


• Coil moving relative to magnetic field

Lenz’s Law

Lenz’s law states that the polarity of the induced voltage is such that the voltagewould produce a current (through an external resistance) that opposes the original change in flux linkages.


Lenz’s Law


US5,975,714 Renewable Energy Flashlight


Voltages Induced inField-Cutting Conductors

BludxBldeBlxBA ===→==λλ


Bludt

Bldt

eBlxBA ===→==λ

Magnetic Field Intensity and g yAmpère’s Law

intensityfieldMagneticH == HB μ7

0 104πμ ×= − AmWb

ytpermeabiliRelativer0μ

μμ =

Ampère’s Law: ∑∫ =⋅ idlH


Ampère’s Law

The line integral of the magnetic field g gintensity around a closed path is equal to the sum of the currents flowing through the


g gsurface bounded by the path.

Magnetic Field Intensity and g yAmpère’s Law

productdotHdld =• lH )cos(θ productdotHdldlH )cos(θ

directionsametheinpointsandmagnitudeconstanta hasHfieldmagnetic the If

iHldlength lincrementa the asdirection sametheinpointsand magnitudeΣ= l


Magnetic Field Around a Long g gStraight Wire

IIrHHl π2 ==

rIHπ2

=

IHB

rμ

π2

rIHB

πμμ

2==


Flux Density in a Toroidal Corey

NIRHHl π2 ==

RNIHπ2

=

NIB

Rμπ2

RB

πμ2

=


Flux Density in a Toroidal Corey

rR

NIBA2

2ππ

μφ ==

NIrR2

2μπ

=

INR2

22

=

RIrNN

2

22μφλ ==


Example 15.7

Σ=•∫ idlH

1path for A10=∫

3pathforA102path for 0

==

3path for A10−=


Example 15.8Find the force between these two wires if theythese two wires if they are 1 m long and separated by 0.1 m:p y

rI

mBπ

μ2

)1.0(

7

101 =

Tm

Axπ

π

20)1.0(2

)10)(104( 7=

−

TmAlBif

T

μθ

μ

)20)(1)(10()sin(

20

1221 ==

→


repulsiveNTmA

μμ

200)20)(1)(10(

==

Magnetic CircuitsMagnetic Circuits

I i i li ti d tIn many engineering applications, we need to compute the magnetic fields for structures that l k ffi i t t f t i ht f dlack sufficient symmetry for straight-forward application of Ampère’s law. Then, we use an

i t th d k ti i itapproximate method known as magnetic-circuitanalysis.


magnetomotive force (mmf) of an N-turn current carrying coilcurrent-carrying coil

IN=F Analog: Voltage (emf)

reluctance of a path for magnetic flux

g g ( )

l=R

Aμ=R

Analog: Resistance

φRF =

g


φRFAnalog: Ohm’s Law

Reluctance ↔Resistance

⎟⎠⎞

⎜⎝⎛=↔⎟

⎠⎞

⎜⎝⎛=

AlR

Al ρ1

R


⎠⎝⎠⎝ AAρ

μ

Magnetic Circuit for Toroidal Coilg

rARl 2 2ππ

⎞⎛⎞⎛⎞⎛

==

rR

rR

Al 21211

22 μππ

μμ ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛=R

IN

NI2

=⎠⎝⎠⎝

F

F

RINr

2

2μφ ==RF


Advantage of theAdvantage of theMagnetic-Circuit Approachg pp

Th d f h i i i hThe advantage of the magnetic-circuit approach is that it can be applied to unsymmetrical

ti ith lti l ilmagnetic cores with multiple coils.


Magnetic Circuit with an Air Gapg

Find what current is required to generate a flux density


of Bgap=0.25 T in the air gap.


)5.064(11 cmxlR −==

20

105.231

)3)(2(

mx

cmcmAR

rcore ==

−

μμμ

4

47

101955

106)104)(6000(

x

mxx

=

=−−π

10195.5 x=

6000=rμ


r

Fringingg gWe approximately account for fringing by adding the length of the gap to the depth and g g g p pwidth in computing effective gap area.

2410758

)5.03()5.02( cmcmxcmcmAgap ++=−

70

24

104

1075.8

x

mx

gap =≈

=−

−

πμμ

24

2

7 1075.8105.0

1041

mxmx

xRgap =

−

−

−π


610547.4 x=


RRR +

xxx

RRR gapcoretotal

10600.410547.410195.524

664 =+=

+=

Wbx

mxTAB gapgap

10188.2

)1075.8)(25.0(4

24

=

==−

−φ

turnsAxxRF

1006)10600.4)(10188.2( 64

=== −φ

AturnsAFi

Ni

012.21006===

=


AturnsN

i 012.2500

Exercise 15.9

Determine the current required to establish a flux


Determine the current required to establish a flux density of 0.5T in the air gap

Exercise 15.9

)12()12( cmcmxcmcmAgap ++=24109

)12()12(

mx

cmcmxcmcmAgap

=

++−

2

70

1011

104

mx

xgap =≈−

−πμμ

6

247

1084281075.8

1011041

mxmx

xRgap =

−−π610842.8 x=


Exercise 15.9)16282(11 cmxxlRcore

−+==

20

10271

)2)(2(

mx

cmcmAR

rcore

−

μμμ

3

47

104107

104)104)(5000( mxx=

−−π3104.107 x=


Exercise 15.9RRR coregaptotal +=

mxTAB

Rxx

gapgap

gap

)109)(5.0(

10107.010842.824

66

==

≈+=−φ

Ri

mWb

total

45.0

=

=φ

xxxN

i

1000)1045.0)(10107.010842.8( 366 +

=−

A027.41000

=


A Magnetic Circuit with ReluctancesA Magnetic Circuit with Reluctances in Series and Parallel

Find the flux density in each gap


Find the flux density in each gap

A Magnetic Circuit with Reluctances

t t l RR +=1

in Series and Parallel

ba

ctotal

NiRR

RR+

+11

b

totalc

dividercurrentR

RNi

φφ

φ

=

=

)(

ca

b

cba

a

RRR

dividercurrentRR

φφ

φφ

=

+= )(

a

aa

cba

b

AB

RRφ

=

+

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.a

aa

a

AB

Aφ

=

Magnetic Materialsg


Magnetic MaterialsMagnetic Materials

The relationship between B and H is not linear for the types of iron used in motors and transformers.yp

B-H curves exhibits “h t i ”


“hysterysis”

Magnetic MaterialsT t l li t

gTotal alignmentResidual

alignment at H=0

Alignment

• Magnetic field of atoms within small domains are aligned

Linear 1-22-3

Magnetic field of atoms within small domains are aligned

• Magnetic fields of the small domains are initially randomly oriented

• As the magnetic field intensity increases, the domains tend to align, leaving a residual alignment even when the applied field


is reduced to zero

Energy Considerationsgy

φ

∫∫∫ ===tt

dNidtidtdNdtviW

000

φφφ

∫∫ ==

==BB

dBHVdBAlHW

AdBdandHlNi φ

∫

∫∫ ==

B

core

dBHWW

dBHVdBAlHW00


∫== V dBHWVW

0

Energy Considerationsgy

W B

dBHAlWW

B

v ∫==0Al 0

The area between the B-H curve and the B axis represents


pthe volumetric energy supplied to the core

Core LossCore Loss Power loss due to hysteresis is proportional to y p pfrequency, assuming constant peak flux.


Eddy-Current Lossy

As the magnetic field changes in a material, it g gcauses “eddy currents” to flow. Power loss due to eddy currents is proportional to the square of y p p qfrequency, assuming constant peak flux.

vP2

RP =


Energy Stored in theEnergy Stored in theMagnetic Fieldg

2B

μμ 2

2BdBBWB

v == ∫ μμ 20∫


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